December 22, 2008 Re: NP=P | ||||
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Andrei Alexandrescu Wrote:
> Tim M wrote:
> > If they really did find proof that p==np wouldn't they be millionaires and probably should have kept it to themselves. (I haven't read that all the way through btw)
> >
> > On Sun, 14 Dec 2008 08:43:48 +1300, BCS <ao@pathlink.com> wrote:
> >
> >> Reply to Knud,
> >>
> >>> Læs lige denne artikel
> >>> http://arxiv.org/abs/0812.1385
> >>>
> >>
> >> If I'm reading that correctly, not exactly, the verbiage seems to imply that they didn't solve P=NP but a related problem.
> >>
> >> "... these problems most of which are not believed to have even a polynomial time sequential algorithm."
>
> The paper shows that #P=FP. I'm not that versed with theory to figure how important that result is.
>
> http://en.wikipedia.org/wiki/Sharp-P
> http://en.wikipedia.org/wiki/FP_(complexity)
>
>
> Andrei
Proving FP=#P is a far more grandiose claim than proving P = NP.
To clarify:
FP is the class of all *functions* that can be computed 'easily' (on a deterministic computer in polynomial time). It is a pretty simple generalization of, P, which is the class of easy *decision problems* (must have a yes/no answer.)
While on the other hand:
#P is the set of all functions which compute the number of solutions for problems in NP. For example, *counting* the number of Hamiltonian circuits in a graph is in #P, while simply *testing* if it has Hamiltonian circuit is in NP.
If this were indeed true, it would have many screwy consequences, such as NP=coNP (but then again pretty much any hierarchy collapse would do the same thing.) Of course, most likely this is just noise.
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