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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!

user108220
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  • 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." – Tassle Aug 04 '19 at 20:51
  • The NP version is, given a graph and an integer $k$, to determine whether the graph has a cut containing at least $k$ edges. – Yuval Filmus Aug 04 '19 at 21:12
  • @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. – user108220 Aug 05 '19 at 07:23

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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.

Yuval Filmus
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  • What you are talking about the decision version? what is the formal way to do it? – user108220 Aug 05 '19 at 07:21
  • 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. – Yuval Filmus Aug 05 '19 at 07:22
  • NP is a category of decision problems. Only decision problems can be in NP. – Yuval Filmus Aug 05 '19 at 07:23
  • The version im talking is this one (as you wrote): "given a graph, find a cut maximizing the number of edges cut" – user108220 Aug 05 '19 at 07:25
  • 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. – Yuval Filmus Aug 05 '19 at 07:25
  • It is enough to say: "Let's look at the decision problem of Max-Cut, and not Max-Cut itself, ..." ??? – user108220 Aug 05 '19 at 07:26
  • Just like you cannot prove that your cat is in NP, so you cannot prove that the optimization version of MAX-CUT is in NP, since NP is a category of decision problems. When we say that an optimization problem is in NP, we really mean that its decision version is in NP. – Yuval Filmus Aug 05 '19 at 07:27
  • See also https://cs.stackexchange.com/questions/67938/decisional-problems-vs-optimization-problems-np-complete-vs-np-hard. – Yuval Filmus Aug 05 '19 at 07:28
  • Ok, that is what i was missing, thanks! – user108220 Aug 05 '19 at 07:29