Complexity classes P and NP/Talk

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This question was asked on a little while ago, and it seems relevant here: What, if any, would be the effects of a proof that P=NP, or that P≠NP, or indeed that the question is undecidable?
- Stuart

The latter two wouldn't have any practical consequences, but it would be a big deal in the world of theoretical computer science (and the last also one in logic and mathematics). If somebody proves P=NP, then the practical consequences depend on the way they prove it. If they explicitly exhibit a polynomial algorithm for an NP complete problem, then we would know polynomial algorithms for all NP problems, and if the degrees of the involved polynomials are reasonable, that would be huge news in many fields, but is considered extremely unlikely. It's much more likely that either the polynomial has a ridiculously high degree, or that the proof wouldn't be constructive and just generally concludes P=NP without exhibiting any algorithm altogether. In those two cases, there wouldn't be any immediate practical consequences either. --AxelBoldt

I agree with your conclusion. However, it isn't really possible for there to be a nonconstructive proof of P=NP with no known algorithm. That's because the construction has already been done. I could easily give you an algorithm for, say, Travelling Salesman, whose correctness is obvious, but whose asymptotic running time is unknown. It's easy to prove that if P=NP, then my algorithm is polytime. If anyone ever publishes a "nonconstructive" proof of P=NP, the construction will already exist. This is all academic, of course. The huge multiplicative constant means none of this would have any practical value. --LC

That's very interesting. How does the construction work? --AxelBoldt

Assume the goal is a program that, given a TSP instance, returns the certificate (if "yes") or fails to halt (if "no"). Assume you already have a program to check certificates. Let X=1. Consider the list of all possible programs, sorted first by size, then lexicographically. Run the first X programs for X steps each, with the TSP instance as an input. If any of them halts and returns a certificate, check it. If it's right, halt and return it. If not, continue. If none of them succeed, then increment X, and repeat. Clearly, this algorithm is correct, and its theta running time will be the optimal running time, plus the time to check a certificate. The constant hidden by the theta is exponential in the length of the first program that solves the problem. Totally impractical. I don't remember offhand which books cover this, though I could look it up if needed. It's been a while since I've visited Wikipedia; it's nice to see that it's still going strong. --LC

The description is clear, I don't think we need a reference, but I think it's nice enough to put it in the main article. --AxelBoldt

I've added it, in a form that's hopefully accessible to a wider audience. --LC

Can you also concoct an algorithm which always gives the correct answer in polynomial time? --AxelBoldt

Yes, if you can tell me the running time of the first algorithm (or give a polynomial upper bound for it). Otherwise, I doubt it's possible now. It's too bad that we can construct this algorithm for NP-complete problems, but not for co-NP-complete problems. --LC

But then my original statement stands: if we find a non-constructive proof of P=NP, then we still wouldn't have a polynomial algorithm for NP problems, since a polynomial algorithm has to stop after p(n) steps. --AxelBoldt