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After removing the leaves of the dfs tree of a random graph , suppose the number of edges left is |S|, can we prove that the matching for that graph will be |S|/2?

justin waugh
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  • By |S|/2 you mean floor or ceiling? – kunigami Jan 27 '11 at 12:15
  • 1)Is |S| the number of edges after removing the leaves of the DFS tree or the original random graph? 2) By matching you mean the maximum matching or any matching is ok? – kunigami Jan 27 '11 at 21:06
  • @kunigami 1) |S| is after removing the leaves of DFS tree 2) any matching which is maximal will do so any matching will do. To correct my question, it is ceiling(|S|/2) instead of just |S|/2. – justin waugh Jan 27 '11 at 21:46
  • Ok, now I understood. So I think Keith Randall answered your question :) – kunigami Jan 28 '11 at 11:34

1 Answers1

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Here's a proof.

Theorem: Let T be any tree with i leaves. There is a (|T|-i)/2 matching in T.

Proof: by induction. If T is a tree with i leaves, let T' be the tree that results when removing all the leaves from T. T' has j <= i leaves. Similarly, let T'' be the tree that results when removing all the leaves from T'. T'' has k <= j leaves.

Apply the theorem by induction to T'', so there exists a matching of size (|T''|-k)/2 = (|T|-i-j-k)/2 in T''. The set of edges T-T' contains at least j edges that are not incident to any edge in T'' or to each other (pick one incident to each leaf in T'), so add those edges to make a matching in T of size (|T|-i+j-k)/2. Since j >= k, this is at least (|T|-i)/2 edges. QED.

I've glossed over the floor/ceiling issues with the /2, but I suspect the proof would still work if you included them.

Keith Randall
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