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How can i show/explain/prove that Max-Cut is in NP? "For a graph, a maximum cut is a cut whose size is at least the size of any other cut. The problem of finding a maximum cut in a graph is known as the Max-Cut Problem."

Thanks!

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  • $\begingroup$ The Wikipedia article you are citing tells you why it is easy to see that the decision version of the problem is in NP. Quote : "It is easy to see that the problem is in NP: a yes answer is easy to prove by presenting a large enough cut." $\endgroup$ – Tassle Aug 4 at 20:51
  • $\begingroup$ The NP version is, given a graph and an integer $k$, to determine whether the graph has a cut containing at least $k$ edges. $\endgroup$ – Yuval Filmus Aug 4 at 21:12
  • $\begingroup$ @Tassle - you are not helping. clearly i saw this cite in wikipedia. the question is why if i want to prove that max-cut is in NP is need to talk about the this decision problem. $\endgroup$ – user108220 Aug 5 at 7:23
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The decision version of MAX-CUT is as follows:

Given a graph $G$ and an integer $k$, is there a cut in $G$ containing at least $k$ edges?

This version is clearly in NP.

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  • $\begingroup$ What you are talking about the decision version? what is the formal way to do it? $\endgroup$ – user108220 Aug 5 at 7:21
  • $\begingroup$ MAX-CUT can be considered both as a decision problem (as in my answer) and as an optimization problem (given a graph, find a cut maximizing the number of edges cut). The version you quote in your OP is unfamiliar to me. $\endgroup$ – Yuval Filmus Aug 5 at 7:22
  • $\begingroup$ NP is a category of decision problems. Only decision problems can be in NP. $\endgroup$ – Yuval Filmus Aug 5 at 7:23
  • $\begingroup$ The version im talking is this one (as you wrote): "given a graph, find a cut maximizing the number of edges cut" $\endgroup$ – user108220 Aug 5 at 7:25
  • $\begingroup$ This is not a decision problem, so it cannot be in NP. Only decision problems can be in NP. A decision problem is one in which the answer is either Yes or No. $\endgroup$ – Yuval Filmus Aug 5 at 7:25

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