Max Min composition for crisp relation

Max Min composition is one way of computing interaction between variables of different relations. The composition of relation R and S is denoted as R ∘ S. Mathematically, it is defined as,

R ∘ S = {(x, z) | (x, y) ∈ R, and (y, z) ∈ S, ∀y ∈ Y}

The composition of the relation is computed in two different ways:

• Max–min composition
• Max–product composition

Although, for crisp relations both are identical, for fuzzy relations, the results of max-min composition and max-product composition would be different.

Example 1: Max min composition

We will try to understand max min composition with multiple examples. Let R = { (x₁, y₁), (x₁, y₃), (x₂, y₄) } and S = { (y₁, z₂), (y₃, z₂) } . let us find the Max-Min composition of these relations.

As we know, the representation of crisp relation could take multiple forms. The above relation R and S we can represent as,

The final composition of relation would look something like this,

Let us see how to fill the cells of the composition matrix T

χ_T(x₁, z₁) = max(min( χ_R(x₁, y₁), χ_S(y₁, z₁)), min( χ_R(x₁, y₂), χ_S(y₂, z₁)), min( χ_R(x₁, y₃), χ_S(y₃, z₁)), min( χ_R(x₁, y₄), χ_S(y₄, z₁))

χ_T(x₁, z₁) = max(min(1, 0),min(0, 0),min(1, 0),min(0, 0))

χ_T(x₁, z₁) = max(0,0,0,0) = 0

Similarly,

χ_T(x₁, z₂) = max(min(1, 1), min(0, 0), min(1, 1), min(0, 0)) = max(1, 0, 1, 0) = 1

χ_T(x₂, z₁) = max(min(0, 0), min(0, 0), min(0, 0), min(1, 0)) = max(0, 0, 0, 0) = 0

χ_T(x₂, z₂) = max(min(0, 1), min(0, 0), min(0, 1), min(1, 0)) = max(0, 0, 0, 0) = 0

χ_T(x₃, z₁) = max(min(0, 0), min(0, 0), min(0, 0), min(0, 0)) = max(0, 0, 0, 0) = 0

χ_T(x₃, z₂) = max(min(0, 1), min(0, 0), min(0, 1), min(0, 0)) = max(0, 0, 0, 0) = 0

Thus, the composition of relation R and S would be,

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Example 2: Max min composition

Given X={1, 3, 5}, Y={1, 3, 5},

R = { (x, y) | y = x + 2 } = { (1, 3), (3, 5) }

S = { (x, y) | x < y } = { (1, 3), (1, 5), (3, 5) }

Find the Max-Min composition of relations R and S

Solution:

X×Y={(1, 1), (1, 3), (1, 5),(3, 1), (3, 3), (3, 5),(5, 1), (5, 3), (5, 5)}

R = { (x, y) | y = x + 2 } = { (1, 3), (3, 5) }

S = { (x, y) | x < y } = { (1, 3), (1, 5), (3, 5) }

In matrix form, we can represent the relation R and S as,

T= R ∘ S = {(x, z) | (x, y) ∈ R, and (y, z) ∈ S, ∀y ∈ Y}

χ_T(1, 1) = max(min(0, 0), min(1, 0), min(0, 0)) = max(0, 0, 0) = 0

χ_T(1, 3) = max(min(0, 1), min(1, 0), min(0, 0)) = max(0, 0, 0) = 0

χ_T(1, 5) = max(min(0, 1), min(1, 1), min(0, 0)) = max(0, 1, 0) = 1

χ_T(3, 1) = max(min(0, 0), min(0, 0), min(1, 0)) = max(0, 0, 0) = 0

χ_T(3, 3) = max(min(0, 1), min(0, 0), min(1, 0)) = max(0, 0, 0) = 0

χ_T(3, 5) = max(min(0, 1), min(0, 1), min(1, 0)) = max(0, 0, 0) = 0

χ_T(5, 1) = max(min(0, 0), min(0, 0), min(0, 0)) = max(0, 0, 0) = 0

χ_T(5, 3) = max(min(0, 1), min(0, 0), min(1, 0)) = max(0, 0, 0) = 0

χ_T(5, 5) = max(min(0, 1), min(1, 0), min(0, 0)) = max(0, 0, 0) = 0

Thus, the composition of relation R and S would be,

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Test your knowledge:

For the given relations,

Find max-min composition.

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