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Let $G$ be an undirected, simple graph, and let $\alpha(G)$ denote the independence number of $G$, i.e., the size of a maximum independent set (stable set) in $G$. A graph is $\alpha$-critical if for every edge $e$ of $G$, we have $\alpha(G - e) > \alpha(G)$.

Conjecture: for every $\alpha$-critical graph $G$ without isolated vertices, for every maximum independent set $S$ in $G$, there is a maximum independent set $S'$ in $G$ that is disjoint from $S$.

The conjecture is true for the simplest classes of $\alpha$-critical graphs, the graphs $K_n$ ($n \geq 2$) and the odd cycles. It also seems to be true for the list of minor-minimal obstructions to having a vertex cover of size at most five, which are also $\alpha$-critical. Is it true in general?

Thanks in advance!

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Bart, I have ``Matching Theory''. To send it to you? –  Boris Novikov Feb 12 '13 at 11:20
    
Thanks, Boris. The theorems in the book do not seem to answer my question, unfortunately. –  Bart Jansen Feb 13 '13 at 9:26
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2 Answers 2

Well, take a graph on $n\ge 3$ vertices with exactly one edge. Clearly, it's $(n-1)$-critical, but does not satisfy your conjecture.

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You're right, of course. I should have added the condition that $G$ contains no isolated vertices, which I just did. (I'm used to thinking about the graphs as minor-minimal obstructions to having a vertex cover of a certain size; $\alpha$-critical graphs are only obstructions if they do not have any isolated vertices.) –  Bart Jansen Feb 13 '13 at 9:28
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I don't think so. You really mean maxiMUM, not maxiMAL, right?

If so, take a complete bipartite graph $K_{m,n}$ with $m$ less than $n$, so the maximum critical set is the big part in the bipartition and is of size n. This graph isn't $\alpha$-critical, clearly, but its maximum independent set is so big that there's a pigeonhole problem with finding a disjoint second one.

It seems to me that if you delete edges from G you can only increase the independence number, so (*) do this iteratively without disconnecting G until the result is $\alpha$-critical.

This should give you a counterexample unless this last step (*) is flawed somehow, which might well be the case, I confess - I haven't thought it through. But rather than agonizing about whether it's possible, you can implement this randomly with a computer program, for explicit small m,n, until you have either generated an explicit counterexample, or waited for such a long time as to convince yourself that it's worth trying to find the error. I might do it if I have a chance today.

EDIT -- I was wrong - step (*) is in fact flawed, because it disconnects the graph before it becomes $\alpha$-critical. Interesting!

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I'm pretty sure (*) fails emphatically. As evidence, consider the case in which the edges form a perfect matching. Here $n$ is the maximum stable set still, but since $n>m$ the graph is disconnected. Moreover it fails for $(m,n)=(1,2)$, and I think you should be able to prove that it fails in general using strong induction and observing the new maximum stable set when you remove an edge. –  Andrew D. King Mar 13 '13 at 17:44
    
I do indeed mean maxiMUM rather than maxiMAL. It is known that the alpha-critical bipartite graphs are exactly the perfect matchings, so these cannot give a counterexample: all perfect matchings have two disjoint maximum independent sets. –  Bart Jansen Mar 14 '13 at 9:09
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