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Q1> All NP problems are polynomial time reducible to an NP-COMPLETE Problem.
So are all NP-COMPLETE problems polynomial time reducible to an NP Problem?

Q2> All NP-COMPLETE problems and NP problems are polynomial time reducible to an NP-HARD Problem.
So are all NP-HARD problems polynomial time reducible to an NP/NP-COMPLETE Problem?

Q3> An NP-HARD problem need not be in NP, so why is NP-Hard definition on NP ?

EDIT :

Not a duplicate of np-vs-np-complete-vs-np-hard-what-does-it-all-mean:

I am asking something else here- like, are the reductions One-way, or both way possible ? And my 1st question on NP-Hard is building over what had been discussed there.

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Somjit
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  • You might find better answers at [cs.se] – pascalhein Apr 16 '15 at 11:48
  • possible duplicate of [NP vs NP-Complete vs NP-Hard -- what does it all mean?](http://stackoverflow.com/questions/1857244/np-vs-np-complete-vs-np-hard-what-does-it-all-mean) –  Apr 16 '15 at 11:48
  • This diagram should help: http://en.wikipedia.org/wiki/NP-hard#/media/File:P_np_np-complete_np-hard.svg – TravisJ Apr 16 '15 at 11:49
  • @Austin : NO its not a duplicate. I have read that already and dint find answers to my question. – Somjit Apr 16 '15 at 11:49
  • @TravisJ : I'm trying to understand if "X can be Reduced to Y in polynomial time" is a both way relation. I fail to understand that from the diagram. – Somjit Apr 16 '15 at 11:55
  • It is not a two way relation. This is because the mapping might be to a "special case" of a problem that is tougher in general. – TravisJ Apr 16 '15 at 12:00
  • So an NP-Hard problem might not be reducible to NP-Complete, but an NP-Complete is always reducible to NP-Hard ? – Somjit Apr 16 '15 at 12:02
  • "Note that NP-hard problems do not have to be in NP (they do not have to be decision problems). The precise definition here is that a problem X is NP-hard if there is an NP-complete problem Y such that Y is reducible to X in polynomial time. But since any NP-complete problem can be reduced to any other NP-complete problem in polynomial time, all NP-complete problems can be reduced to any NP-hard problem in polynomial time. " From the accepted answer in @Austin 's link. – TravisJ Apr 16 '15 at 12:08

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No, reductions are one-way, not both-ways. If you can solve problem A using a solution to problem B, this does not always mean that you can solve problem B using a solution to problem A.

This is very well illustrated by your first question,

Q1. So are all NP-COMPLETE problems polynomial time reducible to an NP Problem?

Definitely not, unless P=NP. Note that even P problems are NP (you do not need any certificate to check the answer for P problems), but you can not reduce an NP-complete problem to a P problem unless P=NP (if a NP-complete problem is reduced to a P-problem, this immediately means that P=NP).

Note that it all stands only unless P=NP. If it turns out that P=NP, than it is easy to see that any NP problem can be reduced to any other NP problem.

BTW, your questions are not quite clear when you say "...polynomial time reducible to an NP Problem". Do you mean any NP-problem or at least one NP-problem? If you mean any, my answer above stands. If you mean "are all NP-COMPLETE problems polynomial time reducible to at least one NP Problem?", that the answer is yes, definitely: every problem can be reduced to itself.

Petr
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  • regarding the last few lines you said... how about "at least one/a few , except itself" ? – Somjit Apr 17 '15 at 21:04
  • @Somjit, then you will first need to define which problems do you consider "equivalent". For most problems, I can slightly alter its text by adding an insignificant requirement or whatever, so that problem will change a tiny bit, but the solution will remain the same. Obviously the old and new problems will be reducable to each other, but I think that's not what you want. – Petr Apr 18 '15 at 07:11
  • @Somjit, and anyway any NP complete problem can be reduced to any other NP complete problem, and *only* to NP complete, so "are all NP-COMPLETE problems polynomial time reducible to *some* NP Problem?" --- yes, to all *NP complete* problems and only to them. – Petr Apr 18 '15 at 07:14