# Intelligent Systems II — all public teaching content Exact source JSON: https://deep-learning-study-room.vercel.app/exports/is2.json --- ## M2 · Start here: your IS2 M2 study order Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/m2-start--how-to-use/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/m2-start/how-to-use M2 scope and pattern confirmed in your 8 October class update. Learn **Fuzzy Logic → Genetic Algorithms → Particle Swarm Optimization**, with the required maths explained just before it is used. The answers stay below the questions. ## The confirmed 10-mark pattern | Question | Marks | Attempt rule | Prepare for | | --- | --- | --- | --- | | Q1(a) | 2 | Compulsory | A short numerical or theory question | | Q1(b) | 2 | Compulsory | A short numerical or theory question, including swarm intelligence | | Q2, Q3, Q4 | 3 each | Attempt any two | GA, PSO and fuzzy logic; the mix can include theory and numericals | Your class update specifically highlights fuzzy union/intersection/difference/complement, GA with supplied operators, two-particle PSO minimisation, fuzzy rules, architecture, advantages and disadvantages. Work GA and PSO through **iteration 2**. The mock question numbering is a practice arrangement; the actual exam can rearrange the topics. ## Read in this order 1. [Essential maths](https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--reading-formulas/): candidate versus score, fractions, min/max, binary and two-coordinate movement. 2. [Fuzzy logic](https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--membership-meaning/): membership meaning → graphs → set operations → rule base → Mamdani → defuzzification. Then the complete 32°C / 7-person fan example. 3. [Genetic algorithms](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--meaning/): terms → encoding → selection → crossover/mutation → complete two-generation solutions. 4. [PSO](https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--meaning/): position, velocity and memory → single updates → two particles → two dimensions → class photograph example. 5. [Three 10-mark mocks](https://deep-learning-study-room.vercel.app/read/is2/m2/index/mocks/), followed by [last-day revision](https://deep-learning-study-room.vercel.app/read/is2/m2/index/revision/). Fuzzy relation matrices are kept as additional Unit 2 practice after the three main topics. ACO and GA-for-TSP are in the TEE route; they are not mixed into this M2 reading path. ## How to write a numerical **Full question → Given → To find → Formula and why → numbered calculation steps → final answer.** Read one worked example, then write the same calculation in your notebook. Keep its intermediate table: in GA and PSO that table is the input to the next iteration. ## If the exam is tomorrow First complete fuzzy operations, triangle membership and the Mamdani architecture. Next finish one GA two-generation solution and one PSO two-iteration solution. Then do the remaining variants, theory answers and mocks. Use the formula page for revision after learning the symbols; it is not a substitute for the worked explanations. **Yaad rakho:** Fuzzy = degree aur rules. GA = select, cross, mutate. PSO = velocity, position, best memories. --- ## M2 · Symbols, brackets and iteration labels: how to read a formula Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--reading-formulas/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/core-maths/reading-formulas Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO. ## Read a formula as a calculation instruction A **variable** is a name for a number that can change. In **f(x)=x²+1**, x is the input, f is the name of the scoring rule, and f(x) is its result. An **equation** says that the expressions on both sides have the same value. A formula tells you how to calculate a quantity. | Read this | Meaning | Concrete example | | --- | --- | --- | | x² | x multiplied by itself | 3² = 3 × 3 = 9 | | 2x or 2(x) | 2 multiplied by x | For x=3: 2x=6, while x²=9 | | f(3) | Apply the rule named f to input 3 | If f(x)=x²+1, then f(3)=10 | | x₁ and x₂ | First and second named entries; the small lower number is a label | If a position is (3,7), x₁=3 and x₂=7 | | x at time t | A version of x at a particular iteration | Old x=3; after a move x=2. The time label is not squaring. | | = and ≈ | Exactly equal; approximately equal after rounding | 1/3 ≈ 0.333333 | | ∈ and { } | Belongs to; a collection of values | x ∈ {2,5} means x may be 2 or 5 | **Evaluate f(x)=2x²+1 at x=−3.** [2 marks · Worked example] #### Given The rule is f(x)=2x²+1. The supplied input is x=−3. #### To find f(−3): the output of this rule for input −3. #### Formula and why it applies $$ f(x)=2x^2+1 $$ The question supplies the rule. Replace x by −3, square that whole input, multiply the square by 2, and add 1. Parentheses keep the negative input together. #### Solution ##### Step 1. Substitute with parentheses $$ f(-3)=2(-3)^2+1 $$ ##### Step 2. Do the square first $$ (-3)^2=(-3)\times(-3)=9 $$ ##### Step 3. Multiply by 2 $$ 2\times9=18 $$ ##### Step 4. Add 1 $$ f(-3)=18+1=19 $$ **Final answer.** f(−3)=19. **Evaluate g(x)=(x−2)/4 at x=10.** [2 marks · Worked example] #### Given The rule is g(x)=(x−2)/4. The supplied input is x=10. #### To find g(10): the output of this rule for input 10. #### Formula and why it applies $$ g(x)=\frac{x-2}{4} $$ The fraction bar groups the numerator x−2. Subtract 2 from the input first, then divide the entire difference by 4. #### Solution ##### Step 1. Replace the input $$ g(10)=\frac{10-2}{4} $$ ##### Step 2. Finish the numerator $$ 10-2=8 $$ ##### Step 3. Divide $$ g(10)=8/4=2 $$ **Final answer.** g(10)=2. **At iteration 0, x=4. The rule is new x = old x − 1. Find x after two iterations.** [2 marks · Worked example] #### Given Initial value x=4 at iteration 0. Every update subtracts 1 from the current value. Perform two updates. #### To find The value after update 1 and after update 2. #### Formula and why it applies $$ x^{(t+1)}=x^{(t)}-1 $$ t labels the current iteration, and t+1 labels the next. The parenthesised upper label records the step; it is not a power. Apply the supplied rule once, then use that new answer as the input to the second update. #### Solution ##### Step 1. Iteration 1 uses the initial value $$ x^{(1)}=4-1=3 $$ ##### Step 2. Iteration 2 uses the updated value $$ x^{(2)}=3-1=2 $$ ##### Step 3. Distinguish a label from a power x at iteration 2 is 2. Squaring the initial value would give 4²=16; that is a different operation. **Final answer.** After update 1: x=3. After update 2: x=2. The sequence is 4 → 3 → 2. **Yaad rakho:** Symbol sirf naam hai. Pehle bolo kaunsa number chahiye, phir uski jagah value likho. --- ## M2 · Position, score, minimisation and maximisation Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--objective/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/core-maths/objective Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO. ## What are we trying to do in IS2? **An intelligent system** uses information and a decision procedure to choose an output or an action. In M2, you study three different procedures. | Topic | Input | What the procedure does | Output | | --- | --- | --- | --- | | Fuzzy logic | Temperature and occupancy | Applies rules such as IF hot AND crowded THEN fast | A fan-speed setting | | Genetic algorithm (GA) | A population of candidate answers | Selects, combines and changes candidate answers | A good solution and its fitness | | Particle swarm optimization (PSO) | Positions and velocities of particles | Moves candidate answers using individual and shared experience | A best position and its objective value | ## Candidate answer and score are different An **objective function** is the calculation used to score an answer. In **minimisation**, a smaller score is better; in **maximisation**, a larger score is better. The letter **x** names a candidate answer. The notation **f(x)**, read “f of x”, means its score. Parentheses here do not mean f multiplied by x. **For the minimisation problem $$ f(x)=(x-3)^2 $$ , find the score at $$ x=2 $$ .** [2 marks · Practice] #### Given $$ x=2 $$ , $$ f(x)=(x-3)^2 $$ #### To find The objective value f(x). #### Formula and why it applies Replace x by the supplied number. Square means multiply the result by itself. #### Solution ##### Step 1. Subtract 3 $$ x-3=2-3=-1 $$ ##### Step 2. Square the difference $$ f(2)=(-1)^2=(-1)\times(-1)=1 $$ **Final answer.** The position is x = 2; its score is 1. These are two different numbers. **For the minimisation problem $$ f(x)=(x-3)^2 $$ , find the score at $$ x=5 $$ .** [2 marks · Practice] #### Given $$ x=5 $$ , $$ f(x)=(x-3)^2 $$ #### To find The objective value f(x). #### Formula and why it applies Replace x by the supplied number. Square means multiply the result by itself. #### Solution ##### Step 1. Subtract 3 $$ x-3=5-3=2 $$ ##### Step 2. Square the difference $$ f(5)=(2)^2=(2)\times(2)=4 $$ **Final answer.** The position is x = 5; its score is 4. These are two different numbers. **For the minimisation problem $$ f(x)=(x-3)^2 $$ , find the score at $$ x=-1 $$ .** [2 marks · Practice] #### Given $$ x=-1 $$ , $$ f(x)=(x-3)^2 $$ #### To find The objective value f(x). #### Formula and why it applies Replace x by the supplied number. Square means multiply the result by itself. #### Solution ##### Step 1. Subtract 3 $$ x-3=(-1)-3=-4 $$ ##### Step 2. Square the difference $$ f(-1)=(-4)^2=(-4)\times(-4)=16 $$ **Final answer.** The position is x = -1; its score is 16. These are two different numbers. Comparing these candidates: x = 2 has score 1, x = 5 has score 4, and x = −1 has score 16. For minimisation, x = 2 is the best of these three. A negative position is allowed unless the question gives a restriction; a low score, not a low coordinate, decides the winner. **Yaad rakho:** x = answer ki jagah; f(x) = us answer ka score. Min mein chhota score jeetega. --- ## M2 · Fractions, brackets, minimum and maximum Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--arithmetic/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/core-maths/arithmetic Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO. ## Fractions, min and max A fraction means division: the top number is the **numerator**, and the bottom is the **denominator**. Calculate each bracket first. **min** selects the smallest listed number; **max** selects the largest. Neither word means addition. **Evaluate $$ \frac{32-25}{35-25} $$ .** [2 marks · Practice] #### Given Top: 32 − 25; bottom: 35 − 25. #### To find The quotient. #### Formula and why it applies Subtract separately in the numerator and denominator, then divide. #### Solution ##### Step 1. Calculate the numerator $$ 32-25=7 $$ ##### Step 2. Calculate the denominator $$ 35-25=10 $$ ##### Step 3. Divide $$ \frac{7}{10}=0.7 $$ **Final answer.** 0.7 **Evaluate $$ \frac{35-32}{35-25} $$ .** [2 marks · Practice] #### Given Top: 35 − 32; bottom: 35 − 25. #### To find The quotient. #### Formula and why it applies Subtract separately in the numerator and denominator, then divide. #### Solution ##### Step 1. Calculate the numerator $$ 35-32=3 $$ ##### Step 2. Calculate the denominator $$ 35-25=10 $$ ##### Step 3. Divide $$ \frac{3}{10}=0.3 $$ **Final answer.** 0.3 **Evaluate $$ \frac{7-6}{8-6} $$ .** [2 marks · Practice] #### Given Top: 7 − 6; bottom: 8 − 6. #### To find The quotient. #### Formula and why it applies Subtract separately in the numerator and denominator, then divide. #### Solution ##### Step 1. Calculate the numerator $$ 7-6=1 $$ ##### Step 2. Calculate the denominator $$ 8-6=2 $$ ##### Step 3. Divide $$ \frac{1}{2}=0.5 $$ **Final answer.** 0.5 **Find $$ \max\{\min(0.7,0.2),\min(0.4,0.8)\} $$ .** [2 marks · Practice] #### Given Two inner min operations followed by one outer max. #### To find One number. #### Formula and why it applies Work from the innermost brackets outward. This exact pattern occurs in fuzzy relation composition. #### Solution ##### Step 1. First pair $$ \min(0.7,0.2)=0.2 $$ ##### Step 2. Second pair $$ \min(0.4,0.8)=0.4 $$ ##### Step 3. Choose the larger result $$ \max(0.2,0.4)=0.4 $$ **Final answer.** 0.4. **Yaad rakho:** Pehle bracket, phir divide. AND ka min; OR ka max. --- ## M2 · Percentages, probabilities and cumulative intervals Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--probability/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/core-maths/probability Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO. ## Probability is a share of a specified total A probability is a number between 0 and 1. In these finite examples, 0 means an event has no probability mass and 1 means it occurs with certainty under the model. In a continuous distribution, a single exact value can have probability zero without being logically impossible; probabilities then describe intervals. 25% means 25 out of 100, so it equals 25÷100=0.25. For equally likely outcomes, probability is favourable outcomes divided by all possible outcomes. For GA selection, it is the candidate’s weight divided by the sum of weights. $$ P(\text{event})=\frac{\text{favourable count}}{\text{total count}} $$ **A bag contains 3 red balls and 7 blue balls. One ball is drawn uniformly at random. Find the probability of red and of blue.** [2 marks · Worked example] #### Given Red=3; blue=7. Each ball is equally likely. #### To find Both probabilities and their percentages. #### Formula and why it applies $$ P(\text{colour})=\frac{\text{balls of that colour}}{\text{all balls}} $$ All balls are equally likely, so counting them gives probability. The denominator counts all balls; the numerator counts the requested colour. #### Solution ##### Step 1. Find the total $$ 3+7=10 $$ ##### Step 2. Red probability $$ P(R)=3/10=0.3=30\% $$ ##### Step 3. Blue probability $$ P(B)=7/10=0.7=70\% $$ **Final answer.** Red 0.3; blue 0.7. Their sum is 1. ## From weights to probabilities to cumulative probabilities A **weight** need not itself be a probability. Weights 1, 2, 3, 4 add to 10. Dividing every weight by 10 makes their total 1: this is **normalisation**. **Cumulative** means “total accumulated so far”. It turns separate shares into boundaries along a line from 0 to 1. **Candidates A, B, C and D have weights 1, 2, 3 and 4. Find probabilities and cumulative probabilities. Which candidate is selected by random draw r=0.55?** [2 marks · Worked example] #### Given Weights [1,2,3,4], draw 0.55. Use intervals including their lower endpoint and excluding their upper endpoint. #### To find Four probabilities, their running totals and the selected candidate. #### Formula and why it applies $$ P_i=F_i/(F_A+F_B+F_C+F_D) $$ F is weight. P is its share. Add P values in row order for the cumulative boundary. #### Solution ##### Step 1. Add all weights $$ 1+2+3+4=10 $$ ##### Step 2. Divide each weight by that same total | Candidate | Division | Probability | | --- | --- | --- | | A | 1÷10 | 0.1 | | B | 2÷10 | 0.2 | | C | 3÷10 | 0.3 | | D | 4÷10 | 0.4 | ##### Step 3. Build the running totals | Candidate | Running addition | Interval | | --- | --- | --- | | A | 0+0.1=0.1 | 0 ≤ r < 0.1 | | B | 0.1+0.2=0.3 | 0.1 ≤ r < 0.3 | | C | 0.3+0.3=0.6 | 0.3 ≤ r < 0.6 | | D | 0.6+0.4=1 | 0.6 ≤ r ≤ 1 | ##### Step 4. Locate the draw 0.55 is at least 0.3 and less than 0.6. It lies in C’s interval. The draw is not compared with C’s weight 3. **Final answer.** Candidate C is selected. Swipe across the diagram to read its labels ↔ ![Each interval width equals its probability. Running totals mark the ends.](https://deep-learning-study-room.vercel.app/is2/fig/depth_roulette.12804497d2.svg) Diagram: Each interval width equals its probability. Running totals mark the ends.. Labels: Probabilities become intervals on the same 0–1 line | A | 0.1 | 0 | B | 0.2 | 0.1 | C | 0.3 | 0.3 | D | 0.4 | 0.6 | 1 | r = 0.55 selects C Each interval width equals its probability. Running totals mark the ends. **Two unnormalised class scores are 0.12 and 0.08. Convert them to probabilities.** [2 marks · Worked example] #### Given Scores A=0.12 and B=0.08; they describe the only two alternatives. #### To find Normalised probabilities. #### Formula and why it applies Divide each score by the sum, preserving their ratio. #### Solution ##### Step 1. Sum the scores $$ 0.12+0.08=0.20 $$ ##### Step 2. Normalise A $$ P(A)=0.12/0.20=0.6 $$ ##### Step 3. Normalise B $$ P(B)=0.08/0.20=0.4 $$ **Final answer.** A=0.6 and B=0.4. The original score 0.12 was not the final probability. **Yaad rakho:** Probability = apna hissa ÷ sabka total. Cumulative = ab tak ke saare hisse jodna. --- ## M2 · Binary from the beginning: positions, place values and bits Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--binary/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/core-maths/binary Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO. ## 1. Position means the digit's place **A position is simply a place occupied by one digit in a number.** A digit is the symbol written there. The **place value** tells us how much one unit in that position is worth. Start with an ordinary number you already know: **23**. Its two digits are 2 and 3. They do not contribute the same kind of unit. | | LEFT: tens place | RIGHT: ones place | | --- | --- | --- | | Value of this place | 10 | 1 | | Digit written here | 2 | 3 | | Contribution | 2 × 10 = 20 | 3 × 1 = 3 | $$ 23=(2\times 10)+(3\times 1)=20+3 $$ The **2** means two tens because of where it sits. In **32**, the same digit 2 sits on the right and means only two ones. Changing a digit’s position can change its contribution. **Yaad rakho:** Position = digit ki jagah. Place value = us jagah par likhe ONE ki keemat. 23 mein 2 ka contribution 20 hai, kyunki woh tens ki jagah par hai. ## 2. Binary changes the size of the groups Our ordinary number system is **decimal**: it has ten possible digits, 0 to 9. Its places are ones, tens, hundreds, and so on. Moving one place left multiplies the place value by 10. **Binary** has only two possible digits: **0 and 1**. One binary digit is called a **bit**. Therefore, a **4-bit number means four digit boxes**, each holding either 0 or 1. In binary, grouping happens in **twos**. The rightmost box counts single units. After one single, a second single cannot be written as digit “2”, because binary has no digit 2. Instead, the two singles become **one group of two** in the next box on the left. **Binary 10 means one group of two and zero single units.** It represents the ordinary number 2. Read its digits as “one-zero” when discussing binary, so you do not accidentally read it as decimal ten. Two groups of two make **4**, so the next box to the left is the fours place. Two groups of four make **8**, so the next box is the eights place. This is why the values double. | Ordinary number | Binary digits | Meaning | | --- | --- | --- | | 0 | 0 | No single units | | 1 | 1 | One single unit | | 2 | 10 | One two + zero ones = 2 | | 3 | 11 | One two + one one = 3 | | 4 | 100 | One four + zero twos + zero ones = 4 | **Yaad rakho:** Decimal mein 10 ka group; binary mein 2 ka group. Do singles = ek pair; do pairs = ek group of four. ## 3. Build the places from the RIGHT First draw four empty boxes. **Start with the box furthest to the right.** Give that place the value 1. Move left one box at a time, doubling the previous value. 1. **1st box from the right:** value 1. 2. **2nd box from the right:** 1 × 2 = 2. 3. **3rd box from the right:** 2 × 2 = 4. 4. **4th box from the right:** 4 × 2 = 8. Place value: what ONE in this box is worth 4th from right**8** 3rd from right**4** 2nd from right**2** 1st from right**1** ← Double each time you move one box LEFT Start at the RIGHT with 1. Then 1 × 2 = 2, 2 × 2 = 4, and 4 × 2 = 8. We built the values by moving right to left: **1 → 2 → 4 → 8**. Once the boxes are on paper, reading them in the usual left-to-right direction shows **8, 4, 2, 1**. These are the same four places, viewed in the opposite direction. **Position number is not place value.** The third box from the right has position number 3, but its place value is 4. “Third” tells you where the box is; “4” tells you what a 1 in it contributes. If you add a fifth box on the left, its value is 8 × 2 = **16**. The rightmost place stays 1; adding a box on the left does not change the existing places. **Yaad rakho:** Sabse RIGHT se 1 likho. LEFT jaate jaate double karo: 1, 2, 4, 8, 16. Question ke digits baad mein in boxes mein rakho. ## 4. The place value stays fixed; the digit chooses it Keep the digit **1** the same and move it into a different place. Put 0 in every other box. Watch how its position changes the number it represents: | Binary digits | Where the 1 sits | Decimal value | | --- | --- | --- | | 0001 | 1st place from the right: ones | 1 × 1 = 1 | | 0010 | 2nd place from the right: twos | 1 × 2 = 2 | | 0100 | 3rd place from the right: fours | 1 × 4 = 4 | | 1000 | 4th place from the right: eights | 1 × 8 = 8 | In all four rows, the written digit is still 1. Its **position** decides whether that 1 contributes one, two, four or eight. Now put **1 0 1 1** in the four boxes. The values 8, 4, 2 and 1 are labels for the boxes. The written digits tell us which values to include: - **Digit 1:** take one of this place value. - **Digit 0:** take zero of this place value. Place value: what ONE in this box is worth 4th from right**8** 3rd from right**4** 2nd from right**2** 1st from right**1** Digits written in those boxes **1**include **0**skip **1**include **1**include Each box contributes 1 × 8**8** 0 × 4**0** 1 × 2**2** 1 × 1**1** Add the contributions: 8 + 0 + 2 + 1 = 11 in ordinary decimal notation. The 0 above does **not** mean the place value changes from 4 to 0. The place is still worth 4 per unit; the digit says to take zero units of it. Its contribution is therefore 0 × 4 = 0. The three 1s are the same digit, but they sit in different places. They contribute **8, 2 and 1**. Therefore, binary 1011 represents decimal 11. **Yaad rakho:** Box ki value fixed. Andar 1 hai toh value lo; 0 hai toh skip karo. Phir li hui values jodo. ## 5. Binary to decimal: four complete examples **Decode** means convert the binary digits into an ordinary decimal number. In exam notation, a small 2 below a number means binary; a small 10 means decimal. These small labels identify the number system. They are not multipliers. $$ (1011)_2=(11)_{10} $$ This line says: “one-zero-one-one in binary and eleven in decimal represent the same quantity.” **Convert the binary number $$ (1011)_2 $$ to decimal.** [2 marks · Practice] #### Given Binary digits: 1011, in this exact order. #### To find The ordinary decimal number represented by these digits. #### Formula and why it applies Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add. $$ \text{Number}=\text{sum of (digit}\times\text{place value)} $$ #### Solution ##### Step 1. Write the place values above the digits | Place value | 8 | 4 | 2 | 1 | | --- | --- | --- | --- | --- | | Binary digit | 1 | 0 | 1 | 1 | ##### Step 2. Calculate the contribution of the 8-place The digit is 1, so include the value of this place. $$ 1\times 8=8 $$ ##### Step 3. Calculate the contribution of the 4-place The digit is 0, so this place contributes zero. Its place value still stays the same. $$ 0\times 4=0 $$ ##### Step 4. Calculate the contribution of the 2-place The digit is 1, so include the value of this place. $$ 1\times 2=2 $$ ##### Step 5. Calculate the contribution of the 1-place The digit is 1, so include the value of this place. $$ 1\times 1=1 $$ ##### Step 6. Add the contributions $$ 8+0+2+1=11 $$ **Final answer.** $$ (1011)_2=(11)_{10} $$ . **Convert the binary number $$ (0110)_2 $$ to decimal.** [2 marks · Practice] #### Given Binary digits: 0110, in this exact order. #### To find The ordinary decimal number represented by these digits. #### Formula and why it applies Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add. $$ \text{Number}=\text{sum of (digit}\times\text{place value)} $$ #### Solution ##### Step 1. Write the place values above the digits | Place value | 8 | 4 | 2 | 1 | | --- | --- | --- | --- | --- | | Binary digit | 0 | 1 | 1 | 0 | ##### Step 2. Calculate the contribution of the 8-place The digit is 0, so this place contributes zero. Its place value still stays the same. $$ 0\times 8=0 $$ ##### Step 3. Calculate the contribution of the 4-place The digit is 1, so include the value of this place. $$ 1\times 4=4 $$ ##### Step 4. Calculate the contribution of the 2-place The digit is 1, so include the value of this place. $$ 1\times 2=2 $$ ##### Step 5. Calculate the contribution of the 1-place The digit is 0, so this place contributes zero. Its place value still stays the same. $$ 0\times 1=0 $$ ##### Step 6. Add the contributions $$ 0+4+2+0=6 $$ **Final answer.** $$ (0110)_2=(6)_{10} $$ . **Convert the binary number $$ (1001)_2 $$ to decimal.** [2 marks · Practice] #### Given Binary digits: 1001, in this exact order. #### To find The ordinary decimal number represented by these digits. #### Formula and why it applies Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add. $$ \text{Number}=\text{sum of (digit}\times\text{place value)} $$ #### Solution ##### Step 1. Write the place values above the digits | Place value | 8 | 4 | 2 | 1 | | --- | --- | --- | --- | --- | | Binary digit | 1 | 0 | 0 | 1 | ##### Step 2. Calculate the contribution of the 8-place The digit is 1, so include the value of this place. $$ 1\times 8=8 $$ ##### Step 3. Calculate the contribution of the 4-place The digit is 0, so this place contributes zero. Its place value still stays the same. $$ 0\times 4=0 $$ ##### Step 4. Calculate the contribution of the 2-place The digit is 0, so this place contributes zero. Its place value still stays the same. $$ 0\times 2=0 $$ ##### Step 5. Calculate the contribution of the 1-place The digit is 1, so include the value of this place. $$ 1\times 1=1 $$ ##### Step 6. Add the contributions $$ 8+0+0+1=9 $$ **Final answer.** $$ (1001)_2=(9)_{10} $$ . **Convert the binary number $$ (10011)_2 $$ to decimal.** [2 marks · Practice] #### Given Binary digits: 10011, in this exact order. #### To find The ordinary decimal number represented by these digits. #### Formula and why it applies Each digit tells us how many of that place value to include. Multiply each digit by its place value, then add. $$ \text{Number}=\text{sum of (digit}\times\text{place value)} $$ #### Solution ##### Step 1. Write the place values above the digits | Place value | 16 | 8 | 4 | 2 | 1 | | --- | --- | --- | --- | --- | --- | | Binary digit | 1 | 0 | 0 | 1 | 1 | ##### Step 2. Calculate the contribution of the 16-place The digit is 1, so include the value of this place. $$ 1\times 16=16 $$ ##### Step 3. Calculate the contribution of the 8-place The digit is 0, so this place contributes zero. Its place value still stays the same. $$ 0\times 8=0 $$ ##### Step 4. Calculate the contribution of the 4-place The digit is 0, so this place contributes zero. Its place value still stays the same. $$ 0\times 4=0 $$ ##### Step 5. Calculate the contribution of the 2-place The digit is 1, so include the value of this place. $$ 1\times 2=2 $$ ##### Step 6. Calculate the contribution of the 1-place The digit is 1, so include the value of this place. $$ 1\times 1=1 $$ ##### Step 7. Add the contributions $$ 16+0+0+2+1=19 $$ **Final answer.** $$ (10011)_2=(19)_{10} $$ . ## 6. Decimal to binary: choose each digit **Encode** means write an ordinary number using binary digits. The direction of the task is now reversed: the number is known, and you have to choose the 0s and 1s. For four bits, write the places 8, 4, 2, 1. Start with the largest place on the left. Use it only if it does not exceed the amount still needed. Whenever you use a place, subtract it before moving to the next one. **Yaad rakho:** Number banana hai. Badi value se start: jitna bacha hai usmein value fit ho toh 1 aur subtract; fit nahi ho toh 0. **Write the decimal number 6 as an unsigned 4-bit binary number.** [2 marks · Practice] #### Given Decimal number: 6. Required length: 4 bits, meaning exactly 4 digit boxes. #### To find The binary digits, including any zero at the beginning needed to keep the required length. #### Formula and why it applies At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged. This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number. #### Solution ##### Step 1. Make 4 boxes and label their values | Place value | 8 | 4 | 2 | 1 | | --- | --- | --- | --- | --- | | Digit to choose | □ | □ | □ | □ | Start with the whole number 6 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT. ##### Step 2. Choose the digit in the 8-place We still need 6. Taking 8 would be too much, so skip this place. Write **0** in this box. We used nothing from this place. The remaining amount stays **6**. ##### Step 3. Choose the digit in the 4-place We still need 6. This is at least 4, so take one 4. Write **1** in this box. $$ \text{Remaining}=6-4=2 $$ ##### Step 4. Choose the digit in the 2-place We still need 2. This is at least 2, so take one 2. Write **1** in this box. $$ \text{Remaining}=2-2=0 $$ ##### Step 5. Choose the digit in the 1-place We still need 0. Taking 1 would be too much, so skip this place. Write **0** in this box. We used nothing from this place. The remaining amount stays **0**. ##### Step 6. Read the digits from LEFT to RIGHT The boxes now contain **0 1 1 0**. Join them in the same order: **0110**. The remaining amount is 0, so we have made the required number exactly. ##### Step 7. Check by converting back $$ 0\times 8+1\times 4+1\times 2+0\times 1=6 $$ **Final answer.** $$ (6)_{10}=(0110)_2 $$ . The answer has 4 bits. **Write the decimal number 11 as an unsigned 4-bit binary number.** [2 marks · Practice] #### Given Decimal number: 11. Required length: 4 bits, meaning exactly 4 digit boxes. #### To find The binary digits, including any zero at the beginning needed to keep the required length. #### Formula and why it applies At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged. This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number. #### Solution ##### Step 1. Make 4 boxes and label their values | Place value | 8 | 4 | 2 | 1 | | --- | --- | --- | --- | --- | | Digit to choose | □ | □ | □ | □ | Start with the whole number 11 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT. ##### Step 2. Choose the digit in the 8-place We still need 11. This is at least 8, so take one 8. Write **1** in this box. $$ \text{Remaining}=11-8=3 $$ ##### Step 3. Choose the digit in the 4-place We still need 3. Taking 4 would be too much, so skip this place. Write **0** in this box. We used nothing from this place. The remaining amount stays **3**. ##### Step 4. Choose the digit in the 2-place We still need 3. This is at least 2, so take one 2. Write **1** in this box. $$ \text{Remaining}=3-2=1 $$ ##### Step 5. Choose the digit in the 1-place We still need 1. This is at least 1, so take one 1. Write **1** in this box. $$ \text{Remaining}=1-1=0 $$ ##### Step 6. Read the digits from LEFT to RIGHT The boxes now contain **1 0 1 1**. Join them in the same order: **1011**. The remaining amount is 0, so we have made the required number exactly. ##### Step 7. Check by converting back $$ 1\times 8+0\times 4+1\times 2+1\times 1=11 $$ **Final answer.** $$ (11)_{10}=(1011)_2 $$ . The answer has 4 bits. **Write the decimal number 19 as an unsigned 5-bit binary number.** [2 marks · Practice] #### Given Decimal number: 19. Required length: 5 bits, meaning exactly 5 digit boxes. #### To find The binary digits, including any zero at the beginning needed to keep the required length. #### Formula and why it applies At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged. This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number. #### Solution ##### Step 1. Make 5 boxes and label their values | Place value | 16 | 8 | 4 | 2 | 1 | | --- | --- | --- | --- | --- | --- | | Digit to choose | □ | □ | □ | □ | □ | Start with the whole number 19 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT. ##### Step 2. Choose the digit in the 16-place We still need 19. This is at least 16, so take one 16. Write **1** in this box. $$ \text{Remaining}=19-16=3 $$ ##### Step 3. Choose the digit in the 8-place We still need 3. Taking 8 would be too much, so skip this place. Write **0** in this box. We used nothing from this place. The remaining amount stays **3**. ##### Step 4. Choose the digit in the 4-place We still need 3. Taking 4 would be too much, so skip this place. Write **0** in this box. We used nothing from this place. The remaining amount stays **3**. ##### Step 5. Choose the digit in the 2-place We still need 3. This is at least 2, so take one 2. Write **1** in this box. $$ \text{Remaining}=3-2=1 $$ ##### Step 6. Choose the digit in the 1-place We still need 1. This is at least 1, so take one 1. Write **1** in this box. $$ \text{Remaining}=1-1=0 $$ ##### Step 7. Read the digits from LEFT to RIGHT The boxes now contain **1 0 0 1 1**. Join them in the same order: **10011**. The remaining amount is 0, so we have made the required number exactly. ##### Step 8. Check by converting back $$ 1\times 16+0\times 8+0\times 4+1\times 2+1\times 1=19 $$ **Final answer.** $$ (19)_{10}=(10011)_2 $$ . The answer has 5 bits. **Write the decimal number 0 as an unsigned 4-bit binary number.** [2 marks · Practice] #### Given Decimal number: 0. Required length: 4 bits, meaning exactly 4 digit boxes. #### To find The binary digits, including any zero at the beginning needed to keep the required length. #### Formula and why it applies At each place, compare its value with the amount still needed. If that amount is at least the place value, write 1 and subtract that value. Otherwise write 0 and keep the remaining amount unchanged. This works because each place can be used either once (1) or not at all (0). The selected places must add up to the given number. #### Solution ##### Step 1. Make 4 boxes and label their values | Place value | 8 | 4 | 2 | 1 | | --- | --- | --- | --- | --- | | Digit to choose | □ | □ | □ | □ | Start with the whole number 0 still to make. Work from the largest place on the LEFT to the smallest on the RIGHT. ##### Step 2. Choose the digit in the 8-place We still need 0. Taking 8 would be too much, so skip this place. Write **0** in this box. We used nothing from this place. The remaining amount stays **0**. ##### Step 3. Choose the digit in the 4-place We still need 0. Taking 4 would be too much, so skip this place. Write **0** in this box. We used nothing from this place. The remaining amount stays **0**. ##### Step 4. Choose the digit in the 2-place We still need 0. Taking 2 would be too much, so skip this place. Write **0** in this box. We used nothing from this place. The remaining amount stays **0**. ##### Step 5. Choose the digit in the 1-place We still need 0. Taking 1 would be too much, so skip this place. Write **0** in this box. We used nothing from this place. The remaining amount stays **0**. ##### Step 6. Read the digits from LEFT to RIGHT The boxes now contain **0 0 0 0**. Join them in the same order: **0000**. The remaining amount is 0, so we have made the required number exactly. ##### Step 7. Check by converting back $$ 0\times 8+0\times 4+0\times 2+0\times 1=0 $$ **Final answer.** $$ (0)_{10}=(0000)_2 $$ . The answer has 4 bits. ## 7. A zero at the start can keep the required length **Leading zeros** are zeros at the beginning, on the left. Binary 110 and 0110 both represent decimal 6: $$ (110)_2=(1\times4)+(1\times2)+(0\times1)=6 $$ $$ (0110)_2=(0\times8)+(1\times4)+(1\times2)+(0\times1)=6 $$ The added leftmost place contributes zero. It changes the length from three bits to four bits, but not the value. If a GA question requires four bits, write **0110**. This rule is specifically about zeros on the **left**. Appending a zero on the right shifts the existing digits into places worth twice as much. For example, binary 1100 is 8 + 4 = 12, not 6. ## 8. How many bits are enough? For these GA conversions, **unsigned** means the bits represent zero or a positive whole number, with no negative sign. The smallest 4-bit value is 0000. The largest is 1111 because all four places are included. **Find the largest unsigned integer that can be represented using 4 bits. Can decimal 19 fit in 4 bits?** [2 marks · Practice] #### Given Four places with values 8, 4, 2 and 1. Each digit is either 0 or 1. #### To find The largest possible value and whether 19 can be represented. #### Formula and why it applies A digit contributes the most when it is 1. Set every digit to 1 and add all the place values. #### Solution ##### Step 1. Use all four places The largest 4-bit string is **1111**. ##### Step 2. Add their values $$ 8+4+2+1=15 $$ ##### Step 3. Compare with 19 19 is greater than 15, so four bits are insufficient. Add the 16-place on the left. Then 19 = 16 + 2 + 1, giving **10011** in five bits. **Final answer.** The maximum is 15. Decimal 19 needs at least 5 bits. The general rule is $$ \text{largest unsigned value}=2^n-1 $$ . Here n means the number of bits. The power $$ 2^n $$ means multiplying n copies of 2; for four bits, $$ 2^4=2\times2\times2\times2=16 $$ , so the maximum is 16 − 1 = 15. There are 16 different values, from 0 through 15, because zero is also one value. | Bits | Place values from left to right | Unsigned values | | --- | --- | --- | | 3 | 4, 2, 1 | 0 to 7 | | 4 | 8, 4, 2, 1 | 0 to 15 | | 5 | 16, 8, 4, 2, 1 | 0 to 31 | If an exam gives a number outside the capacity of the stated bit length, state that inconsistency. Do not silently cut off a digit: removing the 16-place from 10011 would change 19 into 3. Also retain any separate allowed range stated in the question. ## 9. Where this fits in a genetic algorithm A **chromosome** is the representation of one candidate solution in a genetic algorithm. With the direct unsigned binary encoding used here, decimal 6 is represented by the four-bit chromosome **0110**. During crossover or mutation, the algorithm changes these bits. Afterwards, **decode the resulting chromosome** to find its new value of x. Then use that x in the given objective or fitness function. The binary string, its decimal value, and its fitness score are different things. **Yaad rakho:** Encode: number se bits. Decode: bits se number. GA mein bits badle toh x dobara decode karo, phir uska score nikalo. --- ## M2 · Vectors, coordinates, matrices and transpose from the first entry Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--coordinates/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/core-maths/coordinates Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO. ## A vector is an ordered list of numbers A **scalar** is one number, such as 5. A **vector** is an ordered list, such as (3,7). Order matters: the first number may be horizontal position and the second vertical position. A **matrix** is a rectangular table of numbers. These words describe how numbers are arranged, not a new kind of arithmetic. | Arrangement | Example | How to read it | | --- | --- | --- | | Scalar | 5 | One value | | Row vector | [3,7] | One row, two entries | | Column vector | $$ \begin{bmatrix}3\\7\end{bmatrix} $$ | Two rows, one entry | | Matrix | $$ \begin{bmatrix}3&7\\4&2\end{bmatrix} $$ | Two rows, two columns | ## Rows go across; columns go down For a matrix A, **A₂₁** means the entry in **row 2, column 1**. The first lower label is the row; the second is the column. Matrix size is written **number of rows × number of columns**. A 2×3 matrix has two rows containing three entries each. Swipe across the diagram to read its labels ↔ ![Read an entry by finding its row first and its column second.](https://deep-learning-study-room.vercel.app/is2/fig/depth_matrix.a8870f4ab6.svg) Diagram: Read an entry by finding its row first and its column second.. Labels: Rows across. Columns down. Entry A₂₃ = 7. | Column 1 | Column 2 | Column 3 | Row 1 | 2 | 5 | 8 | Row 2 | 1 | 4 | 7 | Two rows × three columns = 2 × 3 matrix Read an entry by finding its row first and its column second. **Let A have first row [2,5,8] and second row [1,4,7]. State its size and find A₂₃ and A₁₂.** [2 marks · Worked example] #### Given $$ A=\begin{bmatrix}2&5&8\\1&4&7\end{bmatrix} $$ #### To find Matrix dimensions and two entries. #### Formula and why it applies Read the lower labels in row–column order. #### Solution ##### Step 1. Count the arrangement There are 2 horizontal rows and 3 vertical columns. Size=2×3. ##### Step 2. Find A₂₃ Row 2 is [1,4,7]. Its third entry is 7. ##### Step 3. Find A₁₂ Row 1 is [2,5,8]. Its second entry is 5. **Final answer.** Size 2×3; A₂₃=7; A₁₂=5. **A particle has position x=(3,7), velocity v=(−1,2), and best position p=(1,8). Find p−x, half of that difference, and x+v.** [2 marks · Worked example] #### Given Position x=(3,7); velocity v=(−1,2); personal-best position p=(1,8). First and second coordinates stay in the same order. #### To find The difference p−x, the scaled difference 0.5(p−x), and the updated position x+v. #### Formula and why it applies Match first with first and second with second. A scalar multiplies every coordinate. #### Solution ##### Step 1. Subtract corresponding coordinates $$ p-x=(1-3,\;8-7)=(-2,1) $$ ##### Step 2. Multiply both coordinates by 0.5 $$ 0.5(p-x)=(0.5(-2),\;0.5(1))=(-1,0.5) $$ ##### Step 3. Add the velocity to the position $$ x+v=(3+(-1),\;7+2)=(2,9) $$ **Final answer.** Difference (−2,1); scaled difference (−1,0.5); new position (2,9). ## Transpose exchanges rows and columns The superscript **T** in Aᵀ means **transpose**. It does not mean multiply or raise A to a power. Row 1 of A becomes column 1 of Aᵀ; row 2 becomes column 2. An m×n arrangement becomes n×m. **Find the transpose of A = [[2,5,8],[1,4,7]].** [2 marks · Worked example] #### Given $$ A=\begin{bmatrix}2&5&8\\1&4&7\end{bmatrix} $$ #### To find Aᵀ. #### Formula and why it applies Turn each old row into a new column, keeping its order. #### Solution ##### Step 1. Move the first row [2,5,8] becomes the first column from top to bottom. ##### Step 2. Move the second row [1,4,7] becomes the second column from top to bottom. ##### Step 3. Write the new arrangement $$ A^T=\begin{bmatrix}2&1\\5&4\\8&7\end{bmatrix} $$ **Final answer.** Aᵀ has 3 rows and 2 columns. No number changed. **Calculate the dot product of [2,3] and [4,5].** [2 marks · Worked example] #### Given First vector a=[2,3]; second vector b=[4,5]. Each has two entries. #### To find The dot product a·b, a single number obtained by adding the two matching products. #### Formula and why it applies $$ a\cdot b=a_1b_1+a_2b_2 $$ Multiply corresponding entries, then add the products. Markov matrix multiplication later repeats this for each destination column. #### Solution ##### Step 1. First product $$ 2\times4=8 $$ ##### Step 2. Second product $$ 3\times5=15 $$ ##### Step 3. Add $$ 8+15=23 $$ **Final answer.** The dot product is 23. It is not the vector [8,15]. **Yaad rakho:** Vector = order wali list. Matrix = rows-columns ka table. Entry padhte waqt pehle row, phir column. --- ## M2 · Graph axes, intervals and the fractions inside membership functions Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--graphs/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/core-maths/graphs Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO. ## Read a graph one input at a time The **horizontal axis** runs left to right and shows input x. The **vertical axis** runs bottom to top and shows the output. A point (3,0.4) means: input 3 gives output 0.4. In a membership graph, the vertical output is a degree between 0 and 1. To read the graph at x=3, start at 3 on the bottom axis, move vertically until you reach the line, then move horizontally to the vertical axis. The height you read is the membership. Swipe across the diagram to read its labels ↔ ![Halfway along a straight rise from height 0 to height 1 gives height 0.5.](https://deep-learning-study-room.vercel.app/is2/fig/depth_slope.c91a1030aa.svg) Diagram: Halfway along a straight rise from height 0 to height 1 gives height 0.5.. Labels: Read x = 4: halfway from 2 to 6, so height = 0.5 | 0 | 0.25 | 0.5 | 1 | 0 | 2 | 3 | 4 | 6 | 8 | μ | Input x | Straight rising membership | x = 4 Halfway along a straight rise from height 0 to height 1 gives height 0.5. | Read this | Meaning | Concrete example | | --- | --- | --- | | x < 5 | Strictly less than 5 | 4.9 qualifies; 5 does not | | x ≤ 5 | Less than or equal to 5 | Both 4.9 and 5 qualify | | 2 ≤ x ≤ 8 | x is between 2 and 8 including both ends | The interval is [2,8] | | [2,8) | Include 2; exclude 8 | Useful for roulette interval boundaries | | Piecewise formula | Choose one calculation according to where x lies | Rising branch, falling branch or outside | **A straight membership line rises from (2,0) to (6,1). Find its height at x=3.** [2 marks · Worked example] #### Given Left input=2; peak input=6; required input=3. #### To find Membership height. #### Formula and why it applies $$ \text{height}=\frac{\text{distance from left}}{\text{total rising width}} $$ A straight line changes by the same proportion of its height as the proportion travelled along its width. #### Solution ##### Step 1. Find distance travelled $$ 3-2=1 $$ ##### Step 2. Find total width $$ 6-2=4 $$ ##### Step 3. Divide distance by width $$ \mu(3)=\frac{1}{4}=0.25 $$ **Final answer.** Membership=0.25. **A straight membership line rises from (2,0) to (6,1). Find its height at x=4.** [2 marks · Worked example] #### Given Left input=2; peak input=6; required input=4. #### To find Membership height. #### Formula and why it applies $$ \text{height}=\frac{\text{distance from left}}{\text{total rising width}} $$ A straight line changes by the same proportion of its height as the proportion travelled along its width. #### Solution ##### Step 1. Find distance travelled $$ 4-2=2 $$ ##### Step 2. Find total width $$ 6-2=4 $$ ##### Step 3. Divide distance by width $$ \mu(4)=\frac{2}{4}=0.5 $$ **Final answer.** Membership=0.5. **A straight membership line rises from (10,0) to (30,1). Find its height at x=25.** [2 marks · Worked example] #### Given Left input=10; peak input=30; required input=25. #### To find Membership height. #### Formula and why it applies $$ \text{height}=\frac{\text{distance from left}}{\text{total rising width}} $$ A straight line changes by the same proportion of its height as the proportion travelled along its width. #### Solution ##### Step 1. Find distance travelled $$ 25-10=15 $$ ##### Step 2. Find total width $$ 30-10=20 $$ ##### Step 3. Divide distance by width $$ \mu(25)=\frac{15}{20}=0.75 $$ **Final answer.** Membership=0.75. **A membership line falls from (5,1) to (8,0). Find its height at x=7.** [2 marks · Worked example] #### Given Peak=5; right zero=8; x=7. #### To find Membership. #### Formula and why it applies $$ \mu(x)=\frac{\text{distance still remaining to right zero}}{\text{total falling width}} $$ #### Solution ##### Step 1. Remaining distance $$ 8-7=1 $$ ##### Step 2. Total falling width $$ 8-5=3 $$ ##### Step 3. Divide $$ \mu(7)=1/3\approx0.333333 $$ **Final answer.** Membership≈0.333333. Using distance already travelled would reverse the falling line. **Yaad rakho:** Chadhti line: left se kitna aage? Utarti line: right tak kitna baaki? Dono mein poori width se divide. --- ## M2 · Sum notation, ordinary averages and weighted averages Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--sums-averages/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/core-maths/sums-averages Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO. ## Σ means add a list of terms The capital Greek letter **Σ** is read “sum”. The lower label says where to start and the upper label where to stop. The small index i is just a counter: first item, second item, third item. $$ \sum_{i=1}^{3}a_i=a_1+a_2+a_3 $$ **For a₁=2, a₂=5 and a₃=8, calculate Σaᵢ and their mean.** [2 marks · Worked example] #### Given Three numbers: 2,5,8. #### To find Sum and ordinary average. #### Formula and why it applies $$ \bar a=\frac{a_1+a_2+a_3}{3} $$ The bar means average. Each number has equal weight. #### Solution ##### Step 1. Expand the sum symbol $$ \sum_{i=1}^3 a_i=2+5+8 $$ ##### Step 2. Add $$ 2+5=7\quad7+8=15 $$ ##### Step 3. Divide by the count of numbers $$ \bar a=15/3=5 $$ **Final answer.** Sum=15; mean=5. ## A weighted average gives unequal influence If values 20 and 80 have weights 3 and 1, this is equivalent to averaging [20,20,20,80]. Therefore you divide by the **sum of weights**, not by the number of distinct values. Weights may also be decimals. Their total need not initially be 1. $$ \bar z=\frac{w_1z_1+w_2z_2+\cdots}{w_1+w_2+\cdots} $$ **Find the weighted average of [20, 80] with corresponding weights [3, 1].** [2 marks · Worked example] #### Given | Value | Its weight | | --- | --- | | 20 | 3 | | 80 | 1 | #### To find The weighted average of the listed values. #### Formula and why it applies $$ \bar z=\frac{\sum_i w_i z_i}{\sum_i w_i} $$ zᵢ is a value and wᵢ is its corresponding weight. Multiply each value by its own weight, add, then divide by total weight. Here all weights are nonnegative and their total is positive. #### Solution ##### Step 1. Multiply value 1 by its weight $$ 20\times3=60 $$ ##### Step 2. Multiply value 2 by its weight $$ 80\times1=80 $$ ##### Step 3. Add the weighted products $$ 60+80=140 $$ ##### Step 4. Add the weights $$ 3+1=4 $$ ##### Step 5. Divide the two totals $$ \bar z=\frac{140}{4}=35 $$ **Final answer.** 35. **Find the weighted average of [20, 50, 80] with corresponding weights [0.2, 0.6, 0.2].** [2 marks · Worked example] #### Given | Value | Its weight | | --- | --- | | 20 | 0.2 | | 50 | 0.6 | | 80 | 0.2 | #### To find The weighted average of the listed values. #### Formula and why it applies $$ \bar z=\frac{\sum_i w_i z_i}{\sum_i w_i} $$ zᵢ is a value and wᵢ is its corresponding weight. Multiply each value by its own weight, add, then divide by total weight. Here all weights are nonnegative and their total is positive. #### Solution ##### Step 1. Multiply value 1 by its weight $$ 20\times0.2=4 $$ ##### Step 2. Multiply value 2 by its weight $$ 50\times0.6=30 $$ ##### Step 3. Multiply value 3 by its weight $$ 80\times0.2=16 $$ ##### Step 4. Add the weighted products $$ 4+30+16=50 $$ ##### Step 5. Add the weights $$ 0.2+0.6+0.2=1 $$ ##### Step 6. Divide the two totals $$ \bar z=\frac{50}{1}=50 $$ **Final answer.** 50. **Find the weighted average of [65, 85] with corresponding weights [0.3, 0.5].** [2 marks · Worked example] #### Given | Value | Its weight | | --- | --- | | 65 | 0.3 | | 85 | 0.5 | #### To find The weighted average of the listed values. #### Formula and why it applies $$ \bar z=\frac{\sum_i w_i z_i}{\sum_i w_i} $$ zᵢ is a value and wᵢ is its corresponding weight. Multiply each value by its own weight, add, then divide by total weight. Here all weights are nonnegative and their total is positive. #### Solution ##### Step 1. Multiply value 1 by its weight $$ 65\times0.3=19.5 $$ ##### Step 2. Multiply value 2 by its weight $$ 85\times0.5=42.5 $$ ##### Step 3. Add the weighted products $$ 19.5+42.5=62 $$ ##### Step 4. Add the weights $$ 0.3+0.5=0.8 $$ ##### Step 5. Divide the two totals $$ \bar z=\frac{62}{0.8}=77.5 $$ **Final answer.** 77.5. ## Π means multiply a list of terms $$ \prod_{i=1}^{3}a_i=a_1a_2a_3 $$ For [0.5,0.2,0.4], first 0.5×0.2=0.1, then 0.1×0.4=0.04. Σ would instead give 1.1. Naive Bayes multiplies likelihood factors; it does not add them. **Yaad rakho:** Sigma = sab jodo. Product Pi = sab multiply. Weighted mean = value×weight ke total ko weights ke total se divide. --- ## M2 · Powers, roots, absolute values and exp on the calculator Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--powers-exponentials/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/core-maths/powers-exponentials Short prerequisites, with worked examples. Read once before Fuzzy → GA → PSO. ## Powers, roots and exponentials used later A positive integer power repeats multiplication. 2³=2×2×2=8; it does not mean 2×3. A nonzero number raised to power 0 equals 1. A negative power means a reciprocal: 2⁻²=1/(2²)=1/4. A square root reverses a square for a nonnegative result: √9=3. | Read this | Meaning | Concrete example | | --- | --- | --- | | 2⁰, 2¹, 2², 2³ | Binary place values | 1,2,4,8 | | |x| | Absolute value: distance from zero | |−3|=3 and |3|=3 | | e ≈ 2.71828 | A fixed mathematical constant | e⁰=1 | | exp(u) | Another way of writing eᵘ | exp(−2)=e⁻²≈0.135335 | | ln(x) | The natural logarithm; the exponent giving x from e | ln(e²)=2; ln(1)=0 | **Calculate exp(−x²/2) at x=-2.** [2 marks · Worked example] #### Given x=-2; exp means e raised to the bracketed exponent. #### To find The exponential value. #### Formula and why it applies Calculate the complete exponent first. Then use the calculator’s e-to-the-power function. #### Solution ##### Step 1. Square x $$ (-2)^2=4 $$ ##### Step 2. Negate and divide by 2 $$ u=-\frac{4}{2}=-2 $$ ##### Step 3. Evaluate the exponential $$ e^u=e^{-2}\approx 0.135335283 $$ **Final answer.** 0.135335283. **Calculate exp(−x²/2) at x=0.** [2 marks · Worked example] #### Given x=0; exp means e raised to the bracketed exponent. #### To find The exponential value. #### Formula and why it applies Calculate the complete exponent first. Then use the calculator’s e-to-the-power function. #### Solution ##### Step 1. Square x $$ (0)^2=0 $$ ##### Step 2. Negate and divide by 2 $$ u=-\frac{0}{2}=0 $$ ##### Step 3. Evaluate the exponential $$ e^u=e^{0}\approx 1 $$ **Final answer.** 1. **Calculate exp(−x²/2) at x=2.** [2 marks · Worked example] #### Given x=2; exp means e raised to the bracketed exponent. #### To find The exponential value. #### Formula and why it applies Calculate the complete exponent first. Then use the calculator’s e-to-the-power function. #### Solution ##### Step 1. Square x $$ (2)^2=4 $$ ##### Step 2. Negate and divide by 2 $$ u=-\frac{4}{2}=-2 $$ ##### Step 3. Evaluate the exponential $$ e^u=e^{-2}\approx 0.135335283 $$ **Final answer.** 0.135335283. On an fx-991ES-style calculator, eˣ is normally the shifted function above ln. Enter the complete exponent in brackets. Use the negative-number key for −2 where required. Confirm against e⁰=1 before a longer calculation. The exact key labels depend on the model. **Yaad rakho:** exp ka matlab e ki power. Pehle exponent ka poora hisaab, phir eˣ. --- ## M2 · Fuzzy sets: what membership actually means Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--membership-meaning/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/membership-meaning Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Start with an input and a label Suppose the input is **temperature 32°C** and the fuzzy label is **Hot**. A membership function is the rule that converts that temperature into a degree of matching the label. If μHot(32)=0.7, read it aloud as “the membership of 32 degrees in Hot is 0.7”. The lower word Hot names the set; 32 is the input; 0.7 is the output. The same temperature can have membership 0.3 in Medium and 0.7 in Hot. These are two descriptions of one input. Fuzzy memberships do not generally have to sum to 1. By contrast, probabilities of mutually exclusive, exhaustive outcomes do. | Read this | Meaning | Concrete example | | --- | --- | --- | | Universe U | All input values being considered | Temperatures from 0 to 40°C | | Crisp value | One definite measured number | 32°C | | Linguistic term | A word naming a fuzzy category | Low, Medium, High | | μA(x) | Degree to which input x fits category A | μHot(32)=0.7 | **Yaad rakho:** 32 temperature hai; 0.7 us temperature ka Hot label mein degree hai. Dono ko mix mat karo. ## Why “true or false” is sometimes too rigid Suppose a controller says a room is hot only when its temperature is at least 30°C. At 29.9°C it says “not hot”; at 30°C it says “hot”. That sharp boundary is a **crisp** rule. Fuzzy logic allows a gradual degree of belonging to the category “Hot”. **Definition.** A fuzzy set A on a universe U assigns every element x a membership degree μA(x) between 0 and 1. The symbol μ is the Greek letter “mu”; read μA(x) as “the degree to which x belongs to set A”. $$ \mu_A:U\rightarrow[0,1] $$ | Term | Meaning | Temperature example | | --- | --- | --- | | Universe U | All allowed input values | Temperatures from 0°C to 40°C | | Element x | The particular input being considered | 32°C | | Fuzzy set A | A category described by a membership function | Hot | | Membership μA(x) | How strongly x fits that category | μHot(32) = 0.7 | ## Read membership correctly Membership 0 means no belonging; 1 means full belonging; 0.7 means partial belonging under the chosen definition. **It does not mean a 70% chance that the measured temperature is 32°C.** Nor is it a 70% chance that the sensor is correct. Fuzziness concerns the gradual meaning of “Hot”; probability concerns uncertainty about which event occurs. | Input | Warm membership | Hot membership | Interpretation | | --- | --- | --- | --- | | 28 | 0.8 | 0.1 | Fits Warm strongly and Hot weakly | | 32 | 0.3 | 0.7 | Fits Hot more than Warm | | 35 | 0 | 1 | Fully Hot in this chosen model | These memberships need not add to 1. Overlapping categories are allowed. A crisp set is the special case in which membership can only be 0 or 1. A crisp set can also contain infinitely many elements; “crisp” does not mean “finite”. **On U = {20, 25, 30, 35}, the Hot memberships are [0, 0.2, 0.6, 1]. Write the fuzzy set Hot and identify the fully and partly belonging inputs.** [2 marks · Practice] #### Given Ordered inputs [20, 25, 30, 35]; memberships [0, 0.2, 0.6, 1]. #### To find The set notation and interpretation. #### Formula and why it applies In the common fuzzy-set notation μ/x, the slash separates membership from element; it is not division. #### Solution ##### Step 1. Pair each input with its membership | x | μHot(x) | | --- | --- | | 20 | 0 | | 25 | 0.2 | | 30 | 0.6 | | 35 | 1 | ##### Step 2. Write the set $$ Hot=\{0/20,\;0.2/25,\;0.6/30,\;1/35\} $$ ##### Step 3. Interpret 35 fully belongs; 25 and 30 partly belong; 20 does not belong. **Final answer.** Hot = {0/20, 0.2/25, 0.6/30, 1/35}. **Distinguish crisp and fuzzy sets.** [2 marks · Practice] #### Answer | Crisp set | Fuzzy set | | --- | --- | | Membership is 0 or 1 | Membership is any value in [0,1] | | A sharp boundary separates members | A membership function permits gradual boundaries | | Example: temperature ≥ 30°C | Example: degree of being Hot increases gradually | **Yaad rakho:** μ ka matlab: is category mein kitna fit hota hai. Chance nahi. --- ## M2 · Triangular membership: choose the side, then calculate Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--triangle/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/triangle Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Why the triangle has two different fractions Tri(2,5,8) means membership starts at zero at x=2, rises to 1 at x=5, then falls to zero at x=8. The first three numbers are **input positions**, not three membership degrees. | Where is x? | Meaning | Calculation | | --- | --- | --- | | x≤2 or x≥8 | Outside, or at a zero-height endpoint | 0 | | 20; core: degree=1; boundary: 0 0. **Core:** inputs with μ = 1. **Boundary:** inputs with 0 < μ < 1. **Height:** largest membership (more generally the supremum). A set is normal when its height is 1; it is subnormal when its height is below 1. A crossover point has membership 0.5. An α-cut contains inputs with μ ≥ α. For Tri(2,5,8), support is (2,8), core is {5}, height is 1, and the 0.5-cut is [3.5,6.5]. Solve (x−2)/3=0.5 → x=3.5 and (8−x)/3=0.5 → x=6.5. A one-dimensional fuzzy set is **convex** if every α-cut is an interval (or empty). The membership between two inputs must not dip below the smaller endpoint membership. A single triangle is convex; two separated peaks with a lower valley may not be. **Yaad rakho:** Fuzzy A aur NOT A dono partial ho sakte hain. Isliye crisp waali 1 aur 0 identities blindly nahi lagani. --- ## M2 · Mamdani architecture: what every block does Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--mamdani-architecture/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/mamdani-architecture Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Follow one measurement through the controller A **controller** turns measurements into an action setting. Consider temperature 32°C and occupancy 7 people. The input numbers first become membership degrees. The rules then combine those degrees. Only the final defuzzification returns one fan-speed number. | Stage | What goes in | Operation | What comes out | | --- | --- | --- | --- | | Fuzzifier | 32°C and 7 people | Use the supplied membership graphs | Temperature Medium=.3, High=.7; occupancy High=.5 | | Knowledge base | Chosen labels and their shapes | Store graphs and IF–THEN rules | A reusable definition of how the controller works | | Inference | Medium=.3 AND High=.5 | Rule strength=min(.3,.5)=.3 | “Fan High” fires at .3 | | Aggregation | All clipped output sets | Take the largest output degree at each speed | One combined fuzzy output shape | | Defuzzifier | Combined shape | Apply the specified MoM, centroid or representative method | One numerical speed | The **antecedent** is the IF part (“temperature is Medium AND occupancy is High”). The **consequent** is the THEN part (“fan speed is High”). A rule strength of 0.3 is a membership height; it does not mean the fan runs at 30%. ## Definition: a fuzzy inference system A **fuzzy inference system (FIS)** maps inputs to an output by evaluating fuzzy IF–THEN rules. A **Mamdani FIS** uses fuzzy sets in the rule consequents, combines their inferred output sets, and defuzzifies the result. Crisp inputs32°C; 7 people → Fuzzificationdegrees of Warm/Hot etc. → Rules + inferencerule strengths and clipped outputs → Aggregationone combined output shape → Defuzzificationone fan-speed number ## Architecture: what each block contributes | Block | What it stores or calculates | Concrete meaning | | --- | --- | --- | | Fuzzifier | Membership values of the measured inputs | 32°C may be Medium to degree 0.3 and High to degree 0.7 | | Knowledge base: database | Ranges and membership functions | Medium = Tri(15,25,35) | | Knowledge base: rule base | IF–THEN rules | IF temperature High AND occupancy High THEN fan Very High | | Inference engine | Combines conditions and applies consequents | min for AND; clips the output triangle at the rule strength | | Aggregator | Combines all rule outputs | At every output location, take the maximum clipped membership | | Defuzzifier | Converts the aggregate to one crisp value | Use centroid or mean of maxima, as requested | ## Parts of an IF–THEN rule In “IF temperature is High AND occupancy is High THEN fan speed is Very High”, the IF part is the **antecedent** (condition), and the THEN part is the **consequent** (output category). The input variable temperature, the label High, and the numerical membership μHigh(32) are different things. ## Why a rule strength is not yet the fan speed A strength such as 0.3 says how strongly a rule fires. It is on the membership scale 0–1. Fan speed might be on a 0–100% scale. Defuzzification is the step that produces a number on that output scale. **Explain the architecture of a Mamdani fuzzy inference system.** [3 marks · Practice] #### Answer A Mamdani FIS contains a fuzzifier, knowledge base, inference engine and defuzzifier. The database specifies membership functions and the rule base stores IF–THEN rules. Crisp inputs are fuzzified. The inference engine calculates rule firing strengths and constructs output fuzzy sets. These sets are aggregated, then defuzzified into a crisp action. For a fan controller, temperature and occupancy are inputs and fan speed is the output. **Yaad rakho:** Input ka degree → rule ka strength → output ka shape → final number. --- ## M2 · Rule firing, clipping and aggregation: four solved cases Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--rules-clipping/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/rules-clipping Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Clipping means limiting a height A rule firing at strength 0.3 cannot give its consequent a membership above 0.3. At every candidate speed z, keep the smaller of 0.3 and the original output-set membership. The speed axis is unchanged. **At speeds 40, 60 and 80, output set A has memberships [0.1,0.8,0.2] and set B has [0,0.4,1]. Rule A fires at 0.3 and rule B at 0.5. Clip and max-aggregate the sets.** [2 marks · Worked example] #### Given | Speed | Original A membership | Original B membership | | --- | --- | --- | | 40 | 0.1 | 0 | | 60 | 0.8 | 0.4 | | 80 | 0.2 | 1 | Rule A strength=0.3; rule B strength=0.5. Compare memberships at the same speed. #### To find The combined membership at each speed. #### Formula and why it applies $$ \mu_{out}(z)=\max(\min(0.3,\mu_A(z)),\min(0.5,\mu_B(z))) $$ Inner min limits a rule’s height; outer max keeps the stronger support at that speed. #### Solution ##### Step 1. Clip A At 40: min(.3,.1)=.1. At 60: min(.3,.8)=.3. At 80: min(.3,.2)=.2. ##### Step 2. Clip B At 40: min(.5,0)=0. At 60: min(.5,.4)=.4. At 80: min(.5,1)=.5. ##### Step 3. Aggregate at each speed | Speed | Clipped A | Clipped B | max(A,B) | | --- | --- | --- | --- | | 40 | 0.1 | 0 | 0.1 | | 60 | 0.3 | 0.4 | 0.4 | | 80 | 0.2 | 0.5 | 0.5 | **Final answer.** Combined degrees [0.1,0.4,0.5] at speeds [40,60,80]. A separate defuzzification step is still required. **Yaad rakho:** Clip = height ki ceiling. Aggregate = har same x par sabse upar wala height. ## Four different calculations happen in inference **AND/OR** combine conditions within a rule. **Implication** uses the resulting strength to modify that rule’s output membership function. **Aggregation** merges different rule outputs. Here we use min-AND, max-OR, min-implication and max-aggregation. $$ \alpha_{AND}=\min(a,b)\qquad\alpha_{OR}=\max(a,b) $$ $$ \mu_{\mathrm{rule}}(z)=\min(\alpha,\mu_{\mathrm{consequent}}(z)) $$ $$ \mu_{\mathrm{aggregate}}(z)=\max_r\mu_{\mathrm{rule}\ r}(z) $$ α (“alpha”) is the rule firing strength. z names a possible output value. The subscript r labels a rule, not a multiplication. **A rule has two condition degrees 0.7 and 0.5, joined by AND. Find its firing strength.** [2 marks · Practice] #### Given a=0.7; b=0.5; connective=AND. #### To find The strength α. #### Formula and why it applies Use the smaller degree because both conditions must hold. #### Solution ##### Step 1. Select the operator $$ \alpha=\min(0.7,0.5) $$ ##### Step 2. Evaluate $$ \alpha=0.5 $$ **Final answer.** 0.5. **A rule has two condition degrees 0.3 and 0, joined by AND. Find its firing strength.** [2 marks · Practice] #### Given a=0.3; b=0; connective=AND. #### To find The strength α. #### Formula and why it applies Use the smaller degree because both conditions must hold. #### Solution ##### Step 1. Select the operator $$ \alpha=\min(0.3,0) $$ ##### Step 2. Evaluate $$ \alpha=0 $$ **Final answer.** 0. **A rule has two condition degrees 0.2 and 0.8, joined by OR. Find its firing strength.** [2 marks · Practice] #### Given a=0.2; b=0.8; connective=OR. #### To find The strength α. #### Formula and why it applies Use the larger degree because either condition is sufficient. #### Solution ##### Step 1. Select the operator $$ \alpha=\max(0.2,0.8) $$ ##### Step 2. Evaluate $$ \alpha=0.8 $$ **Final answer.** 0.8. **Rule 1 fires at 0.3 and its original output membership at z=60 is 0.8. Rule 2 fires at 0.6 and its original membership at z=60 is 0.2. Find the aggregate membership at z=60 using min-implication and max-aggregation.** [3 marks · Practice] #### Given Rule 1: α₁=0.3, μ₁(60)=0.8. Rule 2: α₂=0.6, μ₂(60)=0.2. #### To find Membership at the single output location z=60 after implication and aggregation. #### Formula and why it applies Clip each original degree at its rule strength; then take the larger clipped degree. #### Solution ##### Step 1. Clip rule 1 $$ \mu_{R1}(60)=\min(0.3,0.8)=0.3 $$ ##### Step 2. Clip rule 2 $$ \mu_{R2}(60)=\min(0.6,0.2)=0.2 $$ ##### Step 3. Aggregate at this output location $$ \mu_{out}(60)=\max(0.3,0.2)=0.3 $$ **Final answer.** The aggregate membership at z=60 is 0.3. This is not a crisp fan-speed answer. Swipe across the diagram to read its labels ↔ ![Above the 0.3 level, the original triangle is cut off. Below 0.3, the original slope remains.](https://deep-learning-study-room.vercel.app/is2/fig/v2_clip.f101b5cfa9.svg) Diagram: Above the 0.3 level, the original triangle is cut off. Below 0.3, the original slope remains.. Labels: Min-implication clips the triangle at strength 0.3 | 0 | 0.3 | 1 | 50 | 54.5 | 65 | 75.5 | 80 | μ | Output z | Original High | clip 0.3 Above the 0.3 level, the original triangle is cut off. Below 0.3, the original slope remains. **Yaad rakho:** AND ke liye min; output clip karne ke liye bhi min; rules ko jodne ke liye max. Har min ka role alag hai. --- ## M2 · Mean of maxima: discrete peaks and clipped plateaus Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--mom/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/mom Definitions → membership graphs → operations → rules → defuzzification → complete controller. **Maximum membership** is a height. **Mean of maxima** averages the horizontal locations attaining that height. If heights at speeds [20,50,80] are [.2,.6,.6], the maximum height is .6 and the qualifying speeds are 50 and 80; MoM=(50+80)/2=65. Do not average [.6,.6], which would return a degree rather than a speed. A continuous flat maximum from 50 to 80 also has midpoint 65. ## Mean of maxima (MoM) **Defuzzification** converts the aggregate membership function into one crisp output. For MoM, first find the largest membership height, then average the output locations attaining that height. Do not average the membership values themselves. $$ z^*_{MoM}=\frac{z_1+\cdots+z_m}{m} $$ For a continuous flat maximum from a to b, the mean is its midpoint: $$ z^*_{MoM}=\frac{a+b}{2} $$ **For the following discrete aggregate, find the mean-of-maxima output.** [2 marks · Practice] #### Given | Output z | Membership μ | | --- | --- | | 10 | 0.2 | | 20 | 0.8 | | 30 | 0.8 | | 40 | 0.3 | #### To find MoM crisp output. #### Formula and why it applies Find all locations at the largest height, then average only those locations. #### Solution ##### Step 1. Find the maximum height $$ \mu_{max}=0.8 $$ ##### Step 2. List the locations attaining it 20, 30 ##### Step 3. Average these locations $$ z^*=\frac{20+30}{2}=25 $$ **Final answer.** 25. **For the following discrete aggregate, find the mean-of-maxima output.** [2 marks · Practice] #### Given | Output z | Membership μ | | --- | --- | | 1 | 0.1 | | 2 | 0.4 | | 3 | 0.9 | | 4 | 0.2 | #### To find MoM crisp output. #### Formula and why it applies Find all locations at the largest height, then average only those locations. #### Solution ##### Step 1. Find the maximum height $$ \mu_{max}=0.9 $$ ##### Step 2. List the locations attaining it 3 ##### Step 3. Average these locations $$ z^*=\frac{3}{1}=3 $$ **Final answer.** 3. **A single triangular output Tri(50,65,80) is clipped at α=0.3. Find its crisp value by mean of maxima.** [3 marks · Practice] #### Given a=50, b=65, c=80; clip height α=0.3. #### To find Both plateau endpoints and their mean. #### Formula and why it applies Set the rising and falling membership fractions equal to 0.3. Every point between those intersections has the maximum height. #### Solution ##### Step 1. Find the left endpoint $$ \frac{z_L-50}{65-50}=0.3 $$ $$ z_L-50=0.3\times15=4.5 $$ $$ z_L=54.5 $$ ##### Step 2. Find the right endpoint $$ \frac{80-z_R}{80-65}=0.3 $$ $$ 80-z_R=4.5 $$ $$ z_R=75.5 $$ ##### Step 3. Average endpoints $$ z^*=(54.5+75.5)/2=65 $$ **Final answer.** MoM = 65. Swipe across the diagram to read its labels ↔ ![The maximum extends over [54.5,75.5], whose midpoint is 65.](https://deep-learning-study-room.vercel.app/is2/fig/v2_clip.f101b5cfa9.svg) Diagram: The maximum extends over [54.5,75.5], whose midpoint is 65.. Labels: Min-implication clips the triangle at strength 0.3 | 0 | 0.3 | 1 | 50 | 54.5 | 65 | 75.5 | 80 | μ | Output z | Original High | clip 0.3 The maximum extends over [54.5,75.5], whose midpoint is 65. For several disjoint continuous maximum plateaus, a continuous MoM uses a length-weighted mean over all maximum intervals; simply averaging two outside endpoints is not generally valid. In tabulated questions, average the listed locations at the maximum. **Yaad rakho:** Sabse upar ka level dekho; us level par x-values ka mean lo. --- ## M2 · Centroid: three table sums and one area calculation Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--centroid/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/centroid Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Centroid means the balance point of the whole shape Imagine the area under the output graph as a thin sheet of uniform material. The centroid is its horizontal balance point. More area on one side pulls the answer toward that side. In a discrete problem, each membership is a weight. In a continuous problem, the entire area supplies the weights. The integral sign ∫ means adding very thin strips across a continuous interval. The denominator ∫μ(z)dz is the total area. The numerator ∫zμ(z)dz is the **first moment**: position multiplied by area, accumulated across the shape. Their ratio has units of the output, such as fan-speed percent. **A non-overlapping aggregate consists of a rectangle of width 4 and height 0.5 centred at z=2, plus a triangle of base 4 and height 0.5 centred horizontally at z=6. Find the area centroid.** [2 marks · Worked example] #### Given Rectangle: width 4, height 0.5, horizontal centre 2. Triangle: base 4, height 0.5, horizontal centroid 6. Their areas do not overlap. #### To find The combined horizontal centroid. #### Formula and why it applies $$ z^*=\frac{A_1c_1+A_2c_2}{A_1+A_2} $$ Each piece contributes area times its own centre. Use this only for a valid decomposition of the aggregate. #### Solution ##### Step 1. Rectangle area $$ A_1=4\times0.5=2 $$ ##### Step 2. Triangle area $$ A_2=\tfrac12\times4\times0.5=1 $$ ##### Step 3. First moments $$ A_1c_1=2\times2=4 $$ $$ A_2c_2=1\times6=6 $$ ##### Step 4. Totals $$ A=2+1=3\qquad M=4+6=10 $$ ##### Step 5. Divide $$ z^*=10/3\approx3.333333 $$ **Final answer.** Centroid≈3.333333, closer to the larger-area piece. For overlapping clipped sets, first form their max-aggregate. Adding the areas of the original overlapping sets would count the overlap twice. ## Centroid: the balance point of the output The centroid uses the whole aggregate membership distribution. For a discrete output table, multiply each output location by its membership, add the products, then divide by total membership. $$ z^*=\frac{\sum_i z_i\mu(z_i)}{\sum_i\mu(z_i)} $$ The index i simply numbers table entries. Σ (“sigma”) means add all listed terms. The denominator must be nonzero. **Find the discrete centroid of the aggregate membership table.** [3 marks · Practice] #### Given | z | μ(z) | | --- | --- | | 10 | 0.2 | | 20 | 0.6 | | 30 | 0.4 | #### To find One centroid value on the output scale. #### Formula and why it applies This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation. $$ z^*=\frac{\sum z_i\mu_i}{\sum\mu_i} $$ #### Solution ##### Step 1. Multiply every location by its membership | z | μ | z × μ | | --- | --- | --- | | 10 | 0.2 | 2 | | 20 | 0.6 | 12 | | 30 | 0.4 | 12 | ##### Step 2. Add the numerator $$ 2+12+12=26 $$ ##### Step 3. Add the denominator $$ 0.2+0.6+0.4=1.2 $$ ##### Step 4. Divide $$ z^*=\frac{26}{1.2}=21.666666667 $$ **Final answer.** 21.666666667. **Find the discrete centroid of the aggregate membership table.** [3 marks · Practice] #### Given | z | μ(z) | | --- | --- | | 0 | 0.5 | | 5 | 1 | | 10 | 0.5 | #### To find One centroid value on the output scale. #### Formula and why it applies This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation. $$ z^*=\frac{\sum z_i\mu_i}{\sum\mu_i} $$ #### Solution ##### Step 1. Multiply every location by its membership | z | μ | z × μ | | --- | --- | --- | | 0 | 0.5 | 0 | | 5 | 1 | 5 | | 10 | 0.5 | 5 | ##### Step 2. Add the numerator $$ 0+5+5=10 $$ ##### Step 3. Add the denominator $$ 0.5+1+0.5=2 $$ ##### Step 4. Divide $$ z^*=\frac{10}{2}=5 $$ **Final answer.** 5. **Find the discrete centroid of the aggregate membership table.** [3 marks · Practice] #### Given | z | μ(z) | | --- | --- | | 20 | 0 | | 40 | 0.3 | | 60 | 0.6 | | 80 | 0.1 | #### To find One centroid value on the output scale. #### Formula and why it applies This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation. $$ z^*=\frac{\sum z_i\mu_i}{\sum\mu_i} $$ #### Solution ##### Step 1. Multiply every location by its membership | z | μ | z × μ | | --- | --- | --- | | 20 | 0 | 0 | | 40 | 0.3 | 12 | | 60 | 0.6 | 36 | | 80 | 0.1 | 8 | ##### Step 2. Add the numerator $$ 0+12+36+8=56 $$ ##### Step 3. Add the denominator $$ 0+0.3+0.6+0.1=1 $$ ##### Step 4. Divide $$ z^*=\frac{56}{1}=56 $$ **Final answer.** 56. ## For a continuous shape, integrate the area $$ z^*=\frac{\int z\mu(z)\,dz}{\int\mu(z)\,dz} $$ The denominator is the area under the aggregate. The numerator is its first moment about zero: area weighted by horizontal location. For a shape made of non-overlapping rectangles and triangles, compute each part’s area A and its own centroid c, then use $$ z^*=(\sum A_jc_j)/(\sum A_j) $$ . Overlapping regions of a max-aggregate must not be counted twice. **An aggregate is a rectangle of height 0.5 from z=0 to z=4, followed by a right triangle decreasing from height 0.5 at z=4 to 0 at z=6. Find its area centroid.** [3 marks · Practice] #### Given Rectangle [0,4], height 0.5; triangle with vertices (4,0),(4,0.5),(6,0). #### To find Centroid of the combined area. #### Formula and why it applies The parts do not overlap except at their boundary. Weight each part’s horizontal centroid by its area. #### Solution ##### Step 1. Rectangle area and centre $$ A_1=4\times0.5=2\qquad c_1=(0+4)/2=2 $$ ##### Step 2. Triangle area and centre $$ A_2=\tfrac12\times2\times0.5=0.5 $$ $$ c_2=(4+4+6)/3=14/3 $$ ##### Step 3. Add first moments $$ A_1c_1+A_2c_2=2(2)+0.5(14/3)=19/3 $$ ##### Step 4. Divide by total area $$ z^*=(19/3)/(2+0.5)=38/15\approx2.533333 $$ **Final answer.** Centroid ≈ 2.533333. The plateau MoM would be 2, so different methods can legitimately give different results. **Yaad rakho:** Centroid = har location × uska weight, total karke total weight se divide. --- ## M2 · Weighted average and centroid are not automatically the same Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--weighted-average/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/weighted-average Definitions → membership graphs → operations → rules → defuzzification → complete controller. This is the same weighted-average calculation taught in Essential maths. Here a representative output value c is weighted by a rule strength α. Keep the pair together: strength .3 belonging to speed 65 contributes .3×65. The denominator is the total strength. It is not automatically the number of rules, and the representative is not automatically the area-centre of a clipped set. ## Weighted average of output representatives Some classroom solutions use one representative (often the peak) per output label and weight it by the firing strength. That is a **weighted-average or height-based calculation**. It is exact for a specified weighted combination of singleton outputs, but is not generally the centroid of a max-aggregated, clipped Mamdani shape. Always name the method. A table of memberships at actual output locations supports the discrete centroid. A list of category peaks and strengths supports the specified representative-weighted calculation. Full continuous output shapes support area centroid. **Using the weighted-average method, combine output representatives [25, 75] mph with respective strengths [0.1, 0.7].** [3 marks · Practice] #### Given | Output representative | Strength | | --- | --- | | 25 | 0.1 | | 75 | 0.7 | #### To find One crisp output by the specified weighted-average method. #### Formula and why it applies Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set. $$ z^*=\frac{\sum_i\alpha_i z_i}{\sum_i\alpha_i} $$ #### Solution ##### Step 1. Multiply each value by its strength | Strength × value | Product | | --- | --- | | 0.1 × 25 | 2.5 | | 0.7 × 75 | 52.5 | ##### Step 2. Add the products $$ 2.5+52.5=55 $$ ##### Step 3. Add the strengths $$ 0.1+0.7=0.8 $$ ##### Step 4. Divide $$ z^*=\frac{55}{0.8}=68.75 $$ **Final answer.** 68.75 mph. **Using the weighted-average method, combine output representatives [40, 65, 85] % fan speed with respective strengths [0, 0.3, 0.5].** [3 marks · Practice] #### Given | Output representative | Strength | | --- | --- | | 40 | 0 | | 65 | 0.3 | | 85 | 0.5 | #### To find One crisp output by the specified weighted-average method. #### Formula and why it applies Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set. $$ z^*=\frac{\sum_i\alpha_i z_i}{\sum_i\alpha_i} $$ #### Solution ##### Step 1. Multiply each value by its strength | Strength × value | Product | | --- | --- | | 0 × 40 | 0 | | 0.3 × 65 | 19.5 | | 0.5 × 85 | 42.5 | ##### Step 2. Add the products $$ 0+19.5+42.5=62 $$ ##### Step 3. Add the strengths $$ 0+0.3+0.5=0.8 $$ ##### Step 4. Divide $$ z^*=\frac{62}{0.8}=77.5 $$ **Final answer.** 77.5 % fan speed. **Using the weighted-average method, combine output representatives [25, 40, 60] minutes with respective strengths [0.6, 0.4, 0.2].** [3 marks · Practice] #### Given | Output representative | Strength | | --- | --- | | 25 | 0.6 | | 40 | 0.4 | | 60 | 0.2 | #### To find One crisp output by the specified weighted-average method. #### Formula and why it applies Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set. $$ z^*=\frac{\sum_i\alpha_i z_i}{\sum_i\alpha_i} $$ #### Solution ##### Step 1. Multiply each value by its strength | Strength × value | Product | | --- | --- | | 0.6 × 25 | 15 | | 0.4 × 40 | 16 | | 0.2 × 60 | 12 | ##### Step 2. Add the products $$ 15+16+12=43 $$ ##### Step 3. Add the strengths $$ 0.6+0.4+0.2=1.2 $$ ##### Step 4. Divide $$ z^*=\frac{43}{1.2}=35.833333333 $$ **Final answer.** 35.833333333 minutes. | Method | What it uses | Limitation | | --- | --- | --- | | MoM | Locations at maximum membership | Ignores lower portions of the shape; can change abruptly | | Area centroid | Entire aggregated output area | Needs an area/moment calculation or numerical integration | | Representative weighted average | Supplied representative values and strengths | Not interchangeable with a clipped Mamdani area centroid | No method is universally “most accurate”. The correct exam calculation is the method specified by the question. If all memberships or all weights are zero, division by zero is undefined; a controller needs an explicit fallback rule. **Yaad rakho:** Formula similar dikhega, par weight kya hai? Membership, area, ya rule strength—pehle woh identify karo. ## Faculty slide example: one temperature input This is a different example from the two-input fan design. With Warm=Tri(20,30,40) and Hot=Tri(30,40,50), temperature 32 gives Warm=(40−32)/10=0.8 and Hot=(32−30)/10=0.2. The slide uses output representatives 50 and 80. **Using the weighted-average method, combine output representatives [50, 80] % fan speed with respective strengths [0.8, 0.2].