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A graph $G$ is said to be $k$-connected (or $k$-vertex connected, or $k$-point connected) if there does not exist a set of $k-1$ vertices whose removal disconnects the graph.

Let $v$ be a vertex of a 2-connected graph $G$. Prove that $v$ has a neighbor $u$ such that $G-v-u$ is connected.

My proof :
G is 2-connected. So, $G-v$ is connected. If we show that $G-v-u$ is connected, the problem is solved.
According to Menger's Theorem, if a graph is $k$-connected, there exists $k$ internally disjoint paths between each two vertices. So, here, in this question, there exists 2 disjoint paths between every two vertex.
Every path in $G$ is one of these two types:

1) The path $P$ between two arbitrary vertices like $a,b$ contains $u$.
2) $P$ doesn't contain $u$.

Case 1 : When we delete $u$, there still exists a path between $a,b$ ( According two Menger's theorem, because our graph is 1-connected) So, we take the path and the graph remains connected.
Case 2 : We don't need $u$ when traveling from $a$ to $b$. So, Nothing happens to $P$ when we remove the vertex $u$.

Is this proof correct ? Or can you prove a better proof?

I also found this proof which i don't understand :proof
Thanks in advance.

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  • $\begingroup$ Your proof seems to show that the removal of any two neighbouring vertices should work, which is not the case $\endgroup$ Apr 29 '16 at 13:01
  • $\begingroup$ @HagenvonEitzen Can you help me with the proof? even a hint is helpful sir :) $\endgroup$
    – Perceptual
    Apr 29 '16 at 13:02
  • $\begingroup$ did you ever figure this out? @ArmanMalekzadeh $\endgroup$ Feb 27 '18 at 5:19
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Think of the blocks of $G-v$. Assume it has a leaf block B (what about the other case?). How do the two internally disjoint paths connecting an "internal" vertex of B with some vertex in another block look like?

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  • $\begingroup$ please explain it more ! i don't understand what you wrote $\endgroup$
    – Perceptual
    Apr 30 '16 at 13:03

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