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Based on the below link , I can know that solving of Satisfiability(NP Complete) in polynomial time means any other NP problem can be solved in polynomial time. But is Vice - Versa true?

Also, If there is a polynimial for any other NP-Complete problemt does it mean , all the other NP-Complete can be solved in polynomial time?

What are the differences between NP, NP-Complete and NP-Hard?

Community
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user10000000
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  • Elaborate what you mean with vice-versa. – orlp Apr 28 '15 at 04:28
  • All NP-complete problems are equivalent in that if any of them is solvable in polynomial time, all problems in NP can be solved in polynomial time. However, this question really is off-topic for Stack Overflow. Better to ask on http://cs.stackexchange.com/ – Ted Hopp Apr 28 '15 at 04:29

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The 'complete' in NP-complete means that if a problem is in NP-complete, a solution for that problem gives a solution to any problem in NP with a polynomial amount of conversion processing.

In layman's terms - if you solve a single NP-complete problem in polynomial time you have proven that NP = P.

orlp
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  • Yes, But if an NP solved in polynomial time, does it mean all the NP complete problems be solved in polynomial time. – user10000000 Apr 28 '15 at 04:33
  • No. Consider for example, the circuit evaluation problem (which is complete for P). The circuit evaluation problem poses the question "does this input satisfy this boolean circuit". The circuit evaluation problem is clearly in NP and contains a polynomial solution. However, this does not imply that Circuit SAT has a polynomial time solution. If you consider another NP-complete problem such as the Hamiltonian Path problem, and show that this problem has a polynomial-time solution, then Satisfiability also has a polytime solution. – Bill Province Apr 29 '15 at 02:24
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Solving SAT does not solve all NP problems. It solves all NP problems that can be reduced to SAT by a P complexity algorithm. There exist problems that do not have a P complexity reduction to SAT. Some are reducible to SAT by NP reductions. Some of those by NP reductions for which the reduction algorithm has a P complexity reduction to SAT.

SAT has a P complexity solution. See arxiv.org/abs/cs/0205064.

george
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