** [3 marks · Practice] #### Given | Output representative | Strength | | --- | --- | | 50 | 0.8 | | 80 | 0.2 | #### To find One crisp output by the specified weighted-average method. #### Formula and why it applies Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set. $$ z^*=\frac{\sum_i\alpha_i z_i}{\sum_i\alpha_i} $$ #### Solution ##### Step 1. Multiply each value by its strength | Strength × value | Product | | --- | --- | | 0.8 × 50 | 40 | | 0.2 × 80 | 16 | ##### Step 2. Add the products $$ 40+16=56 $$ ##### Step 3. Add the strengths $$ 0.8+0.2=1 $$ ##### Step 4. Divide $$ z^*=\frac{56}{1}=56 $$ **Final answer.** 56 % fan speed. --- ## M2 · Class fan question: 32°C, 7 people, every rule and step Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--class-fan/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/class-fan Definitions → membership graphs → operations → rules → defuzzification → complete controller. There are 3 temperature labels and 3 occupancy labels. Pair each temperature label with each occupancy label: 3×3=9 possible AND rules. At the specified input, any rule containing a zero input membership has strength min(0,other)=0 and contributes no output. The rules with nonzero strength remain; their output labels come from the supplied rule table, not from the magnitude of the strength. This is the fan-controller design question shared in class: temperature 32°C and occupancy 7. The photos contain inconsistent graph breakpoints and arithmetic. A design question permits a stated design; below, every membership function is specified explicitly. These choices make the solution reproducible and should not be read as a uniquely recoverable answer from the photos. **Design a Mamdani fan controller for temperature 0–40°C and occupancy 0–10, with three input labels each and five fan-speed labels on 0–100%. For T=32°C and occupancy=7, use the membership functions and nine rules below. Calculate all memberships, firing strengths, and active output sets. Find fan speed by (a) mean of maxima and (b) the specified representative-weighted average.** [10 marks · Class-question adaptation; full design] #### Given | Variable | Low / Very Low | Medium / Low | High / Medium | High output | Very High output | | --- | --- | --- | --- | --- | --- | | Temperature | Tri(0,10,20) | Tri(15,25,35) | Tri(25,35,40) | — | — | | Occupancy | Tri(0,2,4) | Tri(3,5,7) | Tri(6,8,10) | — | — | | Fan speed | Tri(0,10,20) | Tri(10,25,40) | Tri(20,40,60) | Tri(50,65,80) | Tri(70,85,100) | Input T=32°C; occupancy=7. Use min-AND, min-clipping and max-aggregation. | Temperature | Occupancy | Fan output | | --- | --- | --- | | Low | Low | Very Low | | Low | Medium | Low | | Low | High | Medium | | Medium | Low | Low | | Medium | Medium | Medium | | Medium | High | High | | High | Low | Medium | | High | Medium | High | | High | High | Very High | #### To find Input memberships → nine rule strengths → clipped output → two named crisp outputs. #### Formula and why it applies Use the appropriate triangle branch for membership. AND takes the smaller degree. Each output is clipped at that strength; rules with the same consequent combine by max. MoM uses only the highest plateau. The requested representative weighted average uses category peaks and strengths. $$ \alpha_r=\min(\mu_{T,r},\mu_{O,r})\quad\mu_{out}(z)=\max_r\min(\alpha_r,\mu_{C_r}(z)) $$ #### Solution ##### Step 1. Read the temperature graph Swipe across the diagram to read its labels ↔ ![At 32, Medium is on its falling side; High is on its rising side.](https://deep-learning-study-room.vercel.app/is2/fig/v2_fan_temp.434c9b6635.svg) Diagram: At 32, Medium is on its falling side; High is on its rising side.. Labels: Chosen temperature membership functions | 0 | 0.5 | 1 | 0 | 10 | 15 | 20 | 25 | 32 | 35 | 40 | μ | Temperature in °C | Low | Medium | High | 32°C At 32, Medium is on its falling side; High is on its rising side. $$ \mu_L(32)=0 $$ $$ \mu_M(32)=\frac{35-32}{35-25}=\frac3{10}=0.3 $$ $$ \mu_H(32)=\frac{32-25}{35-25}=\frac7{10}=0.7 $$ ##### Step 2. Read the occupancy graph Swipe across the diagram to read its labels ↔ ![At 7, Medium reaches its right zero and High is halfway up its rising edge.](https://deep-learning-study-room.vercel.app/is2/fig/v2_fan_people.1861531dd1.svg) Diagram: At 7, Medium reaches its right zero and High is halfway up its rising edge.. Labels: Chosen occupancy membership functions | 0 | 0.5 | 1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | μ | Number of people | Low | Medium | High | 7 people At 7, Medium reaches its right zero and High is halfway up its rising edge. $$ \mu_L(7)=0\qquad\mu_M(7)=0 $$ $$ \mu_H(7)=\frac{7-6}{8-6}=\frac12=0.5 $$ ##### Step 3. Evaluate all nine rules | T label / degree | O label / degree | min = strength | Output | | --- | --- | --- | --- | | Low / 0 | Low / 0 | 0 | Very Low | | Low / 0 | Medium / 0 | 0 | Low | | Low / 0 | High / 0.5 | 0 | Medium | | Medium / 0.3 | Low / 0 | 0 | Low | | Medium / 0.3 | Medium / 0 | 0 | Medium | | Medium / 0.3 | High / 0.5 | 0.3 | High | | High / 0.7 | Low / 0 | 0 | Medium | | High / 0.7 | Medium / 0 | 0 | High | | High / 0.7 | High / 0.5 | 0.5 | Very High | ##### Step 4. Clip and aggregate $$ \mu_{out}(z)=\max\{\min(0.3,\mu_H(z)),\min(0.5,\mu_{VH}(z))\} $$ Swipe across the diagram to read its labels ↔ ![High is clipped at 0.3; Very High is clipped at 0.5. Use the upper envelope where the two shapes overlap.](https://deep-learning-study-room.vercel.app/is2/fig/v2_fan_output.8708d0ea77.svg) Diagram: High is clipped at 0.3; Very High is clipped at 0.5. Use the upper envelope where the two shapes overlap.. Labels: Two active output sets | 0 | 0.5 | 1 | 50 | 54.5 | 65 | 70 | 75 | 77.5 | 85 | 92.5 | 100 | μ | Fan speed (%) | High | Very High | clip 0.3 | clip 0.5 High is clipped at 0.3; Very High is clipped at 0.5. Use the upper envelope where the two shapes overlap. ##### Step 5. Find the highest plateau for MoM The maximum is 0.5, belonging to Very High. Its triangle is (70,85,100). $$ \frac{z_L-70}{15}=0.5\Rightarrow z_L=77.5 $$ $$ \frac{100-z_R}{15}=0.5\Rightarrow z_R=92.5 $$ $$ z^*_{MoM}=\frac{77.5+92.5}{2}=85 $$ ##### Step 6. Multiply the representatives by strengths $$ 0.3\times65=19.5 $$ $$ 0.5\times85=42.5 $$ ##### Step 7. Add, then divide for the weighted average $$ z^*_{WA}=\frac{19.5+42.5}{0.3+0.5}=\frac{62}{0.8}=77.5 $$ **Final answer.** For this explicitly stated design: MoM = 85%; representative-weighted output = 77.5%. These are different methods. 77.5% is not claimed to be the true continuous Mamdani area centroid. **How to adapt it in the exam:** if breakpoints or output representatives differ, recompute the memberships from those data. The sequence of operations stays unchanged. If the question explicitly asks for the area centroid, integrate the max-aggregated shape; do not silently substitute the weighted average above. **Yaad rakho:** Temperature aur people ka degree nikalo → 9 min → active outputs clip → jo method poocha hai woh lagao. ## Why the Very High plateau has midpoint 85 For the stated Very High triangle (70,85,100) clipped at height 0.5, solve the rising edge: (z−70)/(85−70)=0.5, so z−70=0.5×15=7.5 and z=77.5. On the falling edge, (100−z)/(100−85)=0.5, so 100−z=7.5 and z=92.5. Thus the highest flat interval is [77.5,92.5]. Its midpoint is (77.5+92.5)/2=170/2=85. Continue with the [complete area-centroid calculation for this same fan design](https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--centroid-comparison/): 77.086956522%, with all overlap breakpoints and area/moment arithmetic shown. --- ## M2 · The same fan controller: true area centroid versus MoM and weighted average Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--centroid-comparison/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/centroid-comparison Definitions → membership graphs → operations → rules → defuzzification → complete controller. The preceding class-fan lesson makes the input shapes and nine rules explicit because the photographs do not specify one internally consistent design. Keep that exact stated design here: temperature 32°C, occupancy 7, High output Tri(50,65,80) firing at 0.3 and Very High output Tri(70,85,100) firing at 0.5. This continuation calculates the actual area centroid of their max-aggregate. It does not reinterpret 77.5 as a centroid. **For the stated two-input class fan controller at T=32°C and occupancy=7, calculate the true continuous Mamdani area centroid of High=Tri(50,65,80) clipped at 0.3 and Very High=Tri(70,85,100) clipped at 0.5. Compare it with mean of maxima and the specified representative-weighted average.** [6 marks] Worked teaching continuation of the class-design adaptation; the mark allocation here is for practice. #### Given All other output rules have strength zero. Use min for implication, max for aggregation and speed z on 0–100%. The two active rule strengths were calculated from input memberships: min(0.3,0.5)=0.3 and min(0.7,0.5)=0.5. #### To find Every visible segment of the aggregate, its area and first moment, the centroid, and the differences between the three crisp answers. #### Formula and why it applies $$ \mu_{out}(z)=\max\{\min(0.3,\mu_H(z)),\min(0.5,\mu_{VH}(z))\} $$ $$ z^*_{COA}=\frac{\int z\mu_{out}(z)\,dz}{\int\mu_{out}(z)\,dz} $$ Centroid balances all the area under the upper envelope. A straight strip with endpoints (a,u) and (b,v) has the following area A and first moment M; adding only disjoint strips counts overlap once. $$ A=\frac{(b-a)(u+v)}2,\qquad M=\frac{b-a}{6}[(2a+b)u+(a+2b)v] $$ #### Solution ##### Step 1. Locate the two clipped plateaus $$ H:\quad z_L=50+0.3(65-50)=54.5,\quad z_R=80-0.3(80-65)=75.5 $$ $$ VH:\quad z_L=70+0.5(85-70)=77.5,\quad z_R=100-0.5(100-85)=92.5 $$ High has a 0.3 plateau from 54.5 to 75.5. Very High has a 0.5 plateau from 77.5 to 92.5. ##### Step 2. Find where the upper envelope changes owner Very High rises through the High plateau at height 0.3. Solve its rising equation: $$ (z-70)/15=0.3\quad\Rightarrow\quad z=74.5 $$ Before 74.5, clipped High supplies the upper envelope. After 74.5, Very High is above it. The descending High edge from 75.5 to 80 lies underneath and contributes no extra aggregate area. ##### Step 3. Trace the aggregate graph Swipe across the diagram to read its labels ↔ ![Fan output: trace the upper envelope before finding its centroid](https://deep-learning-study-room.vercel.app/is2/fig/deep_fan_centroid.98bf1c217d.svg) Diagram: Fan output: trace the upper envelope before finding its centroid. Labels: Fan output: trace the upper envelope before finding its centroid | High clipped at 0.3; Very High clipped at 0.5. Hidden overlap is counted once. | 0 | 0.3 | 0.5 | 50 | 54.5 | 74.5 | 77.5 | 92.5 | 100 | Centroid ≈77.087% | Fan speed z (%) | Area = 17.25; first moment = 1329.75; centroid = moment ÷ area. Read left to right: (50,0) → (54.5,0.3) → (74.5,0.3) → (77.5,0.5) → (92.5,0.5) → (100,0). Membership is zero outside this support. Each adjacent pair is one straight strip in the table. ##### Step 4. Work the rising first strip explicitly $$ A_1=(54.5-50)(0+0.3)/2=0.675 $$ $$ M_1=\frac{4.5}{6}[(100+54.5)0+(50+109)0.3]=35.775 $$ The first moment has units of speed multiplied by area. Dividing total moment by total area returns speed. ##### Step 5. Calculate all disjoint strips | a | b | u | v | Area A | First moment M | | --- | --- | --- | --- | --- | --- | | 50 | 54.5 | 0 | 0.3 | 0.675 | 35.775 | | 54.5 | 74.5 | 0.3 | 0.3 | 6 | 387 | | 74.5 | 77.5 | 0.3 | 0.5 | 1.2 | 91.35 | | 77.5 | 92.5 | 0.5 | 0.5 | 7.5 | 637.5 | | 92.5 | 100 | 0.5 | 0 | 1.875 | 178.125 | $$ \sum A=0.675+6+1.2+7.5+1.875=17.25 $$ $$ \sum M=35.775+387+91.35+637.5+178.125=1329.75 $$ ##### Step 6. Divide for the actual centroid $$ z^*_{COA}=1329.75/17.25=1773/23\approx77.086956522\% $$ ##### Step 7. Compare methods on the same aggregate | Method | Calculation | Answer | | --- | --- | --- | | Area centroid | 1329.75 ÷ 17.25 | 77.086956522% | | Mean of maxima | (77.5+92.5) ÷ 2 | 85% | | Specified representative weighted average | (0.3×65+0.5×85) ÷ (0.3+0.5) | 77.5% | MoM uses only the highest plateau. The representative method uses two chosen peaks and two rule strengths. The area centroid uses the full max-aggregate, so these are three legitimate but different calculations. **Final answer.** Area centroid ≈77.086956522%; MoM=85%; representative-weighted output=77.5%. Report the method with the value. The centroid is inside the support [50,100], as a balance point must be. **Boundary coverage of this teaching design.** With the ordinary triangles stated in the class adaptation, every temperature degree is zero at T=0 or T=40, and every occupancy degree is zero at occupancy=0 or 10. Then no rule fires and an area centroid is undefined. To make a controller cover the entire declared input range, explicitly redesign outer sets as shoulders or specify an out-of-coverage rule; do not silently divide by zero. The worked T=32, occupancy=7 case has positive area and is unaffected. **Yaad rakho:** Centroid mein poore shaded area ka balance chahiye. Peak × strength wala shortcut same calculation nahi hai. --- ## M2 · Fuzzy theory: definitions, architecture, advantages and applications Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--exam-theory/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/exam-theory Definitions → membership graphs → operations → rules → defuzzification → complete controller. **Explain fuzzy logic with one application.** [2 marks · Practice] #### Answer Fuzzy logic represents gradual truth or category membership using values between 0 and 1. A temperature controller can use Low, Medium and High temperature sets. It evaluates rules such as IF temperature is High THEN fan speed is High, combines the rule outputs and defuzzifies them into a numerical fan speed. This permits gradual control around category boundaries. **Give the advantages and limitations of a Mamdani fuzzy controller.** [3 marks · Practice] #### Answer | Advantages | Limitations | | --- | --- | | Human-readable rules express expert knowledge | Membership functions and rules require careful design | | Can model gradual boundaries without an exact analytic plant model | Rule count can grow rapidly with more inputs and categories | | Handles overlapping linguistic descriptions | A poorly chosen rule base can produce poor or uncovered outputs | | Suitable for many control applications | Does not automatically learn or guarantee optimal control | **Describe the design of a fuzzy fan controller.** [3 marks · Practice] #### Answer Choose temperature and occupancy as inputs and fan speed as output. Define their ranges and Low/Medium/High membership functions. Construct a rule table linking input labels to output labels. Fuzzify measured inputs, use min to obtain AND-rule strengths, clip the consequent sets and max-aggregate them. Apply the specified defuzzification method to obtain a fan speed. Test the design at normal inputs and range boundaries. ## Other applications you can explain | Application | Inputs | Example rule | Output | | --- | --- | --- | --- | | Washing machine | Dirtiness, load | IF dirt High AND load High THEN wash time Long | Wash duration | | Loan decision support | Income, credit score | IF income High AND credit Good THEN approval suitability High | Suitability score (not an automatic legal lending decision) | | Fire warning | Temperature, smoke | IF temperature High OR smoke High THEN risk High | Alarm-risk score | | Vehicle control | Distance, relative speed | IF distance Small AND closing speed High THEN braking Strong | Braking command | **Yaad rakho:** Theory answer mein definition → blocks/steps → ek concrete example. Sirf headings se marks nahi milte. --- ## M2 · Gaussian, singleton and bell memberships Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--smooth-memberships/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-core/smooth-memberships Definitions → membership graphs → operations → rules → defuzzification → complete controller. For a Gaussian membership, c is the input at the centre and σ controls the width. Compute x−c, square it, divide by 2σ², put a minus sign in front, and apply eˣ. The maximum membership is 1 at x=c because the exponent becomes 0. Increasing σ makes the same distance from the centre count as a smaller relative distance, so the curve becomes wider. A Gaussian membership is not the same formula as a normal probability density: there is no 1/(√(2π)σ) multiplier here. ## A smooth membership function A Gaussian membership function reaches 1 at its centre c and decreases smoothly on both sides. Its width σ must be positive. This is a fuzzy membership curve, not a normalised probability density: there is no 1/(√(2π)σ) factor. $$ \mu(x)=\exp\left(-\frac{(x-c)^2}{2\sigma^2}\right) $$ exp(u) means eᵘ. The calculator evaluates it using eˣ; calculate the entire negative exponent first. A squared distance treats equal left and right offsets equally. **For Gaussian membership with c=10 and σ=2, find μ(10).** [2 marks · Practice] #### Given x=10; centre c=10; width σ=2. #### To find Membership degree. #### Formula and why it applies Use squared distance from the centre in the Gaussian membership formula. #### Solution ##### Step 1. Distance from the centre $$ 10-10=0 $$ ##### Step 2. Square the distance and width $$ (0)^2=0\qquad2\sigma^2=2(2^2)=8 $$ ##### Step 3. Exponent $$ -0/8=0 $$ ##### Step 4. Evaluate e to this power $$ \mu(10)=e^{0}\approx1 $$ **Final answer.** 1. **For Gaussian membership with c=10 and σ=2, find μ(12).** [2 marks · Practice] #### Given x=12; centre c=10; width σ=2. #### To find Membership degree. #### Formula and why it applies Use squared distance from the centre in the Gaussian membership formula. #### Solution ##### Step 1. Distance from the centre $$ 12-10=2 $$ ##### Step 2. Square the distance and width $$ (2)^2=4\qquad2\sigma^2=2(2^2)=8 $$ ##### Step 3. Exponent $$ -4/8=-0.5 $$ ##### Step 4. Evaluate e to this power $$ \mu(12)=e^{-0.5}\approx0.60653066 $$ **Final answer.** 0.60653066. **For Gaussian membership with c=10 and σ=2, find μ(14).** [2 marks · Practice] #### Given x=14; centre c=10; width σ=2. #### To find Membership degree. #### Formula and why it applies Use squared distance from the centre in the Gaussian membership formula. #### Solution ##### Step 1. Distance from the centre $$ 14-10=4 $$ ##### Step 2. Square the distance and width $$ (4)^2=16\qquad2\sigma^2=2(2^2)=8 $$ ##### Step 3. Exponent $$ -16/8=-2 $$ ##### Step 4. Evaluate e to this power $$ \mu(14)=e^{-2}\approx0.135335283 $$ **Final answer.** 0.135335283. ## Singleton A singleton fuzzy set has membership 1 at exactly one specified point and 0 elsewhere. For a singleton at c=50: μ(50)=1, μ(49)=0 and μ(51)=0. A singleton consequent is represented by one numerical output; this is why a specified weighted average of singleton outputs is a legitimate calculation. ## Generalised bell (when its formula is supplied) $$ \mu(x)=\frac1{1+\left|\frac{x-c}{a}\right|^{2b}} $$ c is the centre, a>0 controls width and b>0 controls the slope. Absolute value |u| is the nonnegative magnitude: |−2|=2. At x=c membership is 1; at x=c±a membership is 1/2. **For a=2, b=1 and c=10, find generalised-bell membership at x=14.** [2 marks · Practice] #### Given a=2, b=1, c=10; x=14. #### To find μ(14). #### Formula and why it applies Subtract the centre, divide by a, take the absolute value, raise to 2b, then use the reciprocal. #### Solution ##### Step 1. Scaled distance $$ |(14-10)/2|=|2|=2 $$ ##### Step 2. Power $$ 2^{2(1)}=4 $$ ##### Step 3. Final reciprocal $$ \mu(14)=1/(1+4)=0.2 $$ **Final answer.** 0.2. **Yaad rakho:** Gaussian MF ka centre 1 hota hai. GMM density ke normalising factor ko yahan mat lagao. --- ## M2 · GA meaning, biological idea and every basic term Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--meaning/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/ga-core/meaning Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. | Object | Concrete value | Do not confuse it with | | --- | --- | --- | | Individual number | Row 2 | The candidate’s value x | | Candidate value x | 11 | Its chromosome 1011 | | Chromosome | 1011 | Its fitness | | Fitness when maximising x² | 11×11=121 | Its row number or bit string | In this guide, “gene 3” counts the third bit from the **left** starting at 1. Binary place values were built from the **right** starting with weight 1. These are two labelling conventions, not contradictory place-value rules. Read the end specified in an exam. ## What a genetic algorithm does A **genetic algorithm (GA)** is a population-based optimisation method inspired by biological evolution. It maintains candidate solutions, evaluates their fitness, selects parents, generates children through crossover and mutation, and repeats the process. In biology, genes carry inherited information. In a GA, an encoding plays the role of inherited information. The analogy motivates the operators; it does not mean that a computer chromosome is literally DNA. | Term | Meaning | Four-bit example | | --- | --- | --- | | Individual | One candidate solution | x=11 | | Chromosome | The complete encoded candidate | 1011 | | Gene | One component of an encoding | The third bit | | Allele | The value at a gene | 0 or 1 for binary encoding | | Population | The collection of current individuals | [3,5,10,12] | | Fitness | A score used to favour useful candidates | For max x², fitness of 11 is 121 | | Generation | One cycle producing a new population | Select → cross → mutate → replace | | Search space | All candidate solutions permitted | 4-bit integers: 0–15 | | Elitism | Preserving best individuals unchanged | Copy the best into the next population | An objective describes what the problem wants. Fitness describes what the selection mechanism favours. For maximising a nonnegative objective, they may be the same. For minimising cost, directly selecting in proportion to cost would favour bad candidates; use a specified transformation or a comparison-based selection method. **For a maximisation GA with objective f(x)=x², find the fitness of x=3.** [2 marks · Practice] #### Given x=3; fitness equals the nonnegative objective x². #### To find Fitness, not the encoded chromosome. #### Formula and why it applies Square x to evaluate its objective; binary encoding is a separate representation step. #### Solution ##### Step 1. Substitute x $$ f(3)=3^2 $$ ##### Step 2. Multiply $$ 3\times3=9 $$ **Final answer.** Fitness = 9. **For a maximisation GA with objective f(x)=x², find the fitness of x=7.** [2 marks · Practice] #### Given x=7; fitness equals the nonnegative objective x². #### To find Fitness, not the encoded chromosome. #### Formula and why it applies Square x to evaluate its objective; binary encoding is a separate representation step. #### Solution ##### Step 1. Substitute x $$ f(7)=7^2 $$ ##### Step 2. Multiply $$ 7\times7=49 $$ **Final answer.** Fitness = 49. **For a maximisation GA with objective f(x)=x², find the fitness of x=12.** [2 marks · Practice] #### Given x=12; fitness equals the nonnegative objective x². #### To find Fitness, not the encoded chromosome. #### Formula and why it applies Square x to evaluate its objective; binary encoding is a separate representation step. #### Solution ##### Step 1. Substitute x $$ f(12)=12^2 $$ ##### Step 2. Multiply $$ 12\times12=144 $$ **Final answer.** Fitness = 144. **Yaad rakho:** Chromosome answer ka code; fitness us code wale answer ka score. --- ## M2 · Encoding: binary, real values and permutations Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--encoding/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/ga-core/encoding Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. ## Expand the compact decoding formula before using it $$ x=b_1 2^3+b_2 2^2+b_3 2^1+b_4 2^0 $$ For four bits, b₁ is the leftmost bit and receives weight 8; b₄ is the rightmost and receives weight 1. In general n−i gives the power belonging to bit i. For 1011, the four terms are 1×8, 0×4, 1×2, 1×1. Add 8+0+2+1=11. For real-range mapping, d/(2ⁿ−1) is the fraction of the way through the possible codes. Multiplying by U−L travels that fraction of the physical range. Adding L shifts the result to start at the lower limit. | 3-bit code | Decoded d | Fraction d/7 | Mapped to [−2,5] | | --- | --- | --- | --- | | 000 | 0 | 0/7=0 | −2+0×7=−2 | | 011 | 3 | 3/7 | −2+(3/7)×7=1 | | 111 | 7 | 7/7=1 | −2+1×7=5 | ## Choose an encoding that can express legal solutions **Binary encoding** uses bit strings; **real-valued encoding** stores real numbers directly; **permutation encoding** stores an order, such as a TSP city tour. An operator valid for bits may create invalid city tours, so encoding and operators must fit the problem. | Representation | Candidate | Appropriate use | | --- | --- | --- | | Binary | 1011 | An unsigned integer or a coded real value | | Real-valued | [0.2, 1.7] | Continuous parameters | | Permutation | [A,C,B,D] | Visit order in TSP | ## Decode an n-bit integer $$ x=b_1 2^{n-1}+b_2 2^{n-2}+\cdots+b_n2^0 $$ b₁ is the leftmost bit, not a coefficient you have to learn. In four bits the weights are 8,4,2,1. The maximum is $$ 2^n-1 $$ . **Decode the 4-bit chromosome 0011.** [2 marks · Practice] #### Given Bits 0011; place values 8,4,2,1. #### To find Integer x. #### Formula and why it applies Multiply each bit by its matching place value, then add. #### Solution ##### Step 1. List contributions | Bit | Place value | Contribution | | --- | --- | --- | | 0 | 8 | 0 | | 0 | 4 | 0 | | 1 | 2 | 2 | | 1 | 1 | 1 | ##### Step 2. Add $$ 0+0+2+1=3 $$ **Final answer.** x=3. **Decode the 4-bit chromosome 1010.** [2 marks · Practice] #### Given Bits 1010; place values 8,4,2,1. #### To find Integer x. #### Formula and why it applies Multiply each bit by its matching place value, then add. #### Solution ##### Step 1. List contributions | Bit | Place value | Contribution | | --- | --- | --- | | 1 | 8 | 8 | | 0 | 4 | 0 | | 1 | 2 | 2 | | 0 | 1 | 0 | ##### Step 2. Add $$ 8+0+2+0=10 $$ **Final answer.** x=10. **Decode the 4-bit chromosome 1111.** [2 marks · Practice] #### Given Bits 1111; place values 8,4,2,1. #### To find Integer x. #### Formula and why it applies Multiply each bit by its matching place value, then add. #### Solution ##### Step 1. List contributions | Bit | Place value | Contribution | | --- | --- | --- | | 1 | 8 | 8 | | 1 | 4 | 4 | | 1 | 2 | 2 | | 1 | 1 | 1 | ##### Step 2. Add $$ 8+4+2+1=15 $$ **Final answer.** x=15. ## Encoding a real interval If the question maps an n-bit integer d to a real interval [L,U], first decode d, then use: $$ x=L+\frac{d}{2^n-1}(U-L) $$ **Map the 3-bit string 101 to the real interval [−2,5] using linear binary decoding.** [3 marks · Practice] #### Given n=3, L=−2, U=5; string 101. #### To find Real x. #### Formula and why it applies The smallest bit code maps to L and the largest to U; the remaining codes are equally spaced. $$ x=L+\frac d{2^n-1}(U-L) $$ #### Solution ##### Step 1. Decode integer d $$ d=1(4)+0(2)+1(1)=5 $$ ##### Step 2. Calculate interval width and maximum code $$ U-L=5-(-2)=7\qquad2^3-1=7 $$ ##### Step 3. Map to the interval $$ x=-2+(5/7)(7)=3 $$ **Final answer.** x=3. **Yaad rakho:** Pehle bits se integer. Real range mapping explicitly diya ho tabhi next formula lagao. --- ## M2 · Roulette selection, cumulative intervals and alternatives Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--selection/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/ga-core/selection Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. Before the three complete sums below, read the running-total example in [Percentages, probabilities and cumulative intervals](https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--probability/). For weights [1,2,3,4], candidate C’s own probability is 0.3, but its interval is [0.3,0.6). A draw of 0.55 selects C because it lies inside that interval; it is not required to be less than C’s individual probability. ## Selection chooses parents, usually with replacement **Roulette-wheel selection** gives each candidate a probability proportional to a nonnegative fitness weight. A high weight occupies a larger interval on [0,1]. A supplied random number selects the interval containing it. $$ P_i=F_i/\sum_jF_j $$ Fᵢ is the fitness weight of individual i. The sum must be positive. Probabilities add to 1. Cumulative probability is the running sum and supplies the interval boundaries. **Fitness weights of successive individuals are [1, 2, 3, 4]. Select parents using random numbers [0.05, 0.25, 0.55, 0.9].** [3 marks · Practice] #### Given Fitness weights [1, 2, 3, 4]; draws [0.05, 0.25, 0.55, 0.9]. #### To find Parent row number for each draw. #### Formula and why it applies Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included. #### Solution ##### Step 1. Sum fitness $$ \sum F=1+2+3+4=10 $$ ##### Step 2. Find probabilities and cumulative bounds | Individual | Fitness | Probability | Cumulative | | --- | --- | --- | --- | | 1 | 1 | 0.1 | 0.1 | | 2 | 2 | 0.2 | 0.3 | | 3 | 3 | 0.3 | 0.6 | | 4 | 4 | 0.4 | 1 | ##### Step 3. Locate each draw | Draw | Selected individual | | --- | --- | | 0.05 | 1 | | 0.25 | 2 | | 0.55 | 3 | | 0.9 | 4 | **Final answer.** Selected row numbers: [1, 2, 3, 4]. Repeated selection is permitted. **Fitness weights of successive individuals are [4, 4, 2]. Select parents using random numbers [0.2, 0.6, 0.85].** [3 marks · Practice] #### Given Fitness weights [4, 4, 2]; draws [0.2, 0.6, 0.85]. #### To find Parent row number for each draw. #### Formula and why it applies Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included. #### Solution ##### Step 1. Sum fitness $$ \sum F=4+4+2=10 $$ ##### Step 2. Find probabilities and cumulative bounds | Individual | Fitness | Probability | Cumulative | | --- | --- | --- | --- | | 1 | 4 | 0.4 | 0.4 | | 2 | 4 | 0.4 | 0.8 | | 3 | 2 | 0.2 | 1 | ##### Step 3. Locate each draw | Draw | Selected individual | | --- | --- | | 0.2 | 1 | | 0.6 | 2 | | 0.85 | 3 | **Final answer.** Selected row numbers: [1, 2, 3]. Repeated selection is permitted. **Fitness weights of successive individuals are [0, 3, 1]. Select parents using random numbers [0.2, 0.8].** [3 marks · Practice] #### Given Fitness weights [0, 3, 1]; draws [0.2, 0.8]. #### To find Parent row number for each draw. #### Formula and why it applies Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included. #### Solution ##### Step 1. Sum fitness $$ \sum F=0+3+1=4 $$ ##### Step 2. Find probabilities and cumulative bounds | Individual | Fitness | Probability | Cumulative | | --- | --- | --- | --- | | 1 | 0 | 0 | 0 | | 2 | 3 | 0.75 | 0.75 | | 3 | 1 | 0.25 | 1 | ##### Step 3. Locate each draw | Draw | Selected individual | | --- | --- | | 0.2 | 2 | | 0.8 | 3 | **Final answer.** Selected row numbers: [2, 3]. Repeated selection is permitted. ## Other selection methods | Method | Procedure | Why it is used | | --- | --- | --- | | Tournament | Sample a group and take its best candidate | Only objective comparisons are needed | | Rank | Sort candidates; assign weights based on rank | Reduces sensitivity to a single huge raw fitness | | Elitist survival | Preserve top individuals unchanged into the next population | Protects good solutions from destructive variation | Selection is not automatically “replace the weakest with the fittest”. That is one possible classroom reproduction rule. If a problem gives selected parents, a rank rule, tournament contestants or roulette draws, follow that exact instruction. **Yaad rakho:** Probability nikalo → cumulative banao → random number ka interval dekho. --- ## M2 · Crossover and mutation: four complete worked variants Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--operators/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/ga-core/operators Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. ## Cutting means splitting the written string The **prefix** is the part before the cut; the **suffix** is the part after it. “10 joined with 00 gives 1000” means writing strings next to each other. It is not the arithmetic addition 10+00. | Operation | Parents / instruction | Construction | Children | | --- | --- | --- | --- | | One-point crossover | 10|11 and 01|00 | First prefix + second tail; second prefix + first tail | 1000 and 0111 | | Two-point crossover | 1|01|10 and 0|10|01 | Exchange only the middle blocks | 11010 and 00101 | | Uniform crossover | A=1010, B=0101, mask=1100 | Child 1 takes A where mask=1 and B where mask=0; child 2 reverses this | 1001 and 0110 | | Mutation | 1000; flip position 3 from left | The third bit is 0; replace it by 1; keep every other bit | 1010 | These crossover operators are alternatives. Apply the operator stated in the question; do not apply all three in one generation. Mutation is a separate step after crossover in the worked exam workflow. ## Crossover combines; mutation changes In **single-point crossover**, cut both parents after the specified bit. Child 1 keeps parent 1’s prefix and takes parent 2’s suffix; child 2 does the reverse. A cut after bit 2 in a four-bit string looks like 10|11. **Cross parents 1011 and 0100 after bit 2. Then flip bit 4 of both children, counting from the left. Decode the final children.** [3 marks · Practice] #### Given Parents 1011, 0100; cut after bit 2; mutation bit 4. #### To find Children after crossover, after mutation and their decimal values. #### Formula and why it applies Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1. #### Solution ##### Step 1. Mark the cut 10 | 11 and 01 | 00 ##### Step 2. Exchange tails 10 + 00 → 1000 01 + 11 → 0111 ##### Step 3. Flip the specified bit | Crossover child | Bit before | Bit after | Mutated child | | --- | --- | --- | --- | | 1000 | 0 | 1 | 1001 | | 0111 | 1 | 0 | 0110 | ##### Step 4. Decode | String | Place-value sum | x | | --- | --- | --- | | 1001 | 8 + 0 + 0 + 1 | 9 | | 0110 | 0 + 4 + 2 + 0 | 6 | **Final answer.** Final children 1001 (9) and 0110 (6). **Cross parents 0011 and 1100 after bit 1. Then flip bit 2 of both children, counting from the left. Decode the final children.** [3 marks · Practice] #### Given Parents 0011, 1100; cut after bit 1; mutation bit 2. #### To find Children after crossover, after mutation and their decimal values. #### Formula and why it applies Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1. #### Solution ##### Step 1. Mark the cut 0 | 011 and 1 | 100 ##### Step 2. Exchange tails 0 + 100 → 0100 1 + 011 → 1011 ##### Step 3. Flip the specified bit | Crossover child | Bit before | Bit after | Mutated child | | --- | --- | --- | --- | | 0100 | 1 | 0 | 0000 | | 1011 | 0 | 1 | 1111 | ##### Step 4. Decode | String | Place-value sum | x | | --- | --- | --- | | 0000 | 0 + 0 + 0 + 0 | 0 | | 1111 | 8 + 4 + 2 + 1 | 15 | **Final answer.** Final children 0000 (0) and 1111 (15). **Cross parents 1110 and 0001 after bit 3. Then flip bit 1 of both children, counting from the left. Decode the final children.** [3 marks · Practice] #### Given Parents 1110, 0001; cut after bit 3; mutation bit 1. #### To find Children after crossover, after mutation and their decimal values. #### Formula and why it applies Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1. #### Solution ##### Step 1. Mark the cut 111 | 0 and 000 | 1 ##### Step 2. Exchange tails 111 + 1 → 1111 000 + 0 → 0000 ##### Step 3. Flip the specified bit | Crossover child | Bit before | Bit after | Mutated child | | --- | --- | --- | --- | | 1111 | 1 | 0 | 0111 | | 0000 | 0 | 1 | 1000 | ##### Step 4. Decode | String | Place-value sum | x | | --- | --- | --- | | 0111 | 0 + 4 + 2 + 1 | 7 | | 1000 | 8 + 0 + 0 + 0 | 8 | **Final answer.** Final children 0111 (7) and 1000 (8). **Cross parents 1010 and 1010 after bit 2. Then flip bit 4 of both children, counting from the left. Decode the final children.** [3 marks · Practice] #### Given Parents 1010, 1010; cut after bit 2; mutation bit 4. #### To find Children after crossover, after mutation and their decimal values. #### Formula and why it applies Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1. #### Solution ##### Step 1. Mark the cut 10 | 10 and 10 | 10 ##### Step 2. Exchange tails 10 + 10 → 1010 10 + 10 → 1010 ##### Step 3. Flip the specified bit | Crossover child | Bit before | Bit after | Mutated child | | --- | --- | --- | --- | | 1010 | 0 | 1 | 1011 | | 1010 | 0 | 1 | 1011 | ##### Step 4. Decode | String | Place-value sum | x | | --- | --- | --- | | 1011 | 8 + 0 + 2 + 1 | 11 | | 1011 | 8 + 0 + 2 + 1 | 11 | **Final answer.** Final children 1011 (11) and 1011 (11). **Two-point crossover** exchanges the segment between two cuts. **Uniform crossover** uses a given mask to choose, at each position, which parent contributes a gene. Mutation may flip a bit, perturb a real value, or swap two positions in a permutation. These are different operators; apply the one named in the question. **Yaad rakho:** Crossover = tail exchange. Mutation = specified bit flip. Flip ka matlab 1 bhi 0 banega. --- ## M2 · GA architecture, two-generation flow, constraints and theory Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--algorithm/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/ga-core/algorithm Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. | Read this | Meaning | Concrete example | | --- | --- | --- | | Feasible | Obeys every stated constraint | x≤10: x=9 qualifies; x=12 does not | | Penalty | A deduction for violating a constraint | For maximisation, subtract 100×excess | | Elitism | Explicitly preserve a best old individual | Only apply if the question specifies it | | Replacement | Choose what population the next generation uses | Our full-replacement examples use the new children | | Stopping condition | A rule ending the repeated process | Stop after two complete generations | ## A complete GA generation Encode + evaluate → Select parents → Crossover → Mutation → Evaluate childrenreplace and repeat A **simple generational GA** replaces the population in generations using a chosen encoding, selection rule and variation operators. A more general GA may use steady-state replacement, different encodings, adaptive operators or explicit elitism. Always state the choices used in your worked answer. ## What “iteration 2” means here Initial population → first generation’s selection/crossover/mutation → new population → second generation’s selection/crossover/mutation → final population. Stopping after the first children and labelling them iteration 2 is not two full generations. ## Stopping conditions Stop after a fixed generation budget, after reaching a specified acceptable objective, or after little improvement for a stated number of generations. A finite run does not prove the global optimum. ## Constraints A **constraint** restricts allowed solutions, such as 0≤x≤15 or “visit every city once”. Common approaches are valid-by-design encoding, repairing an invalid child, rejecting it, or using a penalty. A penalty must reduce attractiveness: subtract it in maximisation or add it to a minimised cost. Do not silently remove the original constraint because you used more bits. **A maximisation problem has objective f(x)=x² and constraint x≤10. The given penalised fitness is F(x)=x²−100 max(0,x−10). Compare x=9 and x=12.** [3 marks · Practice] #### Given Penalty coefficient 100; bound 10; candidates 9 and 12. #### To find Penalised fitness of both candidates. #### Formula and why it applies Only excess above the bound receives a penalty. #### Solution ##### Step 1. Feasible candidate 9 $$ \max(0,9-10)=0 $$ $$ F(9)=81-100(0)=81 $$ ##### Step 2. Infeasible candidate 12 $$ \max(0,12-10)=2 $$ $$ F(12)=144-100(2)=-56 $$ ##### Step 3. Compare For this maximisation fitness, 81 > −56, so x=9 is preferred. Raw objective alone would misleadingly favour 12. **Final answer.** F(9)=81; F(12)=−56. | Advantages | Limitations | | --- | --- | | Derivative-free; supports many encodings | Many objective evaluations may be costly | | Population explores several candidates | Premature convergence is possible | | Crossover and mutation can create new candidates | Parameter and encoding choices affect results | | Applicable to discontinuous or combinatorial searches | No guaranteed global optimum; variation can worsen fitness | **Yaad rakho:** Do generation = do baar poora cycle. Sirf children banana aur decode karna do alag iterations nahi. --- ## M2 · Two generations with given selection, cut and mutation Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--two-generations/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/ga-core/two-generations Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. Swipe across the diagram to read its labels ↔ ![This diagram traces the first pair in generation 1. The full solution also processes the second pair and the entire second generation.](https://deep-learning-study-room.vercel.app/is2/fig/v2_ga_generation.fa7bf86c05.svg) Diagram: This diagram traces the first pair in generation 1. The full solution also processes the second pair and the entire second generation.. Labels: One complete GA generation: read from top to bottom | Selected parents | 1010 × 1100 | Cut after bit 2 | Mark the cut | 10 | 10 11 | 00 | Keep prefix, exchange suffix | After crossover | 1000 1110 | Children are 8 and 14 | Flip the last bit | 1001 1111 | 0 becomes 1 in each child | Decode and evaluate | 9² = 81 15² = 225 | These are the children’s fitness values This diagram traces the first pair in generation 1. The full solution also processes the second pair and the entire second generation. **Maximise $$ f(x)=x^2 $$ over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness.** [3 marks · Practice; complete two-generation solution] #### Given Initial ordered population: [3, 5, 10, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [1, 2, 3, 4] in every generation. Replace the whole population with the four children; no elitism. | Generation | Selection instruction | | --- | --- | | 1 | Current individual row numbers [3, 4, 4, 3] | | 2 | Current individual row numbers [2, 3, 2, 3] | #### To find Both generations: evaluation, selection, crossover, mutation, decoding and comparison. #### Formula and why it applies Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step. #### Solution ##### Step 1. Generation 1: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 3 | 0011 | 3 × 3 = 9 | | 2 | 5 | 0101 | 5 × 5 = 25 | | 3 | 10 | 1010 | 10 × 10 = 100 | | 4 | 12 | 1100 | 12 × 12 = 144 | ##### Step 2. Generation 1: apply the supplied selection order The question directly supplies individual numbers [3, 4, 4, 3]. These refer to the current generation’s row numbers, not their x values. | Mating-pool slot | Chosen individual | x | Chromosome | | --- | --- | --- | --- | | 1 | 3 | 10 | 1010 | | 2 | 4 | 12 | 1100 | | 3 | 4 | 12 | 1100 | | 4 | 3 | 10 | 1010 | ##### Step 3. Generation 1: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 10 | 10 | 11 | 00 | 1000 | 1110 | | 3 & 4 | 11 | 00 | 10 | 10 | 1110 | 1000 | ##### Step 4. Generation 1: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1000 | Flip bit 4 | 1001 | | 2 | 1110 | Flip bit 4 | 1111 | | 3 | 1110 | Flip bit 4 | 1111 | | 4 | 1000 | Flip bit 4 | 1001 | ##### Step 5. Generation 1: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 | | 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 | | 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 | | 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 | ##### Step 6. Generation 1: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{9+25+100+144}{4}=\frac{278}{4}=69.5 $$ $$ \overline f_{new}=\frac{81+225+225+81}{4}=\frac{612}{4}=153 $$ Best objective: 144 → 225. Higher is better. The next population is [9, 15, 15, 9]. No elitism is imposed, so the best previous individual is not automatically protected. ##### Step 7. Generation 2: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 9 | 1001 | 9 × 9 = 81 | | 2 | 15 | 1111 | 15 × 15 = 225 | | 3 | 15 | 1111 | 15 × 15 = 225 | | 4 | 9 | 1001 | 9 × 9 = 81 | ##### Step 8. Generation 2: apply the supplied selection order The question directly supplies individual numbers [2, 3, 2, 3]. These refer to the current generation’s row numbers, not their x values. | Mating-pool slot | Chosen individual | x | Chromosome | | --- | --- | --- | --- | | 1 | 2 | 15 | 1111 | | 2 | 3 | 15 | 1111 | | 3 | 2 | 15 | 1111 | | 4 | 3 | 15 | 1111 | ##### Step 9. Generation 2: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 11 | 11 | 11 | 11 | 1111 | 1111 | | 3 & 4 | 11 | 11 | 11 | 11 | 1111 | 1111 | ##### Step 10. Generation 2: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1111 | Flip bit 4 | 1110 | | 2 | 1111 | Flip bit 4 | 1110 | | 3 | 1111 | Flip bit 4 | 1110 | | 4 | 1111 | Flip bit 4 | 1110 | ##### Step 11. Generation 2: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | ##### Step 12. Generation 2: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{81+225+225+81}{4}=\frac{612}{4}=153 $$ $$ \overline f_{new}=\frac{196+196+196+196}{4}=\frac{784}{4}=196 $$ Best objective: 225 → 196. Higher is better. The next population is [14, 14, 14, 14]. No elitism is imposed, so the best previous individual is not automatically protected. **Final answer.** After two generations the ordered population is [14, 14, 14, 14]. Best final objective = 196 at x = [14]. --- ## M2 · Two generations including roulette selection Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--roulette-generations/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/ga-core/roulette-generations Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. **Maximise $$ f(x)=x(15-x)+1 $$ over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness.** [3 marks · Practice; complete two-generation solution] #### Given Initial ordered population: [2, 5, 9, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [2, 4] in every generation. Replace the whole population with the four children; no elitism. | Generation | Selection instruction | | --- | --- | | 1 | Random draws [0.1, 0.4, 0.65, 0.9] | | 2 | Random draws [0.2, 0.4, 0.6, 0.95] | #### To find Both generations: evaluation, selection, crossover, mutation, decoding and comparison. #### Formula and why it applies Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step. #### Solution ##### Step 1. Generation 1: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 2 | 0010 | 2 × (15 − 2) + 1 = 27 | | 2 | 5 | 0101 | 5 × (15 − 5) + 1 = 51 | | 3 | 9 | 1001 | 9 × (15 − 9) + 1 = 55 | | 4 | 12 | 1100 | 12 × (15 − 12) + 1 = 37 | ##### Step 2. Generation 1: calculate roulette intervals Fitness weights are nonnegative. The nonnegative objective itself is the fitness in this maximisation example. $$ \sum F=27+51+55+37\approx170 $$ | Individual | Probability: weight ÷ total | Cumulative: old boundary + probability | | --- | --- | --- | | 1 | 27 ÷ 170 ≈ 0.158823529 | 0 + 0.158823529 ≈ 0.158823529 | | 2 | 51 ÷ 170 ≈ 0.3 | 0.158823529 + 0.3 ≈ 0.458823529 | | 3 | 55 ÷ 170 ≈ 0.323529412 | 0.458823529 + 0.323529412 ≈ 0.782352941 | | 4 | 37 ÷ 170 ≈ 0.217647059 | 0.782352941 + 0.217647059 ≈ 1 | ##### Step 3. Generation 1: use the supplied random draws | Random draw | Interval containing it | Selected individual | Selected x | | --- | --- | --- | --- | | 0.1 | 0 ≤ 0.1 < 0.158823529 | 1 | 2 | | 0.4 | 0.158823529 ≤ 0.4 < 0.458823529 | 2 | 5 | | 0.65 | 0.458823529 ≤ 0.65 < 0.782352941 | 3 | 9 | | 0.9 | 0.782352941 ≤ 0.9 < 1 | 4 | 12 | Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers. ##### Step 4. Generation 1: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 00 | 10 | 01 | 01 | 0001 | 0110 | | 3 & 4 | 10 | 01 | 11 | 00 | 1000 | 1101 | ##### Step 5. Generation 1: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 0001 | No mutation | 0001 | | 2 | 0110 | Flip bit 4 | 0111 | | 3 | 1000 | No mutation | 1000 | | 4 | 1101 | Flip bit 4 | 1100 | ##### Step 6. Generation 1: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 0001 | 0 + 0 + 0 + 1 = 1 | 1(15−1)+1 = 15 | | 0111 | 0 + 4 + 2 + 1 = 7 | 7(15−7)+1 = 57 | | 1000 | 8 + 0 + 0 + 0 = 8 | 8(15−8)+1 = 57 | | 1100 | 8 + 4 + 0 + 0 = 12 | 12(15−12)+1 = 37 | ##### Step 7. Generation 1: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{27+51+55+37}{4}=\frac{170}{4}=42.5 $$ $$ \overline f_{new}=\frac{15+57+57+37}{4}=\frac{166}{4}=41.5 $$ Best objective: 55 → 57. Higher is better. The next population is [1, 7, 8, 12]. No elitism is imposed, so the best previous individual is not automatically protected. ##### Step 8. Generation 2: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 1 | 0001 | 1 × (15 − 1) + 1 = 15 | | 2 | 7 | 0111 | 7 × (15 − 7) + 1 = 57 | | 3 | 8 | 1000 | 8 × (15 − 8) + 1 = 57 | | 4 | 12 | 1100 | 12 × (15 − 12) + 1 = 37 | ##### Step 9. Generation 2: calculate roulette intervals Fitness weights are nonnegative. The nonnegative objective itself is the fitness in this maximisation example. $$ \sum F=15+57+57+37\approx166 $$ | Individual | Probability: weight ÷ total | Cumulative: old boundary + probability | | --- | --- | --- | | 1 | 15 ÷ 166 ≈ 0.090361446 | 0 + 0.090361446 ≈ 0.090361446 | | 2 | 57 ÷ 166 ≈ 0.343373494 | 0.090361446 + 0.343373494 ≈ 0.43373494 | | 3 | 57 ÷ 166 ≈ 0.343373494 | 0.43373494 + 0.343373494 ≈ 0.777108434 | | 4 | 37 ÷ 166 ≈ 0.222891566 | 0.777108434 + 0.222891566 ≈ 1 | ##### Step 10. Generation 2: use the supplied random draws | Random draw | Interval containing it | Selected individual | Selected x | | --- | --- | --- | --- | | 0.2 | 0.090361446 ≤ 0.2 < 0.43373494 | 2 | 7 | | 0.4 | 0.090361446 ≤ 0.4 < 0.43373494 | 2 | 7 | | 0.6 | 0.43373494 ≤ 0.6 < 0.777108434 | 3 | 8 | | 0.95 | 0.777108434 ≤ 0.95 < 1 | 4 | 12 | Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers. ##### Step 11. Generation 2: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 01 | 11 | 01 | 11 | 0111 | 0111 | | 3 & 4 | 10 | 00 | 11 | 00 | 1000 | 1100 | ##### Step 12. Generation 2: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 0111 | No mutation | 0111 | | 2 | 0111 | Flip bit 4 | 0110 | | 3 | 1000 | No mutation | 1000 | | 4 | 1100 | Flip bit 4 | 1101 | ##### Step 13. Generation 2: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 0111 | 0 + 4 + 2 + 1 = 7 | 7(15−7)+1 = 57 | | 0110 | 0 + 4 + 2 + 0 = 6 | 6(15−6)+1 = 55 | | 1000 | 8 + 0 + 0 + 0 = 8 | 8(15−8)+1 = 57 | | 1101 | 8 + 4 + 0 + 1 = 13 | 13(15−13)+1 = 27 | ##### Step 14. Generation 2: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{15+57+57+37}{4}=\frac{166}{4}=41.5 $$ $$ \overline f_{new}=\frac{57+55+57+27}{4}=\frac{196}{4}=49 $$ Best objective: 57 → 57. Higher is better. The next population is [7, 6, 8, 13]. No elitism is imposed, so the best previous individual is not automatically protected. **Final answer.** After two generations the ordered population is [7, 6, 8, 13]. Best final objective = 57 at x = [7, 8]. --- ## M2 · GA minimisation: transform cost before roulette selection Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--minimisation/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/ga-core/minimisation Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. Roulette favours a larger weight. A minimisation objective instead favours a smaller cost. The stated transformation F=1/(1+cost) reverses their order for nonnegative costs: cost 0 → weight 1/1=1; cost 3 → 1/4=.25; cost 24 → 1/25=.04. The +1 prevents division by zero. This is one specified transformation, not a universal rule for every minimisation problem. **Minimise $$ f(x)=(x-7)^2 $$ over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness.** [3 marks · Practice; complete two-generation solution] #### Given Initial ordered population: [2, 4, 8, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [2, 4] in every generation. Replace the whole population with the four children; no elitism. | Generation | Selection instruction | | --- | --- | | 1 | Random draws [0.1, 0.4, 0.65, 0.9] | | 2 | Random draws [0.2, 0.4, 0.6, 0.95] | #### To find Both generations: evaluation, selection, crossover, mutation, decoding and comparison. #### Formula and why it applies Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Roulette uses Fᵢ/ΣF. For minimisation here, use Fᵢ=1/(1+f(xᵢ)); do not directly favour large costs. #### Solution ##### Step 1. Generation 1: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 2 | 0010 | (2 − 7)² = 25 | | 2 | 4 | 0100 | (4 − 7)² = 9 | | 3 | 8 | 1000 | (8 − 7)² = 1 | | 4 | 12 | 1100 | (12 − 7)² = 25 | ##### Step 2. Generation 1: calculate roulette intervals Fitness weights are nonnegative. For this minimisation example use F=1/(1+objective), so a smaller cost receives a larger weight. | Individual | Cost | Fitness calculation | | --- | --- | --- | | 1 | 25 | 1 / (1 + 25) = 1 / 26 ≈ 0.038461538 | | 2 | 9 | 1 / (1 + 9) = 1 / 10 ≈ 0.1 | | 3 | 1 | 1 / (1 + 1) = 1 / 2 ≈ 0.5 | | 4 | 25 | 1 / (1 + 25) = 1 / 26 ≈ 0.038461538 | $$ \sum F=0.038461538+0.1+0.5+0.038461538\approx0.676923077 $$ | Individual | Probability: weight ÷ total | Cumulative: old boundary + probability | | --- | --- | --- | | 1 | 0.038461538 ÷ 0.676923077 ≈ 0.056818182 | 0 + 0.056818182 ≈ 0.056818182 | | 2 | 0.1 ÷ 0.676923077 ≈ 0.147727273 | 0.056818182 + 0.147727273 ≈ 0.204545455 | | 3 | 0.5 ÷ 0.676923077 ≈ 0.738636364 | 0.204545455 + 0.738636364 ≈ 0.943181818 | | 4 | 0.038461538 ÷ 0.676923077 ≈ 0.056818182 | 0.943181818 + 0.056818182 ≈ 1 | ##### Step 3. Generation 1: use the supplied random draws | Random draw | Interval containing it | Selected individual | Selected x | | --- | --- | --- | --- | | 0.1 | 0.056818182 ≤ 0.1 < 0.204545455 | 2 | 4 | | 0.4 | 0.204545455 ≤ 0.4 < 0.943181818 | 3 | 8 | | 0.65 | 0.204545455 ≤ 0.65 < 0.943181818 | 3 | 8 | | 0.9 | 0.204545455 ≤ 0.9 < 0.943181818 | 3 | 8 | Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers. ##### Step 4. Generation 1: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 01 | 00 | 10 | 00 | 0100 | 1000 | | 3 & 4 | 10 | 00 | 10 | 00 | 1000 | 1000 | ##### Step 5. Generation 1: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 0100 | No mutation | 0100 | | 2 | 1000 | Flip bit 4 | 1001 | | 3 | 1000 | No mutation | 1000 | | 4 | 1000 | Flip bit 4 | 1001 | ##### Step 6. Generation 1: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 0100 | 0 + 4 + 0 + 0 = 4 | (4−7)² = 9 | | 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 | | 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 | | 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 | ##### Step 7. Generation 1: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{25+9+1+25}{4}=\frac{60}{4}=15 $$ $$ \overline f_{new}=\frac{9+4+1+4}{4}=\frac{18}{4}=4.5 $$ Best objective: 1 → 1. Lower is better. The next population is [4, 9, 8, 9]. No elitism is imposed, so the best previous individual is not automatically protected. ##### Step 8. Generation 2: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 4 | 0100 | (4 − 7)² = 9 | | 2 | 9 | 1001 | (9 − 7)² = 4 | | 3 | 8 | 1000 | (8 − 7)² = 1 | | 4 | 9 | 1001 | (9 − 7)² = 4 | ##### Step 9. Generation 2: calculate roulette intervals Fitness weights are nonnegative. For this minimisation example use F=1/(1+objective), so a smaller cost receives a larger weight. | Individual | Cost | Fitness calculation | | --- | --- | --- | | 1 | 9 | 1 / (1 + 9) = 1 / 10 ≈ 0.1 | | 2 | 4 | 1 / (1 + 4) = 1 / 5 ≈ 0.2 | | 3 | 1 | 1 / (1 + 1) = 1 / 2 ≈ 0.5 | | 4 | 4 | 1 / (1 + 4) = 1 / 5 ≈ 0.2 | $$ \sum F=0.1+0.2+0.5+0.2\approx1 $$ | Individual | Probability: weight ÷ total | Cumulative: old boundary + probability | | --- | --- | --- | | 1 | 0.1 ÷ 1 ≈ 0.1 | 0 + 0.1 ≈ 0.1 | | 2 | 0.2 ÷ 1 ≈ 0.2 | 0.1 + 0.2 ≈ 0.3 | | 3 | 0.5 ÷ 1 ≈ 0.5 | 0.3 + 0.5 ≈ 0.8 | | 4 | 0.2 ÷ 1 ≈ 0.2 | 0.8 + 0.2 ≈ 1 | ##### Step 10. Generation 2: use the supplied random draws | Random draw | Interval containing it | Selected individual | Selected x | | --- | --- | --- | --- | | 0.2 | 0.1 ≤ 0.2 < 0.3 | 2 | 9 | | 0.4 | 0.3 ≤ 0.4 < 0.8 | 3 | 8 | | 0.6 | 0.3 ≤ 0.6 < 0.8 | 3 | 8 | | 0.95 | 0.8 ≤ 0.95 < 1 | 4 | 9 | Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers. ##### Step 11. Generation 2: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 10 | 01 | 10 | 00 | 1000 | 1001 | | 3 & 4 | 10 | 00 | 10 | 01 | 1001 | 1000 | ##### Step 12. Generation 2: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1000 | No mutation | 1000 | | 2 | 1001 | Flip bit 4 | 1000 | | 3 | 1001 | No mutation | 1001 | | 4 | 1000 | Flip bit 4 | 1001 | ##### Step 13. Generation 2: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 | | 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 | | 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 | | 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 | ##### Step 14. Generation 2: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{9+4+1+4}{4}=\frac{18}{4}=4.5 $$ $$ \overline f_{new}=\frac{1+1+4+4}{4}=\frac{10}{4}=2.5 $$ Best objective: 1 → 1. Lower is better. The next population is [8, 8, 9, 9]. No elitism is imposed, so the best previous individual is not automatically protected. **Final answer.** After two generations the ordered population is [8, 8, 9, 9]. Best final objective = 1 at x = [8]. --- ## M2 · Swarm intelligence, particles, velocity, pbest and gbest Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--meaning/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/pso-core/meaning First understand position, velocity and memory; then solve complete two-iteration minimisation examples. ## Swarm intelligence **Swarm intelligence** is collective problem-solving behaviour that emerges from interactions among many relatively simple agents and their environment. It is inspired by systems such as bird flocks, fish schools and ant colonies. **Particle swarm optimization (PSO)** is a population-based search method inspired by social movement. Each particle represents a candidate solution. It moves through the search space using a velocity influenced by its own best previous position and the best position known to the swarm. ## Translate every PSO word into a concrete object | Term | Meaning | Example | | --- | --- | --- | | Particle | One candidate solution | Particle A | | Position x | The candidate’s coordinates | xA = 3 | | Objective f(x) | The problem’s score for that candidate | f(3) = 3² = 9 | | Velocity v | Proposed change of position | vA = −1 means move left by 1 | | pbest | Best position that this particle has visited | A once visited x=1 with score 1 | | gbest | Best position found by any particle | Among all personal bests, x=0.5 scores 0.25 | Swipe across the diagram to read its labels ↔ ![A best memory changes only when the new score improves.](https://deep-learning-study-room.vercel.app/is2/fig/v2_pso_memory.fcc9783a95.svg) Diagram: A best memory changes only when the new score improves.. Labels: Best memory survives a worse move | Visit 1 | Position x = 3 | Score x² = 9 | Remember pbest = 3 | Visit 2 | Position x = 1 | Score x² = 1 | Remember pbest = 1 | Visit 3 | Position x = 2 | Score x² = 4 | Remember pbest = 1 | Minimise x²: after visiting x = 2, the best past position is still x = 1. | Current position and remembered best position are separate records. A best memory changes only when the new score improves. ## Current position is not necessarily personal best For minimising x², suppose a particle visits x=3, then x=1, then x=2. Its current position is 2, but its personal best stays 1 because 1² is less than 2². The particle remembers a better past result even after a bad move. **A particle visits x=3, x=1, x=2 in that order while minimising f(x)=x². Find pbest after each visit.** [2 marks · Practice] #### Given Positions in chronological order: 3 → 1 → 2. #### To find The best-so-far position and score. #### Formula and why it applies Keep the location with the lowest objective encountered so far. #### Solution ##### Step 1. First visit $$ f(3)=9 $$ No history: pbest=3, best score=9. ##### Step 2. Second visit $$ f(1)=1<9 $$ Replace pbest by 1. ##### Step 3. Third visit $$ f(2)=4>1 $$ Keep pbest=1 even though the current position is 2. **Final answer.** Final current position=2; pbest=1; personal-best score=1. **Define swarm intelligence and briefly explain one nature-inspired algorithm.** [2 marks · Practice] #### Answer Swarm intelligence is collective problem-solving that emerges from interactions among simple agents. Particle swarm optimization is inspired by flocking behaviour. Each particle is a candidate solution with a position and velocity. It remembers its own best position and uses the swarm’s best position to guide movement. Repeated evaluation and movement search for a low-cost or high-fitness solution. **Yaad rakho:** Position abhi kahan ho; pbest apna best past; gbest sabka best past. --- ## M2 · Read the PSO equation and work three individual updates Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--equation/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/pso-core/equation First understand position, velocity and memory; then solve complete two-iteration minimisation examples. ## First read the update in words **New velocity = retained old velocity + pull toward personal best + pull toward the group best.** After finding that velocity, add it to the old position. You are calculating a movement, then applying it. Swipe across the diagram to read its labels ↔ ![A negative difference pulls toward a best position on the left.](https://deep-learning-study-room.vercel.app/is2/fig/depth_pso.88e9c2a672.svg) Diagram: A negative difference pulls toward a best position on the left.. Labels: Read the direction before multiplying the pull | Best = 1 | Current = 4 | difference = 1 − 4 = −3 | coefficient 0.5 → pull 0.5 × (−3) = −1.5 (left) A negative difference pulls toward a best position on the left. Example: current x=4 and personal best=1 gives the difference 1−4=−3. A coefficient of .5 gives personal pull .5×(−3)=−1.5. The negative sign says “move left”; it does not mean a negative probability. If current x=−2 and best=1, the difference is 1−(−2)=1+2=3, which pulls right. In vi,jt, i chooses the particle, j chooses its coordinate, and t chooses the recorded iteration. For particle B, coordinate 2, iteration 0, look at row B and the second velocity entry in the initial table. The letters identify a stored number; they are not extra multiplication factors. ## The velocity equation is three contributions added together $$ v_{i,j}^{t+1}=w v_{i,j}^{t}+c_1r_1(pbest_{i,j}^{t}-x_{i,j}^{t})+c_2r_2(gbest_j^{t}-x_{i,j}^{t}) $$ $$ x_{i,j}^{t+1}=x_{i,j}^{t}+v_{i,j}^{t+1} $$ | Symbol | Meaning | | --- | --- | | i | Which particle | | j | Which coordinate of that particle | | t / t+1 | Before / after this iteration; these are labels, not powers | | w | Inertia weight: how much old velocity is kept | | c₁r₁ | Combined strength of the pull to personal best | | c₂r₂ | Combined strength of the pull to global best | | r₁,r₂ | Random factors, normally in [0,1]; use the values supplied in an exam | A best position to the left gives a negative difference and pulls left; a best position to the right gives a positive difference and pulls right. The multiplication sets the strength of each pull. **For one coordinate, x=4, v=0, pbest=4, gbest=2. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position.** [3 marks · Practice] #### Given x=4, v=0, pbest=4, gbest=2; w=.5; c₁r₁=.5; c₂r₂=.5. #### To find New velocity followed by new position. #### Formula and why it applies Use the three-term velocity equation, then add the new velocity to x. $$ v_{new}=0.5v+0.5(pbest-x)+0.5(gbest-x) $$ #### Solution ##### Step 1. Inertia $$ 0.5\times0=0 $$ ##### Step 2. Personal pull $$ 0.5(4-(4))=0 $$ ##### Step 3. Social pull $$ 0.5(2-(4))=-1 $$ ##### Step 4. New velocity $$ v_{new}=0+0+(-1)=-1 $$ ##### Step 5. New position $$ x_{new}=4+(-1)=3 $$ **Final answer.** vnew=-1; xnew=3. **For one coordinate, x=4, v=-2, pbest=3, gbest=1. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position.** [3 marks · Practice] #### Given x=4, v=-2, pbest=3, gbest=1; w=.5; c₁r₁=.5; c₂r₂=.5. #### To find New velocity followed by new position. #### Formula and why it applies Use the three-term velocity equation, then add the new velocity to x. $$ v_{new}=0.5v+0.5(pbest-x)+0.5(gbest-x) $$ #### Solution ##### Step 1. Inertia $$ 0.5\times(-2)=-1 $$ ##### Step 2. Personal pull $$ 0.5(3-(4))=-0.5 $$ ##### Step 3. Social pull $$ 0.5(1-(4))=-1.5 $$ ##### Step 4. New velocity $$ v_{new}=(-1)+(-0.5)+(-1.5)=-3 $$ ##### Step 5. New position $$ x_{new}=4+(-3)=1 $$ **Final answer.** vnew=-3; xnew=1. **For one coordinate, x=-2, v=1, pbest=-1, gbest=1. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position.** [3 marks · Practice] #### Given x=-2, v=1, pbest=-1, gbest=1; w=.5; c₁r₁=.5; c₂r₂=.5. #### To find New velocity followed by new position. #### Formula and why it applies Use the three-term velocity equation, then add the new velocity to x. $$ v_{new}=0.5v+0.5(pbest-x)+0.5(gbest-x) $$ #### Solution ##### Step 1. Inertia $$ 0.5\times1=0.5 $$ ##### Step 2. Personal pull $$ 0.5(-1-(-2))=0.5 $$ ##### Step 3. Social pull $$ 0.5(1-(-2))=1.5 $$ ##### Step 4. New velocity $$ v_{new}=0.5+0.5+1.5=2.5 $$ ##### Step 5. New position $$ x_{new}=(-2)+2.5=0.5 $$ **Final answer.** vnew=2.5; xnew=0.5. **First-iteration shortcut:** if pbest was just initialised to x, the personal difference is zero. This follows from the initialisation; it is not a permanent property of PSO. A gbest particle can still move because its inertia term can be nonzero. **Yaad rakho:** Old speed ka hissa + apne best ki pull + group best ki pull; phir new speed ko position mein add. --- ## M2 · Two particles, two iterations: start with the simplest minimisation Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--two-particles/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/pso-core/two-particles First understand position, velocity and memory; then solve complete two-iteration minimisation examples. A synchronous iteration is a class photograph of the old swarm: every particle uses that same snapshot of gbest. First calculate all new positions, then compare all new scores and update memories. If particle A discovers a better position halfway through the calculations, B still uses the previous snapshot during this iteration. B can use the improved gbest in the next iteration. Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values. **Minimise $$ f(x)=x^2 $$ using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [3 marks · Practice; expanded teaching solution] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (-2) | (0) | | B | (4) | (0) | w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(-2)=(-2)^2=4=4 $$ **Particle B** $$ f(4)=4^2=16=16 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(-2) $$ Its score is 4. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (-2). pbest values are A: (-2), B: (4). ##### Step 4. Iteration 1: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=-2-(-2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=-2-(-2)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=(-2)+0=-2 $$ ##### Step 5. Iteration 1: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=4-(4)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=-2-(4)=-6 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-6)=-3 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-3)=-3 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=4+(-3)=1 $$ ##### Step 6. Iteration 1: evaluate the new positions **Particle A** $$ f(-2)=(-2)^2=4=4 $$ **Particle B** $$ f(1)=1^2=1=1 $$ ##### Step 7. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 4 | 4 | Keep old: tie | (-2) | | B | 16 | 1 | Replace: new score is lower | (1) | Compare the retained pbest scores. gbest = **(1)**, with objective **1**. ##### Step 8. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (-2) | (0) | (-2) | | B | (1) | (-3) | (1) | ##### Step 9. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (1). pbest values are A: (-2), B: (1). ##### Step 10. Iteration 2: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=-2-(-2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(-2)=3 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(3)=1.5 $$ Add the three velocity contributions. $$ v^{(2)}=0+0+1.5=1.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=(-2)+1.5=-0.5 $$ ##### Step 11. Iteration 2: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-3)=-1.5 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(2)}=(-1.5)+0+0=-1.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1+(-1.5)=-0.5 $$ ##### Step 12. Iteration 2: evaluate the new positions **Particle A** $$ f(-0.5)=(-0.5)^2=0.25=0.25 $$ **Particle B** $$ f(-0.5)=(-0.5)^2=0.25=0.25 $$ ##### Step 13. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 4 | 0.25 | Replace: new score is lower | (-0.5) | | B | 1 | 0.25 | Replace: new score is lower | (-0.5) | Compare the retained pbest scores. gbest = **(-0.5)**, with objective **0.25**. ##### Step 14. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (-0.5) | (1.5) | (-0.5) | | B | (-0.5) | (-1.5) | (-0.5) | **Final answer.** After iteration 2, gbest = (-0.5), objective = 0.25. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. --- ## M2 · A worse move: pbest stays in the past Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--memory-matters/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/pso-core/memory-matters First understand position, velocity and memory; then solve complete two-iteration minimisation examples. Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values. **Minimise $$ f(x)=x^2 $$ using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [3 marks · Practice; expanded teaching solution] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (1) | (4) | | B | (3) | (-1) | w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(1)=1^2=1=1 $$ **Particle B** $$ f(3)=3^2=9=9 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(1) $$ Its score is 1. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (1). pbest values are A: (1), B: (3). ##### Step 4. Iteration 1: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times4=2 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=2+0+0=2 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=1+2=3 $$ ##### Step 5. Iteration 1: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-1)=-0.5 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3-(3)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(3)=-2 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-2)=-1 $$ Add the three velocity contributions. $$ v^{(1)}=(-0.5)+0+(-1)=-1.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=3+(-1.5)=1.5 $$ ##### Step 6. Iteration 1: evaluate the new positions **Particle A** $$ f(3)=3^2=9=9 $$ **Particle B** $$ f(1.5)=1.5^2=2.25=2.25 $$ ##### Step 7. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 1 | 9 | Keep old: new score is worse | (1) | | B | 9 | 2.25 | Replace: new score is lower | (1.5) | Compare the retained pbest scores. gbest = **(1)**, with objective **1**. ##### Step 8. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (3) | (2) | (1) | | B | (1.5) | (-1.5) | (1.5) | ##### Step 9. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (1). pbest values are A: (1), B: (1.5). ##### Step 10. Iteration 2: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times2=1 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(3)=-2 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(-2)=-1 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(3)=-2 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-2)=-1 $$ Add the three velocity contributions. $$ v^{(2)}=1+(-1)+(-1)=-1 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=3+(-1)=2 $$ ##### Step 11. Iteration 2: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-1.5)=-0.75 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1.5-(1.5)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1.5)=-0.5 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-0.5)=-0.25 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.75)+0+(-0.25)=-1 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1.5+(-1)=0.5 $$ ##### Step 12. Iteration 2: evaluate the new positions **Particle A** $$ f(2)=2^2=4=4 $$ **Particle B** $$ f(0.5)=0.5^2=0.25=0.25 $$ ##### Step 13. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 1 | 4 | Keep old: new score is worse | (1) | | B | 2.25 | 0.25 | Replace: new score is lower | (0.5) | Compare the retained pbest scores. gbest = **(0.5)**, with objective **0.25**. ##### Step 14. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (2) | (-1) | (1) | | B | (0.5) | (-1) | (0.5) | **Final answer.** After iteration 2, gbest = (0.5), objective = 0.25. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. **Yaad rakho:** Naya result kharab ho toh position toh badlegi, pbest nahi. --- ## M2 · Two-dimensional PSO: keep the coordinates separate Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--two-coordinates/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/pso-core/two-coordinates First understand position, velocity and memory; then solve complete two-iteration minimisation examples. A two-dimensional particle has two entries in its position and two in its velocity. Apply the scalar velocity calculation once to coordinate 1 and once to coordinate 2. Keep them in the same order to form the new pair. Only after both entries are ready do you insert the pair into f(x₁,x₂). Adding x₁ and x₂ into a single position would lose information. Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values. **Minimise $$ f(x)=x_1^2+x_2^2 $$ using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [3 marks · Practice; expanded teaching solution] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (1, 1) | (0, 0) | | B | (3, 2) | (0, 0) | w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(1, 1)=1^2+1^2=1+1=2 $$ **Particle B** $$ f(3, 2)=3^2+2^2=9+4=13 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(1, 1) $$ Its score is 2. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (1, 1). pbest values are A: (1, 1), B: (3, 2). ##### Step 4. Iteration 1: update particle A **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=1+0=1 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=1+0=1 $$ ##### Step 5. Iteration 1: update particle B **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3-(3)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(3)=-2 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-2)=-1 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-1)=-1 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=3+(-1)=2 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=2-(2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(2)=-1 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-1)=-0.5 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-0.5)=-0.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=2+(-0.5)=1.5 $$ ##### Step 6. Iteration 1: evaluate the new positions **Particle A** $$ f(1, 1)=1^2+1^2=1+1=2 $$ **Particle B** $$ f(2, 1.5)=2^2+1.5^2=4+2.25=6.25 $$ ##### Step 7. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 2 | 2 | Keep old: tie | (1, 1) | | B | 13 | 6.25 | Replace: new score is lower | (2, 1.5) | Compare the retained pbest scores. gbest = **(1, 1)**, with objective **2**. ##### Step 8. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (1, 1) | (0, 0) | (1, 1) | | B | (2, 1.5) | (-1, -0.5) | (2, 1.5) | ##### Step 9. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (1, 1). pbest values are A: (1, 1), B: (2, 1.5). ##### Step 10. Iteration 2: update particle A **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(2)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1+0=1 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(2)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1+0=1 $$ ##### Step 11. Iteration 2: update particle B **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-1)=-0.5 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=2-(2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(2)=-1 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-1)=-0.5 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.5)+0+(-0.5)=-1 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=2+(-1)=1 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-0.5)=-0.25 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1.5-(1.5)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1.5)=-0.5 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-0.5)=-0.25 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.25)+0+(-0.25)=-0.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1.5+(-0.5)=1 $$ ##### Step 12. Iteration 2: evaluate the new positions **Particle A** $$ f(1, 1)=1^2+1^2=1+1=2 $$ **Particle B** $$ f(1, 1)=1^2+1^2=1+1=2 $$ ##### Step 13. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 2 | 2 | Keep old: tie | (1, 1) | | B | 6.25 | 2 | Replace: new score is lower | (1, 1) | Compare the retained pbest scores. gbest = **(1, 1)**, with objective **2**. ##### Step 14. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (1, 1) | (0, 0) | (1, 1) | | B | (1, 1) | (-1, -0.5) | (1, 1) | **Final answer.** After iteration 2, gbest = (1, 1), objective = 2. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. --- ## M2 · Class PSO photograph: all three particles through iteration 2 Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--class-three-particles/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/pso-core/class-three-particles First understand position, velocity and memory; then solve complete two-iteration minimisation examples. Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values. The positions, velocities and coefficients are transcribed from the class photo. We use f(x₁,x₂)=x₁²+x₂²−x₁x₂, minimisation, initial pbest=x, fixed supplied random factors and synchronous updates. The photo does not itself specify all these conventions; they are stated here to make the two-iteration extension unambiguous. **Minimise $$ f(x)=x_1^2+x_2^2-x_1x_2 $$ using 3 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [10 marks · Class data; two-iteration minimisation extension] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (3.14, 7.21) | (0.1, 0.3) | | B | (4.15, 7.31) | (0.5, 0.1) | | C | (6.17, 8.27) | (0.2, 0) | w=0.5, c₁=1.2, c₂=1.2, r₁=0.5, r₂=0.1. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(3.14, 7.21)=3.14^2+7.21^2-(3.14)(7.21) $$ $$ =9.8596+51.9841-22.6394=39.2043 $$ **Particle B** $$ f(4.15, 7.31)=4.15^2+7.31^2-(4.15)(7.31) $$ $$ =17.2225+53.4361-30.3365=40.3221 $$ **Particle C** $$ f(6.17, 8.27)=6.17^2+8.27^2-(6.17)(8.27) $$ $$ =38.0689+68.3929-51.0259=55.4359 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(3.14, 7.21) $$ Its score is 39.2043. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (3.14, 7.21). pbest values are A: (3.14, 7.21), B: (4.15, 7.31), C: (6.17, 8.27). ##### Step 4. Iteration 1: update particle A **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.1=0.05 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3.14-(3.14)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(3.14)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0.05+0+0=0.05 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=3.14+0.05=3.19 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.3=0.15 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=7.21-(7.21)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(7.21)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0.15+0+0=0.15 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=7.21+0.15=7.36 $$ ##### Step 5. Iteration 1: update particle B **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.5=0.25 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=4.15-(4.15)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(4.15)=-1.01 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-1.01)=-0.1212 $$ Add the three velocity contributions. $$ v^{(1)}=0.25+0+(-0.1212)=0.1288 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=4.15+0.1288=4.2788 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.1=0.05 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=7.31-(7.31)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(7.31)=-0.1 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.1)=-0.012 $$ Add the three velocity contributions. $$ v^{(1)}=0.05+0+(-0.012)=0.038 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=7.31+0.038=7.348 $$ ##### Step 6. Iteration 1: update particle C **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.2=0.1 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=6.17-(6.17)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(6.17)=-3.03 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-3.03)=-0.3636 $$ Add the three velocity contributions. $$ v^{(1)}=0.1+0+(-0.3636)=-0.2636 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=6.17+(-0.2636)=5.9064 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=8.27-(8.27)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(8.27)=-1.06 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-1.06)=-0.1272 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-0.1272)=-0.1272 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=8.27+(-0.1272)=8.1428 $$ ##### Step 7. Iteration 1: evaluate the new positions **Particle A** $$ f(3.19, 7.36)=3.19^2+7.36^2-(3.19)(7.36) $$ $$ =10.1761+54.1696-23.4784=40.8673 $$ **Particle B** $$ f(4.2788, 7.348)=4.2788^2+7.348^2-(4.2788)(7.348) $$ $$ =18.30812944+53.993104-31.4406224=40.86061104 $$ **Particle C** $$ f(5.9064, 8.1428)=5.9064^2+8.1428^2-(5.9064)(8.1428) $$ $$ =34.88556096+66.30519184-48.09463392=53.09611888 $$ ##### Step 8. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 39.2043 | 40.8673 | Keep old: new score is worse | (3.14, 7.21) | | B | 40.3221 | 40.86061104 | Keep old: new score is worse | (4.15, 7.31) | | C | 55.4359 | 53.09611888 | Replace: new score is lower | (5.9064, 8.1428) | Compare the retained pbest scores. gbest = **(3.14, 7.21)**, with objective **39.2043**. ##### Step 9. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (3.19, 7.36) | (0.05, 0.15) | (3.14, 7.21) | | B | (4.2788, 7.348) | (0.1288, 0.038) | (4.15, 7.31) | | C | (5.9064, 8.1428) | (-0.2636, -0.1272) | (5.9064, 8.1428) | ##### Step 10. