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by MauranKilom
1509 days ago
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> Consider the complete graph on n vertices; there are roughly n-factorial-many Hamiltonian cycles. Any permutation of the vertices corresponds to one such cycle. Different permutations can correspond to the same cycle, so the number is not exactly n-factorial. Isn't it just (n-1)!/2 as each cycle can be cyclically permuted (n possibilities) and (for n>=3) also be reversed? (Cases n=2 and n=1 have only one cycle.) (That wasn't the point you were trying to get to, I know. I'm just thinking about the precise number.) |
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