This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
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....
View Full Document
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