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (3.14, 7.21). pbest values are A: (3.14, 7.21), B: (4.15, 7.31), C: (5.9064, 8.1428). ##### Step 11. Iteration 2: update particle A **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.05=0.025 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3.14-(3.19)=-0.05 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(-0.05)=-0.03 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(3.19)=-0.05 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.05)=-0.006 $$ Add the three velocity contributions. $$ v^{(2)}=0.025+(-0.03)+(-0.006)=-0.011 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=3.19+(-0.011)=3.179 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.15=0.075 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=7.21-(7.36)=-0.15 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(-0.15)=-0.09 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(7.36)=-0.15 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.15)=-0.018 $$ Add the three velocity contributions. $$ v^{(2)}=0.075+(-0.09)+(-0.018)=-0.033 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=7.36+(-0.033)=7.327 $$ ##### Step 12. Iteration 2: update particle B **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.1288=0.0644 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=4.15-(4.2788)=-0.1288 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(-0.1288)=-0.07728 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(4.2788)=-1.1388 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-1.1388)=-0.136656 $$ Add the three velocity contributions. $$ v^{(2)}=0.0644+(-0.07728)+(-0.136656)=-0.149536 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=4.2788+(-0.149536)=4.129264 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.038=0.019 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=7.31-(7.348)=-0.038 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(-0.038)=-0.0228 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(7.348)=-0.138 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.138)=-0.01656 $$ Add the three velocity contributions. $$ v^{(2)}=0.019+(-0.0228)+(-0.01656)=-0.02036 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=7.348+(-0.02036)=7.32764 $$ ##### Step 13. Iteration 2: update particle C **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-0.2636)=-0.1318 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=5.9064-(5.9064)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(5.9064)=-2.7664 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-2.7664)=-0.331968 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.1318)+0+(-0.331968)=-0.463768 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=5.9064+(-0.463768)=5.442632 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-0.1272)=-0.0636 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=8.1428-(8.1428)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(8.1428)=-0.9328 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.9328)=-0.111936 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.0636)+0+(-0.111936)=-0.175536 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=8.1428+(-0.175536)=7.967264 $$ ##### Step 14. Iteration 2: evaluate the new positions **Particle A** $$ f(3.179, 7.327)=3.179^2+7.327^2-(3.179)(7.327) $$ $$ =10.106041+53.684929-23.292533=40.498437 $$ **Particle B** $$ f(4.129264, 7.32764)=4.129264^2+7.32764^2-(4.129264)(7.32764) $$ $$ =17.050821182+53.69430797-30.257760057=40.487369094 $$ **Particle C** $$ f(5.442632, 7.967264)=5.442632^2+7.967264^2-(5.442632)(7.967264) $$ $$ =29.622243087+63.477295646-43.362885999=49.736652734 $$ ##### Step 15. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 39.2043 | 40.498437 | Keep old: new score is worse | (3.14, 7.21) | | B | 40.3221 | 40.487369094 | Keep old: new score is worse | (4.15, 7.31) | | C | 53.09611888 | 49.736652734 | Replace: new score is lower | (5.442632, 7.967264) | Compare the retained pbest scores. gbest = **(3.14, 7.21)**, with objective **39.2043**. ##### Step 16. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (3.179, 7.327) | (-0.011, -0.033) | (3.14, 7.21) | | B | (4.129264, 7.32764) | (-0.149536, -0.02036) | (4.15, 7.31) | | C | (5.442632, 7.967264) | (-0.463768, -0.175536) | (5.442632, 7.967264) | **Final answer.** After iteration 2, gbest = (3.14, 7.21), objective = 39.2043. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. Swipe across the diagram to read its labels ↔ ![Motion alone does not imply improvement. Evaluate the objective at the new location before updating a best memory.](https://deep-learning-study-room.vercel.app/is2/fig/v2_pso_class_paths.961831cda1.svg) Diagram: Motion alone does not imply improvement. Evaluate the objective at the new location before updating a best memory.. Labels: 3 | 3.5 | 4 | 4.5 | 5 | 5.5 | 6 | 6.5 | 7 | 7.5 | 8 | 8.5 | Class PSO: positions after each iteration | 0 = initial, 1 = iteration 1, 2 = iteration 2 | A0 | A1 | A2 | B0 | B1 | B2 | C0 | C1 | C2 | First coordinate x₁ | The best-so-far location remains A0 = (3.14, 7.21) after both iterations. Motion alone does not imply improvement. Evaluate the objective at the new location before updating a best memory. --- ## M2 · PSO algorithm, architecture, advantages and limitations Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--theory/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/pso-core/theory First understand position, velocity and memory; then solve complete two-iteration minimisation examples. ## Algorithm you can write in an exam 1. Define the objective and whether to minimise or maximise it. Initialise particle positions and velocities. 2. Evaluate every particle. Set each pbest to its initial position and gbest to the best among them. 3. Use the previous state to calculate each new velocity. 4. Update each position using that new velocity. Apply bounds or a velocity limit only according to the stated policy. 5. Evaluate the new positions. Update a pbest only when the new objective is better; choose gbest from the personal bests. 6. Repeat until the iteration budget or another stopping criterion is reached. Initialise → Evaluate + memories → Velocity → Position → Evaluate + update memoriesrepeat until stopping | Advantages | Limitations | | --- | --- | | Simple position/velocity updates; does not require objective derivatives | Can converge prematurely to a local region | | Uses both individual and shared experience | Results depend on coefficients, initialisation and random factors | | Useful for many continuous optimisation problems | Constraints and discrete encodings need suitable handling | | Keeps best-so-far memories | Does not guarantee the global optimum in a finite run | **Explain one application of PSO.** [3 marks · Practice] #### Answer In controller tuning, a particle can represent the gain values of a controller. The objective is a performance cost, such as tracking error plus an overshoot penalty. Each proposed gain vector is evaluated in the model. PSO updates candidate gains using individual and swarm bests. The best feasible gains found are selected and then validated on the actual system. A lower training cost alone does not prove safe real-world behaviour. | Exam situation | Action | | --- | --- | | Minimisation | Lower f is better | | Maximisation | Higher f is better; do not reverse velocity differences | | Tie in best score | Keep the previous best unless a tie rule is supplied | | Different random factors each iteration | Read the relevant row; do not reuse old factors | | Synchronous update | Freeze gbest for all particles within one iteration | | Position/velocity outside a limit | Use the limit-handling rule explicitly supplied; state an assumption if absent | **Yaad rakho:** Velocity → position → score → pbest → gbest. Ye order har iteration mein same. --- ## M2 · Cartesian product: every cell explained Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-relations--cartesian/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-relations/cartesian Longer Unit 2 numerical practice, also useful for TEE. ## How a row membership and a column membership meet Let A₁=.6 and A₂=.2 label two rows. Let B₁=.3, B₂=.8 and B₃=.5 label three columns. Cell R₂₃ belongs to row 2 and column 3, so use A₂=.2 and B₃=.5. Their min is .2. You do not multiply 2×3; those subscripts only locate the cell. $$ R_{23}=\min(A_2,B_3)=\min(0.2,0.5)=0.2 $$ There are two choices of row and three of column, so fill six cells in a 2×3 rectangle. Keep the row and column labels when reporting a relation. ## What is a fuzzy relation? A fuzzy set gives a membership to one element. A **fuzzy relation** gives a degree to an ordered pair. A matrix is a convenient table: rows name the first element and columns name the second. For the standard min-based Cartesian product R=A×B, put the A memberships down the rows and the B memberships across the columns. Each cell is the smaller of its row and column memberships. $$ R_{ij}=\min(\mu_A(a_i),\mu_B(b_j)) $$ **Find the fuzzy Cartesian product A×B for memberships A=[0.6, 0.2] and B=[0.3, 0.8, 0.5].** [3 marks · Practice] #### Given Ordered A memberships [0.6, 0.2]; B memberships [0.3, 0.8, 0.5]. #### To find A 2×3 relation matrix. #### Formula and why it applies Every row corresponds to an A element and every column to a B element; apply min in every cell. $$ R_{ij}=\min(A_i,B_j) $$ #### Solution ##### Step 1. Determine dimensions 2 row elements × 3 column elements. ##### Step 2. Compute each cell | A ↓ / B → | 0.3 | 0.8 | 0.5 | | --- | --- | --- | --- | | 0.6 | min(0.6,0.3) = 0.3 | min(0.6,0.8) = 0.6 | min(0.6,0.5) = 0.5 | | 0.2 | min(0.2,0.3) = 0.2 | min(0.2,0.8) = 0.2 | min(0.2,0.5) = 0.2 | ##### Step 3. Write the matrix $$ R=\begin{bmatrix}0.3&0.6&0.5\\0.2&0.2&0.2\end{bmatrix} $$ **Final answer.** R is the matrix in Step 3. **Find the fuzzy Cartesian product A×B for memberships A=[1, 0.4] and B=[0.7, 0].** [3 marks · Practice] #### Given Ordered A memberships [1, 0.4]; B memberships [0.7, 0]. #### To find A 2×2 relation matrix. #### Formula and why it applies Every row corresponds to an A element and every column to a B element; apply min in every cell. $$ R_{ij}=\min(A_i,B_j) $$ #### Solution ##### Step 1. Determine dimensions 2 row elements × 2 column elements. ##### Step 2. Compute each cell | A ↓ / B → | 0.7 | 0 | | --- | --- | --- | | 1 | min(1,0.7) = 0.7 | min(1,0) = 0 | | 0.4 | min(0.4,0.7) = 0.4 | min(0.4,0) = 0 | ##### Step 3. Write the matrix $$ R=\begin{bmatrix}0.7&0\\0.4&0\end{bmatrix} $$ **Final answer.** R is the matrix in Step 3. **Find the fuzzy Cartesian product A×B for memberships A=[0.2, 0.9, 0.5] and B=[0.6, 0.1].** [3 marks · Practice] #### Given Ordered A memberships [0.2, 0.9, 0.5]; B memberships [0.6, 0.1]. #### To find A 3×2 relation matrix. #### Formula and why it applies Every row corresponds to an A element and every column to a B element; apply min in every cell. $$ R_{ij}=\min(A_i,B_j) $$ #### Solution ##### Step 1. Determine dimensions 3 row elements × 2 column elements. ##### Step 2. Compute each cell | A ↓ / B → | 0.6 | 0.1 | | --- | --- | --- | | 0.2 | min(0.2,0.6) = 0.2 | min(0.2,0.1) = 0.1 | | 0.9 | min(0.9,0.6) = 0.6 | min(0.9,0.1) = 0.1 | | 0.5 | min(0.5,0.6) = 0.5 | min(0.5,0.1) = 0.1 | ##### Step 3. Write the matrix $$ R=\begin{bmatrix}0.2&0.1\\0.6&0.1\\0.5&0.1\end{bmatrix} $$ **Final answer.** R is the matrix in Step 3. **Yaad rakho:** Row ka degree aur column ka degree—dono ka min us cell mein. --- ## M2 · Max–min composition: three complete matrix examples Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-relations--composition/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/read/fuzzy-relations/composition Longer Unit 2 numerical practice, also useful for TEE. ## One result cell combines alternative two-link paths For T₁₁, start at u₁ and end at w₁. If the middle domain has v₁ and v₂, there are two candidate paths. Their strengths come from matching row 1 of R with column 1 of S. | Path | First link from R | Second link from S | Path strength | | --- | --- | --- | --- | | u₁ → v₁ → w₁ | .2 | .5 | min(.2,.5)=.2 | | u₁ → v₂ → w₁ | .8 | .7 | min(.8,.7)=.7 | Choose the stronger path: max(.2,.7)=.7, so T₁₁=.7. The middle label j runs through v₁ and v₂. The labels i and k stay fixed at 1 while this one cell is calculated. Then move to the next result cell. **Yaad rakho:** Ek path mein dono links chahiye, isliye min. Alag paths mein best chalega, isliye max. ## Composition connects through a middle element Suppose R relates U to V and S relates V to W. To find how strongly u connects to w through each middle v, take min of the two links. Then choose the strongest such path using max. $$ T_{ik}=\max_j\min(R_{ij},S_{jk}) $$ We call this matrix operation R composed with S here. Some books write the symbolic order in reverse. State the formula and row/column labels, especially when a question says S∘R. Composition requires the intermediate domains and their ordering to match; having two 3×3 matrices alone does not prove their labels match. **Calculate the max–min composition using $$ T_{ik}=\max_j\min(R_{ij},S_{jk}) $$ .** [3 marks · Practice] #### Given $$ R=\begin{bmatrix}0.2&0.8\\0.6&0.3\end{bmatrix} $$ $$ S=\begin{bmatrix}0.5&0.9\\0.7&0.1\end{bmatrix} $$ #### To find Every entry of T. #### Formula and why it applies Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication. #### Solution ##### Step 1. Cell (1,1): row 1 of R with column 1 of S $$ T_{11}=\max\{\min(0.2,0.5),\min(0.8,0.7)\} $$ $$ =\max\{0.2,0.7\}=0.7 $$ ##### Step 2. Cell (1,2): row 1 of R with column 2 of S $$ T_{12}=\max\{\min(0.2,0.9),\min(0.8,0.1)\} $$ $$ =\max\{0.2,0.1\}=0.2 $$ ##### Step 3. Cell (2,1): row 2 of R with column 1 of S $$ T_{21}=\max\{\min(0.6,0.5),\min(0.3,0.7)\} $$ $$ =\max\{0.5,0.3\}=0.5 $$ ##### Step 4. Cell (2,2): row 2 of R with column 2 of S $$ T_{22}=\max\{\min(0.6,0.9),\min(0.3,0.1)\} $$ $$ =\max\{0.6,0.1\}=0.6 $$ **Final answer.** $$ T=\begin{bmatrix}0.7&0.2\\0.5&0.6\end{bmatrix} $$ . **Calculate the max–min composition using $$ T_{ik}=\max_j\min(R_{ij},S_{jk}) $$ .** [3 marks · Practice] #### Given $$ R=\begin{bmatrix}1&0\\0.4&0.7\end{bmatrix} $$ $$ S=\begin{bmatrix}0.2&0.6\\0.8&0.3\end{bmatrix} $$ #### To find Every entry of T. #### Formula and why it applies Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication. #### Solution ##### Step 1. Cell (1,1): row 1 of R with column 1 of S $$ T_{11}=\max\{\min(1,0.2),\min(0,0.8)\} $$ $$ =\max\{0.2,0\}=0.2 $$ ##### Step 2. Cell (1,2): row 1 of R with column 2 of S $$ T_{12}=\max\{\min(1,0.6),\min(0,0.3)\} $$ $$ =\max\{0.6,0\}=0.6 $$ ##### Step 3. Cell (2,1): row 2 of R with column 1 of S $$ T_{21}=\max\{\min(0.4,0.2),\min(0.7,0.8)\} $$ $$ =\max\{0.2,0.7\}=0.7 $$ ##### Step 4. Cell (2,2): row 2 of R with column 2 of S $$ T_{22}=\max\{\min(0.4,0.6),\min(0.7,0.3)\} $$ $$ =\max\{0.4,0.3\}=0.4 $$ **Final answer.** $$ T=\begin{bmatrix}0.2&0.6\\0.7&0.4\end{bmatrix} $$ . **Calculate the max–min composition using $$ T_{ik}=\max_j\min(R_{ij},S_{jk}) $$ .** [3 marks · Practice] #### Given $$ R=\begin{bmatrix}0.3&0.6&0.9\end{bmatrix} $$ $$ S=\begin{bmatrix}0.8&0.1\\0.2&0.7\\0.5&0.4\end{bmatrix} $$ #### To find Every entry of T. #### Formula and why it applies Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication. #### Solution ##### Step 1. Cell (1,1): row 1 of R with column 1 of S $$ T_{11}=\max\{\min(0.3,0.8),\min(0.6,0.2),\min(0.9,0.5)\} $$ $$ =\max\{0.3,0.2,0.5\}=0.5 $$ ##### Step 2. Cell (1,2): row 1 of R with column 2 of S $$ T_{12}=\max\{\min(0.3,0.1),\min(0.6,0.7),\min(0.9,0.4)\} $$ $$ =\max\{0.1,0.6,0.4\}=0.6 $$ **Final answer.** $$ T=\begin{bmatrix}0.5&0.6\end{bmatrix} $$ . **Yaad rakho:** Har path mein min, phir saare paths mein max. Multiply-and-add nahi. --- ## M2 · Fuzzy formulas, with meanings and worked-example links Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/formulas/m2-formulas-v2--fuzzy/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/formulas/m2-formulas-v2/fuzzy One place for the formulas, meanings, conditions and links to full worked examples. ## Fuzzy formulas ### Triangular membership $$ \mu(x)=\max\left(0,\min\left(\frac{x-a}{b-a},\frac{c-x}{c-b}\right)\right) $$ a=left zero, b=peak, c=right zero, with a0 sets the width. At x=c the exponent is zero and membership is 1. [Worked exponential calculations →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--smooth-memberships/) ### Generalised bell membership $$ \mu(x)=\frac{1}{1+|(x-c)/a|^{2b}} $$ a>0 controls width; b>0 controls steepness; c is the centre. First subtract c, divide by a, take absolute value, apply the power, add 1, then take its reciprocal. ### Alpha-cut $$ A_\alpha=\{x:\mu_A(x)\ge\alpha\} $$ α is the chosen membership threshold. For 0<α≤1 keep inputs meeting or exceeding it, including threshold-equality endpoints. [Discrete and triangular examples →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--properties/) --- ## M2 · GA and PSO formulas, symbols and memory rules Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/formulas/m2-formulas-v2--ga-pso/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/formulas/m2-formulas-v2/ga-pso One place for the formulas, meanings, conditions and links to full worked examples. ### Binary decoding $$ x=\sum_{i=1}^{n}b_i2^{n-i} $$ bᵢ is a bit; the maximum n-bit unsigned integer is 2ⁿ−1. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--encoding/) ### Roulette selection $$ P_i=F_i/\sum_jF_j\quad C_i=\sum_{j=1}^iP_j $$ Fitness weights must be nonnegative with a positive total. For minimisation use the specified transformation or selection method. C is cumulative probability. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--selection/) ### Population average $$ \bar f=\frac{f_1+\cdots+f_N}{N} $$ N is the number of individuals. Compare both best and average; mutation need not improve either. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--two-generations/) ### PSO velocity $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ w weights inertia; c₁r₁ weights personal pull; c₂r₂ weights social pull. Apply the same equation separately to each coordinate. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--equation/) ### PSO position $$ x^{t+1}=x^t+v^{t+1} $$ Use the new velocity, not the old one. Then reevaluate the objective. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--two-particles/) **Memory updates:** minimise → replace a pbest only for a lower score; maximise → only for a higher score. gbest is a position selected from the personal bests. Store the corresponding objective score alongside it. Keep previous best on ties unless instructed otherwise. **Yaad rakho:** Formula ke pehle objective direction aur old state likho. Iteration 2 mein fresh positions aur retained memories use karo. ## Encoding and minimisation conventions $$ x=L+\frac{d}{2^n-1}(U-L) $$ For the specified uniform real-range encoding: decode n bits into unsigned integer d first. L and U are the physical limits. [Endpoint examples →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--encoding/) $$ F=\frac{1}{1+f(x)} $$ This is the stated roulette transformation in our nonnegative-cost minimisation example. It is not required for every GA. It converts a smaller cost into a larger selection weight. [Complete minimisation example →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--minimisation/) --- ## M2 · Q1(c) Fuzzy set operations [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-final-2025--q1c/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-final-2025/q1c Longer past-paper questions on Fuzzy, GA and PSO. Original TEE marks retained; not the M2 paper pattern. **Question. Consider two fuzzy sets X = {(c1, 0.6), (c2, 0.8), (c3, 0.3), (c4, 0.5), (c5, 0.7)} and Y = {(c1, 0.4), (c2, 0.5), (c3, 0.9), (c4, 0.6), (c5, 0.2)}. Find 1) X ∪ Y 2) X ∩ Y 3) ¬X ∪ ¬Y 4) ¬X ∩ ¬Y.** [5 marks] ### Answer #### Given The aligned X and Y memberships printed in this question. #### To find The four requested membership lists in the original element order. #### Formula and why it applies Use max for OR, min for AND, and 1−μ for NOT. $$ \mu_{X\cup Y}=\max(\mu_X,\mu_Y),\quad\mu_{X\cap Y}=\min(\mu_X,\mu_Y) $$ Complement both operands before applying the operator in ¬X∪¬Y or ¬X∩¬Y. #### Solution ##### Step 1. Complements ¬X = {0.4, 0.2, 0.7, 0.5, 0.3}; ¬Y = {0.6, 0.5, 0.1, 0.4, 0.8}. ##### Step 2. Table | | c1 | c2 | c3 | c4 | c5 | | --- | --- | --- | --- | --- | --- | | X | 0.6 | 0.8 | 0.3 | 0.5 | 0.7 | | Y | 0.4 | 0.5 | 0.9 | 0.6 | 0.2 | | **X ∪ Y = max** | 0.6 | 0.8 | 0.9 | 0.6 | 0.7 | | **X ∩ Y = min** | 0.4 | 0.5 | 0.3 | 0.5 | 0.2 | | **¬X ∪ ¬Y** | 0.6 | 0.5 | 0.7 | 0.5 | 0.8 | | **¬X ∩ ¬Y** | 0.4 | 0.2 | 0.1 | 0.4 | 0.3 | ##### Step 3. Check with De Morgan's law ¬X ∪ ¬Y must equal 1 − (X ∩ Y) = {0.6, 0.5, 0.7, 0.5, 0.8} ✓, and ¬X ∩ ¬Y must equal 1 − (X ∪ Y) = {0.4, 0.2, 0.1, 0.4, 0.3} ✓. **Final answer.** X ∪ Y = {(c1,0.6),(c2,0.8),(c3,0.9),(c4,0.6),(c5,0.7)}; X ∩ Y = {(c1,0.4),(c2,0.5),(c3,0.3),(c4,0.5),(c5,0.2)}; ¬X ∪ ¬Y = {(c1,0.6),(c2,0.5),(c3,0.7),(c4,0.5),(c5,0.8)}; ¬X ∩ ¬Y = {(c1,0.4),(c2,0.2),(c3,0.1),(c4,0.4),(c5,0.3)}. Official marking: 1 + 1 + 1.5 + 1.5. ## Calculate every element separately ### Element c1 $$ \neg X:1-0.6=0.4\qquad\neg Y:1-0.4=0.6 $$ $$ X\cup Y:\max(0.6,0.4)=0.6 $$ $$ X\cap Y:\min(0.6,0.4)=0.4 $$ $$ \neg X\cup\neg Y:\max(0.4,0.6)=0.6 $$ $$ \neg X\cap\neg Y:\min(0.4,0.6)=0.4 $$ ### Element c2 $$ \neg X:1-0.8=0.2\qquad\neg Y:1-0.5=0.5 $$ $$ X\cup Y:\max(0.8,0.5)=0.8 $$ $$ X\cap Y:\min(0.8,0.5)=0.5 $$ $$ \neg X\cup\neg Y:\max(0.2,0.5)=0.5 $$ $$ \neg X\cap\neg Y:\min(0.2,0.5)=0.2 $$ ### Element c3 $$ \neg X:1-0.3=0.7\qquad\neg Y:1-0.9=0.1 $$ $$ X\cup Y:\max(0.3,0.9)=0.9 $$ $$ X\cap Y:\min(0.3,0.9)=0.3 $$ $$ \neg X\cup\neg Y:\max(0.7,0.1)=0.7 $$ $$ \neg X\cap\neg Y:\min(0.7,0.1)=0.1 $$ ### Element c4 $$ \neg X:1-0.5=0.5\qquad\neg Y:1-0.6=0.4 $$ $$ X\cup Y:\max(0.5,0.6)=0.6 $$ $$ X\cap Y:\min(0.5,0.6)=0.5 $$ $$ \neg X\cup\neg Y:\max(0.5,0.4)=0.5 $$ $$ \neg X\cap\neg Y:\min(0.5,0.4)=0.4 $$ ### Element c5 $$ \neg X:1-0.7=0.3\qquad\neg Y:1-0.2=0.8 $$ $$ X\cup Y:\max(0.7,0.2)=0.7 $$ $$ X\cap Y:\min(0.7,0.2)=0.2 $$ $$ \neg X\cup\neg Y:\max(0.3,0.8)=0.8 $$ $$ \neg X\cap\neg Y:\min(0.3,0.8)=0.3 $$ --- ## M2 · Q2(a) PSO, one iteration [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-final-2025--q2a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-final-2025/q2a Longer past-paper questions on Fuzzy, GA and PSO. Original TEE marks retained; not the M2 paper pattern. **Question. Maximize $$ F(x_1,x_2)=2x_1^2-3x_1x_2+x_2^2+5x_1+2x_2+7 $$ using Particle Swarm Optimization. $$ X=\begin{bmatrix}2.5&-3.0\\3.5&-1.5\end{bmatrix},\;V=\begin{bmatrix}0.2&0.3\\0.1&0.25\end{bmatrix} $$ . Take $$ W=0.9,\;c_1=c_2=1.5,\;r_1=0.5949,\;r_2=0.0855 $$ . Optimize and give the solution after 1 iteration showing the step-by-step process.** [10 marks] ### Answer #### Given Particle 1: $$ x=(2.5,\,-3.0) $$ , $$ v=(0.2,\,0.3) $$ . Particle 2: $$ x=(3.5,\,-1.5) $$ , $$ v=(0.1,\,0.25) $$ . $$ c_1r_1=1.5\times0.5949=0.89235,\qquad c_2r_2=1.5\times0.0855=0.12825 $$ #### To find Fitness, pbest, gbest; new velocities and positions; updated pbest and gbest. #### Formula and why it applies $$ v_{ij}\leftarrow wv_{ij}+c_1r_1(p^b_{ij}-x_{ij})+c_2r_2(p^g_j-x_{ij}),\qquad x_{ij}\leftarrow x_{ij}+v_{ij} $$ Each particle is pulled by its own memory and by the swarm's best; we maximise, so the larger $$ F $$ is better. #### Solution ##### Step 1. Fitness of particle 1 $$ F(2.5,-3)=2(6.25)-3(2.5)(-3)+9+5(2.5)+2(-3)+7=12.5+22.5+9+12.5-6+7=57.5 $$ ##### Step 2. Fitness of particle 2 $$ F(3.5,-1.5)=2(12.25)-3(3.5)(-1.5)+2.25+17.5-3+7=24.5+15.75+2.25+17.5-3+7=64 $$ ##### Step 3. pbest and gbest | Particle | Position | F | pbest | | --- | --- | --- | --- | | 1 | (2.5, −3.0) | 57.5 | (2.5, −3.0) | | 2 | (3.5, −1.5) | 64 | (3.5, −1.5) | No previous iteration, so pbest = current position. Largest fitness is 64, so **gbest = (3.5, −1.5)**. ##### Step 4. Velocity and position of particle 1 $$ v_{11}=0.9(0.2)+0.89235(2.5-2.5)+0.12825(3.5-2.5)=0.18+0+0.12825=0.30825 $$ $$ x_{11}=2.5+0.30825=2.80825 $$ $$ v_{12}=0.9(0.3)+0.89235(0)+0.12825(-1.5-(-3.0))=0.27+0.192375=0.462375 $$ $$ x_{12}=-3.0+0.462375=-2.537625 $$ ##### Step 5. Velocity and position of particle 2 (it is gbest, so both pull terms are 0) $$ v_{21}=0.9(0.1)+0+0=0.09,\qquad x_{21}=3.5+0.09=3.59 $$ $$ v_{22}=0.9(0.25)+0+0=0.225,\qquad x_{22}=-1.5+0.225=-1.275 $$ ##### Step 6. New fitness $$ F(2.80825,-2.537625)=15.772536+21.378856+6.439541+14.04125-5.07525+7=59.5569 $$ $$ F(3.59,-1.275)=25.7762+13.73175+1.625625+17.95-2.55+7=63.5336 $$ ##### Step 7. Update pbest and gbest | Particle | New position | New F | Old pbest F | pbest after iteration | | --- | --- | --- | --- | --- | | 1 | (2.80825, −2.537625) | 59.5569 | 57.5 | (2.80825, −2.537625) — improved | | 2 | (3.59, −1.275) | 63.5336 | 64 | (3.5, −1.5) — kept | The best pbest fitness is still 64, so gbest stays (3.5, −1.5). **Final answer.** $$ V=\begin{bmatrix}0.30825&0.462375\\0.09&0.225\end{bmatrix} $$ , $$ X=\begin{bmatrix}2.80825&-2.537625\\3.59&-1.275\end{bmatrix} $$ . After one iteration particle 1 improved to F = 59.5569; the best solution so far is gbest = (3.5, −1.5) with F = 64. Official key: fitness and gbest (5 marks), velocity and position updates (5 marks). The calculations above use the question data. The key's summary matrix writes 0.462 in place of 2.808 for $$ x_{11} $$ ; that is a copying slip. --- ## M2 · Q3(b) Fuzzy loan-approval system [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-final-2025--q3b/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-final-2025/q3b Longer past-paper questions on Fuzzy, GA and PSO. Original TEE marks retained; not the M2 paper pattern. **A bank uses fuzzy decision support with Credit Score and Income Stability, both on 1–100. Loan Approval is a score on 1–100 with labels Reject, Consider and Approve. Define triangular membership functions, construct nine rules, explain Mamdani inference and defuzzify for Credit Score=55 and Income Stability=35.** [10 marks] ### Answer #### Given | Variable | Crisp value | | --- | --- | | Credit Score | 55 | | Income Stability | 35 | The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens. #### To find A complete stated design and its continuous Mamdani centroid. #### Formula and why it applies Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier. $$ z^*=\frac{\int z\mu_{out}(z)\,dz}{\int\mu_{out}(z)\,dz} $$ #### Solution ##### Step 1. Specify the membership functions | Input / output position | Triangle | Peak | | --- | --- | --- | | Low / Reject | (1,25,50) | 25 | | Medium / Consider | (25,50,75) | 50 | | High / Approve | (50,75,100) | 75 | Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values. ##### Step 2. Write all nine rules | Credit Score / Income Stability | Low | Medium | High | | --- | --- | --- | --- | | Low | Reject | Reject | Consider | | Medium | Reject | Consider | Approve | | High | Consider | Approve | Approve | Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium. ##### Step 3. Fuzzify Credit Score = 55 $$ \mu_{Low}(55)=0 $$ $$ \mu_{Medium}(55)=\frac{75-55}{75-50}=0.8 $$ $$ \mu_{High}(55)=\frac{55-50}{75-50}=0.2 $$ ##### Step 4. Fuzzify Income Stability = 35 $$ \mu_{Low}(35)=\frac{50-35}{50-25}=0.6 $$ $$ \mu_{Medium}(35)=\frac{35-25}{50-25}=0.4 $$ $$ \mu_{High}(35)=0 $$ ##### Step 5. Evaluate all rules and aggregate matching consequents | Credit Score | Income Stability | Firing strength | Consequent | | --- | --- | --- | --- | | Low | Low | min(0,0.6) = 0 | Reject | | Low | Medium | min(0,0.4) = 0 | Reject | | Low | High | min(0,0) = 0 | Consider | | Medium | Low | min(0.8,0.6) = 0.6 | Reject | | Medium | Medium | min(0.8,0.4) = 0.4 | Consider | | Medium | High | min(0.8,0) = 0 | Approve | | High | Low | min(0.2,0.6) = 0.2 | Consider | | High | Medium | min(0.2,0.4) = 0.2 | Approve | | High | High | min(0.2,0) = 0 | Approve | Aggregated label strengths: Reject = 0.6, Consider = 0.4, Approve = 0.2. $$ \mu_{out}(z)=\max_r\min(\alpha_r,\mu_{C_r}(z)) $$ ##### Step 6. Draw the aggregated output Swipe across the diagram to read its labels ↔ ![The shaded upper envelope counts any overlap only once.](https://deep-learning-study-room.vercel.app/is2/fig/v2_loan_aggregate.2c5ffa32a7.svg) Diagram: The shaded upper envelope counts any overlap only once.. Labels: Chosen triangular output sets and their max-aggregate | 0 | 0.5 | 1 | 1 | 25 | 50 | 75 | 100 | μ | Output score | Reject | Consider | Approve | centroid ≈ 42.78 The shaded upper envelope counts any overlap only once. ##### Step 7. Calculate area and moment of each straight-line strip For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height. $$ A=(b-a)(u+v)/2 $$ $$ M=\frac{b-a}{6}\big[(2a+b)u+(a+2b)v\big] $$ | a | b | u | v | Area A | Moment M | | --- | --- | --- | --- | --- | --- | | 1 | 15.4 | 0 | 0.6 | 4.32 | 45.792 | | 15.4 | 25 | 0.6 | 0.6 | 5.76 | 116.352 | | 25 | 35 | 0.6 | 0.6 | 6 | 180 | | 35 | 40 | 0.6 | 0.4 | 2.5 | 93.333333333 | | 40 | 50 | 0.4 | 0.4 | 4 | 180 | | 50 | 55 | 0.4 | 0.4 | 2 | 105 | | 55 | 65 | 0.4 | 0.4 | 4 | 240 | | 65 | 70 | 0.4 | 0.2 | 1.5 | 100.833333333 | | 70 | 75 | 0.2 | 0.2 | 1 | 72.5 | | 75 | 95 | 0.2 | 0.2 | 4 | 340 | | 95 | 100 | 0.2 | 0 | 0.5 | 48.333333333 | ##### Step 8. Divide total moment by total area $$ \sum A=35.58 $$ $$ \sum M=1522.144 $$ $$ z^*=\frac{1522.144}{35.58}\approx42.780888139 $$ **Final answer.** For these stated triangles and rules, the centroid is 42.780888139. This is a fuzzy suitability score; the category memberships can be evaluated at this score. The model alone is not a complete loan-decision policy. Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised. ## How the centroid table’s breakpoints are obtained For each output triangle, solve “line height = clip height”. Reject is clipped at .6: rising endpoint 1+.6(25−1)=15.4 and falling endpoint 50−.6(50−25)=35. Consider is clipped at .4: endpoints 25+.4(50−25)=35 and 75−.4(75−50)=65. Approve is clipped at .2: endpoints 50+.2(75−50)=55 and 100−.2(100−75)=95. Now take the maximum of the three clipped curves. The declining Reject edge reaches the Consider height .4 at (50−z)/25=.4, hence z=40. The declining Consider edge reaches the Approve height .2 at (75−z)/25=.2, hence z=70. These crossover locations must be included when splitting the aggregate. Original peaks and other clip endpoints may add extra splits without changing the total area. **For the first aggregate strip from a=1 to b=15.4, endpoint heights are u=0 and v=0.6. Find its area and first moment.** [3 marks · Worked example] #### Given Strip endpoints a=1 and b=15.4; left height u=0 and right height v=0.6. This is a straight rising edge of the clipped Reject set. #### To find A and M for one table row. #### Formula and why it applies $$ A=(b-a)(u+v)/2 $$ $$ M=\frac{b-a}{6}[(2a+b)u+(a+2b)v] $$ A is trapezoid area. M weights every small area by its horizontal position; this formula integrates a straight line. #### Solution ##### Step 1. Width and height sum $$ b-a=15.4-1=14.4\qquad u+v=0.6 $$ ##### Step 2. Area $$ A=14.4\times0.6/2=8.64/2=4.32 $$ ##### Step 3. First bracket term $$ (2a+b)u=(2+15.4)\times0=0 $$ ##### Step 4. Second bracket term $$ (a+2b)v=(1+30.8)\times0.6=31.8\times0.6=19.08 $$ ##### Step 5. Moment $$ M=(14.4/6)(0+19.08)=2.4\times19.08=45.792 $$ **Final answer.** Area=4.32; moment=45.792. Repeat the same arithmetic for each row, then sum the areas and moments separately. ### Interpret the loan result using an explicit decision rule Choose the output label having the greatest membership at the centroid; this is an additional stated interpretation rule. At z=42.780888139: $$ \mu_{Reject}=(50-z)/25\approx0.288764474,\quad\mu_{Consider}=(z-25)/25\approx0.711235526,\quad\mu_{Approve}=0 $$ The resulting recommendation is **Consider / manual review**; this design does not recommend automatic approval. If a binary approve/reject answer is required, the question must supply a threshold or you must state the threshold you adopt. A different decision threshold is separate from defuzzification. --- ## M2 · Q4(a) Cartesian product and max–min composition [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-final-2025--q4a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-final-2025/q4a Longer past-paper questions on Fuzzy, GA and PSO. Original TEE marks retained; not the M2 paper pattern. **Question. Consider fuzzy sets A = {0.4/small + 0.3/medium + 0.1/large} and B = {0.2/slow + 0.7/moderate + 0.3/fast}. 1. Find R = A × B. 2. Given C = {0.3/small + 0.1/medium + 0.4/large}, find S = C × B. 3. Calculate the max–min composition S ∘ R.** [10 marks] ### Answer #### Given The fuzzy sets or relation matrices and their listed row/column order. #### To find The relation matrix/matrices and the composition specified in the question. #### Formula and why it applies For Cartesian products take min of row and column membership. For the ordered matrix composition shown here: $$ T_{ik}=\max_j\min(S_{ij},R_{jk}) $$ If max–product is asked instead, replace the inner min by multiplication. #### Solution ##### Step 1. R = A × B (rows small, medium, large; columns slow, moderate, fast) $$ R=\begin{bmatrix}\min(.4,.2)&\min(.4,.7)&\min(.4,.3)\\\min(.3,.2)&\min(.3,.7)&\min(.3,.3)\\\min(.1,.2)&\min(.1,.7)&\min(.1,.3)\end{bmatrix}=\begin{bmatrix}0.2&0.4&0.3\\0.2&0.3&0.3\\0.1&0.1&0.1\end{bmatrix} $$ ##### Step 2. S = C × B $$ S=\begin{bmatrix}0.2&0.3&0.3\\0.1&0.1&0.1\\0.2&0.4&0.3\end{bmatrix} $$ ##### Step 3. S ∘ R by max–min (row of S with column of R) The printed relation domains do not directly compose: both relate size to speed. For the numerical exercise, explicitly assume positional alignment and compute row of S with column of R. Equal dimensions alone are not a justification for semantic relation composition. $$ (1,1)=\max[\min(.2,.2),\min(.3,.2),\min(.3,.1)]=\max(.2,.2,.1)=0.2 $$ $$ (1,2)=\max[\min(.2,.4),\min(.3,.3),\min(.3,.1)]=\max(.2,.3,.1)=0.3 $$ $$ (1,3)=\max[\min(.2,.3),\min(.3,.3),\min(.3,.1)]=0.3 $$ Row 2 of S is all 0.1, so every entry of row 2 is 0.1. $$ (3,1)=\max[\min(.2,.2),\min(.4,.2),\min(.3,.1)]=0.2,\;(3,2)=\max[.2,.3,.1]=0.3,\;(3,3)=\max[.2,.3,.1]=0.3 $$ **Final answer.** $$ S\circ R=\begin{bmatrix}0.2&0.3&0.3\\0.1&0.1&0.1\\0.2&0.3&0.3\end{bmatrix} $$ . **Official key note.** The key first printed a wrong matrix with row 2 = (0.2, 0.3, 0.3), then corrected it by hand to (0.1, 0.1, 0.1). The corrected matrix above is right. Marking: R 3, S 3, composition 4. **A careful remark you can add.** Strictly, composition needs S: X → Y and R: Y → Z, while here both relate size to speed. The numerical matrix calculation above assumes positional identification of the three middle labels; that is a declared repair, not a valid size-to-speed relation composition without an extra mapping. ## Every cell without the row shortcuts Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries. ### R: Cartesian-product cells | Cell | Row degree | Column degree | min result | | --- | --- | --- | --- | | R11 | 0.4 | 0.2 | 0.2 | | R12 | 0.4 | 0.7 | 0.4 | | R13 | 0.4 | 0.3 | 0.3 | | R21 | 0.3 | 0.2 | 0.2 | | R22 | 0.3 | 0.7 | 0.3 | | R23 | 0.3 | 0.3 | 0.3 | | R31 | 0.1 | 0.2 | 0.1 | | R32 | 0.1 | 0.7 | 0.1 | | R33 | 0.1 | 0.3 | 0.1 | ### S: Cartesian-product cells | Cell | Row degree | Column degree | min result | | --- | --- | --- | --- | | S11 | 0.3 | 0.2 | 0.2 | | S12 | 0.3 | 0.7 | 0.3 | | S13 | 0.3 | 0.3 | 0.3 | | S21 | 0.1 | 0.2 | 0.1 | | S22 | 0.1 | 0.7 | 0.1 | | S23 | 0.1 | 0.3 | 0.1 | | S31 | 0.4 | 0.2 | 0.2 | | S32 | 0.4 | 0.7 | 0.4 | | S33 | 0.4 | 0.3 | 0.3 | **Result cell (1,1)** $$ \max\{\min(0.2,0.2),\min(0.3,0.2),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (1,2)** $$ \max\{\min(0.2,0.4),\min(0.3,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.3,0.1\}=0.3 $$ **Result cell (1,3)** $$ \max\{\min(0.2,0.3),\min(0.3,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.3,0.1\}=0.3 $$ **Result cell (2,1)** $$ \max\{\min(0.1,0.2),\min(0.1,0.2),\min(0.1,0.1)\} $$ $$ =\max\{0.1,0.1,0.1\}=0.1 $$ **Result cell (2,2)** $$ \max\{\min(0.1,0.4),\min(0.1,0.3),\min(0.1,0.1)\} $$ $$ =\max\{0.1,0.1,0.1\}=0.1 $$ **Result cell (2,3)** $$ \max\{\min(0.1,0.3),\min(0.1,0.3),\min(0.1,0.1)\} $$ $$ =\max\{0.1,0.1,0.1\}=0.1 $$ **Result cell (3,1)** $$ \max\{\min(0.2,0.2),\min(0.4,0.2),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (3,2)** $$ \max\{\min(0.2,0.4),\min(0.4,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.3,0.1\}=0.3 $$ **Result cell (3,3)** $$ \max\{\min(0.2,0.3),\min(0.4,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.3,0.1\}=0.3 $$ --- ## M2 · Q5(b) Genetic algorithm [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-final-2025--q5b/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-final-2025/q5b Longer past-paper questions on Fuzzy, GA and PSO. Original TEE marks retained; not the M2 paper pattern. **Implement a GA to maximise $$ f(x)=2x^3-5x^2+7x $$ for x in [0,15]. Take 4-bit representations of 13, 22, 18, 23. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness.** [10 marks] ### Answer **The printed data are inconsistent.** 22, 18 and 23 are outside 0–15 and cannot fit into four bits. Increasing the bit count would encode the numbers but would still violate the stated range. There is no single exact solution satisfying all printed conditions. The complete worked variant below explicitly changes the invalid parents to their last four bits (values modulo 16), producing [13,6,2,7]. This is one declared repair, not a silent interpretation of the printed question. If the instructor clarifies different corrected data, follow those data. #### Given Printed parents: [13, 22, 18, 23]. Worked variant: [13, 6, 2, 7] with 4 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2. #### To find Two valid children under the declared assumptions and their fitness. #### Formula and why it applies Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation. $$ f(x)=2x^3-5x^2+7x $$ #### Solution ##### Step 1. Encode and evaluate the declared parent population | x | Binary | Fitness substitution | | --- | --- | --- | | 13 | 1101 | 2(13³) − 5(13²) + 7(13) = 3640 | | 6 | 0110 | 2(6³) − 5(6²) + 7(6) = 294 | | 2 | 0010 | 2(2³) − 5(2²) + 7(2) = 10 | | 7 | 0111 | 2(7³) − 5(7²) + 7(7) = 490 | ##### Step 2. Select the two fittest under the stated assumption [13, 7] with fitness [3640, 490]. ##### Step 3. Cut after bit 2 and exchange tails | Parent 1 | Parent 2 | Child 1 | Child 2 | | --- | --- | --- | --- | | 11 | 01 | 01 | 11 | 1111 | 0101 | ##### Step 4. Flip the last bit of each child | Before | After | Decoded x | | --- | --- | --- | | 1111 | 1110 | 14 | | 0101 | 0100 | 4 | ##### Step 5. Evaluate both children | x | Fitness calculation | | --- | --- | | 14 | 2(2744) − 5(196) + 7(14) = 4606 | | 4 | 2(64) − 5(16) + 7(4) = 76 | ##### Step 6. Compare with the selected parents Best parent score = 3640; best child score = 4606. Parent average = 2065; child average = 2341. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator. **Final answer.** Under these declared repairs and operator choices, children are [14, 4] with fitness [4606, 76]. These are not unique answers to the inconsistent and under-specified original question. The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2. ## Evaluate the polynomial without skipping the powers Calculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper. ### x=13 $$ x^2=13\times13=169\qquad x^3=169\times13=2197 $$ $$ 2x^3=4394\quad 5x^2=845\quad 7x=91 $$ $$ f(13)=4394-845+91=3640 $$ ### x=6 $$ x^2=6\times6=36\qquad x^3=36\times6=216 $$ $$ 2x^3=432\quad 5x^2=180\quad 7x=42 $$ $$ f(6)=432-180+42=294 $$ ### x=2 $$ x^2=2\times2=4\qquad x^3=4\times2=8 $$ $$ 2x^3=16\quad 5x^2=20\quad 7x=14 $$ $$ f(2)=16-20+14=10 $$ ### x=7 $$ x^2=7\times7=49\qquad x^3=49\times7=343 $$ $$ 2x^3=686\quad 5x^2=245\quad 7x=49 $$ $$ f(7)=686-245+49=490 $$ ### x=15 $$ x^2=15\times15=225\qquad x^3=225\times15=3375 $$ $$ 2x^3=6750\quad 5x^2=1125\quad 7x=105 $$ $$ f(15)=6750-1125+105=5730 $$ ### x=5 $$ x^2=5\times5=25\qquad x^3=25\times5=125 $$ $$ 2x^3=250\quad 5x^2=125\quad 7x=35 $$ $$ f(5)=250-125+35=160 $$ ### x=14 $$ x^2=14\times14=196\qquad x^3=196\times14=2744 $$ $$ 2x^3=5488\quad 5x^2=980\quad 7x=98 $$ $$ f(14)=5488-980+98=4606 $$ ### x=4 $$ x^2=4\times4=16\qquad x^3=16\times4=64 $$ $$ 2x^3=128\quad 5x^2=80\quad 7x=28 $$ $$ f(4)=128-80+28=76 $$ “Modulo 16” means keep the remainder after dividing by 16: 22=1×16+6, so its remainder is 6; 18=1×16+2; 23=1×16+7. This is only the declared repair used here, not an unspoken general encoding rule. --- ## M2 · Q6(a) Applications of fuzzy logic [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-final-2025--q6a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-final-2025/q6a Longer past-paper questions on Fuzzy, GA and PSO. Original TEE marks retained; not the M2 paper pattern. **List the applications of fuzzy logic. Explain any one application in detail.** [10 marks] ### Answer ### Definition Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”. ### Applications | Area | Examples | | --- | --- | | Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners | | Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control | | Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) | | Consumer electronics | Camera autofocus and auto-exposure, image stabilisation | | Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants | | Aerospace | Satellite and spacecraft attitude control | | Medicine | Diagnosis support, anaesthesia dosage control | | Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules | | AI and robotics | Robot navigation, natural-language interfaces, expert systems | ### One application in detail: fuzzy washing machine 1. **Problem.** Wash time depends on how dirty and how greasy the clothes are. These are vague (“a bit dirty”), and a fixed timer wastes water and power. 2. **Inputs and outputs.** Sensors measure dirt (turbidity of water) and grease; the output is wash time. Each is described by fuzzy sets: dirt Small/Medium/High, grease Small/Medium/High, time Very Short … Very Long. 3. **Fuzzification.** Sensor readings are converted to membership degrees, e.g. dirt 60% → Medium 0.8, High 0.2. 4. **Rule base.** 9 expert rules, e.g. IF dirt is High AND grease is High THEN time is Very Long; IF dirt is Small AND grease is Small THEN time is Very Short. 5. **Inference.** Mamdani min–max: rule strength = min of the two memberships; outputs combined by max. 6. **Defuzzification.** Use the specified defuzzifier. For a separate representative-weighted illustration, strengths [.6,.4,.2] at times [25,40,60] give (.6×25+.4×40+.2×60)/(.6+.4+.2)=(15+16+12)/1.2≈35.833 minutes. These illustrative strengths are supplied here, not derived from unspecified input graphs. A Mamdani area centroid requires the full aggregate shape. 7. **Benefits.** Smooth adjustment to every load, saves water and electricity, easy to design from expert knowledge without an exact mathematical model. ### Advantages and disadvantages (bonus) **Advantages:** handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. **Disadvantages:** rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required. **Yaad rakho:** Application explain karte waqt 5 cheezein: inputs, fuzzy sets, rules, inference (min/max), defuzzification + ek number ka example. --- ## M2 · Q7(a) Applications of PSO [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-final-2025--q7a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-final-2025/q7a Longer past-paper questions on Fuzzy, GA and PSO. Original TEE marks retained; not the M2 paper pattern. **List the applications of Particle Swarm Optimization and explain any one application in detail.** [10 marks] ### Answer ### Definition (1 line) Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions. ### Applications 1. **Function optimisation**: maximising or minimising complex non-linear, multimodal functions where derivatives are unavailable. 2. **Neural-network training**: optimising weights and biases (your Lab 8 uses the pyswarms library for this). 3. **Hyper-parameter tuning** of ML models (learning rate, number of hidden units, SVM C and γ). 4. **Feature selection**: choosing the most useful subset of features (binary PSO). 5. **Image processing**: threshold selection for segmentation, edge detection, image enhancement. 6. **Path planning** for robots and drones: shortest collision-free path. 7. **Power systems**: economic load dispatch, optimal placement of capacitors. 8. **Control engineering**: tuning PID controller gains. 9. **Scheduling and wireless sensor networks**: job scheduling, sensor node placement and clustering. ### One application in detail: training a neural network with PSO Training data → Neural network with weights = particle position → Error (fitness) → PSO moves the weights → Best weights = gbest 1. **Encoding.** Flatten all weights and biases of the network into one vector. Each particle is one complete set of weights; a network with 50 weights gives a 50-dimensional search space. 2. **Fitness.** Run the training data through the network with that particle's weights and compute the error (for example mean squared error or cross-entropy). Lower error = better particle (minimisation). 3. **Search.** Apply the velocity and position updates. Each particle is pulled towards the weight set that gave its own lowest error (pbest) and towards the lowest-error weights found by the swarm (gbest). 4. **Stop** after a fixed number of iterations or when the error stops improving; gbest is the trained network. 5. **Why PSO here:** it needs no gradients, so it works with non-differentiable activations and can explore candidate weights without differentiating the objective. It can still stagnate; comparison with gradient methods depends on the problem. **Limitation:** large networks create a high-dimensional search space, and evaluating many complete candidate networks can be computationally expensive. Gradient-based training is often more efficient when useful gradients are available; there is no universal speed ordering for every problem and implementation. ### Advantages and disadvantages (add for 10 marks) | Advantages | Disadvantages | | --- | --- | | Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum | | No derivatives needed | Sensitive to parameter choice | | Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum | | Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification | Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version). **Yaad rakho:** Application likhte waqt teen cheezein: particle kya hai, fitness kya hai, gbest ka matlab kya hai. --- ## M2 · Q1(b) Applications of PSO [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-reexam-2026--q1b/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-reexam-2026/q1b Longer past-paper practice restricted to Fuzzy, GA and PSO. **List the applications of Particle Swarm Optimization and explain any one application in detail.** [5 marks] ### Answer ### Definition (1 line) Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions. ### Applications 1. **Function optimisation**: maximising or minimising complex non-linear, multimodal functions where derivatives are unavailable. 2. **Neural-network training**: optimising weights and biases (your Lab 8 uses the pyswarms library for this). 3. **Hyper-parameter tuning** of ML models (learning rate, number of hidden units, SVM C and γ). 4. **Feature selection**: choosing the most useful subset of features (binary PSO). 5. **Image processing**: threshold selection for segmentation, edge detection, image enhancement. 6. **Path planning** for robots and drones: shortest collision-free path. 7. **Power systems**: economic load dispatch, optimal placement of capacitors. 8. **Control engineering**: tuning PID controller gains. 9. **Scheduling and wireless sensor networks**: job scheduling, sensor node placement and clustering. ### One application in detail: training a neural network with PSO Training data → Neural network with weights = particle position → Error (fitness) → PSO moves the weights → Best weights = gbest 1. **Encoding.** Flatten all weights and biases of the network into one vector. Each particle is one complete set of weights; a network with 50 weights gives a 50-dimensional search space. 2. **Fitness.** Run the training data through the network with that particle's weights and compute the error (for example mean squared error or cross-entropy). Lower error = better particle (minimisation). 3. **Search.** Apply the velocity and position updates. Each particle is pulled towards the weight set that gave its own lowest error (pbest) and towards the lowest-error weights found by the swarm (gbest). 4. **Stop** after a fixed number of iterations or when the error stops improving; gbest is the trained network. 5. **Why PSO here:** it needs no gradients, so it works with non-differentiable activations and can explore candidate weights without differentiating the objective. It can still stagnate; comparison with gradient methods depends on the problem. **Limitation:** large networks create a high-dimensional search space, and evaluating many complete candidate networks can be computationally expensive. Gradient-based training is often more efficient when useful gradients are available; there is no universal speed ordering for every problem and implementation. ### Advantages and disadvantages (add for 10 marks) | Advantages | Disadvantages | | --- | --- | | Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum | | No derivatives needed | Sensitive to parameter choice | | Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum | | Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification | Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version). **Yaad rakho:** Application likhte waqt teen cheezein: particle kya hai, fitness kya hai, gbest ka matlab kya hai. --- ## M2 · Q1(c) Fuzzy set operations [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-reexam-2026--q1c/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-reexam-2026/q1c Longer past-paper practice restricted to Fuzzy, GA and PSO. **Question. X = {(c1, 0.7), (c2, 0.9), (c3, 0.2), (c4, 0.4), (c5, 0.9)}, Y = {(c1, 0.6), (c2, 0.3), (c3, 0.8), (c4, 0.7), (c5, 0.1)}. Find X ∪ Y, X ∩ Y, ¬X ∪ ¬Y, ¬X ∩ ¬Y.** [5 marks] ### Answer #### Given The aligned X and Y memberships printed in this question. #### To find The four requested membership lists in the original element order. #### Formula and why it applies Use max for OR, min for AND, and 1−μ for NOT. $$ \mu_{X\cup Y}=\max(\mu_X,\mu_Y),\quad\mu_{X\cap Y}=\min(\mu_X,\mu_Y) $$ Complement both operands before applying the operator in ¬X∪¬Y or ¬X∩¬Y. #### Solution | | c1 | c2 | c3 | c4 | c5 | | --- | --- | --- | --- | --- | --- | | ¬X | 0.3 | 0.1 | 0.8 | 0.6 | 0.1 | | ¬Y | 0.4 | 0.7 | 0.2 | 0.3 | 0.9 | | **X ∪ Y** | 0.7 | 0.9 | 0.8 | 0.7 | 0.9 | | **X ∩ Y** | 0.6 | 0.3 | 0.2 | 0.4 | 0.1 | | **¬X ∪ ¬Y** | 0.4 | 0.7 | 0.8 | 0.6 | 0.9 | | **¬X ∩ ¬Y** | 0.3 | 0.1 | 0.2 | 0.3 | 0.1 | **Final answer.** As in the last four rows. De Morgan check: ¬X ∪ ¬Y = 1 − (X ∩ Y) and ¬X ∩ ¬Y = 1 − (X ∪ Y) ✓. ## Calculate every element separately ### Element c1 $$ \neg X:1-0.7=0.3\qquad\neg Y:1-0.6=0.4 $$ $$ X\cup Y:\max(0.7,0.6)=0.7 $$ $$ X\cap Y:\min(0.7,0.6)=0.6 $$ $$ \neg X\cup\neg Y:\max(0.3,0.4)=0.4 $$ $$ \neg X\cap\neg Y:\min(0.3,0.4)=0.3 $$ ### Element c2 $$ \neg X:1-0.9=0.1\qquad\neg Y:1-0.3=0.7 $$ $$ X\cup Y:\max(0.9,0.3)=0.9 $$ $$ X\cap Y:\min(0.9,0.3)=0.3 $$ $$ \neg X\cup\neg Y:\max(0.1,0.7)=0.7 $$ $$ \neg X\cap\neg Y:\min(0.1,0.7)=0.1 $$ ### Element c3 $$ \neg X:1-0.2=0.8\qquad\neg Y:1-0.8=0.2 $$ $$ X\cup Y:\max(0.2,0.8)=0.8 $$ $$ X\cap Y:\min(0.2,0.8)=0.2 $$ $$ \neg X\cup\neg Y:\max(0.8,0.2)=0.8 $$ $$ \neg X\cap\neg Y:\min(0.8,0.2)=0.2 $$ ### Element c4 $$ \neg X:1-0.4=0.6\qquad\neg Y:1-0.7=0.3 $$ $$ X\cup Y:\max(0.4,0.7)=0.7 $$ $$ X\cap Y:\min(0.4,0.7)=0.4 $$ $$ \neg X\cup\neg Y:\max(0.6,0.3)=0.6 $$ $$ \neg X\cap\neg Y:\min(0.6,0.3)=0.3 $$ ### Element c5 $$ \neg X:1-0.9=0.1\qquad\neg Y:1-0.1=0.9 $$ $$ X\cup Y:\max(0.9,0.1)=0.9 $$ $$ X\cap Y:\min(0.9,0.1)=0.1 $$ $$ \neg X\cup\neg Y:\max(0.1,0.9)=0.9 $$ $$ \neg X\cap\neg Y:\min(0.1,0.9)=0.1 $$ --- ## M2 · Q2(a) Fuzzy fire-hazard system [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-reexam-2026--q2a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-reexam-2026/q2a Longer past-paper practice restricted to Fuzzy, GA and PSO. **A building safety department wants a fuzzy logic system to predict fire hazards. Inputs are Temperature (1–100°C) and Smoke Level (1–100). The output Fire Risk (1–100) has labels Low, Moderate and High. Define triangular membership functions, construct at least nine rules, explain Mamdani inference and defuzzify for Temperature=65°C and Smoke=80.** [10 marks] ### Answer #### Given | Variable | Crisp value | | --- | --- | | Temperature | 65 | | Smoke Level | 80 | The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens. #### To find A complete stated design and its continuous Mamdani centroid. #### Formula and why it applies Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier. $$ z^*=\frac{\int z\mu_{out}(z)\,dz}{\int\mu_{out}(z)\,dz} $$ #### Solution ##### Step 1. Specify the membership functions | Input / output position | Triangle | Peak | | --- | --- | --- | | Low / Low | (1,25,50) | 25 | | Medium / Moderate | (25,50,75) | 50 | | High / High | (50,75,100) | 75 | Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values. ##### Step 2. Write all nine rules | Temperature / Smoke Level | Low | Medium | High | | --- | --- | --- | --- | | Low | Low | Low | Moderate | | Medium | Low | Moderate | High | | High | Moderate | High | High | Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium. ##### Step 3. Fuzzify Temperature = 65 $$ \mu_{Low}(65)=0 $$ $$ \mu_{Medium}(65)=\frac{75-65}{75-50}=0.4 $$ $$ \mu_{High}(65)=\frac{65-50}{75-50}=0.6 $$ ##### Step 4. Fuzzify Smoke Level = 80 $$ \mu_{Low}(80)=0 $$ $$ \mu_{Medium}(80)=0 $$ $$ \mu_{High}(80)=\frac{100-80}{100-75}=0.8 $$ ##### Step 5. Evaluate all rules and aggregate matching consequents | Temperature | Smoke Level | Firing strength | Consequent | | --- | --- | --- | --- | | Low | Low | min(0,0) = 0 | Low | | Low | Medium | min(0,0) = 0 | Low | | Low | High | min(0,0.8) = 0 | Moderate | | Medium | Low | min(0.4,0) = 0 | Low | | Medium | Medium | min(0.4,0) = 0 | Moderate | | Medium | High | min(0.4,0.8) = 0.4 | High | | High | Low | min(0.6,0) = 0 | Moderate | | High | Medium | min(0.6,0) = 0 | High | | High | High | min(0.6,0.8) = 0.6 | High | Aggregated label strengths: Low = 0, Moderate = 0, High = 0.6. $$ \mu_{out}(z)=\max_r\min(\alpha_r,\mu_{C_r}(z)) $$ ##### Step 6. Draw the aggregated output Swipe across the diagram to read its labels ↔ ![The shaded upper envelope counts any overlap only once.](https://deep-learning-study-room.vercel.app/is2/fig/v2_fire_aggregate.e0cab4ad42.svg) Diagram: The shaded upper envelope counts any overlap only once.. Labels: Chosen triangular output sets and their max-aggregate | 0 | 0.5 | 1 | 1 | 25 | 50 | 75 | 100 | μ | Output score | Low | Moderate | High | centroid ≈ 75 The shaded upper envelope counts any overlap only once. ##### Step 7. Calculate area and moment of each straight-line strip For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height. $$ A=(b-a)(u+v)/2 $$ $$ M=\frac{b-a}{6}\big[(2a+b)u+(a+2b)v\big] $$ | a | b | u | v | Area A | Moment M | | --- | --- | --- | --- | --- | --- | | 50 | 65 | 0 | 0.6 | 4.5 | 270 | | 65 | 85 | 0.6 | 0.6 | 12 | 900 | | 85 | 100 | 0.6 | 0 | 4.5 | 405 | ##### Step 8. Divide total moment by total area $$ \sum A=21 $$ $$ \sum M=1575 $$ $$ z^*=\frac{1575}{21}\approx75 $$ **Final answer.** For these stated triangles and rules, the centroid is 75. This is in the high-risk region of this illustrative fuzzy design. Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised. ## Why the clipped High shape is symmetric about 75 Only the High output survives, at height .6. Its triangle is (50,75,100). Solve its rising edge: (z−50)/25=.6, so z=50+15=65. Solve its falling edge: (100−z)/25=.6, so z=100−15=85. | Piece | Area calculation | Centre | Moment | | --- | --- | --- | --- | | Left triangle [50,65] | ½×15×.6=4.5 | (50+65+65)/3=60 | 4.5×60=270 | | Rectangle [65,85] | 20×.6=12 | (65+85)/2=75 | 12×75=900 | | Right triangle [85,100] | ½×15×.6=4.5 | (85+85+100)/3=90 | 4.5×90=405 | The horizontal centroid of a triangle is the average of the three vertex x-coordinates. Add areas: 4.5+12+4.5=21. Add moments:270+900+405=1575. Divide 1575/21=75, agreeing with symmetry. --- ## M2 · Q3(b) Fuzzy relations and composition [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-reexam-2026--q3b/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-reexam-2026/q3b Longer past-paper practice restricted to Fuzzy, GA and PSO. **Question. A = {0.6/small + 0.4/medium + 0.1/large}, B = {0.4/slow + 0.8/moderate + 0.3/fast}, C = {0.5/small + 0.2/medium + 0.4/large}. Find R = A × B, S = C × B and the max–min composition S ∘ R.** [10 marks] ### Answer #### Given The fuzzy sets or relation matrices and their listed row/column order. #### To find The relation matrix/matrices and the composition specified in the question. #### Formula and why it applies For Cartesian products take min of row and column membership. For the ordered matrix composition shown here: $$ T_{ik}=\max_j\min(S_{ij},R_{jk}) $$ If max–product is asked instead, replace the inner min by multiplication. #### Solution ##### Step 1. R and S $$ R=\begin{bmatrix}0.4&0.6&0.3\\0.4&0.4&0.3\\0.1&0.1&0.1\end{bmatrix},\qquad S=\begin{bmatrix}0.4&0.5&0.3\\0.2&0.2&0.2\\0.4&0.4&0.3\end{bmatrix} $$ ##### Step 2. S ∘ R $$ (1,1)=\max[\min(.4,.4),\min(.5,.4),\min(.3,.1)]=\max(.4,.4,.1)=0.4 $$ $$ (1,2)=\max[\min(.4,.6),\min(.5,.4),\min(.3,.1)]=\max(.4,.4,.1)=0.4 $$ $$ (1,3)=\max[\min(.4,.3),\min(.5,.3),\min(.3,.1)]=0.3 $$ $$ \text{row 2 (all 0.2)}=(0.2,0.2,0.2);\quad(3,1)=\max(.4,.4,.1)=0.4,\;(3,2)=\max(.4,.4,.1)=0.4,\;(3,3)=0.3 $$ **Final answer.** $$ R=\begin{bmatrix}.4&.6&.3\\.4&.4&.3\\.1&.1&.1\end{bmatrix} $$ , $$ S=\begin{bmatrix}.4&.5&.3\\.2&.2&.2\\.4&.4&.3\end{bmatrix} $$ , $$ S\circ R=\begin{bmatrix}0.4&0.4&0.3\\0.2&0.2&0.2\\0.4&0.4&0.3\end{bmatrix} $$ . (The key gives R and S only.) ## Every cell without the row shortcuts Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries. ### R: Cartesian-product cells | Cell | Row degree | Column degree | min result | | --- | --- | --- | --- | | R11 | 0.6 | 0.4 | 0.4 | | R12 | 0.6 | 0.8 | 0.6 | | R13 | 0.6 | 0.3 | 0.3 | | R21 | 0.4 | 0.4 | 0.4 | | R22 | 0.4 | 0.8 | 0.4 | | R23 | 0.4 | 0.3 | 0.3 | | R31 | 0.1 | 0.4 | 0.1 | | R32 | 0.1 | 0.8 | 0.1 | | R33 | 0.1 | 0.3 | 0.1 | ### S: Cartesian-product cells | Cell | Row degree | Column degree | min result | | --- | --- | --- | --- | | S11 | 0.5 | 0.4 | 0.4 | | S12 | 0.5 | 0.8 | 0.5 | | S13 | 0.5 | 0.3 | 0.3 | | S21 | 0.2 | 0.4 | 0.2 | | S22 | 0.2 | 0.8 | 0.2 | | S23 | 0.2 | 0.3 | 0.2 | | S31 | 0.4 | 0.4 | 0.4 | | S32 | 0.4 | 0.8 | 0.4 | | S33 | 0.4 | 0.3 | 0.3 | **Result cell (1,1)** $$ \max\{\min(0.4,0.4),\min(0.5,0.4),\min(0.3,0.1)\} $$ $$ =\max\{0.4,0.4,0.1\}=0.4 $$ **Result cell (1,2)** $$ \max\{\min(0.4,0.6),\min(0.5,0.4),\min(0.3,0.1)\} $$ $$ =\max\{0.4,0.4,0.1\}=0.4 $$ **Result cell (1,3)** $$ \max\{\min(0.4,0.3),\min(0.5,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.3,0.3,0.1\}=0.3 $$ **Result cell (2,1)** $$ \max\{\min(0.2,0.4),\min(0.2,0.4),\min(0.2,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (2,2)** $$ \max\{\min(0.2,0.6),\min(0.2,0.4),\min(0.2,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (2,3)** $$ \max\{\min(0.2,0.3),\min(0.2,0.3),\min(0.2,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (3,1)** $$ \max\{\min(0.4,0.4),\min(0.4,0.4),\min(0.3,0.1)\} $$ $$ =\max\{0.4,0.4,0.1\}=0.4 $$ **Result cell (3,2)** $$ \max\{\min(0.4,0.6),\min(0.4,0.4),\min(0.3,0.1)\} $$ $$ =\max\{0.4,0.4,0.1\}=0.4 $$ **Result cell (3,3)** $$ \max\{\min(0.4,0.3),\min(0.4,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.3,0.3,0.1\}=0.3 $$ --- ## M2 · Q4(a) PSO, one iteration [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-reexam-2026--q4a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-reexam-2026/q4a Longer past-paper practice restricted to Fuzzy, GA and PSO. **Question. Maximize $$ F(x)=x_1^2-x_1x_2+x_2^2+3x_1+4x_2+5 $$ using Particle Swarm Optimization. $$ X=\begin{bmatrix}3.1472&-4.0246\\4.0579&-2.2150\end{bmatrix},\;V=\begin{bmatrix}0.2194&0.2449\\0.1908&0.2228\end{bmatrix} $$ . Take $$ W=0.9,\;c_1=c_2=1.5,\;r_1=0.5949,\;r_2=0.0855 $$ . Optimize and give the solution after 1 iteration showing the step-by-step process.** [10 marks] ### Answer #### Given P1: $$ x=(3.1472,\,-4.0246) $$ , $$ v=(0.2194,\,0.2449) $$ . P2: $$ x=(4.0579,\,-2.2150) $$ , $$ v=(0.1908,\,0.2228) $$ . $$ c_1r_1=0.89235 $$ , $$ c_2r_2=0.12825 $$ . #### To find Updated velocities, positions, pbest and gbest after one iteration. #### Formula and why it applies $$ v^{new}=wv+c_1r_1(pbest-x)+c_2r_2(gbest-x),\quad x^{new}=x+v^{new} $$ Initial personal bests equal the starting positions. Use one frozen global best for all particles. This question maximises fitness, so keep a memory only when its score is larger. #### Solution ##### Step 1. Fitness of P1 (term by term) $$ x_1^2=9.904868,\;-x_1x_2=+12.666221,\;x_2^2=16.197405,\;3x_1=9.4416,\;4x_2=-16.0984 $$ $$ F_1=9.904868+12.666221+16.197405+9.4416-16.0984+5=37.1117 $$ ##### Step 2. Fitness of P2 $$ F_2=16.466552+8.988249+4.906225+12.1737-8.86+5=38.6747 $$ ##### Step 3. pbest and gbest pbest = current positions. $$ 38.6747>37.1117 $$ , so **gbest = P2 = (4.0579, −2.2150)**. ##### Step 4. Particle 1 updates $$ v_{11}=0.9(0.2194)+0+0.12825(4.0579-3.1472)=0.19746+0.116797=0.314257 $$ $$ x_{11}=3.1472+0.314257=3.461457 $$ $$ v_{12}=0.9(0.2449)+0+0.12825(-2.2150+4.0246)=0.22041+0.232081=0.452491 $$ $$ x_{12}=-4.0246+0.452491=-3.572109 $$ ##### Step 5. Particle 2 updates (gbest particle: only inertia acts) $$ v_{21}=0.9(0.1908)=0.17172,\quad x_{21}=4.22962;\qquad v_{22}=0.9(0.2228)=0.20052,\quad x_{22}=-2.01448 $$ ##### Step 6. New fitness and bests Use the same objective at each new position. Keep full precision internally; displayed coordinates are rounded. $$ F(3.461457275,-3.5721088)=(3.461457275)^2-(3.461457275)(-3.5721088)+(-3.5721088)^2+3(3.461457275)+4(-3.5721088)+5 $$ $$ =11.981686467+12.364701993+12.759961279+10.384371825+(-14.2884352)+5\approx38.202286364 $$ $$ F(4.22962,-2.01448)=(4.22962)^2-(4.22962)(-2.01448)+(-2.01448)^2+3(4.22962)+4(-2.01448)+5 $$ $$ =17.889685344+8.520484898+4.05812967+12.68886+(-8.05792)+5\approx40.099239912 $$ | Particle | New position | New F | Old F | pbest | | --- | --- | --- | --- | --- | | 1 | (3.461457, −3.572109) | 38.2023 | 37.1117 | updated | | 2 | (4.22962, −2.01448) | 40.0992 | 38.6747 | updated | The best pbest is now P2's with F = 40.0992, so gbest moves to (4.22962, −2.01448). **Final answer.** $$ V=\begin{bmatrix}0.314257&0.452491\\0.17172&0.20052\end{bmatrix} $$ , $$ X=\begin{bmatrix}3.461457&-3.572109\\4.22962&-2.01448\end{bmatrix} $$ . Both particles improved; new gbest = (4.22962, −2.01448) with F = 40.0992. **Official key warning.** The re-exam answer key solves a *different* problem: particles (0.4, 0.3) and (0.3, 0.4) with velocities (0.1, −0.2) and (−0.2, 0.1), giving fitness 7.53 and 7.86. Do not copy it. The numbers above solve the question as printed. This X and V are particles 1 and 2 of your lecture's five-particle example; only the function differs (+3x₁ and +5 here instead of +2x₁ and +3). --- ## M2 · Q5(a) Genetic algorithm [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-reexam-2026--q5a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-reexam-2026/q5a Longer past-paper practice restricted to Fuzzy, GA and PSO. **Implement a GA to maximise $$ f(x)=2x^3-3x^2+8x $$ for x in [0,50]. Take 4-bit representations of 12, 20, 17, 13. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness.** [10 marks] ### Answer **The printed data are inconsistent.** 20 and 17 do not fit into four bits, and covering 0–50 needs six bits. We use six-bit encoding to preserve the supplied parent values and the stated range. #### Given Printed parents: [12, 20, 17, 13]. Worked variant: [12, 20, 17, 13] with 6 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2. #### To find Two valid children under the declared assumptions and their fitness. #### Formula and why it applies Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation. $$ f(x)=2x^3-3x^2+8x $$ #### Solution ##### Step 1. Encode and evaluate the declared parent population | x | Binary | Fitness substitution | | --- | --- | --- | | 12 | 001100 | 2(12³) − 3(12²) + 8(12) = 3120 | | 20 | 010100 | 2(20³) − 3(20²) + 8(20) = 14960 | | 17 | 010001 | 2(17³) − 3(17²) + 8(17) = 9095 | | 13 | 001101 | 2(13³) − 3(13²) + 8(13) = 3991 | ##### Step 2. Select the two fittest under the stated assumption [20, 17] with fitness [14960, 9095]. ##### Step 3. Cut after bit 2 and exchange tails | Parent 1 | Parent 2 | Child 1 | Child 2 | | --- | --- | --- | --- | | 01 | 0100 | 01 | 0001 | 010001 | 010100 | ##### Step 4. Flip the last bit of each child | Before | After | Decoded x | | --- | --- | --- | | 010001 | 010000 | 16 | | 010100 | 010101 | 21 | ##### Step 5. Evaluate both children | x | Fitness calculation | | --- | --- | | 16 | 2(4096) − 3(256) + 8(16) = 7552 | | 21 | 2(9261) − 3(441) + 8(21) = 17367 | ##### Step 6. Compare with the selected parents Best parent score = 14960; best child score = 17367. Parent average = 12027.5; child average = 12459.5. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator. **Final answer.** Under these declared repairs and operator choices, children are [16, 21] with fitness [7552, 17367]. These are not unique answers to the inconsistent and under-specified original question. The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2. ## Evaluate the polynomial without skipping the powers Calculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper. ### x=12 $$ x^2=12\times12=144\qquad x^3=144\times12=1728 $$ $$ 2x^3=3456\quad 3x^2=432\quad 8x=96 $$ $$ f(12)=3456-432+96=3120 $$ ### x=20 $$ x^2=20\times20=400\qquad x^3=400\times20=8000 $$ $$ 2x^3=16000\quad 3x^2=1200\quad 8x=160 $$ $$ f(20)=16000-1200+160=14960 $$ ### x=17 $$ x^2=17\times17=289\qquad x^3=289\times17=4913 $$ $$ 2x^3=9826\quad 3x^2=867\quad 8x=136 $$ $$ f(17)=9826-867+136=9095 $$ ### x=13 $$ x^2=13\times13=169\qquad x^3=169\times13=2197 $$ $$ 2x^3=4394\quad 3x^2=507\quad 8x=104 $$ $$ f(13)=4394-507+104=3991 $$ ### x=16 $$ x^2=16\times16=256\qquad x^3=256\times16=4096 $$ $$ 2x^3=8192\quad 3x^2=768\quad 8x=128 $$ $$ f(16)=8192-768+128=7552 $$ ### x=21 $$ x^2=21\times21=441\qquad x^3=441\times21=9261 $$ $$ 2x^3=18522\quad 3x^2=1323\quad 8x=168 $$ $$ f(21)=18522-1323+168=17367 $$ --- ## M2 · Q7(a) Applications of fuzzy logic [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/papers/paper-reexam-2026--q7a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/papers/paper-reexam-2026/q7a Longer past-paper practice restricted to Fuzzy, GA and PSO. **List the applications of fuzzy logic. Explain any one application in detail.** [10 marks] ### Answer ### Definition Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”. ### Applications | Area | Examples | | --- | --- | | Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners | | Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control | | Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) | | Consumer electronics | Camera autofocus and auto-exposure, image stabilisation | | Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants | | Aerospace | Satellite and spacecraft attitude control | | Medicine | Diagnosis support, anaesthesia dosage control | | Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules | | AI and robotics | Robot navigation, natural-language interfaces, expert systems | ### One application in detail: fuzzy washing machine 1. **Problem.** Wash time depends on how dirty and how greasy the clothes are. These are vague (“a bit dirty”), and a fixed timer wastes water and power. 2. **Inputs and outputs.** Sensors measure dirt (turbidity of water) and grease; the output is wash time. Each is described by fuzzy sets: dirt Small/Medium/High, grease Small/Medium/High, time Very Short … Very Long. 3. **Fuzzification.** Sensor readings are converted to membership degrees, e.g. dirt 60% → Medium 0.8, High 0.2. 4. **Rule base.** 9 expert rules, e.g. IF dirt is High AND grease is High THEN time is Very Long; IF dirt is Small AND grease is Small THEN time is Very Short. 5. **Inference.** Mamdani min–max: rule strength = min of the two memberships; outputs combined by max. 6. **Defuzzification.** Use the specified defuzzifier. For a separate representative-weighted illustration, strengths [.6,.4,.2] at times [25,40,60] give (.6×25+.4×40+.2×60)/(.6+.4+.2)=(15+16+12)/1.2≈35.833 minutes. These illustrative strengths are supplied here, not derived from unspecified input graphs. A Mamdani area centroid requires the full aggregate shape. 7. **Benefits.** Smooth adjustment to every load, saves water and electricity, easy to design from expert knowledge without an exact mathematical model. ### Advantages and disadvantages (bonus) **Advantages:** handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. **Disadvantages:** rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required. **Yaad rakho:** Application explain karte waqt 5 cheezein: inputs, fuzzy sets, rules, inference (min/max), defuzzification + ek number ka example. --- ## M2 · Q1(a) · Fuzzy operations [2] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-1--q1a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-1/q1a Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q1(a). On U = {u1, u2, u3}, A has memberships [0.2, 0.8, 0.6] and B has memberships [0.7, 0.3, 0.5]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations.** [2 marks · Practice] #### Given | Element | u1 | u2 | u3 | | --- | --- | --- | --- | | A | 0.2 | 0.8 | 0.6 | | B | 0.7 | 0.3 | 0.5 | #### To find Four membership lists in exactly the same element order. #### Formula and why it applies Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A. $$ \mu_{A\cup B}=\max(\mu_A,\mu_B)\qquad\mu_{A\cap B}=\min(\mu_A,\mu_B) $$ $$ \mu_{B^c}=1-\mu_B\qquad\mu_{A\setminus B}=\min(\mu_A,1-\mu_B) $$ #### Solution ##### Step 1. Align the same element All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂. ##### Step 2. Calculate one element completely $$ \mu_{B^c}(u_1)=1-0.7=0.3 $$ $$ \mu_{A\setminus B}(u_1)=\min(0.2,0.3)=0.2 $$ ##### Step 3. Repeat for each element | Element | A | B | Union | Intersection | Bᶜ | A ∖ B | | --- | --- | --- | --- | --- | --- | --- | | u1 | 0.2 | 0.7 | max(0.2, 0.7) = 0.7 | min(0.2, 0.7) = 0.2 | 1 − 0.7 = 0.3 | min(0.2, 0.3) = 0.2 | | u2 | 0.8 | 0.3 | max(0.8, 0.3) = 0.8 | min(0.8, 0.3) = 0.3 | 1 − 0.3 = 0.7 | min(0.8, 0.7) = 0.7 | | u3 | 0.6 | 0.5 | max(0.6, 0.5) = 0.6 | min(0.6, 0.5) = 0.5 | 1 − 0.5 = 0.5 | min(0.6, 0.5) = 0.5 | **Final answer.** A ∪ B = [0.7, 0.8, 0.6]; A ∩ B = [0.2, 0.3, 0.5]; Bᶜ = [0.3, 0.7, 0.5]; A ∖ B = [0.2, 0.7, 0.5]. --- ## M2 · Q1(b) · Swarm intelligence [2] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-1--q1b/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-1/q1b Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q1(b). Define swarm intelligence. Briefly explain one nature-inspired algorithm.** [2 marks · Practice] #### Answer Swarm intelligence is collective problem-solving arising from interactions among simple agents. PSO is inspired by social movement in flocks. Each particle represents a candidate solution and has a position and velocity. It moves using its own best visited position and the swarm’s best position. Repeated objective evaluation guides the search. --- ## M2 · Q2 · GA through two generations [3] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-1--q2/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-1/q2 Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q2. Maximise $$ f(x)=x^2 $$ over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness.** [3 marks · Practice; complete two-generation solution] #### Given Initial ordered population: [3, 5, 10, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [1, 2, 3, 4] in every generation. Replace the whole population with the four children; no elitism. | Generation | Selection instruction | | --- | --- | | 1 | Current individual row numbers [3, 4, 4, 3] | | 2 | Current individual row numbers [2, 3, 2, 3] | #### To find Both generations: evaluation, selection, crossover, mutation, decoding and comparison. #### Formula and why it applies Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step. #### Solution ##### Step 1. Generation 1: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 3 | 0011 | 3 × 3 = 9 | | 2 | 5 | 0101 | 5 × 5 = 25 | | 3 | 10 | 1010 | 10 × 10 = 100 | | 4 | 12 | 1100 | 12 × 12 = 144 | ##### Step 2. Generation 1: apply the supplied selection order The question directly supplies individual numbers [3, 4, 4, 3]. These refer to the current generation’s row numbers, not their x values. | Mating-pool slot | Chosen individual | x | Chromosome | | --- | --- | --- | --- | | 1 | 3 | 10 | 1010 | | 2 | 4 | 12 | 1100 | | 3 | 4 | 12 | 1100 | | 4 | 3 | 10 | 1010 | ##### Step 3. Generation 1: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 10 | 10 | 11 | 00 | 1000 | 1110 | | 3 & 4 | 11 | 00 | 10 | 10 | 1110 | 1000 | ##### Step 4. Generation 1: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1000 | Flip bit 4 | 1001 | | 2 | 1110 | Flip bit 4 | 1111 | | 3 | 1110 | Flip bit 4 | 1111 | | 4 | 1000 | Flip bit 4 | 1001 | ##### Step 5. Generation 1: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 | | 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 | | 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 | | 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 | ##### Step 6. Generation 1: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{9+25+100+144}{4}=\frac{278}{4}=69.5 $$ $$ \overline f_{new}=\frac{81+225+225+81}{4}=\frac{612}{4}=153 $$ Best objective: 144 → 225. Higher is better. The next population is [9, 15, 15, 9]. No elitism is imposed, so the best previous individual is not automatically protected. ##### Step 7. Generation 2: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 9 | 1001 | 9 × 9 = 81 | | 2 | 15 | 1111 | 15 × 15 = 225 | | 3 | 15 | 1111 | 15 × 15 = 225 | | 4 | 9 | 1001 | 9 × 9 = 81 | ##### Step 8. Generation 2: apply the supplied selection order The question directly supplies individual numbers [2, 3, 2, 3]. These refer to the current generation’s row numbers, not their x values. | Mating-pool slot | Chosen individual | x | Chromosome | | --- | --- | --- | --- | | 1 | 2 | 15 | 1111 | | 2 | 3 | 15 | 1111 | | 3 | 2 | 15 | 1111 | | 4 | 3 | 15 | 1111 | ##### Step 9. Generation 2: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 11 | 11 | 11 | 11 | 1111 | 1111 | | 3 & 4 | 11 | 11 | 11 | 11 | 1111 | 1111 | ##### Step 10. Generation 2: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1111 | Flip bit 4 | 1110 | | 2 | 1111 | Flip bit 4 | 1110 | | 3 | 1111 | Flip bit 4 | 1110 | | 4 | 1111 | Flip bit 4 | 1110 | ##### Step 11. Generation 2: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | ##### Step 12. Generation 2: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{81+225+225+81}{4}=\frac{612}{4}=153 $$ $$ \overline f_{new}=\frac{196+196+196+196}{4}=\frac{784}{4}=196 $$ Best objective: 225 → 196. Higher is better. The next population is [14, 14, 14, 14]. No elitism is imposed, so the best previous individual is not automatically protected. **Final answer.** After two generations the ordered population is [14, 14, 14, 14]. Best final objective = 196 at x = [14]. --- ## M2 · Q3 · Two-particle PSO through iteration 2 [3] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-1--q3/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-1/q3 Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q3. Minimise $$ f(x)=x^2 $$ using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [3 marks · Practice; expanded teaching solution] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (-1) | (0) | | B | (3) | (0) | w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(-1)=(-1)^2=1=1 $$ **Particle B** $$ f(3)=3^2=9=9 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(-1) $$ Its score is 1. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (-1). pbest values are A: (-1), B: (3). ##### Step 4. Iteration 1: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=-1-(-1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=-1-(-1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=(-1)+0=-1 $$ ##### Step 5. Iteration 1: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3-(3)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=-1-(3)=-4 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-4)=-2 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-2)=-2 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=3+(-2)=1 $$ ##### Step 6. Iteration 1: evaluate the new positions **Particle A** $$ f(-1)=(-1)^2=1=1 $$ **Particle B** $$ f(1)=1^2=1=1 $$ ##### Step 7. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 1 | 1 | Keep old: tie | (-1) | | B | 9 | 1 | Replace: new score is lower | (1) | Compare the retained pbest scores. gbest = **(-1)**, with objective **1**. ##### Step 8. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (-1) | (0) | (-1) | | B | (1) | (-2) | (1) | ##### Step 9. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (-1). pbest values are A: (-1), B: (1). ##### Step 10. Iteration 2: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=-1-(-1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=-1-(-1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(2)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=(-1)+0=-1 $$ ##### Step 11. Iteration 2: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-2)=-1 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=-1-(1)=-2 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-2)=-1 $$ Add the three velocity contributions. $$ v^{(2)}=(-1)+0+(-1)=-2 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1+(-2)=-1 $$ ##### Step 12. Iteration 2: evaluate the new positions **Particle A** $$ f(-1)=(-1)^2=1=1 $$ **Particle B** $$ f(-1)=(-1)^2=1=1 $$ ##### Step 13. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 1 | 1 | Keep old: tie | (-1) | | B | 1 | 1 | Keep old: tie | (1) | Compare the retained pbest scores. gbest = **(-1)**, with objective **1**. ##### Step 14. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (-1) | (0) | (-1) | | B | (-1) | (-2) | (1) | **Final answer.** After iteration 2, gbest = (-1), objective = 1. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. --- ## M2 · Q4 · Fuzzy architecture [3] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-1--q4/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-1/q4 Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q4. Explain the architecture of a Mamdani fuzzy inference system with a fan-controller example.** [3 marks · Practice] #### Answer The fuzzifier maps crisp temperature and occupancy inputs to membership degrees. The knowledge base contains membership functions and IF–THEN rules. The inference engine combines antecedent degrees, applies each rule’s consequent and aggregates the inferred outputs. The defuzzifier converts this aggregate to a numerical fan speed. For example, IF temperature is High AND occupancy is High THEN fan speed is Very High. --- ## M2 · Q1(a) · Fuzzy operations [2] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-2--q1a/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-2/q1a Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q1(a). On U = {u1, u2, u3}, A has memberships [0.1, 0.5, 1] and B has memberships [0.9, 0.4, 0]. Find A ∪ B, A ∩ B, Bᶜ and A ∖ B using standard max–min fuzzy operations.** [2 marks · Practice] #### Given | Element | u1 | u2 | u3 | | --- | --- | --- | --- | | A | 0.1 | 0.5 | 1 | | B | 0.9 | 0.4 | 0 | #### To find Four membership lists in exactly the same element order. #### Formula and why it applies Union is OR → max. Intersection is AND → min. Complement is NOT → 1 minus membership. Difference A ∖ B means A AND NOT B, so first complement B, then intersect with A. $$ \mu_{A\cup B}=\max(\mu_A,\mu_B)\qquad\mu_{A\cap B}=\min(\mu_A,\mu_B) $$ $$ \mu_{B^c}=1-\mu_B\qquad\mu_{A\setminus B}=\min(\mu_A,1-\mu_B) $$ #### Solution ##### Step 1. Align the same element All operations compare the two memberships of the same u. Do not compare A at u₁ with B at u₂. ##### Step 2. Calculate one element completely $$ \mu_{B^c}(u_1)=1-0.9=0.1 $$ $$ \mu_{A\setminus B}(u_1)=\min(0.1,0.1)=0.1 $$ ##### Step 3. Repeat for each element | Element | A | B | Union | Intersection | Bᶜ | A ∖ B | | --- | --- | --- | --- | --- | --- | --- | | u1 | 0.1 | 0.9 | max(0.1, 0.9) = 0.9 | min(0.1, 0.9) = 0.1 | 1 − 0.9 = 0.1 | min(0.1, 0.1) = 0.1 | | u2 | 0.5 | 0.4 | max(0.5, 0.4) = 0.5 | min(0.5, 0.4) = 0.4 | 1 − 0.4 = 0.6 | min(0.5, 0.6) = 0.5 | | u3 | 1 | 0 | max(1, 0) = 1 | min(1, 0) = 0 | 1 − 0 = 1 | min(1, 1) = 1 | **Final answer.** A ∪ B = [0.9, 0.5, 1.0]; A ∩ B = [0.1, 0.4, 0.0]; Bᶜ = [0.1, 0.6, 1.0]; A ∖ B = [0.1, 0.5, 1.0]. --- ## M2 · Q1(b) · Swarm intelligence [2] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-2--q1b/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-2/q1b Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q1(b). Define swarm intelligence. Briefly explain one nature-inspired algorithm.** [2 marks · Practice] #### Answer Swarm intelligence is collective problem-solving arising from interactions among simple agents. PSO is inspired by social movement in flocks. Each particle represents a candidate solution and has a position and velocity. It moves using its own best visited position and the swarm’s best position. Repeated objective evaluation guides the search. --- ## M2 · Q2 · GA through two generations [3] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-2--q2/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-2/q2 Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q2. Maximise $$ f(x)=x^2 $$ over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness.** [3 marks · Practice; complete two-generation solution] #### Given Initial ordered population: [1, 6, 9, 14]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 3 of children [1, 2, 3, 4] in every generation. Replace the whole population with the four children; no elitism. | Generation | Selection instruction | | --- | --- | | 1 | Current individual row numbers [3, 4, 4, 3] | | 2 | Current individual row numbers [2, 3, 2, 3] | #### To find Both generations: evaluation, selection, crossover, mutation, decoding and comparison. #### Formula and why it applies Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step. #### Solution ##### Step 1. Generation 1: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 1 | 0001 | 1 × 1 = 1 | | 2 | 6 | 0110 | 6 × 6 = 36 | | 3 | 9 | 1001 | 9 × 9 = 81 | | 4 | 14 | 1110 | 14 × 14 = 196 | ##### Step 2. Generation 1: apply the supplied selection order The question directly supplies individual numbers [3, 4, 4, 3]. These refer to the current generation’s row numbers, not their x values. | Mating-pool slot | Chosen individual | x | Chromosome | | --- | --- | --- | --- | | 1 | 3 | 9 | 1001 | | 2 | 4 | 14 | 1110 | | 3 | 4 | 14 | 1110 | | 4 | 3 | 9 | 1001 | ##### Step 3. Generation 1: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 10 | 01 | 11 | 10 | 1010 | 1101 | | 3 & 4 | 11 | 10 | 10 | 01 | 1101 | 1010 | ##### Step 4. Generation 1: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1010 | Flip bit 3 | 1000 | | 2 | 1101 | Flip bit 3 | 1111 | | 3 | 1101 | Flip bit 3 | 1111 | | 4 | 1010 | Flip bit 3 | 1000 | ##### Step 5. Generation 1: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1000 | 8 + 0 + 0 + 0 = 8 | 8² = 64 | | 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 | | 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 | | 1000 | 8 + 0 + 0 + 0 = 8 | 8² = 64 | ##### Step 6. Generation 1: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{1+36+81+196}{4}=\frac{314}{4}=78.5 $$ $$ \overline f_{new}=\frac{64+225+225+64}{4}=\frac{578}{4}=144.5 $$ Best objective: 196 → 225. Higher is better. The next population is [8, 15, 15, 8]. No elitism is imposed, so the best previous individual is not automatically protected. ##### Step 7. Generation 2: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 8 | 1000 | 8 × 8 = 64 | | 2 | 15 | 1111 | 15 × 15 = 225 | | 3 | 15 | 1111 | 15 × 15 = 225 | | 4 | 8 | 1000 | 8 × 8 = 64 | ##### Step 8. Generation 2: apply the supplied selection order The question directly supplies individual numbers [2, 3, 2, 3]. These refer to the current generation’s row numbers, not their x values. | Mating-pool slot | Chosen individual | x | Chromosome | | --- | --- | --- | --- | | 1 | 2 | 15 | 1111 | | 2 | 3 | 15 | 1111 | | 3 | 2 | 15 | 1111 | | 4 | 3 | 15 | 1111 | ##### Step 9. Generation 2: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 11 | 11 | 11 | 11 | 1111 | 1111 | | 3 & 4 | 11 | 11 | 11 | 11 | 1111 | 1111 | ##### Step 10. Generation 2: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1111 | Flip bit 3 | 1101 | | 2 | 1111 | Flip bit 3 | 1101 | | 3 | 1111 | Flip bit 3 | 1101 | | 4 | 1111 | Flip bit 3 | 1101 | ##### Step 11. Generation 2: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 | | 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 | | 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 | | 1101 | 8 + 4 + 0 + 1 = 13 | 13² = 169 | ##### Step 12. Generation 2: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{64+225+225+64}{4}=\frac{578}{4}=144.5 $$ $$ \overline f_{new}=\frac{169+169+169+169}{4}=\frac{676}{4}=169 $$ Best objective: 225 → 169. Higher is better. The next population is [13, 13, 13, 13]. No elitism is imposed, so the best previous individual is not automatically protected. **Final answer.