8.4 Closures of Relations

8.4 Closures of Relations

Intro

Consider the following example (telephone line, bus route,…)

a b c

d

Is R, defined above on the set A={a, b, c, d}, transitive?If not, is there a (possibly indirect) link between each of the

cities?To answer, we want to find the Transitive Closure

Closures, in general

Def: Let R be a relation on a set A that may or may not have some property P. (Ex: Reflexive,…) If there is a relation S with property P containing R such that S is a subset of every relation with property P containing R, then S is called the closure of R with respect to P.

Note: the closure may or may not exist

Reflexive Closures- Idea, Example

Reflexive Closure of R—the smallest reflexive relation that contains R

Consider R={(1,2),(2,3),(3,2)} on A={1,2,3}

1 23

Using both ordered pairs and digraphs, find the reflexive closure.

Reflexive Closures

Reflexive Closure of R—the smallest reflexive relation that contains R

Reflexive Closure = R

Where ={(a,a)| a A} is the diagonal relation on A.

More examples

• Find the reflexive closures for:– R={(a,b)|a<b} on the integers Z

– R={(a,b)|a ≠ b} on Z

Symmetric Example

• Find the symmetric closure of R={(1,1), (1,2),(2,2),(2,3),(3,1),(3,2)} on A={1,2,3}

1 2 3

Symmetric Closures

Symmetric Closure of R = R R-1

Where R-1= {(b,a) | (a,b) R}

Example:R={(a,b)|a>b} on the integers ZSymmetric closure:

Transitive Theory- example

1 2

4 3

Add all (a,c) such that (a,b), (b,c) R.

Keep going. (Why?)

Transitive Closure Theory, and Def of Path

Def: A path from a to b in a directed graph G is a

sequence of edges (x0,x1), (x1,x2)… (xn-1, xn) in G where x0=a and xn=b. It is denoted x1, x2,…xn and has length n.

When a=b, the path is called a circuit or cycle.

Find Transitive Closure- see worksheet

Do Worksheet 1 2

4 3 Find the transitive closure Find circuits and paths of length 2, 3, 4

Example- in matrices

Using the idea that R n+1 = Rn°R and

MS°R = MR MS , Find the matrices for R R 2 R 3 R 4

The find paths of length 2, 3, 4

Example

=

Next step

In order to come up with a theory for the transitive closure, we will first study paths….

Theorem 1

Theorem 1: Let R be a relation on a set A.There is a path of length n from a to b iff

(a,b)Rn

Proof method?

Proof of Thm. 1By induction:N=1: true by definition (path from a to b of l=1 iff (a,b) R).

Induction step: Assume: There exists a path of length __ from ___iff ______Show: There exists a path of length __ from ___iff ______Assuming the IH (Inductive Hypothesis), There is a path of length __ from ___ Iff There exists an element c with a path from a to c in R and a

path of length n from c to b in ___ Iff There exists an element c with (a,c) ___ and (c,b) ___Iff (a,b) ____ = _______

Def 2: Connectivity relation

Def. 2: Let R be a relation on set A.The connectivity relation R* consists of the pairs

(a,b) such that there is a path between a and b in R.

R* =

Examples• R={(a,b)| a has met b}– 6 degrees – Erdos number– R* include (you,__)

• R={(a,b)| it is possible to travel from stop a to b directly} on set A of all subway stops– R*=

• R={(a,b)|state a and b have a common border” on the set A of states. – R*=

Thm. 2: Transitive closure is the connectivity relation

Theorem 2: The transitive closure of a relation R equals the connectivity relation

R* =

Elements of the Proof:Note that R R*To show R* is the transitive closure of R, show:1) R* is ________2) Whenever S is a transitive relation that contains R, then

R* ______

Proof of Thm 21) Assume (a,b) R* and (b,c) R*So (a,b) ___ and (b,c) ___By Thm. 1, there exists paths…

2 paths:

In conclusion ________

Thm 2 proof…

2) Suppose S is a transitive relation containing RIt can be shown by induction that Sn is transitive.By a previous theorem in sec. 8.1, S n ___ S.Since S* = S k and S k __ S , the S* ___ S.Since R ___S, the R* ____ S*.Therefore R* ___ S* ___ S.