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### Chapter 4 Relations and Digraphs

Weiqi Luo (骆伟祺)

School of Software

Sun Yat-Sen University

Email：weiqi.luo@yahoo.com Office：A309

Chapter Four: Relations and Digraphs

- 4.1 Product Sets and Partitions
- 4.2 Relations and Digraphs
- 4.3 Paths in Relations and Digraphs
- 4.4 Properties of Relations
- 4.5 Equivalence Relations
- 4.6 Data structures for Relations and Digraphs
- 4.7 Operations on Relations
- 4.8 Transitive Closure and Warshall’s Algorithm

4.1. Product Sets and Partitions

- Ordered pair

An order pair (a, b) is a listing of objects a and b in a prescribed order, which a appearing first and b appearing second

(a,b) = (c,d) <=> a=c and b=d

- Product set

If A and B are two nonempty sets, we define the product set or Cartesian product A × B as the set of all ordered pairs (a, b) with a ∈ A and b ∈ B. Thus

A × B = {(a, b) | a ∈ A and b ∈ B}

4.1. Product Sets and Partitions

- Example 1 & 2

Let A={1, 2, 3} and B={r, s}

then

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

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

Note: A × B is not equal to B × A

4.1. Product Sets and Partitions

- Theorem 1: For any two finite, nonempty sets A and B, |A×B| = |A||B|

Proof:

Suppose |A| = m, |B|=n. To form an ordered pair (a, b), a ∈ A, and b∈B, we must perform two successive tasks:

1) choose a first element from A (m ways)

2) choose a second element from B (n ways)

By multiplication principle (Section 3.1). There are totally m×n ways to form an ordered pair (a, b), which means that |A×B| = |A||B| = m×n

4.1. Product Sets and Partitions

- Example 3

If A=B=R, the set of all real numbers, then R×R, also denoted by R2, is the set of all points in the plane. The ordered pair (a, b) gives the coordinates of a point in the plane

Y

(a, b)

O

X

4.1. Product Sets and Partitions

- Cartesian product

A1 × A2 × … × Am = {(a1,a2, …,am) | ai ∈ Ai}

where A1, A2 , … , Am are non-empty sets.

- | A1× A2× … × Am | = |A1| |A2| … |Am|

where A1, A2 , … , Am are finite, nonempty sets

4.1. Product Sets and Partitions

- Example 5

A manufacture offers the following options for its refrigerators

Doors: side-by-side(s) over-under(u) , three (t)

Icemaker: freezer(f), door(d)

Finish: standard(r), metallic(m), custom(c)

Let D={s,u,t} I={f,d} and F={r, m, c}

then the Cartesian product D ×I ×F contains all the categories that describe refigerator options. There are 3 ×2 ×3=18 categories.

4.1. Product Sets and Partitions

- Employees
- Select D[ai1=required1, ai2=required2,…, aik=requiredk]

4.1. Product Sets and Partitions

- Partition

A partition or quotient set of a nonempty set A is a collection B of nonempty subsets of A such that

1. Each element of A belongs to one of the set in B

2. If A1 and A2 are distinct elements of B, then

A1∩ A2 = { }

A

B={A1, A2, A3, A4, A5, A6}

A1

A2

A3

A4

A6

A5

Blocks or cells

4.1. Product Sets and Partitions

- Example 6

Let A={a, b, c, d, e, f, g, h}, consider the following subsets of A

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

A3= {a, c, e, g} A4={b, d} A5={f, h}

Then {A1, A2} is not a partition, since A1 ∩ A2 is not an empty set. Also {A1 , A5} is not a partition since e is not in A1 and A5 . The collection B={A3 A4 A5} is a partition of A

4.2 Relations and Digraphs

- Relation

Let A and B be nonempty sets. A relation R from A to B is a subset of A × B. If R ⊆ A× B and (a, b) ∈ R, we say that a is related to b by R, write a R b, if a is not related to b by R, we write a b.

If A and B are equal. We say that If R ⊆ A× A is a related on A, instead of a relation from A to A.

4.2 Relations and Digraphs

- Example 1

Let A ={1, 2, 3} and B={r, s}. Then we define R={(1, r), (2, s), (3, r)} is a relation from A to B.

- Example 2

Let A and B be sets of real numbers. We define the following relation R(equals) from A to B

Y

O

X

4.2 Relations and Digraphs

- Example 3

Let A={1, 2, 3, 4}. Define the following relation R (less than) on A

a R b if and only if a<b

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

- Example 4

Let A=Z+, the set of all positive integers. Define the following relation R on A:

a R b if and only if a divides b

Then 4 R 12, 6 R 30, but 5 7 .

4.2 Relations and Digraphs

- Example 5

Let A be the set of all people in the world. We define the following relation R on A: a R b if any only if there is a sequence a0, a1, a2, …,an of people such that a0=a, an=b and ai-1 knows ai, i=1,2,…,n (n will depend on a and b)

Small-world network

http://en.wikipedia.org/wiki/Small-world_network

4.2 Relations and Digraphs

- Example 6

Let A=R, the set of real numbers. We define the following relation R on A

x R y if and only if x and y satisfy the equation

x2/4+y2/9=1

Y

(0,3)

O

(2,0)

X

4.2 Relations and Digraphs

- Example 9

Ci R Cj if and only if the cost of going from Ci to Cj defined and less than or equal to $180.

