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Suppose P = NP, would that mean that Hamiltonian-Cycle is no longer NP-Hard? Hamiltonian-Cycle is a language where a given graph G contains a Ham-Cycle.

  • I’m voting to close this question because it is not a computer programming question. There is no computer program here. It is a question about the theory of computing. – Raymond Chen Dec 16 '21 at 01:09

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The definition of NP-Hard is that it allows you to solve a NP-complete problem, which is a fancy way of saying "solving this problem is at least as hard as solving any NP problem".

If P = NP, a NP-hard problem stays NP-hard, in the sense that "solving this problem is still at least as hard as solving any NP problem". But since P = NP, this is equivalent to "solving this problem is at least as hard as solving any P problem". Which is generally not considered hard

tbrugere
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  • But honestly, most of the world’s cryptography is based on the assumption that P != NP, and it has never failed. So for your applications, there’s a good chance you can assume that too. And if it fails, look at the bright side of things, you’ve just solved a problem of the millenium – tbrugere Dec 16 '21 at 00:20
  • So, if true, Ham-Cycle would still be NP-Hard? –  Dec 16 '21 at 00:23
  • Yes. But NP-hard would be equivalent to P-hard (which doesn’t mean very hard) – tbrugere Dec 16 '21 at 00:24
  • Btw hamiltonian-cycle is not only NP-hard, it is actually NP-complete (which means it is NP-hard but also NP). This means that if P=NP, then Hamiltonian-cycle is P. – tbrugere Dec 16 '21 at 00:26