On Tue, Feb 25, 2014 at 9:53 PM, Ben Goertzel <[email protected]> wrote:

> Intuitively, I suspect that P!=NP will eventually be demonstrated by some
> sort of counting argument...
>
> That is, I suspect that if
>
> X = the number of polynomial-time algorithms
>

"Number of Turing machines of size N that
execute
 in polynomial time" seems to be an incomputable number -- due to the
halting problem... no?

Y = the number of problems in NP (i.e. for which potential solutions can be
> checked in polynomial time)
>
> one can show somehow that
>
> X < Y
>

It would have to be a carefully managed limiting argument; since in full
> generality, of course, both of those classes are infinite....  That is, one
> would have to talk about, given an appropriate Turing-complete language L,
>
> X = the number of distinct polynomial time algorithms describable in N
> bits, in language L
>
> Y = the number of distinct problems for which there is a solution
> describable in N bits in language L, for which potential solutions can be
> checked in polynomial times
>
> (where it's assumed that each "algorithm" can solve exactly one problem,
> in the presupposed formalism).
>

There may exist just a small number of Turing machines that can solve an
NP-hard problem in polynomial time, with the problem-size N as an input
parameter.  That would allow P=NP without violating your counting
argument....

Good algorithms are hard to find, but when they do exist, they prove that
it is possible =)



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