Date: Nov 10, 2011 11:29 AM
Author: Craig Feinstein
Subject: Re: an interesting observation about Natural Proofs

On Nov 9, 11:58 am, Craig Feinstein <cafei...@msn.com> wrote:
> Unless it is an axiom that P!=NP, any proof of P!=NP has to be
> essentially be of the form:
>
> f solves SAT -> f in C
> f in C -> f cannot be computed in polynomial time.
> Therefore, f solves SAT -> f cannot be computed in polynomial time.
>
> The Razborov-Rudich paper on "Natural Proofs" essentially shows that
> if (1) C is a property that is common (Largeness) and (2) it is easy
> to check whether a function f has property C (Constructivity or Low
> Complexity), then the above proof cannot be valid unless "pigs can
> fly".
>
> Suppose C is an uncommon property, so (1) is broken. In the most
> extreme case, C=the class of functions which solve SAT. Then the above
> proof would be of the form:
>
> f solves SAT -> f solves SAT
> f solves SAT -> f cannot be computed in polynomial time.
> Therefore, f solves SAT -> f cannot be computed in polynomial time.
>
> So this proof would be of the form "P!=NP because P!=NP".
>
> Now, suppose it is difficult to check whether a function f has
> property C, so (2) is broken. In the most extreme case, C = the class
> of functions f that cannot be computed in polynomial-time. Then the
> above proof would be of the form:
>
> f solves SAT -> f cannot be computed in polynomial time
> f cannot be computed in polynomial time -> f cannot be computed in
> polynomial time.
> Therefore, f solves SAT -> f cannot be computed in polynomial time.
>
> So this proof would be of the form "P!=NP because P!=NP".
>
> So we see that breaking one of the two conditions for a natural proof,
> largeness or constructivity, leads one in the direction of a proof of
> the flavor "P!=NP because P!=NP". And the two conditions, largeness
> and constructivity, are "natural" because they lead a proof away from
> a proof of the flavor "P!=NP because P!=NP". From this observation, we
> can conclude that a proof that P!=NP cannot involve deep mathematics,
> since deep mathematics is "natural", at least "natural" to
> mathematicians. For example, here is one such proof:
>
> SUBSET-SUM (where the target integer is zero) is equivalent to the
> problem of determining whether the set of subset-sums of {s1,...,sk}
> and the set of subset-sums of {-s_{k+1},...,-s_n} intersect
> nontrivially. There are 2^k members of the set of subset-sums of
> {s1,...,sk} and 2^{n-k} members of the set of subset-sums of {-s_{k
> +1},...,-s_n}, and it is impossible to simplify the problem further.
>
> Hence, since the combined size of the sets that the SUBSET-SUM problem
> is about is 2^k+2^{n-k}, for k in {1,...,n}, we can get a lower bound
> for the worst-case running-time of an algorithm that solves SUBSET-SUM
> by minimizing 2^k+2^{n-k}, s.t. k in {1,...,n}. We get 2^{n/2}, so
> Omega(2^{n/2}) is an exponential lower bound for algorithms which
> solve SUBSET-SUM.
>
> This is an example of a "supernatural proof". It is not natural,
> because in this case, C=the set of functions that cannot be computed
> in o(2^{n/2}) time, so it violates the constructivity property above.
> And it also does not involve deep mathematics. This supernatural proof
> is almost of the form "P!=NP because P!=NP", but not quite or else it
> would be a circular argument.
>
> I believe that the reason why the P vs NP problem has been considered
> so difficult is because of human psychology. Most people either want
> there to be some deep reason why P!=NP (some almost magic secret using
> some deep mathematics like the proof of FLT had) or they refuse to
> believe that P!=NP. They don't want to accept that there is no deep
> reason why P!=NP. It essentially just is.


Not everything I said in this post is technically correct. But I think
the idea is correct.