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I'm dealing with the following problem in graph's matchings. Let G = (V,E) be an undirected graph, and let S,T subgroups of V be two sets of vertices with no common neighbours. That is, there exist no vertices s in S, t in T, and v in V such that (s,v) (t,v) in E (note that there may still be edges between S and T or within S or T).

I wanna show that if there exists a matching in G in which all vertices in S are covered, and also a (possibly different) matching in which all vertices in T are covered, then there also exists a matching in G in which all vertices in S union T are covered.

An idea: dividing M1 and M2 (the two matchings for S,T accordingly) to groups of different edges: 1 - Edges from S to S 2 - Edges from S to T 3 - Edges from S to an unmatched vertex (harmless edges).

Same for T.

I think the first thing to do in order to show such matching exists, is take edges from group 3 first, that will make our problem smaller. Or perhaps there's a counter example? Any ideas?

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Hint: Try to cover $S$ first, and then cover $T$. The first step must be done as minimally as possible so that the second step works.

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  • $\begingroup$ @Jojo Ignore my previous ideas. This is the correct way to go... $\endgroup$ Feb 14 '15 at 16:07
  • $\begingroup$ Not sure I understand. What is the correct way to go? Because I haven't succeeded yet with my idea. $\endgroup$
    – itaymeller
    Feb 14 '15 at 19:05
  • $\begingroup$ @Jojo If you saw the previous revisions of the post, ignore them. This latest one is correct. One of the groups need to be covered first before the other. $\endgroup$ Feb 15 '15 at 3:30

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