# sol6 - 2 = 6 pred 3 = 2 pred 4 = 3 pred 5 = 4 pred 6 = 1...

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AMS 301.2 (Spring, 2011) Estie Arkin Homework Set # 6 – Solution Notes A: (a) Kruskal’s algorithms would insert the edges in the following order: (E,F) (A,B) (C,D) (C,E) (C,B). Cost of the MST 2 + 5 + 4 + 1 + 3 = 15 (b). Prim’s algorithm would insert the edges in the following order: (A,B), (B,C), (C,E), (C,D), (E,F). Cost of the MST 2 + 5 + 4 + 1 + 3 = 15 (c). Edge (A,B) is currently part of the minimum spanning tree. Suppose its cost is increased from 2 to 11. Will it still be part of the minimum spanning tree? Explain. No. If (A,B) remains in the tree, its cost is now 11 + 5 + 4 + 1 + 3 = 24, but removing (A,B) and including (A,D) yields a tree of smaller cost, 9 + 5 + 4 + 1 + 3 = 22. B: The edges in the shortest path tree are: (A,B) (B,C) (A,D) (C,E) (A,F). Here is the entire execution of the algorithm: u A u B u C u D u E u F node becoming perm 0 2 9 12 node B - - 7 9 12 12 node C - - - 9 11 12 node D - - - - 11 12 node E - - - - - 12 node F The final pred labels are: pred
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Unformatted text preview: 2 = 6, pred 3 = 2, pred 4 = 3, pred 5 = 4, pred 6 = 1, pred 7 = 6. C: Quick TSP: T 1 = 1 T 2 = 1 , 4 , 1 T 3 = 1 , 3 , 4 , 1 T 4 = 1 , 2 , 3 , 4 , 1 T 5 = 1 , 2 , 3 , 5 , 4 , 1 T 6 = 1 , 2 , 3 , 6 , 5 , 4 , 1, at cost 4 + 2 + 6 + 2 + 2 + 2 = 18 Using the double the MST method we get: The minimum spanning tree contains edges (3,4) (2,3) (1,4) (4,5) (5,6). Doubling it, we get an Euler cycle 1-4-3-2-3-4-5-6-5-4-1. Shortcutting this by eliminating repeated nodes, we get the nal tour: 1-4-3-2-5-6-1 of length 2+1+2+5+2+4 = 16. Note that this solution is dierent, neither method gaurantees that we nd an optimal tour! D: Let each red edge have a cost 0 and each blue edge have cost 1. Now nd a mimimum cost spanning tree by any of the algorithms from class. It will contain as few blue edges as possible, and therefore as many red edges as possible....
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