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CS 173:

Discrete Mathematical Structures

Transitive Closure

• Let c(R ) denote the transitive

closure of relation R.

Then c(R ) = R U {

(a,c): ∃b (a,b),(b,c) ∈ R

Example: A={1,2,3,4}, R={(1,2),(2,3),(3,4)}.

Apply definition to get:

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

Which of the following is true: a) This set is transitive, but we added too much. b) This set is the transitive closure of R. c) This set is not transitive, one pair is missing. d) This set is not transitive, more than 1 pair ismissing.

Transitive Closure

Formally:

If t(R) is the transitive closure of R, and if

R contains a path from a to b, then (a,b)

∈ t(R)

Notes:

– Later classes will give you efficient

algorithms for determining if there is a

path between two vertices in a graph

(graph connectivity problem)

– Read about Warshall’s algorithm in the

text.

Equivalence Relations

Example:

Let S = {people in this classroom},

and let

R = {(a,b): a has same # of coins in his/her bag as b}

Quiz time:

Is R reflexive?

Is R symmetric?

Is R transitive?

Yes

Yes

Yes

This is a special

kind of relation,

characterized by

the properties it

has.

What’s special

about it?

Everyone with the same

of coins as

you is just like you.

Equivalence Relations

Example:

S = Z (integers), R = {(a,b) : a ≡ b mod

Is this relation an equivalence

relation on S?

Let a be any integer. aRa since a = 4⋅0 + a.

Consider aRb. Then a = 4k + b. But b = -4k + a.

Consider aRb and bRc. Then a = 4k + b and b = 4j + c.

So, a = 4k + 4j +c = 4(k+j) + c.

Start by thinking of R a different way: aRb iff

there is an int k so that a = 4k + b. Your quest

becomes one of finding ks.

Equivalence Relations

Example:

S = people in this room,

R = {(a,b) : total $ on a is within $1.00 of total $ on b}

Is this relation an equivalence

relation on S?

Clearly reflexive and symmetric. Is it transitive?

a) Yes, I can give a proof. b) Yes, I think so, but I can’t prove it. c) No, I can give a proof. d) No, I don’t think so, but I can’t prove it.

Equivalence Classes

What equivalence relation we’ve seen

recently has representatives

[244], [7], [58], [1]?

Equivalence Classes

Definition: Let R be an equivalence

relation on S. The equivalence class

of a ∈ S, [a] R , is

[a] R = {b: aRb}

What does the set of equivalence classes on S look

like?

Notice this is just a subset of S.

To answer, think about the relation from before:

S = {people in this room}

R = {(a,b) : a has the same # of coins in his/her bag as b}

In how many different equivalence classes can each

person fall?

Equivalence Classes

Lemma: Let R be an equivalence

relation on S. Then

1. If aRb, then [a] R = [b] R

2. If not aRb, then [a] R ∩ [b] R = ∅

Proof:

2. Suppose to the contrary that ∃ x ∈ [a] (^) R ∩ [b] (^) R.

x ∈ [a]R x ∈ [b]R ⇔ aRx and bRx

⇔ aRb, contradicting “not aRb”

⇔ aRx and xRb

Thus, [a]R and [b] (^) R are either identical or disjoint.

S = [ a ] R

a ∈ S

Equivalence Classes

So S is the union of disjoint

equivalence classes of R.

A partition of a set S is a (perhaps infinite…or

uncountably infinite) collection of sets {Ai} with

Each Ai non-empty Each Ai ⊆ S For all i, j, A (^) i ∩ Aj = ∅ S = ∪A (^) i Each Ai is called a block of the partition.

Equivalence Classes

Theorem: if R is a _____ S, then {[a] (^) R : a ∈ S} is a _____ S.

A. Partition of, equivalence relation on B. Subset of, equivalence class of C. Relation on, partition of D. Equivalence relation on, partition of E. I have no clue.

Theorem: if R is an equivalence relation on S, then {[a] (^) R : a ∈ S} is a partition of S.

Proof: we need to show that an equivalence relation R satisfies the definition of a partition. (we’ve spent the whole day doing this!)

Equivalence Classes

Theorem: if {Ai} is any partition of S, then there exists an

equivalence relation R, whose equivalence classes are exactly the blocks Ai.

Proof If {Ai} partitions S then define relation R on S to be R = {(a,b) : ∃ i, a ∈ Ai and b ∈ Ai} Next show that R is an equivalence relation. Reflexive and symmetric. Transitive? Suppose aRb and bRc. Then a and b are in A (^) i, and b and c are in A (^) j. But b ∈ Ai ∩ A (^) j , so A (^) i = Aj. So, a, b, c ∈ Ai, thus aRc.

Partially Ordered Sets (POSets)

Ex. (Z +, | ), the relation “divides” on positive integers.

Reflexive? Yes, x|x since x=1x (k=1)

Transitive?

Antisymmetric?

a|b means b=ak, b|c means c=bj. Does c=am for some m?

c = bj = akj (m=kj)

a|b means b=ak, b|a means a=bj. But b = bjk (subst) only if jk=1.

jk=1 means j=k=1, and we have b=a1, or b=a

Yes, or No?

Partially Ordered Sets (POSets)

Ex. (Z, | ), the relation “divides” on integers.

Reflexive? Yes, x|x since x=1x (k=1)

Transitive?

Antisymmetric?

a|b means b=ak, b|c means c=bj. Does c=am for some m?

c = bj = akj (m=kj)

3|-3, and -3|3, but 3 ≠ -3.

Not a poset.

Yes, or No?