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Discrete Mathematics
Chapter 7
Relations
感謝 大葉大學 資訊工程系 黃鈴玲老師 提供
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7.1 Relations and their properties.※The most direct way to express a
relationship between elements of two sets
is to use ordered pairs.
For this reason, sets of ordered pairs are
called binary relations.
Example 1.A : the set of students in your school.B : the set of courses.R = { (a, b) : aA, bB, a is enrolled in course b }
7.1.1
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Def 1. Let A and B be sets. A binary relation
from A to B is a subset R of AB.
( Note AB = { (a,b) : aA and bB } )
Def 1’. We use the notation aRb to denote
that (a, b)R, and aRb to denote that
(a,b)R.
Moreover, a is said to be related to b
by R if aRb.
7.1.2
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Example 3. Let A={0, 1, 2} and B={a, b},
then {(0,a),(0,b),(1,a),(2,b)} is a relation R
from A to B. This means, for instance, that
0Ra, but that 1Rb
0
1
2
a
b
A B
R
R AB = { (0,a) , (0,b) , (1,a) (1,b) , (2,a) , (2,b)}
R R
7.1.3
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Example ( 上例 ) : A : 男生 , B : 女生 , R : 夫妻關系 A : 城市 , B : 州 , 省 R : 屬於 (Example
2)
Note. Relations vs. Functions
A relation can be used to express a 1-to-many
relationship between the elements of the sets
A and B.
( function 不可一對多,只可多對一 )
Def 2. A relation on the set A is a subset of A A ( i.e., a relation from A to A ).
7.1.4
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Example 4. Let A be the set {1, 2, 3, 4}. Which ordered pairs are in the relation R = { (a, b)| a divides b }?
Sol :
1
2
3
4
12
3
4
R = { (1,1), (1,2), (1,3), (1,4), (2,2), (2,4), (3,3), (4,4) }
7.1.5
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Example 5. Consider the following relations on Z.R1 = { (a, b) | a b }
R2 = { (a, b) | a > b }
R3 = { (a, b) | a = b or a = b }
R4 = { (a, b) | a = b }
R5 = { (a, b) | a = b+1 }
R6 = { (a, b) | a + b 3 }
Which of these relationscontain each of the pairs(1,1), (1,2), (2,1), (1,1),and (2,2)?
Sol : (1,1) (1,2) (2,1) (1,1) (2,2)
R1
R2
R3
R4
R5
R6
● ● ●
● ●
● ● ●
● ● ● ●
●
● ●
7.1.6
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Def 3. A relation R on a set A is called reflexive
if (a,a)R for every aA.Example 7. Consider the following relations on
{1, 2, 3, 4} : R2 = { (1,1), (1,2), (2,1) }
R3 = { (1,1), (1,2), (1,4), (2,1), (2,2), (3,3), (4,1), (4,4) }
R4 = { (2,1), (3,1), (3,2), (4,1), (4,2), (4,3) }
which of them are reflexive ?
反身性
Sol : R3
7.1.7
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Example 8. Which of the relations from
Example 5 are reflexive ?
Sol :
R1, R3 and R4 7.1.8
R1 = { (a, b) | a b }
R2 = { (a, b) | a > b }
R3 = { (a, b) | a = b or a = b }
R4 = { (a, b) | a = b }
R5 = { (a, b) | a = b+1 }
R6 = { (a, b) | a + b 3 }
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Def 4.
(1) A relation R on a set A is called symmetric
if for a, bA, (a, b)R (b, a)R.
(2) A relation R on a set A is called antisymmetric ( 反對稱 ) if for a, bA,
(a, b)R and (b, a)R a = b.
即若 a≠b 且 (a,b)R (b, a)R
7.1.9
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Example 10. Which of the relations from Example 7 are symmetric or antisymmetric ? R2 = { (1,1), (1,2), (2,1) }
R3 = { (1,1), (1,2), (1,4), (2,1), (2,2), (3,3), (4,1), (4,4) }
R4 = { (2,1), (3,1), (3,2), (4,1), (4,2), (4,3) }
Sol : R2, R3 are symmetricR4 are antisymmetric.
7.1.10
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Def 5. A relation R on a set A is called transitive( 遞移 ) if for a, b, c A, (a, b)R and (b, c)R (a, c)R.
7.1.11
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Example 13. Which of the relations in Example 7 are transitive ? R2 = { (1,1), (1,2), (2,1) }
R3 = { (1,1), (1,2), (1,4), (2,1), (2,2), (3,3), (4,1), (4,4) }
R4 = { (2,1), (3,1), (3,2), (4,1), (4,2), (4,3) }
Sol : R2 is not transitive since (2,1) R2 and (1,2) R2 but (2,2) R2. R3 is not transitive since (2,1) R3 and (1,4) R3 but (2,4) R3. R4 is transitive.
