# Q3(a) Fuzzy relation and max-product comparison [10]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-2024-reexam/q3a

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-2024-reexam--q3a/

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(a)** [10 marks · 2024-06-25]

D={0.3/0 + 0.7/1 + 1/2}, I={0.5/20 + 0.6/30 + 0.8/40}, V={0.6/20 + 0.5/30 + 0.4/40}. Determine (1) R=D×I, (2) max–min composition of V∘R, (3) max–product composition of V∘R.

### Answer

#### Given

D labels 0,1,2; I and V labels 20,30,40.

#### To find

The 3×3 relation and the two resulting memberships on D’s universe.

#### Formula and why it applies

R connects a D label to an I label. V supplies a membership on the I side. For each D row, combine its three relation memberships with V at the same I labels, then take the maximum. Max–product changes only the inner operation.



$$
R_{ij}=\min(D_i,I_j),\quad t_i=\max_j\min(R_{ij},V_j),\quad u_i=\max_j(R_{ij}V_j)
$$



#### Solution

##### Step 1. Create each Cartesian-product row

| D label | D membership | Min with.5 | Min with.6 | Min with.8 |
| --- | --- | --- | --- | --- |
| 0 | 0.3 | 0.3 | 0.3 | 0.3 |
| 1 | 0.7 | 0.5 | 0.6 | 0.7 |
| 2 | 1 | 0.5 | 0.6 | 0.8 |



$$
R=\begin{bmatrix}0.3&0.3&0.3\\0.5&0.6&0.7\\0.5&0.6&0.8\end{bmatrix}
$$



##### Step 2. Max–min: pair each row with V

| Output cell | Compare every matching middle index | Take the largest result |
| --- | --- | --- |
| (1, 1) | min(0.3, 0.6) = 0.3, min(0.3, 0.5) = 0.3, min(0.3, 0.4) = 0.3 | max(0.3, 0.3, 0.3) = 0.3 |
| (2, 1) | min(0.5, 0.6) = 0.5, min(0.6, 0.5) = 0.5, min(0.7, 0.4) = 0.4 | max(0.5, 0.5, 0.4) = 0.5 |
| (3, 1) | min(0.5, 0.6) = 0.5, min(0.6, 0.5) = 0.5, min(0.8, 0.4) = 0.4 | max(0.5, 0.5, 0.4) = 0.5 |

##### Step 3. Max–product: multiply instead of taking the minimum

| Output cell | Compare every matching middle index | Take the largest result |
| --- | --- | --- |
| (1, 1) | 0.3 × 0.6 = 0.18, 0.3 × 0.5 = 0.15, 0.3 × 0.4 = 0.12 | max(0.18, 0.15, 0.12) = 0.18 |
| (2, 1) | 0.5 × 0.6 = 0.3, 0.6 × 0.5 = 0.3, 0.7 × 0.4 = 0.28 | max(0.3, 0.3, 0.28) = 0.3 |
| (3, 1) | 0.5 × 0.6 = 0.3, 0.6 × 0.5 = 0.3, 0.8 × 0.4 = 0.32 | max(0.3, 0.3, 0.32) = 0.32 |

##### Step 4. Attach the original D labels

The max–min output is {0.3/0 + 0.5/1 + 0.5/2}. The max–product output is {0.18/0 + 0.30/1 + 0.32/2}. These are fuzzy sets, so the slash notation means membership/label, not division.

**Final answer.** R=[[.3,.3,.3],[.5,.6,.7],[.5,.6,.8]]; max–min=[.3,.5,.5]; max–product=[.18,.30,.32].