问题
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 calculated using this formula, K3 has only one cycle and K4 has 5. Was my calculation incorrect?
回答1:
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.
回答2:
In answer to your Google Code Jam comment, see this SO question
回答3:
I think when we have a Hamiltonian cycle since each vertex lies in the Hamiltonian cycle if we consider one vertex as starting and ending cycle . we should use 2 edges of this vertex.So we have (n-1)(n-2)/2 Hamiltonian cycle because we should select 2 edges of n-1 edges which linked to this vertex.
来源：https://stackoverflow.com/questions/1387523/how-can-i-find-the-number-of-hamiltonian-cycles-in-a-complete-undirected-graph