Discrete Mathematicscis.catholic.ac.kr/sunoh/Courses/DiscreteM/DMChapter4.pdf · Chapter 4. Relations and Digraphs Sanguk Noh Discrete Mathematics . Table Product sets and partitions

Chapter 4. Relations and Digraphs

Sanguk Noh

Discrete Mathematics

Table

Product sets and partitions

Relations and digraphs

Paths in relations and digraphs

Properties of relations

Equivalence relations

Operations on relations

Product sets and partitions Def.) product set or Cartesian product A×B

A×B={(a,b)|a∈A and b∈B}

e.g.) A={1,2,3} and B={r,s}

A×B={(1,r),(1,s),(2,r),(2,s),(3,r),(3,s)}

Def.) partition or quotient set P of a nonempty set A

Each element of A belongs to one of the sets in P.

If A1 and A2 are distinct elements of P, then A1∩A2=∅.

e.g.) A={a, b, c, d, e, f, g, h}

A1={a,b,c,d}, A2={a,c,e,f,g,h}, A3={a,c,e,g}, A4={b,d}

A5={f,h}

{A1, A2} : not a partition ∵A1∩A2≠∅

P={A3,A4,A5} is a partition of A.

Relations and digraphs Def.) Let A and B be nonempty sets. Relation R from A to B ⊆ A×B.

If R⊆A×B and (a,b)∈R, then a is related to b by R, a R b.

When R⊆A×A , R is a relation on A.

e.g.) A: the set of positive integers

a R b iff a divides b, a|b.

Then, 4R12, 5R7.

Def.) Let R ⊆ A×B be a relation from A to B.

the domain of R, Dom(R) : the set of elements in A

the range of R, Ran(R) : the set of elements in B

the R-relative set of x, R(x)={y∈B|x R y},

if R is a relation from A to B and x∈A.

e.g.) A={a,b,c,d} and R⊆A×A.

R={(a,a),(a,b),(b,c),(c,a),(d,c),(c,b)}

If A1={c,d}, then R(A1)={a,b,c}

Relations and digraphs Theorem : Let R be a relation from A to B, and let A1⊆A and A2⊆A.

If A1⊆ A2, then R(A1) ⊆ R(A2).

R(A1∪ A2)=R(A1) ∪ R(A2). R(A1 ∩ A2)⊆R(A1) ∩ R(A2)

e.g.) A={x|x is an integer}

R:≤

A1={0,1,2}, A2={9,13}

R(A1)={0,1,2,…}, if x≤y.

R(A2)={9,10,11,…}, if x≤y.

∴R(A1) ∩ R(A2)={9,10,11,…}

A1∩ A2=∅, R(A1∩ A2) = ∅

0≤n or 1≤n or 2≤n


e.g.) Let A= {1,2,3} and B={x,y,z,w,p,q}, and let R ⊆ A×B.

Then, consider R={(1,x),(1,z),(2,w),(2,p),(2,q),(3,y)}.

Let A1={1,2} and A2={2,3}.

(1) R(A1)={x,z,w,p,q}

R(A2)={w,p,q,y}

R(A1)∪R(A2)={x,y,z,w,p,q}=B

Since A1∪A2=A, R(A1∪A2)=R(A)=B

(2) R(A1)∩R(A2)={w,p,q}=R(A1∩A2)

∴ R(A1)∩R(A2)⊇ R(A1∩A2)


Def.) If A={a1,a2…am}and B={b1,b2…bn} are finite sets,

and R is a relation from A to B, then

R: m×n matrix MR=[mij],

mij= 1 if (ai,bj)∈R

0 if (ai,bj)∉R

MR : the matrix of R.

e.g.) A={1,2,3}, B={r,s}, R={(1,r),(2,s),(3,r)}

the matrix of R, m×n,

MR = 1 0 1

0 1 2

1 0 3


Def.) digraph (or directed graph) of R

e.g.) A={1,2}, R={(1,1),(1,2),(2,1),(2,2)}

R is a relation on A.

