How can I find the number of Hamiltonian cycles in a complete undirected graph?

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再見小時候
再見小時候 2021-02-05 22:03

Can someone explain how to find the number of Hamiltonian cycles in a complete undirected graph?

Wikipedia says that the formula is (n-1)!/2, but when I cal

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  •  借酒劲吻你
    2021-02-05 23:03

    Since the graph is complete, any permutation starting with a fixed vertex gives an (almost) unique cycle (the last vertex in the permutation will have an edge back to the first, fixed vertex. Except for one thing: if you visit the vertices in the cycle in reverse order, then that's really the same cycle (because of this, the number is half of what permutations of (n-1) vertices would give you).

    e.g. for vertices 1,2,3, fix "1" and you have:

    123 132

    but 123 reversed (321) is a rotation of (132), because 32 is 23 reversed.

    There are (n-1)! permutations of the non-fixed vertices, and half of those are the reverse of another, so there are (n-1)!/2 distinct Hamiltonian cycles in the complete graph of n vertices.

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