[–] SyzygyRhythm 3 days ago
The Exponential Time Hypothesis in effect says that 3SAT can only be solved by brute force--trying every possible combination until you find the right one. If your 3SAT expression has 100 variables, then this will take about 2^100 tries (exponential time).
What they found was an algorithm faster than brute force. 3SAT is still potentially one of these hard problems, but finding anything faster than brute force is a major advance.
[–] pcstl 3 days ago
NP-complete problems are essentially those where you need to make some kind of choice over a superlinearly-growing set of options where each option can be solved in polynomial time. Like, for example, finding a route between cities on a map under a certain total distance. You might need to go over all choices of routes.
The ETH essentially assumes that the best solution (in terms of complexity, i.e. how it grows) you can get for these kinds of problems is exponential. OpenAI just proved that this is false, you can have solutions that are below exponential complexity (but still above polynomial - if anyone finds a polynomial solution then we'd have proven that P=NP)
[–] eru 3 days ago
Seems like a weird way to describe NP, and I'm not even sure it's true?
NP is the class of decision problems where if the answer is YES and someone slips you a hint, you can solve them quickly (or quickly figure out that the hint was useless and report that). Where quickly means in polynomial time.
There's nothing about linear time in the classic definition.
[–] pcstl 3 days ago
NP = nondeterministically polynomial. i.e. if you could evaluate all choices in parallel (or just "know" the right choice), then you'd get a polynomial runtime. That's what I was trying to informally hint at, but I ended up using the wrong term. I'll edit my original post.
That's equivalent to your description - the "hint" is knowing which of the options to choose.
The reason why superlinearity comes into play is that if you had a linearly growing amount of solutions to test then you'd just have a sum of polynomials, which is itself polynomial.
[–] eru 3 days ago
One nitpick: I think your set of options needs to grow a lot faster than just superlinearly, I suspect it needs to grew superpolynomially?
However otherwise what you describe is essentially true of all problems in NP, not just the NP hard problems:
Sorting is (trivially) in NP, and you can say that to sort you need to pick one out of n! many permutations (which certainly grows superlinearly and even superpolynomially).
NP hard problems have the distinction that if you solve one of them efficiently, you can solve all of them efficiently, because you can reduce them to each other. (Basically because NP hard problems allow you to sort-of describe a computer, and reducing them to each other is equivalent to writing an interpreter or compiler.)
[–] dnautics 3 days ago
P means "this problem can be solved in polynomial # of steps relative to the problem size"
NP means "we don't know if the problem can be solved in polynomial # of steps relative to problem size, but we know we can check if the result is correct in a polynomial # of steps relative to the problem size"
!ETH: it was widely suspected that the flagship problem in NP, 3SAT, requires an exponential # of steps relative to problem size. We now know that 3SAT requires AT MOST something more than polynomial but strictly less than exponential to solve.
"P = NP" means "if you can check it in polynomial time, there's some way to solve it in polynomial time". "P != NP" means "there are some problems checkable in polynomial time for which there is no polynomial way to obtain the result"
Today's new proof makes "P = NP" "feel more likely", because we know that the flagship problem in NP is not exponential.
[–] tfpgh 3 days ago
Solving an arbitrary NP problem looks like first reducing that problem to SAT and then solving the equivalent SAT problem. We know (via the Cook–Levin theorem) that any NP problem can be reduced to SAT in polynomial time, and we now know that solving SAT is subexponential in terms of the size of the SAT problem, but it's possible for the conversion to SAT to grow the problem enough that it's exponential in terms of the original problem.
[–] dnautics 3 days ago
[–] jibalt 3 days ago
It seems to me that the logic only goes the other way ... if ETH had been proven then it would follow that P != NP. With !ETH we really don't know any more about P ?= NP then we did before. Reasoning that P = NP is "more likely" strikes me as a form of a fallacy of affirming the consequent.
[–] dnautics 2 days ago
[–] caughtinthought 3 days ago
Fwiw never thought it was a well-named conjecture
[–] pcstl 3 days ago
This proves that EXP != NP
[–] jibalt 3 days ago
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
[–] ehershey 3 days ago
[–] gberger 3 days ago
[–] newsy-combi 3 days ago
[–] pvab3 3 days ago
[–] nullc 3 days ago
A common complexity theory misunderstanding is that statements like this applies to all or even most instances-- and it leads to objectively bad engineering of the form "oh look this is actually 3sat, it's unsolvable".
(Great fodder for a midwit meme: idiot and genius say "my heuristics go brrr!"-- midwit says "No no no! NP problems can't be efficiently solved!")
Any specific instance may be efficiently solvable (and perhaps all instances coming from your particular problem set -- a subset of all possible 3sats).
So I would add to that paragraph "at least for some 3sat instances".
[–] marshray 3 days ago