1

2 3

vertices

Edge: a3 R a1

1 2

Relations and digraphs Def.) in-degree of a: the no. of b∈A s.t. (b,a)∈R

out-degree of a: the no. of b∈A s.t. (a,b)∈R

e.g.) The vertex 1 in the previous figure has in-degree 2.

e.g.) A={1,4,5}

Sol.) MR =

R={(1,4),(1,5),(4,1),(4,4),(5,4),(5,5)}

1 4

5

R

1

4

5

1

0

1

0

4

1

1

1

5

1

0

1


Def.) the restriction of R to B : R∩(B×B)

if R is a relation on a set A and B⊆A.

e.g.) R ∩(B×B)

A={a,b,c}, B={a,b}

R={(a,a),(a,c),(b,c),(b,a),(c,c)}

B×B={(a,a),(a,b),(b,a),(b,b)}

R ∩(B×B) = {(a,a),(b,a)}

Paths in relations and digraphs Def.) a path of length n in R from a to b:

π: a, x1,x2,…,xn-1, b such that

a finite sequence aRx1, x1Rx2, … , xn-1Rb

e.g.)

π1:1,2,3 a path of length 2 from vertex 1 to vertex 3

π2: 1,2,5,1 π3:2,3

1

5 4

2

3

n+1 elements

Paths in relations and digraphs Def.)

Cycle : a path that begins and ends at the same vertex.

x Rn y: There is a path of length n from x to y in R

x R∞ y: There is some path in R from x to y.

⇒ connectivity relation for R.

e.g.) A={a,b,c,d,e}, R={(a,a),(a,b),(b,c),(c,e),(c,d),(d,e)}

(a) R2 (b) R∞

(a) R2 ={(a,a),(a,b),(a,c),(b,d),(b,e),(c,e)}

(b) R∞={(a,a),(a,b),(a,c),(a,d),(a,e),(b,c,),(b,d),(b,e),(c,d),(c,e),(d,e)}

a R2 a a R2 b a R2 c b R2 d b R2 e c R2 e

a b

c

d e

Paths in relations and digraphs Theorem

R is a relation on A ={a1,a2,…an}.

MR2=MR⊙MR

Proof) MR=[mij] MR2=[nij]

By the definition of MR⊙MR, the i, jth element of MR⊙MR is l iff the row i of MR and the column j of MR have a 1 in the same relative position, say k.

⇒mik=1 and mkj=1 for some k, 1≤k≤n.

By the definition of MR, this means that ai R ak and akR aj.

Thus, ai R2 aj , so nij=1

∴ MR⊙MR = MR2

MRn

Paths in relations and digraphs Theorem

For n≥2 and R a relation on a finite set A, we have

= MR⊙ MR⊙ … ⊙ MR (n factors).

Proof by induction

Basis step Let n=2. MR2 = MR⊙ MR

Induction hypothesis

n=k for some k≥2, MRk = MR⊙.. ⊙ MR (k factors)

Induction step

n=k+1. MRk+1 =[xij], MR

k =[yij], and MR =[mij]

if xij =1, we must have a path of length k+1 from ai to aj.

ai as aj

k 1


⇒ yis=1 and msj=1, so MRk ⊙ MR has a 1 in position i,j.

Similarly, if MRk ⊙ MR has a l in position i,j, then xij=1.

⇒ MRk+1 = MR

k ⊙ MR

MRk+1 = MR

k ⊙ MR

= (MR⊙.. ⊙ MR )⊙ MR (k+1 factors)

Thus, by the induction, MRn = MR⊙ MR⊙ … ⊙ MR (n

factors) is true for all n≥2.


Page 140, #26

A={1,2,3,4,5}, R: aRb iff a<b

(a) R2 and R3

R2=

R3=

Then, R=?

(b) a R2 b iff ?

(c) a R3 b iff ?


Def.) reflexive

a relation R on a set A(≡ R⊆A×A) is reflexive if (a,a)∈R

for all a∈A.

e.g.) A={1,2,3}, R={(1,1),(2,2),(3,3)}:reflexive

cf.) “irreflexive” if (a,a)∉R for all a∈A.

e.g.) R={(1,1),(2,3)}

Is the empty relation reflexive?


