While this has no immediate bearing on P-vs-NP, it is interesting to note:
- Early CS researchers like Donald Knuth entered CS in the days when regularly someone would come up with a surprising faster algorithm for some problem, and their intuitions were that most problems had faster algorithms than you'd expect.
- Later generations of CS researchers, as all the easy algorithms were discovered, and post Karp's definition of NP-completeness (and the Garey & Johnson book, say), began to specialize in proving hardness, and complexity theory with a “zoo” of complexity classes became more about separation and lower bounds, e.g. I've seen algorithms like Arora's PTAS for Euclidean TSP being described as a “failed attempt to prove a lower bound”. So an entire generation of researchers' intuition was mostly in the direction of an elegant universe with neat separation results (even if we couldn't get them).
- So e.g. surveys of computer scientists showed most of them believing that P≠NP, and someone like Knuth has always seemed a bit of an outlier for saying that maybe P=NP: his argument was simply that there are so many possible polynomial-time algorithms that we haven't explored, and maybe one of them with a large exponent could solve NP-complete problems. (In response to arguments that many researchers have failed to find any such algorithms, he would say that many researchers have failed to prove that P≠NP.) E.g. in 2017 when Scott Aaronson wrote his survey (https://www.scottaaronson.com/papers/pnp.pdf = https://www.scottaaronson.com/papers/pnp-kindle.pdf) he wrote:
> Admittedly, it can be hard to tell whether declarations that P = NP are meant seriously, or are merely attempts to be contrarian. However, we can surely agree with Knuth and Lipton that we’re far from understanding the limits of efficient computation, and that there are further surprises in store.
- What we've seen in this month is surprising unexpected algorithms, like o(n log n) algorithms for FFT and integer multiplication, and now this sub–exponential-time algorithm for 3-SAT. (Yes the ETH has had less support than P≠NP but I still gather it's unexpected.)
So (though I don't expect intuitions on P≠NP to change, and probably that's still the right intuition… but I'm just an interested amateur), maybe there will be a bit of a swing of the pendulum back to the idea that maybe clever algorithms exist that we just haven't thought of.
There are hard problems where the worst case may be exponentially hard, but the average case is much faster. Linear programming and the traveling salesman problem are like that. The faster approaches are considered "probabilistic solutions".
In cryptography, most of the problems believed to be exponentially hard, such as factoring, don't have proofs of an exponential lower bound. There's some risk of a breakthrough that makes those problems easier. A probabilistic solution would be good enough to break many cryptosystems.
I suspect that people at the cryptanalysis agencies worldwide are staying up late tonight, trying to figure out the impacts and risks of this new result.
Funny you mention probabilistic algorithms. If I’m not mistaken one of the OpenAI results is that most of the probabilistic methods (specifically those that use a logarithmic amount of space)-4 have non-probabilistic versions that are as efficient in time and space (L = RL = BPL)
> In fact, one of the first things I wanted OpenAI to check is that no existing cryptosystems are compromised by the result. Apparently they are safe...
so if you trust his trust of OpenAI, then this has no (immediately obvious) bearing on cryptosystems for now.
Seeing the panicked state of the PHP ecosystem at the moment, and having worked in the financial system in the past, I dread the thought of what would happen to global banking if a frontier model broke a maths problem that underpins our crypto systems...
Part of the problem with the current centralization of the world's math capability within AI labs is that if they do find that P=NP via an efficient algorithm, it could have implications on all modern public key based cryptography. So it's not completely clear that they will (be allowed to) release these results.
> if AI frontier labs do find that P=NP via an efficient algorithm, it could have implications on all modern public key based cryptography.
The fact that other math problems are bulk solved yet nothing from cryptography was released/published as solved could be a strong indicator that "all modern public key based cryptography" could have been at risk.
It seems rather unlikely that solving the Navier-Stokes problem is harder than e.g. a better attack on a reduced-round AES. Perhaps they think an epsilon improvement in the exponent isn't worth publishing, but it seems more likely that they're intentionally withholding any results they have.
It's a reasonable assumption that the cases where the NSA, etc, had a role in designing cryptographic systems, and reportedly made specific suggestions that could be forms of a backdoor, would all be revealed by AI being tasked to find it. Maybe even indirectly as part of the broader defensive security research by models unreleased to the public. If algorithmic complexity or a really obscure backdoor has existed in cryptographic systems and put there by the Gov't, it seems that current frontier models would have the best chance of revealing it.
Either that or they've not put the model(s) on those problems. Remember that even the Navier-Stokes solution did use a large amount of compute. I'm not sure they'd even want to waste it on problems that they can't get PR from.
I think a proof of P=NP via a galactic algorithm is far more likely than a proof of P=NP via an efficient algorithm.
I mean if someone develops an O(n^100) algorithm for solving an NP-complete problem, that would prove P=NP, but pose no immediate threat to cryptography.
It increases risk, in that a galactic algorithm increases the odds that an efficient algorithm exists waiting to be found, but it is entirely (epistemically) possible that the galactic algorithm exists but the efficient algorithm doesn't.
Maybe, but that feels significantly more exotic and unlikely to me. You have to basically negate the natural cost scaling of brute force in the thousand to million n range but lose that speed benefit later.
