t04s_y2008 - The University of Sydney MATH2969/2069 Graph...

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The University of Sydney MATH2969/2069 Graph Theory Tutorial 4 (Week 11) Solutions 2008 1. ( i ) How many Hamiltonian cycles are there in this graph? X 7 6 7 9 9 5 6 9 7 8 ( ii ) Delete the vertex labelled X (and its incident edges). How many spanning trees are there on the remaining subgraph? How many of these spanning trees are part of a Hamiltonian cycle? ( iii ) Find a minimum weight spanning tree on the remaining subgraph. Hence show that the weight (or length) of a solution to the travelling salesman problem is at least 32. Solution. ( i ) The graph is K 5 , and so a Hamiltonian cycle is formed by starting at any vertex, taking the remaining 4 vertices in any order, and returning to the ±rst vertex. Note that which vertex is considered as the initial (and ±nal) vertex is not signi±cant. For example, v 1 v 2 v 3 v 4 v 5 v 1 and v 2 v 3 v 4 v 5 v 1 v 2 are the same cycles. So, if the vertices are v 1 , . . . , v 5 , we may without loss of generality assume v 1 is the start (and end). There are then 4! arrangements of the other 4 vertices. But each cycle has been counted twice in these 4! arrangements – once forwards, and once backwards. Therefore the number of di²erent cycles is 4! 2 = 12. ( ii ) The remaining subgraph is K 4 . The number of spanning trees is equal to the number of labelled trees on 4 vertices, which is 4 2 = 16, by Cayley’s Theorem. There are 4 spanning trees on K 4 with degree sequence (1 , 1 , 1 , 3), and such trees cannot be part of a Hamiltonian cycle. The remaining 12 spanning trees have degree sequence (1 , 1 , 2 , 2) and are each part of some Hamiltonian cycle. ( iii
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t04s_y2008 - The University of Sydney MATH2969/2069 Graph...

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