4.2 Relations and Digraphs

- Sets arising from relations

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

a) The domain of R

Denoted by Dom(R), is the set of all first elements in the pairs that make up R. (Dom(R) ⊆ A )

b) The range of R

Denoted by Ran(R), is the set of all second elements in the pairs that make up R. (Ran(R) ⊆ B )

4.2 Relations and Digraphs

- Example 12

Let R be the relation of Example 6

then

Dom(R) = [-2, +2] Ran(R) = [-3, +3]

Y

(0,3)

O

(2,0)

X

4.2 Relations and Digraphs

- R-relative set of x

If R is a relation from A to B and x ∈ A, we define R(x), the R-relative set of x, to be the set of all y in B with the property that x is R-related to y.

R(x) = {y ∈ B | x R y}

If A1 ⊆A, then R(A1), the R-relative set of A1, is the set of all y in B with the property that x is R-related to y for some x in A1.

R(A1) = {y ∈ B | x R y for some x in A1}

4.2 Relations and Digraphs

- Example 13

Let A={a, b, c, d} and let R={(a,a), (a,b), (b,c), (c,a), (d,c), (c,b)}. Then

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

If A1={c, d}, Then

R(A1)= {a, b, c}

4.2 Relations and Digraphs

- Let R be the relation of Example 6

A1=[-∞, -2) U (+2, +∞], A2=[-2, +2] then

R(A1)= ф , R(A2)= [-3, +3]

R(-2)={0}, R(+2)={0} , R(0) = {-3, +3}

Y

(0,3)

O

(2,0)

X

4.2 Relations and Digraphs

- Theorem 1

Let R be a relation from A to B, and let A1 and A2 be subsets of A, then

(a) If A1⊆ A2, then R(A1) ⊆ R(A2)

(b) R(A1 U A2 ) = R(A1) U R(A2)

(c) R (A1 ∩ A2 ) ⊆R(A1 ) ∩ R(A2)

4.2 Relations and Digraphs

- (a) If A1⊆ A2, then R(A1) ⊆ R(A2)

Proof:

If y ∈ R(A1), then x R y for some x in A1, since A1⊆ A2, x ∈A2, thus y ∈ R(A2) and therefore

R(A1) ⊆ R(A2)

4.2 Relations and Digraphs

- (b) R(A1 U A2 ) = R(A1) U R(A2)

Proof :

(1) R(A1 U A2 ) ⊆ R(A1) U R(A2)

If y ∈ R(A1 U A2 ), then x R y for some x in A1 U A2, then x in A1 or A2. If x in A1 , y ∈ R(A1); if x in A2 . y ∈ R(A2). In either cases, y ∈ R(A1) U R(A2) , therefore

R(A1 U A2 ) ⊆ R(A1) U R(A2)

(2) R(A1) U R(A2) ⊆ R(A1 U A2 )

A1⊆ A1 U A2 , then R(A1) ⊆ R(A1 U A2 )

A2⊆ A1 U A2 , then R(A2) ⊆ R(A1 U A2 )

Thus R(A1) U R(A2) ⊆ R(A1 U A2 )

Therefore, we obtain R(A1 U A2 ) = R(A1) U R(A2)

4.2 Relations and Digraphs

- (c) R (A1 ∩ A2) ⊆ R(A1) ∩ R(A2)

Proof:

If y ∈ R (A1 ∩ A2) , then x R y for some x in A1 ∩ A2, since x is in both A1 and A2, it follows that y is in both R(A1) and R(A2) ; that is, y ∈ R(A1 ) ∩ R(A2). Therefore

R (A1 ∩ A2) ⊆ R(A1) ∩ R(A2)

Note: R(A1) ∩ R(A2) ⊆ R (A1 ∩ A2)

4.2 Relations and Digraphs

- Example 15

Let A=Z, R be “≤” A1={0,1,2} and A2={9,13}. Then R(A1) consists of all integers n such that 0 ≤n, or 1 ≤n, or 2 ≤ n, thus R(A1)={0, 1, 2,…}.

Similarly R(A2)={9, 10, 11,…}, so

R(A1) ∩ R(A2)= R(A2)={9,10,11,…}

On the other hand, A1 ∩ A2 = ф , therefore,

R (A1 ∩ A2 ) = ф

This shows that

R(A1 ) ∩ R(A2) ⊆ R (A1 ∩ A2 )

4.2 Relations and Digraphs

- Theorem 2

Let R and S be relations from A to B, If R(a) =S(a) for all a in A, then R=S

Proof:

If a R b, then b ∈ R(a). Therefore, b ∈ S(a) and a S b. A completely similar argument shows that, if a S b, then a R b. Thus R=S.

4.2 Relations and Digraphs

- The matrix of a Relation

If A={a1,a2, …am} and B={b1,b2,…bn} are finite sets containing m and n elements, respectively, and R is a relation from A to B, we represent R by the m×n matrix MR =[mij], which is defined by

mij = 1 if (ai bj) ∈ R

= 0 otherwise

The matrix MR is called the matrix of R.

4.2 Relations and Digraphs

- Example 17

Let A ={1, 2, 3} and B={r, s}. Then we define R={(1, r), (2, s), (3, r)} is a relation from A to B.

