# 55PS11 - SOLUTION SET 11—DUE Problem 1#8.4.2 Let R be the...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: SOLUTION SET 11—DUE 4/30/2008 Problem 1 (#8.4.2) . Let R be the relation { ( a,b ) | a 6 = b } on the set of integers. What is the reflexive closure of R ? Solution. Since R already includes all ( a,b ) for a 6 = b , and we now must add the ( a,a ), the reflexive closure must include every ordered pair in Z × Z . Problem 2 (#8.4.4) . How can the directed graph representing the reflexive closure of a relation on a finite set be constructed from the directed graph of the relation? Solution. The reflexive closure of a relation is just the relation together with any of the elements ( a,a ) that were not already in the relation. When represented as a directed graph, the ( a,a ) are the self-loops, so the reflexive closure of a relation is represented by adding all possible self-loops to the graph. Problem 3 (#8.4.8) . How can the directed graph representing the symmetric clo- sure of a relation on a finite set be constructed from the directed graph for this relation? Solution. The symmetric closure of a relation is formed by adding all ( b,a ) for any ( a,b ) in the relation. In directed graph terms, this is equivalent to adding the reverse of every edge to the graph, whenever it isn’t already in the graph. Problem 4 (#8.4.14) . Show that the closure of a relation R with respect to a property P , if it exists, is the intersection of all the relations with property P that contain R . Proof. We may assume that property P is preserved under intersection. (This may be proved if we define P formally; for now it suffices to note that all the properties that we have seen thus far are preserved under intersection.) Let C P ( R ) be the P-closure of R . Let the R α be the supersets of R with property P , where α ∈ I for some indexing set I . Since by definition C P ( R ) ⊂ R α for all α ∈ I , we have C P ( R ) ⊂ \ α ∈ I R α . But since C P ( R ) is itself a superset of R with property P , it must be one of the R α . Since an intersection is a subset of all the intersected sets, \ α ∈ I R α ⊂ C P ( R ) . Thus ∩ α ∈ I R α = C P ( R ), as desired. Problem 5 (#8.4.16) . Determine whether these sequences of vertices are paths in this directed graph. 1 2 SOLUTION SET 11—DUE 4/30/2008 a b c d e a) a,b,c,e b) b,e,c,b,e c) a,a,b,e,d,e d) b,c,e,d,a,a,b e) b,c,c,b,e,d,e,d f) a,a,b,b,c,c,b,e,d Solution. The vertices in (b) don’t form a path because there’s no edge from e to c on this directed graph. Similarly, path (f) tried to move from b to b , but there’s no self-loop at that vertex. The remaining vertex sequences are paths. Problem 6 (#8.4.20) . Let R be the relation that contains the pair ( a,b ) if a and b are cities such that there is a direct non-stop airline flight from a to b . When is ( a,b ) in a) R 2 ?...
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

55PS11 - SOLUTION SET 11—DUE Problem 1#8.4.2 Let R be the...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online