Graph

Graph - 1) Summer 2008 Q7 Prove or give a counterexample:...

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1) Summer 2008 Q7 Prove or give a counterexample: Let G be an undirected, connected, bipartite, weighted graph. If the weight of each edge in G is +1, and for every pair of vertices (u,v) in G there is exactly one shortest path, then G is a tree. If G is not a tree, we show that there must be some pair of vertices (u, v) in G for whom there are at least two shortest paths. Since G is not a tree, it must have some circle. Since G is a bipartite graph, that circle must have even number of nodes. Assume there are 2n nodes in that circle as shown in the below graph, there are two paths from node 1 to node n+1 with the same length n. If the path from node 1 to node n+1 in that circle is the shortest path, then we have two shortest paths, which is a contradiction. If not, there must be another path from node 1 to node n+1 whose length is shorter than n. Thus we can find a circle with 2m nodes and m<n. We can check in this new circle whether the two paths between the node 1 and node m+1 are the shortest paths, which either returns a contradiction or a new circle of smaller size. We keep doing this.
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This note was uploaded on 01/24/2011 for the course CS 570 at USC.

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Graph - 1) Summer 2008 Q7 Prove or give a counterexample:...

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