chapter 4 relations and digraphs weiqi luo ( 骆伟祺 ) school of software sun yat-sen university...

Chapter 4 Relations and Digraphs

Weiqi Luo (骆伟祺 )School of Software

Sun Yat-Sen UniversityEmail ： [email protected] Office ： A309

mailto:[email protected]

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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

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Chapter Four: Relations and Digraphs

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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

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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


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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


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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


6

O X

Y

(a, b)

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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


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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.


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Employees

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


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Employee ID Last Name Department Years with company

8341 Croft Front Office 2

7984 Cottongim Sales 2

2086 King Human Resources 4

0340 Boswell Research 3

…

…

3465 Harris Sales 4

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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}


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A1 A2

A3

A4

A5

A6

Blocks or cells

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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


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Homework

Ex. 2, Ex. 10, Ex. 17, Ex. 32, Ex. 36, Ex. 37


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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

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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


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Y

O X

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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 .


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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)


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

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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


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O

X

Y(0,3)

(2,0)

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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.


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C1 C2 C3 C4 C5

C1 140 100 150 200

C2 190 200 160 220

C3 110 180 190 250

C4 190 200 120 150

C5 200 100 200 150

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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 )


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Example 12

Let R be the relation of Example 6

then

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


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O

X

Y(0,3)

(2,0)

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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 A∈ 1}


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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}


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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}


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O

X

Y(0,3)

(2,0)

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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)


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(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)


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(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)


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(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)


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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 )


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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.


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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.


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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


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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}


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1 0 0 1

0 1 1 0

1 0 1 0

M

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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.


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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)}


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Example 20 Find the relation determined by the following Figure.

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


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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


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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


37

1 0 0 0

0 1 0 0

1 1 1 0

0 1 0 1

RM

a b c d

2 3 1 1

1 1 3 2

In-degree

Out-degree

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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)}


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Homework

Ex. 2, Ex. 8, Ex.12, Ex.20

Ex.24, Ex. 28, Ex. 32


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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

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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 )


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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


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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∞


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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


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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 ∞


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R

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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)}


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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


47

R

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

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

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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


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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


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1 1 1 , 1

0

ik kj

ij

if a and b for some k k pc

otherwise

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Example

Let

then


50

1 1 0

0 1 0

1 1 0

0 0 1

A

1 0 0 0

0 1 1 0

1 0 1 1

B

1 1 0 1 1 1 01 0 0 0

0 1 0 0 1 1 00 1 1 0

1 1 0 1 1 1 01 0 1 1

0 0 1 1 0 1 1

A B

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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)⊙ ⊙ ⊙ ⊙


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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 m ik=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 M⊙ R is 1 if and only if nij=1.


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2 R RRM M M

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Example 6 Let A and R be as in Example 5. Then


53

1 1 0 0 0

0 0 1 0 0

0 0 0 1 1

0 0 0 0 1

0 0 0 0 0

RM

1 1 0 0 0 1 1 0 0 0 1 1 1 0 0

0 0 1 0 0 0 0 1 0 0 0 0 0 1 1

0 0 0 1 1 0 0 0 1 1 0 0 0 0 1

0 0 0 0 1 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

R RM M

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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)


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... ( )n R R R RRM M M M M n factors

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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

2 3 n1

R =R R R .. Rn

2 3RR R R

2 3R R R

M =M ...

M (M ) (M ) ...

M M

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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


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* nR R

2 3n R R R

M = I M

I M (M ) (M ) ...

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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 O

п1


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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 O

п1 : 1, 2, 3, 5, 6, 2, 4 from 1 to 4 of length 6


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Homework

Ex. 2, Ex. 6, Ex. 12, Ex. 18, Ex. 20, Ex. 26


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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

4.4 Properties of Relations

60

1 * ... *

* 1 ... *

...

* ... 1

reflexive

0 * ... *

* 0 ... *

...