** After two generations the ordered population is [13, 13, 13, 13]. Best final objective = 169 at x = [13]. --- ## M2 · Q3 · Two-particle PSO through iteration 2 [3] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-2--q3/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-2/q3 Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q3. Minimise $$ f(x)=x^2 $$ using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [3 marks · Practice; expanded teaching solution] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (2) | (0) | | B | (5) | (0) | w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(2)=2^2=4=4 $$ **Particle B** $$ f(5)=5^2=25=25 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(2) $$ Its score is 4. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (2). pbest values are A: (2), B: (5). ##### Step 4. Iteration 1: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=2-(2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=2-(2)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=2+0=2 $$ ##### Step 5. Iteration 1: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=5-(5)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=2-(5)=-3 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-3)=-1.5 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-1.5)=-1.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=5+(-1.5)=3.5 $$ ##### Step 6. Iteration 1: evaluate the new positions **Particle A** $$ f(2)=2^2=4=4 $$ **Particle B** $$ f(3.5)=3.5^2=12.25=12.25 $$ ##### Step 7. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 4 | 4 | Keep old: tie | (2) | | B | 25 | 12.25 | Replace: new score is lower | (3.5) | Compare the retained pbest scores. gbest = **(2)**, with objective **4**. ##### Step 8. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (2) | (0) | (2) | | B | (3.5) | (-1.5) | (3.5) | ##### Step 9. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (2). pbest values are A: (2), B: (3.5). ##### Step 10. Iteration 2: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=2-(2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=2-(2)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(2)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=2+0=2 $$ ##### Step 11. Iteration 2: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-1.5)=-0.75 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3.5-(3.5)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=2-(3.5)=-1.5 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-1.5)=-0.75 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.75)+0+(-0.75)=-1.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=3.5+(-1.5)=2 $$ ##### Step 12. Iteration 2: evaluate the new positions **Particle A** $$ f(2)=2^2=4=4 $$ **Particle B** $$ f(2)=2^2=4=4 $$ ##### Step 13. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 4 | 4 | Keep old: tie | (2) | | B | 12.25 | 4 | Replace: new score is lower | (2) | Compare the retained pbest scores. gbest = **(2)**, with objective **4**. ##### Step 14. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (2) | (0) | (2) | | B | (2) | (-1.5) | (2) | **Final answer.** After iteration 2, gbest = (2), objective = 4. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. --- ## M2 · Q4 · Fuzzy membership [3] Canonical: https://deep-learning-study-room.vercel.app/read/is2/m2/mocks/m2-paper-v2-2--q4/ Reader: https://deep-learning-study-room.vercel.app/is2/#m2/mocks/m2-paper-v2-2/q4 Q1(a)+Q1(b) compulsory; any two of Q2–Q4. Answers immediately below each question. **Confirmed M2 pattern:** Q1(a) [2] and Q1(b) [2] are compulsory. Attempt any two of Q2, Q3 and Q4, each worth 3 marks. The paper offers 13 marks; you attempt 10. These are newly written practice questions, not claims about the exact upcoming paper. Solutions are directly below each question; the teaching working is deliberately longer than a concise exam answer. **Q4. A fuzzy set A has triangular membership Tri(15, 25, 35). Find membership at x = 32.** [3 marks · Practice] #### Given Left endpoint a = 15; peak b = 25; right endpoint c = 35; input x = 32. #### To find The membership degree μA(x), a number from 0 to 1. #### Formula and why it applies Locate x relative to a, b, c before choosing the branch. $$ \mu_A(x)=\begin{cases}0,&x\le a\text{ or }x\ge c\\\dfrac{x-a}{b-a},&a0 $$ #### Solution ##### Step 1. Choose numerator and denominator Numerator=140; denominator=160. The quantity after the bar identifies the denominator. ##### Step 2. Divide $$ \frac{140}{160}=0.875 $$ **Final answer.** 0.875. **Given P(A and B)=0.12 and P(B)=0.30, find P(A | B).** [2 marks · Practice] #### Given Joint event probability=0.12; evidence probability=0.30. #### To find The conditional probability. #### Formula and why it applies Keep only the evidence group. The denominator measures that group; the numerator measures the part satisfying both events. $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)>0 $$ #### Solution ##### Step 1. Choose numerator and denominator Numerator=0.12; denominator=0.3. The quantity after the bar identifies the denominator. ##### Step 2. Divide $$ \frac{0.12}{0.3}=0.4 $$ **Final answer.** 0.4. --- ## TEE · Bayes' rule with a tree of 1000 people Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-zero--bayes-tree/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-zero/bayes-tree Start here if probability is new: counting, AND/OR, conditional probability and Bayes with pictures A probability tree branches by successive conditions. Multiply along one complete path because all its events must occur together. Add different complete paths when either path qualifies. Counts such as 0.99 out of 1,000 are **expected counts** in the illustration, not fractions of an actual person. Scaling the same model to 100,000 gives 100 diseased people, 99 positive among them, and 1,998 false positives among 99,900 others. The posterior is 99/(99+1998)=99/2097≈0.0472103. **In one line:** to reverse a “given”, imagine a crowd, split it by the cause, then by the evidence, and count. ## The question A disease affects 1 person in 1000. A test is positive for 99% of sick people, and wrongly positive for 2% of healthy people. Your test is positive. How likely is it that you are sick? Swipe across the diagram to read its labels ↔ ![Imagine 1000 people (disease 1 in 1000, test catches 99%, false alarm 2%)](https://deep-learning-study-room.vercel.app/is2/fig/prob_tree.201d46f630.svg) Diagram: Imagine 1000 people (disease 1 in 1000, test catches 99%, false alarm 2%). Labels: Imagine 1000 people (disease 1 in 1000, test catches 99%, false alarm 2%) | 1000 people | 1 has disease | 0.001 × 1000 | 999 healthy | 0.999 × 1000 | 0.99 test + | 99% of 1 | 0.01 test − | 19.98 test + | 2% of 999 (false alarm) | 979.02 test − | All positives = 0.99 + 19.98 = 20.97 people; only 0.99 of them are ill → 0.99 / 20.97 = 0.047 (4.7%) ## Counting on the tree 1. Of 1000 people, 1 is sick and 999 are healthy. 2. The sick person tests positive with 99%: 0.99 people. 3. Healthy people test positive with 2%: 0.02 × 999 = 19.98 people. 4. Everyone who tests positive: 0.99 + 19.98 = 20.97 people. 5. Of those, sick: 0.99. So P(sick | positive) = 0.99 / 20.97 = 0.047. Only about 4.7%, because the healthy group is so much bigger that its small false-alarm rate still produces most positives. ## The same thing as a formula $$ P(\text{sick}\mid+)=\frac{P(+\mid\text{sick})\,P(\text{sick})}{P(+\mid\text{sick})P(\text{sick})+P(+\mid\text{healthy})P(\text{healthy})}=\frac{0.99\times0.001}{0.99\times0.001+0.02\times0.999}=0.047 $$ Each part of the formula is one branch of the tree. Names: P(sick) is the **prior** (before the test), P(+ | sick) the **likelihood**, the denominator the **evidence** (all positives), and the answer the **posterior** (after the test). **Yaad rakho:** Bayes = tree banao: pehle cause se baanto, phir evidence se. Answer = “jo humein chahiye” ÷ “saare jo evidence dikhate hain”. Next: the formal version and more examples in [Bayes' rule](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-prob--bayes-rule/). --- ## TEE · Why agents need probability (and the basic rules) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-prob--uncertainty/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-prob/uncertainty Unit 1 · Quantifying uncertainty, inference using the full joint distribution, Bayes rule **Action A pays 10 points with probability 0.6 and −2 points with probability 0.4. Action B always pays 4 points. Which has greater expected reward?** [2 marks · Worked example] #### Given Action A: reward 10 with probability 0.6, reward −2 with probability 0.4. Action B: reward 4 with probability 1. The two probabilities for A add to 1. #### To find Expected score for each action. #### Formula and why it applies $$ \mathbb E[R]=\sum_i P_iR_i $$ Expectation is a probability-weighted average. It need not equal any one actual outcome. #### Solution ##### Step 1. A: weight each outcome $$ 0.6\times10=6\qquad0.4\times(-2)=-0.8 $$ ##### Step 2. A: add $$ \mathbb E[R_A]=6-0.8=5.2 $$ ##### Step 3. B: certain outcome $$ \mathbb E[R_B]=1\times4=4 $$ ##### Step 4. Compare 5.2>4, so A has the greater expected reward. A single attempt can still return −2. **Final answer.** A:5.2; B:4. Choose A under an expected-reward criterion. Ω names the sample space (all outcomes). ω names one outcome. The symbol E or 𝔼 means expectation here; it does not mean an observed reward. A rational probabilistic decision uses both how likely outcomes are and how desirable they are. **In one line:** when an agent cannot be certain, it should hold a degree of belief between 0 and 1 for each possibility and update it with evidence; probability theory is the tool for that. ## Acting under uncertainty An agent is uncertain because of **partial observability** (it cannot see everything) and **non-determinism** (actions do not always have the same result). Pure logic fails in such domains. Example: “Toothache ⇒ Cavity” is false (gum disease also causes toothache), and listing every possible cause is impossible. Logic fails for three reasons: 1. **Laziness**: listing every antecedent and consequent is too much work. 2. **Theoretical ignorance**: science has no complete theory of the domain. 3. **Practical ignorance**: we cannot run every test on every patient. So the agent assigns a **degree of belief** (probability), e.g. “80% of toothache patients have cavities”. **Decision theory = probability theory + utility theory**: a rational agent chooses the action with the **maximum expected utility** (e.g. a taxi choosing when to leave for the airport balances the chance of missing the flight against waiting time). Sources of uncertainty in AI: incomplete data, noise, ambiguity, stochastic environments, model and algorithm limits. ## Basic notation | Term | Meaning / rule | | --- | --- | | Sample space Ω | All possible worlds; $$ 0\le P(\omega)\le1 $$ , $$ \sum_\omega P(\omega)=1 $$ | | Prior (unconditional) | $$ P(a) $$ : belief before evidence, e.g. $$ P(cavity)=0.2 $$ | | Posterior (conditional) | $$ P(a\mid b)=\dfrac{P(a\land b)}{P(b)} $$ , $$ P(b)>0 $$ | | Product rule | $$ P(a\land b)=P(a\mid b)P(b)=P(b\mid a)P(a) $$ | | Complement | $$ P(\lnot a)=1-P(a) $$ | | Inclusion–exclusion | $$ P(a\lor b)=P(a)+P(b)-P(a\land b) $$ | | Marginalisation | $$ P(Y)=\sum_z P(Y,z) $$ | | Total probability | $$ P(b)=P(b\mid a)P(a)+P(b\mid\lnot a)P(\lnot a) $$ | | Independence | $$ P(a\land b)=P(a)P(b) $$ , equivalently $$ P(a\mid b)=P(a) $$ | | Distribution | Bold $$ \mathbf P(Weather)=\langle0.6,0.1,0.29,0.01\rangle $$ lists all values | **Yaad rakho:** Logic “haan/na” bolta hai; probability “kitna yakeen” batati hai. Conditional = joint ÷ evidence. --- ## TEE · Inference using the full joint distribution (5-mark PYQs) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-prob--full-joint/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-prob/full-joint Unit 1 · Quantifying uncertainty, inference using the full joint distribution, Bayes rule A joint-probability table is a list of complete combinations. Three Yes/No variables have 2×2×2=8 combinations. “Marginalise” means ignore a variable by adding its possible cases. “Condition” means first restrict attention to the evidence cases, then divide by their total. For unnormalised weights .12 and .08, the total is .20; α=1/.20=5, so α×.12=.6 and α×.08=.4. α is only a rescaling constant. **In one line:** a full joint table lists the probability of every combination; to answer any question, add the cells where it is true, and for “given” questions divide by the cells of the evidence. ## Method 1. **Marginal** $$ P(a) $$ : add every cell in which a is true. 2. **Conditional** $$ P(a\mid e)=\dfrac{P(a\land e)}{P(e)} $$ : add cells where both hold, divide by the sum of cells where e holds. 3. **Normalisation**: $$ \mathbf P(A\mid e)=\alpha\,\mathbf P(A,e) $$ with $$ \alpha=1/P(e) $$ : compute the unnormalised values for every value of A, then scale them to sum to 1. **Question. For the dentist table below find P(cavity ∨ toothache), P(cavity | toothache) and P(¬cavity | toothache).** [class example · 3 marks] #### Given The complete joint table or cell values printed in this question. The rows and columns identify the event combinations. #### To find All requested marginal and conditional probabilities. #### Formula and why it applies Marginal: add cells satisfying the event. Conditional: retain the evidence cells and divide the joint-event sum by the evidence sum. $$ P(A\mid E)=\frac{P(A\cap E)}{P(E)},\quad P(E)>0 $$ #### Solution | | toothache, catch | toothache, ¬catch | ¬toothache, catch | ¬toothache, ¬catch | | --- | --- | --- | --- | --- | | **cavity** | 0.108 | 0.012 | 0.072 | 0.008 | | **¬cavity** | 0.016 | 0.064 | 0.144 | 0.576 | ##### Step 1. Disjunction $$ P(cavity\lor toothache)=0.108+0.012+0.072+0.008+0.016+0.064=0.28 $$ ##### Step 2. Conditionals $$ P(toothache)=0.108+0.012+0.016+0.064=0.2,\qquad P(cavity\mid toothache)=\frac{0.12}{0.2}=0.6,\qquad P(\lnot cavity\mid toothache)=\frac{0.08}{0.2}=0.4 $$ With α: $$ \mathbf P(Cavity\mid toothache)=\alpha\langle0.12,\,0.08\rangle=\langle0.6,\,0.4\rangle $$ , $$ \alpha=1/0.2=5 $$ . **Final answer.** P(cavity OR toothache)=0.28; P(cavity | toothache)=0.6; P(no cavity | toothache)=0.4. **Question. A supermarket recorded the joint distribution of Product type (Y: premium = 1, regular = 0), Payment mode (X: credit = 1, debit = 0) and Gender (Z: female = 1, male = 0). (a) If a customer bought a premium item, find the probability that (1) they are female, (2) they paid by credit card. (b) Find (1) P(premium) and (2) P(credit card). (c) Given a male customer paid by debit card, find the probability he bought a regular item.** [5 marks · Final Nov 2025 Q1a] #### Given The printed eight entries are reproduced below. Their total is 1.002, not 1. Therefore the table is inconsistent as a probability distribution. | | Credit, female | Credit, male | Debit, female | Debit, male | | --- | --- | --- | --- | --- | | **Premium (Y=1)** | 0.08 | 0.10 | 0.072 | 0.05 | | **Regular (Y=0)** | 0.10 | 0.15 | 0.25 | 0.20 | #### To find The five requested probabilities, stating how the inconsistent data are interpreted. #### Formula and why it applies For this worked solution, treat all printed entries as nonnegative weights and normalise by their total 1.002. This preserves the printed ratios. It is a declared interpretation, not a claim about the examiner’s intended correction. $$ P(E)=\frac{\text{sum of weights for }E}{\text{sum of all weights}} $$ $$ P(A\mid E)=\frac{\text{sum of weights for }A\cap E}{\text{sum of weights for }E} $$ #### Solution ##### Step 1. Validate the full table $$ 0.08+0.10+0.072+0.05+0.10+0.15+0.25+0.20=1.002 $$ Use total weight 1.002 consistently for unconditional probabilities. ##### Step 2. Premium row total $$ w(premium)=0.08+0.10+0.072+0.05=0.302 $$ ##### Step 3. Female given premium $$ P(female\mid premium)=\frac{0.08+0.072}{0.302}=\frac{0.152}{0.302}\approx0.503311 $$ ##### Step 4. Credit given premium $$ P(credit\mid premium)=\frac{0.08+0.10}{0.302}=\frac{0.18}{0.302}\approx0.596026 $$ ##### Step 5. Normalised premium marginal $$ P(premium)=\frac{0.302}{1.002}\approx0.301397 $$ ##### Step 6. Normalised credit marginal $$ P(credit)=\frac{0.08+0.10+0.10+0.15}{1.002}=\frac{0.43}{1.002}\approx0.429142 $$ ##### Step 7. Regular given male and debit $$ P(regular\mid male,debit)=\frac{0.20}{0.05+0.20}=0.8 $$ **Final answer.** Under common normalisation: (a) 0.503311 and 0.596026; (b) 0.301397 and 0.429142; (c) 0.8. **Alternative correction:** if 0.072 is corrected to 0.070, the total becomes 1 and the corresponding answers are (a) 0.50 and 0.60; (b) 0.30 and 0.43; (c) 0.8. This agrees with the interpretation used in the supplied key. State the correction; do not call raw totals 0.302 and 0.43 valid marginals of the original inconsistent table. **Question. For a disease Y with fever X as a symptom, a test Z is conducted. Joint probabilities: Disease (Y=1): fever & test+ 0.576, fever & test− 0.144, no fever & test+ 0.072, no fever & test− 0.016; No disease (Y=0): 0.008, 0.064, 0.012, 0.108. (1) If fever occurs and the test is positive, what is the probability of disease? (2) What is the probability that a random person tests positive? (3) Given no fever, what is the probability of still having the disease?** [5 marks · Re-exam Jan 2026 Q1a] #### Given The complete joint table or cell values printed in this question. The rows and columns identify the event combinations. #### To find All requested marginal and conditional probabilities. #### Formula and why it applies Marginal: add cells satisfying the event. Conditional: retain the evidence cells and divide the joint-event sum by the evidence sum. $$ P(A\mid E)=\frac{P(A\cap E)}{P(E)},\quad P(E)>0 $$ #### Solution ##### Step 1. (1) $$ P(Y{=}1\mid X{=}1,Z{=}1)=\frac{0.576}{0.576+0.008}=\frac{0.576}{0.584}=0.986 $$ ##### Step 2. (2) Add every test-positive cell $$ P(Z{=}1)=0.576+0.072+0.008+0.012=0.668 $$ ##### Step 3. (3) $$ P(X{=}0)=0.072+0.016+0.012+0.108=0.208,\qquad P(Y{=}1\mid X{=}0)=\frac{0.072+0.016}{0.208}=\frac{0.088}{0.208}=0.423 $$ **Final answer.** (1) 0.986; (2) 0.668; (3) 0.423. **Key correction.** The key writes 0.088/0.076 = 1.158 for part (3). A probability cannot exceed 1; the denominator must be P(no fever) = 0.208. **Yaad rakho:** “Given” wali cheez denominator mein jaati hai. Pehle evidence ke saare cells jodo, phir upar wale. --- ## TEE · Bayes’ rule: meaning, derivation and three complete examples Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-prob--bayes-rule/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-prob/bayes-rule Unit 1 · Quantifying uncertainty, inference using the full joint distribution, Bayes rule ## What Bayes’ rule reverses A **prior** is the probability of a cause before the new evidence. A **likelihood** is the probability of that evidence if the cause is true. A **posterior** is the probability of the cause after observing the evidence. These are different conditional directions: P(positive | disease) is not P(disease | positive). ## Derivation $$ P(A\cap B)=P(B\mid A)P(A)=P(A\mid B)P(B) $$ Divide both sides by P(B), provided that P(B)>0. $$ P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B)} $$ If A and its complement are the only two possibilities, evidence B can arise by either path. Add both paths in the denominator. $$ P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid\neg A)P(\neg A)} $$ | Symbol | Read it as | | --- | --- | | P(A) | Prior: how common is A? | | P(B | A) | Likelihood: if A is true, how likely is B? | | P(B) | Evidence: how common is B through all possible causes? | | P(A | B) | Posterior: among B cases, what fraction come from A? | **A hypothetical population has a disease prevalence of 1 in 1000. A test is positive in 99% of diseased cases and 2% of healthy cases. Given a positive result, calculate the probability of disease.** [4 marks · Worked probability example] #### Given A = disease; B = positive test. P(A)=0.001; P(B|A)=0.99; P(B|not A)=0.02. #### To find The reversed conditional probability P(A | B). #### Formula and why it applies The question gives evidence given each cause, but asks for cause given evidence. Bayes’ rule combines the prior with that likelihood and divides by all ways of obtaining the evidence. $$ P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid\neg A)P(\neg A)} $$ #### Solution ##### Step 1. Find the other prior $$ P(\neg A)=1-0.001=0.999 $$ ##### Step 2. Evidence through the requested cause Multiply the prior for disease by the likelihood within that group. $$ P(A\cap B)=0.001\times0.99=0.00099 $$ ##### Step 3. Evidence through the other cause Now include the path through healthy. $$ P(\neg A\cap B)=0.999\times0.02=0.01998 $$ ##### Step 4. Add both paths for total evidence $$ P(B)=0.00099+0.01998=0.02097 $$ ##### Step 5. Normalise the requested path $$ P(A\mid B)=\frac{0.00099}{0.02097}\approx0.0472103 $$ **Final answer.** 0.0472103 ≈ 4.721030043%. **In a hypothetical population, 49% are female and 51% male. 8% of females and 12% of males are classified high-risk in the supplied data. Given that a person is high-risk, find the probability that the person is male.** [3 marks · Worked probability example] #### Given A = male; B = high-risk category. P(A)=0.51; P(B|A)=0.12; P(B|not A)=0.08. #### To find The reversed conditional probability P(A | B). #### Formula and why it applies The question gives evidence given each cause, but asks for cause given evidence. Bayes’ rule combines the prior with that likelihood and divides by all ways of obtaining the evidence. $$ P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid\neg A)P(\neg A)} $$ #### Solution ##### Step 1. Find the other prior $$ P(\neg A)=1-0.51=0.49 $$ ##### Step 2. Evidence through the requested cause Multiply the prior for male by the likelihood within that group. $$ P(A\cap B)=0.51\times0.12=0.0612 $$ ##### Step 3. Evidence through the other cause Now include the path through female. $$ P(\neg A\cap B)=0.49\times0.08=0.0392 $$ ##### Step 4. Add both paths for total evidence $$ P(B)=0.0612+0.0392=0.1004 $$ ##### Step 5. Normalise the requested path $$ P(A\mid B)=\frac{0.0612}{0.1004}\approx0.609561753 $$ **Final answer.** 0.609561753 ≈ 60.956175299%. **Machine A makes 60% of items and Machine B makes 40%. Defect rates are 4% for A and 1% for B. A randomly chosen item is defective. Find the probability that Machine A made it.** [3 marks · Worked probability example] #### Given Cause A = Machine A; evidence B = defective item. P(A)=0.60; P(B|A)=0.04; P(B|not A)=0.01. #### To find The reversed conditional probability P(A | B). #### Formula and why it applies The question gives evidence given each cause, but asks for cause given evidence. Bayes’ rule combines the prior with that likelihood and divides by all ways of obtaining the evidence. $$ P(A\mid B)=\frac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid\neg A)P(\neg A)} $$ #### Solution ##### Step 1. Find the other prior $$ P(\neg A)=1-0.6=0.4 $$ ##### Step 2. Evidence through the requested cause Multiply the prior for Machine A by the likelihood within that group. $$ P(A\cap B)=0.6\times0.04=0.024 $$ ##### Step 3. Evidence through the other cause Now include the path through Machine B. $$ P(\neg A\cap B)=0.4\times0.01=0.004 $$ ##### Step 4. Add both paths for total evidence $$ P(B)=0.024+0.004=0.028 $$ ##### Step 5. Normalise the requested path $$ P(A\mid B)=\frac{0.024}{0.028}\approx0.857142857 $$ **Final answer.** 0.857142857 ≈ 85.714285714%. A rare prior can make a positive result much less conclusive than its sensitivity suggests. The posterior still depends on the actual prior and both test rates; rarity alone does not determine the answer. **Yaad rakho:** Pehle requested cause ka path multiply. Neeche evidence tak pahunchne wale saare paths add. --- ## TEE · The naive Bayes model (worked on PlayTennis) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-nb--model/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-nb/model Unit 1 · Naive Bayes model · 10-mark PYQ in both papers ## What the data and the question represent A **training row** is one past case whose correct class is known. Its **features** are the information used to predict that class. In the flu exercise, Fever, Cough and Temperature are features; Flu Yes/No is the class. A **query** is the new case whose features are known but whose class must be predicted. A **prior** is the class share before using the query features. A **likelihood** asks how common one query feature is inside that class. A **posterior** is the class probability after accounting for the query. The naive assumption makes the feature likelihoods multiply once the class is fixed; it does not assert that all features are independent in the whole population. | Read this | Meaning | Concrete example | | --- | --- | --- | | Π likelihoods | Multiply every feature factor | With factors .5,.2,.4: .5×.2=.1; .1×.4=.04 | | score(C) | An unnormalised class weight | Prior .3 times .04 gives .012 | | max | The largest numerical score | max(.012,.008)=.012 | | argmax | The label attaining the largest score | The winning label is the class with score .012 | | Laplace denominator N꜀+k | Class count plus number of possible values for this feature | Temperature has three values, so k=3; a Yes/No feature has k=2 | **In one line:** assume the features are independent given the class; then score each class as prior × product of feature likelihoods and pick the largest. ## Definition **Naive Bayes** is a supervised probabilistic classifier based on Bayes' rule with the **naive assumption** that all features are conditionally independent given the class: $$ P(C\mid a_1,\dots,a_n)\;\propto\;P(C)\prod_{i=1}^{n}P(a_i\mid C),\qquad v_{NB}=\arg\max_{c}\,P(c)\prod_iP(a_i\mid c) $$ “Naive” because real features are often correlated; it still works very well for spam filtering, sentiment analysis and medical screening. ## Steps (as taught) 1. **Priors**: $$ P(c)= $$ (rows of class c) / (total rows). 2. **Likelihood tables**: for each attribute value, $$ P(a_i\mid c)= $$ (rows of class c with that value) / (rows of class c). 3. **Score** each class for the new instance and choose the larger. Optionally normalise: divide each score by their sum. **Zero-frequency problem.** If a value never occurs with a class, that factor is 0 and the whole score is 0. The main unsmoothed solutions below leave a zero count as zero. Apply smoothing when it is specified, consistently in every class. Textbooks fix it with **Laplace smoothing**: $$ P(a\mid c)=\dfrac{\text{count}+1}{N_c+k} $$ , k = number of values of the attribute. Mention it in one line; it can change the answer. **Question. Using the 14-day PlayTennis data (9 Yes, 5 No), classify ⟨Outlook = Sunny, Temperature = Cool, Humidity = High, Wind = Strong⟩.** [class example · 6 marks] #### Given Class counts: Yes=9, No=5. Within Yes, counts for Sunny/Cool/High/Strong are 2/3/3/3; within No they are 3/1/4/3. Classify Sunny, Cool, High, Strong. #### To find Both class scores and the predicted class. #### Formula and why it applies Conditional independence given the class allows the product of feature likelihoods. $$ score(C)=P(C)\prod_jP(a_j\mid C) $$ Divide each score by their sum only after computing both. #### Solution ##### Step 1. Priors $$ P(yes)=\tfrac9{14}\approx0.643,\qquad P(no)=\tfrac5{14}\approx0.357 $$ ##### Step 2. Likelihoods needed | | Sunny | Cool | High | Strong | | --- | --- | --- | --- | --- | | Yes | 2/9 | 3/9 | 3/9 | 3/9 | | No | 3/5 | 1/5 | 4/5 | 3/5 | ##### Step 3. Scores $$ v(yes)=\tfrac9{14}\cdot\tfrac29\cdot\tfrac39\cdot\tfrac39\cdot\tfrac39\approx0.005291005,\qquad v(no)=\tfrac5{14}\cdot\tfrac35\cdot\tfrac15\cdot\tfrac45\cdot\tfrac35\approx0.020571429 $$ ##### Step 4. Normalise $$ P(yes\mid x)=\frac{1/189}{1/189+18/875}\approx0.204582651,\qquad P(no\mid x)\approx0.795417349 $$ **Final answer.** PlayTennis = **No** (0.795). **Yaad rakho:** Naive Bayes = prior × (har feature ka fraction) ka guna. Jo bada, wohi class. --- ## TEE · Solved PYQ: naive Bayes for flu (Final 2025, 10 marks) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-nb--flu/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-nb/flu Unit 1 · Naive Bayes model · 10-mark PYQ in both papers **Question. A hospital wants to predict flu from Fever (Yes/No), Cough (Yes/No) and Temperature Level (Low/Normal/High). Training data: (Yes, Yes, High, Yes), (No, No, Normal, No), (Yes, No, Low, No), (Yes, Yes, High, Yes), (No, Yes, Normal, No), (Yes, Yes, High, Yes), (No, No, Low, No), (Yes, No, Normal, No). Using naive Bayes, predict whether a patient with Fever = Yes, Cough = Yes, Temperature = Normal has flu.** [10 marks · Final Nov 2025 Q4b] #### Given The full training data are in Step 1. Query: Fever=Yes, Cough=Yes, Temperature=Normal. The main calculation uses unsmoothed counts. #### To find Both class scores and the predicted class. #### Formula and why it applies $$ score(C)=P(C)\prod_jP(a_j\mid C) $$ C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods. #### Solution ##### Step 1. Write the training rows as a table | Row | Fever | Cough | Temperature | Class | | --- | --- | --- | --- | --- | | 1 | Yes | Yes | High | Yes | | 2 | No | No | Normal | No | | 3 | Yes | No | Low | No | | 4 | Yes | Yes | High | Yes | | 5 | No | Yes | Normal | No | | 6 | Yes | Yes | High | Yes | | 7 | No | No | Low | No | | 8 | Yes | No | Normal | No | ##### Step 2. Separate the two class groups | Class | Rows in this group | Class count | Prior | | --- | --- | --- | --- | | Yes | 1, 4, 6 | 3 | 3/8 = 0.375 | | No | 2, 3, 5, 7, 8 | 5 | 5/8 = 0.625 | ##### Step 3. For class Yes, count Fever=Yes Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6. $$ P(\text{Fever}=\text{Yes}\mid C=\text{Yes})=\frac{3}{3}=1 $$ ##### Step 4. For class Yes, count Cough=Yes Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6. $$ P(\text{Cough}=\text{Yes}\mid C=\text{Yes})=\frac{3}{3}=1 $$ ##### Step 5. For class Yes, count Temperature=Normal Look only at class Yes rows 1, 4, 6. No row in this class has that feature value. $$ P(\text{Temperature}=\text{Normal}\mid C=\text{Yes})=\frac{0}{3}=0 $$ ##### Step 6. Multiply the Yes score one factor at a time $$ P(Yes)=3/8=0.375 $$ $$ 0.375\times1=0.375 $$ $$ 0.375\times1=0.375 $$ $$ 0.375\times0=0 $$ ##### Step 7. For class No, count Fever=Yes Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 3, 8. $$ P(\text{Fever}=\text{Yes}\mid C=\text{No})=\frac{2}{5}=0.4 $$ ##### Step 8. For class No, count Cough=Yes Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 5. $$ P(\text{Cough}=\text{Yes}\mid C=\text{No})=\frac{1}{5}=0.2 $$ ##### Step 9. For class No, count Temperature=Normal Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 2, 5, 8. $$ P(\text{Temperature}=\text{Normal}\mid C=\text{No})=\frac{3}{5}=0.6 $$ ##### Step 10. Multiply the No score one factor at a time $$ P(No)=5/8=0.625 $$ $$ 0.625\times0.4=0.25 $$ $$ 0.25\times0.2=0.05 $$ $$ 0.05\times0.6=0.03 $$ ##### Step 11. Compare the scores $$ score(Yes)=0\qquad score(No)=0.03 $$ The larger score belongs to No. Normalise only if a posterior probability is requested: divide each score by their total 0.03. **Final answer.** Predict No. Yes score=0; No score=0.03. The zero Yes score is caused by an absent training combination. If Laplace smoothing is explicitly required, apply it to all feature likelihoods in both classes. That is a different calculation and may change the prediction; do not mix smoothed and unsmoothed sides. **Yaad rakho:** Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo. --- ## TEE · Solved PYQ: naive Bayes for a job offer (Re-exam 2026, 10 marks) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-nb--job-offer/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-nb/job-offer Unit 1 · Naive Bayes model · 10-mark PYQ in both papers **Question. Predict whether a candidate accepts a job offer from Experience (Fresher/Mid/Senior), Expected Salary (Low/Medium/High) and Company Reputation (Good/Average). Data: (Fresher, Low, Good, Yes), (Mid, High, Good, Yes), (Senior, High, Average, No), (Fresher, Medium, Average, Yes), (Senior, Low, Good, No), (Mid, Medium, Good, Yes), (Fresher, High, Average, No), (Senior, Medium, Average, No). New candidate: Mid, Low, Good.** [10 marks · Re-exam Jan 2026 Q4b] #### Given The full training data are in Step 1. Query: Experience=Mid, Salary=Low, Reputation=Good. The main calculation uses unsmoothed counts. #### To find Both class scores and the predicted class. #### Formula and why it applies $$ score(C)=P(C)\prod_jP(a_j\mid C) $$ C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods. #### Solution ##### Step 1. Write the training rows as a table | Row | Experience | Salary | Reputation | Class | | --- | --- | --- | --- | --- | | 1 | Fresher | Low | Good | Yes | | 2 | Mid | High | Good | Yes | | 3 | Senior | High | Average | No | | 4 | Fresher | Medium | Average | Yes | | 5 | Senior | Low | Good | No | | 6 | Mid | Medium | Good | Yes | | 7 | Fresher | High | Average | No | | 8 | Senior | Medium | Average | No | ##### Step 2. Separate the two class groups | Class | Rows in this group | Class count | Prior | | --- | --- | --- | --- | | Yes | 1, 2, 4, 6 | 4 | 4/8 = 0.5 | | No | 3, 5, 7, 8 | 4 | 4/8 = 0.5 | ##### Step 3. For class Yes, count Experience=Mid Look only at class Yes rows 1, 2, 4, 6. The matching rows are 2, 6. $$ P(\text{Experience}=\text{Mid}\mid C=\text{Yes})=\frac{2}{4}=0.5 $$ ##### Step 4. For class Yes, count Salary=Low Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1. $$ P(\text{Salary}=\text{Low}\mid C=\text{Yes})=\frac{1}{4}=0.25 $$ ##### Step 5. For class Yes, count Reputation=Good Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1, 2, 6. $$ P(\text{Reputation}=\text{Good}\mid C=\text{Yes})=\frac{3}{4}=0.75 $$ ##### Step 6. Multiply the Yes score one factor at a time $$ P(Yes)=4/8=0.5 $$ $$ 0.5\times0.5=0.25 $$ $$ 0.25\times0.25=0.0625 $$ $$ 0.0625\times0.75=0.046875 $$ ##### Step 7. For class No, count Experience=Mid Look only at class No rows 3, 5, 7, 8. No row in this class has that feature value. $$ P(\text{Experience}=\text{Mid}\mid C=\text{No})=\frac{0}{4}=0 $$ ##### Step 8. For class No, count Salary=Low Look only at class No rows 3, 5, 7, 8. The matching rows are 5. $$ P(\text{Salary}=\text{Low}\mid C=\text{No})=\frac{1}{4}=0.25 $$ ##### Step 9. For class No, count Reputation=Good Look only at class No rows 3, 5, 7, 8. The matching rows are 5. $$ P(\text{Reputation}=\text{Good}\mid C=\text{No})=\frac{1}{4}=0.25 $$ ##### Step 10. Multiply the No score one factor at a time $$ P(No)=4/8=0.5 $$ $$ 0.5\times0=0 $$ $$ 0\times0.25=0 $$ $$ 0\times0.25=0 $$ ##### Step 11. Compare the scores $$ score(Yes)=0.046875\qquad score(No)=0 $$ The larger score belongs to Yes. Normalise only if a posterior probability is requested: divide each score by their total 0.046875. **Final answer.** Predict Yes. Yes score=0.046875; No score=0. **Official key note.** The key gives No = 0.0068 and P(Yes) = 0.873. That No value comes from Laplace smoothing (1/7 × 2/7 × 2/6 × 0.5), but the key did not smooth the Yes side. If you smooth, smooth both: Yes = 0.5 × 3/7 × 2/7 × 4/6 = 0.0408, No = 0.0068, P(Yes) = 0.857. The prediction is Yes either way. **Yaad rakho:** Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo. --- ## TEE · Bayesian networks: structure, CPTs and the chain rule Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-bn--semantics/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-bn/semantics Unit 1 · Semantics of Bayesian networks and exact inference · 10-mark PYQ in both papers ## A Bayesian network starts as a diagram of variables A **node** is a variable, such as Rain. An **edge** is an arrow representing a direct dependency in the model. A **parent** has an arrow into a node. A **root** has no parents. **Directed acyclic graph (DAG)** means arrows have directions and you cannot follow them around a loop back to the starting node. The graph alone does not establish real-world causation. A **CPT** is a conditional probability table: for each parent configuration, it states the probabilities of the child’s values. A **Boolean** variable has two values, true and false. If a CPT gives P(Wet=true|Rain=true)=.9, its complementary entry is P(Wet=false|Rain=true)=1−.9=.1. **In Rain → Wet, P(Rain)=0.2, P(Wet|Rain)=0.9 and P(Wet|no Rain)=0.1. Find P(Rain AND Wet), P(Wet), and P(Rain|Wet).** [2 marks · Worked example] #### Given P(Rain)=0.2; P(Wet|Rain)=0.9; P(Wet|no Rain)=0.1. Each variable has two possible values: true or false. Rain is the root; Wet depends on Rain. #### To find A joint probability, a marginal probability and a posterior. #### Formula and why it applies Multiply down one complete case, add the alternative ways to get Wet, then divide by the evidence probability. #### Solution ##### Step 1. Rain and wet $$ P(R,W)=0.2\times0.9=0.18 $$ ##### Step 2. No rain and wet $$ P(\neg R,W)=(1-0.2)\times0.1=0.8\times0.1=0.08 $$ ##### Step 3. Total wet $$ P(W)=0.18+0.08=0.26 $$ ##### Step 4. Restrict to the wet cases $$ P(R\mid W)=0.18/0.26=9/13\approx0.692307692 $$ **Final answer.** Joint .18; marginal .26; posterior≈.692308. **In one line:** a Bayesian network is a DAG of random variables in which each node stores P(node | its parents); multiplying those tables gives the full joint distribution. ## Definition A **Bayesian network** (belief network) is a directed acyclic graph in which: 1. each **node** is a random variable; 2. a directed **edge** X → Y means the model conditions Y on X (X is a parent of Y); a causal interpretation needs additional assumptions; there are no directed cycles; 3. each node has a **conditional probability table (CPT)** $$ \mathbf P(X_i\mid Parents(X_i)) $$ . A root node stores its prior. Swipe across the diagram to read its labels ↔ ![Burglary network (Russell and Norvig)](https://deep-learning-study-room.vercel.app/is2/fig/bn_burglary.9802d6d53d.svg) Diagram: Burglary network (Russell and Norvig). Labels: Burglary network (Russell and Norvig) | Burglary | P(B) = 0.001 | Earthquake | P(E) = 0.002 | Alarm | P(A|B,E): .95 · .94 · .29 · .001 | (TT · TF · FT · FF) | JohnCalls | P(J|A) = .90, P(J|¬A) = .05 | MaryCalls | P(M|A) = .70, P(M|¬A) = .01 ## Semantics: the chain rule for Bayesian networks $$ P(x_1,\dots,x_n)=\prod_{i=1}^{n}P\big(x_i\mid parents(X_i)\big) $$ So the network *is* a compact description of the full joint distribution. For the burglary network: $$ P(J,M,A,B,E)=P(J\mid A)\,P(M\mid A)\,P(A\mid B,E)\,P(B)\,P(E) $$ **Conditional independence:** each node is conditionally independent of its non-descendants given its parents (e.g. JohnCalls is independent of Burglary once Alarm is known). **Compactness:** a Boolean node with k Boolean parents needs $$ 2^k $$ numbers. The burglary network needs 1 + 1 + 4 + 2 + 2 = 10 numbers instead of $$ 2^5-1=31 $$ for the full joint. ## Building a network 1. Choose the variables. 2. Choose an acyclic variable order; when a causal model is justified, cause-before-effect is often useful. 3. For each variable, choose as parents a suitable set of earlier variables that captures its conditional dependencies. 4. Write its CPT. **Yaad rakho:** Har node sirf apne parents se baat karta hai. Joint = sab nodes ka P(node | parents) ka guna. --- ## TEE · Inference in Bayesian networks: enumeration, α and variable elimination Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-bn--inference/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-bn/inference Unit 1 · Semantics of Bayesian networks and exact inference · 10-mark PYQ in both papers ## Expand the hidden-variable calculation before the compact formula For burglary given both calls, B is the query, J=true and M=true are evidence, and E and A are hidden. “Hidden” only means their values are not supplied in the query. A sum over a Boolean variable adds its true and false cases. | Calculation | Burglary true | Burglary false | | --- | --- | --- | | Average alarm probability over E | .95×.002 + .94×.998 = .94002 | .29×.002 + .001×.998 = .001578 | | No-alarm probability | 1−.94002=.05998 | 1−.001578=.998422 | | Both calls if alarm | .9×.7=.63 | .9×.7=.63 | | Both calls if no alarm | .05×.01=.0005 | .05×.01=.0005 | | Calls and this B value | .001×(.63×.94002 + .0005×.05998)=.00059224259 | .999×(.63×.001578 + .0005×.998422)=.001491857649 | Add the final two weights: .00059224259+.001491857649=.002084100239. Divide the burglary weight by that total: .00059224259/.002084100239≈.284171835. No step equates P(B|calls) with P(calls|B); the prior P(B) is included explicitly. **In one line:** to answer P(query | evidence), sum the chain-rule product over every hidden variable, do it for each value of the query, and normalise with α. ## Three kinds of question | Question type | Method | | --- | --- | | Every variable's value given | Multiply the CPT entries (chain rule) | | One marginal, e.g. P(J) | Sum over hidden variables; no α | | Query given evidence, e.g. P(B | j, m) | Enumerate for each value of B, then $$ \alpha=1/\text{(sum)} $$ | **Question. In the burglary network find (a) P(j, m, a, ¬b, ¬e), (b) P(JohnCalls), (c) P(Burglary | John and Mary call).** [class examples · 10 marks] For the burglary example: B and E are independent roots with P(B)=0.001 and P(E)=0.002. Alarm A has parents B,E. John J and Mary M each have parent A. P(A|B,E)=0.95, P(A|B,¬E)=0.94, P(A|¬B,E)=0.29, P(A|¬B,¬E)=0.001. P(J|A)=0.9, P(J|¬A)=0.05; P(M|A)=0.7, P(M|¬A)=0.01. False probabilities are the complements of the listed true probabilities. #### Given The network, root probabilities and conditional probability tables stated in the question. #### To find The requested joint, marginal or evidence-conditioned probability. #### Formula and why it applies A Bayesian network factorises the joint using each node conditioned on its parents. Sum over hidden values; normalise for evidence. $$ P(x_1,\ldots,x_n)=\prod_iP(x_i\mid parents_i) $$ #### Solution ##### (a) Chain rule $$ P=P(j\mid a)P(m\mid a)P(a\mid\lnot b,\lnot e)P(\lnot b)P(\lnot e)=0.90\times0.70\times0.001\times0.999\times0.998=0.000628 $$ ##### (b) Marginal: first P(a) $$ P(a)=0.95(0.001)(0.002)+0.94(0.001)(0.998)+0.29(0.999)(0.002)+0.001(0.999)(0.998)=0.002516 $$ $$ P(j)=P(j\mid a)P(a)+P(j\mid\lnot a)P(\lnot a)=0.90(0.002516)+0.05(0.997484)=0.0521 $$ ##### (c) Enumeration with α (hidden: E, A) $$ P(b\mid j,m)=\alpha\,P(b)\sum_aP(j\mid a)P(m\mid a)\sum_eP(a\mid b,e)P(e) $$ $$ =\alpha\,(0.001)\big[0.9(0.7)(0.95\times0.002+0.94\times0.998)+0.05(0.01)(0.05\times0.002+0.06\times0.998)\big]=\alpha\times0.000592 $$ $$ P(\lnot b\mid j,m)=\alpha\,(0.999)\big[0.9(0.7)(0.29\times0.002+0.001\times0.998)+0.05(0.01)(0.71\times0.002+0.999\times0.998)\big]=\alpha\times0.001492 $$ $$ \alpha=\frac1{0.000592+0.001492}=479.8\;\Rightarrow\;P(b\mid j,m)=0.284,\;P(\lnot b\mid j,m)=0.716 $$ **Final answer.** (a) 0.000628; (b) 0.0521; (c) 0.284. Even when both neighbours call, a burglary is only about 28% likely because burglaries are rare. **Slide note.** One slide prints .71 for $$ P(\lnot a\mid b,\lnot e) $$ ; it is 1 − 0.94 = 0.06 (corrected by hand on the slide). ## Variable elimination (why it is faster) Enumeration repeats the same products many times (exponential, $$ O(2^n) $$ ). **Variable elimination** pushes each sum as far inside as possible, computes it once and stores the result as a **factor**; hidden variables are “summed out” one at a time: $$ P(J)=\sum_AP(J\mid A)\sum_MP(M\mid A)\sum_BP(B)\underbrace{\sum_EP(A\mid B,E)P(E)}_{f_1(A,B)} $$ Any sum of a CPT over its own variable is 1 (e.g. $$ \sum_MP(M\mid A)=1 $$ ), so irrelevant (barren) nodes drop out. **Question. Heart-disease network: Exercise → HD ← Diet; HD → BP. P(E) = 0.7, P(D = healthy) = 0.25; P(HD | E, D) = 0.25 (E, healthy), 0.45 (E, unhealthy), 0.55 (¬E, healthy), 0.75 (¬E, unhealthy); P(BP high | HD) = 0.85, P(BP high | ¬HD) = 0.2. A patient has high BP. Find P(HD | BP).** [class example · 6 marks] #### Given The network, root probabilities and conditional probability tables stated in the question. #### To find The requested joint, marginal or evidence-conditioned probability. #### Formula and why it applies A Bayesian network factorises the joint using each node conditioned on its parents. Sum over hidden values; normalise for evidence. $$ P(x_1,\ldots,x_n)=\prod_iP(x_i\mid parents_i) $$ #### Solution ##### Step 1. P(HD) by summing out E and D $$ P(hd)=0.7(0.25\times0.25+0.45\times0.75)+0.3(0.55\times0.25+0.75\times0.75)=0.7(0.4)+0.3(0.7)=0.49 $$ ##### Step 2. Unnormalised $$ \alpha\,P(bp\mid hd)P(hd)=\alpha(0.85)(0.49)=\alpha\,0.4165,\qquad\alpha\,P(bp\mid\lnot hd)P(\lnot hd)=\alpha(0.2)(0.51)=\alpha\,0.102 $$ ##### Step 3. Normalise $$ \alpha=\frac1{0.5185},\quad P(hd\mid bp)=0.803,\quad P(\lnot hd\mid bp)=0.197 $$ **Final answer.** 0.803. **Yaad rakho:** Hidden variables par sum, query ki dono values nikaalo, α se 1 banao. Sirf marginal (P(J)) ho to α nahi. --- ## TEE · Solved PYQ: wet grass given no rain (Final 2025, 10 marks) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-bn--sprinkler/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-bn/sprinkler Unit 1 · Semantics of Bayesian networks and exact inference · 10-mark PYQ in both papers The following table enumerates the values of Cloudy and Sprinkler that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome. **Question. Cloudy → Sprinkler, Cloudy → Rain, Sprinkler and Rain → Wet Grass. P(Cloudy) = 0.5; P(Sprinkler | C) = 0.10, P(Sprinkler | ¬C) = 0.50; P(Rain | C) = 0.80, P(Rain | ¬C) = 0.20; P(Wet | S, R) = 0.99, P(Wet | S, ¬R) = 0.90, P(Wet | ¬S, R) = 0.90, P(Wet | ¬S, ¬R) = 0.01. Calculate the probability of the grass being wet given it doesn't rain.** [10 marks · Final Nov 2025 Q3a] #### Given C→S, C→R, (S,R)→W. Root and conditional probabilities are listed in the question. Evidence R=false. #### To find P(W=true | R=false). Hidden variables C and S must be summed out. #### Formula and why it applies Use the network joint and divide by the evidence probability. $$ P(W\mid\neg R)=\frac{\sum_{c,s}P(c)P(s\mid c)P(\neg R\mid c)P(W\mid s,\neg R)}{\sum_cP(c)P(\neg R\mid c)} $$ #### Solution Swipe across the diagram to read its labels ↔ ![Final 2025 Q3a: sprinkler network](https://deep-learning-study-room.vercel.app/is2/fig/bn_sprinkler.a251583272.svg) Diagram: Final 2025 Q3a: sprinkler network. Labels: Final 2025 Q3a: sprinkler network | Cloudy | P(C) = 0.5 | Sprinkler | P(S|C) = 0.1, P(S|¬C) = 0.5 | Rain | P(R|C) = 0.8, P(R|¬C) = 0.2 | Wet grass | P(W|S,R): .99 · .90 · .90 · .01 | (TT · TF · FT · FF) ##### Step 1. Joint by the chain rule $$ P(C,S,R,W)=P(C)\,P(S\mid C)\,P(R\mid C)\,P(W\mid S,R) $$ ##### Step 2. Sum over C and S with R = false, W = true | C | S | P(C) | P(S|C) | P(¬R|C) | P(W|S,¬R) | Product | | --- | --- | --- | --- | --- | --- | --- | | T | T | 0.5 | 0.1 | 0.2 | 0.90 | 0.0090 | | T | F | 0.5 | 0.9 | 0.2 | 0.01 | 0.0009 | | F | T | 0.5 | 0.5 | 0.8 | 0.90 | 0.1800 | | F | F | 0.5 | 0.5 | 0.8 | 0.01 | 0.0020 | | **P(W, ¬R)** | | | | | | **0.1919** | ##### Step 3. Evidence $$ P(\lnot R)=P(C)P(\lnot R\mid C)+P(\lnot C)P(\lnot R\mid\lnot C)=0.5(0.2)+0.5(0.8)=0.5 $$ ##### Step 4. Divide $$ P(W\mid\lnot R)=\frac{0.1919}{0.5}=0.3838 $$ **Final answer.** P(grass wet | no rain) ≈ **0.384**. Without rain, the grass is wet mainly when the sprinkler runs (more likely on non-cloudy days). **Keep the requested conditional probability separate.** The supplied key discusses P(Wet) and P(Rain | Wet), whereas this question asks P(Wet | no rain). The answer to the printed question is 0.3838. For comparison, the exact marginal from the printed CPT is calculated below; it is not 0.6505. #### Optional comparison: calculate P(Wet) from the same network The four no-rain cases already total 0.1919. Now add the four rain cases, keeping the same Cloudy/Sprinkler order. $$ P(W,R)=0.5(0.1)(0.8)(0.99)+0.5(0.9)(0.8)(0.90)+0.5(0.5)(0.2)(0.99)+0.5(0.5)(0.2)(0.90) $$ $$ P(W,R)=0.0396+0.324+0.0495+0.045=0.4581 $$ $$ P(W)=P(W,R)+P(W,\neg R)=0.4581+0.1919=0.6500 $$ $$ P(R\mid W)=\frac{0.4581}{0.6500}\approx0.704769231 $$ **Meaning:** the unconditional chance of Wet is 0.65, the chance of Rain among wet cases is about 0.704769, and the requested chance of Wet among no-rain cases is 0.3838. These are three different questions. **Trap.** Do not plug in the marginal P(Sprinkler) = 0.3; sprinkler and rain both depend on Cloudy, so you must sum over Cloudy (given no rain, P(S) = 0.42). --- ## TEE · Solved PYQ: fever given no flu (Re-exam 2026, 10 marks) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-bn--flu-network/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-bn/flu-network Unit 1 · Semantics of Bayesian networks and exact inference · 10-mark PYQ in both papers The following table enumerates the values of Weather and Cough that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome. **Question. Weather → Flu, Weather → Cough, Flu and Cough → Fever. P(Weather) = 0.75; P(Flu | W) = 0.2, P(Flu | ¬W) = 0.5; P(Cough | W) = 0.70, P(Cough | ¬W) = 0.25; P(Fever | Flu, Cough) = 0.60, P(Fever | Flu, ¬Cough) = 0.02, P(Fever | ¬Flu, Cough) = 0.60, P(Fever | ¬Flu, ¬Cough) = 0.01. Calculate the probability of a person having fever given he doesn't have flu.** [10 marks · Re-exam Jan 2026 Q3a] #### Given W→F, W→C, (F,C)→V. The complete probabilities are in the question. Evidence F=false. #### To find P(V=true | F=false), summing hidden W and C. #### Formula and why it applies Use the network joint and normalise by the no-Flu evidence. $$ P(V\mid\neg F)=\frac{\sum_{w,c}P(w)P(\neg F\mid w)P(c\mid w)P(V\mid\neg F,c)}{\sum_wP(w)P(\neg F\mid w)} $$ #### Solution Swipe across the diagram to read its labels ↔ ![Re-exam 2026 Q3a: weather–flu–cough–fever network](https://deep-learning-study-room.vercel.app/is2/fig/bn_flu.1d732f0fb6.svg) Diagram: Re-exam 2026 Q3a: weather–flu–cough–fever network. Labels: Re-exam 2026 Q3a: weather–flu–cough–fever network | Weather | P(W) = 0.75 | Flu | P(F|W) = 0.2, P(F|¬W) = 0.5 | Cough | P(C|W) = 0.7, P(C|¬W) = 0.25 | Fever | P(V|F,C): .60 · .02 · .60 · .01 | (TT · TF · FT · FF) ##### Step 1. Chain rule $$ P(W,F,C,V)=P(W)\,P(F\mid W)\,P(C\mid W)\,P(V\mid F,C) $$ ##### Step 2. Sum over W and C with F = false, V = true | W | C | P(W) | P(¬F|W) | P(C|W) | P(V|¬F,C) | Product | | --- | --- | --- | --- | --- | --- | --- | | T | T | 0.75 | 0.8 | 0.70 | 0.60 | 0.252 | | T | F | 0.75 | 0.8 | 0.30 | 0.01 | 0.0018 | | F | T | 0.25 | 0.5 | 0.25 | 0.60 | 0.01875 | | F | F | 0.25 | 0.5 | 0.75 | 0.01 | 0.0009375 | | **P(V, ¬F)** | | | | | | **0.2734875** | ##### Step 3. Evidence $$ P(\lnot F)=0.75(0.8)+0.25(0.5)=0.6+0.125=0.725 $$ ##### Step 4. Divide $$ P(V\mid\lnot F)=\frac{0.2734875}{0.725}=0.3772 $$ **Final answer.** P(fever | no flu) ≈ **0.377** (fever without flu comes mostly from cough). **Key correction.** The key uses P(Flu) = 0.05 and P(Fever | ¬Flu) = 0.2, which are not in this network, and answers 0.2. Marking scheme (still applies): write the equation with hidden variables (2), take true/false cases (3), substitute (2), answer (3). --- ## TEE · Before GMM: centre, spread, density and likelihood Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-gmm--centre-spread/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-gmm/centre-spread Unit 1 · Mixture models, GMM, expectation maximization, clustering using GMM · 10-mark PYQ in both papers ## Mean, variance and standard deviation A Gaussian component describes a cluster using its **mean** (centre) and **variance** (spread). The **standard deviation** is the square root of variance. The Greek letters μ and σ name these two quantities; σ² is the variance. **Calculate the mean, population variance and standard deviation of [1, 3, 5].** [2 marks · Worked example] #### Given Data values: [1, 3, 5]. Number of observations n=3. Use population variance, dividing by n. #### To find The mean μ, population variance σ², and standard deviation σ. #### Formula and why it applies $$ \mu=\frac{\sum_i x_i}{n}\quad\sigma^2=\frac{\sum_i(x_i-\mu)^2}{n}\quad\sigma=\sqrt{\sigma^2} $$ This descriptive/maximum-likelihood convention divides by n. The n−1 sample-variance estimator is a different convention. #### Solution ##### Step 1. Add and divide for the mean $$ \mu=\frac{1+3+5}{3}=3 $$ ##### Step 2. Subtract the mean, then square | x | x−mean | Squared deviation | | --- | --- | --- | | 1 | -2 | 4 | | 3 | 0 | 0 | | 5 | 2 | 4 | ##### Step 3. Average the squared deviations $$ \sigma^2=\frac{4+0+4}{3}=2.666666667 $$ ##### Step 4. Take the square root $$ \sigma=\sqrt{2.666666667}\approx1.632993162 $$ **Final answer.** Mean=3; variance=2.666666667; standard deviation≈1.632993162. **Calculate the mean, population variance and standard deviation of [2, 4].** [2 marks · Worked example] #### Given Data values: [2, 4]. Number of observations n=2. Use population variance, dividing by n. #### To find The mean μ, population variance σ², and standard deviation σ. #### Formula and why it applies $$ \mu=\frac{\sum_i x_i}{n}\quad\sigma^2=\frac{\sum_i(x_i-\mu)^2}{n}\quad\sigma=\sqrt{\sigma^2} $$ This descriptive/maximum-likelihood convention divides by n. The n−1 sample-variance estimator is a different convention. #### Solution ##### Step 1. Add and divide for the mean $$ \mu=\frac{2+4}{2}=3 $$ ##### Step 2. Subtract the mean, then square | x | x−mean | Squared deviation | | --- | --- | --- | | 2 | -1 | 1 | | 4 | 1 | 1 | ##### Step 3. Average the squared deviations $$ \sigma^2=\frac{1+1}{2}=1 $$ ##### Step 4. Take the square root $$ \sigma=\sqrt{1}\approx1 $$ **Final answer.** Mean=3; variance=1; standard deviation≈1. **Calculate the mean, population variance and standard deviation of [4, 4, 4].** [2 marks · Worked example] #### Given Data values: [4, 4, 4]. Number of observations n=3. Use population variance, dividing by n. #### To find The mean μ, population variance σ², and standard deviation σ. #### Formula and why it applies $$ \mu=\frac{\sum_i x_i}{n}\quad\sigma^2=\frac{\sum_i(x_i-\mu)^2}{n}\quad\sigma=\sqrt{\sigma^2} $$ This descriptive/maximum-likelihood convention divides by n. The n−1 sample-variance estimator is a different convention. #### Solution ##### Step 1. Add and divide for the mean $$ \mu=\frac{4+4+4}{3}=4 $$ ##### Step 2. Subtract the mean, then square | x | x−mean | Squared deviation | | --- | --- | --- | | 4 | 0 | 0 | | 4 | 0 | 0 | | 4 | 0 | 0 | ##### Step 3. Average the squared deviations $$ \sigma^2=\frac{0+0+0}{3}=0 $$ ##### Step 4. Take the square root $$ \sigma=\sqrt{0}\approx0 $$ **Final answer.** Mean=4; variance=0; standard deviation≈0. The constant data [4,4,4] have variance zero. A standard Gaussian density formula requires positive variance; do not divide by zero when fitting a component. ## Density, likelihood and logarithm have different jobs A **density** is the height of a continuous distribution. Probability over an interval is area, so density height can exceed 1 without violating probability rules. A **likelihood** treats the observed data as fixed and scores candidate model parameters. For independent observations, their densities multiply. **A model assigns density 0.2 to one observation and 0.5 to a second independent observation. Find likelihood and natural log-likelihood.** [2 marks · Worked example] #### Given Two density values .2 and .5. #### To find Product and its natural logarithm. #### Formula and why it applies $$ L=p(x_1)p(x_2)\qquad\ln L=\ln p(x_1)+\ln p(x_2) $$ The logarithm turns a product into a sum. #### Solution ##### Step 1. Multiply the densities $$ L=0.2\times0.5=0.1 $$ ##### Step 2. Take ln of that product $$ \ln L=\ln(0.1)\approx-2.302585093 $$ ##### Step 3. Show the equivalent sum $$ \ln(0.2)+\ln(0.5)\approx-1.609437912-0.693147181=-2.302585093 $$ **Final answer.** Likelihood .1; log-likelihood≈−2.302585093. Do not turn a sum inside a log into a sum of logs: ln(.2+.5)=ln(.7)≈−.356675, whereas ln(.2)+ln(.5)≈−2.302585. A GMM has an inner sum over components, which must stay inside each observation’s logarithm. **Yaad rakho:** Mean = centre. Variance = squared distance ka average. Likelihood = fixed data ko model kitna support karta hai. --- ## TEE · GMM prerequisites: mean, variance, density and mixture Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-gmm--mixture-models/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-gmm/mixture-models Unit 1 · Mixture models, GMM, expectation maximization, clustering using GMM · 10-mark PYQ in both papers | Read this | Meaning | Concrete example | | --- | --- | --- | | Component k | One Gaussian group in the mixture | Component 1 may have mean 2; component 2 mean 4 | | Mixing weight πₖ | The prior share assigned to component k | π₁=.4 and π₂=.6 sum to 1 | | N(x|μ,σ²) | The Gaussian density evaluated at x | The vertical bar supplies parameters; it is not division | | Σ over k | Add contributions of every component | p(x)=.4N₁(x)+.6N₂(x) | ## Mean, spread and a Gaussian component A **mean** is the centre, found by adding observations and dividing by their count. A **variance** measures squared spread around the mean. In one dimension, a Gaussian component is described by mean μ and positive variance σ². Its standard deviation σ is the square root of that variance. A **density** is a height, not the probability of one exact real-valued observation. Areas under a density give probabilities. Density heights can exceed 1; their total area must be 1. $$ \mathcal N(x\mid\mu,\sigma^2)=\frac1{\sqrt{2\pi}\,\sigma}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right) $$ Here π inside 2π is the constant 3.14159…; the separately written πₖ in a mixture denotes a component weight. exp(a)=eᵃ, and the calculator eˣ key evaluates it. The minus sign in the exponent reduces density away from the mean. **For a standard Gaussian with μ=0 and σ²=1, find the density at x=0.** [3 marks · Practice] #### Given x=0; μ=0; σ=1. #### To find Density N(x|0,1). #### Formula and why it applies Use the Gaussian density with the supplied mean and variance. #### Solution ##### Step 1. Squared distance $$ (0-0)^2=0 $$ ##### Step 2. Exponent $$ -0/(2\times1)=0 $$ ##### Step 3. Exponential $$ e^{0}\approx1 $$ ##### Step 4. Divide by the normalising factor $$ \frac{1}{\sqrt{2\pi}}\approx0.39894228 $$ **Final answer.** 0.39894228 (a density, not an exact-point probability). **For a standard Gaussian with μ=0 and σ²=1, find the density at x=1.** [3 marks · Practice] #### Given x=1; μ=0; σ=1. #### To find Density N(x|0,1). #### Formula and why it applies Use the Gaussian density with the supplied mean and variance. #### Solution ##### Step 1. Squared distance $$ (1-0)^2=1 $$ ##### Step 2. Exponent $$ -1/(2\times1)=-0.5 $$ ##### Step 3. Exponential $$ e^{-0.5}\approx0.60653066 $$ ##### Step 4. Divide by the normalising factor $$ \frac{0.60653066}{\sqrt{2\pi}}\approx0.241970725 $$ **Final answer.** 0.241970725 (a density, not an exact-point probability). **For a standard Gaussian with μ=0 and σ²=1, find the density at x=2.** [3 marks · Practice] #### Given x=2; μ=0; σ=1. #### To find Density N(x|0,1). #### Formula and why it applies Use the Gaussian density with the supplied mean and variance. #### Solution ##### Step 1. Squared distance $$ (2-0)^2=4 $$ ##### Step 2. Exponent $$ -4/(2\times1)=-2 $$ ##### Step 3. Exponential $$ e^{-2}\approx0.135335283 $$ ##### Step 4. Divide by the normalising factor $$ \frac{0.135335283}{\sqrt{2\pi}}\approx0.053990967 $$ **Final answer.** 0.053990967 (a density, not an exact-point probability). ## Mixture model and hidden component A **mixture model** combines component densities. A Gaussian mixture model (GMM) uses K Gaussian components. A hidden component label says which component generated a point; it is not observed in ordinary clustering. A weight πₖ is the prior probability of choosing component k. All weights are nonnegative and sum to 1. $$ p(x)=\sum_{k=1}^K\pi_k\mathcal N(x\mid\mu_k,\sigma_k^2) $$ For two components this means π₁×density₁ + π₂×density₂. A GMM can represent several concentrations of data rather than forcing all points around one mean. Swipe across the diagram to read its labels ↔ ![The total density is a weighted sum of component densities.](https://deep-learning-study-room.vercel.app/is2/fig/gmm_mix.84e959698a.svg) Diagram: The total density is a weighted sum of component densities.. Labels: A 1-D Gaussian mixture: three weighted bells add up to p(x) | -8 | -4 | 0 | 4 | 8 | p | x The total density is a weighted sum of component densities. **At x, two component densities are 0.2 and 0.05, with weights 0.6 and 0.4. Find the mixture density.** [2 marks · Practice] #### Given π₁=.6, π₂=.4; densities=.2,.05. #### To find p(x). #### Formula and why it applies Weight each component density, then add. $$ p(x)=\pi_1N_1+\pi_2N_2 $$ #### Solution ##### Step 1. First contribution $$ 0.6(0.2)=0.12 $$ ##### Step 2. Second contribution $$ 0.4(0.05)=0.02 $$ ##### Step 3. Add $$ p(x)=0.12+0.02=0.14 $$ **Final answer.** Mixture density = 0.14. **Yaad rakho:** Har Gaussian ek group ka model; mixture = weights ke saath sab densities ka sum. --- ## TEE · EM for GMM: every responsibility and parameter update Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-gmm--em/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-gmm/em Unit 1 · Mixture models, GMM, expectation maximization, clustering using GMM · 10-mark PYQ in both papers **At one observation, the two weighted component densities are q₁=0.12 and q₂=0.03. Find the responsibilities. Then use responsibilities [0.8,0.2] for data [1,5] to find a component’s new mean.** [2 marks · Worked example] #### Given Part A: weighted component densities q₁=0.12 and q₂=0.03 for one observation. Part B: data x₁=1 and x₂=5, with component responsibilities 0.8 and 0.2 respectively. Part A compares components for one point; Part B combines points for one component. #### To find Two responsibilities and a weighted mean. #### Formula and why it applies $$ \gamma_{ik}=q_{ik}/\sum_jq_{ij}\qquad\mu_k^{new}=\frac{\sum_i\gamma_{ik}x_i}{\sum_i\gamma_{ik}} $$ i identifies an observation; k identifies a component. Notice which index is being added in each formula. #### Solution ##### Step 1. Normalise one observation $$ q_1+q_2=0.12+0.03=0.15 $$ $$ \gamma_1=0.12/0.15=0.8\quad\gamma_2=0.03/0.15=0.2 $$ ##### Step 2. Find the effective component count $$ N_k=0.8+0.2=1 $$ ##### Step 3. Weight each observation $$ 0.8\times1=0.8\qquad0.2\times5=1 $$ ##### Step 4. Add and divide $$ \mu_k^{new}=(0.8+1)/1=1.8 $$ **Final answer.** Responsibilities [.8,.2]; new mean 1.8. A point contributes a fraction of a count instead of being assigned wholly to one component. ## What EM is estimating **Expectation–maximisation (EM)** alternates two calculations when component assignments are hidden. The E-step estimates how responsible each component is for each observation. The M-step uses those responsibilities as weights to update the model parameters. A **responsibility** γᵢₖ is a posterior component probability: for observation i, how likely is component k under the current model? For each observation, the responsibilities sum to 1. Unlike fuzzy memberships, these probabilities are explicitly normalised across mixture components. | Symbol | Meaning | | --- | --- | | i / k | Observation number / component number | | γᵢₖ | Responsibility of component k for observation i | | Nₖ | Sum of responsibilities for component k: an effective count | | μₖ | Component mean | | σ²ₖ | Component variance | | πₖ | Component mixing weight | $$ \gamma_{ik}=\frac{\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)}{\sum_j\pi_j\mathcal N(x_i\mid\mu_j,\sigma_j^2)} $$ $$ N_k=\sum_i\gamma_{ik},\quad\mu_k^{new}=\frac{\sum_i\gamma_{ik}x_i}{N_k} $$ $$ (\sigma_k^2)^{new}=\frac{\sum_i\gamma_{ik}(x_i-\mu_k^{new})^2}{N_k},\quad\pi_k^{new}=N_k/n $$ The E-step uses all old parameters. The M-step mean uses those fixed responsibilities; the variance uses the newly calculated mean. One exact EM update cannot lower the likelihood when its optimisation steps are performed correctly; numerical approximations and regularisation need their own interpretation. EM can still reach a local solution. **For observations {1,3,5}, initialise a two-component GMM with means 2 and 4, both variances 1, and both weights 0.5. Perform one complete EM iteration.** [6 marks · Practice] #### Given n=3, K=2, μ₁=2, μ₂=4; σ₁²=σ₂²=1; π₁=π₂=.5. #### To find Six responsibilities and the new weights, means and variances. #### Formula and why it applies Apply the E-step and M-step equations above. Calculate with full precision; decimals below are displayed to nine places. #### Solution ##### Step 1. E-step: weighted densities for x=1 $$ q_1=0.5\frac{e^{-(1-2)^2/2}}{\sqrt{2\pi}}\approx0.120985362 $$ $$ q_2=0.5\frac{e^{-(1-4)^2/2}}{\sqrt{2\pi}}\approx0.002215924 $$ ##### Step 2. E-step: weighted densities for x=3 $$ q_1=0.5\frac{e^{-(3-2)^2/2}}{\sqrt{2\pi}}\approx0.120985362 $$ $$ q_2=0.5\frac{e^{-(3-4)^2/2}}{\sqrt{2\pi}}\approx0.120985362 $$ ##### Step 3. E-step: weighted densities for x=5 $$ q_1=0.5\frac{e^{-(5-2)^2/2}}{\sqrt{2\pi}}\approx0.002215924 $$ $$ q_2=0.5\frac{e^{-(5-4)^2/2}}{\sqrt{2\pi}}\approx0.120985362 $$ ##### Step 4. E-step: normalise responsibilities for x=1 $$ \gamma_1=\frac{0.120985362}{0.120985362+0.002215924}\approx0.98201379 $$ $$ \gamma_2=1-\gamma_1\approx0.01798621 $$ ##### Step 5. E-step: normalise responsibilities for x=3 $$ \gamma_1=\frac{0.120985362}{0.120985362+0.120985362}\approx0.5 $$ $$ \gamma_2=1-\gamma_1\approx0.5 $$ ##### Step 6. E-step: normalise responsibilities for x=5 $$ \gamma_1=\frac{0.002215924}{0.002215924+0.120985362}\approx0.01798621 $$ $$ \gamma_2=1-\gamma_1\approx0.98201379 $$ ##### Step 7. M-step: effective count for component 1 $$ N_1=0.98201379+0.5+0.01798621=1.5 $$ ##### Step 8. M-step: mean for component 1 $$ \mu_1^{new}=\frac{0.98201379(1)+0.5(3)+0.01798621(5)}{1.5}\approx1.714629893 $$ ##### Step 9. M-step: variance for component 1 | x | Responsibility | Squared deviation from new mean | Weighted squared deviation | | --- | --- | --- | --- | | 1 | 0.98201379 | 0.510695884 | 0.501510401 | | 3 | 0.5 | 1.652176311 | 0.826088156 | | 5 | 0.01798621 | 10.793656738 | 0.194136976 | $$ (\sigma_1^2)^{new}=\frac{1.521735533}{1.5}\approx1.014490355 $$ ##### Step 10. M-step: mixture weight for component 1 $$ \pi_1^{new}=1.5/3=0.5 $$ ##### Step 11. M-step: effective count for component 2 $$ N_2=0.01798621+0.5+0.98201379=1.5 $$ ##### Step 12. M-step: mean for component 2 $$ \mu_2^{new}=\frac{0.01798621(1)+0.5(3)+0.98201379(5)}{1.5}\approx4.285370107 $$ ##### Step 13. M-step: variance for component 2 | x | Responsibility | Squared deviation from new mean | Weighted squared deviation | | --- | --- | --- | --- | | 1 | 0.01798621 | 10.793656738 | 0.194136976 | | 3 | 0.5 | 1.652176311 | 0.826088156 | | 5 | 0.98201379 | 0.510695884 | 0.501510401 | $$ (\sigma_2^2)^{new}=\frac{1.521735533}{1.5}\approx1.014490355 $$ ##### Step 14. M-step: mixture weight for component 2 $$ \pi_2^{new}=1.5/3=0.5 $$ **Final answer.** Means ≈ [1.714629893, 4.285370107]; variances ≈ [1.014490355, 1.014490355]; weights [0.5,0.5]. **Yaad rakho:** E: har point ki responsibility. M: responsibilities se weighted mean, variance, weight. --- ## TEE · 10-mark answer: GMM vs k-means, log-likelihood derivation, parameters for K = 5 Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-gmm--ten-mark-answer/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-gmm/ten-mark-answer Unit 1 · Mixture models, GMM, expectation maximization, clustering using GMM · 10-mark PYQ in both papers ## Read the derivation’s symbols first “i.i.d.” means the observations are modelled as independent draws from the same distribution. The product Π runs over observations; the inner sum Σ runs over components. λki in some notes is the same responsibility that other sections call γik. A superscript i here labels the observation, not an exponent. Jensen’s inequality gives a lower bound on the log of the component sum. EM chooses responsibilities making that bound touch the current likelihood, then updates the parameters to maximise the bound with those responsibilities fixed. This explains the two alternating steps; λ values are not arbitrary new model parameters. | Parameter | Number in a 1D K-component GMM | Why | | --- | --- | --- | | Means | K | One centre for each component | | Variances | K | One spread for each component | | Independent mixing weights | K−1 | The last weight is 1 minus the others | | Total | 3K−1 | K+K+(K−1) | For K=2, choose μ₁,μ₂,σ₁²,σ₂² and π₁: five free numbers. Then π₂=1−π₁. For K=5, 5+5+4=14. In 2D with a full covariance matrix, each component instead has two means and three independent covariance entries (two variances and one shared off-diagonal covariance). Five such components have 5×(2+3)+4=29 free parameters. **Asked twice as:** “With reference to GMM: compare GMM with the k-means algorithm; derive the equation for the log-likelihood function; what will be the number of parameters to be computed if K = 5?” [10 marks · Final 2025 Q2b and Re-exam 2026 Q2b; key: up to the maximum-likelihood equation 5, estimation 5] ### 1. GMM versus k-means | Aspect | k-means | GMM | | --- | --- | --- | | Assignment | Hard: each point in exactly one cluster | Soft: probability $$ \lambda_k^i $$ of belonging to each cluster | | Model | Geometric: nearest centroid | Probabilistic generative model $$ \sum_k\pi_k\mathcal N(\mu_k,\Sigma_k) $$ | | Cluster shape | Spherical, similar size | Elliptical, any size/orientation; overlapping clusters | | Parameters | K centroids | Weights, means, covariances (1-D: 3K − 1) | | Algorithm / objective | Assign–update; minimise squared distances | EM; maximise log-likelihood | | Output | Hard labels and fitted centroids | Labels, membership probabilities and a density (useful for anomaly detection) | | Relation | The hard-assignment, equal spherical variance, small-variance limit of a Gaussian-mixture objective leads to the k-means objective. Ordinary finite-variance EM is not the same algorithm as k-means. Both need K in advance and depend on initialisation. | | ### 2. Derivation of the log-likelihood **Step 1. Model.** Each point is generated by choosing component k with probability $$ \pi_k $$ and drawing $$ x_i\sim\mathcal N(\mu_k,\sigma_k^2) $$ , so its density is $$ f_{mix}(x_i)=\sum_k\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2) $$ . **Step 2. Likelihood.** Points are i.i.d., so the likelihood is a product (θ = all parameters): $$ L(\theta)=\prod_{i=1}^{n}\Big[\sum_{k=1}^{K}\pi_k\frac{1}{\sqrt{2\pi}\,\sigma_k}e^{-\frac{(x_i-\mu_k)^2}{2\sigma_k^2}}\Big] $$ **Step 3. Log-likelihood.** Logs turn the product into a sum: $$ \log L(\theta)=\sum_{i=1}^{n}\log\Big(\sum_{k=1}^{K}\pi_k\frac{1}{\sqrt{2\pi}\,\sigma_k}e^{-\frac{(x_i-\mu_k)^2}{2\sigma_k^2}}\Big) $$ **Step 4. Why it cannot be solved directly.** The sum sits inside the log, so setting derivatives to zero gives no closed-form solution. **Step 5. Jensen's inequality.** log is concave, so for weights $$ \lambda_k\ge0,\;\sum\lambda_k=1 $$ : $$ \log\big(\sum_k\lambda_ka_k\big)\ge\sum_k\lambda_k\log a_k $$ . Introduce $$ \lambda_k^i $$ for every point ( $$ \sum_k\lambda_k^i=1 $$ ), multiply and divide inside the log: $$ \log L(\theta)=\sum_i\log\Big(\sum_k\lambda_k^i\,\frac{\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)}{\lambda_k^i}\Big)\;\ge\;\sum_i\sum_k\lambda_k^i\log\frac{\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)}{\lambda_k^i} $$ **Step 6. Estimation (EM).** Maximise the lower bound alternately. Fixing θ gives the E-step $$ \lambda_k^i=\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)/\sum_l\pi_l\mathcal N(x_i\mid\mu_l,\sigma_l^2) $$ . Fixing λ and differentiating (with a Lagrange multiplier for $$ \sum\pi_k=1 $$ ) gives the M-step: $$ \hat\mu_k=\frac{\sum_i\lambda_k^ix_i}{\sum_i\lambda_k^i},\qquad\hat\sigma_k^2=\frac{\sum_i\lambda_k^i(x_i-\hat\mu_k)^2}{\sum_i\lambda_k^i},\qquad\hat\pi_k=\frac{\sum_i\lambda_k^i}{n} $$ ### 3. Number of parameters for K = 5 The dimension and covariance structure must also be specified. For the one-dimensional, independent-variance model used in the derivation: $$ 2K+(K-1)=3K-1=3(5)-1=\mathbf{14} $$ (5 means + 5 variances + 4 free weights, since the 5 weights must add to 1.) In two dimensions with full covariance it would be 29. **Slide note.** One slide puts the convex inequality under the “concave” heading. Correct: log is *concave*, so $$ f(\sum\lambda a)\ge\sum\lambda f(a) $$ (≥, not >). --- ## TEE · Clustering with GMM: uses, advantages and limits Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-gmm--clustering-applications/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u1-gmm/clustering-applications Unit 1 · Mixture models, GMM, expectation maximization, clustering using GMM · 10-mark PYQ in both papers ## Clustering using GMM 1. Choose K (domain knowledge, or a criterion such as BIC). 2. Fit the GMM with EM. 3. For each point compute $$ \lambda_k^i $$ ; the largest gives its cluster, and the values show how confident the assignment is. 4. Points with very low $$ p(x) $$ under every component are anomalies. **Example (class).** Marks 45, 48, 50, 52, 80, 82, 85, 88, 90 form two groups, “average” near 50 and “high performers” near 85. A two-component GMM gives each student a probability of belonging to each group; students in between (none here) would get split probabilities instead of a forced label. ## Applications - Customer segmentation in marketing - Image segmentation (pixel colours) and background subtraction in video - Speaker identification and speech recognition - Anomaly and fraud detection (low-density points) - Density estimation and generating new samples ## Advantages and limitations | Advantages | Limitations | | --- | --- | | Soft assignments with probabilities | K must be chosen in advance | | Flexible elliptical, overlapping clusters | Sensitive to initialisation; can stop at a local optimum | | Interpretable parameters (weights, means, covariances) | Assumes each cluster is Gaussian | | Gives a full density model | Costly for many dimensions (covariance matrices) | **Probability vs likelihood** (lab question): probability fixes the parameters and asks how likely the data is; likelihood fixes the observed data and asks how plausible each parameter value is. Maximum-likelihood estimation chooses the parameters that make the observed data most probable. ### Why an unrestricted GMM can collapse A Gaussian centred on one training observation can make its variance approach zero, causing its density there and the mixture likelihood to grow without a finite bound. This is a singularity, not evidence of a useful perfect cluster. Practical fitting therefore uses a positive variance/covariance floor or a prior, sensible initialisation, and checks for nearly empty components. Different starts can reach different local solutions. These qualifications do not change the stated one-iteration classroom calculation. Primary reference: [Bishop, Pattern Recognition and Machine Learning, chapter 9](https://www.microsoft.com/en-us/research/wp-content/uploads/2006/01/Bishop-Pattern-Recognition-and-Machine-Learning-2006.pdf). --- ## TEE · Fuzzy sets: what membership actually means Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--membership-meaning/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/membership-meaning Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Start with an input and a label Suppose the input is **temperature 32°C** and the fuzzy label is **Hot**. A membership function is the rule that converts that temperature into a degree of matching the label. If μHot(32)=0.7, read it aloud as “the membership of 32 degrees in Hot is 0.7”. The lower word Hot names the set; 32 is the input; 0.7 is the output. The same temperature can have membership 0.3 in Medium and 0.7 in Hot. These are two descriptions of one input. Fuzzy memberships do not generally have to sum to 1. By contrast, probabilities of mutually exclusive, exhaustive outcomes do. | Read this | Meaning | Concrete example | | --- | --- | --- | | Universe U | All input values being considered | Temperatures from 0 to 40°C | | Crisp value | One definite measured number | 32°C | | Linguistic term | A word naming a fuzzy category | Low, Medium, High | | μA(x) | Degree to which input x fits category A | μHot(32)=0.7 | **Yaad rakho:** 32 temperature hai; 0.7 us temperature ka Hot label mein degree hai. Dono ko mix mat karo. ## Why “true or false” is sometimes too rigid Suppose a controller says a room is hot only when its temperature is at least 30°C. At 29.9°C it says “not hot”; at 30°C it says “hot”. That sharp boundary is a **crisp** rule. Fuzzy logic allows a gradual degree of belonging to the category “Hot”. **Definition.** A fuzzy set A on a universe U assigns every element x a membership degree μA(x) between 0 and 1. The symbol μ is the Greek letter “mu”; read μA(x) as “the degree to which x belongs to set A”. $$ \mu_A:U\rightarrow[0,1] $$ | Term | Meaning | Temperature example | | --- | --- | --- | | Universe U | All allowed input values | Temperatures from 0°C to 40°C | | Element x | The particular input being considered | 32°C | | Fuzzy set A | A category described by a membership function | Hot | | Membership μA(x) | How strongly x fits that category | μHot(32) = 0.7 | ## Read membership correctly Membership 0 means no belonging; 1 means full belonging; 0.7 means partial belonging under the chosen definition. **It does not mean a 70% chance that the measured temperature is 32°C.** Nor is it a 70% chance that the sensor is correct. Fuzziness concerns the gradual meaning of “Hot”; probability concerns uncertainty about which event occurs. | Input | Warm membership | Hot membership | Interpretation | | --- | --- | --- | --- | | 28 | 0.8 | 0.1 | Fits Warm strongly and Hot weakly | | 32 | 0.3 | 0.7 | Fits Hot more than Warm | | 35 | 0 | 1 | Fully Hot in this chosen model | These memberships need not add to 1. Overlapping categories are allowed. A crisp set is the special case in which membership can only be 0 or 1. A crisp set can also contain infinitely many elements; “crisp” does not mean “finite”. **On U = {20, 25, 30, 35}, the Hot memberships are [0, 0.2, 0.6, 1]. Write the fuzzy set Hot and identify the fully and partly belonging inputs.** [2 marks · Practice] #### Given Ordered inputs [20, 25, 30, 35]; memberships [0, 0.2, 0.6, 1]. #### To find The set notation and interpretation. #### Formula and why it applies In the common fuzzy-set notation μ/x, the slash separates membership from element; it is not division. #### Solution ##### Step 1. Pair each input with its membership | x | μHot(x) | | --- | --- | | 20 | 0 | | 25 | 0.2 | | 30 | 0.6 | | 35 | 1 | ##### Step 2. Write the set $$ Hot=\{0/20,\;0.2/25,\;0.6/30,\;1/35\} $$ ##### Step 3. Interpret 35 fully belongs; 25 and 30 partly belong; 20 does not belong. **Final answer.** Hot = {0/20, 0.2/25, 0.6/30, 1/35}. **Distinguish crisp and fuzzy sets.** [2 marks · Practice] #### Answer | Crisp set | Fuzzy set | | --- | --- | | Membership is 0 or 1 | Membership is any value in [0,1] | | A sharp boundary separates members | A membership function permits gradual boundaries | | Example: temperature ≥ 30°C | Example: degree of being Hot increases gradually | **Yaad rakho:** μ ka matlab: is category mein kitna fit hota hai. Chance nahi. --- ## TEE · Triangular membership: choose the side, then calculate Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--triangle/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/triangle Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Why the triangle has two different fractions Tri(2,5,8) means membership starts at zero at x=2, rises to 1 at x=5, then falls to zero at x=8. The first three numbers are **input positions**, not three membership degrees. | Where is x? | Meaning | Calculation | | --- | --- | --- | | x≤2 or x≥8 | Outside, or at a zero-height endpoint | 0 | | 20; core: degree=1; boundary: 0 0. **Core:** inputs with μ = 1. **Boundary:** inputs with 0 < μ < 1. **Height:** largest membership (more generally the supremum). A set is normal when its height is 1; it is subnormal when its height is below 1. A crossover point has membership 0.5. An α-cut contains inputs with μ ≥ α. For Tri(2,5,8), support is (2,8), core is {5}, height is 1, and the 0.5-cut is [3.5,6.5]. Solve (x−2)/3=0.5 → x=3.5 and (8−x)/3=0.5 → x=6.5. A one-dimensional fuzzy set is **convex** if every α-cut is an interval (or empty). The membership between two inputs must not dip below the smaller endpoint membership. A single triangle is convex; two separated peaks with a lower valley may not be. **Yaad rakho:** Fuzzy A aur NOT A dono partial ho sakte hain. Isliye crisp waali 1 aur 0 identities blindly nahi lagani. --- ## TEE · Mamdani architecture: what every block does Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--mamdani-architecture/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/mamdani-architecture Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Follow one measurement through the controller A **controller** turns measurements into an action setting. Consider temperature 32°C and occupancy 7 people. The input numbers first become membership degrees. The rules then combine those degrees. Only the final defuzzification returns one fan-speed number. | Stage | What goes in | Operation | What comes out | | --- | --- | --- | --- | | Fuzzifier | 32°C and 7 people | Use the supplied membership graphs | Temperature Medium=.3, High=.7; occupancy High=.5 | | Knowledge base | Chosen labels and their shapes | Store graphs and IF–THEN rules | A reusable definition of how the controller works | | Inference | Medium=.3 AND High=.5 | Rule strength=min(.3,.5)=.3 | “Fan High” fires at .3 | | Aggregation | All clipped output sets | Take the largest output degree at each speed | One combined fuzzy output shape | | Defuzzifier | Combined shape | Apply the specified MoM, centroid or representative method | One numerical speed | The **antecedent** is the IF part (“temperature is Medium AND occupancy is High”). The **consequent** is the THEN part (“fan speed is High”). A rule strength of 0.3 is a membership height; it does not mean the fan runs at 30%. ## Definition: a fuzzy inference system A **fuzzy inference system (FIS)** maps inputs to an output by evaluating fuzzy IF–THEN rules. A **Mamdani FIS** uses fuzzy sets in the rule consequents, combines their inferred output sets, and defuzzifies the result. Crisp inputs32°C; 7 people → Fuzzificationdegrees of Warm/Hot etc. → Rules + inferencerule strengths and clipped outputs → Aggregationone combined output shape → Defuzzificationone fan-speed number ## Architecture: what each block contributes | Block | What it stores or calculates | Concrete meaning | | --- | --- | --- | | Fuzzifier | Membership values of the measured inputs | 32°C may be Medium to degree 0.3 and High to degree 0.7 | | Knowledge base: database | Ranges and membership functions | Medium = Tri(15,25,35) | | Knowledge base: rule base | IF–THEN rules | IF temperature High AND occupancy High THEN fan Very High | | Inference engine | Combines conditions and applies consequents | min for AND; clips the output triangle at the rule strength | | Aggregator | Combines all rule outputs | At every output location, take the maximum clipped membership | | Defuzzifier | Converts the aggregate to one crisp value | Use centroid or mean of maxima, as requested | ## Parts of an IF–THEN rule In “IF temperature is High AND occupancy is High THEN fan speed is Very High”, the IF part is the **antecedent** (condition), and the THEN part is the **consequent** (output category). The input variable temperature, the label High, and the numerical membership μHigh(32) are different things. ## Why a rule strength is not yet the fan speed A strength such as 0.3 says how strongly a rule fires. It is on the membership scale 0–1. Fan speed might be on a 0–100% scale. Defuzzification is the step that produces a number on that output scale. **Explain the architecture of a Mamdani fuzzy inference system.