Intelligent Systems II · TEE · papers
Q1(c) Fuzzy union, intersection and complements [5]
Final / re-examination · 28 Apr 2025
Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.
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
LaTeX: \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))
LaTeX: \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.