# Q3(b) Fuzzy Cartesian relations and composition-domain ambiguity [10]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2025-april/q3b

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2025-april--q3b/

Actual historical paper ·100 marks attempted /140 offered · Q1 compulsory + any four of Q2–Q7. Source defects and solution assumptions are explicitly identified.

**Question. Q3(b)** [10 marks · 2025-04-28]

Consider 2 fuzzy sets

A={0.8/small+0.5/medium+0.2/large}

B={0.6/slow+0.9/moderate+0.4/fast}

- Find the fuzzy relation R=A×B (Cartesian Product)
- Given another fuzzy set C={0.7/small+0.3/medium+0.5/large}. Find the fuzzy relation S=C×B
- Calculate the max-min composition S∘R

### Answer

### Given

A and C use size universe X={small, medium, large}; B uses speed universe Y={slow, moderate, fast}. Use the listed orders for rows and columns.

### To find

R and S; then explain and resolve the composition convention explicitly.

### Formula and why



$$
\mu_{A\times B}(x,y)=\min(\mu_A(x),\mu_B(y)),\quad T_{ij}=\max_k\min(U_{ik},V_{kj})
$$



Fuzzy Cartesian product uses minimum here, not multiplication. A relation composition additionally needs the first relation's output universe to match the second's input universe.

### Step 1. Calculate all entries of R

| Size / speed | slow 0.6 | moderate 0.9 | fast 0.4 |
| --- | --- | --- | --- |
| small 0.8 | min(0.8,0.6)=0.6 | min(0.8,0.9)=0.8 | min(0.8,0.4)=0.4 |
| medium 0.5 | min(0.5,0.6)=0.5 | min(0.5,0.9)=0.5 | min(0.5,0.4)=0.4 |
| large 0.2 | min(0.2,0.6)=0.2 | min(0.2,0.9)=0.2 | min(0.2,0.4)=0.2 |



$$
R=\begin{bmatrix}0.6&0.8&0.4\\0.5&0.5&0.4\\0.2&0.2&0.2\end{bmatrix}
$$



### Step 2. Calculate all entries of S

| Size / speed | slow 0.6 | moderate 0.9 | fast 0.4 |
| --- | --- | --- | --- |
| small 0.7 | min(0.7,0.6)=0.6 | min(0.7,0.9)=0.7 | min(0.7,0.4)=0.4 |
| medium 0.3 | min(0.3,0.6)=0.3 | min(0.3,0.9)=0.3 | min(0.3,0.4)=0.3 |
| large 0.5 | min(0.5,0.6)=0.5 | min(0.5,0.9)=0.5 | min(0.5,0.4)=0.4 |



$$
S=\begin{bmatrix}0.6&0.7&0.4\\0.3&0.3&0.3\\0.5&0.5&0.4\end{bmatrix}
$$



### Step 3. State the source-domain problem

**As typed relations, both R and S map X→Y.** S cannot directly follow R because S expects a size while R ends in a speed. Equal matrix dimensions do not make the universes equal. Thus the requested composition has no unique meaningful typed interpretation as written.

### Step 4. If the examiner intends an indexed matrix operation, state it first

**Conditional classroom calculation:** temporarily identify intermediate labels by position 1,2,3 and define the requested notation here as T=S⊙R, where⊙ is max–min matrix multiplication. This is an index convention, not a claim that “slow” equals “small”.

| Entry | Three minimum terms | Maximum |
| --- | --- | --- |
| T11 | min(0.6,0.6)=0.6, min(0.7,0.5)=0.5, min(0.4,0.2)=0.2 | 0.6 |
| T12 | min(0.6,0.8)=0.6, min(0.7,0.5)=0.5, min(0.4,0.2)=0.2 | 0.6 |
| T13 | min(0.6,0.4)=0.4, min(0.7,0.4)=0.4, min(0.4,0.2)=0.2 | 0.4 |
| T21 | min(0.3,0.6)=0.3, min(0.3,0.5)=0.3, min(0.3,0.2)=0.2 | 0.3 |
| T22 | min(0.3,0.8)=0.3, min(0.3,0.5)=0.3, min(0.3,0.2)=0.2 | 0.3 |
| T23 | min(0.3,0.4)=0.3, min(0.3,0.4)=0.3, min(0.3,0.2)=0.2 | 0.3 |
| T31 | min(0.5,0.6)=0.5, min(0.5,0.5)=0.5, min(0.4,0.2)=0.2 | 0.5 |
| T32 | min(0.5,0.8)=0.5, min(0.5,0.5)=0.5, min(0.4,0.2)=0.2 | 0.5 |
| T33 | min(0.5,0.4)=0.4, min(0.5,0.4)=0.4, min(0.4,0.2)=0.2 | 0.4 |



$$
T=S\odot R=\begin{bmatrix}0.6&0.6&0.4\\0.3&0.3&0.3\\0.5&0.5&0.4\end{bmatrix}
$$



If the course instead writes S∘R to mean first R then S, its untyped matrix computation is R⊙S, which yields rows(0.6,0.6,0.4), (0.5,0.5,0.4), (0.2,0.2,0.2) for these values. The convention must be declared; silently reversing the factors changes the answer.

### Step 5. Show a mathematically valid typed repair

One possible corrected task is to compose R:X→Y with the inverse relation Sᵀ:Y→X. Then compare size elements through a common speed:



$$
U=R\odot S^T=\begin{bmatrix}0.7&0.3&0.5\\0.5&0.3&0.5\\0.2&0.2&0.2\end{bmatrix}
$$





$$
U_{11}=\max\{\min(0.6,0.6),\min(0.8,0.7),\min(0.4,0.4)\}=0.7
$$



This last matrix answers the explicitly repaired task, not an unannounced correction to the printed question.