Relations (1) Rosen 6 th ed., ch. 8. 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

Relations (1)Relations (1)

Rosen 6th ed., ch. 8

Binary RelationsBinary 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 PR:A×B→{T,F} defined over the 2 sets A,B; e.g., “eats” :≡ {(a,b)| organism a eats food b}

Complementary RelationsComplementary Relations

• Let R:A↔B be any binary relation.• Then, R:A↔B, the complement of R, is the

binary relation defined byR :≡ {(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.Example: < = {(a,b) | (a,b)<<} = {(} = {(aa,,bb) | ) | ¬¬aa<<bb} = } = ≥≥

Inverse RelationsInverse 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., <−1 = {(b,a) | a<b} = {(b,a) | b>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.)

Relations on a SetRelations 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)}

ReflexivityReflexivity

• A relation R on A is reflexive if aA, aRa.– E.g., the relation ≥ :≡ {(a,b) | a≥b} is reflexive.

• A relation is irreflexive iff its complementary relation is reflexive. (for every aA, (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.)

Symmetry & AntisymmetrySymmetry & Antisymmetry

• A binary relation R on A is symmetric iff R = R−1, that is, if (a,b)R ↔ (b,a)R.

i.e, ab((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 ab(((a, b) R (b, a) R) → (a = b))– < is antisymmetric, “likes” is not.

AsymmetryAsymmetry

• 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.

TransitivityTransitivity

• 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… ?

Examples of Properties of RelationsExamples of Properties of Relations

• Example 7. Relations on {1, 2, 3, 4}– R1 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 4), (4, 1), (4, 4)},

R2 = {(1, 1), (1, 2), (2, 1)},

R3 = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (3, 3), (4, 1),

(4, 4)}, R4 = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)},

R5 = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4),

(3, 3), (3, 4), (4, 4)}, R6 = {(3, 4)}.

Examples Cont.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


• Example 5. Relations on the set of integers– R1 = {(a, b) | a ≤ b},

R2 = {(a, b) | a > b},

R3 = {(a, b) | a = b or a = -b},

R4 = {(a, b) | a = b},

R5 = {(a, b) | a = b + 1},

R6 = {(a, b) | a + b ≤ 3},


• 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

Composition of RelationsComposition 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.

Composition of Relations Cont.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

wx

y

u

R SAB C

x

Az

w

CS°R

ExampleExample

1

2

3

1

2

3

4

1

2

3

4

1

2

R S

S°R

1

2

3

1

2

Example Cont.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)}

Power of RelationsPower of Relations

• DEFINITION: Let R be a relation on the set A. The powers Rn, n = 1, 2, 3, …, are defined recursively by R1 = R and Rn+1 = Rn◦R.

• EXAMPLE: Let R = {(1, 1), (2, 1), (3, 2), (4, 3)}. Find the powers Rn, n = 2, 3, 4, ….

• Solution: R2 = R◦R = {(1, 1), (2, 1), (3, 1), (4, 2)}, R3 = R2◦R = {(1, 1), (2, 1), (3, 1), (4, 1)}, R4 = R2◦R = {(1, 1), (2, 1), (3, 1), (4, 1)} => Rn = R3

Power of Relations Cont.Power of Relations Cont.

• THEOREM: The relation R on a set A is transitive if and only if Rn 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: R2 = ?, R3 = ?, R4 = ? => maybe a tedious task^^

Representing RelationsRepresenting Relations

• Some special ways to represent binary relations:– With a zero-one matrix.– With a directed graph.

Using Zero-One MatricesUsing Zero-One Matrices

• To represent a relation R by a matrix MR = [mij], let mij = 1 if (ai,bj)R, else 0.

• E.g., Joe likes Susan and Mary, Fred likes Mary, and Mark likes Sally.

• The 0-1 matrix representationof that “Likes”relation:

1 00

010

0 1 1

Mark

Fred

JoeSallyMarySusan

Zero-One Reflexive, SymmetricZero-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.

0

0

101

0

0

01

1

0

0

0

0

1

1

1

1

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

any-thing

any-thing

any-thing anything

anything

Finding Composite MatrixFinding Composite Matrix

• Let the zero-one matrices for S◦R, R, S be M S◦R, MR, MS. Then we can find the matrix representing the relation S◦R by:

• M S◦R = MR M⊙ S

• EXAMPLE: MR MS M S◦R

⊙ =

101

010

101

010

010

010

010

010

010

Examples of matrix representationExamples 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).

101

010

101

010

010

010

111

101

111


• Determine properties of these relations on {1, 2, 3, 4}.

R1 R2 R3

1101

1110

0101

1011

1001

1100

0010

0111

0101

1010

0101

1010


• Solution – Reflective : X, O, X – Irreflective : O, X, O– Symmetric : O, X, O– Antisymmetric : X, X, X– Asymmetric : X, X, X


– Transitive :

– R12 = = ⊙

– Not Transitive – Notice: (1, 4) and (4, 3) are in R1 but not (1, 3)

1101

1110

0101

1011

1101

1110

0101

1011

1111

1111

1111

1111


– R22 = = ⊙

– R23 = = ⊙

1111

1101

0010

1111

1001

1100

0010

0111

1001

1100

0010

0111

1111

1101

0010

1111

1001

1100

0010

0111

1111

1111

0010

1111


– R24 = = ⊙

– Not transitive – Notice: (1, 3) and (3, 4) are in R1 but not (1, 4)

1111

1111

0010

1111

1001

1100

0010

0111

1111

1111

0010

1111

Examples Cont.Examples Cont.• EXAMPLE: Let R = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}. Is R

transitive?• Solution: transitive (see below)

– R2 = ⊙

– R3 = ⊙

– R33 = = ⊙

0111

0011

0001

0000

0111

0011

0001

0000

0011

0001

0000

0000

0111

0011

0001

0000

0001

0000

0000

0000

0011

0001

0000

0000

Using Directed GraphsUsing Directed Graphs

• A directed graph or digraph G=(VG,EG) is a set VG of vertices (nodes) with a set EGVG×VG 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 GR=(VG=AB, EG=R).

1 00

010

0 1 1

Mark

Fred

JoeSallyMarySusan

MR

GRJoe

Fred

Mark

Susan

Mary

Sally

Node set VG(black dots)

Edge set EG(blue arrows)

Digraph Reflexive, SymmetricDigraph 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 isbidirectional

Antisymmetric:

No link isbidirectional

Asymmetric, non-antisymmetric Non-reflexive, non-irreflexive

ExampleExample

• Determine which are reflexive, irreflexive, symmetric, antisymmetric, and transitive.

a b

c

a b

ca b

c

transitive ReflexiveSymmetricNot transitive

Reflexiveantisymmetric

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Relations (1) Rosen 6 th ed., ch. 8. 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