From: Pavitra (firstname.lastname@example.org)
Date: Tue Oct 20 2009 - 22:00:58 MDT
Matt Mahoney wrote:
> Pavitra wrote:
>> Just to check: I think you mean "...even if it turns out that P =
> No, I mean P != NP. Suppose it were proven. You would know that some
> instances of, say, SAT or traveling salesman required exponential
> time to solve, but you wouldn't know which ones. There are heuristics
> that can solve lot of NP-complete problems quickly, just not all of
> them. You don't know that any particular instance is hard because
> there might be another heuristic that makes it easy.
I thought the definition of NP-complete was that if any single
NP-complete problem is solvable in polynomial time (i.e, is in P), then
any problem in NP is solvable in polynomial time.
Thus, "there exists at least one NP-complete problem solvable in
polynomial time" is equivalent to "P = NP", and "there exists at least
one NP-complete problem not solvable in polynomial time" is equivalent
to "P != NP".
That is, I believe that "there exists at least one NP-complete problem
solvable in polynomial time, and at least one other NP-complete problem
not solvable in polynomial time" has been mathematically disproven.
Am I completely missing the point?
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