h2_sol - CSE 202 Homework 2 Solution TA: Nan Zang Problem 1...

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CSE 202 Homework 2 Solution TA: Nan Zang Problem 1 Suppose you are given a connect graph G, with edge costs that are all distinct. Prove that G has a unique MST. Proof: Suppose the graph has two different MSTs, T and T1. Then there must be some edge (u, v) which is present in T but not in T1. The graph obtained by removing the edge (u, v) from T consists of two connected components, which define a cut on G. There must be some edge (u1, v1) across this cut in T1. Also, one of these two edges must be lighter than the other, since all edge weights are distinct. Without loss of generality, assume that w(u, v) < w(u1, v1). But then, removing (u1, v1) from T1 and adding (u, v) gives a tree of lower cost than T1. However, this is a contradiction with that T1 is an MST. Hence, the MST of the graph must be unique. Problem 2 Design a spanning network for which the most expensive edge is as cheap as possible. A spanning tree T of G is a minimum-bottleneck spanning tree if there is no spanning tree T’ of G with a cheaper bottleneck edge. (a) Is every minimum bottleneck tree of G a minimum spanning tree of G. No. Every spanning tree of the following graph is a MBT, but only one of them is a MST. (b) Is every minimum spanning tree of G a minimum bottleneck tree of G. Yes, Proof by Contradiction. Suppose T is a MST but not a MBT. T’ is a MBT. The bottleneck edge in T is e. the weight of e is larger than any edge in T’, since T’ is a MBT. Removing edge e from T, T will have two connected components S and M. Since T’ is a spanning tree, in T’, there must be an edge connecting these two components S and M, denoting this edge e’. So, T-{e} +{e’} is a new spanning tree with smaller weight than T. This is a contradiction with the assumption that T is a MST.
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This note was uploaded on 05/21/2008 for the course CS CO453 taught by Professor Graham during the Winter '08 term at UCSD.

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h2_sol - CSE 202 Homework 2 Solution TA: Nan Zang Problem 1...

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