** [3 marks · Practice] #### Answer A Mamdani FIS contains a fuzzifier, knowledge base, inference engine and defuzzifier. The database specifies membership functions and the rule base stores IF–THEN rules. Crisp inputs are fuzzified. The inference engine calculates rule firing strengths and constructs output fuzzy sets. These sets are aggregated, then defuzzified into a crisp action. For a fan controller, temperature and occupancy are inputs and fan speed is the output. **Yaad rakho:** Input ka degree → rule ka strength → output ka shape → final number. --- ## TEE · Rule firing, clipping and aggregation: four solved cases Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--rules-clipping/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/rules-clipping Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Clipping means limiting a height A rule firing at strength 0.3 cannot give its consequent a membership above 0.3. At every candidate speed z, keep the smaller of 0.3 and the original output-set membership. The speed axis is unchanged. **At speeds 40, 60 and 80, output set A has memberships [0.1,0.8,0.2] and set B has [0,0.4,1]. Rule A fires at 0.3 and rule B at 0.5. Clip and max-aggregate the sets.** [2 marks · Worked example] #### Given | Speed | Original A membership | Original B membership | | --- | --- | --- | | 40 | 0.1 | 0 | | 60 | 0.8 | 0.4 | | 80 | 0.2 | 1 | Rule A strength=0.3; rule B strength=0.5. Compare memberships at the same speed. #### To find The combined membership at each speed. #### Formula and why it applies $$ \mu_{out}(z)=\max(\min(0.3,\mu_A(z)),\min(0.5,\mu_B(z))) $$ Inner min limits a rule’s height; outer max keeps the stronger support at that speed. #### Solution ##### Step 1. Clip A At 40: min(.3,.1)=.1. At 60: min(.3,.8)=.3. At 80: min(.3,.2)=.2. ##### Step 2. Clip B At 40: min(.5,0)=0. At 60: min(.5,.4)=.4. At 80: min(.5,1)=.5. ##### Step 3. Aggregate at each speed | Speed | Clipped A | Clipped B | max(A,B) | | --- | --- | --- | --- | | 40 | 0.1 | 0 | 0.1 | | 60 | 0.3 | 0.4 | 0.4 | | 80 | 0.2 | 0.5 | 0.5 | **Final answer.** Combined degrees [0.1,0.4,0.5] at speeds [40,60,80]. A separate defuzzification step is still required. **Yaad rakho:** Clip = height ki ceiling. Aggregate = har same x par sabse upar wala height. ## Four different calculations happen in inference **AND/OR** combine conditions within a rule. **Implication** uses the resulting strength to modify that rule’s output membership function. **Aggregation** merges different rule outputs. Here we use min-AND, max-OR, min-implication and max-aggregation. $$ \alpha_{AND}=\min(a,b)\qquad\alpha_{OR}=\max(a,b) $$ $$ \mu_{\mathrm{rule}}(z)=\min(\alpha,\mu_{\mathrm{consequent}}(z)) $$ $$ \mu_{\mathrm{aggregate}}(z)=\max_r\mu_{\mathrm{rule}\ r}(z) $$ α (“alpha”) is the rule firing strength. z names a possible output value. The subscript r labels a rule, not a multiplication. **A rule has two condition degrees 0.7 and 0.5, joined by AND. Find its firing strength.** [2 marks · Practice] #### Given a=0.7; b=0.5; connective=AND. #### To find The strength α. #### Formula and why it applies Use the smaller degree because both conditions must hold. #### Solution ##### Step 1. Select the operator $$ \alpha=\min(0.7,0.5) $$ ##### Step 2. Evaluate $$ \alpha=0.5 $$ **Final answer.** 0.5. **A rule has two condition degrees 0.3 and 0, joined by AND. Find its firing strength.** [2 marks · Practice] #### Given a=0.3; b=0; connective=AND. #### To find The strength α. #### Formula and why it applies Use the smaller degree because both conditions must hold. #### Solution ##### Step 1. Select the operator $$ \alpha=\min(0.3,0) $$ ##### Step 2. Evaluate $$ \alpha=0 $$ **Final answer.** 0. **A rule has two condition degrees 0.2 and 0.8, joined by OR. Find its firing strength.** [2 marks · Practice] #### Given a=0.2; b=0.8; connective=OR. #### To find The strength α. #### Formula and why it applies Use the larger degree because either condition is sufficient. #### Solution ##### Step 1. Select the operator $$ \alpha=\max(0.2,0.8) $$ ##### Step 2. Evaluate $$ \alpha=0.8 $$ **Final answer.** 0.8. **Rule 1 fires at 0.3 and its original output membership at z=60 is 0.8. Rule 2 fires at 0.6 and its original membership at z=60 is 0.2. Find the aggregate membership at z=60 using min-implication and max-aggregation.** [3 marks · Practice] #### Given Rule 1: α₁=0.3, μ₁(60)=0.8. Rule 2: α₂=0.6, μ₂(60)=0.2. #### To find Membership at the single output location z=60 after implication and aggregation. #### Formula and why it applies Clip each original degree at its rule strength; then take the larger clipped degree. #### Solution ##### Step 1. Clip rule 1 $$ \mu_{R1}(60)=\min(0.3,0.8)=0.3 $$ ##### Step 2. Clip rule 2 $$ \mu_{R2}(60)=\min(0.6,0.2)=0.2 $$ ##### Step 3. Aggregate at this output location $$ \mu_{out}(60)=\max(0.3,0.2)=0.3 $$ **Final answer.** The aggregate membership at z=60 is 0.3. This is not a crisp fan-speed answer. Swipe across the diagram to read its labels ↔ ![Above the 0.3 level, the original triangle is cut off. Below 0.3, the original slope remains.](https://deep-learning-study-room.vercel.app/is2/fig/v2_clip.f101b5cfa9.svg) Diagram: Above the 0.3 level, the original triangle is cut off. Below 0.3, the original slope remains.. Labels: Min-implication clips the triangle at strength 0.3 | 0 | 0.3 | 1 | 50 | 54.5 | 65 | 75.5 | 80 | μ | Output z | Original High | clip 0.3 Above the 0.3 level, the original triangle is cut off. Below 0.3, the original slope remains. **Yaad rakho:** AND ke liye min; output clip karne ke liye bhi min; rules ko jodne ke liye max. Har min ka role alag hai. --- ## TEE · Mean of maxima: discrete peaks and clipped plateaus Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--mom/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/mom Definitions → membership graphs → operations → rules → defuzzification → complete controller. **Maximum membership** is a height. **Mean of maxima** averages the horizontal locations attaining that height. If heights at speeds [20,50,80] are [.2,.6,.6], the maximum height is .6 and the qualifying speeds are 50 and 80; MoM=(50+80)/2=65. Do not average [.6,.6], which would return a degree rather than a speed. A continuous flat maximum from 50 to 80 also has midpoint 65. ## Mean of maxima (MoM) **Defuzzification** converts the aggregate membership function into one crisp output. For MoM, first find the largest membership height, then average the output locations attaining that height. Do not average the membership values themselves. $$ z^*_{MoM}=\frac{z_1+\cdots+z_m}{m} $$ For a continuous flat maximum from a to b, the mean is its midpoint: $$ z^*_{MoM}=\frac{a+b}{2} $$ **For the following discrete aggregate, find the mean-of-maxima output.** [2 marks · Practice] #### Given | Output z | Membership μ | | --- | --- | | 10 | 0.2 | | 20 | 0.8 | | 30 | 0.8 | | 40 | 0.3 | #### To find MoM crisp output. #### Formula and why it applies Find all locations at the largest height, then average only those locations. #### Solution ##### Step 1. Find the maximum height $$ \mu_{max}=0.8 $$ ##### Step 2. List the locations attaining it 20, 30 ##### Step 3. Average these locations $$ z^*=\frac{20+30}{2}=25 $$ **Final answer.** 25. **For the following discrete aggregate, find the mean-of-maxima output.** [2 marks · Practice] #### Given | Output z | Membership μ | | --- | --- | | 1 | 0.1 | | 2 | 0.4 | | 3 | 0.9 | | 4 | 0.2 | #### To find MoM crisp output. #### Formula and why it applies Find all locations at the largest height, then average only those locations. #### Solution ##### Step 1. Find the maximum height $$ \mu_{max}=0.9 $$ ##### Step 2. List the locations attaining it 3 ##### Step 3. Average these locations $$ z^*=\frac{3}{1}=3 $$ **Final answer.** 3. **A single triangular output Tri(50,65,80) is clipped at α=0.3. Find its crisp value by mean of maxima.** [3 marks · Practice] #### Given a=50, b=65, c=80; clip height α=0.3. #### To find Both plateau endpoints and their mean. #### Formula and why it applies Set the rising and falling membership fractions equal to 0.3. Every point between those intersections has the maximum height. #### Solution ##### Step 1. Find the left endpoint $$ \frac{z_L-50}{65-50}=0.3 $$ $$ z_L-50=0.3\times15=4.5 $$ $$ z_L=54.5 $$ ##### Step 2. Find the right endpoint $$ \frac{80-z_R}{80-65}=0.3 $$ $$ 80-z_R=4.5 $$ $$ z_R=75.5 $$ ##### Step 3. Average endpoints $$ z^*=(54.5+75.5)/2=65 $$ **Final answer.** MoM = 65. Swipe across the diagram to read its labels ↔ ![The maximum extends over [54.5,75.5], whose midpoint is 65.](https://deep-learning-study-room.vercel.app/is2/fig/v2_clip.f101b5cfa9.svg) Diagram: The maximum extends over [54.5,75.5], whose midpoint is 65.. Labels: Min-implication clips the triangle at strength 0.3 | 0 | 0.3 | 1 | 50 | 54.5 | 65 | 75.5 | 80 | μ | Output z | Original High | clip 0.3 The maximum extends over [54.5,75.5], whose midpoint is 65. For several disjoint continuous maximum plateaus, a continuous MoM uses a length-weighted mean over all maximum intervals; simply averaging two outside endpoints is not generally valid. In tabulated questions, average the listed locations at the maximum. **Yaad rakho:** Sabse upar ka level dekho; us level par x-values ka mean lo. --- ## TEE · Centroid: three table sums and one area calculation Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--centroid/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/centroid Definitions → membership graphs → operations → rules → defuzzification → complete controller. ## Centroid means the balance point of the whole shape Imagine the area under the output graph as a thin sheet of uniform material. The centroid is its horizontal balance point. More area on one side pulls the answer toward that side. In a discrete problem, each membership is a weight. In a continuous problem, the entire area supplies the weights. The integral sign ∫ means adding very thin strips across a continuous interval. The denominator ∫μ(z)dz is the total area. The numerator ∫zμ(z)dz is the **first moment**: position multiplied by area, accumulated across the shape. Their ratio has units of the output, such as fan-speed percent. **A non-overlapping aggregate consists of a rectangle of width 4 and height 0.5 centred at z=2, plus a triangle of base 4 and height 0.5 centred horizontally at z=6. Find the area centroid.** [2 marks · Worked example] #### Given Rectangle: width 4, height 0.5, horizontal centre 2. Triangle: base 4, height 0.5, horizontal centroid 6. Their areas do not overlap. #### To find The combined horizontal centroid. #### Formula and why it applies $$ z^*=\frac{A_1c_1+A_2c_2}{A_1+A_2} $$ Each piece contributes area times its own centre. Use this only for a valid decomposition of the aggregate. #### Solution ##### Step 1. Rectangle area $$ A_1=4\times0.5=2 $$ ##### Step 2. Triangle area $$ A_2=\tfrac12\times4\times0.5=1 $$ ##### Step 3. First moments $$ A_1c_1=2\times2=4 $$ $$ A_2c_2=1\times6=6 $$ ##### Step 4. Totals $$ A=2+1=3\qquad M=4+6=10 $$ ##### Step 5. Divide $$ z^*=10/3\approx3.333333 $$ **Final answer.** Centroid≈3.333333, closer to the larger-area piece. For overlapping clipped sets, first form their max-aggregate. Adding the areas of the original overlapping sets would count the overlap twice. ## Centroid: the balance point of the output The centroid uses the whole aggregate membership distribution. For a discrete output table, multiply each output location by its membership, add the products, then divide by total membership. $$ z^*=\frac{\sum_i z_i\mu(z_i)}{\sum_i\mu(z_i)} $$ The index i simply numbers table entries. Σ (“sigma”) means add all listed terms. The denominator must be nonzero. **Find the discrete centroid of the aggregate membership table.** [3 marks · Practice] #### Given | z | μ(z) | | --- | --- | | 10 | 0.2 | | 20 | 0.6 | | 30 | 0.4 | #### To find One centroid value on the output scale. #### Formula and why it applies This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation. $$ z^*=\frac{\sum z_i\mu_i}{\sum\mu_i} $$ #### Solution ##### Step 1. Multiply every location by its membership | z | μ | z × μ | | --- | --- | --- | | 10 | 0.2 | 2 | | 20 | 0.6 | 12 | | 30 | 0.4 | 12 | ##### Step 2. Add the numerator $$ 2+12+12=26 $$ ##### Step 3. Add the denominator $$ 0.2+0.6+0.4=1.2 $$ ##### Step 4. Divide $$ z^*=\frac{26}{1.2}=21.666666667 $$ **Final answer.** 21.666666667. **Find the discrete centroid of the aggregate membership table.** [3 marks · Practice] #### Given | z | μ(z) | | --- | --- | | 0 | 0.5 | | 5 | 1 | | 10 | 0.5 | #### To find One centroid value on the output scale. #### Formula and why it applies This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation. $$ z^*=\frac{\sum z_i\mu_i}{\sum\mu_i} $$ #### Solution ##### Step 1. Multiply every location by its membership | z | μ | z × μ | | --- | --- | --- | | 0 | 0.5 | 0 | | 5 | 1 | 5 | | 10 | 0.5 | 5 | ##### Step 2. Add the numerator $$ 0+5+5=10 $$ ##### Step 3. Add the denominator $$ 0.5+1+0.5=2 $$ ##### Step 4. Divide $$ z^*=\frac{10}{2}=5 $$ **Final answer.** 5. **Find the discrete centroid of the aggregate membership table.** [3 marks · Practice] #### Given | z | μ(z) | | --- | --- | | 20 | 0 | | 40 | 0.3 | | 60 | 0.6 | | 80 | 0.1 | #### To find One centroid value on the output scale. #### Formula and why it applies This question supplies memberships at discrete output locations; use the discrete centroid, not a peak-only approximation. $$ z^*=\frac{\sum z_i\mu_i}{\sum\mu_i} $$ #### Solution ##### Step 1. Multiply every location by its membership | z | μ | z × μ | | --- | --- | --- | | 20 | 0 | 0 | | 40 | 0.3 | 12 | | 60 | 0.6 | 36 | | 80 | 0.1 | 8 | ##### Step 2. Add the numerator $$ 0+12+36+8=56 $$ ##### Step 3. Add the denominator $$ 0+0.3+0.6+0.1=1 $$ ##### Step 4. Divide $$ z^*=\frac{56}{1}=56 $$ **Final answer.** 56. ## For a continuous shape, integrate the area $$ z^*=\frac{\int z\mu(z)\,dz}{\int\mu(z)\,dz} $$ The denominator is the area under the aggregate. The numerator is its first moment about zero: area weighted by horizontal location. For a shape made of non-overlapping rectangles and triangles, compute each part’s area A and its own centroid c, then use $$ z^*=(\sum A_jc_j)/(\sum A_j) $$ . Overlapping regions of a max-aggregate must not be counted twice. **An aggregate is a rectangle of height 0.5 from z=0 to z=4, followed by a right triangle decreasing from height 0.5 at z=4 to 0 at z=6. Find its area centroid.** [3 marks · Practice] #### Given Rectangle [0,4], height 0.5; triangle with vertices (4,0),(4,0.5),(6,0). #### To find Centroid of the combined area. #### Formula and why it applies The parts do not overlap except at their boundary. Weight each part’s horizontal centroid by its area. #### Solution ##### Step 1. Rectangle area and centre $$ A_1=4\times0.5=2\qquad c_1=(0+4)/2=2 $$ ##### Step 2. Triangle area and centre $$ A_2=\tfrac12\times2\times0.5=0.5 $$ $$ c_2=(4+4+6)/3=14/3 $$ ##### Step 3. Add first moments $$ A_1c_1+A_2c_2=2(2)+0.5(14/3)=19/3 $$ ##### Step 4. Divide by total area $$ z^*=(19/3)/(2+0.5)=38/15\approx2.533333 $$ **Final answer.** Centroid ≈ 2.533333. The plateau MoM would be 2, so different methods can legitimately give different results. **Yaad rakho:** Centroid = har location × uska weight, total karke total weight se divide. --- ## TEE · Weighted average and centroid are not automatically the same Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--weighted-average/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/weighted-average Definitions → membership graphs → operations → rules → defuzzification → complete controller. This is the same weighted-average calculation taught in Essential maths. Here a representative output value c is weighted by a rule strength α. Keep the pair together: strength .3 belonging to speed 65 contributes .3×65. The denominator is the total strength. It is not automatically the number of rules, and the representative is not automatically the area-centre of a clipped set. ## Weighted average of output representatives Some classroom solutions use one representative (often the peak) per output label and weight it by the firing strength. That is a **weighted-average or height-based calculation**. It is exact for a specified weighted combination of singleton outputs, but is not generally the centroid of a max-aggregated, clipped Mamdani shape. Always name the method. A table of memberships at actual output locations supports the discrete centroid. A list of category peaks and strengths supports the specified representative-weighted calculation. Full continuous output shapes support area centroid. **Using the weighted-average method, combine output representatives [25, 75] mph with respective strengths [0.1, 0.7].** [3 marks · Practice] #### Given | Output representative | Strength | | --- | --- | | 25 | 0.1 | | 75 | 0.7 | #### To find One crisp output by the specified weighted-average method. #### Formula and why it applies Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set. $$ z^*=\frac{\sum_i\alpha_i z_i}{\sum_i\alpha_i} $$ #### Solution ##### Step 1. Multiply each value by its strength | Strength × value | Product | | --- | --- | | 0.1 × 25 | 2.5 | | 0.7 × 75 | 52.5 | ##### Step 2. Add the products $$ 2.5+52.5=55 $$ ##### Step 3. Add the strengths $$ 0.1+0.7=0.8 $$ ##### Step 4. Divide $$ z^*=\frac{55}{0.8}=68.75 $$ **Final answer.** 68.75 mph. **Using the weighted-average method, combine output representatives [40, 65, 85] % fan speed with respective strengths [0, 0.3, 0.5].** [3 marks · Practice] #### Given | Output representative | Strength | | --- | --- | | 40 | 0 | | 65 | 0.3 | | 85 | 0.5 | #### To find One crisp output by the specified weighted-average method. #### Formula and why it applies Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set. $$ z^*=\frac{\sum_i\alpha_i z_i}{\sum_i\alpha_i} $$ #### Solution ##### Step 1. Multiply each value by its strength | Strength × value | Product | | --- | --- | | 0 × 40 | 0 | | 0.3 × 65 | 19.5 | | 0.5 × 85 | 42.5 | ##### Step 2. Add the products $$ 0+19.5+42.5=62 $$ ##### Step 3. Add the strengths $$ 0+0.3+0.5=0.8 $$ ##### Step 4. Divide $$ z^*=\frac{62}{0.8}=77.5 $$ **Final answer.** 77.5 % fan speed. **Using the weighted-average method, combine output representatives [25, 40, 60] minutes with respective strengths [0.6, 0.4, 0.2].** [3 marks · Practice] #### Given | Output representative | Strength | | --- | --- | | 25 | 0.6 | | 40 | 0.4 | | 60 | 0.2 | #### To find One crisp output by the specified weighted-average method. #### Formula and why it applies Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set. $$ z^*=\frac{\sum_i\alpha_i z_i}{\sum_i\alpha_i} $$ #### Solution ##### Step 1. Multiply each value by its strength | Strength × value | Product | | --- | --- | | 0.6 × 25 | 15 | | 0.4 × 40 | 16 | | 0.2 × 60 | 12 | ##### Step 2. Add the products $$ 15+16+12=43 $$ ##### Step 3. Add the strengths $$ 0.6+0.4+0.2=1.2 $$ ##### Step 4. Divide $$ z^*=\frac{43}{1.2}=35.833333333 $$ **Final answer.** 35.833333333 minutes. | Method | What it uses | Limitation | | --- | --- | --- | | MoM | Locations at maximum membership | Ignores lower portions of the shape; can change abruptly | | Area centroid | Entire aggregated output area | Needs an area/moment calculation or numerical integration | | Representative weighted average | Supplied representative values and strengths | Not interchangeable with a clipped Mamdani area centroid | No method is universally “most accurate”. The correct exam calculation is the method specified by the question. If all memberships or all weights are zero, division by zero is undefined; a controller needs an explicit fallback rule. **Yaad rakho:** Formula similar dikhega, par weight kya hai? Membership, area, ya rule strength—pehle woh identify karo. ## Faculty slide example: one temperature input This is a different example from the two-input fan design. With Warm=Tri(20,30,40) and Hot=Tri(30,40,50), temperature 32 gives Warm=(40−32)/10=0.8 and Hot=(32−30)/10=0.2. The slide uses output representatives 50 and 80. **Using the weighted-average method, combine output representatives [50, 80] % fan speed with respective strengths [0.8, 0.2].** [3 marks · Practice] #### Given | Output representative | Strength | | --- | --- | | 50 | 0.8 | | 80 | 0.2 | #### To find One crisp output by the specified weighted-average method. #### Formula and why it applies Each representative is multiplied by its strength; add these products and divide by the total strength. Here α is a strength and z is a representative value. These are not automatically the full area of a clipped Mamdani set. $$ z^*=\frac{\sum_i\alpha_i z_i}{\sum_i\alpha_i} $$ #### Solution ##### Step 1. Multiply each value by its strength | Strength × value | Product | | --- | --- | | 0.8 × 50 | 40 | | 0.2 × 80 | 16 | ##### Step 2. Add the products $$ 40+16=56 $$ ##### Step 3. Add the strengths $$ 0.8+0.2=1 $$ ##### Step 4. Divide $$ z^*=\frac{56}{1}=56 $$ **Final answer.** 56 % fan speed. --- ## TEE · Class fan question: 32°C, 7 people, every rule and step Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--class-fan/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/class-fan Definitions → membership graphs → operations → rules → defuzzification → complete controller. There are 3 temperature labels and 3 occupancy labels. Pair each temperature label with each occupancy label: 3×3=9 possible AND rules. At the specified input, any rule containing a zero input membership has strength min(0,other)=0 and contributes no output. The rules with nonzero strength remain; their output labels come from the supplied rule table, not from the magnitude of the strength. This is the fan-controller design question shared in class: temperature 32°C and occupancy 7. The photos contain inconsistent graph breakpoints and arithmetic. A design question permits a stated design; below, every membership function is specified explicitly. These choices make the solution reproducible and should not be read as a uniquely recoverable answer from the photos. **Design a Mamdani fan controller for temperature 0–40°C and occupancy 0–10, with three input labels each and five fan-speed labels on 0–100%. For T=32°C and occupancy=7, use the membership functions and nine rules below. Calculate all memberships, firing strengths, and active output sets. Find fan speed by (a) mean of maxima and (b) the specified representative-weighted average.** [10 marks · Class-question adaptation; full design] #### Given | Variable | Low / Very Low | Medium / Low | High / Medium | High output | Very High output | | --- | --- | --- | --- | --- | --- | | Temperature | Tri(0,10,20) | Tri(15,25,35) | Tri(25,35,40) | — | — | | Occupancy | Tri(0,2,4) | Tri(3,5,7) | Tri(6,8,10) | — | — | | Fan speed | Tri(0,10,20) | Tri(10,25,40) | Tri(20,40,60) | Tri(50,65,80) | Tri(70,85,100) | Input T=32°C; occupancy=7. Use min-AND, min-clipping and max-aggregation. | Temperature | Occupancy | Fan output | | --- | --- | --- | | Low | Low | Very Low | | Low | Medium | Low | | Low | High | Medium | | Medium | Low | Low | | Medium | Medium | Medium | | Medium | High | High | | High | Low | Medium | | High | Medium | High | | High | High | Very High | #### To find Input memberships → nine rule strengths → clipped output → two named crisp outputs. #### Formula and why it applies Use the appropriate triangle branch for membership. AND takes the smaller degree. Each output is clipped at that strength; rules with the same consequent combine by max. MoM uses only the highest plateau. The requested representative weighted average uses category peaks and strengths. $$ \alpha_r=\min(\mu_{T,r},\mu_{O,r})\quad\mu_{out}(z)=\max_r\min(\alpha_r,\mu_{C_r}(z)) $$ #### Solution ##### Step 1. Read the temperature graph Swipe across the diagram to read its labels ↔ ![At 32, Medium is on its falling side; High is on its rising side.](https://deep-learning-study-room.vercel.app/is2/fig/v2_fan_temp.434c9b6635.svg) Diagram: At 32, Medium is on its falling side; High is on its rising side.. Labels: Chosen temperature membership functions | 0 | 0.5 | 1 | 0 | 10 | 15 | 20 | 25 | 32 | 35 | 40 | μ | Temperature in °C | Low | Medium | High | 32°C At 32, Medium is on its falling side; High is on its rising side. $$ \mu_L(32)=0 $$ $$ \mu_M(32)=\frac{35-32}{35-25}=\frac3{10}=0.3 $$ $$ \mu_H(32)=\frac{32-25}{35-25}=\frac7{10}=0.7 $$ ##### Step 2. Read the occupancy graph Swipe across the diagram to read its labels ↔ ![At 7, Medium reaches its right zero and High is halfway up its rising edge.](https://deep-learning-study-room.vercel.app/is2/fig/v2_fan_people.1861531dd1.svg) Diagram: At 7, Medium reaches its right zero and High is halfway up its rising edge.. Labels: Chosen occupancy membership functions | 0 | 0.5 | 1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | μ | Number of people | Low | Medium | High | 7 people At 7, Medium reaches its right zero and High is halfway up its rising edge. $$ \mu_L(7)=0\qquad\mu_M(7)=0 $$ $$ \mu_H(7)=\frac{7-6}{8-6}=\frac12=0.5 $$ ##### Step 3. Evaluate all nine rules | T label / degree | O label / degree | min = strength | Output | | --- | --- | --- | --- | | Low / 0 | Low / 0 | 0 | Very Low | | Low / 0 | Medium / 0 | 0 | Low | | Low / 0 | High / 0.5 | 0 | Medium | | Medium / 0.3 | Low / 0 | 0 | Low | | Medium / 0.3 | Medium / 0 | 0 | Medium | | Medium / 0.3 | High / 0.5 | 0.3 | High | | High / 0.7 | Low / 0 | 0 | Medium | | High / 0.7 | Medium / 0 | 0 | High | | High / 0.7 | High / 0.5 | 0.5 | Very High | ##### Step 4. Clip and aggregate $$ \mu_{out}(z)=\max\{\min(0.3,\mu_H(z)),\min(0.5,\mu_{VH}(z))\} $$ Swipe across the diagram to read its labels ↔ ![High is clipped at 0.3; Very High is clipped at 0.5. Use the upper envelope where the two shapes overlap.](https://deep-learning-study-room.vercel.app/is2/fig/v2_fan_output.8708d0ea77.svg) Diagram: High is clipped at 0.3; Very High is clipped at 0.5. Use the upper envelope where the two shapes overlap.. Labels: Two active output sets | 0 | 0.5 | 1 | 50 | 54.5 | 65 | 70 | 75 | 77.5 | 85 | 92.5 | 100 | μ | Fan speed (%) | High | Very High | clip 0.3 | clip 0.5 High is clipped at 0.3; Very High is clipped at 0.5. Use the upper envelope where the two shapes overlap. ##### Step 5. Find the highest plateau for MoM The maximum is 0.5, belonging to Very High. Its triangle is (70,85,100). $$ \frac{z_L-70}{15}=0.5\Rightarrow z_L=77.5 $$ $$ \frac{100-z_R}{15}=0.5\Rightarrow z_R=92.5 $$ $$ z^*_{MoM}=\frac{77.5+92.5}{2}=85 $$ ##### Step 6. Multiply the representatives by strengths $$ 0.3\times65=19.5 $$ $$ 0.5\times85=42.5 $$ ##### Step 7. Add, then divide for the weighted average $$ z^*_{WA}=\frac{19.5+42.5}{0.3+0.5}=\frac{62}{0.8}=77.5 $$ **Final answer.** For this explicitly stated design: MoM = 85%; representative-weighted output = 77.5%. These are different methods. 77.5% is not claimed to be the true continuous Mamdani area centroid. **How to adapt it in the exam:** if breakpoints or output representatives differ, recompute the memberships from those data. The sequence of operations stays unchanged. If the question explicitly asks for the area centroid, integrate the max-aggregated shape; do not silently substitute the weighted average above. **Yaad rakho:** Temperature aur people ka degree nikalo → 9 min → active outputs clip → jo method poocha hai woh lagao. ## Why the Very High plateau has midpoint 85 For the stated Very High triangle (70,85,100) clipped at height 0.5, solve the rising edge: (z−70)/(85−70)=0.5, so z−70=0.5×15=7.5 and z=77.5. On the falling edge, (100−z)/(100−85)=0.5, so 100−z=7.5 and z=92.5. Thus the highest flat interval is [77.5,92.5]. Its midpoint is (77.5+92.5)/2=170/2=85. Continue with the [complete area-centroid calculation for this same fan design](https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--centroid-comparison/): 77.086956522%, with all overlap breakpoints and area/moment arithmetic shown. --- ## TEE · The same fan controller: true area centroid versus MoM and weighted average Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--centroid-comparison/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/centroid-comparison Definitions → membership graphs → operations → rules → defuzzification → complete controller. The preceding class-fan lesson makes the input shapes and nine rules explicit because the photographs do not specify one internally consistent design. Keep that exact stated design here: temperature 32°C, occupancy 7, High output Tri(50,65,80) firing at 0.3 and Very High output Tri(70,85,100) firing at 0.5. This continuation calculates the actual area centroid of their max-aggregate. It does not reinterpret 77.5 as a centroid. **For the stated two-input class fan controller at T=32°C and occupancy=7, calculate the true continuous Mamdani area centroid of High=Tri(50,65,80) clipped at 0.3 and Very High=Tri(70,85,100) clipped at 0.5. Compare it with mean of maxima and the specified representative-weighted average.** [6 marks] Worked teaching continuation of the class-design adaptation; the mark allocation here is for practice. #### Given All other output rules have strength zero. Use min for implication, max for aggregation and speed z on 0–100%. The two active rule strengths were calculated from input memberships: min(0.3,0.5)=0.3 and min(0.7,0.5)=0.5. #### To find Every visible segment of the aggregate, its area and first moment, the centroid, and the differences between the three crisp answers. #### Formula and why it applies $$ \mu_{out}(z)=\max\{\min(0.3,\mu_H(z)),\min(0.5,\mu_{VH}(z))\} $$ $$ z^*_{COA}=\frac{\int z\mu_{out}(z)\,dz}{\int\mu_{out}(z)\,dz} $$ Centroid balances all the area under the upper envelope. A straight strip with endpoints (a,u) and (b,v) has the following area A and first moment M; adding only disjoint strips counts overlap once. $$ A=\frac{(b-a)(u+v)}2,\qquad M=\frac{b-a}{6}[(2a+b)u+(a+2b)v] $$ #### Solution ##### Step 1. Locate the two clipped plateaus $$ H:\quad z_L=50+0.3(65-50)=54.5,\quad z_R=80-0.3(80-65)=75.5 $$ $$ VH:\quad z_L=70+0.5(85-70)=77.5,\quad z_R=100-0.5(100-85)=92.5 $$ High has a 0.3 plateau from 54.5 to 75.5. Very High has a 0.5 plateau from 77.5 to 92.5. ##### Step 2. Find where the upper envelope changes owner Very High rises through the High plateau at height 0.3. Solve its rising equation: $$ (z-70)/15=0.3\quad\Rightarrow\quad z=74.5 $$ Before 74.5, clipped High supplies the upper envelope. After 74.5, Very High is above it. The descending High edge from 75.5 to 80 lies underneath and contributes no extra aggregate area. ##### Step 3. Trace the aggregate graph Swipe across the diagram to read its labels ↔ ![Fan output: trace the upper envelope before finding its centroid](https://deep-learning-study-room.vercel.app/is2/fig/deep_fan_centroid.98bf1c217d.svg) Diagram: Fan output: trace the upper envelope before finding its centroid. Labels: Fan output: trace the upper envelope before finding its centroid | High clipped at 0.3; Very High clipped at 0.5. Hidden overlap is counted once. | 0 | 0.3 | 0.5 | 50 | 54.5 | 74.5 | 77.5 | 92.5 | 100 | Centroid ≈77.087% | Fan speed z (%) | Area = 17.25; first moment = 1329.75; centroid = moment ÷ area. Read left to right: (50,0) → (54.5,0.3) → (74.5,0.3) → (77.5,0.5) → (92.5,0.5) → (100,0). Membership is zero outside this support. Each adjacent pair is one straight strip in the table. ##### Step 4. Work the rising first strip explicitly $$ A_1=(54.5-50)(0+0.3)/2=0.675 $$ $$ M_1=\frac{4.5}{6}[(100+54.5)0+(50+109)0.3]=35.775 $$ The first moment has units of speed multiplied by area. Dividing total moment by total area returns speed. ##### Step 5. Calculate all disjoint strips | a | b | u | v | Area A | First moment M | | --- | --- | --- | --- | --- | --- | | 50 | 54.5 | 0 | 0.3 | 0.675 | 35.775 | | 54.5 | 74.5 | 0.3 | 0.3 | 6 | 387 | | 74.5 | 77.5 | 0.3 | 0.5 | 1.2 | 91.35 | | 77.5 | 92.5 | 0.5 | 0.5 | 7.5 | 637.5 | | 92.5 | 100 | 0.5 | 0 | 1.875 | 178.125 | $$ \sum A=0.675+6+1.2+7.5+1.875=17.25 $$ $$ \sum M=35.775+387+91.35+637.5+178.125=1329.75 $$ ##### Step 6. Divide for the actual centroid $$ z^*_{COA}=1329.75/17.25=1773/23\approx77.086956522\% $$ ##### Step 7. Compare methods on the same aggregate | Method | Calculation | Answer | | --- | --- | --- | | Area centroid | 1329.75 ÷ 17.25 | 77.086956522% | | Mean of maxima | (77.5+92.5) ÷ 2 | 85% | | Specified representative weighted average | (0.3×65+0.5×85) ÷ (0.3+0.5) | 77.5% | MoM uses only the highest plateau. The representative method uses two chosen peaks and two rule strengths. The area centroid uses the full max-aggregate, so these are three legitimate but different calculations. **Final answer.** Area centroid ≈77.086956522%; MoM=85%; representative-weighted output=77.5%. Report the method with the value. The centroid is inside the support [50,100], as a balance point must be. **Boundary coverage of this teaching design.** With the ordinary triangles stated in the class adaptation, every temperature degree is zero at T=0 or T=40, and every occupancy degree is zero at occupancy=0 or 10. Then no rule fires and an area centroid is undefined. To make a controller cover the entire declared input range, explicitly redesign outer sets as shoulders or specify an out-of-coverage rule; do not silently divide by zero. The worked T=32, occupancy=7 case has positive area and is unaffected. **Yaad rakho:** Centroid mein poore shaded area ka balance chahiye. Peak × strength wala shortcut same calculation nahi hai. --- ## TEE · Fuzzy theory: definitions, architecture, advantages and applications Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--exam-theory/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/exam-theory Definitions → membership graphs → operations → rules → defuzzification → complete controller. **Explain fuzzy logic with one application.** [2 marks · Practice] #### Answer Fuzzy logic represents gradual truth or category membership using values between 0 and 1. A temperature controller can use Low, Medium and High temperature sets. It evaluates rules such as IF temperature is High THEN fan speed is High, combines the rule outputs and defuzzifies them into a numerical fan speed. This permits gradual control around category boundaries. **Give the advantages and limitations of a Mamdani fuzzy controller.** [3 marks · Practice] #### Answer | Advantages | Limitations | | --- | --- | | Human-readable rules express expert knowledge | Membership functions and rules require careful design | | Can model gradual boundaries without an exact analytic plant model | Rule count can grow rapidly with more inputs and categories | | Handles overlapping linguistic descriptions | A poorly chosen rule base can produce poor or uncovered outputs | | Suitable for many control applications | Does not automatically learn or guarantee optimal control | **Describe the design of a fuzzy fan controller.** [3 marks · Practice] #### Answer Choose temperature and occupancy as inputs and fan speed as output. Define their ranges and Low/Medium/High membership functions. Construct a rule table linking input labels to output labels. Fuzzify measured inputs, use min to obtain AND-rule strengths, clip the consequent sets and max-aggregate them. Apply the specified defuzzification method to obtain a fan speed. Test the design at normal inputs and range boundaries. ## Other applications you can explain | Application | Inputs | Example rule | Output | | --- | --- | --- | --- | | Washing machine | Dirtiness, load | IF dirt High AND load High THEN wash time Long | Wash duration | | Loan decision support | Income, credit score | IF income High AND credit Good THEN approval suitability High | Suitability score (not an automatic legal lending decision) | | Fire warning | Temperature, smoke | IF temperature High OR smoke High THEN risk High | Alarm-risk score | | Vehicle control | Distance, relative speed | IF distance Small AND closing speed High THEN braking Strong | Braking command | **Yaad rakho:** Theory answer mein definition → blocks/steps → ek concrete example. Sirf headings se marks nahi milte. --- ## TEE · Gaussian, singleton and bell memberships Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-core--smooth-memberships/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-core/smooth-memberships Definitions → membership graphs → operations → rules → defuzzification → complete controller. For a Gaussian membership, c is the input at the centre and σ controls the width. Compute x−c, square it, divide by 2σ², put a minus sign in front, and apply eˣ. The maximum membership is 1 at x=c because the exponent becomes 0. Increasing σ makes the same distance from the centre count as a smaller relative distance, so the curve becomes wider. A Gaussian membership is not the same formula as a normal probability density: there is no 1/(√(2π)σ) multiplier here. ## A smooth membership function A Gaussian membership function reaches 1 at its centre c and decreases smoothly on both sides. Its width σ must be positive. This is a fuzzy membership curve, not a normalised probability density: there is no 1/(√(2π)σ) factor. $$ \mu(x)=\exp\left(-\frac{(x-c)^2}{2\sigma^2}\right) $$ exp(u) means eᵘ. The calculator evaluates it using eˣ; calculate the entire negative exponent first. A squared distance treats equal left and right offsets equally. **For Gaussian membership with c=10 and σ=2, find μ(10).** [2 marks · Practice] #### Given x=10; centre c=10; width σ=2. #### To find Membership degree. #### Formula and why it applies Use squared distance from the centre in the Gaussian membership formula. #### Solution ##### Step 1. Distance from the centre $$ 10-10=0 $$ ##### Step 2. Square the distance and width $$ (0)^2=0\qquad2\sigma^2=2(2^2)=8 $$ ##### Step 3. Exponent $$ -0/8=0 $$ ##### Step 4. Evaluate e to this power $$ \mu(10)=e^{0}\approx1 $$ **Final answer.** 1. **For Gaussian membership with c=10 and σ=2, find μ(12).** [2 marks · Practice] #### Given x=12; centre c=10; width σ=2. #### To find Membership degree. #### Formula and why it applies Use squared distance from the centre in the Gaussian membership formula. #### Solution ##### Step 1. Distance from the centre $$ 12-10=2 $$ ##### Step 2. Square the distance and width $$ (2)^2=4\qquad2\sigma^2=2(2^2)=8 $$ ##### Step 3. Exponent $$ -4/8=-0.5 $$ ##### Step 4. Evaluate e to this power $$ \mu(12)=e^{-0.5}\approx0.60653066 $$ **Final answer.** 0.60653066. **For Gaussian membership with c=10 and σ=2, find μ(14).** [2 marks · Practice] #### Given x=14; centre c=10; width σ=2. #### To find Membership degree. #### Formula and why it applies Use squared distance from the centre in the Gaussian membership formula. #### Solution ##### Step 1. Distance from the centre $$ 14-10=4 $$ ##### Step 2. Square the distance and width $$ (4)^2=16\qquad2\sigma^2=2(2^2)=8 $$ ##### Step 3. Exponent $$ -16/8=-2 $$ ##### Step 4. Evaluate e to this power $$ \mu(14)=e^{-2}\approx0.135335283 $$ **Final answer.** 0.135335283. ## Singleton A singleton fuzzy set has membership 1 at exactly one specified point and 0 elsewhere. For a singleton at c=50: μ(50)=1, μ(49)=0 and μ(51)=0. A singleton consequent is represented by one numerical output; this is why a specified weighted average of singleton outputs is a legitimate calculation. ## Generalised bell (when its formula is supplied) $$ \mu(x)=\frac1{1+\left|\frac{x-c}{a}\right|^{2b}} $$ c is the centre, a>0 controls width and b>0 controls the slope. Absolute value |u| is the nonnegative magnitude: |−2|=2. At x=c membership is 1; at x=c±a membership is 1/2. **For a=2, b=1 and c=10, find generalised-bell membership at x=14.** [2 marks · Practice] #### Given a=2, b=1, c=10; x=14. #### To find μ(14). #### Formula and why it applies Subtract the centre, divide by a, take the absolute value, raise to 2b, then use the reciprocal. #### Solution ##### Step 1. Scaled distance $$ |(14-10)/2|=|2|=2 $$ ##### Step 2. Power $$ 2^{2(1)}=4 $$ ##### Step 3. Final reciprocal $$ \mu(14)=1/(1+4)=0.2 $$ **Final answer.** 0.2. **Yaad rakho:** Gaussian MF ka centre 1 hota hai. GMM density ke normalising factor ko yahan mat lagao. --- ## TEE · Cartesian product: every cell explained Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-relations--cartesian/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-relations/cartesian Longer Unit 2 numerical practice, also useful for TEE. ## How a row membership and a column membership meet Let A₁=.6 and A₂=.2 label two rows. Let B₁=.3, B₂=.8 and B₃=.5 label three columns. Cell R₂₃ belongs to row 2 and column 3, so use A₂=.2 and B₃=.5. Their min is .2. You do not multiply 2×3; those subscripts only locate the cell. $$ R_{23}=\min(A_2,B_3)=\min(0.2,0.5)=0.2 $$ There are two choices of row and three of column, so fill six cells in a 2×3 rectangle. Keep the row and column labels when reporting a relation. ## What is a fuzzy relation? A fuzzy set gives a membership to one element. A **fuzzy relation** gives a degree to an ordered pair. A matrix is a convenient table: rows name the first element and columns name the second. For the standard min-based Cartesian product R=A×B, put the A memberships down the rows and the B memberships across the columns. Each cell is the smaller of its row and column memberships. $$ R_{ij}=\min(\mu_A(a_i),\mu_B(b_j)) $$ **Find the fuzzy Cartesian product A×B for memberships A=[0.6, 0.2] and B=[0.3, 0.8, 0.5].** [3 marks · Practice] #### Given Ordered A memberships [0.6, 0.2]; B memberships [0.3, 0.8, 0.5]. #### To find A 2×3 relation matrix. #### Formula and why it applies Every row corresponds to an A element and every column to a B element; apply min in every cell. $$ R_{ij}=\min(A_i,B_j) $$ #### Solution ##### Step 1. Determine dimensions 2 row elements × 3 column elements. ##### Step 2. Compute each cell | A ↓ / B → | 0.3 | 0.8 | 0.5 | | --- | --- | --- | --- | | 0.6 | min(0.6,0.3) = 0.3 | min(0.6,0.8) = 0.6 | min(0.6,0.5) = 0.5 | | 0.2 | min(0.2,0.3) = 0.2 | min(0.2,0.8) = 0.2 | min(0.2,0.5) = 0.2 | ##### Step 3. Write the matrix $$ R=\begin{bmatrix}0.3&0.6&0.5\\0.2&0.2&0.2\end{bmatrix} $$ **Final answer.** R is the matrix in Step 3. **Find the fuzzy Cartesian product A×B for memberships A=[1, 0.4] and B=[0.7, 0].** [3 marks · Practice] #### Given Ordered A memberships [1, 0.4]; B memberships [0.7, 0]. #### To find A 2×2 relation matrix. #### Formula and why it applies Every row corresponds to an A element and every column to a B element; apply min in every cell. $$ R_{ij}=\min(A_i,B_j) $$ #### Solution ##### Step 1. Determine dimensions 2 row elements × 2 column elements. ##### Step 2. Compute each cell | A ↓ / B → | 0.7 | 0 | | --- | --- | --- | | 1 | min(1,0.7) = 0.7 | min(1,0) = 0 | | 0.4 | min(0.4,0.7) = 0.4 | min(0.4,0) = 0 | ##### Step 3. Write the matrix $$ R=\begin{bmatrix}0.7&0\\0.4&0\end{bmatrix} $$ **Final answer.** R is the matrix in Step 3. **Find the fuzzy Cartesian product A×B for memberships A=[0.2, 0.9, 0.5] and B=[0.6, 0.1].** [3 marks · Practice] #### Given Ordered A memberships [0.2, 0.9, 0.5]; B memberships [0.6, 0.1]. #### To find A 3×2 relation matrix. #### Formula and why it applies Every row corresponds to an A element and every column to a B element; apply min in every cell. $$ R_{ij}=\min(A_i,B_j) $$ #### Solution ##### Step 1. Determine dimensions 3 row elements × 2 column elements. ##### Step 2. Compute each cell | A ↓ / B → | 0.6 | 0.1 | | --- | --- | --- | | 0.2 | min(0.2,0.6) = 0.2 | min(0.2,0.1) = 0.1 | | 0.9 | min(0.9,0.6) = 0.6 | min(0.9,0.1) = 0.1 | | 0.5 | min(0.5,0.6) = 0.5 | min(0.5,0.1) = 0.1 | ##### Step 3. Write the matrix $$ R=\begin{bmatrix}0.2&0.1\\0.6&0.1\\0.5&0.1\end{bmatrix} $$ **Final answer.** R is the matrix in Step 3. **Yaad rakho:** Row ka degree aur column ka degree—dono ka min us cell mein. --- ## TEE · Max–min composition: three complete matrix examples Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/fuzzy-relations--composition/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/fuzzy-relations/composition Longer Unit 2 numerical practice, also useful for TEE. ## One result cell combines alternative two-link paths For T₁₁, start at u₁ and end at w₁. If the middle domain has v₁ and v₂, there are two candidate paths. Their strengths come from matching row 1 of R with column 1 of S. | Path | First link from R | Second link from S | Path strength | | --- | --- | --- | --- | | u₁ → v₁ → w₁ | .2 | .5 | min(.2,.5)=.2 | | u₁ → v₂ → w₁ | .8 | .7 | min(.8,.7)=.7 | Choose the stronger path: max(.2,.7)=.7, so T₁₁=.7. The middle label j runs through v₁ and v₂. The labels i and k stay fixed at 1 while this one cell is calculated. Then move to the next result cell. **Yaad rakho:** Ek path mein dono links chahiye, isliye min. Alag paths mein best chalega, isliye max. ## Composition connects through a middle element Suppose R relates U to V and S relates V to W. To find how strongly u connects to w through each middle v, take min of the two links. Then choose the strongest such path using max. $$ T_{ik}=\max_j\min(R_{ij},S_{jk}) $$ We call this matrix operation R composed with S here. Some books write the symbolic order in reverse. State the formula and row/column labels, especially when a question says S∘R. Composition requires the intermediate domains and their ordering to match; having two 3×3 matrices alone does not prove their labels match. **Calculate the max–min composition using $$ T_{ik}=\max_j\min(R_{ij},S_{jk}) $$ .** [3 marks · Practice] #### Given $$ R=\begin{bmatrix}0.2&0.8\\0.6&0.3\end{bmatrix} $$ $$ S=\begin{bmatrix}0.5&0.9\\0.7&0.1\end{bmatrix} $$ #### To find Every entry of T. #### Formula and why it applies Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication. #### Solution ##### Step 1. Cell (1,1): row 1 of R with column 1 of S $$ T_{11}=\max\{\min(0.2,0.5),\min(0.8,0.7)\} $$ $$ =\max\{0.2,0.7\}=0.7 $$ ##### Step 2. Cell (1,2): row 1 of R with column 2 of S $$ T_{12}=\max\{\min(0.2,0.9),\min(0.8,0.1)\} $$ $$ =\max\{0.2,0.1\}=0.2 $$ ##### Step 3. Cell (2,1): row 2 of R with column 1 of S $$ T_{21}=\max\{\min(0.6,0.5),\min(0.3,0.7)\} $$ $$ =\max\{0.5,0.3\}=0.5 $$ ##### Step 4. Cell (2,2): row 2 of R with column 2 of S $$ T_{22}=\max\{\min(0.6,0.9),\min(0.3,0.1)\} $$ $$ =\max\{0.6,0.1\}=0.6 $$ **Final answer.** $$ T=\begin{bmatrix}0.7&0.2\\0.5&0.6\end{bmatrix} $$ . **Calculate the max–min composition using $$ T_{ik}=\max_j\min(R_{ij},S_{jk}) $$ .** [3 marks · Practice] #### Given $$ R=\begin{bmatrix}1&0\\0.4&0.7\end{bmatrix} $$ $$ S=\begin{bmatrix}0.2&0.6\\0.8&0.3\end{bmatrix} $$ #### To find Every entry of T. #### Formula and why it applies Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication. #### Solution ##### Step 1. Cell (1,1): row 1 of R with column 1 of S $$ T_{11}=\max\{\min(1,0.2),\min(0,0.8)\} $$ $$ =\max\{0.2,0\}=0.2 $$ ##### Step 2. Cell (1,2): row 1 of R with column 2 of S $$ T_{12}=\max\{\min(1,0.6),\min(0,0.3)\} $$ $$ =\max\{0.6,0\}=0.6 $$ ##### Step 3. Cell (2,1): row 2 of R with column 1 of S $$ T_{21}=\max\{\min(0.4,0.2),\min(0.7,0.8)\} $$ $$ =\max\{0.2,0.7\}=0.7 $$ ##### Step 4. Cell (2,2): row 2 of R with column 2 of S $$ T_{22}=\max\{\min(0.4,0.6),\min(0.7,0.3)\} $$ $$ =\max\{0.4,0.3\}=0.4 $$ **Final answer.** $$ T=\begin{bmatrix}0.2&0.6\\0.7&0.4\end{bmatrix} $$ . **Calculate the max–min composition using $$ T_{ik}=\max_j\min(R_{ij},S_{jk}) $$ .** [3 marks · Practice] #### Given $$ R=\begin{bmatrix}0.3&0.6&0.9\end{bmatrix} $$ $$ S=\begin{bmatrix}0.8&0.1\\0.2&0.7\\0.5&0.4\end{bmatrix} $$ #### To find Every entry of T. #### Formula and why it applies Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication. #### Solution ##### Step 1. Cell (1,1): row 1 of R with column 1 of S $$ T_{11}=\max\{\min(0.3,0.8),\min(0.6,0.2),\min(0.9,0.5)\} $$ $$ =\max\{0.3,0.2,0.5\}=0.5 $$ ##### Step 2. Cell (1,2): row 1 of R with column 2 of S $$ T_{12}=\max\{\min(0.3,0.1),\min(0.6,0.7),\min(0.9,0.4)\} $$ $$ =\max\{0.1,0.6,0.4\}=0.6 $$ **Final answer.** $$ T=\begin{bmatrix}0.5&0.6\end{bmatrix} $$ . **Yaad rakho:** Har path mein min, phir saare paths mein max. Multiply-and-add nahi. --- ## TEE · GA meaning, biological idea and every basic term Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/ga-core--meaning/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/ga-core/meaning Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. | Object | Concrete value | Do not confuse it with | | --- | --- | --- | | Individual number | Row 2 | The candidate’s value x | | Candidate value x | 11 | Its chromosome 1011 | | Chromosome | 1011 | Its fitness | | Fitness when maximising x² | 11×11=121 | Its row number or bit string | In this guide, “gene 3” counts the third bit from the **left** starting at 1. Binary place values were built from the **right** starting with weight 1. These are two labelling conventions, not contradictory place-value rules. Read the end specified in an exam. ## What a genetic algorithm does A **genetic algorithm (GA)** is a population-based optimisation method inspired by biological evolution. It maintains candidate solutions, evaluates their fitness, selects parents, generates children through crossover and mutation, and repeats the process. In biology, genes carry inherited information. In a GA, an encoding plays the role of inherited information. The analogy motivates the operators; it does not mean that a computer chromosome is literally DNA. | Term | Meaning | Four-bit example | | --- | --- | --- | | Individual | One candidate solution | x=11 | | Chromosome | The complete encoded candidate | 1011 | | Gene | One component of an encoding | The third bit | | Allele | The value at a gene | 0 or 1 for binary encoding | | Population | The collection of current individuals | [3,5,10,12] | | Fitness | A score used to favour useful candidates | For max x², fitness of 11 is 121 | | Generation | One cycle producing a new population | Select → cross → mutate → replace | | Search space | All candidate solutions permitted | 4-bit integers: 0–15 | | Elitism | Preserving best individuals unchanged | Copy the best into the next population | An objective describes what the problem wants. Fitness describes what the selection mechanism favours. For maximising a nonnegative objective, they may be the same. For minimising cost, directly selecting in proportion to cost would favour bad candidates; use a specified transformation or a comparison-based selection method. **For a maximisation GA with objective f(x)=x², find the fitness of x=3.** [2 marks · Practice] #### Given x=3; fitness equals the nonnegative objective x². #### To find Fitness, not the encoded chromosome. #### Formula and why it applies Square x to evaluate its objective; binary encoding is a separate representation step. #### Solution ##### Step 1. Substitute x $$ f(3)=3^2 $$ ##### Step 2. Multiply $$ 3\times3=9 $$ **Final answer.** Fitness = 9. **For a maximisation GA with objective f(x)=x², find the fitness of x=7.** [2 marks · Practice] #### Given x=7; fitness equals the nonnegative objective x². #### To find Fitness, not the encoded chromosome. #### Formula and why it applies Square x to evaluate its objective; binary encoding is a separate representation step. #### Solution ##### Step 1. Substitute x $$ f(7)=7^2 $$ ##### Step 2. Multiply $$ 7\times7=49 $$ **Final answer.** Fitness = 49. **For a maximisation GA with objective f(x)=x², find the fitness of x=12.** [2 marks · Practice] #### Given x=12; fitness equals the nonnegative objective x². #### To find Fitness, not the encoded chromosome. #### Formula and why it applies Square x to evaluate its objective; binary encoding is a separate representation step. #### Solution ##### Step 1. Substitute x $$ f(12)=12^2 $$ ##### Step 2. Multiply $$ 12\times12=144 $$ **Final answer.** Fitness = 144. **Yaad rakho:** Chromosome answer ka code; fitness us code wale answer ka score. --- ## TEE · Encoding: binary, real values and permutations Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/ga-core--encoding/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/ga-core/encoding Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. ## Expand the compact decoding formula before using it $$ x=b_1 2^3+b_2 2^2+b_3 2^1+b_4 2^0 $$ For four bits, b₁ is the leftmost bit and receives weight 8; b₄ is the rightmost and receives weight 1. In general n−i gives the power belonging to bit i. For 1011, the four terms are 1×8, 0×4, 1×2, 1×1. Add 8+0+2+1=11. For real-range mapping, d/(2ⁿ−1) is the fraction of the way through the possible codes. Multiplying by U−L travels that fraction of the physical range. Adding L shifts the result to start at the lower limit. | 3-bit code | Decoded d | Fraction d/7 | Mapped to [−2,5] | | --- | --- | --- | --- | | 000 | 0 | 0/7=0 | −2+0×7=−2 | | 011 | 3 | 3/7 | −2+(3/7)×7=1 | | 111 | 7 | 7/7=1 | −2+1×7=5 | ## Choose an encoding that can express legal solutions **Binary encoding** uses bit strings; **real-valued encoding** stores real numbers directly; **permutation encoding** stores an order, such as a TSP city tour. An operator valid for bits may create invalid city tours, so encoding and operators must fit the problem. | Representation | Candidate | Appropriate use | | --- | --- | --- | | Binary | 1011 | An unsigned integer or a coded real value | | Real-valued | [0.2, 1.7] | Continuous parameters | | Permutation | [A,C,B,D] | Visit order in TSP | ## Decode an n-bit integer $$ x=b_1 2^{n-1}+b_2 2^{n-2}+\cdots+b_n2^0 $$ b₁ is the leftmost bit, not a coefficient you have to learn. In four bits the weights are 8,4,2,1. The maximum is $$ 2^n-1 $$ . **Decode the 4-bit chromosome 0011.** [2 marks · Practice] #### Given Bits 0011; place values 8,4,2,1. #### To find Integer x. #### Formula and why it applies Multiply each bit by its matching place value, then add. #### Solution ##### Step 1. List contributions | Bit | Place value | Contribution | | --- | --- | --- | | 0 | 8 | 0 | | 0 | 4 | 0 | | 1 | 2 | 2 | | 1 | 1 | 1 | ##### Step 2. Add $$ 0+0+2+1=3 $$ **Final answer.** x=3. **Decode the 4-bit chromosome 1010.** [2 marks · Practice] #### Given Bits 1010; place values 8,4,2,1. #### To find Integer x. #### Formula and why it applies Multiply each bit by its matching place value, then add. #### Solution ##### Step 1. List contributions | Bit | Place value | Contribution | | --- | --- | --- | | 1 | 8 | 8 | | 0 | 4 | 0 | | 1 | 2 | 2 | | 0 | 1 | 0 | ##### Step 2. Add $$ 8+0+2+0=10 $$ **Final answer.** x=10. **Decode the 4-bit chromosome 1111.** [2 marks · Practice] #### Given Bits 1111; place values 8,4,2,1. #### To find Integer x. #### Formula and why it applies Multiply each bit by its matching place value, then add. #### Solution ##### Step 1. List contributions | Bit | Place value | Contribution | | --- | --- | --- | | 1 | 8 | 8 | | 1 | 4 | 4 | | 1 | 2 | 2 | | 1 | 1 | 1 | ##### Step 2. Add $$ 8+4+2+1=15 $$ **Final answer.** x=15. ## Encoding a real interval If the question maps an n-bit integer d to a real interval [L,U], first decode d, then use: $$ x=L+\frac{d}{2^n-1}(U-L) $$ **Map the 3-bit string 101 to the real interval [−2,5] using linear binary decoding.** [3 marks · Practice] #### Given n=3, L=−2, U=5; string 101. #### To find Real x. #### Formula and why it applies The smallest bit code maps to L and the largest to U; the remaining codes are equally spaced. $$ x=L+\frac d{2^n-1}(U-L) $$ #### Solution ##### Step 1. Decode integer d $$ d=1(4)+0(2)+1(1)=5 $$ ##### Step 2. Calculate interval width and maximum code $$ U-L=5-(-2)=7\qquad2^3-1=7 $$ ##### Step 3. Map to the interval $$ x=-2+(5/7)(7)=3 $$ **Final answer.** x=3. **Yaad rakho:** Pehle bits se integer. Real range mapping explicitly diya ho tabhi next formula lagao. --- ## TEE · Roulette selection, cumulative intervals and alternatives Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/ga-core--selection/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/ga-core/selection Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. Before the three complete sums below, read the running-total example in [Percentages, probabilities and cumulative intervals](https://deep-learning-study-room.vercel.app/read/is2/m2/read/core-maths--probability/). For weights [1,2,3,4], candidate C’s own probability is 0.3, but its interval is [0.3,0.6). A draw of 0.55 selects C because it lies inside that interval; it is not required to be less than C’s individual probability. ## Selection chooses parents, usually with replacement **Roulette-wheel selection** gives each candidate a probability proportional to a nonnegative fitness weight. A high weight occupies a larger interval on [0,1]. A supplied random number selects the interval containing it. $$ P_i=F_i/\sum_jF_j $$ Fᵢ is the fitness weight of individual i. The sum must be positive. Probabilities add to 1. Cumulative probability is the running sum and supplies the interval boundaries. **Fitness weights of successive individuals are [1, 2, 3, 4]. Select parents using random numbers [0.05, 0.25, 0.55, 0.9].** [3 marks · Practice] #### Given Fitness weights [1, 2, 3, 4]; draws [0.05, 0.25, 0.55, 0.9]. #### To find Parent row number for each draw. #### Formula and why it applies Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included. #### Solution ##### Step 1. Sum fitness $$ \sum F=1+2+3+4=10 $$ ##### Step 2. Find probabilities and cumulative bounds | Individual | Fitness | Probability | Cumulative | | --- | --- | --- | --- | | 1 | 1 | 0.1 | 0.1 | | 2 | 2 | 0.2 | 0.3 | | 3 | 3 | 0.3 | 0.6 | | 4 | 4 | 0.4 | 1 | ##### Step 3. Locate each draw | Draw | Selected individual | | --- | --- | | 0.05 | 1 | | 0.25 | 2 | | 0.55 | 3 | | 0.9 | 4 | **Final answer.** Selected row numbers: [1, 2, 3, 4]. Repeated selection is permitted. **Fitness weights of successive individuals are [4, 4, 2]. Select parents using random numbers [0.2, 0.6, 0.85].** [3 marks · Practice] #### Given Fitness weights [4, 4, 2]; draws [0.2, 0.6, 0.85]. #### To find Parent row number for each draw. #### Formula and why it applies Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included. #### Solution ##### Step 1. Sum fitness $$ \sum F=4+4+2=10 $$ ##### Step 2. Find probabilities and cumulative bounds | Individual | Fitness | Probability | Cumulative | | --- | --- | --- | --- | | 1 | 4 | 0.4 | 0.4 | | 2 | 4 | 0.4 | 0.8 | | 3 | 2 | 0.2 | 1 | ##### Step 3. Locate each draw | Draw | Selected individual | | --- | --- | | 0.2 | 1 | | 0.6 | 2 | | 0.85 | 3 | **Final answer.** Selected row numbers: [1, 2, 3]. Repeated selection is permitted. **Fitness weights of successive individuals are [0, 3, 1]. Select parents using random numbers [0.2, 0.8].** [3 marks · Practice] #### Given Fitness weights [0, 3, 1]; draws [0.2, 0.8]. #### To find Parent row number for each draw. #### Formula and why it applies Divide by the fitness sum, form cumulative intervals, then locate each random draw. Use half-open intervals [lower,upper); the final upper endpoint 1 is included. #### Solution ##### Step 1. Sum fitness $$ \sum F=0+3+1=4 $$ ##### Step 2. Find probabilities and cumulative bounds | Individual | Fitness | Probability | Cumulative | | --- | --- | --- | --- | | 1 | 0 | 0 | 0 | | 2 | 3 | 0.75 | 0.75 | | 3 | 1 | 0.25 | 1 | ##### Step 3. Locate each draw | Draw | Selected individual | | --- | --- | | 0.2 | 2 | | 0.8 | 3 | **Final answer.** Selected row numbers: [2, 3]. Repeated selection is permitted. ## Other selection methods | Method | Procedure | Why it is used | | --- | --- | --- | | Tournament | Sample a group and take its best candidate | Only objective comparisons are needed | | Rank | Sort candidates; assign weights based on rank | Reduces sensitivity to a single huge raw fitness | | Elitist survival | Preserve top individuals unchanged into the next population | Protects good solutions from destructive variation | Selection is not automatically “replace the weakest with the fittest”. That is one possible classroom reproduction rule. If a problem gives selected parents, a rank rule, tournament contestants or roulette draws, follow that exact instruction. **Yaad rakho:** Probability nikalo → cumulative banao → random number ka interval dekho. --- ## TEE · Crossover and mutation: four complete worked variants Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/ga-core--operators/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/ga-core/operators Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. ## Cutting means splitting the written string The **prefix** is the part before the cut; the **suffix** is the part after it. “10 joined with 00 gives 1000” means writing strings next to each other. It is not the arithmetic addition 10+00. | Operation | Parents / instruction | Construction | Children | | --- | --- | --- | --- | | One-point crossover | 10|11 and 01|00 | First prefix + second tail; second prefix + first tail | 1000 and 0111 | | Two-point crossover | 1|01|10 and 0|10|01 | Exchange only the middle blocks | 11010 and 00101 | | Uniform crossover | A=1010, B=0101, mask=1100 | Child 1 takes A where mask=1 and B where mask=0; child 2 reverses this | 1001 and 0110 | | Mutation | 1000; flip position 3 from left | The third bit is 0; replace it by 1; keep every other bit | 1010 | These crossover operators are alternatives. Apply the operator stated in the question; do not apply all three in one generation. Mutation is a separate step after crossover in the worked exam workflow. ## Crossover combines; mutation changes In **single-point crossover**, cut both parents after the specified bit. Child 1 keeps parent 1’s prefix and takes parent 2’s suffix; child 2 does the reverse. A cut after bit 2 in a four-bit string looks like 10|11. **Cross parents 1011 and 0100 after bit 2. Then flip bit 4 of both children, counting from the left. Decode the final children.** [3 marks · Practice] #### Given Parents 1011, 0100; cut after bit 2; mutation bit 4. #### To find Children after crossover, after mutation and their decimal values. #### Formula and why it applies Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1. #### Solution ##### Step 1. Mark the cut 10 | 11 and 01 | 00 ##### Step 2. Exchange tails 10 + 00 → 1000 01 + 11 → 0111 ##### Step 3. Flip the specified bit | Crossover child | Bit before | Bit after | Mutated child | | --- | --- | --- | --- | | 1000 | 0 | 1 | 1001 | | 0111 | 1 | 0 | 0110 | ##### Step 4. Decode | String | Place-value sum | x | | --- | --- | --- | | 1001 | 8 + 0 + 0 + 1 | 9 | | 0110 | 0 + 4 + 2 + 0 | 6 | **Final answer.** Final children 1001 (9) and 0110 (6). **Cross parents 0011 and 1100 after bit 1. Then flip bit 2 of both children, counting from the left. Decode the final children.** [3 marks · Practice] #### Given Parents 0011, 1100; cut after bit 1; mutation bit 2. #### To find Children after crossover, after mutation and their decimal values. #### Formula and why it applies Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1. #### Solution ##### Step 1. Mark the cut 0 | 011 and 1 | 100 ##### Step 2. Exchange tails 0 + 100 → 0100 1 + 011 → 1011 ##### Step 3. Flip the specified bit | Crossover child | Bit before | Bit after | Mutated child | | --- | --- | --- | --- | | 0100 | 1 | 0 | 0000 | | 1011 | 0 | 1 | 1111 | ##### Step 4. Decode | String | Place-value sum | x | | --- | --- | --- | | 0000 | 0 + 0 + 0 + 0 | 0 | | 1111 | 8 + 4 + 2 + 1 | 15 | **Final answer.** Final children 0000 (0) and 1111 (15). **Cross parents 1110 and 0001 after bit 3. Then flip bit 1 of both children, counting from the left. Decode the final children.** [3 marks · Practice] #### Given Parents 1110, 0001; cut after bit 3; mutation bit 1. #### To find Children after crossover, after mutation and their decimal values. #### Formula and why it applies Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1. #### Solution ##### Step 1. Mark the cut 111 | 0 and 000 | 1 ##### Step 2. Exchange tails 111 + 1 → 1111 000 + 0 → 0000 ##### Step 3. Flip the specified bit | Crossover child | Bit before | Bit after | Mutated child | | --- | --- | --- | --- | | 1111 | 1 | 0 | 0111 | | 0000 | 0 | 1 | 1000 | ##### Step 4. Decode | String | Place-value sum | x | | --- | --- | --- | | 0111 | 0 + 4 + 2 + 1 | 7 | | 1000 | 8 + 0 + 0 + 0 | 8 | **Final answer.** Final children 0111 (7) and 1000 (8). **Cross parents 1010 and 1010 after bit 2. Then flip bit 4 of both children, counting from the left. Decode the final children.** [3 marks · Practice] #### Given Parents 1010, 1010; cut after bit 2; mutation bit 4. #### To find Children after crossover, after mutation and their decimal values. #### Formula and why it applies Exchange only suffixes. Mutation means 0→1 or 1→0; it does not mean set the bit to 1. #### Solution ##### Step 1. Mark the cut 10 | 10 and 10 | 10 ##### Step 2. Exchange tails 10 + 10 → 1010 10 + 10 → 1010 ##### Step 3. Flip the specified bit | Crossover child | Bit before | Bit after | Mutated child | | --- | --- | --- | --- | | 1010 | 0 | 1 | 1011 | | 1010 | 0 | 1 | 1011 | ##### Step 4. Decode | String | Place-value sum | x | | --- | --- | --- | | 1011 | 8 + 0 + 2 + 1 | 11 | | 1011 | 8 + 0 + 2 + 1 | 11 | **Final answer.** Final children 1011 (11) and 1011 (11). **Two-point crossover** exchanges the segment between two cuts. **Uniform crossover** uses a given mask to choose, at each position, which parent contributes a gene. Mutation may flip a bit, perturb a real value, or swap two positions in a permutation. These are different operators; apply the one named in the question. **Yaad rakho:** Crossover = tail exchange. Mutation = specified bit flip. Flip ka matlab 1 bhi 0 banega. --- ## TEE · GA architecture, two-generation flow, constraints and theory Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/ga-core--algorithm/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/ga-core/algorithm Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. | Read this | Meaning | Concrete example | | --- | --- | --- | | Feasible | Obeys every stated constraint | x≤10: x=9 qualifies; x=12 does not | | Penalty | A deduction for violating a constraint | For maximisation, subtract 100×excess | | Elitism | Explicitly preserve a best old individual | Only apply if the question specifies it | | Replacement | Choose what population the next generation uses | Our full-replacement examples use the new children | | Stopping condition | A rule ending the repeated process | Stop after two complete generations | ## A complete GA generation Encode + evaluate → Select parents → Crossover → Mutation → Evaluate childrenreplace and repeat A **simple generational GA** replaces the population in generations using a chosen encoding, selection rule and variation operators. A more general GA may use steady-state replacement, different encodings, adaptive operators or explicit elitism. Always state the choices used in your worked answer. ## What “iteration 2” means here Initial population → first generation’s selection/crossover/mutation → new population → second generation’s selection/crossover/mutation → final population. Stopping after the first children and labelling them iteration 2 is not two full generations. ## Stopping conditions Stop after a fixed generation budget, after reaching a specified acceptable objective, or after little improvement for a stated number of generations. A finite run does not prove the global optimum. ## Constraints A **constraint** restricts allowed solutions, such as 0≤x≤15 or “visit every city once”. Common approaches are valid-by-design encoding, repairing an invalid child, rejecting it, or using a penalty. A penalty must reduce attractiveness: subtract it in maximisation or add it to a minimised cost. Do not silently remove the original constraint because you used more bits. **A maximisation problem has objective f(x)=x² and constraint x≤10. The given penalised fitness is F(x)=x²−100 max(0,x−10). Compare x=9 and x=12.** [3 marks · Practice] #### Given Penalty coefficient 100; bound 10; candidates 9 and 12. #### To find Penalised fitness of both candidates. #### Formula and why it applies Only excess above the bound receives a penalty. #### Solution ##### Step 1. Feasible candidate 9 $$ \max(0,9-10)=0 $$ $$ F(9)=81-100(0)=81 $$ ##### Step 2. Infeasible candidate 12 $$ \max(0,12-10)=2 $$ $$ F(12)=144-100(2)=-56 $$ ##### Step 3. Compare For this maximisation fitness, 81 > −56, so x=9 is preferred. Raw objective alone would misleadingly favour 12. **Final answer.** F(9)=81; F(12)=−56. | Advantages | Limitations | | --- | --- | | Derivative-free; supports many encodings | Many objective evaluations may be costly | | Population explores several candidates | Premature convergence is possible | | Crossover and mutation can create new candidates | Parameter and encoding choices affect results | | Applicable to discontinuous or combinatorial searches | No guaranteed global optimum; variation can worsen fitness | **Yaad rakho:** Do generation = do baar poora cycle. Sirf children banana aur decode karna do alag iterations nahi. --- ## TEE · Two generations with given selection, cut and mutation Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/ga-core--two-generations/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/ga-core/two-generations Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. Swipe across the diagram to read its labels ↔ ![This diagram traces the first pair in generation 1. The full solution also processes the second pair and the entire second generation.](https://deep-learning-study-room.vercel.app/is2/fig/v2_ga_generation.fa7bf86c05.svg) Diagram: This diagram traces the first pair in generation 1. The full solution also processes the second pair and the entire second generation.. Labels: One complete GA generation: read from top to bottom | Selected parents | 1010 × 1100 | Cut after bit 2 | Mark the cut | 10 | 10 11 | 00 | Keep prefix, exchange suffix | After crossover | 1000 1110 | Children are 8 and 14 | Flip the last bit | 1001 1111 | 0 becomes 1 in each child | Decode and evaluate | 9² = 81 15² = 225 | These are the children’s fitness values This diagram traces the first pair in generation 1. The full solution also processes the second pair and the entire second generation. **Maximise $$ f(x)=x^2 $$ over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness.** [3 marks · Practice; complete two-generation solution] #### Given Initial ordered population: [3, 5, 10, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [1, 2, 3, 4] in every generation. Replace the whole population with the four children; no elitism. | Generation | Selection instruction | | --- | --- | | 1 | Current individual row numbers [3, 4, 4, 3] | | 2 | Current individual row numbers [2, 3, 2, 3] | #### To find Both generations: evaluation, selection, crossover, mutation, decoding and comparison. #### Formula and why it applies Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step. #### Solution ##### Step 1. Generation 1: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 3 | 0011 | 3 × 3 = 9 | | 2 | 5 | 0101 | 5 × 5 = 25 | | 3 | 10 | 1010 | 10 × 10 = 100 | | 4 | 12 | 1100 | 12 × 12 = 144 | ##### Step 2. Generation 1: apply the supplied selection order The question directly supplies individual numbers [3, 4, 4, 3]. These refer to the current generation’s row numbers, not their x values. | Mating-pool slot | Chosen individual | x | Chromosome | | --- | --- | --- | --- | | 1 | 3 | 10 | 1010 | | 2 | 4 | 12 | 1100 | | 3 | 4 | 12 | 1100 | | 4 | 3 | 10 | 1010 | ##### Step 3. Generation 1: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 10 | 10 | 11 | 00 | 1000 | 1110 | | 3 & 4 | 11 | 00 | 10 | 10 | 1110 | 1000 | ##### Step 4. Generation 1: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1000 | Flip bit 4 | 1001 | | 2 | 1110 | Flip bit 4 | 1111 | | 3 | 1110 | Flip bit 4 | 1111 | | 4 | 1000 | Flip bit 4 | 1001 | ##### Step 5. Generation 1: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 | | 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 | | 1111 | 8 + 4 + 2 + 1 = 15 | 15² = 225 | | 1001 | 8 + 0 + 0 + 1 = 9 | 9² = 81 | ##### Step 6. Generation 1: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{9+25+100+144}{4}=\frac{278}{4}=69.5 $$ $$ \overline f_{new}=\frac{81+225+225+81}{4}=\frac{612}{4}=153 $$ Best objective: 144 → 225. Higher is better. The next population is [9, 15, 15, 9]. No elitism is imposed, so the best previous individual is not automatically protected. ##### Step 7. Generation 2: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 9 | 1001 | 9 × 9 = 81 | | 2 | 15 | 1111 | 15 × 15 = 225 | | 3 | 15 | 1111 | 15 × 15 = 225 | | 4 | 9 | 1001 | 9 × 9 = 81 | ##### Step 8. Generation 2: apply the supplied selection order The question directly supplies individual numbers [2, 3, 2, 3]. These refer to the current generation’s row numbers, not their x values. | Mating-pool slot | Chosen individual | x | Chromosome | | --- | --- | --- | --- | | 1 | 2 | 15 | 1111 | | 2 | 3 | 15 | 1111 | | 3 | 2 | 15 | 1111 | | 4 | 3 | 15 | 1111 | ##### Step 9. Generation 2: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 11 | 11 | 11 | 11 | 1111 | 1111 | | 3 & 4 | 11 | 11 | 11 | 11 | 1111 | 1111 | ##### Step 10. Generation 2: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1111 | Flip bit 4 | 1110 | | 2 | 1111 | Flip bit 4 | 1110 | | 3 | 1111 | Flip bit 4 | 1110 | | 4 | 1111 | Flip bit 4 | 1110 | ##### Step 11. Generation 2: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | | 1110 | 8 + 4 + 2 + 0 = 14 | 14² = 196 | ##### Step 12. Generation 2: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{81+225+225+81}{4}=\frac{612}{4}=153 $$ $$ \overline f_{new}=\frac{196+196+196+196}{4}=\frac{784}{4}=196 $$ Best objective: 225 → 196. Higher is better. The next population is [14, 14, 14, 14]. No elitism is imposed, so the best previous individual is not automatically protected. **Final answer.** After two generations the ordered population is [14, 14, 14, 14]. Best final objective = 196 at x = [14]. --- ## TEE · Two generations including roulette selection Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/ga-core--roulette-generations/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/ga-core/roulette-generations Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. **Maximise $$ f(x)=x(15-x)+1 $$ over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness.** [3 marks · Practice; complete two-generation solution] #### Given Initial ordered population: [2, 5, 9, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [2, 4] in every generation. Replace the whole population with the four children; no elitism. | Generation | Selection instruction | | --- | --- | | 1 | Random draws [0.1, 0.4, 0.65, 0.9] | | 2 | Random draws [0.2, 0.4, 0.6, 0.95] | #### To find Both generations: evaluation, selection, crossover, mutation, decoding and comparison. #### Formula and why it applies Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Use exactly the supplied selection rule; replacing the weakest by the strongest is not a universal GA step. #### Solution ##### Step 1. Generation 1: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 2 | 0010 | 2 × (15 − 2) + 1 = 27 | | 2 | 5 | 0101 | 5 × (15 − 5) + 1 = 51 | | 3 | 9 | 1001 | 9 × (15 − 9) + 1 = 55 | | 4 | 12 | 1100 | 12 × (15 − 12) + 1 = 37 | ##### Step 2. Generation 1: calculate roulette intervals Fitness weights are nonnegative. The nonnegative objective itself is the fitness in this maximisation example. $$ \sum F=27+51+55+37\approx170 $$ | Individual | Probability: weight ÷ total | Cumulative: old boundary + probability | | --- | --- | --- | | 1 | 27 ÷ 170 ≈ 0.158823529 | 0 + 0.158823529 ≈ 0.158823529 | | 2 | 51 ÷ 170 ≈ 0.3 | 0.158823529 + 0.3 ≈ 0.458823529 | | 3 | 55 ÷ 170 ≈ 0.323529412 | 0.458823529 + 0.323529412 ≈ 0.782352941 | | 4 | 37 ÷ 170 ≈ 0.217647059 | 0.782352941 + 0.217647059 ≈ 1 | ##### Step 3. Generation 1: use the supplied random draws | Random draw | Interval containing it | Selected individual | Selected x | | --- | --- | --- | --- | | 0.1 | 0 ≤ 0.1 < 0.158823529 | 1 | 2 | | 0.4 | 0.158823529 ≤ 0.4 < 0.458823529 | 2 | 5 | | 0.65 | 0.458823529 ≤ 0.65 < 0.782352941 | 3 | 9 | | 0.9 | 0.782352941 ≤ 0.9 < 1 | 4 | 12 | Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers. ##### Step 4. Generation 1: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 00 | 10 | 01 | 01 | 0001 | 0110 | | 3 & 4 | 10 | 01 | 11 | 00 | 1000 | 1101 | ##### Step 5. Generation 1: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 0001 | No mutation | 0001 | | 2 | 0110 | Flip bit 4 | 0111 | | 3 | 1000 | No mutation | 1000 | | 4 | 1101 | Flip bit 4 | 1100 | ##### Step 6. Generation 1: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 0001 | 0 + 0 + 0 + 1 = 1 | 1(15−1)+1 = 15 | | 0111 | 0 + 4 + 2 + 1 = 7 | 7(15−7)+1 = 57 | | 1000 | 8 + 0 + 0 + 0 = 8 | 8(15−8)+1 = 57 | | 1100 | 8 + 4 + 0 + 0 = 12 | 12(15−12)+1 = 37 | ##### Step 7. Generation 1: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{27+51+55+37}{4}=\frac{170}{4}=42.5 $$ $$ \overline f_{new}=\frac{15+57+57+37}{4}=\frac{166}{4}=41.5 $$ Best objective: 55 → 57. Higher is better. The next population is [1, 7, 8, 12]. No elitism is imposed, so the best previous individual is not automatically protected. ##### Step 8. Generation 2: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 1 | 0001 | 1 × (15 − 1) + 1 = 15 | | 2 | 7 | 0111 | 7 × (15 − 7) + 1 = 57 | | 3 | 8 | 1000 | 8 × (15 − 8) + 1 = 57 | | 4 | 12 | 1100 | 12 × (15 − 12) + 1 = 37 | ##### Step 9. Generation 2: calculate roulette intervals Fitness weights are nonnegative. The nonnegative objective itself is the fitness in this maximisation example. $$ \sum F=15+57+57+37\approx166 $$ | Individual | Probability: weight ÷ total | Cumulative: old boundary + probability | | --- | --- | --- | | 1 | 15 ÷ 166 ≈ 0.090361446 | 0 + 0.090361446 ≈ 0.090361446 | | 2 | 57 ÷ 166 ≈ 0.343373494 | 0.090361446 + 0.343373494 ≈ 0.43373494 | | 3 | 57 ÷ 166 ≈ 0.343373494 | 0.43373494 + 0.343373494 ≈ 0.777108434 | | 4 | 37 ÷ 166 ≈ 0.222891566 | 0.777108434 + 0.222891566 ≈ 1 | ##### Step 10. Generation 2: use the supplied random draws | Random draw | Interval containing it | Selected individual | Selected x | | --- | --- | --- | --- | | 0.2 | 0.090361446 ≤ 0.2 < 0.43373494 | 2 | 7 | | 0.4 | 0.090361446 ≤ 0.4 < 0.43373494 | 2 | 7 | | 0.6 | 0.43373494 ≤ 0.6 < 0.777108434 | 3 | 8 | | 0.95 | 0.777108434 ≤ 0.95 < 1 | 4 | 12 | Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers. ##### Step 11. Generation 2: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 01 | 11 | 01 | 11 | 0111 | 0111 | | 3 & 4 | 10 | 00 | 11 | 00 | 1000 | 1100 | ##### Step 12. Generation 2: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 0111 | No mutation | 0111 | | 2 | 0111 | Flip bit 4 | 0110 | | 3 | 1000 | No mutation | 1000 | | 4 | 1100 | Flip bit 4 | 1101 | ##### Step 13. Generation 2: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 0111 | 0 + 4 + 2 + 1 = 7 | 7(15−7)+1 = 57 | | 0110 | 0 + 4 + 2 + 0 = 6 | 6(15−6)+1 = 55 | | 1000 | 8 + 0 + 0 + 0 = 8 | 8(15−8)+1 = 57 | | 1101 | 8 + 4 + 0 + 1 = 13 | 13(15−13)+1 = 27 | ##### Step 14. Generation 2: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{15+57+57+37}{4}=\frac{166}{4}=41.5 $$ $$ \overline f_{new}=\frac{57+55+57+27}{4}=\frac{196}{4}=49 $$ Best objective: 57 → 57. Higher is better. The next population is [7, 6, 8, 13]. No elitism is imposed, so the best previous individual is not automatically protected. **Final answer.** After two generations the ordered population is [7, 6, 8, 13]. Best final objective = 57 at x = [7, 8]. --- ## TEE · GA minimisation: transform cost before roulette selection Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/ga-core--minimisation/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/ga-core/minimisation Candidate → fitness → encoding → selection → crossover → mutation → two complete generations. Roulette favours a larger weight. A minimisation objective instead favours a smaller cost. The stated transformation F=1/(1+cost) reverses their order for nonnegative costs: cost 0 → weight 1/1=1; cost 3 → 1/4=.25; cost 24 → 1/25=.04. The +1 prevents division by zero. This is one specified transformation, not a universal rule for every minimisation problem. **Minimise $$ f(x)=(x-7)^2 $$ over unsigned 4-bit integers using the supplied GA settings. Perform two complete generations and compare the population fitness.