1. 8.4 Closures of Relations

2. Intro Consider the following example (telephone line, bus route,…) abc 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

3. 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

4. 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} 12 3 Using both ordered pairs and digraphs, find the reflexive closure.

5. 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.

6. More examples Find the reflexive closures for: – R={(a,b)|a<b} on the integers Z – R={(a,b)|a ≠ b} on Z

7. 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} 12 3

8. 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 Z Symmetric closure:

9. Transitive Theory- example 12 43 Add all (a,c) such that (a,b), (b,c)  R. Keep going. (Why?)

10. Transitive Closure Theory, and Def of Path Def: A path from a to b in a directed graph G is a sequence of edges (x 0,x 1 ), (x 1,x 2 )… (x n-1, x n ) in G where x 0 =a and x n =b. It is denoted x 1, x 2,…x n and has length n. When a=b, the path is called a circuit or cycle.

11. Find Transitive Closure- see worksheet Do Worksheet 12 43 Find the transitive closure Find circuits and paths of length 2, 3, 4

12. Example- in matrices Using the idea that R n+1 = R n °R and M S°R = M R   M S, Find the matrices for RR 2 R 3 R 4 The find paths of length 2, 3, 4

13. Example =

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

15. 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)  R n Proof method?

16. 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 ___ IffThere exists an element c with (a,c)  ___ and (c,b)  ___ Iff (a,b)  ____ = _______

17. 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* =

18. 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*=

19. 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* ______

20. Proof of Thm 2 1)Assume (a,b)  R* and (b,c)  R* So (a,b)  ___ and (b,c)  ___ By Thm. 1, there exists paths… 2 paths: In conclusion ________

21. Thm 2 proof… 2) Suppose S is a transitive relation containing R It can be shown by induction that S n 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.