> Even a P=NP result doesn't tell us that NP problems have efficient solutions.

Yes it does. That is literally exactly what it means. The class P is the class of problems which are considered theoretically "tractable"/"efficiently solvable"/"feasibly solvable" (Cobham-Edmonds thesis). Hence, if NP=P, then that same definition extends to all problems in NP.

There are very obviously algorithms in P which are not "efficient". For example, an algorithm in O(n^10^10^10) is not efficient in any reasonable sense. It is in fact much much much less efficient than O(e^n) for any n low enough that the whole thing would finish in under a year or so.

In practical terms, the class of efficient algorithms is probably O(n^3) at best, and even then assuming no huge constant factors.

I am not quite the mathematician to find out but I would love to know the relationship between the constant factor and the exponent in matrix multiplication research papers.

P vs NP is not an "in practical terms" question. It is a theoretical question with theoretical definitions of theoretical terms, including "efficient", which directly corresponds to the class P by definition.

Ok. When I say efficient, I mean "produces efficient code on near-term hardware". I understand that complexity theorists have a different definition of "efficient"-- they also have a different definition of "important" too.

The question being asked was "what would proving P=NP mean for us in practical terms". The fact that mathematicians call all polynomial-time algorithms efficient is irrelevant to this question.

> The question being asked

Where was that question asked?

It wasn't exactly a question, but the thread started by discussing practical implications:

> That said, there is definitely potential practical implications for this. Even if it means we can know np problems do not have efficient solutions [emphasis mine]

So, this was about efficiency in the practical sense, not some largely useless definition of efficiency by which galactic algorithms are "efficient".

That's not Cobham-Edmonds thesis. The assertion is that problems are feasible _only if_ they are in P, not _if and only if_ they are in P.

Sorry, but no. Edmonds clearly maps "efficiently-solvable problems" onto the notion of P in a complete way, not only as a subset of P.

I chased some links from Wikipedia and you're right that Edmonds uses "efficiently solvable" to mean P. However, he does not take "efficiently solvable" to mean "feasible" for basically the same reasons as I've said in this thread. From section 2 of Edmonds' "Paths, Trees, and Flowers" (1965):

> An explanation is due on the use of the words "efficient algorithm." First, what I present is a conceptual description of an algorithm and not a particular formalized algorithm or "code." For practical purposes computational details are vital. However, my purpose is only to show as attractively as I can that there is an efficient algorithm. According to the dictionary, "efficient" means "adequate in operation or performance." This is roughly the meaning I want—in the sense that it is conceivable for maximum matching to have no efficient algorithm. Perhaps a better word is "good."

> It is by no means obvious whether or not there exists an algorithm whose difficulty increases only algebraically with the size of the graph. The mathematical significance of this paper rests largely on the assumption that the two preceding sentences have mathematical meaning.

> ...

> When the measure of problem-size is reasonable and when the sizes assume values arbitrarily large, an asymptotic estimate of FA(N) (let us call it the order of difficulty of algorithm A) is theoretically important. It cannot be rigged by making the algorithm artificially difficult for smaller sizes. It is one criterion showing how good the algorithm is—not merely in comparison with other given algorithms for the same class of problems, but also on the whole how good in comparison with itself. There are, of course, other equally valuable criteria. And in practice this one is rough, one reason being that the size of a problem which would ever be considered is bounded.

You should read the rest of section 2, it's short very clear. Calling P the class of "efficiently solvable" problems is a completely reasonable framing for this paper, considering that this was written at a time when we had fewer tools to formally compare algorithms. Edmonds correctly does not claim that all P algorithms are feasible, and my opinion that P=NP is not important is based on the 60 years of feasible non-P computations we've had since.

I think even a negative solution could have practical implications. For starters you can now know for sure the problem can't be solved efficiently, which means you can stop looking and focus on other things. Possibly you can use it in situations you need a hard problem lkke in crypto (although there is more to it then that, since just because a problem is NP doesnt mean that a specific instance is)

I am not a complexity theorist, but ive always doubted the n^1000 solution to p=np being likely. Think about it - that essentially means you have to loop through the data 1000 times, but no more than that, no matter how much data. It just doesn't seem natural that looping 999 wouldn't be enough, but 1000 would be precisely enough. Its a sort of middle ground that seems like it would be extremely odd to me. Much more so than p=np with a low exponent or p!=np.

Apologies for nitpicking, but n^m doesn't mean m loops (ie n^1000 dient mean 1000 passes): that would be mn (1000n in your example). I think your intuition argument kind of breaks down there.

I meant to say 1000 nested loops. You're right to call me out on it though, as that is a huge difference in meaning.

I still think it would be bizarre to have to nest a loop n layers deep, where n-1 is not enough, but n layers is sufficient for really large n. Like what extra info would you get on the 1000th nested loop that you didn't on the first 999.

Of course there is nothing formal about this, it just feels like it would be wrong to me (and hence personally i would consider it the most interesting result). Of course gut feelings dont really count for much and i have nothing more than that.

I suppose my intuition is that increasing the exponent gives diminishing returns to how much more power you really get , so it doesn't make sense for problems to be in the n^1000 range. My gut feeling is they should either be easier or harder. I certainly can't think of very many non-exponential algorithms in the > n^50 range.

Not 1000 iterations of 1 loop, but 1000 loops nested inside each other.

> - having no other mitigating factors (requiring absurd amounts of space, for example)

I think you can't require absurd amounts of space, because you only have P steps to access that space, therefore the space is bounded to P anyway. (Memory you can't access can't help you.)