Form of the smallest vertex cover in a bipartite graph I'm trying to write a proof of Konig's theor

dream13rxs

dream13rxs

Answered question

2022-07-14

Form of the smallest vertex cover in a bipartite graph
I'm trying to write a proof of Konig's theorem using Menger's theorem. However, I got stuck along the way. In order to move forward I'd (apparently) need to show the following fact.
Let G be a bipartite graph with partite sets X and Y. The smallest vertex cover of G is of the form ( X A ) N ( A ) for some A X.
I tried doing a proof by contradiction but to no avail. Maybe the thesis I'm trying to show is false in the first place? Any help would be greatly appreciated.

Answer & Explanation

zlepljalz2

zlepljalz2

Beginner2022-07-15Added 22 answers

Step 1
Let U be any vertex cover; let A = X U (the smallest set for which U contains X A).
For every vertex b N ( A ), there is some edge ab with a A which needs to be covered by U somehow. By construction a U, so ab is not covered by a. Therefore ab must be covered by b: we must have b U. Therefore N ( A ) U.
Step 2
We have shown that U contains ( X A ) N ( A ). We can check that ( X A ) N ( A ) is a vertex cover all by itself. Therefore if U is a minimal vertex cover (if it has no proper subset which is a vertex cover) we must have U = ( X A ) N ( A ). In particular, this is true of the smallest vertex cover.
I admit I do not see how you need this fact to prove Konig's theorem from Menger's theorem. Neither one deals in N(A) as a concept. Maybe you are trying to prove Hall's theorem from Konig's theorem?

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