Intelligent Systems II · TEE · read

Max–min composition: three complete matrix examples

Fuzzy relations and matrices

Longer Unit 2 numerical practice, also useful for TEE.

One result cell combines alternative two-link paths

For T₁₁, start at u₁ and end at w₁. If the middle domain has v₁ and v₂, there are two candidate paths. Their strengths come from matching row 1 of R with column 1 of S.

PathFirst link from RSecond link from SPath strength
u₁ → v₁ → w₁.2.5min(.2,.5)=.2
u₁ → v₂ → w₁.8.7min(.8,.7)=.7

Choose the stronger path: max(.2,.7)=.7, so T₁₁=.7. The middle label j runs through v₁ and v₂. The labels i and k stay fixed at 1 while this one cell is calculated. Then move to the next result cell.

Yaad rakho: Ek path mein dono links chahiye, isliye min. Alag paths mein best chalega, isliye max.

Composition connects through a middle element

Suppose R relates U to V and S relates V to W. To find how strongly u connects to w through each middle v, take min of the two links. Then choose the strongest such path using max.

T_{ik}=\max_j\min(R_{ij},S_{jk})LaTeX: T_{ik}=\max_j\min(R_{ij},S_{jk})

We call this matrix operation R composed with S here. Some books write the symbolic order in reverse. State the formula and row/column labels, especially when a question says S∘R. Composition requires the intermediate domains and their ordering to match; having two 3×3 matrices alone does not prove their labels match.

Calculate the max–min composition using T_{ik}=\max_j\min(R_{ij},S_{jk})LaTeX: T_{ik}=\max_j\min(R_{ij},S_{jk}). [3 marks · Practice]

Given

R=\begin{bmatrix}0.2&0.8\\0.6&0.3\end{bmatrix}LaTeX: R=\begin{bmatrix}0.2&0.8\\0.6&0.3\end{bmatrix}
S=\begin{bmatrix}0.5&0.9\\0.7&0.1\end{bmatrix}LaTeX: S=\begin{bmatrix}0.5&0.9\\0.7&0.1\end{bmatrix}

To find

Every entry of T.

Formula and why it applies

Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication.

Solution

Step 1. Cell (1,1): row 1 of R with column 1 of S
T_{11}=\max\{\min(0.2,0.5),\min(0.8,0.7)\}LaTeX: T_{11}=\max\{\min(0.2,0.5),\min(0.8,0.7)\}
=\max\{0.2,0.7\}=0.7LaTeX: =\max\{0.2,0.7\}=0.7
Step 2. Cell (1,2): row 1 of R with column 2 of S
T_{12}=\max\{\min(0.2,0.9),\min(0.8,0.1)\}LaTeX: T_{12}=\max\{\min(0.2,0.9),\min(0.8,0.1)\}
=\max\{0.2,0.1\}=0.2LaTeX: =\max\{0.2,0.1\}=0.2
Step 3. Cell (2,1): row 2 of R with column 1 of S
T_{21}=\max\{\min(0.6,0.5),\min(0.3,0.7)\}LaTeX: T_{21}=\max\{\min(0.6,0.5),\min(0.3,0.7)\}
=\max\{0.5,0.3\}=0.5LaTeX: =\max\{0.5,0.3\}=0.5
Step 4. Cell (2,2): row 2 of R with column 2 of S
T_{22}=\max\{\min(0.6,0.9),\min(0.3,0.1)\}LaTeX: T_{22}=\max\{\min(0.6,0.9),\min(0.3,0.1)\}
=\max\{0.6,0.1\}=0.6LaTeX: =\max\{0.6,0.1\}=0.6
Final answer. T=\begin{bmatrix}0.7&0.2\\0.5&0.6\end{bmatrix}LaTeX: T=\begin{bmatrix}0.7&0.2\\0.5&0.6\end{bmatrix}.

