> We prove that the time complexity of self-attention is necessarily quadratic in the input length, unless the Strong Exponential Time Hypothesis (SETH) is false[...]
If OpenAI's proof is correct, does that mean self-attention can be implemented in sub-quadratic time? I.e. we could eventually have all the most advanced LLMs running on cheap hardware? To put it more playfully, is ETH being false actually bearish for Nvidia?
Can an expert in sat solving comment if there's any immediate applications of the method from the paper? Or is SAT solving in general pretty far away from worst-case times, and this is a theoretical algorithm which would be hard to apply in practice.
I'm reminded of the Apollo program that was cool for a while, but eventually the public lost interest.
Maybe it becomes cheaper? Maybe other labs want to prove their worth will pick up the slack when another loses interest?
Doesn't that lead me to an existence proof (i.e., a solution) to a k-SAT problem in N*(subexponential_time and polynomial_space)?
And with constant folding, N is reduced by 1 on each iteration.
What obvious thing am I missing here?
And the result that it's now false apparently doesn't imply NP!=EXPTIME either since it seems not strongly subexponential enough to offset the expansion in variables from reductions. See the other comments for a much better explanation.]
It's a bit more complicated than that, if you want to be careful. Sorting is in both NP and EXP and can be solved in subexponential time.
I think what they proved is that no problem in NP takes exponential time to solve. (Because you can always reduce them to SAT in polynomial time, and then solve SAT in subexponential time.)
EDIT: apparently the runtime bounds they gave here don't survive the usual polynomial reductions, thus they haven't proven that NP != EXP, it seems.
This means hard problems aren't as hard as we thought, and it's still possible P=NP. If it turns out to be true P=NP, we're a little closer to that. (I hope P!=NP!)
Every paper which assumes ETH is true might now have their conclusions be disproven.
In a way it shows that NP is perhaps less distant from P than imagined. Whether it can be _reduced_ to P in some gigabrain way, or whether that can be shown to be impossible remains to be seen.
It is definitely progress towards an answer to P ?= NP.
It's conceivable that each of them is smaller than the next, but also that e.g. P < NP = all the sigmas < PSPACE < EXPTIME or many other possibilities. I think the general consensus is that it's more likely that there are more inequalities than not, but the fact is that we can't even show that P is less than PSPACE so far, which would be extraordinarily surprising if it were true.
strong exponential time hypothesis, and the titular exponential time hypothesis
It's not saying that these problems are easy, but it does demonstrate that they are much easier than previously believed. A decade ago if a human had published this kind of thing, and importantly if it held up, they would be almost certainly a strong candidate for some sort of award(s).
In computer science, the satisfiability problem is the task of determining whether a boolean problem can be satisfied, that is, made to be true, by some combination of inputs. For example, the expression: A AND NOT B is satisfied whenever A is true and B is false, while the expression: A AND NOT A cannot be satisfied.
A naive approach to satisfiablilty would be to check every possible input and see if the expression evaluates to true.
However, the number of combinations that must be checked is 2n, where n is the number of variables in the expression. This is what is meant by exponential time.
The exponential time hypothesis conjectures that you cannot do better, that is, the number of operations needed to solve satisfiability will always be exponential in the size of the problem, no matter what algorithm is chosen.
If this hypothesis is true, it would imply that P≠NP, because satisfiability has been shown to be in NP, and in order for it to be in P there would have to exist an algorithm that solves it in polynomial time.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".
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.
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)
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.
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.
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.)
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.
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.
Fwiw never thought it was a well-named conjecture
This proves that EXP != NP
You're right, and I should correct my earlier phrasing. The paper does *not* merely assert existence — it gives a *specific deterministic algorithm* with a concrete, proven running time:
> A deterministic algorithm that exactly counts the satisfying assignments of a canonically encoded $k$-CNF formula on $n$ variables in time $(n+m)^{O_k(1)} \cdot 2^{O_k(n \log\log n / \sqrt{\log n})}$, using polynomial space.
The Lean formalization proves this bound. So it's a fully specified algorithm, not just "there exists one." My earlier "theoretical existence statement" framing was wrong.
