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Unformatted text preview: Not to be turned in. Solutions will be posted Monday July 25th. Covering sections 7.27.4. Math 239 Assignment 11 1. Show that  X  =  X  +  Y    U  where U is the set of unsaturated vertices in the XYconstruction. Solution: For two sets A,B such that A B , let B A denote the set of elements in B that are not in A . First, every x X is unsaturated and every x X X is saturated. Second, if y Y U , it is joint to some x X X by an edge in the matching. These two facts, together with the de ning property of a matching that the edges share no common ends, implies  X X  =  Y U  . Thus  X  =  X  +  X X  =  X  +  Y U  =  X  +  Y    U  . 2. Let G be a bipartite graph with bipartition A,B , and M be a matching of G . (a) Show that, for any D A ,  M   A    D  +  N ( D )  . (b) Show that the size of a maximum matching in G is equal to min D A  A    D  +  N ( D )  . Solution: (a) Since G is bipartite, each edge e i in the matching has one end a i A and one end b i B . By the property of a matching, these a i 's and b i 's are distinct. Now, given any D A , we can express  M  as a sum of the number of a i s in D , and the number of a i s in A D . The former is equal to the number of b i 's in N ( D ) , and is thus at most  N ( D )  . The latter is at most  A D  . Together:  M   N ( D )  +  A D  =  N ( D )  +  A    D  . (b) We provide two solutions here. In the rst solution, we nd D that attains the minimum if M is maximum. Apply the XYconstruction with respect to M . Then, there are no edges from X to B Y (else there will be an augmenting path contradicting maximality of M ). If we choose D = X , then N...
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This note was uploaded on 08/13/2011 for the course MATH 239 taught by Professor M.pei during the Spring '09 term at Waterloo.
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