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Unformatted text preview: tree). The remaining weights can be assigned arbitrarily. b) assign the weights 1, 1, 2 and 2 to the rim, and the remaining weights arbitrarily. A minimal spanning tree will contain exactly one of the edges with weight 2, and both choices will give at least one minimal spanning tree. 13.2.3: Not necessarily. Consider for instance a cycle of length 4 with edge weights 1, 1, 1 and 2, and let u and v be the vertices joined by the edge of weight 2. The minimum spanning tree is the path of three edges of weight 1 from u to v (weight 3), the path of minimum weight between u and v consists of the edge uv (weight 2). 13.2.9: The proof of the correctness of Kruskals theorem implies that in this case, any spanning tree other than the one found by the algorithm has strictly larger weight....
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This note was uploaded on 07/15/2010 for the course MACM 201 taught by Professor Marnimishna during the Spring '09 term at Simon Fraser.
 Spring '09
 MarniMishna

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