From: Johnicholas Hines (johnicholas.hines@gmail.com)
Date: Wed Oct 21 2009 - 12:11:48 MDT
ALL instances of a problem DO have shortcuts. Specifically, given an
instance, consider the solver that checks for that instance, and
outputs a pre-computed answer for that instance. That's why we talk
about the hardness of problem families rather than problems.
On Wed, Oct 21, 2009 at 11:57 AM, Matt Mahoney <matmahoney@yahoo.com> wrote:
> Pavitra wrote:
>> 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".
>
> No. "Traveling Salesman" is an NP-complete problem. But suppose I have n cities equally spaced at 1 mile intervals on a circle. Then I know (and can prove in polynomial time) that the shortest path for these instances is n miles.
>
> Other instances may have more subtle shortcuts, but you don't know which ones. A proof of P != NP would only say that not all instances of a problem do.
> -- Matt Mahoney, matmahoney@yahoo.com
>
>
>
> ----- Original Message ----
> From: Pavitra <celestialcognition@gmail.com>
> To: sl4@sl4.org
> Sent: Wed, October 21, 2009 12:00:58 AM
> Subject: Re: [sl4] to-do list for strong, nice AI
>
> Matt Mahoney wrote:
>> Pavitra wrote:
>>> Just to check: I think you mean "...even if it turns out that P =
>>> NP"?
>>
>> 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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