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Dec 1, 2013 at 12:58 vote accept Mohammad Al-Turkistany
Dec 1, 2013 at 12:55 history edited nvcleemp CC BY-SA 3.0
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Dec 1, 2013 at 12:51 comment added nvcleemp No problem, I have added a drawing.
Dec 1, 2013 at 12:41 history edited nvcleemp CC BY-SA 3.0
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Dec 1, 2013 at 12:16 comment added Mohammad Al-Turkistany I guess I am missing something obvious. Can you illustrate you answer using a drawing?
Dec 1, 2013 at 7:40 comment added nvcleemp Well, that minimum is 2. Take a $K_4$ and repeat the operation described to the same edge and the minimum hamming distance remains equal to 2.
Dec 1, 2013 at 4:12 comment added Mohammad Al-Turkistany Yes. I am looking for asymptotic tight minimum.
Nov 30, 2013 at 18:46 comment added nvcleemp Have you read the construction? Give me any even number greater than 2 and I can construct a cubic graph with that number of vertices and minimum hamming distance 2.
Nov 30, 2013 at 15:58 comment added Mohammad Al-Turkistany I conjecture that it is at least $\Omega(\log N)$ and I even conjecture a tighter lower bound of $\Omega( \epsilon N)$ for some constant $\epsilon \gt 0$.
Nov 30, 2013 at 12:20 comment added nvcleemp There is no such function except maybe the constant function 2, as is proven by the construction above. You can have $N$ arbitrarily large and still have the minimum hamming distance 2.
Nov 30, 2013 at 11:32 comment added Mohammad Al-Turkistany Thanks for the answer. However, I am not interested in trivial lower bound. I am looking for a tight lower bound in cubic Hamiltonian graphs (as a function of the number of nodes $N$).
Nov 30, 2013 at 8:56 history answered nvcleemp CC BY-SA 3.0