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Relations (1) Rosen 6 th ed., ch. 8
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Binary Relations Let A, B be any two sets. A binary relation R from A to B, written (with signature) R:A↔B, is a subset of A×B. – E.g., let < : N↔N :≡ {(n,m) | n < m} The notation a R b or aRb means (a,b) R. – E.g., a < b means (a,b) < If aRb we may say “a is related to b (by relation R)”, or “a relates to b (under relation R)”. A binary relation R corresponds to a predicate function P R :A×B→{T,F} defined over the 2 sets A,B; e.g., “eats” :≡ {(a,b)| organism a eats food b}
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Complementary Relations Let R:A↔B be any binary relation. Then, R:A↔B, the complement of R, is the binary relation defined by R :≡ {(a,b) | (a,b) R} = (A×B) − R Note this is just R if the universe of discourse is U = A×B; thus the name complement. Note the complement of R is R. <} = {(a,b) | ¬a<b} = ≥ Example: < = {(a,b) | (a,b) <} = {(a,b) | ¬a<b} = ≥
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Inverse Relations Any binary relation R:A↔B has an inverse relation R −1 :B↔A, defined by R −1 :≡ {(b,a) | (a,b) R}. E.g., a} = >. E.g., if R:People→Foods is defined by aRb a eats b, then: b R −1 a b is eaten by a. (Passive voice.)
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Relations on a Set A relation on the set A is a relation from A to A. In other words, a relation on a set A is a subset of A х A. Example 4. Let A be the set {1, 2, 3, 4}. Which ordered pairs are in the relation R = {(a, b) | a divides b}? Solution: R = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)}
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Reflexivity A relation R on A is reflexive if a A, aRa. – E.g., the relation ≥ :≡ {(a,b) | a≥b} is reflexive. A relation is irreflexive iff its complementary relation is reflexive. (for every a A, (a, a) R) – Note “irreflexive” ≠ “not reflexive”! – Example: < is irreflexive. – Note: “likes” between people is not reflexive, but not irreflexive either. (Not everyone likes themselves, but not everyone dislikes themselves either.)
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Symmetry & Antisymmetry A binary relation R on A is symmetric iff R = R −1, that is, if (a,b) R ↔ (b,a) R. i.e, a b((a, b) R → (b, a) R)) – E.g., = (equality) is symmetric. < is not. – “is married to” is symmetric, “likes” is not. A binary relation R is antisymmetric if a b(((a, b) R (b, a) R) → (a = b)) – < is antisymmetric, “likes” is not.
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Asymmetry A relation R is called asymmetric if (a,b) R implies that (b,a) R. – ‘=’ is antisymmetric, but not asymmetric. – < is antisymmetric and asymmetric. – “likes” is not antisymmetric and asymmetric.
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Transitivity A relation R is transitive iff (for all a,b,c) (a,b) R (b,c) R → (a,c) R. A relation is intransitive if it is not transitive. Examples: “is an ancestor of” is transitive. “likes” is intransitive. “is within 1 mile of” is… ?
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Examples of Properties of Relations Example 7. Relations on {1, 2, 3, 4} – R 1 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 4), (4, 1), (4, 4)}, R 2 = {(1, 1), (1, 2), (2, 1)}, R 3 = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1), (4, 4)}, R 4 = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}, R 5 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)}, R 6 = {(3, 4)}.
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Examples Cont. Solution – Reflective : X, X, O, X, O, X – Irreflective : X, X, X, O, X, O – Symmetric : X, O, O, X, X, X – Antisymmetric : X, X, X, O, O, O – Asymmetric : X, X, X, O, X, O – Transitive : X, X, X, O, O, O
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Examples Cont. Example 5. Relations on the set of integers – R 1 = {(a, b) | a ≤ b}, R 2 = {(a, b) | a > b}, R 3 = {(a, b) | a = b or a = -b}, R 4 = {(a, b) | a = b}, R 5 = {(a, b) | a = b + 1}, R 6 = {(a, b) | a + b ≤ 3},
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Examples Cont. Solution – Reflective : O, X, O, O, X, X – Irreflective : X, O, X, X, O, X – Symmetric : X, X, O, O, X, O – Antisymmetric : O, O, X, O, O, X – Asymmetric : X, O, X, X, O, X – Transitive : O, O, O, O, X, X
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Composition of Relations DEFINITION: Let R be a relation from a set A to a set B and S a relation from B to a set C. The composite of R and S is the relation consisting of ordered pairs (a, c), where a A, c C, and for which there exists an element b B such that (a, b) R and (b, c) S. We denote the composite of R and S by S ◦ R.