We had O(n lg n) since 2019 already; now we (supposedly) know the limit is less than that (but of course at least linear).
> In 2019, Harvey and van der Hoeven came up with a galactic algorithm with complexity O ( n log n ) {\displaystyle O(n\log n)}. This matches a guess by Schönhage and Strassen that this would be the optimal bound.
> In September 2026, OpenAI released a preprint claiming an algorithm with complexity O ( n ( log n ) α ) {\displaystyle O(n(\log n)^{\alpha })} with α < 1 {\displaystyle \alpha <1}
I believe this stems from an incomplete understanding of the problem space on my part (excuse the multiple puns); but, I am skeptical Knuth really said P=NP. There are so many mathemtical examples of equations that clearly show P≠NP. Computing in itself was originally formed from using iterative methods to solve hard problems. Least-squares methods, sin() cos(), traveling salesman, so many to list? To say P=NP for all problems seems like a fools errand to me.
> Don Knuth: As you say, I've come to believe that P = N P, namely that there does exist an integer M and an algorithm that will solve every n-bit problem belonging to the class N P in n**M elementary steps.
This is an amazing and unexpected result. To be clear though, it does not imply that NP is subexponential: "Note that runtime bounds of the form 2^o(n) do not transfer under general polynomial-time reductions (2^n^o(1) would be needed for this), so our result does not imply that general NP problems
admit subexponential time algorithms."
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.
> 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.
Wouldn't that also include functions that can be evaluated in hardware using a fixed number of gates? E.g., symmetric cryptography?
A circuit with a fixed number of (binary logical) gates can be solved in constant time, as it will have a fixed number of input bits, so brute force will have a fixed number of input options to consider. This does apply to cryptographic algorithms with a fixed key size, like the ones used in practice, as well.
The whole point of computational complexity is to look at the scaling behavior when you increase the inputs/key size/size of the circuit.
From my reading no. Because brute forcing anything symmetrical is already O(n). What protects us is the assumption that there is not enough energy in the observable universe to brute force 2^256 key.
It's NP-complete problems if "subexponential" means O(2^{n^{\epsilon}}) for every \epsilon>0.
If subexponential just means 2^{o(n)}, (as it is here, O(2^{n/\sqrt{\log n}})), then no. E.g. if the cost of the reduction took you from n to \sqrt{\log n}n, then you'd be back to exponential time. The Subexponential Reduction Family is the class of problems that preserve subexponential time in the second sense.
After some back and forth with AI, I think I understand:
1. The Exponential Time Hypothesis being false shows that 3SAT specifically is sub-exponential time
2. Any NP complete problem can be changed into 3SAT in polynomial time
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.
The halting problem is NP hard, technically speaking, but it's much harder than that - it's uncomputable. I don't think it's common to refer to it as NP-hard because it's confusing.
"Any NP complete problem can be changed into 3SAT in polynomial time" - if I remember correctly, turning an NP-complete problem into 3SAT is not part of the problem statement - it just means that when working with NP-complete problems, they are always translated to 3SAT first, and whenever someone is trying to prove something they are working against 3SAT.
No I get that, but I don't get how proving 3SAT can be proved in subexponential time does not imply all NP-complete problems can be as well, since they can all be reduced to 3SAT in polynomial time.
Suppose 3sat is solved in 2^sqrt(n). That's subexponential. An n^2 reduction from sat to 3sat is polynomial. However, an algorithm using this reduction with the discovered solution, would run in (2^sqrt(n))^2 which is just 2^n.
That last statement is not correct. (2^sqrt(n))^2 equals 2^(2 * sqrt(n)). If you just take n=1 in your expression you can see the error, your (2^sqrt(n))^2 yields 4, while 2^n is just 2.
Suppose 3SAT can be solved in 2^(n^0.9) time (this is not the actual bound in paper). And another NP problem can be reduced to 3SAT in polynomial time, but the reduction increases the problem size from n to n^2. Then after the reduction the other NP problem can solved in 2^(n^1.8) time, so it does not mean all np problems are sub exponential.
The algorithm's runtime is in faster growing than O(exp(n^a)) for any a < 1. It's not subexponential in the sense that factoring is, which I believe has a secondary exponent of 1/3 under the best algorithms.
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.
For example, a 2^√n subexponential algorithm then used on an input after a n^2 polynomial reduction from some other NP-complete language incurs a total running time of 2^√(n^2) = 2^n, which again is exponential.
Wow. Complexity class separations are really really rare.
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.
SAT-solving expert here. No immediate applications (the results in the paper are "galactic"). The main question is whether some ideas here will snowball into other ideas, one of which could eventually be very practical.
This is a complete non-sequitur, but I checked out your web page. I love the Curious George image. My daughter has been deeply into the Curious George episodes for the past couple of months, and I don't think I've seen him show up in many other places besides kids' shows :D
Galactic algorithms have such huge constants they figuratively need a galaxy-sized computer and/or galaxy-sized problem, to have any advantage over the normal solutions. Imagine a generic sorting algorithm in O(n) that had a constant of 10^92 **. Technically superior but in practice unusable.
There are realistic situations where galactic O(n) sorting algorithms exist, right? 32-bit integers can be sorted in O(n) using 2^32 words of memory, using bucket sort[1].