** [3 marks · Practice; complete two-generation solution] #### Given Initial ordered population: [2, 4, 8, 12]. Use 4 bits, crossover after bit 2, pair mating-pool slots 1–2 and 3–4. Flip bit 4 of children [2, 4] in every generation. Replace the whole population with the four children; no elitism. | Generation | Selection instruction | | --- | --- | | 1 | Random draws [0.1, 0.4, 0.65, 0.9] | | 2 | Random draws [0.2, 0.4, 0.6, 0.95] | #### To find Both generations: evaluation, selection, crossover, mutation, decoding and comparison. #### Formula and why it applies Encoding represents x; it does not change the objective. Crossover exchanges tails. Mutation flips specified bits. Fitness must be recomputed after mutation. Roulette uses Fᵢ/ΣF. For minimisation here, use Fᵢ=1/(1+f(xᵢ)); do not directly favour large costs. #### Solution ##### Step 1. Generation 1: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 2 | 0010 | (2 − 7)² = 25 | | 2 | 4 | 0100 | (4 − 7)² = 9 | | 3 | 8 | 1000 | (8 − 7)² = 1 | | 4 | 12 | 1100 | (12 − 7)² = 25 | ##### Step 2. Generation 1: calculate roulette intervals Fitness weights are nonnegative. For this minimisation example use F=1/(1+objective), so a smaller cost receives a larger weight. | Individual | Cost | Fitness calculation | | --- | --- | --- | | 1 | 25 | 1 / (1 + 25) = 1 / 26 ≈ 0.038461538 | | 2 | 9 | 1 / (1 + 9) = 1 / 10 ≈ 0.1 | | 3 | 1 | 1 / (1 + 1) = 1 / 2 ≈ 0.5 | | 4 | 25 | 1 / (1 + 25) = 1 / 26 ≈ 0.038461538 | $$ \sum F=0.038461538+0.1+0.5+0.038461538\approx0.676923077 $$ | Individual | Probability: weight ÷ total | Cumulative: old boundary + probability | | --- | --- | --- | | 1 | 0.038461538 ÷ 0.676923077 ≈ 0.056818182 | 0 + 0.056818182 ≈ 0.056818182 | | 2 | 0.1 ÷ 0.676923077 ≈ 0.147727273 | 0.056818182 + 0.147727273 ≈ 0.204545455 | | 3 | 0.5 ÷ 0.676923077 ≈ 0.738636364 | 0.204545455 + 0.738636364 ≈ 0.943181818 | | 4 | 0.038461538 ÷ 0.676923077 ≈ 0.056818182 | 0.943181818 + 0.056818182 ≈ 1 | ##### Step 3. Generation 1: use the supplied random draws | Random draw | Interval containing it | Selected individual | Selected x | | --- | --- | --- | --- | | 0.1 | 0.056818182 ≤ 0.1 < 0.204545455 | 2 | 4 | | 0.4 | 0.204545455 ≤ 0.4 < 0.943181818 | 3 | 8 | | 0.65 | 0.204545455 ≤ 0.65 < 0.943181818 | 3 | 8 | | 0.9 | 0.204545455 ≤ 0.9 < 0.943181818 | 3 | 8 | Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers. ##### Step 4. Generation 1: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 01 | 00 | 10 | 00 | 0100 | 1000 | | 3 & 4 | 10 | 00 | 10 | 00 | 1000 | 1000 | ##### Step 5. Generation 1: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 0100 | No mutation | 0100 | | 2 | 1000 | Flip bit 4 | 1001 | | 3 | 1000 | No mutation | 1000 | | 4 | 1000 | Flip bit 4 | 1001 | ##### Step 6. Generation 1: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 0100 | 0 + 4 + 0 + 0 = 4 | (4−7)² = 9 | | 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 | | 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 | | 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 | ##### Step 7. Generation 1: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{25+9+1+25}{4}=\frac{60}{4}=15 $$ $$ \overline f_{new}=\frac{9+4+1+4}{4}=\frac{18}{4}=4.5 $$ Best objective: 1 → 1. Lower is better. The next population is [4, 9, 8, 9]. No elitism is imposed, so the best previous individual is not automatically protected. ##### Step 8. Generation 2: evaluate and encode the current population | Individual | x | Binary | Objective calculation | | --- | --- | --- | --- | | 1 | 4 | 0100 | (4 − 7)² = 9 | | 2 | 9 | 1001 | (9 − 7)² = 4 | | 3 | 8 | 1000 | (8 − 7)² = 1 | | 4 | 9 | 1001 | (9 − 7)² = 4 | ##### Step 9. Generation 2: calculate roulette intervals Fitness weights are nonnegative. For this minimisation example use F=1/(1+objective), so a smaller cost receives a larger weight. | Individual | Cost | Fitness calculation | | --- | --- | --- | | 1 | 9 | 1 / (1 + 9) = 1 / 10 ≈ 0.1 | | 2 | 4 | 1 / (1 + 4) = 1 / 5 ≈ 0.2 | | 3 | 1 | 1 / (1 + 1) = 1 / 2 ≈ 0.5 | | 4 | 4 | 1 / (1 + 4) = 1 / 5 ≈ 0.2 | $$ \sum F=0.1+0.2+0.5+0.2\approx1 $$ | Individual | Probability: weight ÷ total | Cumulative: old boundary + probability | | --- | --- | --- | | 1 | 0.1 ÷ 1 ≈ 0.1 | 0 + 0.1 ≈ 0.1 | | 2 | 0.2 ÷ 1 ≈ 0.2 | 0.1 + 0.2 ≈ 0.3 | | 3 | 0.5 ÷ 1 ≈ 0.5 | 0.3 + 0.5 ≈ 0.8 | | 4 | 0.2 ÷ 1 ≈ 0.2 | 0.8 + 0.2 ≈ 1 | ##### Step 10. Generation 2: use the supplied random draws | Random draw | Interval containing it | Selected individual | Selected x | | --- | --- | --- | --- | | 0.2 | 0.1 ≤ 0.2 < 0.3 | 2 | 9 | | 0.4 | 0.3 ≤ 0.4 < 0.8 | 3 | 8 | | 0.6 | 0.3 ≤ 0.6 < 0.8 | 3 | 8 | | 0.95 | 0.8 ≤ 0.95 < 1 | 4 | 9 | Intervals are left-closed and right-open, except that a draw of exactly 1 selects the last interval. No draw here is on a boundary. Draws select parents with replacement: selecting a parent once does not remove it. Calculations use unrounded weights; the table displays rounded numbers. ##### Step 11. Generation 2: crossover after bit 2 Keep each prefix and exchange the suffixes. Bit positions are counted from the left starting at 1. | Pool slots paired | Parent A | Parent B | Child A | Child B | | --- | --- | --- | --- | --- | | 1 & 2 | 10 | 01 | 10 | 00 | 1000 | 1001 | | 3 & 4 | 10 | 00 | 10 | 01 | 1001 | 1000 | ##### Step 12. Generation 2: mutate only the specified bits | Child | Before mutation | Instruction | After mutation | | --- | --- | --- | --- | | 1 | 1000 | No mutation | 1000 | | 2 | 1001 | Flip bit 4 | 1000 | | 3 | 1001 | No mutation | 1001 | | 4 | 1000 | Flip bit 4 | 1001 | ##### Step 13. Generation 2: decode and evaluate every child | Chromosome | Decode using place values | Objective | | --- | --- | --- | | 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 | | 1000 | 8 + 0 + 0 + 0 = 8 | (8−7)² = 1 | | 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 | | 1001 | 8 + 0 + 0 + 1 = 9 | (9−7)² = 4 | ##### Step 14. Generation 2: compare and form the next population Add the objective scores in each population, then divide by its number of individuals. $$ \overline f_{old}=\frac{9+4+1+4}{4}=\frac{18}{4}=4.5 $$ $$ \overline f_{new}=\frac{1+1+4+4}{4}=\frac{10}{4}=2.5 $$ Best objective: 1 → 1. Lower is better. The next population is [8, 8, 9, 9]. No elitism is imposed, so the best previous individual is not automatically protected. **Final answer.** After two generations the ordered population is [8, 8, 9, 9]. Best final objective = 1 at x = [8]. --- ## TEE · GA approach to the travelling salesman problem Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u3-ga--ga-tsp/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u3-ga/ga-tsp Unit 3 · Evolutionary intelligence and genetic algorithms · 10-mark GA numerical in both 2025–26 papers ## Read the route before encoding it A **city** is a location. An **edge** connects two cities and has a travel distance. A **tour** visits each city exactly once and returns to its start. A **permutation** is an ordered arrangement without omissions or repetitions: [A,B,D,C] is valid for four cities; [A,B,B,C] is not. A distance matrix uses row=departure city and column=arrival city. Entry dAB=5 means the A→B leg costs 5. In a symmetric problem dBA=5 too. A diagonal 0 means no distance from a city to itself, not a free trip to another city. For A→B→D→C→A, add dAB+dBD+dDC+dCA. Factorial counts arrangements: 3!=3×2×1=6. With one of four cities fixed as the start, the other three can be ordered in six ways. In a symmetric TSP reverse-direction tours have equal length, although they are different written orders. **In one line:** encode a tour as a permutation of cities, use fitness = 1/(tour length), and use crossover and mutation operators that always produce valid tours. ## The problem **Travelling salesman problem (TSP):** given n cities and the distance between every pair, find the shortest route that visits each city exactly once and returns to the start. For a symmetric TSP with a fixed starting city, reversal-equivalent tours counted once, and n ≥ 3, the number of distinct tours is $$ (n-1)!/2 $$ (12 for 5 cities, about $$ 6\times10^{16} $$ for 20 cities), so exhaustive search quickly becomes impossible. ## How the GA is adapted | GA part | For TSP | | --- | --- | | Encoding | **Permutation**: C A E B D means C → A → E → B → D → back to C. | | Fitness | $$ F=1/L $$ , where $$ L=\sum d(c_i,c_{i+1})+d(c_n,c_1) $$ includes the return edge. Shorter tour = higher fitness. | | Selection | Roulette wheel on F, or tournament. | | Crossover | Order crossover (OX), partially mapped crossover (PMX) or cycle crossover (CX). | | Mutation | Swap two cities, or invert a segment (inversion), with small probability. | **Why ordinary crossover fails.** Single-point crossover of A B C | D E and C E A | B D gives A B C B D: city B twice and E missing, which is not a tour. Permutation operators avoid this. ## Order-preserving crossover: the stated left-to-right variant 1. Copy a randomly chosen segment from Parent 1 into the child, in the same positions. 2. Fill the remaining positions from left to right with the cities of Parent 2 in their order, skipping cities already copied. This variant fills empty positions from left to right; standard cyclic OX instead starts scanning and filling after the second cut. Do not mix the two conventions. Example: P1 = 2 4 [1 8 3] 5 7 6, P2 = 7 5 6 8 3 4 2 1. Copy 1 8 3; P2 order without them is 7 5 6 4 2 → child = **7 5 1 8 3 6 4 2**. (One GA slide prints 4 2 1 8 3 7 5 6, which uses a different fill order; state your rule in the exam.) **Question. Five cities with distances AB = 2, AC = 9, AD = 10, AE = 7, BC = 6, BD = 4, BE = 3, CD = 8, CE = 5, DE = 1. Parents are ABECD and ADBEC. (a) Find their lengths and fitness. (b) Apply the left-to-right order-preserving crossover variant described above with the segment at positions 2–3. (c) Apply swap mutation to positions 3 and 4 of the child. Comment.** [practice · 10 marks] #### Given All ten symmetric distances in the question; parents ABECD and ADBEC. Use the explicitly stated left-to-right order-preserving crossover at positions 2–3, then swap positions 3 and 4. #### To find Parent lengths/fitness, the crossover child, and the mutated child. #### Formula and why it applies Add all edges including the return to the start. A shorter positive length gives a larger reciprocal fitness. $$ L=\sum_{i=1}^{n-1}d(c_i,c_{i+1})+d(c_n,c_1),\quad F=1/L $$ #### Solution ##### Step 1. Parent lengths (include the return edge) $$ L(ABECD)=2+3+5+8+10=28,\quad F=1/28=0.0357 $$ $$ L(ADBEC)=10+4+3+5+9=31,\quad F=1/31=0.0323 $$ ##### Step 2. Order crossover Copy positions 2–3 from P1: \_ B E \_ \_. P2's order without B, E: A, D, C → fill positions 1, 4, 5. Swipe across the diagram to read its labels ↔ ![Order crossover keeps a valid tour: copy B E from P1, fill A, D, C in P2's order](https://deep-learning-study-room.vercel.app/is2/fig/ga_ox.8fd21cbc51.svg) Diagram: Order crossover keeps a valid tour: copy B E from P1, fill A, D, C in P2's order. Labels: Order crossover keeps a valid tour: copy B E from P1, fill A, D, C in P2's order | Parent 1 | A | B | E | C | D | length 28 | Parent 2 | A | D | B | E | C | length 31 | Child (OX) | A | B | E | D | C | length 23 | After swap | A | B | D | E | C | length 21 (shortest) $$ \text{Child}=ABEDC,\quad L=2+3+1+8+9=23,\quad F=0.0435 $$ ##### Step 3. Swap mutation (positions 3 and 4) $$ ABEDC\to ABDEC,\quad L=2+4+1+5+9=21,\quad F=0.0476 $$ **Final answer.** Child ABEDC (23) is shorter than both parents (28, 31) because it kept P1's short edge B–E and gained D–E (1). The mutated child ABDEC (21) is the shortest possible tour for this graph (checked against all 12 tours). ## Two more permutation crossovers (textbook, not in your slides) **PMX** (partially mapped crossover): swap a middle segment and use the position mapping to repair duplicates. With the slide parents and segment 3–5 the mapping is 1↔6, giving children 7 5 1 8 3 4 2 6 and 2 4 6 8 3 5 7 1. **Cycle crossover (CX)**: find cycles of positions between the parents and take alternate cycles from each parent; every city keeps a position it had in one of the parents. ## Algorithm (from your slides) 1. Initialise random tours (permutations). 2. Evaluate fitness = 1/L. 3. Select parents (roulette or tournament). 4. Apply OX to each pair. 5. Apply swap mutation with small probability. 6. Replace the population. 7. Repeat until the stopping condition; return the shortest tour. **Yaad rakho:** TSP mein chromosome = shehron ka order. Fitness = 1/length. Normal crossover se shehar repeat honge, isliye OX. --- ## TEE · How pheromone feedback guides ant-colony search Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u3-aco--how-ants-work/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u3-aco/how-ants-work Unit 3 · Ant colony optimization for TSP · asked as a 10-mark numerical in both 2025–26 papers **In one line:** ants leave pheromone as they walk; shorter paths get refreshed more often, so their scent stays stronger, so more ants choose them: positive feedback can favour shorter paths. ## Definition **Ant colony optimization (ACO)** is a population-based, probabilistic metaheuristic, introduced by **Marco Dorigo in 1992**, that finds good paths through a graph by imitating how ants communicate with **pheromone** trails. It is especially suited to routing problems such as the **travelling salesman problem (TSP)**. Swipe across the diagram to read its labels ↔ ![Why ants find the shorter path](https://deep-learning-study-room.vercel.app/is2/fig/aco_bridge.3fc4d22f6d.svg) Diagram: Why ants find the shorter path. Labels: Why ants find the shorter path | Nest | Food | short path: ants return sooner, pheromone builds up | long path: fewer trips per minute, pheromone evaporates ## How shorter paths can gain an advantage 1. **Random exploration.** At first, ants take all paths roughly equally and drop pheromone as they walk. 2. **Shorter path → faster round trip.** Ants on the short path return sooner, so that path receives pheromone more often per minute. 3. **Evaporation.** Pheromone evaporates. On the long path it has more time to fade between visits. 4. **Positive feedback.** Later ants prefer stronger scent, add even more pheromone, which can reinforce a useful route. A finite run is not guaranteed to discover the global shortest path. This indirect communication through changes in the environment is called **stigmergy**: no ant tells another where to go; the trail does. ## From real ants to artificial ants | Real ants | Artificial ants in ACO | | --- | --- | | Walk from nest to food | Build a solution by moving from node to node in a graph | | Drop pheromone on the path used | Increase the pheromone value $$ \tau_{ij} $$ on every edge used | | Pheromone evaporates | All $$ \tau_{ij} $$ shrink by the evaporation rate $$ \rho $$ each iteration | | Prefer stronger scent | Choose the next node with a probability based on pheromone and distance | | Colony reinforces frequently used paths | The algorithm searches for low-cost tours; it may stagnate on a suboptimal tour | ## TSP as a graph | TSP | ACO | | --- | --- | | City | Node | | Distance between two cities | Edge weight $$ d_{ij} $$ | | Round trip visiting every city once | One ant's path | | Tour length | Sum of the edges used, **including the edge back to the start** | **Yaad rakho:** Chhota raasta = jaldi wapas = zyada pheromone = aur cheentiyan. Lamba raasta = pheromone ud jaata hai. --- ## TEE · ACO formulas: transition probability, city choice and pheromone update Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u3-aco--formulas/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u3-aco/formulas Unit 3 · Ant colony optimization for TSP · asked as a 10-mark numerical in both 2025–26 papers **From city A an ant may choose B or C. Distances are AB=2 and AC=4. Pheromones are both 1 and α=β=1. Find the next-city probabilities.** [2 marks · Worked example] #### Given Current city A; allowed next cities B and C. Distances dAB=2 and dAC=4. Pheromones τAB=τAC=1, and exponents α=β=1. Visited cities are excluded. #### To find P(A→B) and P(A→C). #### Formula and why it applies $$ \eta_{ij}=1/d_{ij}\qquad P_{ij}=\tau_{ij}\eta_{ij}/\sum_{l\in allowed}\tau_{il}\eta_{il} $$ Shorter distance gives a larger reciprocal. Normalise the desirability weights to probabilities. #### Solution ##### Step 1. Take reciprocal distances $$ \eta_{AB}=1/2=0.5\qquad\eta_{AC}=1/4=0.25 $$ ##### Step 2. Multiply by pheromone $$ w_B=1\times0.5=0.5\qquad w_C=1\times0.25=0.25 $$ ##### Step 3. Add allowed weights $$ 0.5+0.25=0.75 $$ ##### Step 4. Divide each weight by their total $$ P_{AB}=0.5/0.75=2/3 $$ $$ P_{AC}=0.25/0.75=1/3 $$ **Final answer.** B has probability 2/3; C has probability 1/3. These are probabilities, not a rule forcing the nearest city every time. Evaporation ρ=.5 means retain 1−.5=.5 of the old pheromone. A completed tour of length 20 deposits Q/20 on each edge it used; with Q=1 this is .05. An unused edge receives no deposit from that ant. Use complete tour length, not the length of the individual edge, in this denominator. **In one line:** an ant picks the next city with probability ∝ pheromone × (1/distance); after all ants finish, pheromone evaporates and each ant adds Q/(its tour length) to every edge it used. ## 1. Transition probability An ant $$ k $$ at city $$ i $$ moves to an unvisited city $$ j $$ with probability $$ P^{k}_{ij}=\frac{\tau_{ij}^{\,\alpha}\,\eta_{ij}^{\,\beta}}{\displaystyle\sum_{l\in\text{allowed}}\tau_{il}^{\,\alpha}\,\eta_{il}^{\,\beta}},\qquad \eta_{ij}=\frac1{d_{ij}} $$ - $$ \tau_{ij} $$ : pheromone on edge $$ i $$ – $$ j $$ ; $$ \eta_{ij} $$ : visibility (heuristic), the inverse of distance. - $$ \alpha $$ : weight of pheromone; $$ \beta $$ : weight of distance. In every class example $$ \alpha=\beta=1 $$ . - “allowed” = cities this ant has not visited yet. With all $$ \tau=1 $$ and $$ \alpha=\beta=1 $$ , the probabilities are simply proportional to $$ 1/d $$ . ## 2. Choosing the city with a random number (cumulative rule) 1. List the allowed cities **in alphabetical (city) order**, not sorted by probability. 2. Write their cumulative probabilities. 3. Pick the **first** city whose cumulative value is ≥ the random number $$ r $$ . 4. When only one city is left, the ant goes there without using a random number, then returns to the start. **Exam trap.** In Final 2025 Q7b, from city C the candidates are B (0.4) and D (0.6). In city order the ranges are B: 0–0.4, D: 0.4–1, so r = 0.8 selects D. If you sort by probability (D first) you would pick B and get the wrong tour. The official key uses city order. ## 3. Pheromone update (after all ants finish) $$ \tau_{ij}\leftarrow(1-\rho)\,\tau_{ij}+\sum_{k=1}^{m}\Delta\tau^{k}_{ij},\qquad \Delta\tau^{k}_{ij}=\begin{cases}Q/L_k&\text{if ant }k\text{ used edge }(i,j)\\0&\text{otherwise}\end{cases} $$ - $$ \rho $$ : evaporation rate (fraction lost). Class examples use $$ \rho=0.5 $$ . $$ Q $$ : deposit constant, usually 1. - $$ L_k $$ : length of ant $$ k $$ 's complete tour **including the return edge**. Shorter tours deposit more. - Edges are undirected: update $$ \tau_{ij} $$ and $$ \tau_{ji} $$ together (the matrix stays symmetric). - The update happens once per iteration, after every ant has finished; within an iteration all ants use the same pheromone values. **When the question gives no ρ or Q** (both 2025–26 papers): write “Assume ρ = 0.5 and Q = 1, as in the class example” and proceed. That is exactly what the Final 2025 key does. ## 4. Algorithm (Ant System) 1. Initialise parameters: number of ants, α, β, ρ, Q, initial pheromone $$ \tau_0 $$ , maximum iterations. 2. Place each ant on its start city. 3. Each ant repeatedly chooses the next city with the transition probability until it has visited all cities, then returns to the start. 4. Compute each tour length $$ L_k $$ and record the best tour. 5. Update pheromone: evaporation, then deposit $$ Q/L_k $$ . 6. Repeat steps 2–5 until the iteration limit; output the best tour. | Parameter | Meaning | Typical value | | --- | --- | --- | | m | Number of ants | number of cities (or 20) | | α | Pheromone importance | 1 | | β | Distance (visibility) importance | 1 (often 2–5 in practice) | | ρ | Evaporation rate | 0.5 in class numericals (0.05–0.5 in practice) | | Q | Deposit constant | 1 (100 in one slide example) | | $$ \tau_0 $$ | Initial pheromone | 1 in exam questions | **Yaad rakho:** Probability = τ × (1/d), phir normalize. City order mein cumulative, pehla ≥ r. Update = ½ purana + Q/L har us edge par jo use hua (wapas wala edge bhi). --- ## TEE · Solved PYQ: ACO for a 4-city TSP (Final 2025, 10 marks) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u3-aco--final-2025/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u3-aco/final-2025 Unit 3 · Ant colony optimization for TSP · asked as a 10-mark numerical in both 2025–26 papers **Question. For the four-city travelling salesman problem, the distance and pheromone matrices (A, B, C, D) are given. Apply Ant Colony Optimization. Calculate the path travelled by 2 ants to cover all cities starting from city A. Random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration.** [10 marks · Final Nov 2025 Q7b] #### Given $$ D=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&4&2&4\\B&4&0&3&6\\C&2&3&0&2\\D&4&6&2&0\end{array}\qquad \tau=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&1&1&1\\B&1&0&1&1\\C&1&1&0&1\\D&1&1&1&0\end{array} $$ Assume $$ \alpha=\beta=1 $$ , $$ \rho=0.5 $$ , $$ Q=1 $$ (class convention; not stated in the question). #### To find Tour of each ant, tour lengths, updated pheromone matrix. Swipe across the diagram to read its labels ↔ ![Final 2025 Q7b: distances (all pheromones start at 1)](https://deep-learning-study-room.vercel.app/is2/fig/aco_final_graph.ac3cd4aeac.svg) Diagram: Final 2025 Q7b: distances (all pheromones start at 1). Labels: Final 2025 Q7b: distances (all pheromones start at 1) | 4 | 2 | 4 | 3 | 6 | 2 | A | B | C | D #### Formula and why it applies $$ P_{ij}=\frac{\tau_{ij}\,\eta_{ij}}{\sum_{l\in\text{allowed}}\tau_{il}\,\eta_{il}},\qquad\eta_{ij}=\frac1{d_{ij}} $$ $$ \tau_{ij}^{new}=(1-\rho)\tau_{ij}+\sum_{k:\,ij\in\mathrm{tour}_k}\frac{Q}{L_k} $$ The sum includes **only ants whose completed tour contains edge ij**. An unused edge receives no deposit but still evaporates. Here distances are symmetric, so a traversed edge updates both symmetric matrix cells. Both ants construct their tours using the old pheromones; update after both finish, including their return to A. #### Solution ##### Step 1. Probabilities from A (both ants start at A) $$ \tau\eta:\;A\to B=1\times\tfrac14=0.25,\quad A\to C=1\times\tfrac12=0.5,\quad A\to D=1\times\tfrac14=0.25;\quad\text{sum}=1 $$ | Move | P | Cumulative | Range | | --- | --- | --- | --- | | A→B | 0.25 | 0.25 | 0–0.25 | | A→C | 0.50 | 0.75 | 0.25–0.75 | | A→D | 0.25 | 1.00 | 0.75–1 | Ant 1: r = 0.3 → **C**. Ant 2: r = 0.5 → **C**. ##### Step 2. Probabilities from C (B and D left) $$ C\to B=\tfrac13=0.3333,\quad C\to D=\tfrac12=0.5;\quad\text{sum}=0.8333\Rightarrow P_{CB}=0.4,\;P_{CD}=0.6 $$ Cumulative in city order: B 0–0.4, D 0.4–1. Ant 1: r = 0.8 → **D**. Ant 2: r = 0.7 → **D**. ##### Step 3. Complete the tours Only B is left, then return to A. $$ \text{Ant 1}=\text{Ant 2}:\;A\to C\to D\to B\to A,\qquad L=2+2+6+4=14 $$ Swipe across the diagram to read its labels ↔ ![Both ants: A → C → D → B → A, length 14](https://deep-learning-study-room.vercel.app/is2/fig/aco_final_tour.1dbfdb8b98.svg) Diagram: Both ants: A → C → D → B → A, length 14. Labels: Both ants: A → C → D → B → A, length 14 | 4 | 2 | 4 | 3 | 6 | 2 | A | B | C | D ##### Step 4. Pheromone update Each ant deposits $$ Q/L=1/14=0.0714 $$ on each edge of its tour; two ants used the same four edges (AC, CD, DB, BA), so each gets $$ 2/14=0.1429 $$ . $$ \text{used edge: }(1-0.5)(1)+0.1429=0.6429\qquad\text{unused edge (AD, BC): }(1-0.5)(1)+0=0.5 $$ $$ \tau_{\text{new}}=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&0.6429&0.6429&0.5\\B&0.6429&0&0.5&0.6429\\C&0.6429&0.5&0&0.6429\\D&0.5&0.6429&0.6429&0\end{array} $$ **Final answer.** Both ants: A → C → D → B → A, length 14. Updated pheromone: 0.6429 on AB, AC, CD, BD; 0.5 on AD and BC. In the next iteration the ants will prefer the edges of this tour. **Official key note.** The key's matrix shows 0.643 on AC, CD and BD but leaves AB at 0.5, i.e. it forgot to deposit on the return edge B → A even though it counted its length (4) in L = 14. Your slides deposit on the return edge, so AB = 0.6429 is correct. If you wish, write one line: “the return edge B–A is part of the tour, so it also receives Q/L.” Official marking: path of Ant 1 (3) + path of Ant 2 (3) + pheromone matrix (4). --- ## TEE · Solved PYQ: ACO for a 4-city TSP (Re-exam 2026, 10 marks) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u3-aco--reexam-2026/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u3-aco/reexam-2026 Unit 3 · Ant colony optimization for TSP · asked as a 10-mark numerical in both 2025–26 papers **Question. For the four-city travelling salesman problem the distance matrix and pheromone matrix are given (all pheromones 1). Apply Ant Colony Optimization. Calculate the path travelled by 2 ants starting from city A; random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration.** [10 marks · Re-exam Jan 2026 Q6a] #### Given $$ D=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&1&2&3\\B&1&0&4&5\\C&2&4&0&6\\D&3&5&6&0\end{array},\qquad\tau_{ij}=1\;(i\ne j) $$ Assume $$ \alpha=\beta=1,\;\rho=0.5,\;Q=1 $$ . #### To find Both tours, their lengths and the pheromone matrix after one iteration. #### Formula and why it applies $$ P_{ij}=\frac{\tau_{ij}^{\alpha}(1/d_{ij})^\beta}{\sum_{k\in allowed}\tau_{ik}^{\alpha}(1/d_{ik})^\beta} $$ $$ \tau_{ij}^{new}=(1-\rho)\tau_{ij}+\sum_{k:\,ij\in tour_k}\frac{Q}{L_k} $$ Use unvisited cities in alphabetical order for cumulative selection. Every ant completes its tour using old pheromones; then evaporate and deposit on each undirected edge, including the return edge. #### Solution ##### Step 1. From A $$ \eta:\;B=1,\;C=0.5,\;D=0.3333;\;\text{sum}=1.8333\Rightarrow P=0.5455,\;0.2727,\;0.1818 $$ Cumulative: B 0–0.5455, C 0.5455–0.8182, D 0.8182–1. Ant 1 (0.3) → **B**; Ant 2 (0.5) → **B**. ##### Step 2. From B (C and D left) $$ \eta:\;C=0.25,\;D=0.2;\;\text{sum}=0.45\Rightarrow P_{BC}=0.5556,\;P_{BD}=0.4444 $$ Cumulative: C 0–0.5556, D 0.5556–1. Ant 1 (0.8) → **D**; Ant 2 (0.7) → **D**. ##### Step 3. Complete the tours $$ A\to B\to D\to C\to A,\qquad L=1+5+6+2=14\;\text{(both ants)} $$ Swipe across the diagram to read its labels ↔ ![Re-exam 2026 Q6a: both ants A → B → D → C → A, length 14](https://deep-learning-study-room.vercel.app/is2/fig/aco_reexam_graph.02195c6ee9.svg) Diagram: Re-exam 2026 Q6a: both ants A → B → D → C → A, length 14. Labels: Re-exam 2026 Q6a: both ants A → B → D → C → A, length 14 | 1 | 2 | 3 | 4 | 5 | 6 | A | B | C | D ##### Step 4. Pheromone update Used edges AB, BD, DC, CA get $$ 0.5(1)+2\times\frac1{14}=0.6429 $$ ; unused edges AD and BC become $$ 0.5 $$ . $$ \tau_{\text{new}}=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&0.6429&0.6429&0.5\\B&0.6429&0&0.5&0.6429\\C&0.6429&0.5&0&0.6429\\D&0.5&0.6429&0.6429&0\end{array} $$ **Final answer.** Both ants: A → B → D → C → A, length 14. Pheromone 0.6429 on AB, BD, CD, AC; 0.5 on AD, BC. **Official key note.** This key applies *no evaporation* ( $$ 1+0.1428=1.1428 $$ ) and its matrix is inconsistent: it increases the unused edge A–D, leaves the used return edge C–A at 1, and gives B–D and D–B different values. If your examiner follows the no-evaporation idea, the consistent version is 1.1429 on AB, BD, CD, AC and 1 on AD, BC. The paper does not specify ρ. State the convention you use; the resulting matrix depends on it. Grading depends on the examiner’s intended convention. --- ## TEE · Practice: the lecture's Ant System example (two different tours) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u3-aco--lecture-example/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u3-aco/lecture-example Unit 3 · Ant colony optimization for TSP · asked as a 10-mark numerical in both 2025–26 papers **Why do this:** in the papers both ants happened to take the same tour. In the class example they take different tours, which shows how the deposit adds up when two lengths differ. **Question. Distances: AB = 5, AC = 6, AD = 7, BC = 2, BD = 9, CD = 3; all pheromones 1; $$ \alpha=\beta=1,\;\rho=0.5,\;Q=1 $$ . Two ants start at A. Ant 1 uses random numbers 0.45 then 0.5; Ant 2 uses 0.83 then 0.5. Find both tours and the updated pheromone matrix.** [lecture example · 10 marks] #### Given Symmetric distances AB=5, AC=6, AD=7, BC=2, BD=9, CD=3. Initial off-diagonal pheromone=1. α=β=1, ρ=0.5, Q=1. Ant draws: [0.45,0.5] and [0.83,0.5]. #### To find Two complete tours and every updated edge pheromone. #### Formula and why it applies Prefer high pheromone and small distance, normalise over unvisited cities, then evaporate and deposit once per iteration. $$ P_{ij}=\frac{\tau_{ij}^{\alpha}(1/d_{ij})^{\beta}}{\sum_{k\in allowed}\tau_{ik}^{\alpha}(1/d_{ik})^{\beta}} $$ $$ \tau_{ij}^{new}=(1-\rho)\tau_{ij}^{old}+\sum_{ants\ using\ ij}Q/L $$ #### Solution Swipe across the diagram to read its labels ↔ ![Lecture example (Ant System): distances](https://deep-learning-study-room.vercel.app/is2/fig/aco_lecture_graph.629672bf56.svg) Diagram: Lecture example (Ant System): distances. Labels: Lecture example (Ant System): distances | 5 | 6 | 7 | 2 | 9 | 3 | A | B | C | D #### Solution ##### Step 1. From A $$ \eta=\tfrac15,\tfrac16,\tfrac17=0.2,\;0.1667,\;0.1429;\;\text{sum}=0.5095\Rightarrow P=0.3925,\;0.3271,\;0.2804 $$ Cumulative: B 0–0.3925, C 0.3925–0.7196, D 0.7196–1. Ant 1 (0.45) → C. Ant 2 (0.83) → D. ##### Step 2. Ant 1 from C (B, D left) $$ \eta=\tfrac12,\tfrac13\Rightarrow P_{CB}=0.6,\;P_{CD}=0.4 $$ r = 0.5 ≤ 0.6 → B; then D; then back to A. $$ \text{Ant 1: }A\to C\to B\to D\to A,\quad L_1=6+2+9+7=24 $$ ##### Step 3. Ant 2 from D (B, C left) $$ \eta=\tfrac19,\tfrac13\Rightarrow P_{DB}=0.25,\;P_{DC}=0.75 $$ Cumulative B 0–0.25, C 0.25–1; r = 0.5 → C; then B; then A. $$ \text{Ant 2: }A\to D\to C\to B\to A,\quad L_2=7+3+2+5=17 $$ ##### Step 4. Deposits | Edge | Used by | New τ | | --- | --- | --- | | AD, BC | both ants | 0.5 + 1/24 + 1/17 = 0.6005 | | AC, BD | Ant 1 only | 0.5 + 1/24 = 0.5417 | | AB, CD | Ant 2 only | 0.5 + 1/17 = 0.5588 | **Final answer.** Ant 1: A-C-B-D-A (24); Ant 2: A-D-C-B-A (17). Ant 2's shorter tour deposits more (1/17 > 1/24), so its edges grow faster. AD and BC, used by both, end highest at 0.6005. **Slide corrections.** The slide gives P(D→B) = 0.33 and P(D→C) = 0.66; the correct values are 0.25 and 0.75 (the choice of C does not change). The slide truncates 0.6005 and 0.5417 to 0.6004 and 0.5416. ## Write the requested final matrix Ant 1 deposits 1/24≈.041666667; ant 2 deposits 1/17≈.058823529. Evaporation leaves .5 on every edge. Their shared edges AD and BC become .600490196. Ant-1-only edges AC and BD become .541666667; ant-2-only edges AB and CD become .558823529. $$ \tau^{new}\approx\begin{bmatrix}0&.558823529&.541666667&.600490196\\.558823529&0&.600490196&.541666667\\.541666667&.600490196&0&.558823529\\.600490196&.541666667&.558823529&0\end{bmatrix} $$ Row and column order is A,B,C,D. The diagonal 0 marks excluded self-edges. Off-diagonal entries mirror each other because this problem is undirected and symmetric. --- ## TEE · 5/10-mark answer: applications of ACO (explain one) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u3-aco--applications/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u3-aco/applications Unit 3 · Ant colony optimization for TSP · asked as a 10-mark numerical in both 2025–26 papers **Asked as:** “List the applications of Ant Colony Optimization and explain any one application in detail.” [5 marks · Final 2025 Q1b; key: listing 3 + explanation 2] ### Definition (1 line) Ant Colony Optimization is a probabilistic, population-based metaheuristic (Dorigo, 1992) in which artificial ants build solutions on a graph and communicate through pheromone trails. In the update rule used here, evaporation reduces all stored trails; deposits then reinforce edges of selected tours according to solution quality. Evaporation does not occur only on poor paths. ### Applications 1. **Travelling salesman problem**: shortest tour visiting every city once. 2. **Vehicle routing**: delivery routes for a fleet with capacity and time windows (minimising distance, time, fuel). 3. **Network routing**: adaptive routing of data packets in communication and sensor networks (AntNet). 4. **Job-shop and project scheduling**: assigning jobs to machines to minimise completion time. 5. **Image processing**: edge detection, with ants moving over pixel-intensity graphs. 6. **Assignment problems**: quadratic assignment, timetabling, graph colouring. 7. **Feature selection and data mining**: Ant-Miner for classification rules. 8. **Load balancing** in distributed systems and cloud task allocation. ### One application in detail: vehicle routing (or TSP) 1. **Graph.** Depot and customers are nodes; travel times or distances are edge weights. 2. **Ants.** Each ant builds a complete set of routes. From its current customer it chooses the next with probability $$ \propto\tau_{ij}\,(1/d_{ij}) $$ , skipping customers already served or that would exceed the vehicle capacity; when the vehicle is full it returns to the depot. 3. **Evaluation.** Total distance (plus penalties for late deliveries). 4. **Learning.** Pheromone evaporates everywhere and is added in proportion to $$ Q/L $$ on the routes of good solutions, so short connections are reinforced. 5. **Result.** Repeated iterations search for low-cost routes; finite-run convergence to the optimum is not guaranteed. Companies use this for parcel delivery, waste collection and school-bus routing. ### Advantages and limitations | Advantages | Limitations | | --- | --- | | Naturally suited to graph and routing problems | Several parameters to tune (α, β, ρ, number of ants) | | Can share useful route information and adapt search to changing graphs | Convergence time is uncertain; may stagnate on a sub-optimal path | | Positive feedback + parallel ants | Theoretical analysis is difficult | **Yaad rakho:** ACO = “raaste wale problems” (TSP, routing, scheduling). PSO = “numbers wale problems” (continuous function, weights). --- ## TEE · Swarm intelligence, particles, velocity, pbest and gbest Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/pso-core--meaning/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/pso-core/meaning First understand position, velocity and memory; then solve complete two-iteration minimisation examples. ## Swarm intelligence **Swarm intelligence** is collective problem-solving behaviour that emerges from interactions among many relatively simple agents and their environment. It is inspired by systems such as bird flocks, fish schools and ant colonies. **Particle swarm optimization (PSO)** is a population-based search method inspired by social movement. Each particle represents a candidate solution. It moves through the search space using a velocity influenced by its own best previous position and the best position known to the swarm. ## Translate every PSO word into a concrete object | Term | Meaning | Example | | --- | --- | --- | | Particle | One candidate solution | Particle A | | Position x | The candidate’s coordinates | xA = 3 | | Objective f(x) | The problem’s score for that candidate | f(3) = 3² = 9 | | Velocity v | Proposed change of position | vA = −1 means move left by 1 | | pbest | Best position that this particle has visited | A once visited x=1 with score 1 | | gbest | Best position found by any particle | Among all personal bests, x=0.5 scores 0.25 | Swipe across the diagram to read its labels ↔ ![A best memory changes only when the new score improves.](https://deep-learning-study-room.vercel.app/is2/fig/v2_pso_memory.fcc9783a95.svg) Diagram: A best memory changes only when the new score improves.. Labels: Best memory survives a worse move | Visit 1 | Position x = 3 | Score x² = 9 | Remember pbest = 3 | Visit 2 | Position x = 1 | Score x² = 1 | Remember pbest = 1 | Visit 3 | Position x = 2 | Score x² = 4 | Remember pbest = 1 | Minimise x²: after visiting x = 2, the best past position is still x = 1. | Current position and remembered best position are separate records. A best memory changes only when the new score improves. ## Current position is not necessarily personal best For minimising x², suppose a particle visits x=3, then x=1, then x=2. Its current position is 2, but its personal best stays 1 because 1² is less than 2². The particle remembers a better past result even after a bad move. **A particle visits x=3, x=1, x=2 in that order while minimising f(x)=x². Find pbest after each visit.** [2 marks · Practice] #### Given Positions in chronological order: 3 → 1 → 2. #### To find The best-so-far position and score. #### Formula and why it applies Keep the location with the lowest objective encountered so far. #### Solution ##### Step 1. First visit $$ f(3)=9 $$ No history: pbest=3, best score=9. ##### Step 2. Second visit $$ f(1)=1<9 $$ Replace pbest by 1. ##### Step 3. Third visit $$ f(2)=4>1 $$ Keep pbest=1 even though the current position is 2. **Final answer.** Final current position=2; pbest=1; personal-best score=1. **Define swarm intelligence and briefly explain one nature-inspired algorithm.** [2 marks · Practice] #### Answer Swarm intelligence is collective problem-solving that emerges from interactions among simple agents. Particle swarm optimization is inspired by flocking behaviour. Each particle is a candidate solution with a position and velocity. It remembers its own best position and uses the swarm’s best position to guide movement. Repeated evaluation and movement search for a low-cost or high-fitness solution. **Yaad rakho:** Position abhi kahan ho; pbest apna best past; gbest sabka best past. --- ## TEE · Read the PSO equation and work three individual updates Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/pso-core--equation/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/pso-core/equation First understand position, velocity and memory; then solve complete two-iteration minimisation examples. ## First read the update in words **New velocity = retained old velocity + pull toward personal best + pull toward the group best.** After finding that velocity, add it to the old position. You are calculating a movement, then applying it. Swipe across the diagram to read its labels ↔ ![A negative difference pulls toward a best position on the left.](https://deep-learning-study-room.vercel.app/is2/fig/depth_pso.88e9c2a672.svg) Diagram: A negative difference pulls toward a best position on the left.. Labels: Read the direction before multiplying the pull | Best = 1 | Current = 4 | difference = 1 − 4 = −3 | coefficient 0.5 → pull 0.5 × (−3) = −1.5 (left) A negative difference pulls toward a best position on the left. Example: current x=4 and personal best=1 gives the difference 1−4=−3. A coefficient of .5 gives personal pull .5×(−3)=−1.5. The negative sign says “move left”; it does not mean a negative probability. If current x=−2 and best=1, the difference is 1−(−2)=1+2=3, which pulls right. In vi,jt, i chooses the particle, j chooses its coordinate, and t chooses the recorded iteration. For particle B, coordinate 2, iteration 0, look at row B and the second velocity entry in the initial table. The letters identify a stored number; they are not extra multiplication factors. ## The velocity equation is three contributions added together $$ v_{i,j}^{t+1}=w v_{i,j}^{t}+c_1r_1(pbest_{i,j}^{t}-x_{i,j}^{t})+c_2r_2(gbest_j^{t}-x_{i,j}^{t}) $$ $$ x_{i,j}^{t+1}=x_{i,j}^{t}+v_{i,j}^{t+1} $$ | Symbol | Meaning | | --- | --- | | i | Which particle | | j | Which coordinate of that particle | | t / t+1 | Before / after this iteration; these are labels, not powers | | w | Inertia weight: how much old velocity is kept | | c₁r₁ | Combined strength of the pull to personal best | | c₂r₂ | Combined strength of the pull to global best | | r₁,r₂ | Random factors, normally in [0,1]; use the values supplied in an exam | A best position to the left gives a negative difference and pulls left; a best position to the right gives a positive difference and pulls right. The multiplication sets the strength of each pull. **For one coordinate, x=4, v=0, pbest=4, gbest=2. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position.** [3 marks · Practice] #### Given x=4, v=0, pbest=4, gbest=2; w=.5; c₁r₁=.5; c₂r₂=.5. #### To find New velocity followed by new position. #### Formula and why it applies Use the three-term velocity equation, then add the new velocity to x. $$ v_{new}=0.5v+0.5(pbest-x)+0.5(gbest-x) $$ #### Solution ##### Step 1. Inertia $$ 0.5\times0=0 $$ ##### Step 2. Personal pull $$ 0.5(4-(4))=0 $$ ##### Step 3. Social pull $$ 0.5(2-(4))=-1 $$ ##### Step 4. New velocity $$ v_{new}=0+0+(-1)=-1 $$ ##### Step 5. New position $$ x_{new}=4+(-1)=3 $$ **Final answer.** vnew=-1; xnew=3. **For one coordinate, x=4, v=-2, pbest=3, gbest=1. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position.** [3 marks · Practice] #### Given x=4, v=-2, pbest=3, gbest=1; w=.5; c₁r₁=.5; c₂r₂=.5. #### To find New velocity followed by new position. #### Formula and why it applies Use the three-term velocity equation, then add the new velocity to x. $$ v_{new}=0.5v+0.5(pbest-x)+0.5(gbest-x) $$ #### Solution ##### Step 1. Inertia $$ 0.5\times(-2)=-1 $$ ##### Step 2. Personal pull $$ 0.5(3-(4))=-0.5 $$ ##### Step 3. Social pull $$ 0.5(1-(4))=-1.5 $$ ##### Step 4. New velocity $$ v_{new}=(-1)+(-0.5)+(-1.5)=-3 $$ ##### Step 5. New position $$ x_{new}=4+(-3)=1 $$ **Final answer.** vnew=-3; xnew=1. **For one coordinate, x=-2, v=1, pbest=-1, gbest=1. With w=0.5, c₁=c₂=1 and r₁=r₂=0.5, find the next velocity and position.** [3 marks · Practice] #### Given x=-2, v=1, pbest=-1, gbest=1; w=.5; c₁r₁=.5; c₂r₂=.5. #### To find New velocity followed by new position. #### Formula and why it applies Use the three-term velocity equation, then add the new velocity to x. $$ v_{new}=0.5v+0.5(pbest-x)+0.5(gbest-x) $$ #### Solution ##### Step 1. Inertia $$ 0.5\times1=0.5 $$ ##### Step 2. Personal pull $$ 0.5(-1-(-2))=0.5 $$ ##### Step 3. Social pull $$ 0.5(1-(-2))=1.5 $$ ##### Step 4. New velocity $$ v_{new}=0.5+0.5+1.5=2.5 $$ ##### Step 5. New position $$ x_{new}=(-2)+2.5=0.5 $$ **Final answer.** vnew=2.5; xnew=0.5. **First-iteration shortcut:** if pbest was just initialised to x, the personal difference is zero. This follows from the initialisation; it is not a permanent property of PSO. A gbest particle can still move because its inertia term can be nonzero. **Yaad rakho:** Old speed ka hissa + apne best ki pull + group best ki pull; phir new speed ko position mein add. --- ## TEE · Two particles, two iterations: start with the simplest minimisation Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/pso-core--two-particles/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/pso-core/two-particles First understand position, velocity and memory; then solve complete two-iteration minimisation examples. A synchronous iteration is a class photograph of the old swarm: every particle uses that same snapshot of gbest. First calculate all new positions, then compare all new scores and update memories. If particle A discovers a better position halfway through the calculations, B still uses the previous snapshot during this iteration. B can use the improved gbest in the next iteration. Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values. **Minimise $$ f(x)=x^2 $$ using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [3 marks · Practice; expanded teaching solution] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (-2) | (0) | | B | (4) | (0) | w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(-2)=(-2)^2=4=4 $$ **Particle B** $$ f(4)=4^2=16=16 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(-2) $$ Its score is 4. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (-2). pbest values are A: (-2), B: (4). ##### Step 4. Iteration 1: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=-2-(-2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=-2-(-2)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=(-2)+0=-2 $$ ##### Step 5. Iteration 1: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=4-(4)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=-2-(4)=-6 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-6)=-3 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-3)=-3 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=4+(-3)=1 $$ ##### Step 6. Iteration 1: evaluate the new positions **Particle A** $$ f(-2)=(-2)^2=4=4 $$ **Particle B** $$ f(1)=1^2=1=1 $$ ##### Step 7. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 4 | 4 | Keep old: tie | (-2) | | B | 16 | 1 | Replace: new score is lower | (1) | Compare the retained pbest scores. gbest = **(1)**, with objective **1**. ##### Step 8. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (-2) | (0) | (-2) | | B | (1) | (-3) | (1) | ##### Step 9. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (1). pbest values are A: (-2), B: (1). ##### Step 10. Iteration 2: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=-2-(-2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(-2)=3 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(3)=1.5 $$ Add the three velocity contributions. $$ v^{(2)}=0+0+1.5=1.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=(-2)+1.5=-0.5 $$ ##### Step 11. Iteration 2: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-3)=-1.5 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(2)}=(-1.5)+0+0=-1.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1+(-1.5)=-0.5 $$ ##### Step 12. Iteration 2: evaluate the new positions **Particle A** $$ f(-0.5)=(-0.5)^2=0.25=0.25 $$ **Particle B** $$ f(-0.5)=(-0.5)^2=0.25=0.25 $$ ##### Step 13. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 4 | 0.25 | Replace: new score is lower | (-0.5) | | B | 1 | 0.25 | Replace: new score is lower | (-0.5) | Compare the retained pbest scores. gbest = **(-0.5)**, with objective **0.25**. ##### Step 14. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (-0.5) | (1.5) | (-0.5) | | B | (-0.5) | (-1.5) | (-0.5) | **Final answer.** After iteration 2, gbest = (-0.5), objective = 0.25. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. --- ## TEE · A worse move: pbest stays in the past Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/pso-core--memory-matters/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/pso-core/memory-matters First understand position, velocity and memory; then solve complete two-iteration minimisation examples. Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values. **Minimise $$ f(x)=x^2 $$ using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [3 marks · Practice; expanded teaching solution] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (1) | (4) | | B | (3) | (-1) | w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(1)=1^2=1=1 $$ **Particle B** $$ f(3)=3^2=9=9 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(1) $$ Its score is 1. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (1). pbest values are A: (1), B: (3). ##### Step 4. Iteration 1: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times4=2 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=2+0+0=2 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=1+2=3 $$ ##### Step 5. Iteration 1: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-1)=-0.5 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3-(3)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(3)=-2 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-2)=-1 $$ Add the three velocity contributions. $$ v^{(1)}=(-0.5)+0+(-1)=-1.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=3+(-1.5)=1.5 $$ ##### Step 6. Iteration 1: evaluate the new positions **Particle A** $$ f(3)=3^2=9=9 $$ **Particle B** $$ f(1.5)=1.5^2=2.25=2.25 $$ ##### Step 7. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 1 | 9 | Keep old: new score is worse | (1) | | B | 9 | 2.25 | Replace: new score is lower | (1.5) | Compare the retained pbest scores. gbest = **(1)**, with objective **1**. ##### Step 8. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (3) | (2) | (1) | | B | (1.5) | (-1.5) | (1.5) | ##### Step 9. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (1). pbest values are A: (1), B: (1.5). ##### Step 10. Iteration 2: update particle A **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times2=1 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(3)=-2 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(-2)=-1 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(3)=-2 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-2)=-1 $$ Add the three velocity contributions. $$ v^{(2)}=1+(-1)+(-1)=-1 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=3+(-1)=2 $$ ##### Step 11. Iteration 2: update particle B **Only coordinate** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-1.5)=-0.75 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1.5-(1.5)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1.5)=-0.5 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-0.5)=-0.25 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.75)+0+(-0.25)=-1 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1.5+(-1)=0.5 $$ ##### Step 12. Iteration 2: evaluate the new positions **Particle A** $$ f(2)=2^2=4=4 $$ **Particle B** $$ f(0.5)=0.5^2=0.25=0.25 $$ ##### Step 13. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 1 | 4 | Keep old: new score is worse | (1) | | B | 2.25 | 0.25 | Replace: new score is lower | (0.5) | Compare the retained pbest scores. gbest = **(0.5)**, with objective **0.25**. ##### Step 14. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (2) | (-1) | (1) | | B | (0.5) | (-1) | (0.5) | **Final answer.** After iteration 2, gbest = (0.5), objective = 0.25. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. **Yaad rakho:** Naya result kharab ho toh position toh badlegi, pbest nahi. --- ## TEE · Two-dimensional PSO: keep the coordinates separate Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/pso-core--two-coordinates/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/pso-core/two-coordinates First understand position, velocity and memory; then solve complete two-iteration minimisation examples. A two-dimensional particle has two entries in its position and two in its velocity. Apply the scalar velocity calculation once to coordinate 1 and once to coordinate 2. Keep them in the same order to form the new pair. Only after both entries are ready do you insert the pair into f(x₁,x₂). Adding x₁ and x₂ into a single position would lose information. Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values. **Minimise $$ f(x)=x_1^2+x_2^2 $$ using 2 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [3 marks · Practice; expanded teaching solution] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (1, 1) | (0, 0) | | B | (3, 2) | (0, 0) | w=0.5, c₁=1, c₂=1, r₁=0.5, r₂=0.5. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(1, 1)=1^2+1^2=1+1=2 $$ **Particle B** $$ f(3, 2)=3^2+2^2=9+4=13 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(1, 1) $$ Its score is 2. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (1, 1). pbest values are A: (1, 1), B: (3, 2). ##### Step 4. Iteration 1: update particle A **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=1+0=1 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=1+0=1 $$ ##### Step 5. Iteration 1: update particle B **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3-(3)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(3)=-2 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-2)=-1 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-1)=-1 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=3+(-1)=2 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=2-(2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(2)=-1 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-1)=-0.5 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-0.5)=-0.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=2+(-0.5)=1.5 $$ ##### Step 6. Iteration 1: evaluate the new positions **Particle A** $$ f(1, 1)=1^2+1^2=1+1=2 $$ **Particle B** $$ f(2, 1.5)=2^2+1.5^2=4+2.25=6.25 $$ ##### Step 7. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 2 | 2 | Keep old: tie | (1, 1) | | B | 13 | 6.25 | Replace: new score is lower | (2, 1.5) | Compare the retained pbest scores. gbest = **(1, 1)**, with objective **2**. ##### Step 8. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (1, 1) | (0, 0) | (1, 1) | | B | (2, 1.5) | (-1, -0.5) | (2, 1.5) | ##### Step 9. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (1, 1). pbest values are A: (1, 1), B: (2, 1.5). ##### Step 10. Iteration 2: update particle A **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(2)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1+0=1 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1-(1)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(0)=0 $$ Add the three velocity contributions. $$ v^{(2)}=0+0+0=0 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1+0=1 $$ ##### Step 11. Iteration 2: update particle B **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-1)=-0.5 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=2-(2)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(2)=-1 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-1)=-0.5 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.5)+0+(-0.5)=-1 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=2+(-1)=1 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-0.5)=-0.25 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=1.5-(1.5)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1\times0.5=0.5 $$ $$ C=0.5\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=1-(1.5)=-0.5 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1\times0.5=0.5 $$ $$ S=0.5\times(-0.5)=-0.25 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.25)+0+(-0.25)=-0.5 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=1.5+(-0.5)=1 $$ ##### Step 12. Iteration 2: evaluate the new positions **Particle A** $$ f(1, 1)=1^2+1^2=1+1=2 $$ **Particle B** $$ f(1, 1)=1^2+1^2=1+1=2 $$ ##### Step 13. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 2 | 2 | Keep old: tie | (1, 1) | | B | 6.25 | 2 | Replace: new score is lower | (1, 1) | Compare the retained pbest scores. gbest = **(1, 1)**, with objective **2**. ##### Step 14. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (1, 1) | (0, 0) | (1, 1) | | B | (1, 1) | (-1, -0.5) | (1, 1) | **Final answer.** After iteration 2, gbest = (1, 1), objective = 2. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. --- ## TEE · Class PSO photograph: all three particles through iteration 2 Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/pso-core--class-three-particles/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/pso-core/class-three-particles First understand position, velocity and memory; then solve complete two-iteration minimisation examples. Work down the page: each particle is fully updated, then all scores and memories are compared. Keep at least six digits internally for longer decimal examples; the final table displays rounded values. The positions, velocities and coefficients are transcribed from the class photo. We use f(x₁,x₂)=x₁²+x₂²−x₁x₂, minimisation, initial pbest=x, fixed supplied random factors and synchronous updates. The photo does not itself specify all these conventions; they are stated here to make the two-iteration extension unambiguous. **Minimise $$ f(x)=x_1^2+x_2^2-x_1x_2 $$ using 3 particles. Use the initial data below and perform exactly two PSO iterations, reporting position, velocity, pbest and gbest after each iteration.** [10 marks · Class data; two-iteration minimisation extension] #### Given | Particle | Initial position x⁰ | Initial velocity v⁰ | | --- | --- | --- | | A | (3.14, 7.21) | (0.1, 0.3) | | B | (4.15, 7.31) | (0.5, 0.1) | | C | (6.17, 8.27) | (0.2, 0) | w=0.5, c₁=1.2, c₂=1.2, r₁=0.5, r₂=0.1. Initialise each pbest at its initial position. Use these random coefficients unchanged in both iterations. No bounds or velocity clamp are imposed in this example. Updates are synchronous: every particle uses the same previous-iteration gbest. #### To find The complete swarm state after iterations 1 and 2, and the best location found. #### Formula and why it applies PSO combines inertia, personal memory and shared memory. Subtract positions coordinate by coordinate. Evaluate the objective only after moving. For minimisation, replace a best only when its score decreases. $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ $$ x^{t+1}=x^t+v^{t+1} $$ #### Solution ##### Step 1. Evaluate every initial position **Particle A** $$ f(3.14, 7.21)=3.14^2+7.21^2-(3.14)(7.21) $$ $$ =9.8596+51.9841-22.6394=39.2043 $$ **Particle B** $$ f(4.15, 7.31)=4.15^2+7.31^2-(4.15)(7.31) $$ $$ =17.2225+53.4361-30.3365=40.3221 $$ **Particle C** $$ f(6.17, 8.27)=6.17^2+8.27^2-(6.17)(8.27) $$ $$ =38.0689+68.3929-51.0259=55.4359 $$ ##### Step 2. Initialise the memories pbest is each initial position because there is no earlier history. The smallest objective decides gbest. $$ gbest^0=(3.14, 7.21) $$ Its score is 39.2043. Store the position, not only the score. ##### Step 3. Iteration 1: freeze the previous memories All particles in this iteration use gbest = (3.14, 7.21). pbest values are A: (3.14, 7.21), B: (4.15, 7.31), C: (6.17, 8.27). ##### Step 4. Iteration 1: update particle A **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.1=0.05 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3.14-(3.14)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(3.14)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0.05+0+0=0.05 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=3.14+0.05=3.19 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.3=0.15 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=7.21-(7.21)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(7.21)=0 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(0)=0 $$ Add the three velocity contributions. $$ v^{(1)}=0.15+0+0=0.15 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=7.21+0.15=7.36 $$ ##### Step 5. Iteration 1: update particle B **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.5=0.25 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=4.15-(4.15)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(4.15)=-1.01 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-1.01)=-0.1212 $$ Add the three velocity contributions. $$ v^{(1)}=0.25+0+(-0.1212)=0.1288 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=4.15+0.1288=4.2788 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.1=0.05 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=7.31-(7.31)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(7.31)=-0.1 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.1)=-0.012 $$ Add the three velocity contributions. $$ v^{(1)}=0.05+0+(-0.012)=0.038 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=7.31+0.038=7.348 $$ ##### Step 6. Iteration 1: update particle C **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.2=0.1 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=6.17-(6.17)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(6.17)=-3.03 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-3.03)=-0.3636 $$ Add the three velocity contributions. $$ v^{(1)}=0.1+0+(-0.3636)=-0.2636 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=6.17+(-0.2636)=5.9064 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0=0 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=8.27-(8.27)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(8.27)=-1.06 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-1.06)=-0.1272 $$ Add the three velocity contributions. $$ v^{(1)}=0+0+(-0.1272)=-0.1272 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(1)}=8.27+(-0.1272)=8.1428 $$ ##### Step 7. Iteration 1: evaluate the new positions **Particle A** $$ f(3.19, 7.36)=3.19^2+7.36^2-(3.19)(7.36) $$ $$ =10.1761+54.1696-23.4784=40.8673 $$ **Particle B** $$ f(4.2788, 7.348)=4.2788^2+7.348^2-(4.2788)(7.348) $$ $$ =18.30812944+53.993104-31.4406224=40.86061104 $$ **Particle C** $$ f(5.9064, 8.1428)=5.9064^2+8.1428^2-(5.9064)(8.1428) $$ $$ =34.88556096+66.30519184-48.09463392=53.09611888 $$ ##### Step 8. Iteration 1: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 39.2043 | 40.8673 | Keep old: new score is worse | (3.14, 7.21) | | B | 40.3221 | 40.86061104 | Keep old: new score is worse | (4.15, 7.31) | | C | 55.4359 | 53.09611888 | Replace: new score is lower | (5.9064, 8.1428) | Compare the retained pbest scores. gbest = **(3.14, 7.21)**, with objective **39.2043**. ##### Step 9. Iteration 1: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (3.19, 7.36) | (0.05, 0.15) | (3.14, 7.21) | | B | (4.2788, 7.348) | (0.1288, 0.038) | (4.15, 7.31) | | C | (5.9064, 8.1428) | (-0.2636, -0.1272) | (5.9064, 8.1428) | ##### Step 10. Iteration 2: freeze the previous memories All particles in this iteration use gbest = (3.14, 7.21). pbest values are A: (3.14, 7.21), B: (4.15, 7.31), C: (5.9064, 8.1428). ##### Step 11. Iteration 2: update particle A **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.05=0.025 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=3.14-(3.19)=-0.05 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(-0.05)=-0.03 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(3.19)=-0.05 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.05)=-0.006 $$ Add the three velocity contributions. $$ v^{(2)}=0.025+(-0.03)+(-0.006)=-0.011 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=3.19+(-0.011)=3.179 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.15=0.075 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=7.21-(7.36)=-0.15 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(-0.15)=-0.09 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(7.36)=-0.15 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.15)=-0.018 $$ Add the three velocity contributions. $$ v^{(2)}=0.075+(-0.09)+(-0.018)=-0.033 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=7.36+(-0.033)=7.327 $$ ##### Step 12. Iteration 2: update particle B **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.1288=0.0644 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=4.15-(4.2788)=-0.1288 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(-0.1288)=-0.07728 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(4.2788)=-1.1388 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-1.1388)=-0.136656 $$ Add the three velocity contributions. $$ v^{(2)}=0.0644+(-0.07728)+(-0.136656)=-0.149536 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=4.2788+(-0.149536)=4.129264 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times0.038=0.019 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=7.31-(7.348)=-0.038 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(-0.038)=-0.0228 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(7.348)=-0.138 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.138)=-0.01656 $$ Add the three velocity contributions. $$ v^{(2)}=0.019+(-0.0228)+(-0.01656)=-0.02036 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=7.348+(-0.02036)=7.32764 $$ ##### Step 13. Iteration 2: update particle C **Coordinate 1** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-0.2636)=-0.1318 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=5.9064-(5.9064)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=3.14-(5.9064)=-2.7664 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-2.7664)=-0.331968 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.1318)+0+(-0.331968)=-0.463768 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=5.9064+(-0.463768)=5.442632 $$ **Coordinate 2** Inertia = keep a fraction of the old velocity. $$ I=0.5\times(-0.1272)=-0.0636 $$ Personal pull: first subtract the current coordinate from the personal-best coordinate. $$ d_p=8.1428-(8.1428)=0 $$ Multiply the two personal coefficients, then multiply by the difference. $$ c_1r_1=1.2\times0.5=0.6 $$ $$ C=0.6\times(0)=0 $$ Social pull: subtract the current coordinate from the frozen global-best coordinate. $$ d_g=7.21-(8.1428)=-0.9328 $$ Multiply the social coefficients, then multiply by that difference. $$ c_2r_2=1.2\times0.1=0.12 $$ $$ S=0.12\times(-0.9328)=-0.111936 $$ Add the three velocity contributions. $$ v^{(2)}=(-0.0636)+0+(-0.111936)=-0.175536 $$ Add the new velocity to the old position. The parenthesised upper label is the iteration number. $$ x^{(2)}=8.1428+(-0.175536)=7.967264 $$ ##### Step 14. Iteration 2: evaluate the new positions **Particle A** $$ f(3.179, 7.327)=3.179^2+7.327^2-(3.179)(7.327) $$ $$ =10.106041+53.684929-23.292533=40.498437 $$ **Particle B** $$ f(4.129264, 7.32764)=4.129264^2+7.32764^2-(4.129264)(7.32764) $$ $$ =17.050821182+53.69430797-30.257760057=40.487369094 $$ **Particle C** $$ f(5.442632, 7.967264)=5.442632^2+7.967264^2-(5.442632)(7.967264) $$ $$ =29.622243087+63.477295646-43.362885999=49.736652734 $$ ##### Step 15. Iteration 2: update pbest, then gbest | Particle | Old pbest score | New score | Decision | pbest position | | --- | --- | --- | --- | --- | | A | 39.2043 | 40.498437 | Keep old: new score is worse | (3.14, 7.21) | | B | 40.3221 | 40.487369094 | Keep old: new score is worse | (4.15, 7.31) | | C | 53.09611888 | 49.736652734 | Replace: new score is lower | (5.442632, 7.967264) | Compare the retained pbest scores. gbest = **(3.14, 7.21)**, with objective **39.2043**. ##### Step 16. Iteration 2: record the complete state | Particle | Position | Velocity | pbest | | --- | --- | --- | --- | | A | (3.179, 7.327) | (-0.011, -0.033) | (3.14, 7.21) | | B | (4.129264, 7.32764) | (-0.149536, -0.02036) | (4.15, 7.31) | | C | (5.442632, 7.967264) | (-0.463768, -0.175536) | (5.442632, 7.967264) | **Final answer.** After iteration 2, gbest = (3.14, 7.21), objective = 39.2043. This is the best found in these two iterations, not a claim that PSO has proved the global optimum. Swipe across the diagram to read its labels ↔ ![Motion alone does not imply improvement. Evaluate the objective at the new location before updating a best memory.](https://deep-learning-study-room.vercel.app/is2/fig/v2_pso_class_paths.961831cda1.svg) Diagram: Motion alone does not imply improvement. Evaluate the objective at the new location before updating a best memory.. Labels: 3 | 3.5 | 4 | 4.5 | 5 | 5.5 | 6 | 6.5 | 7 | 7.5 | 8 | 8.5 | Class PSO: positions after each iteration | 0 = initial, 1 = iteration 1, 2 = iteration 2 | A0 | A1 | A2 | B0 | B1 | B2 | C0 | C1 | C2 | First coordinate x₁ | The best-so-far location remains A0 = (3.14, 7.21) after both iterations. Motion alone does not imply improvement. Evaluate the objective at the new location before updating a best memory. --- ## TEE · PSO algorithm, architecture, advantages and limitations Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/pso-core--theory/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/pso-core/theory First understand position, velocity and memory; then solve complete two-iteration minimisation examples. ## Algorithm you can write in an exam 1. Define the objective and whether to minimise or maximise it. Initialise particle positions and velocities. 2. Evaluate every particle. Set each pbest to its initial position and gbest to the best among them. 3. Use the previous state to calculate each new velocity. 4. Update each position using that new velocity. Apply bounds or a velocity limit only according to the stated policy. 5. Evaluate the new positions. Update a pbest only when the new objective is better; choose gbest from the personal bests. 6. Repeat until the iteration budget or another stopping criterion is reached. Initialise → Evaluate + memories → Velocity → Position → Evaluate + update memoriesrepeat until stopping | Advantages | Limitations | | --- | --- | | Simple position/velocity updates; does not require objective derivatives | Can converge prematurely to a local region | | Uses both individual and shared experience | Results depend on coefficients, initialisation and random factors | | Useful for many continuous optimisation problems | Constraints and discrete encodings need suitable handling | | Keeps best-so-far memories | Does not guarantee the global optimum in a finite run | **Explain one application of PSO.** [3 marks · Practice] #### Answer In controller tuning, a particle can represent the gain values of a controller. The objective is a performance cost, such as tracking error plus an overshoot penalty. Each proposed gain vector is evaluated in the model. PSO updates candidate gains using individual and swarm bests. The best feasible gains found are selected and then validated on the actual system. A lower training cost alone does not prove safe real-world behaviour. | Exam situation | Action | | --- | --- | | Minimisation | Lower f is better | | Maximisation | Higher f is better; do not reverse velocity differences | | Tie in best score | Keep the previous best unless a tie rule is supplied | | Different random factors each iteration | Read the relevant row; do not reuse old factors | | Synchronous update | Freeze gbest for all particles within one iteration | | Position/velocity outside a limit | Use the limit-handling rule explicitly supplied; state an assumption if absent | **Yaad rakho:** Velocity → position → score → pbest → gbest. Ye order har iteration mein same. --- ## TEE · Reinforcement learning: learning from rewards Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-rl--what-is-rl/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-rl/what-is-rl Unit 4 · Introduction to reinforcement learning · TEE ## One interaction, with every word attached to a value | Time | Current state | Chosen action | Observed reward | Next state | | --- | --- | --- | --- | --- | | 0 | Robot in room A | Move right | −1 for the move | Room B | | 1 | Room B | Pick up parcel | +5 for pickup | Carrying parcel | | 2 | Carrying parcel | Finish | +2 for delivery | Terminal | The agent is the robot; the environment supplies the next state and reward after an action. A policy tells the robot which action to take in each state. An episode is one run up to termination. The immediate reward at time 0 is −1; the undiscounted return for the whole displayed run is −1+5+2=6. A value estimates future return, not just the next reward. For ε-greedy with two actions, ε=.1 and a unique greedy action: exploit that action with probability .9; during the remaining .1, select uniformly from both actions. Thus the greedy action has probability .9+.1/2=.95 and the other .05. If a question uses a different exploration convention, follow that definition. **In one line:** an agent learns what to do by trying actions, receiving rewards, and adjusting its behaviour so that the total reward over time is as large as possible. ## Definition (write this in the exam) **Reinforcement learning (RL)** is a type of machine learning in which an **agent** learns a **policy**, a rule that maps situations (**states**) to **actions**, by interacting with an **environment**. After every action the environment returns a numerical **reward** and a new state. The agent's goal is to maximise the **expected cumulative (discounted) reward**, called the **return**. No labelled “correct action” is given; the agent discovers good actions by trial and error. Swipe across the diagram to read its labels ↔ ![At time t the agent sees state S t , chooses action A t ; the environment replies with reward R t+1 and next state S t+1 .](https://deep-learning-study-room.vercel.app/is2/fig/rl_loop_full.a9937cb961.svg) Diagram: At time t the agent sees state S t , chooses action A t ; the environment replies with reward R t+1 and next state S t+1 .. Labels: One time-step of reinforcement learning | Agent | policy π(a | s) | Environment | P(s′ | s, a), R | action Aₜ | next state Sₜ₊₁ and reward Rₜ₊₁ | The loop repeats: S₀, A₀, R₁, S₁, A₁, R₂, S₂, … At time t the agent sees state St, chooses action At; the environment replies with reward Rt+1 and next state St+1. ## The vocabulary you must know | Term | Meaning | Example: a robot vacuum | | --- | --- | --- | | **Agent** | The learner and decision maker. | The vacuum's controller. | | **Environment** | Everything outside the agent that responds to its actions. | The room, furniture, dust. | | **State $$ s $$ ** | A description of the current situation. | Position + battery level. | | **Action $$ a $$ ** | A choice the agent can make in a state. | Forward, turn left, return to dock. | | **Reward $$ r $$ ** | Immediate scalar feedback after an action. | +1 per dust patch, −10 for hitting a wall. | | **Policy $$ \pi $$ ** | The agent's behaviour: $$ \pi(a\mid s) $$ = probability of action $$ a $$ in state $$ s $$ . | “If battery < 15%, go to dock.” | | **Return $$ G_t $$ ** | Total discounted reward from time $$ t $$ onward. | All future dust collected, minus penalties. | | **Value $$ v(s) $$ ** | Expected return starting from state $$ s $$ . | How good it is to be at the centre with a full battery. | | **Model** | The agent's knowledge of $$ P(s'\mid s,a) $$ and $$ R $$ (optional). | A map of the room. | ## Four elements of an RL system (Sutton and Barto) 1. **Policy**: defines behaviour (state → action). 2. **Reward signal**: defines the goal; the immediate desirability of a step. 3. **Value function**: long-term desirability of a state (what reward is expected later). 4. **Model of the environment** (optional): predicts the next state and reward, used for planning. Methods with a model are *model-based*; without one, *model-free* (for example, Q-learning). ## How RL differs from other learning | | Supervised | Unsupervised | Reinforcement | | --- | --- | --- | --- | | Feedback | Correct label for every input | None | A reward, often delayed | | Goal | Predict labels | Find structure (clusters) | Maximise long-term return | | Data | Fixed dataset (i.i.d.) | Fixed dataset | Generated by the agent's own actions (sequential) | | Example | Spam classifier | Customer segments (GMM) | Game-playing agent | ## Two ideas that make RL special **Delayed reward.** A move in chess earns nothing now but may win the game 30 moves later. RL must assign credit to earlier actions. **Exploration versus exploitation.** To earn reward the agent should *exploit* the best action it knows, but to find better actions it must *explore* untried ones. A common rule is **ε-greedy**: with probability $$ 1-\varepsilon $$ take the best-known action; with probability $$ \varepsilon $$ take a random one. **Yaad rakho:** Agent karta hai, environment batata hai (reward + next state), policy decide karti hai, value future ka hisaab rakhti hai. --- ## TEE · Markov property, transition matrix and state diagram Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-markov--markov-chain/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-markov/markov-chain Unit 4 · Markov processes · asked in every recent paper (5 and 10 marks) A state is the current category, such as Savings or Current. A transition is one step from one category to another. The Markov assumption says that, once the current state is known, older states provide no additional information about the next-state probabilities in this model. It does not claim that the past never mattered; the current state must capture the relevant information. **In one line:** in a Markov process the next state depends only on the present state, so the whole system is described by one table of “from → to” probabilities. ## Definitions **Markov property.** A sequence of states $$ S_1, S_2, \dots $$ has the Markov property if the future is independent of the past given the present: $$ P(S_{t+1}\mid S_t, S_{t-1},\dots,S_1)=P(S_{t+1}\mid S_t) $$ **Markov process (Markov chain).** A memoryless random process given by a pair $$ \langle \mathcal S, P\rangle $$ : a finite set of states $$ \mathcal S $$ and a **transition probability matrix** $$ P $$ with entries $$ P_{ij}=P(S_{t+1}=j\mid S_t=i) $$ Row $$ i $$ is the state you are **in now** (“from”); column $$ j $$ is the state you go **to next**. Two rules always hold: 1. Every entry is between 0 and 1: $$ 0\le P_{ij}\le 1 $$ . 2. Every row sums to 1: $$ \sum_j P_{ij}=1 $$ (from any state you must go somewhere, possibly staying). A **state transition diagram** draws each state as a circle and each non-zero $$ P_{ij} $$ as an arrow from $$ i $$ to $$ j $$ labelled with the probability. Staying in the same state is a self-loop. **Exam trap.** Rows are “from”, columns are “to”. “20% of Savings switch to Current” goes in row Savings, column Current. Always check that each row adds to 1 before moving on; an inconsistent row needs clarification before it can represent a probability distribution. **Question. A survey shows that users switch between Brand X and Brand Y each year: 25% of Brand X users switch to Brand Y and 75% remain with X; 40% of Brand Y users switch to X and 60% remain with Y. (a) Write the transition probability matrix. (b) Draw the state transition diagram.** [5 marks · Final Nov 2025 Q1d] #### Given Two states: X, Y. From X: stay 0.75, to Y 0.25. From Y: to X 0.40, stay 0.60. #### To find Matrix $$ P $$ and its diagram. #### Formula and why it applies $$ P_{ij}=P(\text{next}=j\mid\text{now}=i) $$ . Each yearly switch depends only on the brand used this year, so the process is Markov. #### Solution ##### Step 1. Order the states and fill row by row $$ P=\begin{array}{c|cc} & X & Y\\\hline X & 0.75 & 0.25\\ Y & 0.40 & 0.60\end{array}\qquad\text{i.e. } P=\begin{bmatrix}0.75&0.25\\0.40&0.60\end{bmatrix} $$ ##### Step 2. Check the rows Row X: 0.75 + 0.25 = 1. Row Y: 0.40 + 0.60 = 1. Valid. ##### Step 3. Draw the diagram Swipe across the diagram to read its labels ↔ ![Self-loops carry the retention probabilities (0.75, 0.6); the arrows between states carry the switching probabilities.](https://deep-learning-study-room.vercel.app/is2/fig/mk_brand.c2c4be3e6a.svg) Diagram: Self-loops carry the retention probabilities (0.75, 0.6); the arrows between states carry the switching probabilities.. Labels: Brand switching each year | 0.25 | 0.4 | 0.75 | 0.6 | X | Y Self-loops carry the retention probabilities (0.75, 0.6); the arrows between states carry the switching probabilities. **Final answer.** $$ P=\begin{bmatrix}0.75&0.25\\0.40&0.60\end{bmatrix} $$ with the diagram above. Marking in the official key: matrix 3 marks, diagram 2 marks. **Question. A bank offers Savings and Current accounts. Each year 20% of Savings holders switch to Current and 80% stay; 10% of Current holders switch to Savings and 90% stay. (a) Construct the transition probability matrix. (b) Draw the state diagram.** [5 marks · Re-exam Jan 2026 Q1d] #### Given For Savings: stay 80%, switch 20%. For Current: switch to Savings 10%, stay 90%. State order: Savings, Current. #### To find The transition matrix and labelled state diagram. #### Formula and why it applies $$ P_{ij}=P(S_{t+1}=j\mid S_t=i) $$ Convert percentages to fractions of 100. A row fixes the present state, while columns list possible destinations. #### Solution ##### Step 1. Matrix (rows = this year, columns = next year) $$ P=\begin{array}{c|cc} & S & C\\\hline S & 0.80 & 0.20\\ C & 0.10 & 0.90\end{array} $$ ##### Step 2. Diagram Swipe across the diagram to read its labels ↔ ![Savings (S) and Current (C) accounts per year](https://deep-learning-study-room.vercel.app/is2/fig/mk_bank.879499f6eb.svg) Diagram: Savings (S) and Current (C) accounts per year. Labels: Savings (S) and Current (C) accounts per year | 0.2 | 0.1 | 0.8 | 0.9 | S | C **Final answer.** $$ P=\begin{bmatrix}0.8&0.2\\0.1&0.9\end{bmatrix} $$ ; diagram as shown. Each part carries 2.5 marks in the key. **Yaad rakho:** Row = “abhi kahan ho”, column = “kal kahan jaoge”. Har row ka total 1. --- ## TEE · Next-step and n-step distributions Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-markov--next-step/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-markov/next-step Unit 4 · Markov processes · asked in every recent paper (5 and 10 marks) **A bank has 60 Savings customers and 40 Current customers. In one period, 80% of Savings stay and 20% switch to Current; 10% of Current switch to Savings and 90% stay. Find expected counts after one and two periods.** [3 marks · Worked example] #### Given $$ P=\begin{bmatrix}0.8&0.2\\0.1&0.9\end{bmatrix} $$ Row/column order is Savings, Current; initial counts [60,40]. #### To find Two successive count vectors. #### Formula and why it applies Each destination receives incoming contributions from every origin. Multiply origin count by the corresponding transition probability, then add. #### Solution ##### Step 1. Period 1: contributions to Savings $$ 60\times0.8=48\qquad40\times0.1=4 $$ $$ S_1=48+4=52 $$ ##### Step 2. Period 1: contributions to Current $$ 60\times0.2=12\qquad40\times0.9=36 $$ $$ C_1=12+36=48 $$ ##### Step 3. Period 2: use the new counts [52,48] $$ S_2=52\times0.8+48\times0.1=41.6+4.8=46.4 $$ ##### Step 4. Period 2: second destination $$ C_2=52\times0.2+48\times0.9=10.4+43.2=53.6 $$ **Final answer.** After one period [52,48]; after two [46.4,53.6]. Each total is 100. Fractional counts are expectations, not split people. Matrix notation [60,40]P is shorthand for these destination-column calculations. P² means P multiplied by P using row-by-column multiplication; it does not mean squaring every individual entry. **In one line:** write today's numbers as a row, multiply by the matrix, and you get tomorrow's numbers. ## The rule Let $$ x_t $$ be a **row vector** of today's shares (or counts) in each state, in the same order as the matrix rows. Then $$ x_{t+1}=x_t\,P,\qquad x_{t+n}=x_t\,P^{\,n} $$ Entry $$ j $$ of $$ x_{t+1} $$ is the **column $$ j $$ ** calculation: everyone who arrives in $$ j $$ , from every state. $$ x_{t+1}(j)=\sum_i x_t(i)\,P_{ij} $$ $$ P^{\,n} $$ is the **n-step transition matrix**: $$ (P^{\,n})_{ij} $$ is the probability of being in $$ j $$ after $$ n $$ steps when starting in $$ i $$ (the Chapman–Kolmogorov result). Counts stay counts (the total is preserved); shares stay shares (they still add to 1). **Exam trap.** Multiply the row vector by the *columns* of $$ P $$ , not its rows. A quick check: the answer's total must equal the starting total (100 guests stay 100 guests). **Question. A hotel chain has properties in Paris (P), London (L) and Rome (R). Daily movement: from Paris 60% stay, 25% to London, 15% to Rome; from London 30% to Paris, 50% stay, 20% to Rome; from Rome 20% to Paris, 30% to London, 50% stay. Today's guest counts are [50, 30, 20] in P, L, R. Find the expected counts tomorrow.** [part of 10 marks · Final Nov 2025 Q5a] #### Given $$ P=\begin{bmatrix}0.60&0.25&0.15\\0.30&0.50&0.20\\0.20&0.30&0.50\end{bmatrix},\qquad x_0=[50,\;30,\;20] $$ #### To find $$ x_1=x_0P $$ . #### Formula and why it applies $$ x_{t+1}=x_tP,\qquad x_{t+2}=(x_tP)P $$ For each destination, multiply each present-state share by its probability of arriving there, then add all arrivals. #### Solution ##### Step 1. Paris tomorrow (column 1) $$ 50(0.60)+30(0.30)+20(0.20)=30+9+4=43 $$ ##### Step 2. London tomorrow (column 2) $$ 50(0.25)+30(0.50)+20(0.30)=12.5+15+6=33.5 $$ ##### Step 3. Rome tomorrow (column 3) $$ 50(0.15)+30(0.20)+20(0.50)=7.5+6+10=23.5 $$ ##### Step 4. Check the total 43 + 33.5 + 23.5 = 100 = 50 + 30 + 20. Correct. **Final answer.** Expected guests tomorrow: Paris 43, London 33.5, Rome 23.5. **Question. Using the brand-switching matrix $$ P=\begin{bmatrix}0.75&0.25\\0.40&0.60\end{bmatrix} $$ , a customer uses Brand X this year. Find the probability that the customer uses each brand after two years.** [practice · 3 marks] #### Given Transition matrix P is printed in the question. For the two-year brand example the initial state is certainly X, so x₀=[1,0]. #### To find The requested next-step or two-step distribution. #### Formula and why it applies $$ x_{t+1}=x_tP,\qquad x_{t+2}=(x_tP)P $$ For each destination, multiply each present-state share by its probability of arriving there, then add all arrivals. #### Solution ##### Step 1. Start vector $$ x_0=[1,\;0] $$ (certainly X now). ##### Step 2. One year $$ x_1=[1,0]P=[0.75,\;0.25] $$ ##### Step 3. Two years $$ x_2=x_1P=[0.75(0.75)+0.25(0.40),\;0.75(0.25)+0.25(0.60)]=[0.6625,\;0.3375] $$ **Final answer.