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| From | Daniel Bastos <dbastos@toledo.com> |
|---|---|
| Newsgroups | comp.programming |
| Subject | on bipartite graphs and matchings |
| Date | 2017-10-21 18:15 -0200 |
| Organization | Aioe.org NNTP Server |
| Message-ID | <86a80kdzpw.fsf@toledo.com> (permalink) |
I've a question about the proof of the converse of Theorem 5.2 in
``Graph Theory with Applications'' by J.A. Bondy and U.S.R. Murty, 1976,
North-Holland.
--8<---------------cut here---------------start------------->8---
Theorem. Let G be a bipartite graph with bipartition (X,Y). Then G
contains a matching that saturates every vertex in X if and only if
|N(S)| >= |S| for all S subset X. (5.2)
Proof. Suppose that G contains a matching M which saturates every
vertex in X, and let S be a subset of X. Since the vertices in S are
matched under M with distinct vertices in N(S), we clearly have
|N(S)| >= |S|.
Conversely, suppose that G is a bipartite graph satisfying (5.2), but
that G contains no matching saturating all the vertices in X. We shall
obtain a contradiction. Let M* be a maximum matching in G. By our
supposition, M* does not saturate all vertices in X. Let u be an
M*-unsaturated vertex in X, and let Z denote the set of all vertices
connected to u by M*-alternating paths. Since M* is a maximum matching,
it follows from theorem 5.1 that u is the only M*-unsaturated vertex in
Z. [...]
--8<---------------cut here---------------end--------------->8---
Comment. Whoa! How come u is in Z? Z is the set of all vertices
connected to u. But u is not connected to itself because G is
bipartite. So there's no way u could be in Z. What's going on there?
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on bipartite graphs and matchings Daniel Bastos <dbastos@toledo.com> - 2017-10-21 18:15 -0200
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