Intelligent Systems II · TEE · papers

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

Final examination · 24 Nov 2025

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})LaTeX: 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}LaTeX: 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}LaTeX: 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)=0.2LaTeX: (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)=0.3LaTeX: (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)]=0.3LaTeX: (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]=0.3LaTeX: (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}LaTeX: 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

CellRow degreeColumn degreemin result
R110.40.20.2
R120.40.70.4
R130.40.30.3
R210.30.20.2
R220.30.70.3
R230.30.30.3
R310.10.20.1
R320.10.70.1
R330.10.30.1

S: Cartesian-product cells

CellRow degreeColumn degreemin result
S110.30.20.2
S120.30.70.3
S130.30.30.3
S210.10.20.1
S220.10.70.1
S230.10.30.1
S310.40.20.2
S320.40.70.4
S330.40.30.3

Result cell (1,1)

\max\{\min(0.2,0.2),\min(0.3,0.2),\min(0.3,0.1)\}LaTeX: \max\{\min(0.2,0.2),\min(0.3,0.2),\min(0.3,0.1)\}
=\max\{0.2,0.2,0.1\}=0.2LaTeX: =\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)\}LaTeX: \max\{\min(0.2,0.4),\min(0.3,0.3),\min(0.3,0.1)\}
=\max\{0.2,0.3,0.1\}=0.3LaTeX: =\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)\}LaTeX: \max\{\min(0.2,0.3),\min(0.3,0.3),\min(0.3,0.1)\}
=\max\{0.2,0.3,0.1\}=0.3LaTeX: =\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)\}LaTeX: \max\{\min(0.1,0.2),\min(0.1,0.2),\min(0.1,0.1)\}
=\max\{0.1,0.1,0.1\}=0.1LaTeX: =\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)\}LaTeX: \max\{\min(0.1,0.4),\min(0.1,0.3),\min(0.1,0.1)\}
=\max\{0.1,0.1,0.1\}=0.1LaTeX: =\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)\}LaTeX: \max\{\min(0.1,0.3),\min(0.1,0.3),\min(0.1,0.1)\}
=\max\{0.1,0.1,0.1\}=0.1LaTeX: =\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)\}LaTeX: \max\{\min(0.2,0.2),\min(0.4,0.2),\min(0.3,0.1)\}
=\max\{0.2,0.2,0.1\}=0.2LaTeX: =\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)\}LaTeX: \max\{\min(0.2,0.4),\min(0.4,0.3),\min(0.3,0.1)\}
=\max\{0.2,0.3,0.1\}=0.3LaTeX: =\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)\}LaTeX: \max\{\min(0.2,0.3),\min(0.4,0.3),\min(0.3,0.1)\}
=\max\{0.2,0.3,0.1\}=0.3LaTeX: =\max\{0.2,0.3,0.1\}=0.3