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 2 3 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} • 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 Z Symmetric 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 R2 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. 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) ____ = _______
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 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 ________
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.
