# Q4(a) Cartesian product and max–min composition [10]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-final-2025/q4a

Canonical: https://deep-learning-study-room.vercel.app/read/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)**



$$
\max\{\min(0.2,0.3),\min(0.4,0.3),\min(0.3,0.1)\}
$$





$$
=\max\{0.2,0.3,0.1\}=0.3
$$