Then the matrix of R is

4.2 Relations and Digraphs

- Consider the matrix

Since M is 3×4, we let

A={a1,a2,a3} and B={b1,b2,b3,b4}

Then (ai,bj) in R if and only if mij=1, thus

R={(a1,b1),(a1,b4),(a2,b2),(a2,b3),(a3,b1),(a3,b3}

4.2 Relations and Digraphs

- The Digraph of a Relation

If A is a finite set, R is a relation on A (from A to A)

1) Draw a small circle for each element of A and label the circle with the corresponding element of A (Vertices)

2) Draw an arrow from vertex ai to vertex aj if and only if ai R aj(Edge)

The resulting pictorial representation of R is called a directed graph or digraphy of R.

4.2 Relations and Digraphs

- Example 19

Let A={1, 2, 3, 4}

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

4.2 Relations and Digraphs

- Example 20

Find the relation determined by the following Figure.

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

4.2 Relations and Digraphs

- In-/Out- Degree

If R is a relation on a set A, and a in A, then

The in-degree of a relative to the relation R is the number of b in A such that (b,a) in R.

The out-degree of a is the number of b in A such that (a,b) in R

4.2 Relations and Digraphs

- Example 22

Let A={a, b, c, d}, and R be the relation on A that has the matrix Construct the digraph of R, and list the in-degrees and out-degrees of all vertices

In-degree

Out-degree

4.2 Relations and Digraphs

- If R is a relation on a set A, and B is a subset of A, the restriction of R to B is

R ∩ (B×B)

- Example 24

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

Let B={a,b,c}, then B×B={(a,a), (a,b), (a,c), (b,a), (b,b), (b,c), (c,a), (c,b), (c,c) }

and the restriction of R to B is

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

4.3 Paths in Relations and Digraphs

- Suppose that R is a relation on a set A. A path of length n in R from a to b is a finite sequence

п : a, x1, x2, …, xn-1, b

beginning with a and ending with b, such that

a R x1, x1 R x2, …,xn-1 R b

Note: A path of length n involves n+1 elements of A, although they are not necessarily distinct.

4.3 Paths in Relations and Digraphs

- Example 1

Consider the digraph in the following figure. Then

п1: 1, 2, 5, 4, 3 is a path of

length 4 from vertex 1 to 3

п2: 1, 2, 5, 1 is a path of

length 3 from vertex 1 to itself

п3: 2, 2 is a path of length 1 from

vertex 2 to itself

Cycle: a path that begins and ends at the same vertex (п2 п3 )

4.3 Paths in Relations and Digraphs

- If n is a fixed positive integer, we define a relation Rn on A as follows: x Rn y means that there is a path of length n from x to y in R
- Define a relation R∞ on A, by letting x R y mean that there is some path in R from x to y. The length of such a path will depend on x and y. The relation R∞ is sometimes called the connectivity relation for R.
- Rn(x) consists of all verities that can be reached form x by means of a path in R of length n. The set R∞ (x) consists of all vertices that can be reached from x by some path in R

4.3 Paths in Relations and Digraphs

- Example 2

Let A be the set of all living human beings

R be the relation of mutual acquaintance

a R b means that a and b know one another

a R2 b means that a and b have an acquaintance in common

a Rn b if a knows someone x1, who knows x2, …, who knows xn-1, who knows b.

a R∞ b means that some chain of acquaintances exists that begins at a and ends at b.

Q: It is interesting (and unknown) whether every two Americans, say, are related by R∞

4.3 Paths in Relations and Digraphs

- Example 3

Let A be a set of cities

x R y if there is a direct flight from x to y on at least one airline.

x Rn y if one can book a flight from x to y having exactly n-1 intermediate stops

x R∞ y if one can get from x to y by plane

4.3 Paths in Relations and Digraphs

- Example 5

Let A={a,b,c,d,e} and R={(a,a), (a,b), (b,c), (c,e), (c,d), (d,e)} Compute (a) R2 (b) R ∞

R

4.3 Paths in Relations and Digraphs

- Compute R2

a R2 a since a R a and a R a

a R2 b since a R a and a R b

a R2 c since a R b and b R c

b R2 e since b R c and c R e

b R2 d since b R c and c R d

c R2 e since c R d and d R e

Hence

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

4.3 Paths in Relations and Digraphs

- Compute R ∞

To compute R ∞ , we need all ordered pairs of vertices for which there is a path of any length from the first vertex to the second. We can see that from the figure

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

R

Note: If |R| is large, it can be tedious and perhaps difficult to compute R ∞

4.3 Paths in Relations and Digraphs

- Boolean Matrix (p.37)

A Boolean Matrix is an m×n matrix whose entries are either 0 or 1.

Let A=[aij] and B=[bij] be m×n matrix Boolean matrix.

The Join of A and B : A V B = C = [cij]

cij=1 if aij=1 or bij=1

cij=0 otherwise

The meet of A and B: A ^ B = D = [dij]

dij=1 if aij and bij are both 1

dij=0 otherwise

4.3 Paths in Relations and Digraphs

- Boolean Product (p.38)

A = [aij] is an m×p Boolean matrix and

B = [bij] is a p×n Boolean matrix.