If objects have a maximum size of 4 GB (= 2^35 bits), then arbitrary objects can be sorted in O(n) using 2^2^35 bits of memory.
You can use Radix sort https://en.wikipedia.org/wiki/Radix_sort to lower the memory requirement for limited bit-width integers. But there are more practical o(n log n) methods for sorting integers.
Well optimized radix sorts are actually pretty fast (assuming you have the proper fallbacks to quicksort/insertion sort to deal with medium/small sizes)
> A galactic algorithm is an algorithm with state-of-the-art theoretical (asymptotic) performance, but which is not used due to practical constraints.
Yes, I think we can pretty safely say that the algorithm in the paper is not used due to practical constraints. This is a subjective judgment, but pretty clear from the paper that the algorithms aren't practical to apply.
The data you put into a SAT solver has substantial effect on it's performance in practice. Real world inputs tend to be much nicer than pathological ones, so I don't think there's much practical application, but I also haven't read the paper yet.
It's surprising that Exponential Time Hypothesis would be false. Quoting a paper from 2022[1]:
> 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?
That makes me wonder if "I see your CS proof, how does this affect LLMs / nvidia stonk price???" being immediately self-described as "playful" ensures criticism looks cranky.
"I see your particle physics paper (on sustainability of neutron-induced chain fission reaction in uranium), how does this affect the world politics???"
Turns out, it affected the global politics tremendously.
I’m shaking, so much has happened in the past few years, I think it’s inevitable we will find the efficient algorithm for P=NP. Not sure if I’m ready to live in that world yet.
I think that statement is too strong -- if ETH was a core belief then its implication P≠NP must also have been a core belief ... but I don't think that those who assumed ETH held it that strongly. As you say, it was commonly thought to be true ... as was P≠NP ... and the latter still is. But is not a "core belief" -- I think virtually everyone believes that they could be wrong, whichever side they're on.
Uh no, reinforcemnt learning doesn't magically pop out math proofs. OpenAI has a ton of mathematicians on staff and is spending millions of dollars in compute to pursue these solutions.
It would be interesting to see if this new counting method has any implications to counting 2SAT or parity-counting of 2SAT. Since problem posed in XORSAT and 2SAT clauses is NP-complete, this could have implications about P vs NP.
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.
> the number of operations needed to solve satisfiability will always be exponential in the size of the problem, no matter what algorithm is chosen
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".
There's a set of problems that are (it is conjectured) easy to verify but hard to solve. One of them, called 3SAT, looks for values (a, b, c, ...) such that an expression like ((a or b or c) and (b or not-c or e) and ...) evaluates to true. And it turns out that all of these supposedly hard problems can be converted to 3SAT, so if you find a solution to one you find a solution to all.
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.
The exponential time hypothesis is essentially that NP-complete problems' best solutions are all exponential.
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)
> 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 linear time.
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.
Sorry, you're correct. I meant "polynomial" rather than linear.
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.
Ok, I read your edited comment. Looks a lot better know.
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.)
the language around complexity domains is really confusing.
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.
This isn't quite true, problems in NP can still require exponential time to solve.
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.
But even if every problem in NP is sub-exponential (which, as tfpgh notes, this result doesn't prove), how does that make it "more likely" that every such problem is solvable in a polynomial # of steps? I feel like it's a bit like claiming that, because Alex Honnold has shown that a human can free solo El Capitan in less than 4 hours, that makes it "more likely" that I could. To me, his success does not budge the latter probability at all.
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.
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
[deleted, originally wrote the hypothesis being true would imply NP=EXPTIME which is wrong, since there can be problems in EXPTIME which aren't in NP.
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.]
> Instead we now know that NP is a strict subset of EXP (there are problems in NP that can be solved in subexponential time).
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.
I mean the result that k-SAT cannot be solved in 2^{o(n)} clearly would imply P!=NP, but how does ETH imply NP = EXP? And how does \neg ETH imply NP != EXP?
If I am understanding this correctly (and I am still reading this!) and if this is true, this means NP is sub-exponential. Sub-exponential does not mean P=NP. But yes, this is huge if true.
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.
See this from the paper: "Note that runtime bounds of the form 2^o(n) do not transfer under general polynomial-time reductions (2^n^o(1) would be needed for this), so our result does not imply that general NP problems
admit subexponential time algorithms."
It's not clear what you're trying to say or the relevance. The fact remains that "NP is always poly time reduceable to 3SAT" and "3SAT is sub-exponential" does not imply that NP is sub-exponential.
right, but anything linearly reducable to kSAT is trivially sub-exponential, and a discovery like this means that linear reductions are a whole lot more important than they used to be.
Also for NP problems that can't linear reduce to kSAT just having the knowledge that SAT is sub-exponential (and the tools), makes it a lot more likely that we see a whole lot more problems become exponential.
practically, a poly-time reduction can increase teh size of hte problem instance. So maybe you transform a 3-color instance (graph on n vertices) into a n^2 sized SAT instance. You can then solve this in time 2^{n^2 / sqrt{log{n^2}}}, but this is not sub-exponential.
