# Q7(b) Max–min relational composition [5]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2024-final/q7b

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2024-final--q7b/

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

**Question. Q7(b)** [5 marks · 2024-04-17]

For the following fuzzy relations R and S, obtain T using max–min composition.



$$
R=\begin{bmatrix}0.6&0.3\\0.2&0.5\end{bmatrix},\quad S=\begin{bmatrix}1&0.4&0.5\\0.8&0.3&0.8\end{bmatrix}
$$



### Answer

#### Given

R is 2×2 and S is 2×3.

#### To find

A 2×3 relation T connecting the first and last universes.

#### Formula and why it applies

Use a row of R and a column of S. The common middle dimension is 2. Take two minima, then the larger of them.



$$
T_{ij}=\max_{k=1,2}\min(R_{ik},S_{kj})
$$



#### Solution

##### Step 1. Verify dimensions

(2×2) composed with (2×3) gives 2×3. Ordinary dot-product addition is not used.

##### Step 2. Calculate every output cell

| Output cell | Compare every matching middle index | Take the largest result |
| --- | --- | --- |
| (1, 1) | min(0.6, 1) = 0.6, min(0.3, 0.8) = 0.3 | max(0.6, 0.3) = 0.6 |
| (1, 2) | min(0.6, 0.4) = 0.4, min(0.3, 0.3) = 0.3 | max(0.4, 0.3) = 0.4 |
| (1, 3) | min(0.6, 0.5) = 0.5, min(0.3, 0.8) = 0.3 | max(0.5, 0.3) = 0.5 |
| (2, 1) | min(0.2, 1) = 0.2, min(0.5, 0.8) = 0.5 | max(0.2, 0.5) = 0.5 |
| (2, 2) | min(0.2, 0.4) = 0.2, min(0.5, 0.3) = 0.3 | max(0.2, 0.3) = 0.3 |
| (2, 3) | min(0.2, 0.5) = 0.2, min(0.5, 0.8) = 0.5 | max(0.2, 0.5) = 0.5 |

##### Step 3. Write the final relation



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



**Final answer.** T=[[0.6,0.4,0.5],[0.5,0.3,0.5]].