7.1.12
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Example 17. Let A = {1, 2, 3} and B = {1, 2, 3, 4}.
The relation R1 = {(1,1), (2,2), (3,3)}
and R2 = {(1,1), (1,2), (1,3), (1,4)} can be
combined to obtain
R1 ∪ R2
R1 ∩ R2 = {(1,1)}
R1 - R2 = {(2,2), (3,3)}
R2 - R1 = {(1,2), (1,3), (1,4)}
R1 R2 = {(2,2), (3,3), (1,2), (1,3), (1,4)}
Exercise : 1, 7, 437.1.13
symmetric difference, 即 (A B) – (A B)
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補充 :
antisymmetric 跟 symmetric 可並存
(a, b)R, a≠b
sym. (b, a)R
antisym. (b,a)R
故若 R 中沒有 (a, b) with a≠b 即可同時滿足
eg. Let A = {1,2,3}, give a relation R on A s.t. R is both symmetric and antisymmetric, but not reflexive.Sol :
R = { (1,1),(2,2) } 7.1.14
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7.3 Representing Relations by matrices and digraphs
Suppose that R is a relation from A={a1, a2, …, am}
to B = {b1, b2,…, bn }.
The relation R can be represented by the matrix
MR = [mij], where
mij =1, if (ai,bj)R
0, if (ai,bj)R
7.3.1
※ Matrices
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Example 1. Suppose that A = {1,2,3} and B = {1,2} Let R = {(a, b) | a > b, aA, bB}. What is MR ?Sol :
0 0
1
1 1
0
11
01
00
RM
7.3.2
1 2
1
2
3
A
B
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※ The relation R is symmetric iff (ai,aj)R (aj,ai)R. This means mij = mji ( 即 MR 是對稱矩陣 ).
1
1
1
RM
tRR MM )(
0
01
1
※ Let A={a1, a2, …,an}. A relation R on A is reflexive iff (ai,ai)R,i.
i.e., a1
…a2 …a1 an
:an
:
a2
對角線上全為 1
7.3.3
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※ The relation R is antisymmetric iff (ai,aj)R and i j (aj,ai)R.
This means that if mij=1 with i≠j, then mji=0.
i.e.,
0
01
00
1
RM
7.3.4
※ transitive 性質不易從矩陣判斷出來
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Example 3. Suppose that the relation R on a set is represented by the matrix
110
111
011
RM
Is R reflexive, symmetric, and / or antisymmetric ?
Sol :reflexive, symmetric, not antisymmetric.
7.3.5
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eg. Suppose that S={0,1,2,3} Let R be a relation
containing (a, b) if a b, where a S and b S. Is R reflexive, symmetric, antisymmetric ?
Sol :
1000
1100
1110
1111
RM
00
1 2 3
1
2
3
∴ R is reflexive and antisymmetric, not symmetric.
7.3.6
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※Representing Relations using Digraphs. (directed graphs)
Example 8. Show the directed graph (digraph)
of the relation R={(1,1),(1,3),(2,1),(2,3),(2,4),
(3,1),(3,2),(4,1)} on the set {1,2,3,4}.
Sol :
vertex( 點 ) : 1, 2, 3, 4edge( 邊 ) : 11, 13, 21 23, 24, 31 32, 41
1 2
4 37.3.7
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※ The relation R is reflexive iff for every vertex,
( 每個點上都有 loop)
※ The relation R is symmetric iff
兩點間若有邊,必只有一條邊
7.3.8
※ The relation R is antisymmetric iff
x y
x y
( 兩點間若有邊,必為一對不同方向的邊 )
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※ The relation R is transitive iff for a, b, c A, (a, b)R and (b, c)R (a, c)R.
This means:
a b
d c
a b
d c
( 只要點 x 有路徑走到點 y , x 必定有邊直接連向 y)
7.3.9
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Example 10. Determine whether the relations R and S are reflexive, symmetric, antisymmetric, and / or transitive
Sol :
R : a
b c
Exercise : 1,13,26, 27,31
irreflexive( 非反身性 ) 的定義在 p.480即 (a,a)R, aA
7.3.10
a b
c d
Sreflexive,not symmetric,not antisymmetric,not transitive (a→b, b→c, a→c)
not reflexive,symmetricnot antisymmetricnot transitive(b→a, a→c, b→c)
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7.4 Closures of Relations
The relation R={(1,1), (1,2), (2,1), (3,2)} on the set A={1, 2, 3} is not reflexive.
Q: How to construct a smallest reflexive relation Rr such that R Rr ?
7.4.1
※ Closures
Sol: Let Rr = R {(2,2), (3,3)}.
i. e., Rr = R {(a, a)| a A}.
Rr is a reflexive relation containing R that is as small as possible. It is called the reflexive closure of R.