Your "No" is wrong. As he said, "Conversions to other problems can still end up exponential", so (NP is poly time reduceable to 3SAT + 3SAT is sub-exponential) does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
This defeats a few hypotheses that were widely relied on.
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).
Assuming P != NP (and that this NP != EXP proof holds up), what's the next most "natural" candidate for how fast NP must grow? This paper appears to show Omega(2^(n/sqrt(log(n))) with a couple additional fudge factors which would be a very disappointing answer.
This is not a proof of NP ≠ EXP. NP ≠ EXP would mean that there is a problem that can be solved in exponential time, but not in non-deterministic polynomial time. That’s completely different from what they have shown.
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
Even if all NP problems could be solved in sub-exponential time, it would not imply that NP ≠ EXP. It seems that many people here are confused about what EXP means – it’s the set of all problems that can be solved in at most exponential time, including those in P.
Predictions markets are terrible mechanisms for this, because they require imposing an arbitrary time limit at which point the prediction will pay out. IOW, even if you religiously believed P=NP, but you think it would take at least two years to prove it, you would vote against the prediction if the deadline were one year.
This proves that EXP != NP, which was a stronger claim than P != NP.
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.
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
There's a chain of classes P, NP(aka Sigma_1), Sigma_2, .., Sigma_n, ... PSPACE, EXPTIME. What we know for sure is that P < EXPTIME, others are unknown (well, maybe this paper adds something to this, time will tell).
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.
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
Double-checking, is this more like the recent DFT improvement where even the word-RAM model of computation doesn't justify the fraction of a log factor they eliminated because of an assumption of arbitrary-precision arithmetic, or has ETH actually fallen with the Lean proof being formulated appropriately? Do we know yet?
That doesn't sound right. Previous results proved that with bounded values you couldn't beat O(nlogn), among other things. The OpenAI result explicitly used arbitrary-precision complex arithmetic. Yes, it's galactic [0], but either it's wrong or the previous results were wrong, no?
[0] Very much not something I care much about in these sorts of conversations. Even if it weren't galactic, a random logn factor here and there usually loses to cache efficiency and whatnot. Their speedups were less than that, and I'd already assumed we were in the realm of applied math rather than real engineering.
> but either it's wrong or the previous results were wrong, no?
Not quite. This result uses polylog(n) precision which isn't bounded, so it escapes the previous bounds. This doesn't noticably affect the runtime because not all of the math needs this extra precision (in this case you need a couple computations in higher precision but less than n operations), so the extra precision can be payed for even in a TM/circuit model.
no. The DFT result as published uses unlimited precision, but the comparison with the int multiplication shows that the unlimited precision just makes the paper shorter/easier rather than being an exploit in the computation model.
No practical implications, we dont use 10^87 variables (a point where the proposed new method surpasses the already existing sat solvers), however a good theoritical point.
This construction splits the space into 'small' models with a quadratic lookup cost and 'large' with sub-exponential cost. The small one is by enumerating all SAT problems with their answers, then looking up the answer. The large one is something in polynomials I don't understand. An answer from openai about where the cutoff is was:
the constructed lookup machine has vastly more states than the estimated total number of physically distinguishable configurations of the observable universe.
I'm not inclined to consider "wrote down all the answers ahead of time, so that's free" as a quadratic time solution to SAT. Nor very convinced by the other half.
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.
If I can "count the satisfying assignments of a k-CNF in subexponential time and polynomial space", could I just count the the solutions for assignments of 0 and 1 to x_1, then proceed with whichever is greater (or nonzero), counting the solutions for x_2=0, and so on.
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 about OpenAI allowed ETH to be determined to be false that couldn't be done by humans? Was there something in particular that OpenAI's AI did that help bridge the gap to a solution?
Perhaps this will be the push we need to finally, seriously consider the possibility that our formalisms of “efficient” and “efficiently verifiable” algorithms correspond to nothing resembling reality, and P \in NP is actually ill-posed.
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.)
> Perhaps this will be the push we need to finally, seriously consider the possibility that our formalisms of “efficient” and “efficiently verifiable” algorithms correspond to nothing resembling reality,
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?
"Ill-posed" here means that the formalisms we've chosen to represent "efficient" and "efficiently verifiable" algorithms could turn out to not correspond to anything humans would consider meaningful ("we posed the problem poorly"). For most people, a solution to an NP-complete problem that's barely-subexponential would be clear evidence of these formalisms being ill-posed.
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).
> "Ill-posed" here means that the formalisms we've chosen to represent "efficient" and "efficiently verifiable" algorithms could turn out to not correspond to anything humans would consider meaningful ("we posed the problem poorly").
OK, that's a reasonable worry and one that practitioners share. Thanks for explaining. Though I don't think ill-posed is a good term for this, the technical meaning of the term points in a different direction.
So far, as far as I can tell this new result gives much of any extra credence to that worry.
It is completely acceptable in math to use "ill-posed" to indicate that something is insufficiently specified, improperly formulated, or ambiguous. Math people do this all the time.
I'm not the kind of programmer that even understands P=NP, so I'll either be way off base, or be blessed with an outsider perspective.
