Date: Nov 9, 2011 11:58 AM Author: Craig Feinstein Subject: an interesting observation about Natural Proofs 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.