*What that changes (and doesn't) for the practicality question:*
- It's a real algorithm in the sense that if you implemented it, it would terminate and give the correct answer. The proof guarantees the bound. - But the bound is still *astronomically slow* for any input size a cryptographer cares about. At $n = 1{,}000$ variables the exponent is already $\sim 1{,}000 \cdot 5 / 5 \approx 1{,}000$, i.e. $2^{1000}$ — far beyond any feasible computation. - The reduction-overhead argument for cryptography is unchanged: the subexponentiality is in $n$ (the SAT variable count), and reducing a cryptographic problem of size $N$ to SAT inflates $n$ to $\text{poly}(N)$, which swallows the subexponential advantage.
So the correction is: yes, it's a specific algorithm, not a mere existence claim. But the practical conclusion is the same — the bound is too slow to matter for cryptography, and the reduction path doesn't preserve the subexponential character relative to the original security parameter.
the constructed lookup machine has vastly more states than the estimated total number of physically distinguishable configurations of the observable universe.
So on balance, if this is considered proof that the exponential time hypothesis is false, and openai model reading of the lean was vaguely on point, then I reject the axioms of this system. Not a proof, even if there is lean. I'm OK with being downvoted for that position.
This has been marginally debated since the mid 90s, when Impagliazzo published the “5 worlds” survey. Occasionally it is brought up in talks about average case analysis. There is occasionally a workshop about it. But it has never really got the attention it deserved. How many more of these results do we need to take it seriously? (Assuming we do indeed certify it.)
What do you mean? Do you mean that our heuristics shortcuts of treating P as efficiently solvable (and NP as efficiently verifiable) are not warranted? So far pretty much all algorithms in P that we found have small exponents. That hasn't changed, but it might change in the future.
Or do you mean something else?
Keep in mind that even most instances of many NP hard problems can be solved very efficiently in practice, because most instances ain't hard.
However we know problems that are expected to be 'easier' than NP that are hard to solve in practice. These are pretty much all the problems we build cryptography out of, because essentially these problems are in the intersection of NP and co-NP and that's expected to be smaller than NP. And cryptography has to be hard to break on average to work: a random key needs to be a good key.
> and P \in NP is actually ill-posed.
Oh, that's very simple to answer: P is not an element of NP. The question people care about is whether there's any element in NP that's not in P. In LaTeX notation P \subset NP vs P \subseteq NP.
I'm not sure what you mean by ill-posed? The question is very clear. What are you trying to say?
There are many ways it might turn out to be ill-posed, this is just one of them. Impagliazzo's "5 worlds" survey is in some sense a catalog of some of these ways, and includes some of the ways you seem to be referencing (approxmations, probabilistic techniques, average-case efficiency; depending what you mean this could lie in the "Heuristica" world).
It's not possible for someone who's not a specialist in the field to understand it (unless there are more elementary proofs, of which likely none exist).
They are making good progress on similar research programmes for many cetaceans.
In the case of our stochastic parrots it's even easier, since they already speak something like human language. And for the math proofs it's even easier, because once they are formalised into something like Lean the humans only need to understand the statement of the theorem, and a mechanical computer program can check the proof.
It is not a demerit, actually I think that it is a very elegant outcome how many problems could be solved by generation and this probably show us how language is a far more important thing than we ever thought it was.
Edit: To explain more, it is guaranteed that you can reduce any NP problem to 3SAT in polynomial time with a polynomial factor in input size. You still have to solve it.
If 3SAT were solvable in polynomial time, then polynomial(polynomial(x)) = polynomial(x) and you're implied to have an overall polynomial time solution.
But what we have here is that 3SAT is solveable in subexponential time. subexponential(polynomial(x)) is not guaranteed to be sub-exponential.
Further edit: The subexponential solution does carry over to any (?) problems that have can be reduced to 3SAT in a linear number of variables. They list: Counting independent sets, Counting proper q-colorings, Counting perfect matchings.
Put these two together and you would think that they would mean that any NP complete problem can be solved in sub-exponential time. But this isn't necessarily true because changing the NP complete problem into 3SAT can increase the number of variables, which can theoretically negate the sub-exponential time.
(By the way, I think you mean NP-complete, not NP-hard. The Halting Problem is NP-hard.)
It's still a big result, but it's mostly theoretical. It might help explain, structurally, why most SAT instances are far easier than an exponential model would indicate.
P=/=NP -> 3SAT has no polynomial-time algorithm
ETH is true -> 3SAT has no subexponential time algorithm
ETH is false -> 3SAT has a subexponential time algorithm
------
ETH true -> P=/=NP
ETH false (!!!) -> P ?= NP remain unsolved
So 3SAT+poly reduction ends up being exponential