Considering that no human will be able to sift through these in no less than a team-decade, to verify that there's no mistake, yet still that will be massively less compute than the one necessary to produce the 'solution', isn't the empirical finding contrary to the theoretical finding being suggested? Namely that in practice we are generating these 'proofs' in a massive amount of compute time, but it will take tiny amounts of time to verify them.
Is it very useful to talk about computing in purely mathematical terms and not in empirical terms? There's very obvious truths that we are seeing here, and it seems like those that go through some sort of rigorous mathematical training kind of fail to see it. Similar to how 70% of physicians or something believe in the Many Worlds Interpretation or that time isn't real, a sort of algebraic psychosis where one is removed from a human reality that we have access to in order to attempt getting close to an 'objective reality' that is far removed from the human experience.
To some extent we are computing entities, so I don't even think that we can conceive of computing as decoupled from ourselves[1]. I know the observer effect is very often mis-cited by laymen, but I'll be damned if some computer scientists like Godel don't sound like they were up to something similar to this.
Also, here's another outsider perspective. This is obviously slop, I'm not sure how but it's obviously nerd-sniping the CS majors that understand this, because it's obviously very hard to find the flaw, but it gives epicycles energy. It's also sufficiently complex that under a bayesian analysis it is more likely that the AI has reached the limit of its processing power until it becomes confused enough to believe it has reached something significant, kind of like how I'm doing now.
> Considering that no human will be able to sift through these in no less than a team-decade
How do you figure that? For ETH, OpenAI is engaging with top experts in this area. It's unlikely that they'll need 10 man-years combined to go through the results:
> This work will be verified and shepherded by a group led by Russell Impagliazzo (UC San Diego), Ramamohan Paturi (UC San Diego), and Ryan Williams (MIT). All three are experts in algorithm design and complexity theory.
> Impagliazzo and Paturi, along with Francis Zane, are the original proposers of the Exponential Time Hypothesis, and have authored a number of influential papers on the topic over the last 30 years.
> Williams has refuted strong versions of the ETH, used ETH for hardness results, and connected algorithms and lower bounds.
I think I misread something, the pdf is like 80 pages long, and there’s a list of like 200 pdfs, each around that length, I thought they were all related to this proof, but they are unrelated.
Not a math major here. But is this paper way too long? So long that it raises the possibility that OpenAI's staff mathematicians passed through AI's raw output as-is?
We can record a lot of parrot speech and train LLMs against it and eventually build a translator.
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 just that we never considered how advanced stochastic parrots could be so powerful when dealing with basically the whole library of acquired human knowledge and guided by experts to generate.
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.
Is this really as interpetable math is fundamentally. Like wtf is this accessibility? who is this for? Do top mathematicians read this like a chapter book? Absolutely ridiculous they have all this compute and creative resources and cant make something more intuitive to help people understand it more natively
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).
Even mathematicians take a long time to read math books and articles, especially if they go into the details. They don't read math like you would read a novel.
> Absolutely ridiculous they have all this compute and creative resources and cant make something more intuitive to help people understand it more natively
I agree with you insofar that the primary motivator for these press releases is public relations and investor confidence, so it should be made as minimally hard to understand as possible just by sheer virtue that it exposes oAI to more cash thrown at them.
I totally disagree with you on the precept that all difficult to understand things need be needlessly dumbed-down/watered-down until palatable for a mass (average) audience. Requiring 'math50' before 'trig101' isn't gate-keeping in any moral sense, it's an effort to save everyones' time involved when the effort is futile.
Similarly I don't demand that haute couture fashion companies explain the style to me until 'I get it' ; I just take it for granted that i'm not going to be able to understand the entirety of the world in one lifetime.
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.
Humans in this thread have already explained that concept with much more straightforward prose.
From the HN comment guidelines:
> Don't post generated text or AI-edited text. HN is for conversation between humans.
Disclosure doesn't address the objection (not that it wouldn't be obvious anyway). If you learned something from an AI convo, tell us what you learned, in your own words, not what the LLM said.
While this has no immediate bearing on P-vs-NP, it is interesting to note:
- Early CS researchers like Donald Knuth entered CS in the days when regularly someone would come up with a surprising faster algorithm for some problem, and their intuitions were that most problems had faster algorithms than you'd expect.
- Later generations of CS researchers, as all the easy algorithms were discovered, and post Karp's definition of NP-completeness (and the Garey & Johnson book, say), began to specialize in proving hardness, and complexity theory with a “zoo” of complexity classes became more about separation and lower bounds, e.g. I've seen algorithms like Arora's PTAS for Euclidean TSP being described as a “failed attempt to prove a lower bound”. So an entire generation of researchers' intuition was mostly in the direction of an elegant universe with neat separation results (even if we couldn't get them).
- So e.g. surveys of computer scientists showed most of them believing that P≠NP, and someone like Knuth has always seemed a bit of an outlier for saying that maybe P=NP: his argument was simply that there are so many possible polynomial-time algorithms that we haven't explored, and maybe one of them with a large exponent could solve NP-complete problems. (In response to arguments that many researchers have failed to find any such algorithms, he would say that many researchers have failed to prove that P≠NP.) E.g. in 2017 when Scott Aaronson wrote his survey (https://www.scottaaronson.com/papers/pnp.pdf = https://www.scottaaronson.com/papers/pnp-kindle.pdf) he wrote:
> Admittedly, it can be hard to tell whether declarations that P = NP are meant seriously, or are merely attempts to be contrarian. However, we can surely agree with Knuth and Lipton that we’re far from understanding the limits of efficient computation, and that there are further surprises in store.
