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Dec 1, 2013 at 12:58 vote accept Mohammad Al-Turkistany
Dec 1, 2013 at 12:12 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Nov 30, 2013 at 16:02 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Nov 30, 2013 at 8:56 answer added nvcleemp timeline score: 4
Nov 30, 2013 at 8:00 comment added nvcleemp @hbm: this just proves the upper bound which was given by the OP, since the hamming distance is $N - |E(H_1)\cap E(H_2)|$. It is easy to construct an arbitrarily large cubic graph in which the smallest hamming distance between two hamiltonian cycles is 2.
Nov 29, 2013 at 13:58 comment added nvcleemp I would say that the lower bound is 2, but maybe you want it expressed in terms of some invariants?
Nov 29, 2013 at 12:26 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Nov 29, 2013 at 12:20 history edited Mohammad Al-Turkistany CC BY-SA 3.0
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Nov 29, 2013 at 12:15 history asked Mohammad Al-Turkistany CC BY-SA 3.0