Intelligent Systems II · TEE · papers

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

Final examination · 17 Apr 2024

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]].