- What we've seen in this month is surprising unexpected algorithms, like o(n log n) algorithms for FFT and integer multiplication, and now this sub–exponential-time algorithm for 3-SAT. (Yes the ETH has had less support than P≠NP but I still gather it's unexpected.)
So (though I don't expect intuitions on P≠NP to change, and probably that's still the right intuition… but I'm just an interested amateur), maybe there will be a bit of a swing of the pendulum back to the idea that maybe clever algorithms exist that we just haven't thought of.
There are hard problems where the worst case may be exponentially hard, but the average case is much faster. Linear programming and the traveling salesman problem are like that. The faster approaches are considered "probabilistic solutions".
In cryptography, most of the problems believed to be exponentially hard, such as factoring, don't have proofs of an exponential lower bound. There's some risk of a breakthrough that makes those problems easier. A probabilistic solution would be good enough to break many cryptosystems.
I suspect that people at the cryptanalysis agencies worldwide are staying up late tonight, trying to figure out the impacts and risks of this new result.
Funny you mention probabilistic algorithms. If I’m not mistaken one of the OpenAI results is that most of the probabilistic methods (specifically those that use a logarithmic amount of space)-4 have non-probabilistic versions that are as efficient in time and space (L = RL = BPL)
For linear programing, the interior point method has a polynomial worst case.
It also not probabilistic, so linear programing is not rely a good example
Incidentally, Ryan Williams says (https://x.com/rrwilliams/status/2114891838377320784):
> In fact, one of the first things I wanted OpenAI to check is that no existing cryptosystems are compromised by the result. Apparently they are safe...
so if you trust his trust of OpenAI, then this has no (immediately obvious) bearing on cryptosystems for now.
Not by this result. But it opens up a new area of problems to hammer on that might get somewhere.
Seeing the panicked state of the PHP ecosystem at the moment, and having worked in the financial system in the past, I dread the thought of what would happen to global banking if a frontier model broke a maths problem that underpins our crypto systems...
Part of the problem with the current centralization of the world's math capability within AI labs is that if they do find that P=NP via an efficient algorithm, it could have implications on all modern public key based cryptography. So it's not completely clear that they will (be allowed to) release these results.
> it could have implications on all modern public key based cryptography
Here's a nice low-budget scifi movie about this:
https://en.wikipedia.org/wiki/Travelling_Salesman_(2012_film...
https://www.youtube.com/watch?v=OgeyHkP6i3g (full movie)
This is a book about this:
https://www.goodreads.com/book/show/37986693-factor-man
That premise is great.
> if AI frontier labs do find that P=NP via an efficient algorithm, it could have implications on all modern public key based cryptography.
The fact that other math problems are bulk solved yet nothing from cryptography was released/published as solved could be a strong indicator that "all modern public key based cryptography" could have been at risk.
Or it could indicate that more resources have been put into cryptography than most other math fields, so AI hasn't yet caught up to our progress.
It seems rather unlikely that solving the Navier-Stokes problem is harder than e.g. a better attack on a reduced-round AES. Perhaps they think an epsilon improvement in the exponent isn't worth publishing, but it seems more likely that they're intentionally withholding any results they have.
It's a reasonable assumption that the cases where the NSA, etc, had a role in designing cryptographic systems, and reportedly made specific suggestions that could be forms of a backdoor, would all be revealed by AI being tasked to find it. Maybe even indirectly as part of the broader defensive security research by models unreleased to the public. If algorithmic complexity or a really obscure backdoor has existed in cryptographic systems and put there by the Gov't, it seems that current frontier models would have the best chance of revealing it.
Either that or they've not put the model(s) on those problems. Remember that even the Navier-Stokes solution did use a large amount of compute. I'm not sure they'd even want to waste it on problems that they can't get PR from.
I think a proof of P=NP via a galactic algorithm is far more likely than a proof of P=NP via an efficient algorithm.
I mean if someone develops an O(n^100) algorithm for solving an NP-complete problem, that would prove P=NP, but pose no immediate threat to cryptography.
It increases risk, in that a galactic algorithm increases the odds that an efficient algorithm exists waiting to be found, but it is entirely (epistemically) possible that the galactic algorithm exists but the efficient algorithm doesn't.
And vise versa, a proof that P!=NP do not exclude an algorithm that is fast enough for practical n.
Maybe, but that feels significantly more exotic and unlikely to me. You have to basically negate the natural cost scaling of brute force in the thousand to million n range but lose that speed benefit later.
The 2012 movie "Travelling Salesman" is a thought provoking film that explores this scenario.
We had O(n lg n) since 2019 already; now we (supposedly) know the limit is less than that (but of course at least linear).
> In 2019, Harvey and van der Hoeven came up with a galactic algorithm with complexity O ( n log n ) {\displaystyle O(n\log n)}. This matches a guess by Schönhage and Strassen that this would be the optimal bound.
