Tutorial :What would a P=NP proof be like, hypothetically?



Question:

Would it be an polynomial time algorithm to a specific NP-complete problem, or just abstract reasonings that demonstrate solutions to NP-complete problems exist?

It seems that the a specific algoithm is much more helpful. With it, all we'll have to do to polynomially solve an NP problem is to convert it into the specific NP-complete problem for which the proof has a solution, and we are done.


Solution:1

P = NP: "The 3SAT problem is a classic NP complete problem. In this proof, we demonstrate an algorithm to solve it that has an asymptotic bound of (n^99 log log n). First we ..."

P != NP: "Assume there was a polynomial algorithm for the 3SAT problem. This would imply that .... which by ..... implies we can do .... and then ... and then ... which is impossible. This was all predicated on a polynomial time algorithm for 3SAT. Thus P != NP."

UPDATE: Perhaps something like this paper (for P != NP).

UPDATE 2: Here's a video of Michael Sipser sketching out a proof for P != NP


Solution:2

Call me pessimistic, but it will be like this:

...

∴, P ≠ NP

QED


Solution:3

There are some meta-results about what a P=NP or P≠NP proof can not look like. The details are quite technical, but it is known that the proof cannot be

  • relativizing, which kind of means that the proof must make use of the exact definition of Turing machine used, because with some modifications ("oracles", like very powerful CISC instructions added to the instruction set) P=NP, and with some other modifications, P≠NP. See also this blog post for a nice explanation of relativization.

  • natural, a property of several classic circuit complexity proofs,

  • or algebrizing, a generalization of relativizing.


Solution:4

It could take the form of demonstrating that assuming P ≠ NP leads to a contradiction.


Solution:5

It might not be connected to P and NP in a straightforward way... Many theorems now are based on P!=NP, so proving one assumed fact to be untrue would make a big difference. Even proving something like constant ratio approximation for TS should be enough IIRC. I think, existence of NPI (GI) and other sets is also based on P!=NP, so making any of them equal to P or NP might change the situation completely.

IMHO everything happens now on a very abstract level. If someone proves anything about P=/!=NP, it doesn't have to mention any of those sets or even a specific problem.


Solution:6

Probably it would be in the form of a reduction from an NP problem to a P problem. See the Wikipedia page on reductions.

OR

Like this proof proposed by Vinay Deolalikar.


Solution:7

Set N equal to the multiplicative identity. Then NP = P. QED. ;-)


Solution:8

The most straightforward way is to prove that there is a polynomial time solution to the problems in the class NP-complete. These are problems that are in NP and are reducable to one of the known np problem. That means you could give a faster algorithm to prove the original problem posed by Stephen Cook or many others which have also been shown to be NP-Complete. See Richard Karp's seminal paper and this book for more interesting problems. It has been shown that if you solve one of these problems the entire complexity class collapses. edit: I have to add that i was talking to my friend who is studying quantum computation. Although I had no clue what it means, he said that a certain proof/experiment? in the quantum world could make the entire complexity class, i mean the whole thing, moot. If anyone here knows more about this, please reply.

There have also been numerous attempts to the problem without giving a formal algorithm. You could try to count the set. Theres the Robert/Seymore proof. People have also tried to solve it using the tried and tested diagonlization proof(also used to show that there are problems that you can never solve). Razborov also showed that if there are certain one-way functions then any proof cannot give a resolution. That means that new techniques will be required in order to solve this question.

Its been 38 years since the original paper has been published and there still is no sign of a proof. Not only that but lot of problems that mathematicians had been posing before the notion of complexity classes came in has been shown to be NP. Therefor many mathematicians and computer scientists believe that some of the problems are so fundamental that a new kind of maths may be needed to solve the problem. You have to keep in mind that the best minds human race has to offer have tackled this problem without any success. I think it should be at least decades before somebody cracks the puzzle. But even if there is a polynomial time solution the constants or the exponent could be so large that it would be useless in our problems.

There is an excellent survey available which should answer most of your questions: http://www.scottaaronson.com/papers/pnp.pdf.


Solution:9

Certainly a descriptive proof is the most useful, but there are other categories of proof: it is possible, for example, to provide 'existence proofs' that demonstrate that it is possible to find an answer without finding (or, sometimes, even suggesting how to find) that answer.


Solution:10

It would likely look almost precisely like one of these


Solution:11

Good question; it could take either form. Obviously, the specific algorithm would be more helpful, yes, but there's no determining that that would be the way that a theoretical P=NP proof would occur. Given that the nature of NP-complete problems and how common they are, it would seem that more effort has been put into solving those problems than has been put into solving the theoretical reasoning side of the equation, but that's just supposition.


Solution:12

Any nonconstructive proof that P=NP really is not. It would imply that the following explicit 3-SAT algorithm runs in polynomial time:

Enumerate all programs. On round i, run all programs numbered less than i for one step. If a program terminates with a satisfying input to the formula, return true. If a program terminates with a formal proof that no such input exists, return false.

If P=NP, then there exists a program which runs in O(poly(N)) and outputs a satisfying input to the formula, if such a formula exists.

If P=coNP, there exists a program which runs in O(poly(N)) and outputs a formal proof that no formula exists, if no formula exists.

If P=NP, then since P is closed under complement NP=coNP. So, there exists a program which runs in O(poly(N)) and does both. That program is the k'th program in the enumeration. k is O(1)! Since it runs in O(poly(N)) our brute force simulation only requires

k*O(poly(N))+O(poly(N))^2

rounds once it reaches the program in question. As such, the brute force simulation runs in polynomial time!

(Note that k is exponential in the size of the program; this approach is not really feasible, but it suggests that it would be hard to do a nonconstructive proof that P=NP, even if it were the case.)


Solution:13

To some extent, the form such a proof needs to have depends on your philosophical point of view (= the axioms you deem to be true) - e.g., as a contructivist you would demand the construction of an actual algorithm that requires polynomial time to solve an NP-complete problem. This could be done by using reduction, but not with an indirect proof. Anyhow, it really seems to be very unlikely :)


Solution:14

The proof would deduce a contradiction from to the assumption that at least one element (problem) of NP isn't also an element of P.


Solution:15

An interesting read that is somewhat related to this


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