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P vs. NP

Name: Anonymous 2009-12-31 21:01

Do you still win the 1 million dollars if you show that P=NP (and thus P!=NP) is unprovable, as opposed to equal or unequal?

Name: Anonymous 2009-12-31 22:19

If P=NP, there is some algorithm that can decide NP-complete problems in P time. If such an algorithm exists, it can easily (trivially) be proven that P=NP. Therefore, if it can't be proven that P=NP, P!=NP.

If you can prove to me that N=NP is unprovable, I'll go claim the 1 million dollars ;)

Name: 4tran 2009-12-31 23:39

>>2
N?  Are you trolling?

Name: Anonymous 2009-12-31 23:45

Therefore, if it can't be proven that P=NP, P!=NP.
If P=NP is undecidable, then P!=NP is also undecidable.

But maybe it's possible that an algorithm could exist which decides NP-complete problems in P time, but no proof would exist for P=NP.

Name: Anonymous 2010-01-01 1:27

>>3
most likely a typo.

Name: Anonymous 2010-01-01 10:40

The most entertaining situation would probably be to find a algorithm that seemed to give answers in P time, but be unable to prove it did.
You could revolutionize the world of practical computer science, and still not be able to claim a single prize.

Name: Anonymous 2010-01-01 10:53

>>6
Or find an algorithm which solves most or all practical instances of an NP-complete problem in P time, and yet there would still be instances which take an exponential amount of time. i.e. prove that all NP-complete problems are NP-hard.

Also, that situation would be extremely disappointing for the person who found the algorithm.

Name: Anonymous 2010-01-01 15:02

>>4
That's only possible in the way as outlined by >>6, which I kind of forgot in >>2.

>>3
Just a typo of course. Not everything is a troll (yes, I am aware I'm posting on 4chan).

>>7
Unless I'm misunderstanding your post, that's actually quite common; it's pretty easy to write a program that solves most but not all SAT problems in polynomial time.

Don't change these.
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