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Disclaimer: This is a homework question.

I would like to reduce vertex cover problem to the following problem:

$$L = \{G \mid G\text{ has a vertex cover of size } |V(G)|/2\}\,.$$

I have divided the problem into three parts:

  • $k = n/2$: This case is trivial.
  • $k > n/2$: Yes in vertex cover problem does not necessarily mean yes in $L$.
  • $k < n/2$: No in vertex cover problem does not necessarily mean no in $L$.

I know that I need to change graph $G$ into $G'$ somehow to map results of VC to $L$.

Any advice on how to do so is appreciated.

xskxzr
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Henry Cho
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3 Answers3

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For $k<n/2$, add an isolated complete graph with $n-2k+2$ vertices (note it takes at least $m=n-2k+1$ vertices to cover this complete graph). Now there is a vertex cover of size $(n+(n-2k+2))/2=n-k+1$ in the new graph iff there is a vertex cover of size $(n-k+1)-m=k$ in the old graph.

For $k>n/2$, add $2k-n$ isolated vertices. Now there is a vertex cover of size $(n+(2k-n))/2=k$ in the new graph iff there is a vertex cover of size $k$ in the old graph.

xskxzr
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For k < n/2, add enough vertices with a single edge out of them (pointing back at themselves) such that the size of the new vertex cover is half the size of the new graph.

For k > n/2, add enough vertices connected to an arbitrary vertex already in the vertex cover such that the size of the original vertex cover is half the size of the new graph.

tomatoBisque
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This question was answered in the following video (By me): https://www.youtube.com/watch?v=rKNasOecTu0

In a nut-shell, the key here is to modify the graph in such a way it wouldn't be profitable to select vertices from the modified graph, and by finding a full vertex cover in the original graph, you could get a half vertex cover in the modified graph.

Hope that helps :-)

CUCUMBER
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