** After two years: X with probability 0.6625, Y with probability 0.3375. (Equivalently, row X of $$ P^2=\begin{bmatrix}0.6625&0.3375\\0.54&0.46\end{bmatrix} $$ .) ## Other next-step results, with full working in Solved papers | Paper question | Start | Next step | | --- | --- | --- | | Learning levels B, I, A (Final Q6b) | [50, 30, 20] | [36, 32, 32] | | Booking channels W, A, C (Re-exam Q5b) | [0.40, 0.35, 0.25] | [0.39, 0.335, 0.275] | | Aircraft at hubs A, B, C (Re-exam Q6b) | [30, 40, 30] | [38.5, 32.5, 29] | The complete working for each is in [Solved papers](https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q6b/). **Yaad rakho:** Kal ka number = har jagah se aane wale log jodo (column ke saath multiply). --- ## TEE · Steady state and long-run gain or loss Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-markov--steady-state/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-markov/steady-state Unit 4 · Markov processes · asked in every recent paper (5 and 10 marks) **For transitions Savings→Current=0.2 and Current→Savings=0.1, derive the stationary shares.** [3 marks · Worked example] #### Given Savings→Current=0.2, so Savings→Savings=0.8. Current→Savings=0.1, so Current→Current=0.9. Let s and c be the Savings and Current shares; s+c=1. #### To find s and c that remain unchanged after a step. #### Formula and why it applies At stationarity total incoming share equals current share. #### Solution ##### Step 1. Write Savings balance $$ s=0.8s+0.1c $$ ##### Step 2. Subtract 0.8s from both sides $$ s-0.8s=0.1c\quad\Rightarrow\quad0.2s=0.1c $$ ##### Step 3. Divide both sides by 0.1 $$ 2s=c $$ ##### Step 4. Use the total share $$ s+c=1\quad\Rightarrow\quad s+2s=1\quad\Rightarrow\quad3s=1 $$ ##### Step 5. Divide and recover the other share $$ s=1/3\qquad c=2s=2/3 $$ **Final answer.** Stationary shares [1/3,2/3]. **In one line:** a stationary distribution is unchanged by one transition; that fixed row $$ \pi $$ satisfies $$ \pi=\pi P $$ , and comparing it with today's shares tells you who gains and who loses in the long run. ## Definition A **steady-state (stationary) distribution** is a row vector $$ \pi=[\pi_1,\dots,\pi_n] $$ such that $$ \pi=\pi P,\qquad \sum_i \pi_i=1,\qquad \pi_i\ge 0 $$ Once the system has these shares, one more step leaves them unchanged. For a chain in which every state can reach every other state and that does not cycle with a fixed period (true for every paper question so far), $$ x_t\to\pi $$ from *any* starting distribution. ## Method for any number of states 1. Write one balance equation per column: $$ \pi_j=\sum_i\pi_iP_{ij} $$ . 2. For the irreducible examples here, keep n−1 independent balance equations and add $$ \sum\pi_i=1 $$ . 3. Express every $$ \pi $$ in terms of one of them, then use the sum to find it. **Two-state shortcut.** When a+b>0, if $$ P=\begin{bmatrix}1-a&a\\b&1-b\end{bmatrix} $$ then $$ \pi=\left[\frac{b}{a+b},\;\frac{a}{a+b}\right] $$ Brand switching: $$ a=0.25,\;b=0.40\Rightarrow\pi=[0.40/0.65,\;0.25/0.65]=[0.6154,\;0.3846] $$ . Bank: $$ a=0.2,\;b=0.1\Rightarrow\pi=[1/3,\;2/3] $$ . ## How to interpret a transition matrix (asked as a 2–3 mark part) - **Diagonal entries $$ P_{ii} $$ ** = **retention**: the fraction that stays. - **Off-diagonal entries $$ P_{ij} $$ ** = **transfer/loss** from $$ i $$ to $$ j $$ ; looking down column $$ j $$ shows **inflow (gain)** into $$ j $$ . - **Long-run gainer/loser**: compare $$ \pi_j $$ with the current share. Larger → gains; smaller → loses. - An **absorbing state** has $$ P_{ii}=1 $$ : once entered it is never left. **Question. Learners move daily between Beginner (B), Intermediate (I) and Advanced (A): from B 60% stay, 30% to I, 10% to A; from I 20% to B, 50% stay, 30% to A; from A 10% drop to I, 80% stay and 10% stop learning (treated as remaining in A). Find the steady-state proportions and the level with the largest long-run share.** [part of 10 marks · Final Nov 2025 Q6b] #### Given $$ P=\begin{bmatrix}0.6&0.3&0.1\\0.2&0.5&0.3\\0&0.1&0.9\end{bmatrix} $$ Row A is 0 + 0.1 + (0.8 + 0.1) = 1. #### To find $$ \pi=[\pi_B,\pi_I,\pi_A] $$ . #### Formula and why it applies $$ \pi=\pi P,\quad\sum_i\pi_i=1 $$ At stationarity each destination receives exactly its current share. Solve the balance equations together with the total-probability equation. #### Solution ##### Step 1. Column B balance $$ \pi_B=0.6\pi_B+0.2\pi_I+0\cdot\pi_A\;\Rightarrow\;0.4\pi_B=0.2\pi_I\;\Rightarrow\;\pi_I=2\pi_B $$ ##### Step 2. Column A balance $$ \pi_A=0.1\pi_B+0.3\pi_I+0.9\pi_A\;\Rightarrow\;0.1\pi_A=0.1\pi_B+0.3(2\pi_B)=0.7\pi_B\;\Rightarrow\;\pi_A=7\pi_B $$ ##### Step 3. Use the sum $$ \pi_B+2\pi_B+7\pi_B=1\;\Rightarrow\;\pi_B=0.1,\quad\pi_I=0.2,\quad\pi_A=0.7 $$ ##### Step 4. Check with column I $$ 0.3(0.1)+0.5(0.2)+0.1(0.7)=0.03+0.10+0.07=0.20=\pi_I\;\checkmark $$ **Final answer.** Long run: 10% Beginner, 20% Intermediate, 70% Advanced. Advanced holds the largest share. It is not strictly absorbing (10% drop back to I each day), but it is the dominant level. **Official key correction.** The key prints 16.3%, 25.5%, 58.2%. Those numbers do not satisfy $$ \pi=\pi P $$ (column B gives $$ 0.6(0.163)+0.2(0.255)=0.149\ne0.163 $$ ). The correct steady state is 0.1, 0.2, 0.7. **Question. For the hotel chain (Paris, London, Rome) with today's counts [50, 30, 20], find the long-run distribution and say which city gains or loses guests.** [part of 10 marks · Final Nov 2025 Q5a] #### Given $$ P=\begin{bmatrix}0.60&0.25&0.15\\0.30&0.50&0.20\\0.20&0.30&0.50\end{bmatrix} $$ #### To find Stationary shares and the requested comparison with the initial distribution. #### Formula and why it applies $$ \pi=\pi P,\quad\sum_i\pi_i=1 $$ At stationarity each destination receives exactly its current share. Solve the balance equations together with the total-probability equation. #### Solution ##### Step 1. Column P balance $$ \pi_P=0.6\pi_P+0.3\pi_L+0.2\pi_R\;\Rightarrow\;\pi_P=0.75\pi_L+0.5\pi_R $$ ##### Step 2. Column L balance $$ \pi_L=0.25\pi_P+0.5\pi_L+0.3\pi_R\;\Rightarrow\;0.5\pi_L=0.25\pi_P+0.3\pi_R $$ Substitute Step 1: $$ 0.5\pi_L=0.1875\pi_L+0.125\pi_R+0.3\pi_R\Rightarrow0.3125\pi_L=0.425\pi_R\Rightarrow\pi_L=1.36\,\pi_R $$ . Then $$ \pi_P=0.75(1.36\pi_R)+0.5\pi_R=1.52\,\pi_R $$ . ##### Step 3. Use the sum $$ (1.52+1.36+1)\pi_R=3.88\,\pi_R=1\Rightarrow\pi_R=0.2577,\;\pi_L=0.3505,\;\pi_P=0.3918 $$ Exactly: $$ \pi=[38/97,\;34/97,\;25/97] $$ . ##### Step 4. Compare with today's shares [0.50, 0.30, 0.20] | City | Today | Long run | Verdict | | --- | --- | --- | --- | | Paris | 0.50 | 0.392 | Loses guests | | London | 0.30 | 0.351 | Gains | | Rome | 0.20 | 0.258 | Gains | **Final answer.** $$ \pi\approx[0.392,\;0.351,\;0.258] $$ : Paris loses share; London and Rome gain. (The key's 0.3928 for Paris is a rounding slip; the three values must add to 1.) ## Other steady states from the papers | Question | Steady state | Compared with now | | --- | --- | --- | | Brand X/Y (Final Q1d) | [8/13, 5/13] = [0.615, 0.385] | — | | Savings/Current (Re-exam Q1d) | [1/3, 2/3] | Current dominates | | Booking W, A, C (Re-exam Q5b) | [8/21, 1/3, 2/7] = [0.381, 0.333, 0.286] | Now [0.40, 0.35, 0.25]: Website and App shrink slightly; Counter grows. | | Aircraft A, B, C (Re-exam Q6b) | [25/63, 20/63, 2/7] = [0.397, 0.317, 0.286] | Now [0.3, 0.4, 0.3]: A gains; B loses; C loses slightly. | **Official key correction (Re-exam Q5b).** The key says the Website “may grow”. Its steady share 0.381 is below today's 0.40, so the Website actually shrinks slightly; Counter is the one that grows. **Yaad rakho:** Steady state = “ab kuch nahi badlega”: $$ \pi = \pi P $$ aur total 1. Gain/loss = steady share vs aaj ka share. --- ## TEE · Markov reward process, return and discount Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-bellman--mrp-return/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-bellman/mrp-return Unit 4 · Markov reward process, Markov decision process, value functions, Bellman equations, applications ## Discounting changes the weight of later rewards γ (gamma) is a discount factor. Immediate reward has weight 1; one-step-later reward has weight γ; two-step-later reward has weight γ². This is repeated multiplication, not subtracting γ from a reward. An expectation averages possible future trajectories. A return G is the discounted total along one realised trajectory. **In one line:** a Markov reward process is a Markov chain that also pays a reward in each state; the return adds those rewards, shrinking later ones by γ. ## Definition A **Markov reward process (MRP)** is a tuple $$ \langle\mathcal S,P,R,\gamma\rangle $$ : - $$ \mathcal S $$ : finite set of states; $$ P $$ : transition matrix, $$ P_{ss'}=P(S_{t+1}=s'\mid S_t=s) $$ ; - $$ R $$ : reward function, $$ R(s)=\mathbb E[R_{t+1}\mid S_t=s] $$ (the expected reward for leaving $$ s $$ ); - $$ \gamma\in[0,1] $$ : the **discount factor**. There are no actions yet; the process simply drifts and collects rewards. (Add actions and you get an MDP.) ## Return The **return** $$ G_t $$ is the total discounted reward from time $$ t $$ : $$ G_t=R_{t+1}+\gamma R_{t+2}+\gamma^2R_{t+3}+\dots=\sum_{k=0}^{\infty}\gamma^kR_{t+k+1} $$ **Why discount?** (1) With bounded rewards, it keeps the infinite discounted sum finite when $$ \gamma<1 $$ . (2) Future rewards are uncertain. (3) Immediate rewards are usually preferred (like interest on money). $$ \gamma=0 $$ makes the agent short-sighted (only the next reward counts); $$ \gamma\to1 $$ makes it far-sighted. **Question. An episode gives rewards $$ R_1=-2,\;R_2=-1,\;R_3=-2,\;R_4=0 $$ and then ends. Find $$ G_0 $$ with $$ \gamma=0.9 $$ .** [practice · 2 marks] #### Given Rewards [−2,−1,−2,0]; discount γ=0.9; episode ends after the fourth reward. #### To find Return G₀. #### Formula and why it applies The first reward is undiscounted. Each later reward receives one extra factor of γ. $$ G_0=R_1+\gamma R_2+\gamma^2R_3+\gamma^3R_4 $$ #### Solution ##### Step 1. Write the discounted sum $$ G_0=R_1+\gamma R_2+\gamma^2R_3+\gamma^3R_4 $$ ##### Step 2. Substitute $$ G_0=-2+0.9(-1)+0.81(-2)+0.729(0)=-2-0.9-1.62=-4.52 $$ **Final answer.** $$ G_0=-4.52 $$ . ## State-value function of an MRP The **value** of a state is the expected return when starting there: $$ v(s)=\mathbb E\,[\,G_t\mid S_t=s\,] $$ The return of one episode is random; the value is its average over all possible futures. **Yaad rakho:** Return = aaj ka reward + γ × kal ka + γ² × parson ka… γ chhota = jaldi chahiye, γ bada = lambi soch. **An episode gives rewards [2, 4, 8] and then ends. Find the return from the start with γ=0.5.** [2 marks · Worked example] #### Given Rewards in order: R₁=2, R₂=4, R₃=8. Discount factor γ=0.5. The episode then ends, so there are no later rewards. #### To find G₀: the discounted sum of all three rewards, measured from the start. #### Formula and why it applies $$ G_0=R_1+\gamma R_2+\gamma^2R_3 $$ Keep the first reward undiscounted; multiply each later reward by the appropriate power. #### Solution ##### Step 1. First reward contribution $$ 1\times(2)=2 $$ ##### Step 2. Second reward contribution $$ 0.5\times(4)=2 $$ ##### Step 3. Third reward contribution $$ 0.5^2\times(8)=2 $$ ##### Step 4. Add contributions $$ G_0=(2)+(2)+(2)=6 $$ **Final answer.** 6. **An episode gives rewards [-1, 5, 2] and then ends. Find the return from the start with γ=0.9.** [2 marks · Worked example] #### Given Rewards in order: R₁=-1, R₂=5, R₃=2. Discount factor γ=0.9. The episode then ends, so there are no later rewards. #### To find G₀: the discounted sum of all three rewards, measured from the start. #### Formula and why it applies $$ G_0=R_1+\gamma R_2+\gamma^2R_3 $$ Keep the first reward undiscounted; multiply each later reward by the appropriate power. #### Solution ##### Step 1. First reward contribution $$ 1\times(-1)=-1 $$ ##### Step 2. Second reward contribution $$ 0.9\times(5)=4.5 $$ ##### Step 3. Third reward contribution $$ 0.9^2\times(2)=1.62 $$ ##### Step 4. Add contributions $$ G_0=(-1)+(4.5)+(1.62)=5.12 $$ **Final answer.** 5.12. **An episode gives rewards [2, 4, 8] and then ends. Find the return from the start with γ=0.** [2 marks · Worked example] #### Given Rewards in order: R₁=2, R₂=4, R₃=8. Discount factor γ=0. The episode then ends, so there are no later rewards. #### To find G₀: the discounted sum of all three rewards, measured from the start. #### Formula and why it applies $$ G_0=R_1+\gamma R_2+\gamma^2R_3 $$ Keep the first reward undiscounted; multiply each later reward by the appropriate power. #### Solution ##### Step 1. First reward contribution $$ 1\times(2)=2 $$ ##### Step 2. Second reward contribution $$ 0\times(4)=0 $$ ##### Step 3. Third reward contribution $$ 0^2\times(8)=0 $$ ##### Step 4. Add contributions $$ G_0=(2)+(0)+(0)=2 $$ **Final answer.** 2. --- ## TEE · Bellman equation for an MRP (with a solved example) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-bellman--bellman-mrp/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-bellman/bellman-mrp Unit 4 · Markov reward process, Markov decision process, value functions, Bellman equations, applications **A state pays immediate reward 2. Next it reaches a state of value 10 with probability 0.6 or a state of value 0 with probability 0.4. With γ=0.5, calculate its Bellman value.** [2 marks · Worked example] #### Given Immediate reward R=2; discount γ=0.5. Next-state possibilities: value 10 with probability 0.6, or value 0 with probability 0.4. These supplied continuation values already include rewards after reaching those states. #### To find Immediate reward plus discounted expected next value. #### Formula and why it applies $$ v(s)=R(s)+\gamma\sum_{s'}P(s'\mid s)v(s') $$ s′ is read “next state”; the prime is not a derivative. The sum is a weighted average of next-state values. #### Solution ##### Step 1. First branch contribution $$ 0.6\times10=6 $$ ##### Step 2. Second branch contribution $$ 0.4\times0=0 $$ ##### Step 3. Average, then discount $$ 6+0=6\qquad0.5\times6=3 $$ ##### Step 4. Add immediate reward $$ v(s)=2+3=5 $$ **Final answer.** v(s)=5. In the matrix expression, I is the identity matrix (ones on the diagonal and zero elsewhere); multiplying by I leaves a vector unchanged. The inverse solves simultaneous equations when it exists. For the small exam examples below, ordinary substitution is sufficient. **In one line:** the value of a state = the reward you get now + γ × the average value of where you land next. ## Derivation (short and examinable) $$ \begin{aligned}v(s)&=\mathbb E[G_t\mid S_t=s]\\&=\mathbb E[R_{t+1}+\gamma(R_{t+2}+\gamma R_{t+3}+\dots)\mid S_t=s]\\&=\mathbb E[R_{t+1}+\gamma G_{t+1}\mid S_t=s]\\&=R(s)+\gamma\sum_{s'\in\mathcal S}P_{ss'}\,v(s')\end{aligned} $$ The key step is $$ G_t=R_{t+1}+\gamma G_{t+1} $$ : the return splits into the immediate reward plus the discounted return from the next state. That recursive split is the whole idea of every Bellman equation. **Matrix form.** Stack the values in a column $$ v $$ : $$ v=R+\gamma Pv\Rightarrow (I-\gamma P)v=R\Rightarrow v=(I-\gamma P)^{-1}R $$ . For small problems just solve the linear equations. Swipe across the diagram to read its labels ↔ ![Done is terminal (it loops to itself with reward 0), so v(Done) = 0.](https://deep-learning-study-room.vercel.app/is2/fig/mrp_student.be69eb4eb6.svg) Diagram: Done is terminal (it loops to itself with reward 0), so v(Done) = 0.. Labels: A three-state Markov reward process (rewards: Class −2, Lab −1, Done 0) | 0.5 | 0.5 | 0.4 | 0.6 | 1 | Class | Lab | Done Done is terminal (it loops to itself with reward 0), so v(Done) = 0. **Question. In the MRP shown, Class pays −2 and Lab pays −1 when left; Done pays 0 and is terminal. From Class: 0.5 to Lab, 0.5 to Done. From Lab: 0.4 to Class, 0.6 to Done. With $$ \gamma=0.9 $$ , find v(Class) and v(Lab).** [practice · 5 marks] #### Given $$ R(C)=-2,\;R(L)=-1,\;v(D)=0,\;\gamma=0.9 $$ . #### To find $$ v(C),\;v(L) $$ . #### Formula and why it applies $$ v(s)=R(s)+\gamma\sum_{s'}P_{ss'}v(s') $$ It holds for every state, so we get one equation per non-terminal state. #### Solution ##### Step 1. Equation for Class $$ v(C)=-2+0.9\,[\,0.5\,v(L)+0.5\,(0)\,]=-2+0.45\,v(L) $$ ##### Step 2. Equation for Lab $$ v(L)=-1+0.9\,[\,0.4\,v(C)+0.6\,(0)\,]=-1+0.36\,v(C) $$ ##### Step 3. Substitute Step 2 into Step 1 $$ v(C)=-2+0.45(-1+0.36\,v(C))=-2.45+0.162\,v(C)\Rightarrow0.838\,v(C)=-2.45 $$ ##### Step 4. Solve $$ v(C)=\frac{-2.45}{0.838}=-2.9236,\qquad v(L)=-1+0.36(-2.9236)=-2.0525 $$ ##### Step 5. Check Class $$ -2+0.45(-2.0525)=-2-0.9236=-2.9236\;\checkmark $$ **Final answer.** $$ v(\text{Class})\approx-2.924,\;v(\text{Lab})\approx-2.053,\;v(\text{Done})=0 $$ . Lab is “less bad” because it is more likely to finish quickly. **Yaad rakho:** Bellman = “abhi ka reward + γ × aage ka average value”. Har state ki ek equation, phir solve. ## The substitution algebra in slow motion $$ v_C=-2+0.45(-1+0.36v_C) $$ Multiply .45 into both bracket terms: .45×(−1)=−.45 and .45×.36=.162. $$ v_C=-2-0.45+0.162v_C=-2.45+0.162v_C $$ Subtract .162vC from both sides; the coefficient on the left is 1−.162=.838. $$ 0.838v_C=-2.45\quad\Rightarrow\quad v_C=-2.45/0.838\approx-2.923627685 $$ Now use the Lab equation: .36×(−2.923627685)≈−1.052505967; adding −1 gives vL≈−2.052505967. --- ## TEE · Markov decision process, policies and value functions Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-bellman--mdp-policy/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-bellman/mdp-policy Unit 4 · Markov reward process, Markov decision process, value functions, Bellman equations, applications | Read this | Meaning | Concrete example | | --- | --- | --- | | π(a|s) | Probability the policy chooses action a in state s | In room A: left .3, right .7 | | P(s′|s,a) | Probability of the environment’s next state after that action | Right action reaches room B with probability .8 | | vπ(s) | Expected return starting in state s and following policy π | A number scoring the state | | qπ(s,a) | Expected return if action a is taken first, then policy π is followed | One score for each possible first action | | π in this chapter | A policy | Unlike π in a Markov-chain stationary distribution; context distinguishes them | The policy chooses an action; the transition model describes what happens after the action. They are two separate probability distributions. The worked examples use stationary Markov policies, depending on the current state and unchanged over time. More general policies may depend on time or history. **In one line:** an MDP is an MRP in which the agent chooses actions; a policy says which action to choose, and value functions score how good a state or a state–action pair is under that policy. ## Markov decision process A **Markov decision process (MDP)** is a tuple $$ \langle\mathcal S,\mathcal A,P,R,\gamma\rangle $$ : - $$ \mathcal S $$ : states; $$ \mathcal A $$ : actions; - $$ P^{a}_{ss'}=P(S_{t+1}=s'\mid S_t=s,A_t=a) $$ : transition probabilities, now depending on the action; - $$ R^{a}_{s}=\mathbb E[R_{t+1}\mid S_t=s,A_t=a] $$ : expected reward for taking $$ a $$ in $$ s $$ ; - $$ \gamma $$ : discount factor. The Markov property still holds: the next state and reward depend only on the current state and action. An MDP is a common model for RL when the chosen state captures the relevant information; partially observed problems require additional treatment. Swipe across the diagram to read its labels ↔ ![In s₁ the agent chooses “stay” or “go” (black dots). The environment then decides the outcome: “go” reaches s₂ with probability 0.8 (reward +5) or falls back to s₁ with 0.2 (reward 0).](https://deep-learning-study-room.vercel.app/is2/fig/mdp_two.331bc03880.svg) Diagram: In s₁ the agent chooses “stay” or “go” (black dots). The environment then decides the outcome: “go” reaches s₂ with probability 0.8 (reward +5) or falls back to s₁ with 0.2 (reward 0).. Labels: A two-state MDP: actions branch, then the environment decides the next state | s₁ | s₂ | a = stay | a = go | p = 1, r = +1 | p = 0.8, r = +5 | p = 0.2, r = 0 In s₁ the agent chooses “stay” or “go” (black dots). The environment then decides the outcome: “go” reaches s₂ with probability 0.8 (reward +5) or falls back to s₁ with 0.2 (reward 0). ## Policy A **policy** $$ \pi $$ is a distribution over actions given states: $$ \pi(a\mid s)=P(A_t=a\mid S_t=s) $$ - **Deterministic** policy: one fixed action per state, $$ a=\pi(s) $$ . - **Stochastic** policy: probabilities, e.g. $$ \pi(\text{stay}\mid s_1)=0.5,\;\pi(\text{go}\mid s_1)=0.5 $$ . The policies used in these examples are stationary Markov policies. The term policy more generally also permits time-dependent or history-dependent decision rules. ## Value functions **State-value function**: expected return from state $$ s $$ when following $$ \pi $$ : $$ v_\pi(s)=\mathbb E_\pi[G_t\mid S_t=s] $$ **Action-value function (Q-function)**: expected return when taking action $$ a $$ in $$ s $$ and following $$ \pi $$ afterwards: $$ q_\pi(s,a)=\mathbb E_\pi[G_t\mid S_t=s,A_t=a] $$ They are linked by averaging over the policy's choices: $$ v_\pi(s)=\sum_{a\in\mathcal A}\pi(a\mid s)\,q_\pi(s,a) $$ **Optimal** value functions take the best policy: $$ v_*(s)=\max_\pi v_\pi(s) $$ and $$ q_*(s,a)=\max_\pi q_\pi(s,a) $$ . An optimal policy acts greedily on $$ q_* $$ : $$ \pi_*(s)=\arg\max_a q_*(s,a) $$ . | | Markov process | MRP | MDP | | --- | --- | --- | --- | | Tuple | $$ \langle\mathcal S,P\rangle $$ | $$ \langle\mathcal S,P,R,\gamma\rangle $$ | $$ \langle\mathcal S,\mathcal A,P,R,\gamma\rangle $$ | | Rewards | No | Yes | Yes | | Decisions | No | No | Yes (actions) | | Value | — | $$ v(s) $$ | $$ v_\pi(s),\;q_\pi(s,a) $$ | **Yaad rakho:** MP → +reward = MRP → +action = MDP. v = “state kitni achhi”, q = “is state mein yeh action kitna achha”. --- ## TEE · Bellman expectation and optimality equations (worked) Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-bellman--bellman-equations/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-bellman/bellman-equations Unit 4 · Markov reward process, Markov decision process, value functions, Bellman equations, applications **At a state, action Left has value 3 and action Right has value 7. A policy chooses Left with probability 0.25 and Right with probability 0.75. Find the value under this policy, the maximum action value and the greedy action.** [2 marks · Worked example] #### Given Action values: Left=3 and Right=7. Policy probabilities: Left=0.25 and Right=0.75. The supplied action values already include subsequent returns. #### To find A weighted average, a maximum and the label attaining that maximum. #### Formula and why it applies $$ v_\pi(s)=\sum_a\pi(a\mid s)q_\pi(s,a) $$ The expectation averages according to policy; maximisation selects a largest action score. #### Solution ##### Step 1. Weight Left $$ 0.25\times3=0.75 $$ ##### Step 2. Weight Right $$ 0.75\times7=5.25 $$ ##### Step 3. Add for the policy value $$ v_\pi=0.75+5.25=6 $$ ##### Step 4. Read max and argmax max(3,7)=7 is a number. argmax is Right, the action with that number. **Final answer.** Policy value 6; maximum action value 7; greedy action Right. **In one line:** the expectation equations score a given policy; the optimality equations replace the policy's average by a max and describe the best possible behaviour. ## Bellman expectation equations Split the return into “reward now + discounted value later”, and average over the policy and the environment: $$ v_\pi(s)=\sum_{a}\pi(a\mid s)\sum_{s'}P(s'\mid s,a)\,\big[\,R(s,a,s')+\gamma\,v_\pi(s')\,\big] $$ $$ q_\pi(s,a)=\sum_{s'}P(s'\mid s,a)\,\Big[\,R(s,a,s')+\gamma\sum_{a'}\pi(a'\mid s')\,q_\pi(s',a')\,\Big] $$ Read the first one aloud: *for each action the policy might take, for each next state the environment might give, add the reward and γ × the next state's value, weighted by both probabilities.* ## Bellman optimality equations $$ v_*(s)=\max_{a}\sum_{s'}P(s'\mid s,a)\,\big[\,R(s,a,s')+\gamma\,v_*(s')\,\big] $$ $$ q_*(s,a)=\sum_{s'}P(s'\mid s,a)\,\big[\,R(s,a,s')+\gamma\max_{a'}q_*(s',a')\,\big] $$ The only change is $$ \sum_a\pi(a\mid s)\to\max_a $$ : the optimal agent picks the best action instead of averaging. These equations are non-linear (because of max), so they are solved iteratively (value iteration, policy iteration, Q-learning). **Question. In the two-state MDP (figure in the previous lesson), s₂ is terminal with $$ v(s_2)=0 $$ and $$ \gamma=0.9 $$ . In s₁, “stay” returns to s₁ with reward +1; “go” reaches s₂ with probability 0.8 and reward +5, or returns to s₁ with probability 0.2 and reward 0. (a) Find $$ v_\pi(s_1) $$ for the random policy $$ \pi(\text{stay})=\pi(\text{go})=0.5 $$ . (b) Find $$ v_*(s_1) $$ and the optimal action.** [practice · 6 marks] #### Given $$ q(s_1,\text{stay})=1+0.9\,v(s_1) $$ ; $$ q(s_1,\text{go})=0.8(5+0.9\cdot0)+0.2(0+0.9\,v(s_1))=4+0.18\,v(s_1) $$ . #### To find The value under the random policy, the optimal value and the optimal action. #### Formula and why it applies $$ v_\pi(s)=\sum_a\pi(a\mid s)q_\pi(s,a),\qquad v_*(s)=\max_a q_*(s,a) $$ The expectation equation averages actions using the given policy. The optimality equation chooses the largest action value. Each action value already includes its expected immediate reward and discounted continuation. #### Solution ##### Step 1. (a) Expectation equation for the random policy $$ v=0.5(1+0.9v)+0.5(4+0.18v)=2.5+0.54v\Rightarrow0.46v=2.5\Rightarrow v_\pi(s_1)=5.4348 $$ ##### Step 2. (b) Try “go” always $$ v=4+0.18v\Rightarrow v=4/0.82=4.878 $$ Check whether switching helps: $$ q(\text{stay})=1+0.9(4.878)=5.390>4.878 $$ . So “go” is not optimal. ##### Step 3. Try “stay” always $$ v=1+0.9v\Rightarrow v=10 $$ Check: $$ q(\text{go})=4+0.18(10)=5.8<10=q(\text{stay}) $$ . No improvement possible. **Final answer.** (a) $$ v_\pi(s_1)\approx5.43 $$ . (b) $$ v_*(s_1)=10 $$ with $$ q_*(s_1,\text{stay})=10 $$ , $$ q_*(s_1,\text{go})=5.8 $$ : the optimal action is **stay**. A steady +1 forever is worth $$ 1/(1-0.9)=10 $$ , more than a one-time +5. ## Seeing the max at work: value iteration Start with $$ v_0(s_1)=0 $$ and apply the optimality equation repeatedly: $$ v_{k+1}=\max(1+0.9v_k,\;4+0.18v_k) $$ . | k | stay = 1 + 0.9vₖ | go = 4 + 0.18vₖ | vₖ₊₁ | best | | --- | --- | --- | --- | --- | | 0 | 1.000 | 4.000 | 4.000 | go | | 1 | 4.600 | 4.720 | 4.720 | go | | 2 | 5.248 | 4.850 | 5.248 | stay | | 3 | 5.723 | 4.945 | 5.723 | stay | | … | converges | | 10 | stay | Early on the agent prefers the quick +5; as it “looks further ahead”, the steady reward wins. This is why γ and long-term value matter. **Yaad rakho:** Expectation = policy ka average (Σπ). Optimality = best action (max). Baaki sab same: reward + γ × next value. ## Expand the random-policy equation For the full MDP question above, “stay” gives 1+.9v. “Go” has two possible transitions: .8×(5+.9×0)=4 and .2×(0+.9v)=.18v, so its total is 4+.18v. The random policy then averages these two action values. $$ v=0.5(1+0.9v)+0.5(4+0.18v) $$ $$ v=0.5+0.45v+2+0.09v=2.5+0.54v $$ Subtract .54v from both sides, then divide by .46. $$ (1-0.54)v=2.5\quad0.46v=2.5\quad v=2.5/0.46\approx5.434782609 $$ --- ## TEE · Using Bellman equations: policy and value iteration Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-bellman--solving/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-bellman/solving Unit 4 · Markov reward process, Markov decision process, value functions, Bellman equations, applications A **backup** means applying the Bellman equation once to improve a value estimate. In value iteration, each sweep uses the previous sweep’s values for all states. In policy iteration, first solve values for the current policy; then choose the action with the largest one-step action value using those solved continuation values. Repeat if the policy changes. The “try go, then improve to stay” calculation in the previous lesson is one complete policy-improvement example. **In one line:** policy/value iteration use Bellman equations with a model; Q-learning uses sampled experience. These are examples of methods, not a claim about every RL algorithm. ## 1. Policy evaluation Given $$ \pi $$ , compute $$ v_\pi $$ by repeatedly applying the expectation equation until values stop changing: $$ v_{k+1}(s)\leftarrow\sum_a\pi(a\mid s)\sum_{s'}P(s'\mid s,a)\,[R+\gamma v_k(s')] $$ ## 2. Policy iteration Start with any policy π → Evaluate: find $$ v_\pi $$ → Improve: $$ \pi'(s)=\arg\max_a q_\pi(s,a) $$ → Repeat until the policy stops changing For a finite discounted MDP with exact policy evaluation and consistent tie handling, policy improvement does not reduce value and policy iteration can terminate at an optimal policy. ## 3. Value iteration Skip full evaluation; apply the optimality equation directly: $$ v_{k+1}(s)\leftarrow\max_a\sum_{s'}P(s'\mid s,a)\,[R+\gamma v_k(s')] $$ Stop when the change is tiny, then act greedily. The table in the previous lesson is value iteration on one state. Optional extension: Q-learning (not an observed PYQ) The two supplied TEE papers do not ask a Q-learning update. The January 2026 paper asks an explanation of the Bellman equation and RL applications. This extra worked example can wait until those requirements are covered. ## Q-learning Q-learning is included to illustrate a Bellman target; it is not separately named in the supplied introductory RL syllabus. When $$ P $$ and $$ R $$ are unknown, the agent learns $$ q_* $$ from its own experience $$ (s,a,r,s') $$ : $$ Q(s,a)\leftarrow Q(s,a)+\alpha\,\big[\,r+\gamma\max_{a'}Q(s',a')-Q(s,a)\,\big] $$ $$ \alpha $$ is the learning rate. The bracket is the **temporal-difference error**: (Bellman target) − (current guess). The agent usually explores with ε-greedy actions while learning. **Question. A Q-table has $$ Q(s,a)=2 $$ . The agent takes $$ a $$ in $$ s $$ , receives $$ r=3 $$ and lands in $$ s' $$ where $$ \max_{a'}Q(s',a')=5 $$ . With $$ \alpha=0.5,\;\gamma=0.9 $$ , find the updated $$ Q(s,a) $$ .** [Optional authored example] #### Given Old Q=2, reward=3, best next-state Q=5, α=0.5, γ=0.9; the next state is nonterminal. #### To find The updated Q estimate. #### Formula and why it applies Move partway from the old estimate toward the one-step Bellman target. $$ Q_{new}=Q_{old}+\alpha[r+\gamma\max Q_{next}-Q_{old}] $$ #### Solution ##### Step 1. Target $$ r+\gamma\max_{a'}Q(s',a')=3+0.9(5)=7.5 $$ ##### Step 2. Update $$ Q\leftarrow2+0.5(7.5-2)=2+2.75=4.75 $$ **Final answer.** $$ Q(s,a)=4.75 $$ . **Yaad rakho:** Model pata hai → policy/value iteration. Model nahi pata → Q-learning (experience se seekho). --- ## TEE · 10-mark answer: Bellman equation and applications of RL Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-bellman--ten-mark-answer/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/read/u4-bellman/ten-mark-answer Unit 4 · Markov reward process, Markov decision process, value functions, Bellman equations, applications **Asked as:** “Explain Bellman equation in reinforcement learning. Also give applications of reinforcement learning.” [10 marks · Re-exam Jan 2026 Q7b; marking: Bellman explanation 5, applications 5] Write the answer below in this order. It fits in about two pages. ### Part A · Bellman equation (5 marks) **1. Setting.** Reinforcement learning problems are modelled as a Markov decision process $$ \langle\mathcal S,\mathcal A,P,R,\gamma\rangle $$ . The agent follows a policy $$ \pi(a\mid s) $$ and wants to maximise the return $$ G_t=\sum_{k\ge0}\gamma^kR_{t+k+1} $$ . **2. Value functions.** $$ v_\pi(s)=\mathbb E_\pi[G_t\mid S_t=s] $$ and $$ q_\pi(s,a)=\mathbb E_\pi[G_t\mid S_t=s,A_t=a] $$ . **3. Key idea.** The return is recursive: $$ G_t=R_{t+1}+\gamma G_{t+1} $$ . Taking expectations gives the **Bellman expectation equation**: $$ v_\pi(s)=\sum_a\pi(a\mid s)\sum_{s'}P(s'\mid s,a)\,[R(s,a,s')+\gamma v_\pi(s')] $$ $$ q_\pi(s,a)=\sum_{s'}P(s'\mid s,a)\,\Big[R(s,a,s')+\gamma\sum_{a'}\pi(a'\mid s')q_\pi(s',a')\Big] $$ **4. Optimality.** For the optimal policy the average over actions becomes a maximum (**Bellman optimality equation**): $$ v_*(s)=\max_a\sum_{s'}P(s'\mid s,a)\,[R+\gamma v_*(s')],\qquad q_*(s,a)=\sum_{s'}P(s'\mid s,a)\,[R+\gamma\max_{a'}q_*(s',a')] $$ **5. Meaning and use.** The value of a state equals the immediate reward plus the discounted value of the successor state. This breaks a long-horizon problem into one-step sub-problems (dynamic programming). It is the basis of policy evaluation, policy iteration, value iteration and Q-learning; the optimal policy is $$ \pi_*(s)=\arg\max_a q_*(s,a) $$ . **6. Small illustration** (adds clarity): if staying in a state gives +1 each step with $$ \gamma=0.9 $$ , then $$ v=1+0.9v\Rightarrow v=10 $$ . ### Part B · Applications of reinforcement learning (5 marks) 1. **Game playing**: AlphaGo (defeated Lee Sedol, 2016), Atari agents (DQN), and the Mario agent built in this course's lab. 2. **Robotics**: walking, grasping and balancing learned by trial and error in simulation. 3. **Autonomous driving**: lane changing, merging and parking decisions. 4. **Recommendation systems**: choosing content to maximise long-term engagement rather than single clicks. 5. **Finance**: trading and portfolio rebalancing as sequential decisions. 6. **Healthcare**: dynamic treatment regimes (dose adjustment over time). 7. **Resource and energy management**: data-centre cooling control, traffic-signal timing, inventory control. 8. **Language models**: reinforcement learning from human feedback (RLHF) for chatbots. #### One application in detail: traffic-signal control | | | | --- | --- | | Agent | Controller at an intersection. | | State | Queue length and waiting time on each approach, current signal phase. | | Actions | Keep the current phase or switch to the next green phase. | | Reward | Negative of total waiting time (or vehicles cleared per cycle). | | Learning | Q-learning updates $$ Q(s,a) $$ from observed traffic; ε-greedy exploration tries alternative timings. | | Benefit | Adapts to rush hours and incidents; fixed timers cannot. | | Challenge | Safety during exploration, so agents are trained in simulation first. | **Yaad rakho:** Bellman ka sar: “value = reward abhi + γ × value aage”. Application likho to S, A, R batao: examiner ko yahi chahiye. --- ## TEE · Fuzzy formulas, with meanings and worked-example links Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/formulas/m2-formulas-v2--fuzzy/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/formulas/m2-formulas-v2/fuzzy One place for the formulas, meanings, conditions and links to full worked examples. ## Fuzzy formulas ### Triangular membership $$ \mu(x)=\max\left(0,\min\left(\frac{x-a}{b-a},\frac{c-x}{c-b}\right)\right) $$ a=left zero, b=peak, c=right zero, with a0 sets the width. At x=c the exponent is zero and membership is 1. [Worked exponential calculations →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--smooth-memberships/) ### Generalised bell membership $$ \mu(x)=\frac{1}{1+|(x-c)/a|^{2b}} $$ a>0 controls width; b>0 controls steepness; c is the centre. First subtract c, divide by a, take absolute value, apply the power, add 1, then take its reciprocal. ### Alpha-cut $$ A_\alpha=\{x:\mu_A(x)\ge\alpha\} $$ α is the chosen membership threshold. For 0<α≤1 keep inputs meeting or exceeding it, including threshold-equality endpoints. [Discrete and triangular examples →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/fuzzy-core--properties/) --- ## TEE · GA and PSO formulas, symbols and memory rules Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/formulas/m2-formulas-v2--ga-pso/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/formulas/m2-formulas-v2/ga-pso One place for the formulas, meanings, conditions and links to full worked examples. ### Binary decoding $$ x=\sum_{i=1}^{n}b_i2^{n-i} $$ bᵢ is a bit; the maximum n-bit unsigned integer is 2ⁿ−1. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--encoding/) ### Roulette selection $$ P_i=F_i/\sum_jF_j\quad C_i=\sum_{j=1}^iP_j $$ Fitness weights must be nonnegative with a positive total. For minimisation use the specified transformation or selection method. C is cumulative probability. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--selection/) ### Population average $$ \bar f=\frac{f_1+\cdots+f_N}{N} $$ N is the number of individuals. Compare both best and average; mutation need not improve either. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--two-generations/) ### PSO velocity $$ v^{t+1}=wv^t+c_1r_1(pbest^t-x^t)+c_2r_2(gbest^t-x^t) $$ w weights inertia; c₁r₁ weights personal pull; c₂r₂ weights social pull. Apply the same equation separately to each coordinate. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--equation/) ### PSO position $$ x^{t+1}=x^t+v^{t+1} $$ Use the new velocity, not the old one. Then reevaluate the objective. [Full explanation and worked solutions →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/pso-core--two-particles/) **Memory updates:** minimise → replace a pbest only for a lower score; maximise → only for a higher score. gbest is a position selected from the personal bests. Store the corresponding objective score alongside it. Keep previous best on ties unless instructed otherwise. **Yaad rakho:** Formula ke pehle objective direction aur old state likho. Iteration 2 mein fresh positions aur retained memories use karo. ## Encoding and minimisation conventions $$ x=L+\frac{d}{2^n-1}(U-L) $$ For the specified uniform real-range encoding: decode n bits into unsigned integer d first. L and U are the physical limits. [Endpoint examples →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--encoding/) $$ F=\frac{1}{1+f(x)} $$ This is the stated roulette transformation in our nonnegative-cost minimisation example. It is not required for every GA. It converts a smaller cost into a larger selection weight. [Complete minimisation example →](https://deep-learning-study-room.vercel.app/read/is2/m2/read/ga-core--minimisation/) --- ## TEE · Unit 1 formulas: probability, Bayes, naive Bayes, Bayesian networks, GMM Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/formulas/tee-formulas--unit1/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/formulas/tee-formulas/unit1 Probability, Bayes, GMM/EM, Markov chains and Bellman equations (Units 2–3 are in the M2 sheet) ## Conditions and notation for this revision sheet Conditional probabilities require nonzero evidence probability. “Prior” means the class probability before the features; “likelihood” is the feature probability given a class; “posterior” is the class probability after evidence. Σ adds alternatives; Π multiplies factors; argmax returns the winning label. For Laplace smoothing, Nc is the class count and k is the number of possible values of the particular feature. For the Jensen expression below, λ weights are nonnegative, sum to 1, and the logged arguments are positive. For a Gaussian, σ>0. GMM weights πk are nonnegative and sum to 1; γik or λki are responsibilities, not the mixing weights themselves. Worked explanations: [Naive Bayes](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-nb--model/) · [CPTs and Bayesian networks](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-bn--semantics/) · [Mean, variance and likelihood](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-gmm--centre-spread/) · [EM update](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u1-gmm--em/). | | | | --- | --- | | Conditional | $$ P(a\mid b)=P(a\land b)/P(b) $$ | | Product rule | $$ P(a\land b)=P(a\mid b)P(b) $$ | | Marginalisation | $$ P(Y)=\sum_zP(Y,z) $$ | | Normalisation | $$ \mathbf P(X\mid e)=\alpha\,\mathbf P(X,e),\;\alpha=1/\sum_xP(x,e) $$ | | Bayes' rule | $$ P(A\mid B)=\dfrac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid\lnot A)P(\lnot A)} $$ | | Naive Bayes | $$ v_{NB}=\arg\max_cP(c)\prod_iP(a_i\mid c) $$ ; Laplace: $$ \frac{n_{ic}+1}{N_c+k} $$ | | BN chain rule | $$ P(x_1..x_n)=\prod_iP(x_i\mid parents(X_i)) $$ | | BN query | $$ P(X\mid e)=\alpha\sum_{\text{hidden}}\prod P(\cdot\mid parents) $$ | | CPT size | Boolean node with k Boolean parents: independent probability parameters = $$ 2^k $$ numbers | | GMM density | $$ p(x)=\sum_k\pi_k\mathcal N(x\mid\mu_k,\sigma_k^2) $$ | | Gaussian | $$ \mathcal N(x\mid\mu,\sigma^2)=\frac1{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$ | | Log-likelihood | $$ \log L=\sum_i\log\sum_k\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2) $$ | | Jensen (log concave) | $$ \log\sum_k\lambda_ka_k\ge\sum_k\lambda_k\log a_k $$ | | E-step | $$ \lambda_k^i=\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)/\sum_l\pi_l\mathcal N(x_i\mid\mu_l,\sigma_l^2) $$ | | M-step | $$ \mu_k=\frac{\sum\lambda_k^ix_i}{\sum\lambda_k^i},\;\sigma_k^2=\frac{\sum\lambda_k^i(x_i-\mu_k)^2}{\sum\lambda_k^i},\;\pi_k=\frac{\sum_i\lambda_k^i}{n} $$ | | GMM parameters | 1-D: $$ 3K-1 $$ (K = 5 → 14) | --- ## TEE · Unit 4 formulas: Markov chains, MRP, MDP, Bellman Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/formulas/tee-formulas--unit4/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/formulas/tee-formulas/unit4 Probability, Bayes, GMM/EM, Markov chains and Bellman equations (Units 2–3 are in the M2 sheet) ## Read these symbols in context P is a transition matrix with origin rows and destination columns. π denotes a stationary distribution in the first equations, but a policy in the MDP equations. s′ is a next state; a′ a next action. R is the specified immediate or expected reward; γ discounts future rewards. E means expectation (probability-weighted average). I is the identity matrix. The two-state stationary shortcut needs a+b>0. The inverse (I−γP)⁻¹ exists for a finite stochastic P with 0≤γ<1; at γ=1, check the problem’s terminal structure instead of assuming the full matrix is invertible. The continuing discounted-return examples use γ<1. A finite episode can also use γ=1. Q-learning is supplementary here; the explicitly listed course focus is the Markov/MDP and Bellman material. α in Q-learning is a learning rate; α in Bayes normalisation was a different constant. Worked explanations: [Matrix multiplication](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-markov--next-step/) · [Stationary algebra](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-markov--steady-state/) · [Discounted return](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-bellman--mrp-return/) · [Bellman equations](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-bellman--bellman-equations/). | | | | --- | --- | | Markov property | $$ P(S_{t+1}\mid S_t,\dots,S_1)=P(S_{t+1}\mid S_t) $$ | | Transition matrix | $$ P_{ij}=P(\text{next}=j\mid\text{now}=i) $$ ; rows sum to 1 | | Next step | $$ x_{t+1}=x_tP $$ (row vector); $$ x_{t+n}=x_tP^n $$ | | Steady state | $$ \pi=\pi P,\;\sum\pi_i=1 $$ ; two states $$ \begin{bmatrix}1-a&a\\b&1-b\end{bmatrix}\Rightarrow\pi=\left[\frac b{a+b},\frac a{a+b}\right] $$ | | Return | $$ G_t=\sum_{k\ge0}\gamma^kR_{t+k+1} $$ | | MRP Bellman | $$ v(s)=R(s)+\gamma\sum_{s'}P_{ss'}v(s') $$ ; $$ v=(I-\gamma P)^{-1}R $$ | | Policy | $$ \pi(a\mid s)=P(A_t=a\mid S_t=s) $$ | | Value functions | $$ v_\pi(s)=\mathbb E_\pi[G_t\mid s] $$ ; $$ q_\pi(s,a)=\mathbb E_\pi[G_t\mid s,a] $$ ; $$ v_\pi=\sum_a\pi(a\mid s)q_\pi(s,a) $$ | | Bellman expectation | $$ v_\pi(s)=\sum_a\pi(a\mid s)\sum_{s'}P(s'\mid s,a)[R+\gamma v_\pi(s')] $$ | | Bellman optimality | $$ v_*(s)=\max_a\sum_{s'}P(s'\mid s,a)[R+\gamma v_*(s')] $$ ; $$ q_*(s,a)=\sum_{s'}P(s'\mid s,a)[R+\gamma\max_{a'}q_*(s',a')] $$ | Optional further reading: the [Q-learning extension](https://deep-learning-study-room.vercel.app/read/is2/tee/read/u4-bellman--solving/) is kept with the Bellman lesson, outside this core formula list. --- ## TEE · TEE-only formulas: ACO and GA for TSP Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/formulas/tee-formulas--aco-tsp/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/formulas/tee-formulas/aco-tsp Probability, Bayes, GMM/EM, Markov chains and Bellman equations (Units 2–3 are in the M2 sheet) ## Ant colony optimisation | | | | --- | --- | | Transition probability | $$ P_{ij}=\dfrac{\tau_{ij}^\alpha\eta_{ij}^\beta}{\sum_{l\in\text{allowed}}\tau_{il}^\alpha\eta_{il}^\beta},\;\eta_{ij}=1/d_{ij} $$ | | Choice rule | cumulative probabilities in **city order**; first city with cumulative ≥ r | | Pheromone update | $$ \tau_{ij}\leftarrow(1-\rho)\tau_{ij}+\sum_{k:\,ij\in tour_k} Q/L_k $$ on every edge ant k used (return edge included); symmetric only for an undirected symmetric TSP | | Values assumed in the worked PYQs | $$ \alpha=\beta=1,\;\rho=0.5,\;Q=1,\;\tau_0=1 $$ | α weights pheromone influence; β weights inverse-distance influence; ρ is the evaporation rate; Q is the deposit constant; L is the complete tour length. These must be supplied or explicitly assumed. Do not treat the example values as universal defaults. $$ L=\sum_{i=1}^{n-1}d(c_i,c_{i+1})+d(c_n,c_1) $$ $$ F_{TSP}=1/L $$ The return edge is part of the tour. Permutation crossover/mutation must keep each city exactly once. For positive lengths, shorter length gives larger reciprocal fitness. --- ## TEE · Q1(a) Joint probability table [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q1a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q1a Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. A supermarket recorded the joint distribution of Product type (Y: premium = 1, regular = 0), Payment mode (X: credit = 1, debit = 0) and Gender (Z: female = 1, male = 0). (a) If a customer bought a premium item, find the probability that (1) they are female, (2) they paid by credit card. (b) Find (1) P(premium) and (2) P(credit card). (c) Given a male customer paid by debit card, find the probability he bought a regular item.** [5 marks] ### Answer #### Given The printed eight entries are reproduced below. Their total is 1.002, not 1. Therefore the table is inconsistent as a probability distribution. | | Credit, female | Credit, male | Debit, female | Debit, male | | --- | --- | --- | --- | --- | | **Premium (Y=1)** | 0.08 | 0.10 | 0.072 | 0.05 | | **Regular (Y=0)** | 0.10 | 0.15 | 0.25 | 0.20 | #### To find The five requested probabilities, stating how the inconsistent data are interpreted. #### Formula and why it applies For this worked solution, treat all printed entries as nonnegative weights and normalise by their total 1.002. This preserves the printed ratios. It is a declared interpretation, not a claim about the examiner’s intended correction. $$ P(E)=\frac{\text{sum of weights for }E}{\text{sum of all weights}} $$ $$ P(A\mid E)=\frac{\text{sum of weights for }A\cap E}{\text{sum of weights for }E} $$ #### Solution ##### Step 1. Validate the full table $$ 0.08+0.10+0.072+0.05+0.10+0.15+0.25+0.20=1.002 $$ Use total weight 1.002 consistently for unconditional probabilities. ##### Step 2. Premium row total $$ w(premium)=0.08+0.10+0.072+0.05=0.302 $$ ##### Step 3. Female given premium $$ P(female\mid premium)=\frac{0.08+0.072}{0.302}=\frac{0.152}{0.302}\approx0.503311 $$ ##### Step 4. Credit given premium $$ P(credit\mid premium)=\frac{0.08+0.10}{0.302}=\frac{0.18}{0.302}\approx0.596026 $$ ##### Step 5. Normalised premium marginal $$ P(premium)=\frac{0.302}{1.002}\approx0.301397 $$ ##### Step 6. Normalised credit marginal $$ P(credit)=\frac{0.08+0.10+0.10+0.15}{1.002}=\frac{0.43}{1.002}\approx0.429142 $$ ##### Step 7. Regular given male and debit $$ P(regular\mid male,debit)=\frac{0.20}{0.05+0.20}=0.8 $$ **Final answer.** Under common normalisation: (a) 0.503311 and 0.596026; (b) 0.301397 and 0.429142; (c) 0.8. **Alternative correction:** if 0.072 is corrected to 0.070, the total becomes 1 and the corresponding answers are (a) 0.50 and 0.60; (b) 0.30 and 0.43; (c) 0.8. This agrees with the interpretation used in the supplied key. State the correction; do not call raw totals 0.302 and 0.43 valid marginals of the original inconsistent table. --- ## TEE · Q1(b) Applications of ACO [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q1b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q1b Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **List the applications of Ant Colony Optimization and explain any one application in detail.** [5 marks] ### Answer ### Definition (1 line) Ant Colony Optimization is a probabilistic, population-based metaheuristic (Dorigo, 1992) in which artificial ants build solutions on a graph and communicate through pheromone trails. In the update rule used here, evaporation reduces all stored trails; deposits then reinforce edges of selected tours according to solution quality. Evaporation does not occur only on poor paths. ### Applications 1. **Travelling salesman problem**: shortest tour visiting every city once. 2. **Vehicle routing**: delivery routes for a fleet with capacity and time windows (minimising distance, time, fuel). 3. **Network routing**: adaptive routing of data packets in communication and sensor networks (AntNet). 4. **Job-shop and project scheduling**: assigning jobs to machines to minimise completion time. 5. **Image processing**: edge detection, with ants moving over pixel-intensity graphs. 6. **Assignment problems**: quadratic assignment, timetabling, graph colouring. 7. **Feature selection and data mining**: Ant-Miner for classification rules. 8. **Load balancing** in distributed systems and cloud task allocation. ### One application in detail: vehicle routing (or TSP) 1. **Graph.** Depot and customers are nodes; travel times or distances are edge weights. 2. **Ants.** Each ant builds a complete set of routes. From its current customer it chooses the next with probability $$ \propto\tau_{ij}\,(1/d_{ij}) $$ , skipping customers already served or that would exceed the vehicle capacity; when the vehicle is full it returns to the depot. 3. **Evaluation.** Total distance (plus penalties for late deliveries). 4. **Learning.** Pheromone evaporates everywhere and is added in proportion to $$ Q/L $$ on the routes of good solutions, so short connections are reinforced. 5. **Result.** Repeated iterations search for low-cost routes; finite-run convergence to the optimum is not guaranteed. Companies use this for parcel delivery, waste collection and school-bus routing. ### Advantages and limitations | Advantages | Limitations | | --- | --- | | Naturally suited to graph and routing problems | Several parameters to tune (α, β, ρ, number of ants) | | Can share useful route information and adapt search to changing graphs | Convergence time is uncertain; may stagnate on a sub-optimal path | | Positive feedback + parallel ants | Theoretical analysis is difficult | **Yaad rakho:** ACO = “raaste wale problems” (TSP, routing, scheduling). PSO = “numbers wale problems” (continuous function, weights). --- ## TEE · Q1(c) Fuzzy set operations [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q1c/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q1c Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Consider two fuzzy sets X = {(c1, 0.6), (c2, 0.8), (c3, 0.3), (c4, 0.5), (c5, 0.7)} and Y = {(c1, 0.4), (c2, 0.5), (c3, 0.9), (c4, 0.6), (c5, 0.2)}. Find 1) X ∪ Y 2) X ∩ Y 3) ¬X ∪ ¬Y 4) ¬X ∩ ¬Y.** [5 marks] ### Answer #### Given The aligned X and Y memberships printed in this question. #### To find The four requested membership lists in the original element order. #### Formula and why it applies Use max for OR, min for AND, and 1−μ for NOT. $$ \mu_{X\cup Y}=\max(\mu_X,\mu_Y),\quad\mu_{X\cap Y}=\min(\mu_X,\mu_Y) $$ Complement both operands before applying the operator in ¬X∪¬Y or ¬X∩¬Y. #### Solution ##### Step 1. Complements ¬X = {0.4, 0.2, 0.7, 0.5, 0.3}; ¬Y = {0.6, 0.5, 0.1, 0.4, 0.8}. ##### Step 2. Table | | c1 | c2 | c3 | c4 | c5 | | --- | --- | --- | --- | --- | --- | | X | 0.6 | 0.8 | 0.3 | 0.5 | 0.7 | | Y | 0.4 | 0.5 | 0.9 | 0.6 | 0.2 | | **X ∪ Y = max** | 0.6 | 0.8 | 0.9 | 0.6 | 0.7 | | **X ∩ Y = min** | 0.4 | 0.5 | 0.3 | 0.5 | 0.2 | | **¬X ∪ ¬Y** | 0.6 | 0.5 | 0.7 | 0.5 | 0.8 | | **¬X ∩ ¬Y** | 0.4 | 0.2 | 0.1 | 0.4 | 0.3 | ##### Step 3. Check with De Morgan's law ¬X ∪ ¬Y must equal 1 − (X ∩ Y) = {0.6, 0.5, 0.7, 0.5, 0.8} ✓, and ¬X ∩ ¬Y must equal 1 − (X ∪ Y) = {0.4, 0.2, 0.1, 0.4, 0.3} ✓. **Final answer.** X ∪ Y = {(c1,0.6),(c2,0.8),(c3,0.9),(c4,0.6),(c5,0.7)}; X ∩ Y = {(c1,0.4),(c2,0.5),(c3,0.3),(c4,0.5),(c5,0.2)}; ¬X ∪ ¬Y = {(c1,0.6),(c2,0.5),(c3,0.7),(c4,0.5),(c5,0.8)}; ¬X ∩ ¬Y = {(c1,0.4),(c2,0.2),(c3,0.1),(c4,0.4),(c5,0.3)}. Official marking: 1 + 1 + 1.5 + 1.5. ## Calculate every element separately ### Element c1 $$ \neg X:1-0.6=0.4\qquad\neg Y:1-0.4=0.6 $$ $$ X\cup Y:\max(0.6,0.4)=0.6 $$ $$ X\cap Y:\min(0.6,0.4)=0.4 $$ $$ \neg X\cup\neg Y:\max(0.4,0.6)=0.6 $$ $$ \neg X\cap\neg Y:\min(0.4,0.6)=0.4 $$ ### Element c2 $$ \neg X:1-0.8=0.2\qquad\neg Y:1-0.5=0.5 $$ $$ X\cup Y:\max(0.8,0.5)=0.8 $$ $$ X\cap Y:\min(0.8,0.5)=0.5 $$ $$ \neg X\cup\neg Y:\max(0.2,0.5)=0.5 $$ $$ \neg X\cap\neg Y:\min(0.2,0.5)=0.2 $$ ### Element c3 $$ \neg X:1-0.3=0.7\qquad\neg Y:1-0.9=0.1 $$ $$ X\cup Y:\max(0.3,0.9)=0.9 $$ $$ X\cap Y:\min(0.3,0.9)=0.3 $$ $$ \neg X\cup\neg Y:\max(0.7,0.1)=0.7 $$ $$ \neg X\cap\neg Y:\min(0.7,0.1)=0.1 $$ ### Element c4 $$ \neg X:1-0.5=0.5\qquad\neg Y:1-0.6=0.4 $$ $$ X\cup Y:\max(0.5,0.6)=0.6 $$ $$ X\cap Y:\min(0.5,0.6)=0.5 $$ $$ \neg X\cup\neg Y:\max(0.5,0.4)=0.5 $$ $$ \neg X\cap\neg Y:\min(0.5,0.4)=0.4 $$ ### Element c5 $$ \neg X:1-0.7=0.3\qquad\neg Y:1-0.2=0.8 $$ $$ X\cup Y:\max(0.7,0.2)=0.7 $$ $$ X\cap Y:\min(0.7,0.2)=0.2 $$ $$ \neg X\cup\neg Y:\max(0.3,0.8)=0.8 $$ $$ \neg X\cap\neg Y:\min(0.3,0.8)=0.3 $$ --- ## TEE · Q1(d) Brand-switching Markov chain [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q1d/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q1d Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Users switch between Brand X and Brand Y each year: 25% of X users switch to Y and 75% stay; 40% of Y users switch to X and 60% stay. (a) Write the transition probability matrix. (b) Draw the state transition diagram.** [5 marks] ### Answer #### Given All transition values are given in the full question. State order: X, Y. #### To find The transition matrix and state diagram. #### Formula and why it applies Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add. #### Solution ##### Step 1. Convert every origin row to probabilities | From / To | X | Y | | --- | --- | --- | | X | 0.75 | 0.25 | | Y | 0.4 | 0.6 | Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25. ##### Step 2. Interpret entries and draw the transitions Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving. Swipe across the diagram to read its labels ↔ ![Each arrow is labelled with the probability of that transition.](https://deep-learning-study-room.vercel.app/is2/fig/mk_brand.c2c4be3e6a.svg) Diagram: Each arrow is labelled with the probability of that transition.. Labels: Brand switching each year | 0.25 | 0.4 | 0.75 | 0.6 | X | Y Each arrow is labelled with the probability of that transition. **Final answer.** The transition matrix and state diagram are given above. --- ## TEE · Q2(a) PSO, one iteration [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q2a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q2a Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Maximize $$ F(x_1,x_2)=2x_1^2-3x_1x_2+x_2^2+5x_1+2x_2+7 $$ using Particle Swarm Optimization. $$ X=\begin{bmatrix}2.5&-3.0\\3.5&-1.5\end{bmatrix},\;V=\begin{bmatrix}0.2&0.3\\0.1&0.25\end{bmatrix} $$ . Take $$ W=0.9,\;c_1=c_2=1.5,\;r_1=0.5949,\;r_2=0.0855 $$ . Optimize and give the solution after 1 iteration showing the step-by-step process.** [10 marks] ### Answer #### Given Particle 1: $$ x=(2.5,\,-3.0) $$ , $$ v=(0.2,\,0.3) $$ . Particle 2: $$ x=(3.5,\,-1.5) $$ , $$ v=(0.1,\,0.25) $$ . $$ c_1r_1=1.5\times0.5949=0.89235,\qquad c_2r_2=1.5\times0.0855=0.12825 $$ #### To find Fitness, pbest, gbest; new velocities and positions; updated pbest and gbest. #### Formula and why it applies $$ v_{ij}\leftarrow wv_{ij}+c_1r_1(p^b_{ij}-x_{ij})+c_2r_2(p^g_j-x_{ij}),\qquad x_{ij}\leftarrow x_{ij}+v_{ij} $$ Each particle is pulled by its own memory and by the swarm's best; we maximise, so the larger $$ F $$ is better. #### Solution ##### Step 1. Fitness of particle 1 $$ F(2.5,-3)=2(6.25)-3(2.5)(-3)+9+5(2.5)+2(-3)+7=12.5+22.5+9+12.5-6+7=57.5 $$ ##### Step 2. Fitness of particle 2 $$ F(3.5,-1.5)=2(12.25)-3(3.5)(-1.5)+2.25+17.5-3+7=24.5+15.75+2.25+17.5-3+7=64 $$ ##### Step 3. pbest and gbest | Particle | Position | F | pbest | | --- | --- | --- | --- | | 1 | (2.5, −3.0) | 57.5 | (2.5, −3.0) | | 2 | (3.5, −1.5) | 64 | (3.5, −1.5) | No previous iteration, so pbest = current position. Largest fitness is 64, so **gbest = (3.5, −1.5)**. ##### Step 4. Velocity and position of particle 1 $$ v_{11}=0.9(0.2)+0.89235(2.5-2.5)+0.12825(3.5-2.5)=0.18+0+0.12825=0.30825 $$ $$ x_{11}=2.5+0.30825=2.80825 $$ $$ v_{12}=0.9(0.3)+0.89235(0)+0.12825(-1.5-(-3.0))=0.27+0.192375=0.462375 $$ $$ x_{12}=-3.0+0.462375=-2.537625 $$ ##### Step 5. Velocity and position of particle 2 (it is gbest, so both pull terms are 0) $$ v_{21}=0.9(0.1)+0+0=0.09,\qquad x_{21}=3.5+0.09=3.59 $$ $$ v_{22}=0.9(0.25)+0+0=0.225,\qquad x_{22}=-1.5+0.225=-1.275 $$ ##### Step 6. New fitness $$ F(2.80825,-2.537625)=15.772536+21.378856+6.439541+14.04125-5.07525+7=59.5569 $$ $$ F(3.59,-1.275)=25.7762+13.73175+1.625625+17.95-2.55+7=63.5336 $$ ##### Step 7. Update pbest and gbest | Particle | New position | New F | Old pbest F | pbest after iteration | | --- | --- | --- | --- | --- | | 1 | (2.80825, −2.537625) | 59.5569 | 57.5 | (2.80825, −2.537625) — improved | | 2 | (3.59, −1.275) | 63.5336 | 64 | (3.5, −1.5) — kept | The best pbest fitness is still 64, so gbest stays (3.5, −1.5). **Final answer.** $$ V=\begin{bmatrix}0.30825&0.462375\\0.09&0.225\end{bmatrix} $$ , $$ X=\begin{bmatrix}2.80825&-2.537625\\3.59&-1.275\end{bmatrix} $$ . After one iteration particle 1 improved to F = 59.5569; the best solution so far is gbest = (3.5, −1.5) with F = 64. Official key: fitness and gbest (5 marks), velocity and position updates (5 marks). The calculations above use the question data. The key's summary matrix writes 0.462 in place of 2.808 for $$ x_{11} $$ ; that is a copying slip. --- ## TEE · Q2(b) GMM: compare with k-means, log-likelihood, parameters [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q2b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q2b Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **With reference to the Gaussian Mixture Model (GMM): compare GMM with the k-means algorithm; derive the equation for the log-likelihood function; state the number of parameters to be computed if K = 5.** [10 marks] ### Answer ## Read the derivation’s symbols first “i.i.d.” means the observations are modelled as independent draws from the same distribution. The product Π runs over observations; the inner sum Σ runs over components. λki in some notes is the same responsibility that other sections call γik. A superscript i here labels the observation, not an exponent. Jensen’s inequality gives a lower bound on the log of the component sum. EM chooses responsibilities making that bound touch the current likelihood, then updates the parameters to maximise the bound with those responsibilities fixed. This explains the two alternating steps; λ values are not arbitrary new model parameters. | Parameter | Number in a 1D K-component GMM | Why | | --- | --- | --- | | Means | K | One centre for each component | | Variances | K | One spread for each component | | Independent mixing weights | K−1 | The last weight is 1 minus the others | | Total | 3K−1 | K+K+(K−1) | For K=2, choose μ₁,μ₂,σ₁²,σ₂² and π₁: five free numbers. Then π₂=1−π₁. For K=5, 5+5+4=14. In 2D with a full covariance matrix, each component instead has two means and three independent covariance entries (two variances and one shared off-diagonal covariance). Five such components have 5×(2+3)+4=29 free parameters. ### 1. GMM versus k-means | Aspect | k-means | GMM | | --- | --- | --- | | Assignment | Hard: each point in exactly one cluster | Soft: probability $$ \lambda_k^i $$ of belonging to each cluster | | Model | Geometric: nearest centroid | Probabilistic generative model $$ \sum_k\pi_k\mathcal N(\mu_k,\Sigma_k) $$ | | Cluster shape | Spherical, similar size | Elliptical, any size/orientation; overlapping clusters | | Parameters | K centroids | Weights, means, covariances (1-D: 3K − 1) | | Algorithm / objective | Assign–update; minimise squared distances | EM; maximise log-likelihood | | Output | Hard labels and fitted centroids | Labels, membership probabilities and a density (useful for anomaly detection) | | Relation | The hard-assignment, equal spherical variance, small-variance limit of a Gaussian-mixture objective leads to the k-means objective. Ordinary finite-variance EM is not the same algorithm as k-means. Both need K in advance and depend on initialisation. | | ### 2. Derivation of the log-likelihood **Step 1. Model.** Each point is generated by choosing component k with probability $$ \pi_k $$ and drawing $$ x_i\sim\mathcal N(\mu_k,\sigma_k^2) $$ , so its density is $$ f_{mix}(x_i)=\sum_k\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2) $$ . **Step 2. Likelihood.** Points are i.i.d., so the likelihood is a product (θ = all parameters): $$ L(\theta)=\prod_{i=1}^{n}\Big[\sum_{k=1}^{K}\pi_k\frac{1}{\sqrt{2\pi}\,\sigma_k}e^{-\frac{(x_i-\mu_k)^2}{2\sigma_k^2}}\Big] $$ **Step 3. Log-likelihood.** Logs turn the product into a sum: $$ \log L(\theta)=\sum_{i=1}^{n}\log\Big(\sum_{k=1}^{K}\pi_k\frac{1}{\sqrt{2\pi}\,\sigma_k}e^{-\frac{(x_i-\mu_k)^2}{2\sigma_k^2}}\Big) $$ **Step 4. Why it cannot be solved directly.** The sum sits inside the log, so setting derivatives to zero gives no closed-form solution. **Step 5. Jensen's inequality.** log is concave, so for weights $$ \lambda_k\ge0,\;\sum\lambda_k=1 $$ : $$ \log\big(\sum_k\lambda_ka_k\big)\ge\sum_k\lambda_k\log a_k $$ . Introduce $$ \lambda_k^i $$ for every point ( $$ \sum_k\lambda_k^i=1 $$ ), multiply and divide inside the log: $$ \log L(\theta)=\sum_i\log\Big(\sum_k\lambda_k^i\,\frac{\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)}{\lambda_k^i}\Big)\;\ge\;\sum_i\sum_k\lambda_k^i\log\frac{\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)}{\lambda_k^i} $$ **Step 6. Estimation (EM).** Maximise the lower bound alternately. Fixing θ gives the E-step $$ \lambda_k^i=\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)/\sum_l\pi_l\mathcal N(x_i\mid\mu_l,\sigma_l^2) $$ . Fixing λ and differentiating (with a Lagrange multiplier for $$ \sum\pi_k=1 $$ ) gives the M-step: $$ \hat\mu_k=\frac{\sum_i\lambda_k^ix_i}{\sum_i\lambda_k^i},\qquad\hat\sigma_k^2=\frac{\sum_i\lambda_k^i(x_i-\hat\mu_k)^2}{\sum_i\lambda_k^i},\qquad\hat\pi_k=\frac{\sum_i\lambda_k^i}{n} $$ ### 3. Number of parameters for K = 5 The dimension and covariance structure must also be specified. For the one-dimensional, independent-variance model used in the derivation: $$ 2K+(K-1)=3K-1=3(5)-1=\mathbf{14} $$ (5 means + 5 variances + 4 free weights, since the 5 weights must add to 1.) In two dimensions with full covariance it would be 29. **Slide note.** One slide puts the convex inequality under the “concave” heading. Correct: log is *concave*, so $$ f(\sum\lambda a)\ge\sum\lambda f(a) $$ (≥, not >). --- ## TEE · Q3(a) Bayesian network: wet grass given no rain [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q3a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q3a Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Cloudy → Sprinkler, Cloudy → Rain, Sprinkler and Rain → Wet Grass. P(Cloudy) = 0.5; P(Sprinkler | C) = 0.10, P(Sprinkler | ¬C) = 0.50; P(Rain | C) = 0.80, P(Rain | ¬C) = 0.20; P(Wet | S, R) = 0.99, P(Wet | S, ¬R) = 0.90, P(Wet | ¬S, R) = 0.90, P(Wet | ¬S, ¬R) = 0.01. Calculate the probability of the grass being wet given it doesn't rain.** [10 marks] ### Answer The following table enumerates the values of Cloudy and Sprinkler that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome. #### Given C→S, C→R, (S,R)→W. Root and conditional probabilities are listed in the question. Evidence R=false. #### To find P(W=true | R=false). Hidden variables C and S must be summed out. #### Formula and why it applies Use the network joint and divide by the evidence probability. $$ P(W\mid\neg R)=\frac{\sum_{c,s}P(c)P(s\mid c)P(\neg R\mid c)P(W\mid s,\neg R)}{\sum_cP(c)P(\neg R\mid c)} $$ #### Solution Swipe across the diagram to read its labels ↔ ![Final 2025 Q3a: sprinkler network](https://deep-learning-study-room.vercel.app/is2/fig/bn_sprinkler.a251583272.svg) Diagram: Final 2025 Q3a: sprinkler network. Labels: Final 2025 Q3a: sprinkler network | Cloudy | P(C) = 0.5 | Sprinkler | P(S|C) = 0.1, P(S|¬C) = 0.5 | Rain | P(R|C) = 0.8, P(R|¬C) = 0.2 | Wet grass | P(W|S,R): .99 · .90 · .90 · .01 | (TT · TF · FT · FF) ##### Step 1. Joint by the chain rule $$ P(C,S,R,W)=P(C)\,P(S\mid C)\,P(R\mid C)\,P(W\mid S,R) $$ ##### Step 2. Sum over C and S with R = false, W = true | C | S | P(C) | P(S|C) | P(¬R|C) | P(W|S,¬R) | Product | | --- | --- | --- | --- | --- | --- | --- | | T | T | 0.5 | 0.1 | 0.2 | 0.90 | 0.0090 | | T | F | 0.5 | 0.9 | 0.2 | 0.01 | 0.0009 | | F | T | 0.5 | 0.5 | 0.8 | 0.90 | 0.1800 | | F | F | 0.5 | 0.5 | 0.8 | 0.01 | 0.0020 | | **P(W, ¬R)** | | | | | | **0.1919** | ##### Step 3. Evidence $$ P(\lnot R)=P(C)P(\lnot R\mid C)+P(\lnot C)P(\lnot R\mid\lnot C)=0.5(0.2)+0.5(0.8)=0.5 $$ ##### Step 4. Divide $$ P(W\mid\lnot R)=\frac{0.1919}{0.5}=0.3838 $$ **Final answer.** P(grass wet | no rain) ≈ **0.384**. Without rain, the grass is wet mainly when the sprinkler runs (more likely on non-cloudy days). **Keep the requested conditional probability separate.** The supplied key discusses P(Wet) and P(Rain | Wet), whereas this question asks P(Wet | no rain). The answer to the printed question is 0.3838. For comparison, the exact marginal from the printed CPT is calculated below; it is not 0.6505. #### Optional comparison: calculate P(Wet) from the same network The four no-rain cases already total 0.1919. Now add the four rain cases, keeping the same Cloudy/Sprinkler order. $$ P(W,R)=0.5(0.1)(0.8)(0.99)+0.5(0.9)(0.8)(0.90)+0.5(0.5)(0.2)(0.99)+0.5(0.5)(0.2)(0.90) $$ $$ P(W,R)=0.0396+0.324+0.0495+0.045=0.4581 $$ $$ P(W)=P(W,R)+P(W,\neg R)=0.4581+0.1919=0.6500 $$ $$ P(R\mid W)=\frac{0.4581}{0.6500}\approx0.704769231 $$ **Meaning:** the unconditional chance of Wet is 0.65, the chance of Rain among wet cases is about 0.704769, and the requested chance of Wet among no-rain cases is 0.3838. These are three different questions. **Trap.** Do not plug in the marginal P(Sprinkler) = 0.3; sprinkler and rain both depend on Cloudy, so you must sum over Cloudy (given no rain, P(S) = 0.42). --- ## TEE · Q3(b) Fuzzy loan-approval system [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q3b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q3b Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **A bank uses fuzzy decision support with Credit Score and Income Stability, both on 1–100. Loan Approval is a score on 1–100 with labels Reject, Consider and Approve. Define triangular membership functions, construct nine rules, explain Mamdani inference and defuzzify for Credit Score=55 and Income Stability=35.** [10 marks] ### Answer #### Given | Variable | Crisp value | | --- | --- | | Credit Score | 55 | | Income Stability | 35 | The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens. #### To find A complete stated design and its continuous Mamdani centroid. #### Formula and why it applies Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier. $$ z^*=\frac{\int z\mu_{out}(z)\,dz}{\int\mu_{out}(z)\,dz} $$ #### Solution ##### Step 1. Specify the membership functions | Input / output position | Triangle | Peak | | --- | --- | --- | | Low / Reject | (1,25,50) | 25 | | Medium / Consider | (25,50,75) | 50 | | High / Approve | (50,75,100) | 75 | Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values. ##### Step 2. Write all nine rules | Credit Score / Income Stability | Low | Medium | High | | --- | --- | --- | --- | | Low | Reject | Reject | Consider | | Medium | Reject | Consider | Approve | | High | Consider | Approve | Approve | Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium. ##### Step 3. Fuzzify Credit Score = 55 $$ \mu_{Low}(55)=0 $$ $$ \mu_{Medium}(55)=\frac{75-55}{75-50}=0.8 $$ $$ \mu_{High}(55)=\frac{55-50}{75-50}=0.2 $$ ##### Step 4. Fuzzify Income Stability = 35 $$ \mu_{Low}(35)=\frac{50-35}{50-25}=0.6 $$ $$ \mu_{Medium}(35)=\frac{35-25}{50-25}=0.4 $$ $$ \mu_{High}(35)=0 $$ ##### Step 5. Evaluate all rules and aggregate matching consequents | Credit Score | Income Stability | Firing strength | Consequent | | --- | --- | --- | --- | | Low | Low | min(0,0.6) = 0 | Reject | | Low | Medium | min(0,0.4) = 0 | Reject | | Low | High | min(0,0) = 0 | Consider | | Medium | Low | min(0.8,0.6) = 0.6 | Reject | | Medium | Medium | min(0.8,0.4) = 0.4 | Consider | | Medium | High | min(0.8,0) = 0 | Approve | | High | Low | min(0.2,0.6) = 0.2 | Consider | | High | Medium | min(0.2,0.4) = 0.2 | Approve | | High | High | min(0.2,0) = 0 | Approve | Aggregated label strengths: Reject = 0.6, Consider = 0.4, Approve = 0.2. $$ \mu_{out}(z)=\max_r\min(\alpha_r,\mu_{C_r}(z)) $$ ##### Step 6. Draw the aggregated output Swipe across the diagram to read its labels ↔ ![The shaded upper envelope counts any overlap only once.](https://deep-learning-study-room.vercel.app/is2/fig/v2_loan_aggregate.2c5ffa32a7.svg) Diagram: The shaded upper envelope counts any overlap only once.. Labels: Chosen triangular output sets and their max-aggregate | 0 | 0.5 | 1 | 1 | 25 | 50 | 75 | 100 | μ | Output score | Reject | Consider | Approve | centroid ≈ 42.78 The shaded upper envelope counts any overlap only once. ##### Step 7. Calculate area and moment of each straight-line strip For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height. $$ A=(b-a)(u+v)/2 $$ $$ M=\frac{b-a}{6}\big[(2a+b)u+(a+2b)v\big] $$ | a | b | u | v | Area A | Moment M | | --- | --- | --- | --- | --- | --- | | 1 | 15.4 | 0 | 0.6 | 4.32 | 45.792 | | 15.4 | 25 | 0.6 | 0.6 | 5.76 | 116.352 | | 25 | 35 | 0.6 | 0.6 | 6 | 180 | | 35 | 40 | 0.6 | 0.4 | 2.5 | 93.333333333 | | 40 | 50 | 0.4 | 0.4 | 4 | 180 | | 50 | 55 | 0.4 | 0.4 | 2 | 105 | | 55 | 65 | 0.4 | 0.4 | 4 | 240 | | 65 | 70 | 0.4 | 0.2 | 1.5 | 100.833333333 | | 70 | 75 | 0.2 | 0.2 | 1 | 72.5 | | 75 | 95 | 0.2 | 0.2 | 4 | 340 | | 95 | 100 | 0.2 | 0 | 0.5 | 48.333333333 | ##### Step 8. Divide total moment by total area $$ \sum A=35.58 $$ $$ \sum M=1522.144 $$ $$ z^*=\frac{1522.144}{35.58}\approx42.780888139 $$ **Final answer.** For these stated triangles and rules, the centroid is 42.780888139. This is a fuzzy suitability score; the category memberships can be evaluated at this score. The model alone is not a complete loan-decision policy. Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised. ## How the centroid table’s breakpoints are obtained For each output triangle, solve “line height = clip height”. Reject is clipped at .6: rising endpoint 1+.6(25−1)=15.4 and falling endpoint 50−.6(50−25)=35. Consider is clipped at .4: endpoints 25+.4(50−25)=35 and 75−.4(75−50)=65. Approve is clipped at .2: endpoints 50+.2(75−50)=55 and 100−.2(100−75)=95. Now take the maximum of the three clipped curves. The declining Reject edge reaches the Consider height .4 at (50−z)/25=.4, hence z=40. The declining Consider edge reaches the Approve height .2 at (75−z)/25=.2, hence z=70. These crossover locations must be included when splitting the aggregate. Original peaks and other clip endpoints may add extra splits without changing the total area. **For the first aggregate strip from a=1 to b=15.4, endpoint heights are u=0 and v=0.6. Find its area and first moment.** [3 marks · Worked example] #### Given Strip endpoints a=1 and b=15.4; left height u=0 and right height v=0.6. This is a straight rising edge of the clipped Reject set. #### To find A and M for one table row. #### Formula and why it applies $$ A=(b-a)(u+v)/2 $$ $$ M=\frac{b-a}{6}[(2a+b)u+(a+2b)v] $$ A is trapezoid area. M weights every small area by its horizontal position; this formula integrates a straight line. #### Solution ##### Step 1. Width and height sum $$ b-a=15.4-1=14.4\qquad u+v=0.6 $$ ##### Step 2. Area $$ A=14.4\times0.6/2=8.64/2=4.32 $$ ##### Step 3. First bracket term $$ (2a+b)u=(2+15.4)\times0=0 $$ ##### Step 4. Second bracket term $$ (a+2b)v=(1+30.8)\times0.6=31.8\times0.6=19.08 $$ ##### Step 5. Moment $$ M=(14.4/6)(0+19.08)=2.4\times19.08=45.792 $$ **Final answer.** Area=4.32; moment=45.792. Repeat the same arithmetic for each row, then sum the areas and moments separately. ### Interpret the loan result using an explicit decision rule Choose the output label having the greatest membership at the centroid; this is an additional stated interpretation rule. At z=42.780888139: $$ \mu_{Reject}=(50-z)/25\approx0.288764474,\quad\mu_{Consider}=(z-25)/25\approx0.711235526,\quad\mu_{Approve}=0 $$ The resulting recommendation is **Consider / manual review**; this design does not recommend automatic approval. If a binary approve/reject answer is required, the question must supply a threshold or you must state the threshold you adopt. A different decision threshold is separate from defuzzification. --- ## TEE · Q4(a) Cartesian product and max–min composition [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q4a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q4a Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Consider fuzzy sets A = {0.4/small + 0.3/medium + 0.1/large} and B = {0.2/slow + 0.7/moderate + 0.3/fast}. 1. Find R = A × B. 2. Given C = {0.3/small + 0.1/medium + 0.4/large}, find S = C × B. 3. Calculate the max–min composition S ∘ R.** [10 marks] ### Answer #### Given The fuzzy sets or relation matrices and their listed row/column order. #### To find The relation matrix/matrices and the composition specified in the question. #### Formula and why it applies For Cartesian products take min of row and column membership. For the ordered matrix composition shown here: $$ T_{ik}=\max_j\min(S_{ij},R_{jk}) $$ If max–product is asked instead, replace the inner min by multiplication. #### Solution ##### Step 1. R = A × B (rows small, medium, large; columns slow, moderate, fast) $$ R=\begin{bmatrix}\min(.4,.2)&\min(.4,.7)&\min(.4,.3)\\\min(.3,.2)&\min(.3,.7)&\min(.3,.3)\\\min(.1,.2)&\min(.1,.7)&\min(.1,.3)\end{bmatrix}=\begin{bmatrix}0.2&0.4&0.3\\0.2&0.3&0.3\\0.1&0.1&0.1\end{bmatrix} $$ ##### Step 2. S = C × B $$ S=\begin{bmatrix}0.2&0.3&0.3\\0.1&0.1&0.1\\0.2&0.4&0.3\end{bmatrix} $$ ##### Step 3. S ∘ R by max–min (row of S with column of R) The printed relation domains do not directly compose: both relate size to speed. For the numerical exercise, explicitly assume positional alignment and compute row of S with column of R. Equal dimensions alone are not a justification for semantic relation composition. $$ (1,1)=\max[\min(.2,.2),\min(.3,.2),\min(.3,.1)]=\max(.2,.2,.1)=0.2 $$ $$ (1,2)=\max[\min(.2,.4),\min(.3,.3),\min(.3,.1)]=\max(.2,.3,.1)=0.3 $$ $$ (1,3)=\max[\min(.2,.3),\min(.3,.3),\min(.3,.1)]=0.3 $$ Row 2 of S is all 0.1, so every entry of row 2 is 0.1. $$ (3,1)=\max[\min(.2,.2),\min(.4,.2),\min(.3,.1)]=0.2,\;(3,2)=\max[.2,.3,.1]=0.3,\;(3,3)=\max[.2,.3,.1]=0.3 $$ **Final answer.** $$ S\circ R=\begin{bmatrix}0.2&0.3&0.3\\0.1&0.1&0.1\\0.2&0.3&0.3\end{bmatrix} $$ . **Official key note.** The key first printed a wrong matrix with row 2 = (0.2, 0.3, 0.3), then corrected it by hand to (0.1, 0.1, 0.1). The corrected matrix above is right. Marking: R 3, S 3, composition 4. **A careful remark you can add.** Strictly, composition needs S: X → Y and R: Y → Z, while here both relate size to speed. The numerical matrix calculation above assumes positional identification of the three middle labels; that is a declared repair, not a valid size-to-speed relation composition without an extra mapping. ## Every cell without the row shortcuts Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries. ### R: Cartesian-product cells | Cell | Row degree | Column degree | min result | | --- | --- | --- | --- | | R11 | 0.4 | 0.2 | 0.2 | | R12 | 0.4 | 0.7 | 0.4 | | R13 | 0.4 | 0.3 | 0.3 | | R21 | 0.3 | 0.2 | 0.2 | | R22 | 0.3 | 0.7 | 0.3 | | R23 | 0.3 | 0.3 | 0.3 | | R31 | 0.1 | 0.2 | 0.1 | | R32 | 0.1 | 0.7 | 0.1 | | R33 | 0.1 | 0.3 | 0.1 | ### S: Cartesian-product cells | Cell | Row degree | Column degree | min result | | --- | --- | --- | --- | | S11 | 0.3 | 0.2 | 0.2 | | S12 | 0.3 | 0.7 | 0.3 | | S13 | 0.3 | 0.3 | 0.3 | | S21 | 0.1 | 0.2 | 0.1 | | S22 | 0.1 | 0.7 | 0.1 | | S23 | 0.1 | 0.3 | 0.1 | | S31 | 0.4 | 0.2 | 0.2 | | S32 | 0.4 | 0.7 | 0.4 | | S33 | 0.4 | 0.3 | 0.3 | **Result cell (1,1)** $$ \max\{\min(0.2,0.2),\min(0.3,0.2),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (1,2)** $$ \max\{\min(0.2,0.4),\min(0.3,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.3,0.1\}=0.3 $$ **Result cell (1,3)** $$ \max\{\min(0.2,0.3),\min(0.3,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.3,0.1\}=0.3 $$ **Result cell (2,1)** $$ \max\{\min(0.1,0.2),\min(0.1,0.2),\min(0.1,0.1)\} $$ $$ =\max\{0.1,0.1,0.1\}=0.1 $$ **Result cell (2,2)** $$ \max\{\min(0.1,0.4),\min(0.1,0.3),\min(0.1,0.1)\} $$ $$ =\max\{0.1,0.1,0.1\}=0.1 $$ **Result cell (2,3)** $$ \max\{\min(0.1,0.3),\min(0.1,0.3),\min(0.1,0.1)\} $$ $$ =\max\{0.1,0.1,0.1\}=0.1 $$ **Result cell (3,1)** $$ \max\{\min(0.2,0.2),\min(0.4,0.2),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (3,2)** $$ \max\{\min(0.2,0.4),\min(0.4,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.3,0.1\}=0.3 $$ **Result cell (3,3)** $$ \max\{\min(0.2,0.3),\min(0.4,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.2,0.3,0.1\}=0.3 $$ --- ## TEE · Q4(b) Naive Bayes: flu [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q4b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q4b Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. A hospital wants to predict flu from Fever (Yes/No), Cough (Yes/No) and Temperature Level (Low/Normal/High). Training data: (Yes, Yes, High, Yes), (No, No, Normal, No), (Yes, No, Low, No), (Yes, Yes, High, Yes), (No, Yes, Normal, No), (Yes, Yes, High, Yes), (No, No, Low, No), (Yes, No, Normal, No). Using naive Bayes, predict whether a patient with Fever = Yes, Cough = Yes, Temperature = Normal has flu.** [10 marks] ### Answer #### Given The full training data are in Step 1. Query: Fever=Yes, Cough=Yes, Temperature=Normal. The main calculation uses unsmoothed counts. #### To find Both class scores and the predicted class. #### Formula and why it applies $$ score(C)=P(C)\prod_jP(a_j\mid C) $$ C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods. #### Solution ##### Step 1. Write the training rows as a table | Row | Fever | Cough | Temperature | Class | | --- | --- | --- | --- | --- | | 1 | Yes | Yes | High | Yes | | 2 | No | No | Normal | No | | 3 | Yes | No | Low | No | | 4 | Yes | Yes | High | Yes | | 5 | No | Yes | Normal | No | | 6 | Yes | Yes | High | Yes | | 7 | No | No | Low | No | | 8 | Yes | No | Normal | No | ##### Step 2. Separate the two class groups | Class | Rows in this group | Class count | Prior | | --- | --- | --- | --- | | Yes | 1, 4, 6 | 3 | 3/8 = 0.375 | | No | 2, 3, 5, 7, 8 | 5 | 5/8 = 0.625 | ##### Step 3. For class Yes, count Fever=Yes Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6. $$ P(\text{Fever}=\text{Yes}\mid C=\text{Yes})=\frac{3}{3}=1 $$ ##### Step 4. For class Yes, count Cough=Yes Look only at class Yes rows 1, 4, 6. The matching rows are 1, 4, 6. $$ P(\text{Cough}=\text{Yes}\mid C=\text{Yes})=\frac{3}{3}=1 $$ ##### Step 5. For class Yes, count Temperature=Normal Look only at class Yes rows 1, 4, 6. No row in this class has that feature value. $$ P(\text{Temperature}=\text{Normal}\mid C=\text{Yes})=\frac{0}{3}=0 $$ ##### Step 6. Multiply the Yes score one factor at a time $$ P(Yes)=3/8=0.375 $$ $$ 0.375\times1=0.375 $$ $$ 0.375\times1=0.375 $$ $$ 0.375\times0=0 $$ ##### Step 7. For class No, count Fever=Yes Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 3, 8. $$ P(\text{Fever}=\text{Yes}\mid C=\text{No})=\frac{2}{5}=0.4 $$ ##### Step 8. For class No, count Cough=Yes Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 5. $$ P(\text{Cough}=\text{Yes}\mid C=\text{No})=\frac{1}{5}=0.2 $$ ##### Step 9. For class No, count Temperature=Normal Look only at class No rows 2, 3, 5, 7, 8. The matching rows are 2, 5, 8. $$ P(\text{Temperature}=\text{Normal}\mid C=\text{No})=\frac{3}{5}=0.6 $$ ##### Step 10. Multiply the No score one factor at a time $$ P(No)=5/8=0.625 $$ $$ 0.625\times0.4=0.25 $$ $$ 0.25\times0.2=0.05 $$ $$ 0.05\times0.6=0.03 $$ ##### Step 11. Compare the scores $$ score(Yes)=0\qquad score(No)=0.03 $$ The larger score belongs to No. Normalise only if a posterior probability is requested: divide each score by their total 0.03. **Final answer.** Predict No. Yes score=0; No score=0.03. The zero Yes score is caused by an absent training combination. If Laplace smoothing is explicitly required, apply it to all feature likelihoods in both classes. That is a different calculation and may change the prediction; do not mix smoothed and unsmoothed sides. **Yaad rakho:** Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo. --- ## TEE · Q5(a) Hotel guest movement (Markov chain) [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q5a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q5a Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. A hotel chain has properties in Paris (P), London (L) and Rome (R). Of guests in Paris, 60% stay, 25% go to London, 15% to Rome; in London 30% go to Paris, 50% stay, 20% go to Rome; in Rome 20% go to Paris, 30% to London, 50% stay. 1. Construct the transition matrix. 2. Explain the diagonal and off-diagonal entries. 3. Draw the state diagram. 4. Today's counts are [50, 30, 20]; find tomorrow's expected counts. 5. Which city gains or loses guests in the long run?** [10 marks] ### Answer #### Given All transition values are given in the full question. State order: P, L, R. Initial vector [50, 30, 20]. #### To find The matrix, diagram, next-step vector and long-run comparison. #### Formula and why it applies Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add. $$ x_{next}=x_{now}P\qquad\pi=\pi P,\quad\sum_i\pi_i=1 $$ #### Solution ##### Step 1. Convert every origin row to probabilities | From / To | P | L | R | | --- | --- | --- | --- | | P | 0.6 | 0.25 | 0.15 | | L | 0.3 | 0.5 | 0.2 | | R | 0.2 | 0.3 | 0.5 | Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25. ##### Step 2. Interpret entries and draw the transitions Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving. Swipe across the diagram to read its labels ↔ ![Each arrow is labelled with the probability of that transition.](https://deep-learning-study-room.vercel.app/is2/fig/mk_hotel.55f5204329.svg) Diagram: Each arrow is labelled with the probability of that transition.. Labels: Daily guest movement: Paris, London, Rome | 0.25 | 0.15 | 0.3 | 0.2 | 0.2 | 0.3 | 0.6 | 0.5 | 0.5 | P | L | R Each arrow is labelled with the probability of that transition. ##### Step 3. Next step: collect all contributions to P | Origin | Current count/share | Transition to P | Contribution | | --- | --- | --- | --- | | P | 50 | 0.6 | 50 × 0.6 = 30 | | L | 30 | 0.3 | 30 × 0.3 = 9 | | R | 20 | 0.2 | 20 × 0.2 = 4 | $$ P_{new}=30+9+4=43 $$ ##### Step 4. Next step: collect all contributions to L | Origin | Current count/share | Transition to L | Contribution | | --- | --- | --- | --- | | P | 50 | 0.25 | 50 × 0.25 = 12.5 | | L | 30 | 0.5 | 30 × 0.5 = 15 | | R | 20 | 0.3 | 20 × 0.3 = 6 | $$ L_{new}=12.5+15+6=33.5 $$ ##### Step 5. Next step: collect all contributions to R | Origin | Current count/share | Transition to R | Contribution | | --- | --- | --- | --- | | P | 50 | 0.15 | 50 × 0.15 = 7.5 | | L | 30 | 0.2 | 30 × 0.2 = 6 | | R | 20 | 0.5 | 20 × 0.5 = 10 | $$ R_{new}=7.5+6+10=23.5 $$ ##### Step 6. Stationarity: write the Paris equation $$ p=0.6p+0.3l+0.2r $$ $$ 0.4p=0.3l+0.2r $$ Divide every term by .4. $$ p=0.75l+0.5r $$ ##### Step 7. Write the London equation and substitute p $$ l=0.25p+0.5l+0.3r $$ $$ 0.5l=0.25(0.75l+0.5r)+0.3r $$ $$ 0.5l=0.1875l+0.125r+0.3r $$ ##### Step 8. Collect like terms $$ (0.5-0.1875)l=(0.125+0.3)r $$ $$ 0.3125l=0.425r\quad\Rightarrow\quad l=1.36r $$ $$ p=0.75(1.36r)+0.5r=1.52r $$ ##### Step 9. Use total probability to get all three shares $$ p+l+r=1\quad\Rightarrow\quad1.52r+1.36r+r=1 $$ $$ 3.88r=1\quad r=1/3.88=25/97 $$ $$ l=1.36(25/97)=34/97\quad p=1.52(25/97)=38/97 $$ ##### Step 10. Compare the stationary shares with the initial shares | State | Initial share | Stationary share | Change in share | Conclusion | | --- | --- | --- | --- | --- | | P | 0.5 | 0.391752577 | -0.108247423 | Loses | | L | 0.3 | 0.350515464 | 0.050515464 | Gains | | R | 0.2 | 0.257731959 | 0.057731959 | Gains | Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged. **Final answer.