
#1
Dec2712, 04:33 PM

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Suppose G is a HC (Hamiltonianconnected) graph on n >= 4 vertices. Show that connectivity of G is 3.
I tried starting by saying that there would be at least 4C2=6 unique hamiltonian paths. But then I'm not sure where to go from here. Any hints would be appreciated. 



#2
Dec2812, 10:15 PM

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I assume this is edge connectivity.
Suppose there's a cutset of two edges. In relation to these edges, find two points for which there can only be a Hamiltonian path between them under very restrictive conditions. 



#3
Dec2812, 10:33 PM

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#4
Dec2812, 10:40 PM

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Hamiltonian graph 



#5
Dec2812, 10:58 PM

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But i'm not sure how there being a path is helpful >.> hm, lemme think about that for a sec... 



#6
Dec2812, 11:19 PM

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#7
Dec2812, 11:21 PM

P: 36

ooooh, there isn't a HP b/w them, i lied ... i was just testing ya XD .... by doodling it out i can kinda see it i think +.+ but i'm not sure how to put it in words ...
like if by removing those two vertices (let's say u,v) we disconnected the graph, that would mean that the only paths from one component to the other were through u and v. But that would mean that if we tried to find a Hamiltonian path from u to v we would not be able to go from one block of G to the other without passing through v, so we would only be able to go through the vertices of one block of G before inevitably ending up at v. gah ... that doesn't sound very convincing >.< ... i need to work on my proof speak :< 



#8
Dec2912, 02:52 PM

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Yes, that's the argument.




#9
Dec2912, 08:46 PM

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