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8.4 Closures of Relations 8.4 Closures of Relations

8.4 Closures of Relations - PowerPoint Presentation

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8.4 Closures of Relations - PPT Presentation

Intro Consider the following example telephone line bus route a b c d Is R defined above on the set Aa b c d transitive If not is there a possibly indirect link between each of the cities ID: 467243

relation closure path transitive closure relation transitive path find reflexive length iff closures set symmetric def exists thm theorem connectivity proof show

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Slide1

8.4 Closures of Relations

Slide2

Intro

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

a b c

dIs 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 ClosureSlide3

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 existSlide4

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 2 3Using both ordered pairs and digraphs, find the reflexive closure.Slide5

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.Slide6

More examples

Find the reflexive closures for:

R={(

a,b)|a<b} on the integers ZR={(a,b)|a ≠ b} on ZSlide7

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}

2 3Slide8

Symmetric Closures

Symmetric Closure of R = R

R-1Where R-1= {(b,a) | (a,b)  R}

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

Symmetric closure: Slide9

Transitive Theory- example

1 2

4 3

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

Keep going. (Why?)Slide10

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 x

n=b. It is denoted x1, x2,…x

n and has length n. When a=b, the path is called a circuit or cycle.

 Slide11

Find Transitive Closure- see worksheet

Do Worksheet

 

1 2  4 3   Find the transitive closure

 Find circuits and paths of length 2, 3, 4 Slide12

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, 4Slide13

Example

= Slide14

Next step

In order to come up with a theory for the transitive closure, we will first study paths….Slide15

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?Slide16

Proof of Thm

. 1

By 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

)  ____ = _______Slide17

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* =  Slide18

Examples

R

={(

a,b)| a has met b} 6 degrees Erdos numberR* include (you,__)R={(

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

)|state a and b have a common border” on the set A of states. R*=Slide19

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* ______Slide20

Proof of Thm

2

Assume (

a,b) R* and (b,c)  R*So (a,b

) ___ and (b,c)  ___

By Thm. 1, there exists paths…2 paths:

In conclusion ________Slide21

Thm 2 proof…

2)

Suppose S is a transitive relation containing R

It 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.