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I know the correct way of proving NP hard of a problem X is to reduce a known NP-Hard problem to X i.e. the direction is from the known, harder problem to the problem we want to prove is NP-Hard. But all NP-Complete problems are polynomially related (one can be transformed into the other in polynomial time), so I would like to ask if it's correct to assert that a problem is NP-Hard when it can be polynomially reduced to 3SAT?

ccying
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  • You can "reduce" trivial problems to 3SAT too. – Damien_The_Unbeliever Dec 15 '17 at 09:21
  • Might be better asked on one of the [CS](https://cs.stackexchange.com/) [Exchange](https://cstheory.stackexchange.com/) sites (not sure which one though - you'd have to read their help and existing questions to see which is a better fit) – Damien_The_Unbeliever Dec 15 '17 at 10:25
  • It doesn't seem intuitive to me. How could a problem X not be as hard as another problem (3SAT) if it could be transformed to 3SAT in polynomial time? Could you explain a bit on this? Thanks! – ccying Dec 15 '17 at 12:59
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    Have you ever overcomplicated a solution to a problem, only to be told that there are far simpler ways to accomplish your goal? In a similar fashion, you can take a problem with a solution in P and transform it *into*, say, a 3SAT problem. Whilst solving the 3SAT instance would solve the original problem, it's doing it in an overcomplicated way. So proving you can transform problems *into* 3SAT problems demonstrates that they're *no harder* than solving 3SAT, but says nothing about whether they are *easier* to solve. – Damien_The_Unbeliever Dec 15 '17 at 13:07
  • Damien, why don't you put your responses into a true answer? (I would do so, but it's you who should do it, I suppose.) – Prof.Chaos Feb 24 '19 at 08:58

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