Def.) Symmetric

A relation R⊆A×A is symmetric if whenever aRb, then bRa.

Def.) asymmetric

If whenever aRb, then bRa

Def.) antisymmetric

If whenever aRb and bRa, then a=b

⇒if whenever a≠b, aRb or bRa


e.g.) A: the set of integers. R={(a,b)∈A×A|a<b}

Sol.)

Symmetric

if a<b, then b≮a, R is not symmetric.

Asymmetric

if a<b, then b≮a, R is asymmetric.

Antisymmetric

if a≠b, then either a≮b or b≮a, R is antisymmetric

Properties of relations e.g.) A={1,2,3,4}, R={(1,2),(2,2),(3,4),(4,1)}

R: not symmetric (1,2)∈R but (2,1)∉R

not asymmetric (2,2) ∈R

antisymmetric!

• Let MR=[mij]

1. symmetric

if mij=1, then mji=1


2. asymmetric


mii=0 for all i if R is asymmetric.

3. antisymmetric

if i≠j, then mij=0 or mji=0.


e.g.) MR1=

MR2=

MR3=

1 0 1

0 0 1

1 1 1

1 1 1

0 1 0

0 0 1

0 1 1 1

0 0 1 0

0 0 0 1

0 0 0 0

MR1 MR2 MR3

Reflexive

Irreflexive

Symmetric

Asymmetric

antisymmetric


Digraph and graph of symmetric relation

Let A= {a,b,c} and R1⊆A×A.

a c

b

a c

b

Digraph of R1

R1 = 0 1 1

1 0 1

1 1 0

“graph” of R1


Def.) Transitive

If whenever a R b and b R c, then a R c.

e.g.) A: the set of integers. R : the realtion “<“

a R b and b R c. a,b,c∈A

⇒a<b and b<c, then a<c, so a R c. Hence R is transitive.

e.g.) A={1,2,3,4} R={(1,2),(1,3),(4,2)} ⇒ transitive!

e.g.) A={1,2,3}

MR =

1 1 1

0 0 1

0 0 1

R={(1,1),(1,2),(1,3),(2,3),(3,3)}

Since (1,2) and (2,3), then (1,3).

Since (1,3) and (3,3), then (1,3).


Theorem R⊆AxA

Reflexivity of R

a∈R(a) for all a in A

Symmetry of R

a∈R(b) iff b∈R(a)

Transitivity of R

if b∈R(a) and c∈R(b), then c∈R(a)


Def.) A relation R on a set A is an equivalence if it is reflexive,

symmetric, and transitive.

e.g.) Let A={1,2,3,4} and

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

reflexive?

symmetric?

transitive?

Equivalence relations Theorem

Let P be a partition of a set A.

Define the relation R on A:

a R b iff a and b are members of the same block.

Then R is an equivalence relation on A.

Proof Reflexive : If a∈A, then a R a. (∵a is in the same block) Symmetric : If a R b, a and b are in the same block. So. b R a Transitive : If a R b and b R c, then a, b, and c must be in the same

block of P. Thus, a R c.

R is the equivalence relation “determined by P.”

Equivalence relations Theorem

Let R be an equivalence relation on A, and let P be the collection

of all distinct relative sets R(a) for a in A. Then P is a partition of

A, and R is the equivalence relation determined by P.

Proof

(a) Every element of A belongs to some relative set.

(b) If R(a) and R(b) are not identical, then R(a)∩R(b)=∅.

(a) is true, since a∈R(a) by reflexivity of R.

(b) ⇒If R(a)∩R(b)≠∅, then R(a)=R(b). Then a R c and b R c

Since R is symmetric, c R b, and a R b by transitivity.

By Lemma, R(a)=R(b), Thus, P is a partition.

By Theorem, P determines R.

Equivalence relations e.g.) Let A={1,2,3,4} and

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

Find partition of A

Sol.) R(1) = {1,2}=R(2)

R(3) = {3,4}=R(4)

Given R, hence, P={{1,2},{3,4}}

Page 152, problem 20.