* ... 0

irreflexive

Dom(R)=Ran(R)=A

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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


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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


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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


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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


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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)


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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


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TR RM M

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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


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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 (ai R aj ) or mji=0 (aj R ai )


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Example 6


69

1 0 1

0 0 1

1 1 1

1 1 1 0

1 1 0 0

1 0 1 1

0 0 1 1

1 1 1

0 1 0

0 0 0

SymmetricNot AsymmeticNot Antisymmetric

SymmetricNot AsymmeticNot Antisymmetric

Not SymmetricNot AsymmeticAntisymmetric

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Example 6


70

0 0 1 1

0 0 1 0

0 0 0 1

1 0 0 0

1 0 0 1

0 1 1 1

0 0 1 0

0 0 0 1

0 1 1 1

0 0 1 0

0 0 0 1

0 0 0 0

Not SymmetricNot AsymmeticNot Antisymmetric

Not SymmetricNot AsymmeticAntisymmetric

Not SymmetricAsymmeticAntisymmetric

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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”.


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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)


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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)}


73

digraphs graphs

Undirected edge

Adjacent vertices

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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.


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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.


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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.


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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.


77

2( )RM 2( )RM

2( )R RM M

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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.


78

111

001

001RM

2( )R RM M

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


79

2( )RM

0110 0110 0000

0000 0000 0000

0000 0000 0000

0100 0100 0000

R R RM M M

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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


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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?


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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).∈


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Homework

Ex. 2, Ex. 14, Ex. 18, Ex. 26, Ex. 32, Ex. 38


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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

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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.


85

1100

1100

0011

0011

RM

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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.


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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


87

(mod 2)a b(mod 2)a a

(mod 2)a b(mod 2)b a

(mod 2) & (mod 2)a b b c

(mod 2)a c

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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


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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)}


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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)


90

A1 A2

A3

A4

A5

A6

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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


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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.


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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


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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)


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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)


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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}}


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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.


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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


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Homework

Ex. 4, Ex. 12, Ex. 14, Ex. 20, Ex. 23, Ex. 28


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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

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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


101

a Rb

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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)


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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) }


103

R

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

R A B R

School of Software

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) }


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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


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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) }


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Example 4


107

101

011

000RM

011

110

010SM

_

010

100

111R

M

1

100

010

110R

M

S

001

010

000RM

S

111

111

010RM

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Operations on Boolean Matrix


108

1 ( )

R S R S

R S R S

TRR

M M M

M M M

M M

_

_

RR

M M

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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


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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


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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)


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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


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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


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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)


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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


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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.


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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 ) ∪


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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)}


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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.


119

A B CR S

SoR

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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)}


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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


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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


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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


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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


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Homework

Ex. 5, Ex. 12, Ex. 16, Ex. 22 Ex. 25, Ex. 29, Ex. 37


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Theorem 1

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

Proof

(a)R ∞ is transitive. a R ∞ b , b R ∞ c, then a R ∞ c

(b)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

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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)}


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Example 1 Method # 2


128

0100

1010

0001

0000

RM

2 3

1111

1111( ) ( )

0001

0000

R R R RM M M M

2

1010

0101( )

0000

0000

RM

3

1010

1010( )

0000

0000

RM

4

1010

0101( )

0000

0000

RM

2 4: ( ) ( )R RNote M M

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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


129

2 ... nR R R R

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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.


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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


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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)


132

P1, P2 are subpaths with all interior vertices from {a1,a2, …, ak-1}

ak

ai aj

P1

P2

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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


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Example 2 Consider the relation R defined in Example 1. Then


134

0

0100

1010

0001

0000

RW M

1

0100

1110

0001

0000

W

2

1110

1110

0001

0000

W

3

1111

1111

0001

0000

W

pi: 2qj: 2

pi: 1, 2qj: 1, 2, 3

pi: 1, 2qj: 4

pi: 1, 2qj: ф

4 3RM W W

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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)


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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:

1. Reflexive. R, S, R U S △ ⊆ △ ⊆ △ ⊆ ⊆ (R U S) ∞

2. 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 ？ )

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


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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


137

1 2 3 4 5

R

1 2 3 4 5

S

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138

11000

11000

00110

00110

00001

RM

10000

01000

00100

00011

00011

SM

11000

11000

00110

00110

00011

R SM

3 2

1 0

W W

W W

5 4

11000

11000

00111

00111

00111

W W

(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 }∪

1 2 3 4 5

(RUS) ∞

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Homework

Ex. 1, Ex. 2, Ex. 7, Ex. 13, Ex. 18, Ex. 25


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chapter 4 relations and digraphs weiqi luo ( 骆伟祺 ) school of software sun yat-sen university...

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