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Example 1. What is the reflexive closure of the relation R={(a,b) | a < b} on the set of integers ?
Sol : Rr = R ∪ { (a, a) | aZ }
= { (a, b) | a b, a, bZ }Example :
The relation R={ (1,1),(1,2),(2,2),(2,3),(3,1),(3,2) }on the set A={1,2,3} is not symmetric.
Sol : Let R1={ (a, b) | (a, b)R } Let Rs= R∪R1={ (1,1),(1,2),(2,1),(2,2),(2,3), (3,1),(1,3),(3,2) }Rs is the smallest symmetric relation containing R, called the symmetric closure of R. 7.4.2
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Def : 1.(reflexive closure of R on A)
Rr=the smallest set containing R and is reflexive.
Rr=R {∪ (a, a) | aA , (a, a)R}
2.(symmetric closure of R on A)
Rs=the smallest set containing R and is symmetric
Rs=R {∪ (b, a) | (a, b)R & (b, a) R}
3.(transitive closure of R on A)
Rt=the smallest set containing R and is transitive.
Rt=R {∪ (a, c) | (a, b)R & (b, c)R, but (a, c) R}Note. 沒有 antisymmetric closure ,因若不是 antisymmetric , 表示有 a≠b, 且 (a, b) 及 (b, a) 都 R ,此時加任何 pair 都不可能變成 antisymmetric. 7.4.3
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Example 2. What is the symmetric closure of the relation R={(a, b) | a > b } on the set of positive integers ?
Sol :
R { (∪ b, a) | a > b }={ (a,b) | a b }
{ (a, b) | a < b }
||
7.4.4
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Example. Let R be a relation on a set A, where A={1,2,3,4,5}, R={(1,2),(2,3),(3,4),(4,5)}. What is the transitive closure Rt of R ?Sol :
1
2
3
4
5
∴Rt = (1,2),(2,3),(3,4),(4,5) (1,3),(1,4),(1,5) (2,4),(2,5) (3,5)
Exercise : 1,9( 改為找三種 closure)7.4.5
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7.5 Equivalence Relations ( 等價關係 )Def 1. A relation R on a set A is called an
equivalence relation if it is reflexive,
symmetric, and transitive.
Example 1. Let L(x) denote the length of the string x. Suppose that the relation R={(a,b) | L(a)=L(b), a,b are strings of English letters } Is R an equivalence relation ?Sol :
(a,a)R string a reflexive (a,b)R (b,a)R symmetric (a,b)R,(b,c)R (a,c)R transitive
Yes.
7.5.1
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Example 4. (Congruence Modulo m)Let m Z and m > 1. Show that the relation
R={ (a,b) | a≡b (mod m) } is an equivalence relation on the set of integers.
Note that a≡b(mod m) iff m | (ab).∵ a≡a (mod m) (a, a)R reflexive If a≡b(mod m), then ab=km, kZ ba≡(k)m b≡a (mod m) symmetric If a≡b(mod m), b≡c(mod m) then ab=km, bc=lm ac=(k+l)m a≡c(mod m) transitive
∴ R is an equivalence relation.
Sol :
7.5.2
( a is congruent to b modulo m )
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Def 2.
Let R be an equivalence relation on a set A.
The equivalence class of the element aA is
[a]R = { s | (a, s)R }
For any b[a]R , b is called a representative of this equivalence class.
Note:If (a, b)R, then [a]R=[b]R.
7.5.3
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Example 6.
What are the equivalence class of 0 and 1
for congruence modulo 4 ?
Sol :
Let R={ (a,b) | a≡b (mod 4) }
Then [0]R = { s | (0,s)R }
= { …, 8, 4, 0, 4, 8, … }
[1]R = { t | (1,t)R } = { …,7, 3, 1, 5, 9,…}
7.5.4
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Def.
A partition ( 分割 ) of a set S is a collection of disjoint nonempty subsets Ai of S that have S as their union.
In other words, we have
Ai ≠ , i,
Ai∩Aj = , when i≠j
and
∪Ai = S.
7.5.6
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Example 7.
Suppose that S={ 1,2,3,4,5,6 }. The collection
of sets A1={1,2,3}, A2={ 4,5 }, and A3={ 6 } form a partition of S.
Thm 2. Let R be an equivalence relation on a set A.
Then the equivalence classes of R form a partition of A.
7.5.7
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Example 9.
The congruence modulo 4 form a partition of the integers.
Sol :
[0]4 = { …, 8, 4, 0, 4, 8, … }
[1]4 = { …, 7, 3, 1, 5, 9, … }
[2]4 = { …, 6, 2, 2, 6, 10, … }
[3]4 = { …, 5, 1, 3, 7, 11, … }
Exercise : 3,17,19,237.5.8