> In September 2026, OpenAI released a preprint claiming an algorithm with complexity O ( n ( log n ) α ) {\displaystyle O(n(\log n)^{\alpha })} with α < 1 {\displaystyle \alpha <1}
https://en.wikipedia.org/wiki/Multiplication_algorithm
That said, it's interesting to contemplate that there is "space between" polynomial and exponential time complexities.
> “We had O(n lg n) since 2019 already”
Yes, svat’s comment accounts for that by using little-o notation* in o(n log n). This was only achieved in the 2026 result.
* https://en.wikipedia.org/wiki/Little_o_notation
I believe this stems from an incomplete understanding of the problem space on my part (excuse the multiple puns); but, I am skeptical Knuth really said P=NP. There are so many mathemtical examples of equations that clearly show P≠NP. Computing in itself was originally formed from using iterative methods to solve hard problems. Least-squares methods, sin() cos(), traveling salesman, so many to list? To say P=NP for all problems seems like a fools errand to me.
See his answer to the seventeenth question in this interview:
https://www.informit.com/articles/article.aspx?p=2213858&WT....
> Don Knuth: As you say, I've come to believe that P = N P, namely that there does exist an integer M and an algorithm that will solve every n-bit problem belonging to the class N P in n**M elementary steps.
Do you understand what P and NP mean?
This is an amazing and unexpected result. To be clear though, it does not imply that NP is subexponential: "Note that runtime bounds of the form 2^o(n) do not transfer under general polynomial-time reductions (2^n^o(1) would be needed for this), so our result does not imply that general NP problems admit subexponential time algorithms."
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.
> 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.
Wouldn't that also include functions that can be evaluated in hardware using a fixed number of gates? E.g., symmetric cryptography?
A circuit with a fixed number of (binary logical) gates can be solved in constant time, as it will have a fixed number of input bits, so brute force will have a fixed number of input options to consider. This does apply to cryptographic algorithms with a fixed key size, like the ones used in practice, as well. The whole point of computational complexity is to look at the scaling behavior when you increase the inputs/key size/size of the circuit.
From my reading no. Because brute forcing anything symmetrical is already O(n). What protects us is the assumption that there is not enough energy in the observable universe to brute force 2^256 key.
> To be clear though, it does not imply that NP is subexponential
Is there a subset of NP problems for which a subexponential algorithm would witness the entire NP class being subexponential?
~~Yeah, that's NP-complete problems.~~
EDIT: DISREGARD THAT I WAS MISTAKEN
It's NP-complete problems if "subexponential" means O(2^{n^{\epsilon}}) for every \epsilon>0.
If subexponential just means 2^{o(n)}, (as it is here, O(2^{n/\sqrt{\log n}})), then no. E.g. if the cost of the reduction took you from n to \sqrt{\log n}n, then you'd be back to exponential time. The Subexponential Reduction Family is the class of problems that preserve subexponential time in the second sense.
That is unfortunately not the case
Can you explain more?
After some back and forth with AI, I think I understand: 1. The Exponential Time Hypothesis being false shows that 3SAT specifically is sub-exponential time 2. Any NP complete problem can be changed into 3SAT in polynomial time
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.
Oh ok, I understand what you mean.
(By the way, I think you mean NP-complete, not NP-hard. The Halting Problem is NP-hard.)
The halting problem is NP hard, technically speaking, but it's much harder than that - it's uncomputable. I don't think it's common to refer to it as NP-hard because it's confusing.
Thanks for the correction! I updated my answer to fix that :)
"Any NP complete problem can be changed into 3SAT in polynomial time" - if I remember correctly, turning an NP-complete problem into 3SAT is not part of the problem statement - it just means that when working with NP-complete problems, they are always translated to 3SAT first, and whenever someone is trying to prove something they are working against 3SAT.
oh that actually helps a lot. I was tying myself in knots trying to think it through.
P=NP -> 3SAT has a polynomial time algorithm
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
No I get that, but I don't get how proving 3SAT can be proved in subexponential time does not imply all NP-complete problems can be as well, since they can all be reduced to 3SAT in polynomial time.
Suppose 3sat is solved in 2^sqrt(n). That's subexponential. An n^2 reduction from sat to 3sat is polynomial. However, an algorithm using this reduction with the discovered solution, would run in (2^sqrt(n))^2 which is just 2^n.
(2^sqrt(n))^2 = 2^sqrt(n) * 2^sqrt(n) = 2^(sqrt(n) + sqrt(n)) = 2^(2 sqrt(n))
but
2^(sqrt(n)^2) = 2^n
That last statement is not correct. (2^sqrt(n))^2 equals 2^(2 * sqrt(n)). If you just take n=1 in your expression you can see the error, your (2^sqrt(n))^2 yields 4, while 2^n is just 2.
Yeah but you're fixing the wrong thing. The problem after reduction to 3sat is (2^sqrt(n^2)). The altered n becomes the input to 3sat.
This also means your other comment here is wrong to suggest any error: https://news.ycombinator.com/item?id=50254142
...that's...not how exponents work
The result is right, they wrote the parentheses in the wrong spot, don't be so shocked.
Suppose 3SAT can be solved in 2^(n^0.9) time (this is not the actual bound in paper). And another NP problem can be reduced to 3SAT in polynomial time, but the reduction increases the problem size from n to n^2. Then after the reduction the other NP problem can solved in 2^(n^1.8) time, so it does not mean all np problems are sub exponential.