Calculate the max–min composition using T_{ik}=\max_j\min(R_{ij},S_{jk})LaTeX: T_{ik}=\max_j\min(R_{ij},S_{jk}). [3 marks · Practice]

Given

R=\begin{bmatrix}1&0\\0.4&0.7\end{bmatrix}LaTeX: R=\begin{bmatrix}1&0\\0.4&0.7\end{bmatrix}
S=\begin{bmatrix}0.2&0.6\\0.8&0.3\end{bmatrix}LaTeX: S=\begin{bmatrix}0.2&0.6\\0.8&0.3\end{bmatrix}

To find

Every entry of T.

Formula and why it applies

Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication.

Solution

Step 1. Cell (1,1): row 1 of R with column 1 of S
T_{11}=\max\{\min(1,0.2),\min(0,0.8)\}LaTeX: T_{11}=\max\{\min(1,0.2),\min(0,0.8)\}
=\max\{0.2,0\}=0.2LaTeX: =\max\{0.2,0\}=0.2
Step 2. Cell (1,2): row 1 of R with column 2 of S
T_{12}=\max\{\min(1,0.6),\min(0,0.3)\}LaTeX: T_{12}=\max\{\min(1,0.6),\min(0,0.3)\}
=\max\{0.6,0\}=0.6LaTeX: =\max\{0.6,0\}=0.6
Step 3. Cell (2,1): row 2 of R with column 1 of S
T_{21}=\max\{\min(0.4,0.2),\min(0.7,0.8)\}LaTeX: T_{21}=\max\{\min(0.4,0.2),\min(0.7,0.8)\}
=\max\{0.2,0.7\}=0.7LaTeX: =\max\{0.2,0.7\}=0.7
Step 4. Cell (2,2): row 2 of R with column 2 of S
T_{22}=\max\{\min(0.4,0.6),\min(0.7,0.3)\}LaTeX: T_{22}=\max\{\min(0.4,0.6),\min(0.7,0.3)\}
=\max\{0.4,0.3\}=0.4LaTeX: =\max\{0.4,0.3\}=0.4
Final answer. T=\begin{bmatrix}0.2&0.6\\0.7&0.4\end{bmatrix}LaTeX: T=\begin{bmatrix}0.2&0.6\\0.7&0.4\end{bmatrix}.

Calculate the max–min composition using T_{ik}=\max_j\min(R_{ij},S_{jk})LaTeX: T_{ik}=\max_j\min(R_{ij},S_{jk}). [3 marks · Practice]

Given

R=\begin{bmatrix}0.3&0.6&0.9\end{bmatrix}LaTeX: R=\begin{bmatrix}0.3&0.6&0.9\end{bmatrix}
S=\begin{bmatrix}0.8&0.1\\0.2&0.7\\0.5&0.4\end{bmatrix}LaTeX: S=\begin{bmatrix}0.8&0.1\\0.2&0.7\\0.5&0.4\end{bmatrix}

To find

Every entry of T.

Formula and why it applies

Pair each row entry with the corresponding column entry; min for a two-link path, max across the possible middle elements. This is not ordinary matrix multiplication.

Solution

Step 1. Cell (1,1): row 1 of R with column 1 of S
T_{11}=\max\{\min(0.3,0.8),\min(0.6,0.2),\min(0.9,0.5)\}LaTeX: T_{11}=\max\{\min(0.3,0.8),\min(0.6,0.2),\min(0.9,0.5)\}
=\max\{0.3,0.2,0.5\}=0.5LaTeX: =\max\{0.3,0.2,0.5\}=0.5
Step 2. Cell (1,2): row 1 of R with column 2 of S
T_{12}=\max\{\min(0.3,0.1),\min(0.6,0.7),\min(0.9,0.4)\}LaTeX: T_{12}=\max\{\min(0.3,0.1),\min(0.6,0.7),\min(0.9,0.4)\}
=\max\{0.1,0.6,0.4\}=0.6LaTeX: =\max\{0.1,0.6,0.4\}=0.6
Final answer. T=\begin{bmatrix}0.5&0.6\end{bmatrix}LaTeX: T=\begin{bmatrix}0.5&0.6\end{bmatrix}.

Yaad rakho: Har path mein min, phir saare paths mein max. Multiply-and-add nahi.