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

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

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