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Unformatted text preview: Fall Semester 2008 CS300 Algorithms Solution #7 1. (a) False. Figure 1: biconnected, not edge biconnected (b) False. Figure 2: edge biconnected, not biconnected 2. (a) False. 1 2 3 4 (a) 1 3 4 (b) 1 2 4 (c) Figure 3: Original graph G (left), Minimum-bottleneck spanning tree(middle) and Minimum spanning tree(right) (b) True; Assume that a minimum spanning tree T is NOT a minimum-bottleneck tree. Then, there is a spanning tree T with a cheaper bottleneck edge than the one e in T . Let us denote the end nodes of e be v and w , i.e., e = ( v,w ). We make a partition of the set of nodes V by A and B that is only connected by the edge e in T . For T , we also make a partition of V by the same set of nodes A and B . But since in T , no edge e is contained (otherwise, e would be the bottleneck edge of T , too) and T is a spanning tree, there must be a path P in T from v to w . Starting at v , suppose we follow the nodes of P in sequence; there is a first node w on P that is in B . Let 1 v ∈ A be the node just before w on P , and let e be the edge joining them. Note that the cost of e is less than the one of e by our assumption. We now delete e from T and insert e . We claim that this new tree T 00 is a spanning tree. Clearly T 00 is connected, since T is connected, and any path in T that used the edge e = ( v,w ) can now be “rerouted” in T 00 using edge e . To see that T 00 is also acyclic, note that the only cycle in T 00 ∪ e is not present in T 00 due to the deletion of e . We noted above that the edge e has one end in A and the other in B . But e is the cheaper with this property. So the total cost of T 00 is less than that of T which is a contradiction that...
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This note was uploaded on 02/04/2010 for the course COMPUTER S cs300 taught by Professor Unkown during the Spring '08 term at Korea Advanced Institute of Science and Technology.
- Spring '08