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Discrete Mathematics with Graph Theory (3rd Edition) 342

# Discrete Mathematics with Graph Theory (3rd Edition) 342 -...

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340 Solutions to Exercises worst, have to compare the old label on each other vertex v with the weight of edge VVk (n - k comparisons) and then find the minimum of n - k numbers (another n - k - 1 comparisons). In all, this implementation could require (n - 2) + [(n - 2) + (n - 3)] + [(n - 3) + (n - 4)] + ... + [2 + 1] + [1 + 0] = (n - 2) + (2n - 5) + (2n - 7) + ... + 3 + 1 = (n - 2) + L~:g(2k + 1) = (n - 2) + (n - 2)2 comparisons. This function is O(n 2 ) as required. 14. The five subgraphs shown are those obtained by successively removing the vertices A, B, C, D, E. In each case, a minimum spanning tree is shown. A B~9 E 5 2 4 8 C 7 D i&t: 3 2 E C 7 D8 o A \$ 3 1 B 9 E 4 2 C @ Vertex removed Weight of minimum tree A 11 B 10
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Unformatted text preview: D 7 E 12 ® A B~:jy ~D @ Minimum sum of weights of two incident edges 4 9 5 12 3 Sum 15 19 16 19 15 As the table shows, any Hamiltonian cycle must have weight at least 19. The Hamiltonian cycle AECBDA has weight 1 + 2 + 4 + 5 + 7 = 19 and hence is a minimum. 15. (a) [BB] The graph is complete, so an obvious approach is to try to choose the lowest weight available edge at each vertex. One such cycle is ADEFBCA, which has weight 2+3+2+5+ 1+5 = 18. (b) [BB] As shown on the left, the minimum weight of a spanning tree after A is removed is 11. The two edges of least weight at A have weights 2 and 3 so we obtain an estimate of 11 + 2 + 3 = 16 as a lower bound for the weight of any Hamiltonian cycle....
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