The Boolean product of A and B, denoted A⊙B, is the m×n Boolean matrix C=[cij] defined by

4.3 Paths in Relations and Digraphs

- Theorem 5 (p. 39)

1. (a) A V B= B V A (b) A ^ B = B ^ A

2. (a) (A V B) V C = A V (B V C)

(b) (A ^ B) ^ C = A ^ (B ^ C)

3. (a) A ^ ( B V C) = (A ^ B) V (A ^ C)

(b) A V (B ^ C) = (A V B) ^ (A V C)

4. (A ⊙ B) ⊙C = A ⊙ (B ⊙C)

4.3 Paths in Relations and Digraphs

- Theorem 1

If R is a relation on A={a1,a2,…an}, then

Proof:

Let MR=[mij] and MR2 =[nij]. By definition, the i, j-th element of MR ⊙ MR is equal to 1 if only if row i of MR and column j of MR has a 1 in the same relative position, say position k. This means that mik=1 and mkj=1 for some k, 1 ≤ k ≤ n.

By definition of matrix MR, the preceding conditions mean that ai R ak, and ak R aj. Thus ai R2 aj, and so nij=1.

Therefore, position i, j of MR ⊙ MR is 1 if and only if nij=1.

4.3 Paths in Relations and Digraphs

- Theorem 2

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

Proof by Mathematical Induction

(refer to p.138-139 for more details)

4.3 Paths in Relations and Digraphs

- If R and S are relations on A. the relation R U S is defined by x (R U S) y if and only if x R y or x R y. It is easy to verify that MRUS =MR V MS

4.3 Paths in Relations and Digraphs

- The Reachability relation R* of a relation on a set A that has n elements is defined as follows:

x R* y means that x=y or x R∞ y

4.3 Paths in Relations and Digraphs

- Composition

Let п1 = a, x1, x2, …, xn-1,b be a path in relation R of length n form a to b,

and п2 = b, y1, y2, …, ym-1,c be a path in relation R of length m form b to c,

then the composition of п1 and п2 is the path

a, x1, x2, …, xn-1,b y1, y2, …, ym-1,c of length n+m

denoted by п2 Oп1

4.3 Paths in Relations and Digraphs

- Consider the relation whose digraph is given in the following figure and the paths

п1 = 1, 2, 3 and п2 = 3, 5, 6, 2, 4

Then the composition of п1 and п2 is the path

п2 Oп1 : 1, 2, 3, 5, 6, 2, 4 from 1 to 4 of length 6

4.4 Properties of Relations

- Reflexive & Irreflexive

A relation R on a set A is reflexive if (a, a)∈ R for all a ∈A. A relation R on a set A is irreflexive if a R a for all a∈ A

Dom(R)

=Ran(R)

=A

reflexive

irreflexive

4.4 Properties of Relations

- Example 1

(a) Δ = {(a,a) | a ∈A}

Reflexive

(b) R = {(a,b)∈A×A | a ≠ b}

Irreflexive

(c) A={1,2,3} and R={(1,1), (1,2)}

not Reflexive ((2,2) not in R), not Irreflexive ((1,1) in R)

(d) A is a nonempty set. R=Ø ⊆ A×A, the empty relation.

not Reflexive, Irreflexive

4.4 Properties of Relations

- Symmetric, Asymmetric & Antisymmetric

Symmetric: if a R b, then b R a

Asymmetric: if a R b, then b R a (must be irreflexive)

Antisymmetric: if a ≠ b, a R b or b R a

( if a R b and b R a, then a=b )

not Symmetric: some a and b with a R b, but b R a

not Asymmetric: some a and b with a R b, but b R a

not Antisymmetric :

some a and b with a ≠ b, but both a R b and b R a

4.4 Properties of Relations

- Example 2

Let A=Z, the set of integers, and let

R={ (a,b) ∈A×A | a < b }

Is R symmetric, asymmetric, or antisymmetric?

Solution:

not symmetric: if a<b then b<a is not true.

asymmetric: if a<b, then b<a must be true

Antisymmetric: if a ≠ b, then a<b or b<a must be true

4.4 Properties of Relations

- Example 4

Let A={1,2,3,4} and let

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

R is not symmetric, since (1,2) ∈ R, but (2,1) ∈ R.

R is not asymmetric, since (2,2) ∈ R

R is antisymmeric, since if a ≠ b, either (a, b) ∈ R or (b,a) ∈ R

4.4 Properties of Relations

- Example 5

Let A=Z+, the set of positive integers, and let

R={(a,b) ∈A×A | a divides b }

Is R symmetric, asymmetric, or antisymmetric?

Solution:

not symmetric: 3 | 9 but 9 | 3

not asymmetric: 2 | 2

antisymmetric: if a | b and b | a, then a=b (Section 1.4)

4.4 Properties of Relations

- The properties of symmetric matrix

The matrix MR=[mij] of a symmetric relation on a finite set A= {a1,a2, …,an} satisfies the following property

if mij=1 (ai R aj) then mji=1 (aj R ai)

Moreover,

if mij=0 (ai R aj) then mji=0 (ai R aj)

Therefore, we have

4.4 Properties of Relations

- The properties of asymmetric matrix

The matrix MR=[mij] of an asymmetric relation on a finite set A= {a1,a2, …,an} satisfies the following property :

if mij=1 (ai R aj) then mji=0 (aj R ai)

Moreover,

if mii=0 (ai R ai) for all i=1,2, …,n

4.4 Properties of Relations

- The properties of antisymmetric matrix

The matrix MR=[mij] of an antisymmetric relation on a finite set A= {a1,a2, …,an} satisfies the following property

if i ≠ j, ( ai ≠ aj )

then mij=0 (aiR aj ) or mji=0 (ajR ai )