Let A={a,b,c,d,e} and R⊆AxA.

MR =

a b c d e

1 1 1 0 1

1 1 1 0 1

1 1 1 0 1

0 0 0 1 0

1 1 1 0 1

a

b

d

d

e

∴partition of A : A/R=?


Page 151, #14. A={1,2,3,4,5}, P={{1,3,5},{2,4}}

equivalence relation R?

Operations on relations Operations R,S ⊆ A×B

1. complementary relation : R

2. intersection : R∩S

3. union : R∪S

4. inverse : R-1 : relation from B to A

R-1⊆ B×A

e.g.) A={1,2,3,4}, B={a,b,c}

R={(1,a), (1,b), (2,b), (2,c), (3,b), (4,a)}

S={(1,b),(2,c),(3,b),(4,b)}


Sol.) A×B = {(1,a),….,(4,c)}

(a) R=?

(b) R∩S

(c) R∪S

(d) R-1


Closures

reflexive closure of R

Symmetric closure of R

If A={a,b,c,d} and R={(a,b),(b,c),(a,c),(c,d)}, then

R-1={(b,a), (c,b),(c,a),(d,c)}

the symmetric closure of R:

R∪ R-1 ={(a,b),(b,a),(b,c),(c,b),(a,c),(c,a),(c,d),(d,c)}

a

b c

d a

b

d

c

R

(does not possess a

reflexive property) The reflexive closure of R

Operations on relations Transitive closure of R

Method 1: Finding transitivity from R

Method 2: computing R∞

Method 3: Warshall’s algorithm

a d

b c

“R is not transitive”

Warshall’s algorithm

e.g.) A= {1,2,3,4}, R={(1,2),(2,3),(3,4),(2,1)}

Method 1

(1,2),(2,3) ⇒(1,3)

(1,2),(2,1) ⇒(1,1)

Method 2

….

1 2

3

4

R∞={(1,1),(1,2),(1,3),(1,4),(2,1)…(3,4)}

Warshall’s algorithm Method 3


How to implement the transitive closure of R!

Boolean matrix Wk

A={a1,a2,…,an}, R⊆A×A, 1≤k≤n

Wk=[tij]

tij=1 iff there is a path from ai to aj in R whose interior vertices come

from the set {a1,a2,…,ak}.

Wn : iff some path in R connects ai with aj .

Wn = MR∞

(since any vertex must come from the set {a1,a2,…,an})

Suppose Wk=[tij] and Wk-1=[sij]

(1) sij=1 or (2) sik=1 and skj=1

⇒ tij=1

ai aj

ak Subpath 1 Subpath 2


Algorithm

Step 1: Copy Wk-1 into Wk. W0=MR.

Step 2: List the locations p1,p2…, in column k of Wk-1 , where

the entry is 1, and the locations q1,q2…, in row k of Wk-1 ,

where the entry is 1.

Step 3: Put 1’s in all the positions pi, qj of Wk.

Warshall’s algorithm e.g.) A={1,2,3,4}, R={(1,2),(2,3),(3,4),(2,1)}

W0=MR= 1

2

3

4

1

0

1

0

0

2

1

0

0

0

3

0

1

0

0

4

0

0

1

0

1≤k≤4(=n)

1. k=1

2-1, 1-2

i k k j

2. k=2

i 1 2,2 1 j

2 2

3 3. k=3

i 1 3,3 4 j

2

w1=

0

1

0

0

1

1

0

0

0

1

0

0

0

0

1

0

w2= 1

1

0

0

1

1

0

0

1

1

0

0

0

0

1

0

1

1

0

0

1

1

0

0

1

1

0

0

1

1

1

0

w3=

4. k=4

1 4, 4-(None of 1’s)

2

3 No new 1’s are added.

∴MR∞=W4=W3

Discrete Mathematicscis.catholic.ac.kr/sunoh/Courses/DiscreteM/DMChapter4.pdf · Chapter 4. Relations and Digraphs Sanguk Noh Discrete Mathematics . Table Product sets and partitions

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