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Composition of Relations Cont. R : relation between A and B S : relation between B and C S°R : composition of relations R and S – A relation between A and C – {(x,z)| x A, z C, and there exists y B such that xRy and ySz } z w x y u R S A B C x A z w C S°R
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Example 123123 12341234 12341234 1212 RS 123123 1212
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Example Cont. What is the composite of the relations R and S, where R is the relation from {1, 2, 3} to {1, 2, 3, 4} with R = {(1, 1), (1, 4), (2, 3), (3, 1), (3, 4)} and S is the relation from {1, 2, 3, 4} to {0, 1, 2} with S = {(1, 0), (2, 0), (3, 1), (3, 2), (4, 1)}. Solution: S ◦ R = {(1, 0), (1, 1), (2, 1), (2, 2), (3, 0), (3, 1)}
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Power of Relations DEFINITION: Let R be a relation on the set A. The powers R n, n = 1, 2, 3, …, are defined recursively by R 1 = R and R n+1 = R n ◦ R. EXAMPLE: Let R = {(1, 1), (2, 1), (3, 2), (4, 3)}. Find the powers R n, n = 2, 3, 4, …. Solution: R 2 = R ◦ R = {(1, 1), (2, 1), (3, 1), (4, 2)}, R 3 = R 2 ◦ R = {(1, 1), (2, 1), (3, 1), (4, 1)}, R 4 = R 2 ◦ R = {(1, 1), (2, 1), (3, 1), (4, 1)} => R n = R 3
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Power of Relations Cont. THEOREM: The relation R on a set A is transitive if and only if R n R for n = 1, 2, 3, …. EXAMPLE: Let R = {(1, 1), (2, 1), (3, 2), (4, 3)}. Is R transitive? => No (see the previous page). EXAMPLE: Let R = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}. Is R transitive? Solution: R 2 = ?, R 3 = ?, R 4 = ? => maybe a tedious task^^
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Representing Relations Some special ways to represent binary relations: – With a zero-one matrix. – With a directed graph.
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Using Zero-One Matrices To represent a relation R by a matrix M R = [m ij ], let m ij = 1 if (a i,b j ) R, else 0. E.g., Joe likes Susan and Mary, Fred likes Mary, and Mark likes Sally. The 0-1 matrix representation of that “Likes” relation:
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Zero-One Reflexive, Symmetric Terms: Reflexive, non-Reflexive, irreflexive, symmetric, asymmetric, and antisymmetric. – These relation characteristics are very easy to recognize by inspection of the zero-one matrix. Reflexive: all 1’s on diagonal Irreflexive: all 0’s on diagonal Symmetric: all identical across diagonal Antisymmetric: all 1’s are across from 0’s any- thing anything
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Finding Composite Matrix Let the zero-one matrices for S ◦ R, R, S be M S ◦ R, M R, M S. Then we can find the matrix representing the relation S ◦ R by: M S ◦ R = M R ⊙ M S EXAMPLE: M R M S M S ◦ R ⊙ =
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Examples of matrix representation List the ordered pairs in the relation on {1, 2, 3} corresponding to these matrices (where the rows and columns correspond to the integers listed in increasing order).
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Examples Cont. Determine properties of these relations on {1, 2, 3, 4}. R 1 R 2 R 3
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Examples Cont. Solution – Reflective : X, O, X – Irreflective : O, X, O – Symmetric : O, X, O – Antisymmetric : X, X, X – Asymmetric : X, X, X
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Examples Cont. – Transitive : – R 1 2 = ⊙ = – Not Transitive – Notice: (1, 4) and (4, 3) are in R 1 but not (1, 3)
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Examples Cont. – R 2 2 = ⊙ = – R 2 3 = ⊙ =
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Examples Cont. – R 2 4 = ⊙ = – Not transitive – Notice: (1, 3) and (3, 4) are in R 1 but not (1, 4)
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Examples Cont. EXAMPLE: Let R = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}. Is R transitive? Solution: transitive (see below) – R 2 ⊙ = – R 3 ⊙ = – R 3 3 = ⊙ =
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Using Directed Graphs A directed graph or digraph G=(V G,E G ) is a set V G of vertices (nodes) with a set E G V G ×V G of edges (arcs,links). Visually represented using dots for nodes, and arrows for edges. Notice that a relation R:A↔B can be represented as a graph G R =(V G =A B, E G =R). MRMR GRGR Joe Fred Mark Susan Mary Sally Node set V G (black dots) Edge set E G (blue arrows)
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Digraph Reflexive, Symmetric It is extremely easy to recognize the reflexive/irreflexive/ symmetric/antisymmetric properties by graph inspection. Reflexive: Every node has a self-loop Irreflexive: No node links to itself Symmetric: Every link is bidirectional Antisymmetric: No link is bidirectional Asymmetric, non-antisymmetricNon-reflexive, non-irreflexive
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Example Determine which are reflexive, irreflexive, symmetric, antisymmetric, and transitive. a b c a b c a b c transitiveReflexive Symmetric Not transitive Reflexive antisymmetric
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