4.4 Properties of Relations

- Example 6

Symmetric

Not Asymmetic

Not Antisymmetric

Symmetric

Not Asymmetic

Not Antisymmetric

Not Symmetric

Not Asymmetic

Antisymmetric

4.4 Properties of Relations

- Example 6

Not Symmetric

Asymmetic

Antisymmetric

Not Symmetric

Not Asymmetic

Not Antisymmetric

Not Symmetric

Not Asymmetic

Antisymmetric

4.4 Properties of Relations

- The digraphs of asymmetric relation

The digraphs cannot simultaneously have an edge from vertex i to vertex j and an edge from vertex j to vertex i. This is true even i equals j (no cycles). All edges are “one-way streets”

- The digraphs of antisymmetric relation

For different vertices i and j, there cannot be an edge from vertex i to vertex j and edge from vertex j to vertex i. When i=j, no condition is imposed. Thus there may be cycles of length 1, but again all edges are “one way”.

4.4 Properties of Relations

- The digraphs of symmetric relation

If there is an edge from vertex i to vertex j, then there is an edge from vertex j to vertex i. Thus, if two vertex are connected by an edge, they must always be connected in both directions.

We keep the vertices as they appear in the digraph, but if two vertices a and b are connected by edges in each direction, we replace these two edges with one undirected edge, “two-way street ” (Graph of the symmetric relation)

4.4 Properties of Relations

- Example 7

Let A={a,b,c,d,e} and R be the symmetric relation given by R={(a,b), (b,a), (a,c), (c,a), (b,c), (c,b), (b,e), (e,b), (e,d), (d,e), (c,d), (d,c)}

Undirected edge

Adjacent vertices

graphs

digraphs

4.4 Properties of Relations

- Transitive Relations

A relation R on a set A is transitive if whenever a R b and b R c, then a R c.

For instance: (Ex. 8)

the relation “<” on the set of integers

If a < b and b < c, then we have a < c

Note: not Transitive if there exist a, b and c in A so that a R b and b R c, but a R c.

4.4 Properties of Relations

- Example 9

Let A=Z+, and R={(a,b) ∈A×A | a divides b }

Is R transitive?

Solution:

Suppose that a R b and b R c, so that a | b and b | c. It then does follow that a | c (See Theorem 2 of Section 1.4). Thus R is transitive.

4.4 Properties of Relations

- Example 10

Let A ={1,2,3,4} and let R={(1,2),(1,3),(4,2)}

Is R transitive?

Solution:

Since there are no elements a, b and c in A such that a R b and b R c, but a R c, R is transitive.

4.4 Properties of Relations

- A sufficient condition of transitive

A relation R is transitive if and only if its matrix MR=[mij] has the property

if mij=1 and mjk=1, then mik=1.

the left-hand side means that has a 1 in position (i,k). Thus the transitivity of R means that if has a 1 in any position, then MR must have a 1 in the same position.

Thus, in particular, if , then R is transitive.

4.4 Properties of Relations

- Example 11

Let A={1,2,3} and let R be the relation on A whose matrix is

Show that R is transitive.

Solution:

By direct computation, , therefore, R is transitive.

4.4 Properties of Relations

- If R is transitive, may not equal to MR (Verify the Example 10)

4.4 Properties of Relations

- If R is a transitive relation, and a R b and b R c mean that there is a path of length 2 in R from a to c, namely a R2 c. Therefore, we may rephrase the definition of transitivity as follows:

If a R2 c, then a R c, that is R2 ⊆ R

4.4 Properties of Relations

- Theorem 1

A relation R is transitive if and only if it satisfies the following property:

If there is a path of length greater than 1 from vertex a to vertex b, there is a path of length 1 from a to b (that is, a is related to b).

Algebraically stated, R is transitive if and only if Rn ⊆ R for all n >=1

How to Proof the Theorem?

4.4 Properties of Relations

- Theorem 2

Let R be a relation on a set A, then

(a) Reflexivity of R means that a∈R(a) for all a in A.

(b) Symmetry of R means that a∈R(b) iff b∈R(a).

(c) Transitivity of R means that if b∈R(a) and c∈R(b), then c∈R(a).

4.5 Equivalence Relations

- Equivalence Relation

A relation R on a set A is called an equivalence relation if it is reflexive, symmetric, and transitive.

Example 1:

Let A be the set of all triangles in the plane, and let R be the relation on A defined as follow

R= { (a, b) ∈A×A | a is congruent to b}

Therefore, R is an equivalence relation.

Q: R= { (a, b) ∈A×A | a is similar to b} ?

4.5 Equivalence Relations

- Example 2

Let A={1,2,3,4} and let

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

R is an equivalence relation.

4.5 Equivalence Relations

- Example 3

Let A=Z, the set of integers, and let R be defined by a R b if and only if a<=b, is R an equivalence relation?

Reflexive? Yes, since a<=a for all a in A

Symmetric? No, (1, 2) in R but (2,1) is not in R

Transitive?

Yes, since a<=b and b<=c, then we have a<=c

Therefore, R is not an equivalence relation.

4.5 Equivalence Relations

- Example 4

Let A=Z and let R= { (a, b) ∈A×A | a and b yield the same remainder when divided by 2}.

