Proving Hamiltonian Graph Connectivity is 3: n >= 4 Vertices

[Solarmew]

Suppose G is a HC (Hamiltonian-connected) 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.

 

[haruspex]

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.

 

[Solarmew]

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.

it's vertex connectivity, sorry >.> so the smallest number of vertices in any vertex cut of G is 3

 

[haruspex]

it's vertex connectivity, sorry >.> so the smallest number of vertices in any vertex cut of G is 3

OK, same deal, only easier. You're trying to prove connectivity >= 3. So assume false. That means there's a pair vertices whose removal would leave a disconnected graph. Is there a Hamiltonian path between them?

 

[Solarmew]

OK, same deal, only easier. You're trying to prove connectivity >= 3. So assume false. That means there's a pair vertices whose removal would leave a disconnected graph. Is there a Hamiltonian path between them?

There is. Since G is HC, there's a Hamiltonian Path b/w every two pairs of vertices.
But I'm not sure how there being a path is helpful >.>
hm, let me think about that for a sec...

 

[haruspex]

There is. Since G is HC, there's a Hamiltonian Path b/w every two pairs of vertices.
But I'm not sure how there being a path is helpful >.>
hm, let me think about that for a sec...

No, I mean take any graph that has a pair of vertices whose removal would render the graph disconnected. Draw a diagram. How could it have a Hamiltonian path between those two vertices?

 

[Solarmew]

ooooh, there isn't a HP b/w them, i lied ... i was just testing you 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 :<

 

[haruspex]

Yes, that's the argument.

 

[Solarmew]

Yes, that's the argument.

= ^.^ = thanks, i appreciate your help!