** Next-step vector [43.0, 33.5, 23.5]. Stationary shares ≈ [0.391752577, 0.350515464, 0.257731959]. P loses, L gains, R gains. --- ## TEE · Q5(b) Genetic algorithm [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q5b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q5b Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Implement a GA to maximise $$ f(x)=2x^3-5x^2+7x $$ for x in [0,15]. Take 4-bit representations of 13, 22, 18, 23. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness.** [10 marks] ### Answer **The printed data are inconsistent.** 22, 18 and 23 are outside 0–15 and cannot fit into four bits. Increasing the bit count would encode the numbers but would still violate the stated range. There is no single exact solution satisfying all printed conditions. The complete worked variant below explicitly changes the invalid parents to their last four bits (values modulo 16), producing [13,6,2,7]. This is one declared repair, not a silent interpretation of the printed question. If the instructor clarifies different corrected data, follow those data. #### Given Printed parents: [13, 22, 18, 23]. Worked variant: [13, 6, 2, 7] with 4 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2. #### To find Two valid children under the declared assumptions and their fitness. #### Formula and why it applies Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation. $$ f(x)=2x^3-5x^2+7x $$ #### Solution ##### Step 1. Encode and evaluate the declared parent population | x | Binary | Fitness substitution | | --- | --- | --- | | 13 | 1101 | 2(13³) − 5(13²) + 7(13) = 3640 | | 6 | 0110 | 2(6³) − 5(6²) + 7(6) = 294 | | 2 | 0010 | 2(2³) − 5(2²) + 7(2) = 10 | | 7 | 0111 | 2(7³) − 5(7²) + 7(7) = 490 | ##### Step 2. Select the two fittest under the stated assumption [13, 7] with fitness [3640, 490]. ##### Step 3. Cut after bit 2 and exchange tails | Parent 1 | Parent 2 | Child 1 | Child 2 | | --- | --- | --- | --- | | 11 | 01 | 01 | 11 | 1111 | 0101 | ##### Step 4. Flip the last bit of each child | Before | After | Decoded x | | --- | --- | --- | | 1111 | 1110 | 14 | | 0101 | 0100 | 4 | ##### Step 5. Evaluate both children | x | Fitness calculation | | --- | --- | | 14 | 2(2744) − 5(196) + 7(14) = 4606 | | 4 | 2(64) − 5(16) + 7(4) = 76 | ##### Step 6. Compare with the selected parents Best parent score = 3640; best child score = 4606. Parent average = 2065; child average = 2341. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator. **Final answer.** Under these declared repairs and operator choices, children are [14, 4] with fitness [4606, 76]. These are not unique answers to the inconsistent and under-specified original question. The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2. ## Evaluate the polynomial without skipping the powers Calculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper. ### x=13 $$ x^2=13\times13=169\qquad x^3=169\times13=2197 $$ $$ 2x^3=4394\quad 5x^2=845\quad 7x=91 $$ $$ f(13)=4394-845+91=3640 $$ ### x=6 $$ x^2=6\times6=36\qquad x^3=36\times6=216 $$ $$ 2x^3=432\quad 5x^2=180\quad 7x=42 $$ $$ f(6)=432-180+42=294 $$ ### x=2 $$ x^2=2\times2=4\qquad x^3=4\times2=8 $$ $$ 2x^3=16\quad 5x^2=20\quad 7x=14 $$ $$ f(2)=16-20+14=10 $$ ### x=7 $$ x^2=7\times7=49\qquad x^3=49\times7=343 $$ $$ 2x^3=686\quad 5x^2=245\quad 7x=49 $$ $$ f(7)=686-245+49=490 $$ ### x=15 $$ x^2=15\times15=225\qquad x^3=225\times15=3375 $$ $$ 2x^3=6750\quad 5x^2=1125\quad 7x=105 $$ $$ f(15)=6750-1125+105=5730 $$ ### x=5 $$ x^2=5\times5=25\qquad x^3=25\times5=125 $$ $$ 2x^3=250\quad 5x^2=125\quad 7x=35 $$ $$ f(5)=250-125+35=160 $$ ### x=14 $$ x^2=14\times14=196\qquad x^3=196\times14=2744 $$ $$ 2x^3=5488\quad 5x^2=980\quad 7x=98 $$ $$ f(14)=5488-980+98=4606 $$ ### x=4 $$ x^2=4\times4=16\qquad x^3=16\times4=64 $$ $$ 2x^3=128\quad 5x^2=80\quad 7x=28 $$ $$ f(4)=128-80+28=76 $$ “Modulo 16” means keep the remainder after dividing by 16: 22=1×16+6, so its remainder is 6; 18=1×16+2; 23=1×16+7. This is only the declared repair used here, not an unspoken general encoding rule. --- ## TEE · Q6(a) Applications of fuzzy logic [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q6a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q6a Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **List the applications of fuzzy logic. Explain any one application in detail.** [10 marks] ### Answer ### Definition Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”. ### Applications | Area | Examples | | --- | --- | | Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners | | Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control | | Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) | | Consumer electronics | Camera autofocus and auto-exposure, image stabilisation | | Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants | | Aerospace | Satellite and spacecraft attitude control | | Medicine | Diagnosis support, anaesthesia dosage control | | Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules | | AI and robotics | Robot navigation, natural-language interfaces, expert systems | ### One application in detail: fuzzy washing machine 1. **Problem.** Wash time depends on how dirty and how greasy the clothes are. These are vague (“a bit dirty”), and a fixed timer wastes water and power. 2. **Inputs and outputs.** Sensors measure dirt (turbidity of water) and grease; the output is wash time. Each is described by fuzzy sets: dirt Small/Medium/High, grease Small/Medium/High, time Very Short … Very Long. 3. **Fuzzification.** Sensor readings are converted to membership degrees, e.g. dirt 60% → Medium 0.8, High 0.2. 4. **Rule base.** 9 expert rules, e.g. IF dirt is High AND grease is High THEN time is Very Long; IF dirt is Small AND grease is Small THEN time is Very Short. 5. **Inference.** Mamdani min–max: rule strength = min of the two memberships; outputs combined by max. 6. **Defuzzification.** Use the specified defuzzifier. For a separate representative-weighted illustration, strengths [.6,.4,.2] at times [25,40,60] give (.6×25+.4×40+.2×60)/(.6+.4+.2)=(15+16+12)/1.2≈35.833 minutes. These illustrative strengths are supplied here, not derived from unspecified input graphs. A Mamdani area centroid requires the full aggregate shape. 7. **Benefits.** Smooth adjustment to every load, saves water and electricity, easy to design from expert knowledge without an exact mathematical model. ### Advantages and disadvantages (bonus) **Advantages:** handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. **Disadvantages:** rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required. **Yaad rakho:** Application explain karte waqt 5 cheezein: inputs, fuzzy sets, rules, inference (min/max), defuzzification + ek number ka example. --- ## TEE · Q6(b) Learner progression (Markov chain) [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q6b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q6b Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Students are Beginner (B), Intermediate (I) or Advanced (A). Of Beginners, 60% remain, 30% become Intermediate, 10% go to Advanced. Of Intermediates, 20% fall back, 50% remain, 30% progress. Of Advanced, 10% drop to Intermediate, 80% remain and 10% stop (treated as remaining in A). 1. Construct the matrix. 2. Interpret retention, progression and regression. 3. Draw the diagram. 4. For [50, 30, 20], find the distribution after one day. 5. Which level has the largest steady-state share?** [10 marks] ### Answer #### Given All transition values are given in the full question. State order: B, I, A. Initial vector [50, 30, 20]. #### To find The matrix, diagram, next-step vector and long-run comparison. #### Formula and why it applies Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add. $$ x_{next}=x_{now}P\qquad\pi=\pi P,\quad\sum_i\pi_i=1 $$ #### Solution ##### Step 1. Convert every origin row to probabilities | From / To | B | I | A | | --- | --- | --- | --- | | B | 0.6 | 0.3 | 0.1 | | I | 0.2 | 0.5 | 0.3 | | A | 0 | 0.1 | 0.9 | Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25. ##### Step 2. Interpret entries and draw the transitions Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving. Swipe across the diagram to read its labels ↔ ![Each arrow is labelled with the probability of that transition.](https://deep-learning-study-room.vercel.app/is2/fig/mk_learning.61a789edc7.svg) Diagram: Each arrow is labelled with the probability of that transition.. Labels: Daily learner movement: Beginner, Intermediate, Advanced | 0.3 | 0.1 | 0.2 | 0.3 | 0.1 | 0.6 | 0.5 | 0.9 | B | I | A Each arrow is labelled with the probability of that transition. ##### Step 3. Next step: collect all contributions to B | Origin | Current count/share | Transition to B | Contribution | | --- | --- | --- | --- | | B | 50 | 0.6 | 50 × 0.6 = 30 | | I | 30 | 0.2 | 30 × 0.2 = 6 | | A | 20 | 0 | 20 × 0 = 0 | $$ B_{new}=30+6+0=36 $$ ##### Step 4. Next step: collect all contributions to I | Origin | Current count/share | Transition to I | Contribution | | --- | --- | --- | --- | | B | 50 | 0.3 | 50 × 0.3 = 15 | | I | 30 | 0.5 | 30 × 0.5 = 15 | | A | 20 | 0.1 | 20 × 0.1 = 2 | $$ I_{new}=15+15+2=32 $$ ##### Step 5. Next step: collect all contributions to A | Origin | Current count/share | Transition to A | Contribution | | --- | --- | --- | --- | | B | 50 | 0.1 | 50 × 0.1 = 5 | | I | 30 | 0.3 | 30 × 0.3 = 9 | | A | 20 | 0.9 | 20 × 0.9 = 18 | $$ A_{new}=5+9+18=32 $$ ##### Step 6. Stationarity: balance the Beginner column $$ b=0.6b+0.2i $$ $$ 0.4b=0.2i\quad\Rightarrow\quad i=2b $$ ##### Step 7. Balance the Advanced column $$ a=0.1b+0.3i+0.9a $$ $$ 0.1a=0.1b+0.3(2b)=0.7b $$ $$ a=0.7b/0.1=7b $$ ##### Step 8. Use the sum of shares $$ b+i+a=1\quad b+2b+7b=1 $$ $$ 10b=1\quad b=0.1\quad i=0.2\quad a=0.7 $$ Advanced is not an absorbing state: its self-transition is .9, not 1. ##### Step 9. Compare the stationary shares with the initial shares | State | Initial share | Stationary share | Change in share | Conclusion | | --- | --- | --- | --- | --- | | B | 0.5 | 0.1 | -0.4 | Loses | | I | 0.3 | 0.2 | -0.1 | Loses | | A | 0.2 | 0.7 | 0.5 | Gains | Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged. **Final answer.** Next-step vector [36.0, 32.0, 32.0]. Stationary shares ≈ [0.1, 0.2, 0.7]. B loses, I loses, A gains. --- ## TEE · Q7(a) Applications of PSO [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q7a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q7a Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **List the applications of Particle Swarm Optimization and explain any one application in detail.** [10 marks] ### Answer ### Definition (1 line) Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions. ### Applications 1. **Function optimisation**: maximising or minimising complex non-linear, multimodal functions where derivatives are unavailable. 2. **Neural-network training**: optimising weights and biases (your Lab 8 uses the pyswarms library for this). 3. **Hyper-parameter tuning** of ML models (learning rate, number of hidden units, SVM C and γ). 4. **Feature selection**: choosing the most useful subset of features (binary PSO). 5. **Image processing**: threshold selection for segmentation, edge detection, image enhancement. 6. **Path planning** for robots and drones: shortest collision-free path. 7. **Power systems**: economic load dispatch, optimal placement of capacitors. 8. **Control engineering**: tuning PID controller gains. 9. **Scheduling and wireless sensor networks**: job scheduling, sensor node placement and clustering. ### One application in detail: training a neural network with PSO Training data → Neural network with weights = particle position → Error (fitness) → PSO moves the weights → Best weights = gbest 1. **Encoding.** Flatten all weights and biases of the network into one vector. Each particle is one complete set of weights; a network with 50 weights gives a 50-dimensional search space. 2. **Fitness.** Run the training data through the network with that particle's weights and compute the error (for example mean squared error or cross-entropy). Lower error = better particle (minimisation). 3. **Search.** Apply the velocity and position updates. Each particle is pulled towards the weight set that gave its own lowest error (pbest) and towards the lowest-error weights found by the swarm (gbest). 4. **Stop** after a fixed number of iterations or when the error stops improving; gbest is the trained network. 5. **Why PSO here:** it needs no gradients, so it works with non-differentiable activations and can explore candidate weights without differentiating the objective. It can still stagnate; comparison with gradient methods depends on the problem. **Limitation:** large networks create a high-dimensional search space, and evaluating many complete candidate networks can be computationally expensive. Gradient-based training is often more efficient when useful gradients are available; there is no universal speed ordering for every problem and implementation. ### Advantages and disadvantages (add for 10 marks) | Advantages | Disadvantages | | --- | --- | | Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum | | No derivatives needed | Sensitive to parameter choice | | Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum | | Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification | Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version). **Yaad rakho:** Application likhte waqt teen cheezein: particle kya hai, fitness kya hai, gbest ka matlab kya hai. --- ## TEE · Q7(b) ACO for TSP [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-final-2025--q7b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q7b Actual TEE paper · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. For the four-city travelling salesman problem, the distance and pheromone matrices (A, B, C, D) are given. Apply Ant Colony Optimization. Calculate the path travelled by 2 ants to cover all cities starting from city A. Random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration.** [10 marks] ### Answer #### Given $$ D=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&4&2&4\\B&4&0&3&6\\C&2&3&0&2\\D&4&6&2&0\end{array}\qquad \tau=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&1&1&1\\B&1&0&1&1\\C&1&1&0&1\\D&1&1&1&0\end{array} $$ Assume $$ \alpha=\beta=1 $$ , $$ \rho=0.5 $$ , $$ Q=1 $$ (class convention; not stated in the question). #### To find Tour of each ant, tour lengths, updated pheromone matrix. Swipe across the diagram to read its labels ↔ ![Final 2025 Q7b: distances (all pheromones start at 1)](https://deep-learning-study-room.vercel.app/is2/fig/aco_final_graph.ac3cd4aeac.svg) Diagram: Final 2025 Q7b: distances (all pheromones start at 1). Labels: Final 2025 Q7b: distances (all pheromones start at 1) | 4 | 2 | 4 | 3 | 6 | 2 | A | B | C | D #### Formula and why it applies $$ P_{ij}=\frac{\tau_{ij}\,\eta_{ij}}{\sum_{l\in\text{allowed}}\tau_{il}\,\eta_{il}},\qquad\eta_{ij}=\frac1{d_{ij}} $$ $$ \tau_{ij}^{new}=(1-\rho)\tau_{ij}+\sum_{k:\,ij\in\mathrm{tour}_k}\frac{Q}{L_k} $$ The sum includes **only ants whose completed tour contains edge ij**. An unused edge receives no deposit but still evaporates. Here distances are symmetric, so a traversed edge updates both symmetric matrix cells. Both ants construct their tours using the old pheromones; update after both finish, including their return to A. #### Solution ##### Step 1. Probabilities from A (both ants start at A) $$ \tau\eta:\;A\to B=1\times\tfrac14=0.25,\quad A\to C=1\times\tfrac12=0.5,\quad A\to D=1\times\tfrac14=0.25;\quad\text{sum}=1 $$ | Move | P | Cumulative | Range | | --- | --- | --- | --- | | A→B | 0.25 | 0.25 | 0–0.25 | | A→C | 0.50 | 0.75 | 0.25–0.75 | | A→D | 0.25 | 1.00 | 0.75–1 | Ant 1: r = 0.3 → **C**. Ant 2: r = 0.5 → **C**. ##### Step 2. Probabilities from C (B and D left) $$ C\to B=\tfrac13=0.3333,\quad C\to D=\tfrac12=0.5;\quad\text{sum}=0.8333\Rightarrow P_{CB}=0.4,\;P_{CD}=0.6 $$ Cumulative in city order: B 0–0.4, D 0.4–1. Ant 1: r = 0.8 → **D**. Ant 2: r = 0.7 → **D**. ##### Step 3. Complete the tours Only B is left, then return to A. $$ \text{Ant 1}=\text{Ant 2}:\;A\to C\to D\to B\to A,\qquad L=2+2+6+4=14 $$ Swipe across the diagram to read its labels ↔ ![Both ants: A → C → D → B → A, length 14](https://deep-learning-study-room.vercel.app/is2/fig/aco_final_tour.1dbfdb8b98.svg) Diagram: Both ants: A → C → D → B → A, length 14. Labels: Both ants: A → C → D → B → A, length 14 | 4 | 2 | 4 | 3 | 6 | 2 | A | B | C | D ##### Step 4. Pheromone update Each ant deposits $$ Q/L=1/14=0.0714 $$ on each edge of its tour; two ants used the same four edges (AC, CD, DB, BA), so each gets $$ 2/14=0.1429 $$ . $$ \text{used edge: }(1-0.5)(1)+0.1429=0.6429\qquad\text{unused edge (AD, BC): }(1-0.5)(1)+0=0.5 $$ $$ \tau_{\text{new}}=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&0.6429&0.6429&0.5\\B&0.6429&0&0.5&0.6429\\C&0.6429&0.5&0&0.6429\\D&0.5&0.6429&0.6429&0\end{array} $$ **Final answer.** Both ants: A → C → D → B → A, length 14. Updated pheromone: 0.6429 on AB, AC, CD, BD; 0.5 on AD and BC. In the next iteration the ants will prefer the edges of this tour. **Official key note.** The key's matrix shows 0.643 on AC, CD and BD but leaves AB at 0.5, i.e. it forgot to deposit on the return edge B → A even though it counted its length (4) in L = 14. Your slides deposit on the return edge, so AB = 0.6429 is correct. If you wish, write one line: “the return edge B–A is part of the tour, so it also receives Q/L.” Official marking: path of Ant 1 (3) + path of Ant 2 (3) + pheromone matrix (4). --- ## TEE · Q1(a) Disease, fever and test joint table [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q1a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q1a Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. For a disease Y with fever X as a symptom, a test Z is conducted. Joint probabilities: Disease (Y=1): fever & test+ 0.576, fever & test− 0.144, no fever & test+ 0.072, no fever & test− 0.016; No disease (Y=0): 0.008, 0.064, 0.012, 0.108. (1) If fever occurs and the test is positive, what is the probability of disease? (2) What is the probability that a random person tests positive? (3) Given no fever, what is the probability of still having the disease?** [5 marks] ### Answer #### Given The complete joint table or cell values printed in this question. The rows and columns identify the event combinations. #### To find All requested marginal and conditional probabilities. #### Formula and why it applies Marginal: add cells satisfying the event. Conditional: retain the evidence cells and divide the joint-event sum by the evidence sum. $$ P(A\mid E)=\frac{P(A\cap E)}{P(E)},\quad P(E)>0 $$ #### Solution ##### Step 1. (1) $$ P(Y{=}1\mid X{=}1,Z{=}1)=\frac{0.576}{0.576+0.008}=\frac{0.576}{0.584}=0.986 $$ ##### Step 2. (2) Add every test-positive cell $$ P(Z{=}1)=0.576+0.072+0.008+0.012=0.668 $$ ##### Step 3. (3) $$ P(X{=}0)=0.072+0.016+0.012+0.108=0.208,\qquad P(Y{=}1\mid X{=}0)=\frac{0.072+0.016}{0.208}=\frac{0.088}{0.208}=0.423 $$ **Final answer.** (1) 0.986; (2) 0.668; (3) 0.423. **Key correction.** The key writes 0.088/0.076 = 1.158 for part (3). A probability cannot exceed 1; the denominator must be P(no fever) = 0.208. --- ## TEE · Q1(b) Applications of PSO [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q1b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q1b Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **List the applications of Particle Swarm Optimization and explain any one application in detail.** [5 marks] ### Answer ### Definition (1 line) Particle Swarm Optimization is a population-based metaheuristic, inspired by bird flocking and fish schooling, in which particles move through the search space using their own best position (pbest) and the swarm's best position (gbest) to find optimal or near-optimal solutions. ### Applications 1. **Function optimisation**: maximising or minimising complex non-linear, multimodal functions where derivatives are unavailable. 2. **Neural-network training**: optimising weights and biases (your Lab 8 uses the pyswarms library for this). 3. **Hyper-parameter tuning** of ML models (learning rate, number of hidden units, SVM C and γ). 4. **Feature selection**: choosing the most useful subset of features (binary PSO). 5. **Image processing**: threshold selection for segmentation, edge detection, image enhancement. 6. **Path planning** for robots and drones: shortest collision-free path. 7. **Power systems**: economic load dispatch, optimal placement of capacitors. 8. **Control engineering**: tuning PID controller gains. 9. **Scheduling and wireless sensor networks**: job scheduling, sensor node placement and clustering. ### One application in detail: training a neural network with PSO Training data → Neural network with weights = particle position → Error (fitness) → PSO moves the weights → Best weights = gbest 1. **Encoding.** Flatten all weights and biases of the network into one vector. Each particle is one complete set of weights; a network with 50 weights gives a 50-dimensional search space. 2. **Fitness.** Run the training data through the network with that particle's weights and compute the error (for example mean squared error or cross-entropy). Lower error = better particle (minimisation). 3. **Search.** Apply the velocity and position updates. Each particle is pulled towards the weight set that gave its own lowest error (pbest) and towards the lowest-error weights found by the swarm (gbest). 4. **Stop** after a fixed number of iterations or when the error stops improving; gbest is the trained network. 5. **Why PSO here:** it needs no gradients, so it works with non-differentiable activations and can explore candidate weights without differentiating the objective. It can still stagnate; comparison with gradient methods depends on the problem. **Limitation:** large networks create a high-dimensional search space, and evaluating many complete candidate networks can be computationally expensive. Gradient-based training is often more efficient when useful gradients are available; there is no universal speed ordering for every problem and implementation. ### Advantages and disadvantages (add for 10 marks) | Advantages | Disadvantages | | --- | --- | | Simple to implement; few parameters (w, c₁, c₂) | Can converge prematurely to a local optimum | | No derivatives needed | Sensitive to parameter choice | | Fast convergence in early iterations; memory (pbest, gbest) | No guarantee of the global optimum | | Objective evaluations can be parallel; particles share information through gbest | Less suited to discrete problems without modification | Marking pattern used for ACO in the same paper: listing 3 marks + explanation 2 marks (5-mark version); listing 5 + explanation 5 (10-mark version). **Yaad rakho:** Application likhte waqt teen cheezein: particle kya hai, fitness kya hai, gbest ka matlab kya hai. --- ## TEE · Q1(c) Fuzzy set operations [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q1c/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q1c Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. X = {(c1, 0.7), (c2, 0.9), (c3, 0.2), (c4, 0.4), (c5, 0.9)}, Y = {(c1, 0.6), (c2, 0.3), (c3, 0.8), (c4, 0.7), (c5, 0.1)}. Find X ∪ Y, X ∩ Y, ¬X ∪ ¬Y, ¬X ∩ ¬Y.** [5 marks] ### Answer #### Given The aligned X and Y memberships printed in this question. #### To find The four requested membership lists in the original element order. #### Formula and why it applies Use max for OR, min for AND, and 1−μ for NOT. $$ \mu_{X\cup Y}=\max(\mu_X,\mu_Y),\quad\mu_{X\cap Y}=\min(\mu_X,\mu_Y) $$ Complement both operands before applying the operator in ¬X∪¬Y or ¬X∩¬Y. #### Solution | | c1 | c2 | c3 | c4 | c5 | | --- | --- | --- | --- | --- | --- | | ¬X | 0.3 | 0.1 | 0.8 | 0.6 | 0.1 | | ¬Y | 0.4 | 0.7 | 0.2 | 0.3 | 0.9 | | **X ∪ Y** | 0.7 | 0.9 | 0.8 | 0.7 | 0.9 | | **X ∩ Y** | 0.6 | 0.3 | 0.2 | 0.4 | 0.1 | | **¬X ∪ ¬Y** | 0.4 | 0.7 | 0.8 | 0.6 | 0.9 | | **¬X ∩ ¬Y** | 0.3 | 0.1 | 0.2 | 0.3 | 0.1 | **Final answer.** As in the last four rows. De Morgan check: ¬X ∪ ¬Y = 1 − (X ∩ Y) and ¬X ∩ ¬Y = 1 − (X ∪ Y) ✓. ## Calculate every element separately ### Element c1 $$ \neg X:1-0.7=0.3\qquad\neg Y:1-0.6=0.4 $$ $$ X\cup Y:\max(0.7,0.6)=0.7 $$ $$ X\cap Y:\min(0.7,0.6)=0.6 $$ $$ \neg X\cup\neg Y:\max(0.3,0.4)=0.4 $$ $$ \neg X\cap\neg Y:\min(0.3,0.4)=0.3 $$ ### Element c2 $$ \neg X:1-0.9=0.1\qquad\neg Y:1-0.3=0.7 $$ $$ X\cup Y:\max(0.9,0.3)=0.9 $$ $$ X\cap Y:\min(0.9,0.3)=0.3 $$ $$ \neg X\cup\neg Y:\max(0.1,0.7)=0.7 $$ $$ \neg X\cap\neg Y:\min(0.1,0.7)=0.1 $$ ### Element c3 $$ \neg X:1-0.2=0.8\qquad\neg Y:1-0.8=0.2 $$ $$ X\cup Y:\max(0.2,0.8)=0.8 $$ $$ X\cap Y:\min(0.2,0.8)=0.2 $$ $$ \neg X\cup\neg Y:\max(0.8,0.2)=0.8 $$ $$ \neg X\cap\neg Y:\min(0.8,0.2)=0.2 $$ ### Element c4 $$ \neg X:1-0.4=0.6\qquad\neg Y:1-0.7=0.3 $$ $$ X\cup Y:\max(0.4,0.7)=0.7 $$ $$ X\cap Y:\min(0.4,0.7)=0.4 $$ $$ \neg X\cup\neg Y:\max(0.6,0.3)=0.6 $$ $$ \neg X\cap\neg Y:\min(0.6,0.3)=0.3 $$ ### Element c5 $$ \neg X:1-0.9=0.1\qquad\neg Y:1-0.1=0.9 $$ $$ X\cup Y:\max(0.9,0.1)=0.9 $$ $$ X\cap Y:\min(0.9,0.1)=0.1 $$ $$ \neg X\cup\neg Y:\max(0.1,0.9)=0.9 $$ $$ \neg X\cap\neg Y:\min(0.1,0.9)=0.1 $$ --- ## TEE · Q1(d) Savings and current accounts (Markov chain) [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q1d/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q1d Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Each year 20% of Savings holders switch to Current and 80% stay; 10% of Current holders switch to Savings and 90% stay. (a) Construct the transition matrix. (b) Draw the state diagram.** [5 marks] ### Answer #### Given All transition values are given in the full question. State order: S, C. #### To find The transition matrix and state diagram. #### Formula and why it applies Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add. #### Solution ##### Step 1. Convert every origin row to probabilities | From / To | S | C | | --- | --- | --- | | S | 0.8 | 0.2 | | C | 0.1 | 0.9 | Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25. ##### Step 2. Interpret entries and draw the transitions Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving. Swipe across the diagram to read its labels ↔ ![Each arrow is labelled with the probability of that transition.](https://deep-learning-study-room.vercel.app/is2/fig/mk_bank.879499f6eb.svg) Diagram: Each arrow is labelled with the probability of that transition.. Labels: Savings (S) and Current (C) accounts per year | 0.2 | 0.1 | 0.8 | 0.9 | S | C Each arrow is labelled with the probability of that transition. **Final answer.** The transition matrix and state diagram are given above. --- ## TEE · Q2(a) Fuzzy fire-hazard system [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q2a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q2a Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **A building safety department wants a fuzzy logic system to predict fire hazards. Inputs are Temperature (1–100°C) and Smoke Level (1–100). The output Fire Risk (1–100) has labels Low, Moderate and High. Define triangular membership functions, construct at least nine rules, explain Mamdani inference and defuzzify for Temperature=65°C and Smoke=80.** [10 marks] ### Answer #### Given | Variable | Crisp value | | --- | --- | | Temperature | 65 | | Smoke Level | 80 | The question asks us to design the membership functions. This solution chooses Low=Tri(1,25,50), Medium=Tri(25,50,75), High=Tri(50,75,100) for both inputs. Output labels use the same triangles in their respective order. These are explicit design choices, not additional printed givens. #### To find A complete stated design and its continuous Mamdani centroid. #### Formula and why it applies Use min for AND and implication, max for aggregation and the true area centroid of the resulting shape. A peak-only weighted average is a different defuzzifier. $$ z^*=\frac{\int z\mu_{out}(z)\,dz}{\int\mu_{out}(z)\,dz} $$ #### Solution ##### Step 1. Specify the membership functions | Input / output position | Triangle | Peak | | --- | --- | --- | | Low / Low | (1,25,50) | 25 | | Medium / Moderate | (25,50,75) | 50 | | High / High | (50,75,100) | 75 | Use 0 outside each triangle, the rising fraction before the peak and the falling fraction after it. This design is evaluated at the requested input; a deployed controller would also need coverage/fallback at all boundary values. ##### Step 2. Write all nine rules | Temperature / Smoke Level | Low | Medium | High | | --- | --- | --- | --- | | Low | Low | Low | Moderate | | Medium | Low | Moderate | High | | High | Moderate | High | High | Read each cell as an IF–AND–THEN rule. For example, the centre cell combines Medium with Medium. ##### Step 3. Fuzzify Temperature = 65 $$ \mu_{Low}(65)=0 $$ $$ \mu_{Medium}(65)=\frac{75-65}{75-50}=0.4 $$ $$ \mu_{High}(65)=\frac{65-50}{75-50}=0.6 $$ ##### Step 4. Fuzzify Smoke Level = 80 $$ \mu_{Low}(80)=0 $$ $$ \mu_{Medium}(80)=0 $$ $$ \mu_{High}(80)=\frac{100-80}{100-75}=0.8 $$ ##### Step 5. Evaluate all rules and aggregate matching consequents | Temperature | Smoke Level | Firing strength | Consequent | | --- | --- | --- | --- | | Low | Low | min(0,0) = 0 | Low | | Low | Medium | min(0,0) = 0 | Low | | Low | High | min(0,0.8) = 0 | Moderate | | Medium | Low | min(0.4,0) = 0 | Low | | Medium | Medium | min(0.4,0) = 0 | Moderate | | Medium | High | min(0.4,0.8) = 0.4 | High | | High | Low | min(0.6,0) = 0 | Moderate | | High | Medium | min(0.6,0) = 0 | High | | High | High | min(0.6,0.8) = 0.6 | High | Aggregated label strengths: Low = 0, Moderate = 0, High = 0.6. $$ \mu_{out}(z)=\max_r\min(\alpha_r,\mu_{C_r}(z)) $$ ##### Step 6. Draw the aggregated output Swipe across the diagram to read its labels ↔ ![The shaded upper envelope counts any overlap only once.](https://deep-learning-study-room.vercel.app/is2/fig/v2_fire_aggregate.e0cab4ad42.svg) Diagram: The shaded upper envelope counts any overlap only once.. Labels: Chosen triangular output sets and their max-aggregate | 0 | 0.5 | 1 | 1 | 25 | 50 | 75 | 100 | μ | Output score | Low | Moderate | High | centroid ≈ 75 The shaded upper envelope counts any overlap only once. ##### Step 7. Calculate area and moment of each straight-line strip For a strip from a to b with endpoint heights u and v, its trapezoid area is A. Its first moment M is the integral of z times the line height. $$ A=(b-a)(u+v)/2 $$ $$ M=\frac{b-a}{6}\big[(2a+b)u+(a+2b)v\big] $$ | a | b | u | v | Area A | Moment M | | --- | --- | --- | --- | --- | --- | | 50 | 65 | 0 | 0.6 | 4.5 | 270 | | 65 | 85 | 0.6 | 0.6 | 12 | 900 | | 85 | 100 | 0.6 | 0 | 4.5 | 405 | ##### Step 8. Divide total moment by total area $$ \sum A=21 $$ $$ \sum M=1575 $$ $$ z^*=\frac{1575}{21}\approx75 $$ **Final answer.** For these stated triangles and rules, the centroid is 75. This is in the high-risk region of this illustrative fuzzy design. Because the paper asks for a design, a different explicitly specified membership/rule system can yield a different numerical output. Explain the choices and use the chosen defuzzification method consistently; no particular examiner mark outcome is promised. ## Why the clipped High shape is symmetric about 75 Only the High output survives, at height .6. Its triangle is (50,75,100). Solve its rising edge: (z−50)/25=.6, so z=50+15=65. Solve its falling edge: (100−z)/25=.6, so z=100−15=85. | Piece | Area calculation | Centre | Moment | | --- | --- | --- | --- | | Left triangle [50,65] | ½×15×.6=4.5 | (50+65+65)/3=60 | 4.5×60=270 | | Rectangle [65,85] | 20×.6=12 | (65+85)/2=75 | 12×75=900 | | Right triangle [85,100] | ½×15×.6=4.5 | (85+85+100)/3=90 | 4.5×90=405 | The horizontal centroid of a triangle is the average of the three vertex x-coordinates. Add areas: 4.5+12+4.5=21. Add moments:270+900+405=1575. Divide 1575/21=75, agreeing with symmetry. --- ## TEE · Q2(b) GMM: compare with k-means, log-likelihood, parameters [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q2b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q2b Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **With reference to the Gaussian Mixture Model (GMM): compare GMM with the k-means algorithm; derive the equation for the log-likelihood function; state the number of parameters to be computed if K = 5.** [10 marks] ### Answer ## Read the derivation’s symbols first “i.i.d.” means the observations are modelled as independent draws from the same distribution. The product Π runs over observations; the inner sum Σ runs over components. λki in some notes is the same responsibility that other sections call γik. A superscript i here labels the observation, not an exponent. Jensen’s inequality gives a lower bound on the log of the component sum. EM chooses responsibilities making that bound touch the current likelihood, then updates the parameters to maximise the bound with those responsibilities fixed. This explains the two alternating steps; λ values are not arbitrary new model parameters. | Parameter | Number in a 1D K-component GMM | Why | | --- | --- | --- | | Means | K | One centre for each component | | Variances | K | One spread for each component | | Independent mixing weights | K−1 | The last weight is 1 minus the others | | Total | 3K−1 | K+K+(K−1) | For K=2, choose μ₁,μ₂,σ₁²,σ₂² and π₁: five free numbers. Then π₂=1−π₁. For K=5, 5+5+4=14. In 2D with a full covariance matrix, each component instead has two means and three independent covariance entries (two variances and one shared off-diagonal covariance). Five such components have 5×(2+3)+4=29 free parameters. ### 1. GMM versus k-means | Aspect | k-means | GMM | | --- | --- | --- | | Assignment | Hard: each point in exactly one cluster | Soft: probability $$ \lambda_k^i $$ of belonging to each cluster | | Model | Geometric: nearest centroid | Probabilistic generative model $$ \sum_k\pi_k\mathcal N(\mu_k,\Sigma_k) $$ | | Cluster shape | Spherical, similar size | Elliptical, any size/orientation; overlapping clusters | | Parameters | K centroids | Weights, means, covariances (1-D: 3K − 1) | | Algorithm / objective | Assign–update; minimise squared distances | EM; maximise log-likelihood | | Output | Hard labels and fitted centroids | Labels, membership probabilities and a density (useful for anomaly detection) | | Relation | The hard-assignment, equal spherical variance, small-variance limit of a Gaussian-mixture objective leads to the k-means objective. Ordinary finite-variance EM is not the same algorithm as k-means. Both need K in advance and depend on initialisation. | | ### 2. Derivation of the log-likelihood **Step 1. Model.** Each point is generated by choosing component k with probability $$ \pi_k $$ and drawing $$ x_i\sim\mathcal N(\mu_k,\sigma_k^2) $$ , so its density is $$ f_{mix}(x_i)=\sum_k\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2) $$ . **Step 2. Likelihood.** Points are i.i.d., so the likelihood is a product (θ = all parameters): $$ L(\theta)=\prod_{i=1}^{n}\Big[\sum_{k=1}^{K}\pi_k\frac{1}{\sqrt{2\pi}\,\sigma_k}e^{-\frac{(x_i-\mu_k)^2}{2\sigma_k^2}}\Big] $$ **Step 3. Log-likelihood.** Logs turn the product into a sum: $$ \log L(\theta)=\sum_{i=1}^{n}\log\Big(\sum_{k=1}^{K}\pi_k\frac{1}{\sqrt{2\pi}\,\sigma_k}e^{-\frac{(x_i-\mu_k)^2}{2\sigma_k^2}}\Big) $$ **Step 4. Why it cannot be solved directly.** The sum sits inside the log, so setting derivatives to zero gives no closed-form solution. **Step 5. Jensen's inequality.** log is concave, so for weights $$ \lambda_k\ge0,\;\sum\lambda_k=1 $$ : $$ \log\big(\sum_k\lambda_ka_k\big)\ge\sum_k\lambda_k\log a_k $$ . Introduce $$ \lambda_k^i $$ for every point ( $$ \sum_k\lambda_k^i=1 $$ ), multiply and divide inside the log: $$ \log L(\theta)=\sum_i\log\Big(\sum_k\lambda_k^i\,\frac{\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)}{\lambda_k^i}\Big)\;\ge\;\sum_i\sum_k\lambda_k^i\log\frac{\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)}{\lambda_k^i} $$ **Step 6. Estimation (EM).** Maximise the lower bound alternately. Fixing θ gives the E-step $$ \lambda_k^i=\pi_k\mathcal N(x_i\mid\mu_k,\sigma_k^2)/\sum_l\pi_l\mathcal N(x_i\mid\mu_l,\sigma_l^2) $$ . Fixing λ and differentiating (with a Lagrange multiplier for $$ \sum\pi_k=1 $$ ) gives the M-step: $$ \hat\mu_k=\frac{\sum_i\lambda_k^ix_i}{\sum_i\lambda_k^i},\qquad\hat\sigma_k^2=\frac{\sum_i\lambda_k^i(x_i-\hat\mu_k)^2}{\sum_i\lambda_k^i},\qquad\hat\pi_k=\frac{\sum_i\lambda_k^i}{n} $$ ### 3. Number of parameters for K = 5 The dimension and covariance structure must also be specified. For the one-dimensional, independent-variance model used in the derivation: $$ 2K+(K-1)=3K-1=3(5)-1=\mathbf{14} $$ (5 means + 5 variances + 4 free weights, since the 5 weights must add to 1.) In two dimensions with full covariance it would be 29. **Slide note.** One slide puts the convex inequality under the “concave” heading. Correct: log is *concave*, so $$ f(\sum\lambda a)\ge\sum\lambda f(a) $$ (≥, not >). --- ## TEE · Q3(a) Bayesian network: fever given no flu [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q3a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q3a Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Weather → Flu, Weather → Cough, Flu and Cough → Fever. P(Weather) = 0.75; P(Flu | W) = 0.2, P(Flu | ¬W) = 0.5; P(Cough | W) = 0.70, P(Cough | ¬W) = 0.25; P(Fever | Flu, Cough) = 0.60, P(Fever | Flu, ¬Cough) = 0.02, P(Fever | ¬Flu, Cough) = 0.60, P(Fever | ¬Flu, ¬Cough) = 0.01. Calculate the probability of a person having fever given he doesn't have flu.** [10 marks] ### Answer The following table enumerates the values of Weather and Cough that were not fixed by the question. Two Boolean choices give four rows: true/true, true/false, false/true, false/false. Each row is one mutually exclusive complete case. Read the CPT entry using that row’s parent values, multiply its factors, and add the resulting row weights. The denominator must include all cases matching the evidence, including both possible values of the queried outcome. #### Given W→F, W→C, (F,C)→V. The complete probabilities are in the question. Evidence F=false. #### To find P(V=true | F=false), summing hidden W and C. #### Formula and why it applies Use the network joint and normalise by the no-Flu evidence. $$ P(V\mid\neg F)=\frac{\sum_{w,c}P(w)P(\neg F\mid w)P(c\mid w)P(V\mid\neg F,c)}{\sum_wP(w)P(\neg F\mid w)} $$ #### Solution Swipe across the diagram to read its labels ↔ ![Re-exam 2026 Q3a: weather–flu–cough–fever network](https://deep-learning-study-room.vercel.app/is2/fig/bn_flu.1d732f0fb6.svg) Diagram: Re-exam 2026 Q3a: weather–flu–cough–fever network. Labels: Re-exam 2026 Q3a: weather–flu–cough–fever network | Weather | P(W) = 0.75 | Flu | P(F|W) = 0.2, P(F|¬W) = 0.5 | Cough | P(C|W) = 0.7, P(C|¬W) = 0.25 | Fever | P(V|F,C): .60 · .02 · .60 · .01 | (TT · TF · FT · FF) ##### Step 1. Chain rule $$ P(W,F,C,V)=P(W)\,P(F\mid W)\,P(C\mid W)\,P(V\mid F,C) $$ ##### Step 2. Sum over W and C with F = false, V = true | W | C | P(W) | P(¬F|W) | P(C|W) | P(V|¬F,C) | Product | | --- | --- | --- | --- | --- | --- | --- | | T | T | 0.75 | 0.8 | 0.70 | 0.60 | 0.252 | | T | F | 0.75 | 0.8 | 0.30 | 0.01 | 0.0018 | | F | T | 0.25 | 0.5 | 0.25 | 0.60 | 0.01875 | | F | F | 0.25 | 0.5 | 0.75 | 0.01 | 0.0009375 | | **P(V, ¬F)** | | | | | | **0.2734875** | ##### Step 3. Evidence $$ P(\lnot F)=0.75(0.8)+0.25(0.5)=0.6+0.125=0.725 $$ ##### Step 4. Divide $$ P(V\mid\lnot F)=\frac{0.2734875}{0.725}=0.3772 $$ **Final answer.** P(fever | no flu) ≈ **0.377** (fever without flu comes mostly from cough). **Key correction.** The key uses P(Flu) = 0.05 and P(Fever | ¬Flu) = 0.2, which are not in this network, and answers 0.2. Marking scheme (still applies): write the equation with hidden variables (2), take true/false cases (3), substitute (2), answer (3). --- ## TEE · Q3(b) Fuzzy relations and composition [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q3b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q3b Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. A = {0.6/small + 0.4/medium + 0.1/large}, B = {0.4/slow + 0.8/moderate + 0.3/fast}, C = {0.5/small + 0.2/medium + 0.4/large}. Find R = A × B, S = C × B and the max–min composition S ∘ R.** [10 marks] ### Answer #### Given The fuzzy sets or relation matrices and their listed row/column order. #### To find The relation matrix/matrices and the composition specified in the question. #### Formula and why it applies For Cartesian products take min of row and column membership. For the ordered matrix composition shown here: $$ T_{ik}=\max_j\min(S_{ij},R_{jk}) $$ If max–product is asked instead, replace the inner min by multiplication. #### Solution ##### Step 1. R and S $$ R=\begin{bmatrix}0.4&0.6&0.3\\0.4&0.4&0.3\\0.1&0.1&0.1\end{bmatrix},\qquad S=\begin{bmatrix}0.4&0.5&0.3\\0.2&0.2&0.2\\0.4&0.4&0.3\end{bmatrix} $$ ##### Step 2. S ∘ R $$ (1,1)=\max[\min(.4,.4),\min(.5,.4),\min(.3,.1)]=\max(.4,.4,.1)=0.4 $$ $$ (1,2)=\max[\min(.4,.6),\min(.5,.4),\min(.3,.1)]=\max(.4,.4,.1)=0.4 $$ $$ (1,3)=\max[\min(.4,.3),\min(.5,.3),\min(.3,.1)]=0.3 $$ $$ \text{row 2 (all 0.2)}=(0.2,0.2,0.2);\quad(3,1)=\max(.4,.4,.1)=0.4,\;(3,2)=\max(.4,.4,.1)=0.4,\;(3,3)=0.3 $$ **Final answer.** $$ R=\begin{bmatrix}.4&.6&.3\\.4&.4&.3\\.1&.1&.1\end{bmatrix} $$ , $$ S=\begin{bmatrix}.4&.5&.3\\.2&.2&.2\\.4&.4&.3\end{bmatrix} $$ , $$ S\circ R=\begin{bmatrix}0.4&0.4&0.3\\0.2&0.2&0.2\\0.4&0.4&0.3\end{bmatrix} $$ . (The key gives R and S only.) ## Every cell without the row shortcuts Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries. ### R: Cartesian-product cells | Cell | Row degree | Column degree | min result | | --- | --- | --- | --- | | R11 | 0.6 | 0.4 | 0.4 | | R12 | 0.6 | 0.8 | 0.6 | | R13 | 0.6 | 0.3 | 0.3 | | R21 | 0.4 | 0.4 | 0.4 | | R22 | 0.4 | 0.8 | 0.4 | | R23 | 0.4 | 0.3 | 0.3 | | R31 | 0.1 | 0.4 | 0.1 | | R32 | 0.1 | 0.8 | 0.1 | | R33 | 0.1 | 0.3 | 0.1 | ### S: Cartesian-product cells | Cell | Row degree | Column degree | min result | | --- | --- | --- | --- | | S11 | 0.5 | 0.4 | 0.4 | | S12 | 0.5 | 0.8 | 0.5 | | S13 | 0.5 | 0.3 | 0.3 | | S21 | 0.2 | 0.4 | 0.2 | | S22 | 0.2 | 0.8 | 0.2 | | S23 | 0.2 | 0.3 | 0.2 | | S31 | 0.4 | 0.4 | 0.4 | | S32 | 0.4 | 0.8 | 0.4 | | S33 | 0.4 | 0.3 | 0.3 | **Result cell (1,1)** $$ \max\{\min(0.4,0.4),\min(0.5,0.4),\min(0.3,0.1)\} $$ $$ =\max\{0.4,0.4,0.1\}=0.4 $$ **Result cell (1,2)** $$ \max\{\min(0.4,0.6),\min(0.5,0.4),\min(0.3,0.1)\} $$ $$ =\max\{0.4,0.4,0.1\}=0.4 $$ **Result cell (1,3)** $$ \max\{\min(0.4,0.3),\min(0.5,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.3,0.3,0.1\}=0.3 $$ **Result cell (2,1)** $$ \max\{\min(0.2,0.4),\min(0.2,0.4),\min(0.2,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (2,2)** $$ \max\{\min(0.2,0.6),\min(0.2,0.4),\min(0.2,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (2,3)** $$ \max\{\min(0.2,0.3),\min(0.2,0.3),\min(0.2,0.1)\} $$ $$ =\max\{0.2,0.2,0.1\}=0.2 $$ **Result cell (3,1)** $$ \max\{\min(0.4,0.4),\min(0.4,0.4),\min(0.3,0.1)\} $$ $$ =\max\{0.4,0.4,0.1\}=0.4 $$ **Result cell (3,2)** $$ \max\{\min(0.4,0.6),\min(0.4,0.4),\min(0.3,0.1)\} $$ $$ =\max\{0.4,0.4,0.1\}=0.4 $$ **Result cell (3,3)** $$ \max\{\min(0.4,0.3),\min(0.4,0.3),\min(0.3,0.1)\} $$ $$ =\max\{0.3,0.3,0.1\}=0.3 $$ --- ## TEE · Q4(a) PSO, one iteration [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q4a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q4a Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Maximize $$ F(x)=x_1^2-x_1x_2+x_2^2+3x_1+4x_2+5 $$ using Particle Swarm Optimization. $$ X=\begin{bmatrix}3.1472&-4.0246\\4.0579&-2.2150\end{bmatrix},\;V=\begin{bmatrix}0.2194&0.2449\\0.1908&0.2228\end{bmatrix} $$ . Take $$ W=0.9,\;c_1=c_2=1.5,\;r_1=0.5949,\;r_2=0.0855 $$ . Optimize and give the solution after 1 iteration showing the step-by-step process.** [10 marks] ### Answer #### Given P1: $$ x=(3.1472,\,-4.0246) $$ , $$ v=(0.2194,\,0.2449) $$ . P2: $$ x=(4.0579,\,-2.2150) $$ , $$ v=(0.1908,\,0.2228) $$ . $$ c_1r_1=0.89235 $$ , $$ c_2r_2=0.12825 $$ . #### To find Updated velocities, positions, pbest and gbest after one iteration. #### Formula and why it applies $$ v^{new}=wv+c_1r_1(pbest-x)+c_2r_2(gbest-x),\quad x^{new}=x+v^{new} $$ Initial personal bests equal the starting positions. Use one frozen global best for all particles. This question maximises fitness, so keep a memory only when its score is larger. #### Solution ##### Step 1. Fitness of P1 (term by term) $$ x_1^2=9.904868,\;-x_1x_2=+12.666221,\;x_2^2=16.197405,\;3x_1=9.4416,\;4x_2=-16.0984 $$ $$ F_1=9.904868+12.666221+16.197405+9.4416-16.0984+5=37.1117 $$ ##### Step 2. Fitness of P2 $$ F_2=16.466552+8.988249+4.906225+12.1737-8.86+5=38.6747 $$ ##### Step 3. pbest and gbest pbest = current positions. $$ 38.6747>37.1117 $$ , so **gbest = P2 = (4.0579, −2.2150)**. ##### Step 4. Particle 1 updates $$ v_{11}=0.9(0.2194)+0+0.12825(4.0579-3.1472)=0.19746+0.116797=0.314257 $$ $$ x_{11}=3.1472+0.314257=3.461457 $$ $$ v_{12}=0.9(0.2449)+0+0.12825(-2.2150+4.0246)=0.22041+0.232081=0.452491 $$ $$ x_{12}=-4.0246+0.452491=-3.572109 $$ ##### Step 5. Particle 2 updates (gbest particle: only inertia acts) $$ v_{21}=0.9(0.1908)=0.17172,\quad x_{21}=4.22962;\qquad v_{22}=0.9(0.2228)=0.20052,\quad x_{22}=-2.01448 $$ ##### Step 6. New fitness and bests Use the same objective at each new position. Keep full precision internally; displayed coordinates are rounded. $$ F(3.461457275,-3.5721088)=(3.461457275)^2-(3.461457275)(-3.5721088)+(-3.5721088)^2+3(3.461457275)+4(-3.5721088)+5 $$ $$ =11.981686467+12.364701993+12.759961279+10.384371825+(-14.2884352)+5\approx38.202286364 $$ $$ F(4.22962,-2.01448)=(4.22962)^2-(4.22962)(-2.01448)+(-2.01448)^2+3(4.22962)+4(-2.01448)+5 $$ $$ =17.889685344+8.520484898+4.05812967+12.68886+(-8.05792)+5\approx40.099239912 $$ | Particle | New position | New F | Old F | pbest | | --- | --- | --- | --- | --- | | 1 | (3.461457, −3.572109) | 38.2023 | 37.1117 | updated | | 2 | (4.22962, −2.01448) | 40.0992 | 38.6747 | updated | The best pbest is now P2's with F = 40.0992, so gbest moves to (4.22962, −2.01448). **Final answer.** $$ V=\begin{bmatrix}0.314257&0.452491\\0.17172&0.20052\end{bmatrix} $$ , $$ X=\begin{bmatrix}3.461457&-3.572109\\4.22962&-2.01448\end{bmatrix} $$ . Both particles improved; new gbest = (4.22962, −2.01448) with F = 40.0992. **Official key warning.** The re-exam answer key solves a *different* problem: particles (0.4, 0.3) and (0.3, 0.4) with velocities (0.1, −0.2) and (−0.2, 0.1), giving fitness 7.53 and 7.86. Do not copy it. The numbers above solve the question as printed. This X and V are particles 1 and 2 of your lecture's five-particle example; only the function differs (+3x₁ and +5 here instead of +2x₁ and +3). --- ## TEE · Q4(b) Naive Bayes: job offer [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q4b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q4b Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Predict whether a candidate accepts a job offer from Experience (Fresher/Mid/Senior), Expected Salary (Low/Medium/High) and Company Reputation (Good/Average). Data: (Fresher, Low, Good, Yes), (Mid, High, Good, Yes), (Senior, High, Average, No), (Fresher, Medium, Average, Yes), (Senior, Low, Good, No), (Mid, Medium, Good, Yes), (Fresher, High, Average, No), (Senior, Medium, Average, No). New candidate: Mid, Low, Good.** [10 marks] ### Answer #### Given The full training data are in Step 1. Query: Experience=Mid, Salary=Low, Reputation=Good. The main calculation uses unsmoothed counts. #### To find Both class scores and the predicted class. #### Formula and why it applies $$ score(C)=P(C)\prod_jP(a_j\mid C) $$ C is a class label. Each aⱼ is one query feature value. First count inside a class, then multiply the prior and all three likelihoods. #### Solution ##### Step 1. Write the training rows as a table | Row | Experience | Salary | Reputation | Class | | --- | --- | --- | --- | --- | | 1 | Fresher | Low | Good | Yes | | 2 | Mid | High | Good | Yes | | 3 | Senior | High | Average | No | | 4 | Fresher | Medium | Average | Yes | | 5 | Senior | Low | Good | No | | 6 | Mid | Medium | Good | Yes | | 7 | Fresher | High | Average | No | | 8 | Senior | Medium | Average | No | ##### Step 2. Separate the two class groups | Class | Rows in this group | Class count | Prior | | --- | --- | --- | --- | | Yes | 1, 2, 4, 6 | 4 | 4/8 = 0.5 | | No | 3, 5, 7, 8 | 4 | 4/8 = 0.5 | ##### Step 3. For class Yes, count Experience=Mid Look only at class Yes rows 1, 2, 4, 6. The matching rows are 2, 6. $$ P(\text{Experience}=\text{Mid}\mid C=\text{Yes})=\frac{2}{4}=0.5 $$ ##### Step 4. For class Yes, count Salary=Low Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1. $$ P(\text{Salary}=\text{Low}\mid C=\text{Yes})=\frac{1}{4}=0.25 $$ ##### Step 5. For class Yes, count Reputation=Good Look only at class Yes rows 1, 2, 4, 6. The matching rows are 1, 2, 6. $$ P(\text{Reputation}=\text{Good}\mid C=\text{Yes})=\frac{3}{4}=0.75 $$ ##### Step 6. Multiply the Yes score one factor at a time $$ P(Yes)=4/8=0.5 $$ $$ 0.5\times0.5=0.25 $$ $$ 0.25\times0.25=0.0625 $$ $$ 0.0625\times0.75=0.046875 $$ ##### Step 7. For class No, count Experience=Mid Look only at class No rows 3, 5, 7, 8. No row in this class has that feature value. $$ P(\text{Experience}=\text{Mid}\mid C=\text{No})=\frac{0}{4}=0 $$ ##### Step 8. For class No, count Salary=Low Look only at class No rows 3, 5, 7, 8. The matching rows are 5. $$ P(\text{Salary}=\text{Low}\mid C=\text{No})=\frac{1}{4}=0.25 $$ ##### Step 9. For class No, count Reputation=Good Look only at class No rows 3, 5, 7, 8. The matching rows are 5. $$ P(\text{Reputation}=\text{Good}\mid C=\text{No})=\frac{1}{4}=0.25 $$ ##### Step 10. Multiply the No score one factor at a time $$ P(No)=4/8=0.5 $$ $$ 0.5\times0=0 $$ $$ 0\times0.25=0 $$ $$ 0\times0.25=0 $$ ##### Step 11. Compare the scores $$ score(Yes)=0.046875\qquad score(No)=0 $$ The larger score belongs to Yes. Normalise only if a posterior probability is requested: divide each score by their total 0.046875. **Final answer.** Predict Yes. Yes score=0.046875; No score=0. **Official key note.** The key gives No = 0.0068 and P(Yes) = 0.873. That No value comes from Laplace smoothing (1/7 × 2/7 × 2/6 × 0.5), but the key did not smooth the Yes side. If you smooth, smooth both: Yes = 0.5 × 3/7 × 2/7 × 4/6 = 0.0408, No = 0.0068, P(Yes) = 0.857. The prediction is Yes either way. **Yaad rakho:** Class ke rows alag karo. Har feature ka count usi class ke total se divide. Saare factors multiply karo. --- ## TEE · Q5(a) Genetic algorithm [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q5a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q5a Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Implement a GA to maximise $$ f(x)=2x^3-3x^2+8x $$ for x in [0,50]. Take 4-bit representations of 12, 20, 17, 13. Crossover after the second bit and mutate by flipping any one bit. Produce two offspring and comment on fitness.** [10 marks] ### Answer **The printed data are inconsistent.** 20 and 17 do not fit into four bits, and covering 0–50 needs six bits. We use six-bit encoding to preserve the supplied parent values and the stated range. #### Given Printed parents: [12, 20, 17, 13]. Worked variant: [12, 20, 17, 13] with 6 bits. For the unspecified selection step, select the two highest-fitness candidates. For the unspecified mutation location, flip the last bit of each child. Crossover is after bit 2. #### To find Two valid children under the declared assumptions and their fitness. #### Formula and why it applies Encoding must respect the candidate range. Crossover swaps tails; mutation flips the chosen bit. Fitness is evaluated after mutation. $$ f(x)=2x^3-3x^2+8x $$ #### Solution ##### Step 1. Encode and evaluate the declared parent population | x | Binary | Fitness substitution | | --- | --- | --- | | 12 | 001100 | 2(12³) − 3(12²) + 8(12) = 3120 | | 20 | 010100 | 2(20³) − 3(20²) + 8(20) = 14960 | | 17 | 010001 | 2(17³) − 3(17²) + 8(17) = 9095 | | 13 | 001101 | 2(13³) − 3(13²) + 8(13) = 3991 | ##### Step 2. Select the two fittest under the stated assumption [20, 17] with fitness [14960, 9095]. ##### Step 3. Cut after bit 2 and exchange tails | Parent 1 | Parent 2 | Child 1 | Child 2 | | --- | --- | --- | --- | | 01 | 0100 | 01 | 0001 | 010001 | 010100 | ##### Step 4. Flip the last bit of each child | Before | After | Decoded x | | --- | --- | --- | | 010001 | 010000 | 16 | | 010100 | 010101 | 21 | ##### Step 5. Evaluate both children | x | Fitness calculation | | --- | --- | | 16 | 2(4096) − 3(256) + 8(16) = 7552 | | 21 | 2(9261) − 3(441) + 8(21) = 17367 | ##### Step 6. Compare with the selected parents Best parent score = 14960; best child score = 17367. Parent average = 12027.5; child average = 12459.5. Both the best score and the average score increased in this worked variant. An individual child can still be worse than a parent; GA does not guarantee improvement after every random operator. **Final answer.** Under these declared repairs and operator choices, children are [16, 21] with fitness [7552, 17367]. These are not unique answers to the inconsistent and under-specified original question. The printed wording about “two offspring” is not an explicit requirement for two complete generations. The M2 examples separately work two full generations because your class update asks for iteration 2. ## Evaluate the polynomial without skipping the powers Calculate x²=x×x, then x³=x²×x. Multiply each by its coefficient and combine with the written signs. A bit-string repair in the worked answer is an explicitly stated response to inconsistent printed givens; it is not a new instruction in the original paper. ### x=12 $$ x^2=12\times12=144\qquad x^3=144\times12=1728 $$ $$ 2x^3=3456\quad 3x^2=432\quad 8x=96 $$ $$ f(12)=3456-432+96=3120 $$ ### x=20 $$ x^2=20\times20=400\qquad x^3=400\times20=8000 $$ $$ 2x^3=16000\quad 3x^2=1200\quad 8x=160 $$ $$ f(20)=16000-1200+160=14960 $$ ### x=17 $$ x^2=17\times17=289\qquad x^3=289\times17=4913 $$ $$ 2x^3=9826\quad 3x^2=867\quad 8x=136 $$ $$ f(17)=9826-867+136=9095 $$ ### x=13 $$ x^2=13\times13=169\qquad x^3=169\times13=2197 $$ $$ 2x^3=4394\quad 3x^2=507\quad 8x=104 $$ $$ f(13)=4394-507+104=3991 $$ ### x=16 $$ x^2=16\times16=256\qquad x^3=256\times16=4096 $$ $$ 2x^3=8192\quad 3x^2=768\quad 8x=128 $$ $$ f(16)=8192-768+128=7552 $$ ### x=21 $$ x^2=21\times21=441\qquad x^3=441\times21=9261 $$ $$ 2x^3=18522\quad 3x^2=1323\quad 8x=168 $$ $$ f(21)=18522-1323+168=17367 $$ --- ## TEE · Q5(b) Hotel booking channels (Markov chain) [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q5b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q5b Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Bookings come via Website (W), App (A) or Counter (C). From W: 50% W, 30% A, 20% C. From A: 40% A, 40% W, 20% C. From C: 20% W, 30% A, 50% C. 1. Develop the transition matrix. 2. Interpret retention and gain/loss. 3. Draw the diagram. 4. If now 40% W, 35% A, 25% C, estimate next week. 5. Which channel grows or shrinks in the long term?** [10 marks] ### Answer #### Given All transition values are given in the full question. State order: W, A, C. Initial vector [0.4, 0.35, 0.25]. #### To find The matrix, diagram, next-step vector and long-run comparison. #### Formula and why it applies Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add. $$ x_{next}=x_{now}P\qquad\pi=\pi P,\quad\sum_i\pi_i=1 $$ #### Solution ##### Step 1. Convert every origin row to probabilities | From / To | W | A | C | | --- | --- | --- | --- | | W | 0.5 | 0.3 | 0.2 | | A | 0.4 | 0.4 | 0.2 | | C | 0.2 | 0.3 | 0.5 | Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25. ##### Step 2. Interpret entries and draw the transitions Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving. Swipe across the diagram to read its labels ↔ ![Each arrow is labelled with the probability of that transition.](https://deep-learning-study-room.vercel.app/is2/fig/mk_booking.a7176dff51.svg) Diagram: Each arrow is labelled with the probability of that transition.. Labels: Weekly booking channel: Website, App, Counter | 0.3 | 0.2 | 0.4 | 0.2 | 0.2 | 0.3 | 0.5 | 0.4 | 0.5 | W | A | C Each arrow is labelled with the probability of that transition. ##### Step 3. Next step: collect all contributions to W | Origin | Current count/share | Transition to W | Contribution | | --- | --- | --- | --- | | W | 0.4 | 0.5 | 0.4 × 0.5 = 0.2 | | A | 0.35 | 0.4 | 0.35 × 0.4 = 0.14 | | C | 0.25 | 0.2 | 0.25 × 0.2 = 0.05 | $$ W_{new}=0.2+0.14+0.05=0.39 $$ ##### Step 4. Next step: collect all contributions to A | Origin | Current count/share | Transition to A | Contribution | | --- | --- | --- | --- | | W | 0.4 | 0.3 | 0.4 × 0.3 = 0.12 | | A | 0.35 | 0.4 | 0.35 × 0.4 = 0.14 | | C | 0.25 | 0.3 | 0.25 × 0.3 = 0.075 | $$ A_{new}=0.12+0.14+0.075=0.335 $$ ##### Step 5. Next step: collect all contributions to C | Origin | Current count/share | Transition to C | Contribution | | --- | --- | --- | --- | | W | 0.4 | 0.2 | 0.4 × 0.2 = 0.08 | | A | 0.35 | 0.2 | 0.35 × 0.2 = 0.07 | | C | 0.25 | 0.5 | 0.25 × 0.5 = 0.125 | $$ C_{new}=0.08+0.07+0.125=0.275 $$ ##### Step 6. Stationarity: find Counter first $$ c=0.2w+0.2a+0.5c $$ Since w+a+c=1, replace w+a with 1−c. $$ c=0.2(1-c)+0.5c=0.2+0.3c $$ $$ 0.7c=0.2\quad c=2/7 $$ ##### Step 7. Find App in terms of Website $$ a=0.3w+0.4a+0.3c $$ $$ 0.6a=0.3w+0.3c\quad a=0.5w+0.5c $$ ##### Step 8. Use the sum to isolate Website $$ w+(0.5w+0.5c)+c=1 $$ $$ 1.5w=1-1.5c=1-3/7=4/7 $$ $$ w=(4/7)/1.5=8/21 $$ $$ a=1-w-c=1-8/21-6/21=7/21=1/3 $$ ##### Step 9. Compare the stationary shares with the initial shares | State | Initial share | Stationary share | Change in share | Conclusion | | --- | --- | --- | --- | --- | | W | 0.4 | 0.380952381 | -0.019047619 | Loses | | A | 0.35 | 0.333333333 | -0.016666667 | Loses | | C | 0.25 | 0.285714286 | 0.035714286 | Gains | Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged. **Final answer.** Next-step vector [0.39, 0.335, 0.275]. Stationary shares ≈ [0.380952381, 0.333333333, 0.285714286]. W loses, A loses, C gains. --- ## TEE · Q6(a) ACO for TSP [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q6a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q6a Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. For the four-city travelling salesman problem the distance matrix and pheromone matrix are given (all pheromones 1). Apply Ant Colony Optimization. Calculate the path travelled by 2 ants starting from city A; random numbers for Ant 1 are 0.3, 0.8 and for Ant 2 are 0.5, 0.7. Also update the pheromone matrix after 1 iteration.** [10 marks] ### Answer #### Given $$ D=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&1&2&3\\B&1&0&4&5\\C&2&4&0&6\\D&3&5&6&0\end{array},\qquad\tau_{ij}=1\;(i\ne j) $$ Assume $$ \alpha=\beta=1,\;\rho=0.5,\;Q=1 $$ . #### To find Both tours, their lengths and the pheromone matrix after one iteration. #### Formula and why it applies $$ P_{ij}=\frac{\tau_{ij}^{\alpha}(1/d_{ij})^\beta}{\sum_{k\in allowed}\tau_{ik}^{\alpha}(1/d_{ik})^\beta} $$ $$ \tau_{ij}^{new}=(1-\rho)\tau_{ij}+\sum_{k:\,ij\in tour_k}\frac{Q}{L_k} $$ Use unvisited cities in alphabetical order for cumulative selection. Every ant completes its tour using old pheromones; then evaporate and deposit on each undirected edge, including the return edge. #### Solution ##### Step 1. From A $$ \eta:\;B=1,\;C=0.5,\;D=0.3333;\;\text{sum}=1.8333\Rightarrow P=0.5455,\;0.2727,\;0.1818 $$ Cumulative: B 0–0.5455, C 0.5455–0.8182, D 0.8182–1. Ant 1 (0.3) → **B**; Ant 2 (0.5) → **B**. ##### Step 2. From B (C and D left) $$ \eta:\;C=0.25,\;D=0.2;\;\text{sum}=0.45\Rightarrow P_{BC}=0.5556,\;P_{BD}=0.4444 $$ Cumulative: C 0–0.5556, D 0.5556–1. Ant 1 (0.8) → **D**; Ant 2 (0.7) → **D**. ##### Step 3. Complete the tours $$ A\to B\to D\to C\to A,\qquad L=1+5+6+2=14\;\text{(both ants)} $$ Swipe across the diagram to read its labels ↔ ![Re-exam 2026 Q6a: both ants A → B → D → C → A, length 14](https://deep-learning-study-room.vercel.app/is2/fig/aco_reexam_graph.02195c6ee9.svg) Diagram: Re-exam 2026 Q6a: both ants A → B → D → C → A, length 14. Labels: Re-exam 2026 Q6a: both ants A → B → D → C → A, length 14 | 1 | 2 | 3 | 4 | 5 | 6 | A | B | C | D ##### Step 4. Pheromone update Used edges AB, BD, DC, CA get $$ 0.5(1)+2\times\frac1{14}=0.6429 $$ ; unused edges AD and BC become $$ 0.5 $$ . $$ \tau_{\text{new}}=\begin{array}{c|cccc}&A&B&C&D\\\hline A&0&0.6429&0.6429&0.5\\B&0.6429&0&0.5&0.6429\\C&0.6429&0.5&0&0.6429\\D&0.5&0.6429&0.6429&0\end{array} $$ **Final answer.** Both ants: A → B → D → C → A, length 14. Pheromone 0.6429 on AB, BD, CD, AC; 0.5 on AD, BC. **Official key note.** This key applies *no evaporation* ( $$ 1+0.1428=1.1428 $$ ) and its matrix is inconsistent: it increases the unused edge A–D, leaves the used return edge C–A at 1, and gives B–D and D–B different values. If your examiner follows the no-evaporation idea, the consistent version is 1.1429 on AB, BD, CD, AC and 1 on AD, BC. The paper does not specify ρ. State the convention you use; the resulting matrix depends on it. Grading depends on the examiner’s intended convention. --- ## TEE · Q6(b) Aircraft movement between hubs (Markov chain) [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q6b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q6b Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Question. Planes at A: 50% stay, 30% to B, 20% to C. At B: 40% to A, 40% stay, 20% to C. At C: 25% to A, 25% to B, 50% stay. 1. Construct the matrix. 2. Interpret retention and movement. 3. Draw the diagram. 4. For [30, 40, 30] find the next night's distribution. 5. Which airport is a long-term net gainer or loser?** [10 marks] ### Answer #### Given All transition values are given in the full question. State order: A, B, C. Initial vector [30, 40, 30]. #### To find The matrix, diagram, next-step vector and long-run comparison. #### Formula and why it applies Rows mean current state; columns mean next state. For each destination, multiply each origin amount by its transition probability, then add. $$ x_{next}=x_{now}P\qquad\pi=\pi P,\quad\sum_i\pi_i=1 $$ #### Solution ##### Step 1. Convert every origin row to probabilities | From / To | A | B | C | | --- | --- | --- | --- | | A | 0.5 | 0.3 | 0.2 | | B | 0.4 | 0.4 | 0.2 | | C | 0.25 | 0.25 | 0.5 | Each row sums to 1 because it lists all possible destinations for that origin. A percentage is divided by 100: 25%=.25. ##### Step 2. Interpret entries and draw the transitions Diagonal entries are the fractions staying at their current state; off-diagonal entries are fractions moving to a different state. Each arrow uses its origin row and destination column. These fractions have different origin populations, so adding raw column probabilities does not give the number arriving. Swipe across the diagram to read its labels ↔ ![Each arrow is labelled with the probability of that transition.](https://deep-learning-study-room.vercel.app/is2/fig/mk_airline.779a8c5fd6.svg) Diagram: Each arrow is labelled with the probability of that transition.. Labels: Nightly aircraft movement between hubs | 0.3 | 0.2 | 0.4 | 0.2 | 0.25 | 0.25 | 0.5 | 0.4 | 0.5 | A | B | C Each arrow is labelled with the probability of that transition. ##### Step 3. Next step: collect all contributions to A | Origin | Current count/share | Transition to A | Contribution | | --- | --- | --- | --- | | A | 30 | 0.5 | 30 × 0.5 = 15 | | B | 40 | 0.4 | 40 × 0.4 = 16 | | C | 30 | 0.25 | 30 × 0.25 = 7.5 | $$ A_{new}=15+16+7.5=38.5 $$ ##### Step 4. Next step: collect all contributions to B | Origin | Current count/share | Transition to B | Contribution | | --- | --- | --- | --- | | A | 30 | 0.3 | 30 × 0.3 = 9 | | B | 40 | 0.4 | 40 × 0.4 = 16 | | C | 30 | 0.25 | 30 × 0.25 = 7.5 | $$ B_{new}=9+16+7.5=32.5 $$ ##### Step 5. Next step: collect all contributions to C | Origin | Current count/share | Transition to C | Contribution | | --- | --- | --- | --- | | A | 30 | 0.2 | 30 × 0.2 = 6 | | B | 40 | 0.2 | 40 × 0.2 = 8 | | C | 30 | 0.5 | 30 × 0.5 = 15 | $$ C_{new}=6+8+15=29 $$ ##### Step 6. Stationarity: find C first $$ c=0.2(a+b)+0.5c $$ $$ c=0.2(1-c)+0.5c=0.2+0.3c $$ $$ 0.7c=0.2\quad c=2/7 $$ ##### Step 7. Find B in terms of A $$ b=0.3a+0.4b+0.25c $$ $$ 0.6b=0.3a+0.25c $$ $$ b=0.5a+\frac5{12}c $$ ##### Step 8. Use a+b=1−c $$ a+0.5a+\frac5{12}\frac27=1-\frac27=\frac57 $$ $$ 1.5a=\frac57-\frac5{42}=\frac{25}{42} $$ $$ a=\frac{25}{42}\div\frac32=\frac{25}{63} $$ $$ b=1-\frac{25}{63}-\frac{18}{63}=\frac{20}{63} $$ ##### Step 9. Compare the stationary shares with the initial shares | State | Initial share | Stationary share | Change in share | Conclusion | | --- | --- | --- | --- | --- | | A | 0.3 | 0.396825397 | 0.096825397 | Gains | | B | 0.4 | 0.317460317 | -0.082539683 | Loses | | C | 0.3 | 0.285714286 | -0.014285714 | Loses | Compare like with like: divide counts by their total before comparing with stationary probabilities. “Stationary” means one more step leaves the shares unchanged. **Final answer.** Next-step vector [38.5, 32.5, 29.0]. Stationary shares ≈ [0.396825397, 0.317460317, 0.285714286]. A gains, B loses, C loses. --- ## TEE · Q7(a) Applications of fuzzy logic [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q7a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q7a Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **List the applications of fuzzy logic. Explain any one application in detail.** [10 marks] ### Answer ### Definition Fuzzy logic is a multi-valued logic, introduced by Lotfi Zadeh (1965), in which truth values range over [0, 1]; it lets machines reason with vague, human-like terms such as “slightly dirty” or “very hot”. ### Applications | Area | Examples | | --- | --- | | Home appliances | Washing machines (wash time from dirt and load), air conditioners, refrigerators, rice cookers, vacuum cleaners | | Automotive | Automatic transmission, anti-lock braking (ABS), cruise control, engine control | | Transportation | Traffic-signal control, Sendai subway train control (smooth braking and acceleration) | | Consumer electronics | Camera autofocus and auto-exposure, image stabilisation | | Industrial control | Cement kilns, pH, temperature and distillation control in chemical plants | | Aerospace | Satellite and spacecraft attitude control | | Medicine | Diagnosis support, anaesthesia dosage control | | Finance and decisions | Loan approval, credit and fire-risk scoring, stock trading rules | | AI and robotics | Robot navigation, natural-language interfaces, expert systems | ### One application in detail: fuzzy washing machine 1. **Problem.** Wash time depends on how dirty and how greasy the clothes are. These are vague (“a bit dirty”), and a fixed timer wastes water and power. 2. **Inputs and outputs.** Sensors measure dirt (turbidity of water) and grease; the output is wash time. Each is described by fuzzy sets: dirt Small/Medium/High, grease Small/Medium/High, time Very Short … Very Long. 3. **Fuzzification.** Sensor readings are converted to membership degrees, e.g. dirt 60% → Medium 0.8, High 0.2. 4. **Rule base.** 9 expert rules, e.g. IF dirt is High AND grease is High THEN time is Very Long; IF dirt is Small AND grease is Small THEN time is Very Short. 5. **Inference.** Mamdani min–max: rule strength = min of the two memberships; outputs combined by max. 6. **Defuzzification.** Use the specified defuzzifier. For a separate representative-weighted illustration, strengths [.6,.4,.2] at times [25,40,60] give (.6×25+.4×40+.2×60)/(.6+.4+.2)=(15+16+12)/1.2≈35.833 minutes. These illustrative strengths are supplied here, not derived from unspecified input graphs. A Mamdani area centroid requires the full aggregate shape. 7. **Benefits.** Smooth adjustment to every load, saves water and electricity, easy to design from expert knowledge without an exact mathematical model. ### Advantages and disadvantages (bonus) **Advantages:** handles imprecise or noisy data; mimics human reasoning; easy to design and modify; no exact model needed; low computational cost. **Disadvantages:** rules and membership functions depend on expert judgement; results are hard to verify formally; tuning can take time; not suited where very high precision is required. **Yaad rakho:** Application explain karte waqt 5 cheezein: inputs, fuzzy sets, rules, inference (min/max), defuzzification + ek number ka example. --- ## TEE · Q7(b) Bellman equation and applications of RL [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q7b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q7b Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged) **Explain the Bellman equation in reinforcement learning. Also give applications of reinforcement learning.** [10 marks] ### Answer Write the answer below in this order. It fits in about two pages. ### Part A · Bellman equation (5 marks) **1. Setting.** Reinforcement learning problems are modelled as a Markov decision process $$ \langle\mathcal S,\mathcal A,P,R,\gamma\rangle $$ . The agent follows a policy $$ \pi(a\mid s) $$ and wants to maximise the return $$ G_t=\sum_{k\ge0}\gamma^kR_{t+k+1} $$ . **2. Value functions.** $$ v_\pi(s)=\mathbb E_\pi[G_t\mid S_t=s] $$ and $$ q_\pi(s,a)=\mathbb E_\pi[G_t\mid S_t=s,A_t=a] $$ . **3. Key idea.** The return is recursive: $$ G_t=R_{t+1}+\gamma G_{t+1} $$ . Taking expectations gives the **Bellman expectation equation**: $$ v_\pi(s)=\sum_a\pi(a\mid s)\sum_{s'}P(s'\mid s,a)\,[R(s,a,s')+\gamma v_\pi(s')] $$ $$ q_\pi(s,a)=\sum_{s'}P(s'\mid s,a)\,\Big[R(s,a,s')+\gamma\sum_{a'}\pi(a'\mid s')q_\pi(s',a')\Big] $$ **4. Optimality.** For the optimal policy the average over actions becomes a maximum (**Bellman optimality equation**): $$ v_*(s)=\max_a\sum_{s'}P(s'\mid s,a)\,[R+\gamma v_*(s')],\qquad q_*(s,a)=\sum_{s'}P(s'\mid s,a)\,[R+\gamma\max_{a'}q_*(s',a')] $$ **5. Meaning and use.** The value of a state equals the immediate reward plus the discounted value of the successor state. This breaks a long-horizon problem into one-step sub-problems (dynamic programming). It is the basis of policy evaluation, policy iteration, value iteration and Q-learning; the optimal policy is $$ \pi_*(s)=\arg\max_a q_*(s,a) $$ . **6. Small illustration** (adds clarity): if staying in a state gives +1 each step with $$ \gamma=0.9 $$ , then $$ v=1+0.9v\Rightarrow v=10 $$ . ### Part B · Applications of reinforcement learning (5 marks) 1. **Game playing**: AlphaGo (defeated Lee Sedol, 2016), Atari agents (DQN), and the Mario agent built in this course's lab. 2. **Robotics**: walking, grasping and balancing learned by trial and error in simulation. 3. **Autonomous driving**: lane changing, merging and parking decisions. 4. **Recommendation systems**: choosing content to maximise long-term engagement rather than single clicks. 5. **Finance**: trading and portfolio rebalancing as sequential decisions. 6. **Healthcare**: dynamic treatment regimes (dose adjustment over time). 7. **Resource and energy management**: data-centre cooling control, traffic-signal timing, inventory control. 8. **Language models**: reinforcement learning from human feedback (RLHF) for chatbots. #### One application in detail: traffic-signal control | | | | --- | --- | | Agent | Controller at an intersection. | | State | Queue length and waiting time on each approach, current signal phase. | | Actions | Keep the current phase or switch to the next green phase. | | Reward | Negative of total waiting time (or vehicles cleared per cycle). | | Learning | Q-learning updates $$ Q(s,a) $$ from observed traffic; ε-greedy exploration tries alternative timings. | | Benefit | Adapts to rush hours and incidents; fixed timers cannot. | | Challenge | Safety during exploration, so agents are trained in simulation first. | **Yaad rakho:** Bellman ka sar: “value = reward abhi + γ × value aage”. Application likho to S, A, R batao: examiner ko yahi chahiye. --- ## TEE · Q1(a) Joint masses, normalization and conditional probability [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2025-april--q1a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2025-april/q1a Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified. **Question. Q1(a)** [5 marks · 2025-04-28] | Machine | No Fault (X = 0) | Fault (X = 1) | | --- | --- | --- | | Machine A | 0.40 | 0.05 | | Machine B | 0.25 | 0.10 | | Machine C | 0.20 | 0.08 | | Machine D | 0.05 | 0.07 | A factory uses four machines (Machine A, Machine B, Machine C, and Machine D) for production. The above table shows the joint probability distribution of a fault occurring (X) and the machine type (Y). - If a fault occurs, what is the probability that it came from Machine C? - What is the probability that a randomly selected product has a fault? - Given the product is from Machine B, what is the probability that it has a fault? ### Answer ### Given The eight printed entries are used exactly as shown. X=1 means fault; Y identifies the machine. ### To find P(C | fault), P(fault), and P(fault | B). ### Formula and why $$ P(A\mid B)=\frac{P(A\cap B)}{P(B)} $$ Conditioning restricts the denominator to the given group. Before treating a whole table as a joint probability distribution, its entries must sum to 1. ### Step 1. Check the printed total $$ \text{No-fault total}=0.40+0.25+0.20+0.05=0.90 $$ $$ \text{Fault total}=0.05+0.10+0.08+0.07=0.30 $$ $$ \text{Grand total}=0.90+0.30=1.20\ne1 $$ **Source inconsistency:** as printed, this is not a valid normalized probability table. Treating the entries as proportional nonnegative masses is one explicit repair; dividing every entry by 1.20 preserves conditional ratios. ### Step 2. Machine C, given a fault $$ P(C\mid X=1)=\frac{0.08/1.20}{0.30/1.20}=\frac{0.08}{0.30}=\frac4{15}\approx0.266667 $$ Among all fault mass, 0.08 belongs to C. The common normalization factor cancels. ### Step 3. Overall fault probability under the stated repair $$ P(X=1)=\frac{0.30}{1.20}=0.25 $$ **Answer under mass normalization: 25%.** The raw column sum 0.30 is only the unnormalized fault mass. If the examiner intended another corrected entry, the unconditional probability depends on that correction; 0.30 cannot be asserted as a valid probability while retaining all eight printed numbers. ### Step 4. Fault, given Machine B $$ P(X=1\mid B)=\frac{0.10}{0.25+0.10}=\frac{0.10}{0.35}=\frac27\approx0.285714 $$ Machine B is the denominator because the product is known to come from B. **Final conditional answers: 26.6667% and 28.5714%.** --- ## TEE · Q1(b) Genetic algorithm applications [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2025-april--q1b/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2025-april/q1b Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified. **Question. Q1(b)** [5 marks · 2025-04-28] List the applications of genetic algorithm and explain any one application in detail. ### Answer **Genetic algorithm (GA)** is a population-based search method that evolves encoded candidate solutions using fitness evaluation, selection, crossover and mutation. - Job-shop scheduling and timetable construction. - Vehicle routing and travelling-salesman-type optimization. - Feature selection and hyperparameter search in machine learning. - Engineering design, such as selecting dimensions subject to strength or cost constraints. - Resource allocation and placement problems. ### Detailed application: examination timetable | Stage | How it is used | | --- | --- | | Representation | A chromosome assigns a time slot and room to every examination. A gene identifies one exam's assignment. | | Fitness | Penalize student clashes, room-capacity violations and unavailable slots; among feasible schedules, reward spread-out exams and efficient room use. | | Initial population | Create several different candidate timetables, using random assignments or constructive rules. | | Selection | Prefer feasible and lower-penalty timetables while retaining diversity. | | Crossover | Combine assignments from two parents; repair conflicts if the operator can violate feasibility. | | Mutation | Move an exam to another permitted slot or swap compatible assignments. Recheck constraints. | | Replacement / stopping | Keep useful candidates, often protecting the best feasible schedule; stop at a generation/time limit or sufficient improvement. | **Illustrative example:** two exams sharing students cannot occupy the same slot. A candidate with that clash receives a strong penalty or is repaired before acceptance. This shows why fitness must reflect the actual constraints, not merely how many rooms are occupied. **Benefit:** explores many combinations without exhaustively listing all timetables. **Limitation:** stochastic search can stagnate and does not guarantee a global optimum; poorly designed penalties can reward unusable schedules. --- ## TEE · Q1(c) Fuzzy union, intersection and complements [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2025-april--q1c/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2025-april/q1c Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified. **Question. Q1(c)** [5 marks · 2025-04-28] Consider two fuzzy sets X={(c1,0.6),(c2,0.8),(c3,0.3),(c4,0.5),(c5,0.7)} Y={(c1,0.4),(c2,0.5),(c3,0.9),(c4,0.6),(c5,0.2)} Find 1) X∪Y 2) X∩Y 3) ¬X∪¬Y 4) ¬X∩¬Y ### Answer ### Given The same universe {c1, c2, c3, c4, c5} is used for both sets. ### To find All four membership vectors, keeping the element order fixed. ### Formula and why $$ \mu_{X\cup Y}(c)=\max(\mu_X(c),\mu_Y(c)),\quad\mu_{X\cap Y}(c)=\min(\mu_X(c),\mu_Y(c)) $$ $$ \mu_{\neg X}(c)=1-\mu_X(c) $$ Union takes the larger membership, intersection the smaller, and standard complement subtracts from 1. Apply each operation separately at the same element. ### Step 1. Find complements | Element | X | Y | 1−X | 1−Y | | --- | --- | --- | --- | --- | | c1 | 0.6 | 0.4 | 1−0.6=0.4 | 1−0.4=0.6 | | c2 | 0.8 | 0.5 | 1−0.8=0.2 | 1−0.5=0.5 | | c3 | 0.3 | 0.9 | 1−0.3=0.7 | 1−0.9=0.1 | | c4 | 0.5 | 0.6 | 1−0.5=0.5 | 1−0.6=0.4 | | c5 | 0.7 | 0.2 | 1−0.7=0.3 | 1−0.2=0.8 | ### Step 2. Compute the four results element by element | Element | X∪Y = max (X, Y) | X∩Y = min (X, Y) | ¬X∪¬Y | ¬X∩¬Y | | --- | --- | --- | --- | --- | | c1 | max(0.6,0.4)=0.6 | min(0.6,0.4)=0.4 | max(0.4,0.6)=0.6 | min(0.4,0.6)=0.4 | | c2 | max(0.8,0.5)=0.8 | min(0.8,0.5)=0.5 | max(0.2,0.5)=0.5 | min(0.2,0.5)=0.2 | | c3 | max(0.3,0.9)=0.9 | min(0.3,0.9)=0.3 | max(0.7,0.1)=0.7 | min(0.7,0.1)=0.1 | | c4 | max(0.5,0.6)=0.6 | min(0.5,0.6)=0.5 | max(0.5,0.4)=0.5 | min(0.5,0.4)=0.4 | | c5 | max(0.7,0.2)=0.7 | min(0.7,0.2)=0.2 | max(0.3,0.8)=0.8 | min(0.3,0.8)=0.3 | ### Step 3. Write the final sets **X∪Y**={(c1,0.6),(c2,0.8),(c3,0.9),(c4,0.6),(c5,0.7)}. **X∩Y**={(c1,0.4),(c2,0.5),(c3,0.3),(c4,0.5),(c5,0.2)}. **¬X∪¬Y**={(c1,0.6),(c2,0.5),(c3,0.7),(c4,0.5),(c5,0.8)}. **¬X∩¬Y**={(c1,0.4),(c2,0.2),(c3,0.1),(c4,0.4),(c5,0.3)}. Yaad rakho: union bada, intersection chhota, complement one minus. --- ## TEE · Q1(d) Cable-company Markov matrix and diagram [5] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2025-april--q1d/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2025-april/q1d Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified. **Question. Q1(d)** [5 marks · 2025-04-28] A city is served by two cable TV companies, BestTV and CableCast. Due to their aggressive sales tactics, each year 40% of BestTV customers switch to CableCast; the other 60% of BestTV customers stay with BestTV. On the other hand, 30% of the CableCast customers switch to Best TV. The two states in this example are BestTV and CableCast. - Express the information above as a transition matrix which displays the probabilities of going from one state into another state. - Draw the State diagram for the above information. ### Answer ### Given Order the states as B=BestTV and C=CableCast. ### To find The one-year transition matrix and all directed state transitions. ### Formula and why $$ P_{ij}=P(\text{next state}=j\mid\text{current state}=i),\qquad\sum_jP_{ij}=1 $$ Rows mean current state; columns mean next state. Each row includes every possible destination. ### Step 1. Complete the missing stay probability BestTV stays =1−0.40=0.60. CableCast stays =1−0.30=0.70. ### Step 2. Place each probability in its row $$ P=\begin{bmatrix}0.60&0.40\\0.30&0.70\end{bmatrix}_{(B,C)} $$ | Current → next | BestTV | CableCast | Row total | | --- | --- | --- | --- | | BestTV | 0.6 | 0.4 | 1 | | CableCast | 0.3 | 0.7 | 1 | ### Step 3. Draw the state diagram Swipe across the diagram to read its labels ↔ ![BestTV → CableCast:0.40; CableCast → BestTV:0.30. Self-loops:0.60 and 0.70.](https://deep-learning-study-room.vercel.app/is2/fig/legacy25-cable.c15b41f741.svg) Diagram: BestTV → CableCast:0.40; CableCast → BestTV:0.30. Self-loops:0.60 and 0.70.. Labels: Cable-company transition probabilities | BestTV | CableCast | 0.40 | 0.30 | 0.60 | 0.70 BestTV → CableCast:0.40; CableCast → BestTV:0.30. Self-loops:0.60 and 0.70. --- ## TEE · Q2(a) Fuzzy investment-controller design and centroid [10] Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2025-april--q2a/ Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2025-april/q2a Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified. **Question. Q2(a)** [10 marks · 2025-04-28] Create a fuzzy logic system to assist in investment decision-making. The system should consider variables such as risk tolerance, and historical performance to provide recommendations on asset allocation and investment strategies. Input variables: Risk Tolerance (range 1-100) and historical performance (range 1-100). Output variable is Asset allocation (range 1-100) (Conservative, Balanced and Aggressive). Use 3 descriptors for input as well as output. Set rules to accurately predict the output. Show defuzzification for risk 40 and historical performance 8. ### Answer ### Given **Printed inputs:** risk=40, historical performance=8, both on 1–100; output allocation score on 1–100. Membership shapes, breakpoints, rules and inference method are design choices, not printed data. ### To find One fully specified fuzzy system and the crisp output for these inputs. ### Formula and why Use a Mamdani system: fuzzify inputs → apply min for AND → clip each output membership at its rule strength → aggregate by max → centroid. The result is a defined model output, not a guaranteed real investment return. $$ z^*=\frac{\int_1^{100}z\mu_{agg}(z)\,dz}{\int_1^{100}\mu_{agg}(z)\,dz} $$ ### Step 1. Specify all membership functions For either input t∈[1,100], choose Low as a left shoulder, Medium as a triangle and High as a right shoulder: $$ \mu_L(t)=\begin{cases}(50-t)/49&1\le t\le50\\0&50