In this case, we call 2 the modulus and write

read “a is congruent to b mod 2”

Reflexive:

Symmetric: if , then a and b yield the same remainder when divided by 2, thus

Transitive: if , then a, b and c yield the same remainder when divided by 2. Thus

4.5 Equivalence Relations

- Theorem 1

Let P be a partition on a set A. Recall that the sets in P are called the blocks of P. Define the relation A as follows:

a R b if and only if a and b are member of the same block

Then R is an equivalence relation on A . Proof:

(a) If a in A, then a is in the same block as itself, so a R a

(b) If a R b, then a and b are in the same block, so b R a

(c) If a R b, and b R c, then a , b and c must all lie in the same block of P, thus a R c.

Note: If P is a partition of a set A, then P can be used to construct an equivalence on A

4.5 Equivalence Relations

- Example 6

Let A={1,2,3,4} and consider the partition P={{1,2,3},{4}} of A. Find the equivalence relation R on A determined by P.

Solution:

Based on the Theorem 1, each element in a block is related to every other element in the same block and only to those elements. Thus,

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

4.5 Equivalence Relations

- P is a partition of A and R is the equivalence relation determined by P

If a in Ai, i=1,2…6

then Ai=R(a)

Lemma 1*:

a R b iff R(a)=R(b)

A1

A2

A3

A4

A6

A5

4.5 Equivalence Relations

- Lemma 1 (Note: Lemma is a theorem whose main purpose is to aid in proving some other theorem)

Let R be an equivalence relation on a set A, and let a ∈ A , b ∈ A. Then

a R b if and only if R(a)=R(b)

Proof:

First, suppose that R(a)=R(b).

Since R is reflexive, b ∈ R(b);

therefore, b in R(a), so a R b

4.5 Equivalence Relations

Conversely, suppose that a R b, then

(1) b in R(a) by definition. Therefore, since R is symmetric

(2) a in R(b) by theorem 2 (b) of Section 4.4.

To prove R(a)=R(b)

First, we choose x in R(b), since R is transitive, the fact x in R(b) with b in R(a) (1) , implies by Theorem 2(c) of Section 4.4 that x in R(a). Thus, R(b) ⊆ R(a)

Second, support y in R(a). This fact and a in R(b) (2) imply. As before, that y in R(b). Thus R(a) ⊆ R(b)

Therefore, we have R(a) = R(b)

The Lemma proved.

4.5 Equivalence Relations

- Theorem 2

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:

According to the definition of a partition, we should show the two following properties

(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) = ф

4.5 Equivalence Relations

- Property (a) is true, since a ∈ R(a) (reflexivity)
- Property (b) is equivalent to the following statement

If R(a) ∩ R(b) ≠ф, then R(a) = R(b) (p59 Theorem 2 b)

Assume c∈ R(a) ∩ R(b), then a R c, b R c.

then we have c R b (symmetric)

a R c, c R b a R b (transitivity)

Therefore, R(a) = R(b) (by lemma 1)

4.5 Equivalence Relations

- Equivalence classes

If R is an equivalence relation on A, then the sets R(a) (or [a]) are traditionally called equivalence classes of R.

The partition P constructed in Theorem 2 consists of all equivalence classes of R, and this partition will be denoted by A/R. (the quotients set of A)

4.5 Equivalence Relations

- Example 7

Let A={1,2,3,4} and let

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

Determine A/R

Solution:

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

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

Hence

A/R = { {1,2} , {3,4}}

4.5 Equivalence Relations

- Example 8

Let A=Z and let R= { (a, b) ∈A×A | a and b yield the same remainder when divided by 2}.

Solution:

First R(0)={…,-4,-2,0,+2,+4,…}, the set of even integers, since the remainder is zero when each of these numbers is divided by 2.

R(1) = {…, -3,-1,0,+1,+3,… }, the set of odd integers, since the remainder is 1 when divided by 2. Hence,

A/R consists of the even integers and the set of odd integers.

4.5 Equivalence Relations

- The procedure of determining A/R is as follows:

Support A={a1,a2,…} (finite or countable)

Step 1: i=0; j=0;

Step 2: i=i+1;

Step 3: if ai ∈ A, then j=j+1; bj=ai

and compute the equivalence class R(bj), A= A- R(bj)

Step 4: if A become an empty set, then we obtain the equivalence classes R(b1), R(b2), … R(bj) (Note: j may be finite or infinite)

otherwise repeat step 2 ~ 4

4.7 Operations on Relations

- Let R and S be relations from a set A to a set B

R and S are subsets of A×B. We can use set operations on the relations R and S

Relations: (three representations)

1. the set of ordered pairs (finite or infinite)

2. digraph (finite)

3. matrix (finite)

4.7 Operations on Relations

- Complementary relation

if and only if a R b

- The intersection R ∩ S

a (R ∩ S) b means that a R b and a S b

- The union R U S

a (R U S) b means that a R b or a S b

4.7 Operations on Relations

- Inverse

Let R be a relation from A to B, the relation R-1 is a relation from B to A (reverse order from R) denoted by

b R-1 a if and only if a R b

Note: (R-1) -1=R

Dom(R-1) = Ran (R)

Ran(R-1) = Dom(R)

4.7 Operations on Relations

- Example 1

Let A={1,2,3,4} and B= {a,b,c}. We Let

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

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

Compute: (a) (b) R ∩ S (c) R U S (d) R-1

Solution:

A×B = { (1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b),

(3,c), (4,a), (4,b), (4,c) }

4.7 Operations on Relations

- Example 1

R∩S = { (1,b), (3,b), (2,c) }

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

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

4.7 Operations on Relations

- Example 2

Let A=R. Let R be the relation ≤ on A and let S be ≥

Then the complement of R is the relation >, since a ≤ b means a>b

Similarly, the complement of S is the relation <.

On the other hand, R-1=S, since for any number a and b

a R-1 b if and only if b R a if and only if b ≤ a

if and only if a ≥ b if and only if a S b

R ∩ S : “=”

R U S : A×A

4.7 Operations on Relations

- Example 3

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

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

R∩S = {(a b), (b e) (c c) }

4.7 Operations on Relations

- Example 4

4.7 Operations on Relations

- Operations on Boolean Matrix

4.7 Operations on Relations

- Theorem 1

Suppose that R and S are relations from A to B

(a) If R ⊆ S, then R-1 ⊆ S-1.

(b) If R ⊆ S, then S ⊆ R

(c) (R∩S) -1 =R -1 ∩ S -1 and (R U S) -1 = R -1 U S -1

(d) (R∩S) = R U S and R U S = R ∩ S

4.7 Operations on Relations

- Theorem 2

Let R and S be relations on a set A

(a) If R is reflexive, so is R-1

(b) If R and S are reflexive, then so are R∩S and R U S

(c) R is reflexive if and only if R is irrflexive

4.7 Operations on Relations

- Example 5

Let A={1,2,3} and consider the two reflexive relations

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

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

then R-1 = { (1,1), (2,1), (3,1), (2,2), (3,3) }. R and R-1

are both reflexive.

R= { (2,1), (2,3), (3,1), (3,2) } is irreflexive while R is

reflexive.

R∩S = {(1,1), (1,2), (2,2), (3,3) } (reflexive)

R∪S = { (1,1), (1,2), (1,3), (2,2), (3,2), (3,3) } (reflexive)

4.7 Operations on Relations

- Theorem 3 (Homework Ex. 37)

Let R be a relation on a set A. Then

(a) R is symmetric if and only if R=R-1

(b) R is antisymmetric if and only if R ∩ R -1 ⊆Δ

(c) R is asymmetric if and only if R ∩ R -1 = Ø

4.7 Operations on Relations

- Theorem 4

Let R and S be relations on A

(a) If R is symmetric, so are R-1 and R

(b) If R and S are symmetric, So R ∩ S and R U S

4.7 Operations on Relations

- Example 6

Let A={1,2,3} and consider the symmetric relations

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

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

then

(a) R-1 = { (1,1), (2,1), (1,2), (3,1), (1,3) } (symmetric)

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

(b) R∩S = { (1,1), (1,2), (2,1) } (symmetric)

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

(symmetric)

4.7 Operations on Relations

- Theorem 5

Let R and S be relations on A

(a) (R ∩ S )2⊆ R2 ∩ S2

(b) If R and S are transitive, so is R ∩ S

(c) If R and S are equivalence relations, so is R ∩ S

4.7 Operations on Relations

- Closure

If R is a relation on a set A, and R lacks some relational properties (e.g. reflexivity, symmetry and transitivity) .

We try to add as few new pairs as possible to R, so that the resulting relation R1 has the property we desired.

If R1 exists, we call it the closure of R with the respect to the property in question.

4.7 Operations on Relations

- Example 8

Support that R is a relation on A, and R is not reflexive.

R1=R∪Δ is the smallest reflexive relation on A containing R. The reflexive closure of R is R∪Δ.

- Example 9

Support that R is a relation on A, and R is not symmetric.

R1=R∪R-1 is the smallest symmetric relation containing R.

(Note: (R∪R-1)-1 = R-1∪R )

4.7 Operations on Relations

- The symmetric closure of R

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

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

4.7 Operations on Relations

- Composition

Support A, B and C are sets, R is a relation from A to B, and S is a relation from B to C, the composition of R and S, written SoR is a relation from A to C

If a is in A and c is in C, then a SoR c if and only if for some b in B, we have a R b, and b S c.

R

S

C

A

B

SoR

4.7 Operations on Relations

- Example 10

Let A= {1,2,3,4} and R={(1,2), (1,1), (1,3), (2,4),(3,2)} and S={(1,4), (1,3), (2,3),(3,1), (4,1)}.

Since (1,2) in R, (2,3) in S, we must have (1,3) in SoR

Similarly, we have

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

4.7 Operations on Relations

- Theorem 6

Let R be a relation from A to B and let S be a relation from B to C. Then if A1 is any subset of A, we have

(SoR)(A1)= S(R(A1) )

Proof: by the definition of composition and relative set

4.7 Operations on Relations

- Theorem 7

Let A, B, C and D be sets, R a relation from A to B, S a relation from B to C and T a relation from C to D. Then

T o (S o R) = (T o S) o R

Proof: by the definition of composition

4.7 Operations on Relations

- Example 13

Let A={a, b}, R={(a, a), (b, a), (b,b)} and S={ (a, b), (b, a), (b, b)}. Then

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

Ro S={(a,a), (a,b), (b,a),(b,b)}

In general, So R does not equal to Ro S

4.7 Operations on Relations

- Theorem 8

Let A, B and C be sets, R a relation from A to B, and S a relation from B to C. Then

(S o R) -1= R-1 o S-1

Proof: by the definition of composition and inverse

4.8 Transitive Closure and Warshall’s Algorithm

- Theorem 1

Let R be a relation on a set A. Then R ∞ is the transitive closure of R

Proof

- R ∞ is transitive. a R ∞ b , b R ∞ c, then a R ∞ c
- R ∞ is the smallest transitive relation containing R.

Support S is any transitive relation on A, and R ⊆ S ,

S is transitive Sn⊆ S, for all n (P.146 Theorem 1)

then S ∞ =S1 U S2… ⊆ S

R ∞⊆ S ∞ ⊆ S (why?)

This means that R ∞is the smallest one.

4.8 Transitive Closure and Warshall’s Algorithm

- Example 1

Let A={1,2,3,4} and let R ={(1,2), (2,3), (3,4), (2,1)}. Find the transitive closure of R

Method # 1:

1 1, 2, 3, 4

2 1, 2, 3, 4

3 4

4 ф

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

4.8 Transitive Closure and Warshall’s Algorithm

- Theorem 2

Let A be a set with |A|=n, and let R be a relation on A. Then

Proof:

If x0 R ∞ xm, then there is a path from x0 to xm with length m

x0, x1, x2, …, xm

case #1: if 1≤ m≤ n, then x0 Rm xm

4.8 Transitive Closure and Warshall’s Algorithm

- Theorem 2

case #2: if m>n, there must have some vertex appearing two times in the path (why?), saying xi = xj, where i<j

x0, x1, …, xi, …, xj, xj+1 … xm

there must exists another path from x0 to xm with length less than m (why?),

x0, x1, …, xi, xj+1 … xm

Therefore, if x0 R ∞ xm , we can find the shortest path from x0 to xm with length less than or equal to n (why?) , namely x0 R k xm 1≤ k≤ n

Therefore, the theorem has been proofed.

4.8 Transitive Closure and Warshall’s Algorithm

- Let R be a relation on a set A={a1,a2,…,an}.

Interior vertices

If x1,x2,…xm is a path in R, then any vertices other than x1 and xm are called interior vertices of the path.

Boolean matrix Wk (n matrices with size n-by-n)

For 1 ≤ k ≤n, Wk has a 1 in position i , j if and only if there is a path from ai to aj in R whose interior vertices, if any, come from the set {a1,a2, …, ak}

Wn=M R ∞

Since any vertex must come from the set {a1,a2,…,an}, it follows that the matrix Wn has a 1 in position i , j if some path in R connects ai with aj

4.8 Transitive Closure and Warshall’s Algorithm

- How to compute Wk k=1,2,…n

Let W0=MR

Support Wk-1=[sij] is known, let Wk=[tij] then

if tij =1, then there must be a path from ai to aj whose interior vertices come from the set {a1,a2, …,ak}.

Case #1: If ak is not an interior vertex of this path, then all interior vertices must actually come from the set {a1,a2,…ak-1}, so sij=1.

Case #2: If ak is an interior vertex of this path, then sik=1 and skj=1 (refer to the proof of Theorem 2)

ak

P1, P2 are subpaths with all interior

vertices from {a1,a2, …, ak-1}

P2

P1

ai

aj

4.8 Transitive Closure and Warshall’s Algorithm

- Warshall’s Algorithm

Warshall’s Algorithm is a more efficient algorithm for computing transitive closure, the procedure of the algorithm is as follows:

Step1: First transfer Wk all 1’s in Wk-1

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

4.8 Transitive Closure and Warshall’s Algorithm

- Example 2

Consider the relation R defined in Example 1. Then

pi: 2

qj: 2

pi: 1, 2

qj: 1, 2, 3

pi: 1, 2

qj: 4

pi: 1, 2

qj: ф

4.8 Transitive Closure and Warshall’s Algorithm

- Algorithm WARSHALL

1. CLOSURE MAT

2. FOR K=1 THRU N

a. FOR I=1 THRU N

1. FOR J=1 THRU N

a. CLOSURE[I,J] CLOSURE[I,J]

V (CLOSURE[I,K] ^ CLOSURE[K,J] )

Warshall’s Algorithm: O(N3)

Boolean Matrix operations: O(N4)

4.8 Transitive Closure and Warshall’s Algorithm

- Theorem 3

If R and S are equivalence relations on a set A, then the smallest equivalence relation containing both R and S is (R U S) ∞

Proof:

- Reflexive. △ ⊆ R, △ ⊆ S, △ ⊆ R U S ⊆ (R U S) ∞
- Symmetric. R=R-1, S=S-1, (RUS)-1=R-1 U S-1=R U S

all paths in RUS are two-way streets, and it follows from the definitions that (R U S) ∞ must be symmetric (why？)

- Transitive. (R U S) ∞ is the transitive closure of R U S (Theorem 1). (the smallest)

4.8 Transitive Closure and Warshall’s Algorithm

- Example 3

Let A={1,2,3,4,5}, R={(1,1), (1,2), (2,1), (2,2), (3,3), (3,4), (4,3), (4,4), (5,5)}, and S= { (1,1), (2,2), (3,3), (4,4), (4,5), (5,4), (5,5) }.

Find the smallest equivalence relation containing R and S, and compute the partition of A that it produces

R

S

1 2 3 4 5

1 2 3 4 5

4.8 Transitive Closure and Warshall’s Algorithm

(RUS) ∞

1 2 3 4 5

(R∪S)∞ = { (1,1), (1,2), (2,1), (2,2), (3,3), (3,4), (3,5), (4,3), (4,4), (4,5), (5,3), (5,4), (5,5 }

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