Hey, this is a common error. (2^(n^0.9))^2 is not 2^(n^1.8) but 2^(2 * n^0.9).
Basically 3SAT algo isn't sub-exponential enough to absorb the arbitrary poly cost for the reduction
So 3SAT+poly reduction ends up being exponential
Gotcha, thanks
To achieve this you would need to show that 3SAT is solvable in 2^{n^c} time for every c>0, not just a single c<1.
In this case they have c = 1-0.5(\log \log n)/(\log n) which is approaching 1 for large n.
Misunderstood your question sorry, I have no idea!
The algorithm's runtime is in faster growing than O(exp(n^a)) for any a < 1. It's not subexponential in the sense that factoring is, which I believe has a secondary exponent of 1/3 under the best algorithms.
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.
For example, a 2^√n subexponential algorithm then used on an input after a n^2 polynomial reduction from some other NP-complete language incurs a total running time of 2^√(n^2) = 2^n, which again is exponential.
Yep! Or for a slightly generalised example, if we pick our sublinear function of n to be n^(1-epsilon) for some small epsilon > 0
then 2^(n^(1-epsilon)) belongs to 2^o(n)
Suppose p(n) is some polyonomial in n, our polynomial reduction. Let's make it a monomial for simplicity, p(n) = n^a
Does 2^(p(n)^(1-epsilon)) necessary belong to 2^o(n)?
No, that only holds if (n^a)^(1-epsilon) = n^(a (1-epsilon) ) < n^1 . So, it fails for p(n) = n^a for any a such that a >= 1/(1-epsilon)
It appears that Anthropic has released a proof as well (same bounds), but with a different approach? https://drive.google.com/drive/u/0/folders/1uaWnPqWkXy1sjfsD...
Wow. Complexity class separations are really really rare.
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.
SAT-solving expert here. No immediate applications (the results in the paper are "galactic"). The main question is whether some ideas here will snowball into other ideas, one of which could eventually be very practical.
This is a complete non-sequitur, but I checked out your web page. I love the Curious George image. My daughter has been deeply into the Curious George episodes for the past couple of months, and I don't think I've seen him show up in many other places besides kids' shows :D
I think of Curious George Goes to a Chocolate Factory when I fire up Claude Code.
Such a good show. My kids grew up on it!
What do you mean by "galactic"?
Galactic algorithms have such huge constants they figuratively need a galaxy-sized computer and/or galaxy-sized problem, to have any advantage over the normal solutions. Imagine a generic sorting algorithm in O(n) that had a constant of 10^92 **. Technically superior but in practice unusable.
** Yes, I know that's proven to be impossible.
There are realistic situations where galactic O(n) sorting algorithms exist, right? 32-bit integers can be sorted in O(n) using 2^32 words of memory, using bucket sort[1].
If objects have a maximum size of 4 GB (= 2^35 bits), then arbitrary objects can be sorted in O(n) using 2^2^35 bits of memory.
[1] https://en.wikipedia.org/wiki/Bucket_sort
Right, that's why I wrote "a generic sorting algorithm". Your case adds the integer limit which allows it to avoid O(nlogn) minimum.
You can use Radix sort https://en.wikipedia.org/wiki/Radix_sort to lower the memory requirement for limited bit-width integers. But there are more practical o(n log n) methods for sorting integers.
Well optimized radix sorts are actually pretty fast (assuming you have the proper fallbacks to quicksort/insertion sort to deal with medium/small sizes)
https://en.wikipedia.org/wiki/Galactic_algorithm
Do we know they are galactic?
> A galactic algorithm is an algorithm with state-of-the-art theoretical (asymptotic) performance, but which is not used due to practical constraints.
Yes, I think we can pretty safely say that the algorithm in the paper is not used due to practical constraints. This is a subjective judgment, but pretty clear from the paper that the algorithms aren't practical to apply.
The data you put into a SAT solver has substantial effect on it's performance in practice. Real world inputs tend to be much nicer than pathological ones, so I don't think there's much practical application, but I also haven't read the paper yet.
I also haven't read ths paper but coding agents absolutely are a part of modern SAT solver coding challenges.
It's surprising that Exponential Time Hypothesis would be false. Quoting a paper from 2022[1]:
> 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?
[1]: https://arxiv.org/abs/2209.04881
No, SETH implies quadratic time does not mean !SETH implies !quadratic time.
but you're telling me.... there's a chance
This but unironically.
put a number on your probabilities. there's a chance you have an aneurysm in the next few minutes and fail to see my comment. if so, oh well, i tried.
SETH is a much stronger claim than ETH
Not relevant, because !ETH => !SETH
What is relevant is that the logic goes the other way: SETH implies quadratic, !SETH implies nothing.
Right, so a refutation of ETH would also refute SETH.
Yes, it's one strong stronger.
It's playful!
But it's also not even wrong, in the Pauli sense.
That makes me wonder if "I see your CS proof, how does this affect LLMs / nvidia stonk price???" being immediately self-described as "playful" ensures criticism looks cranky.
"I see your particle physics paper (on sustainability of neutron-induced chain fission reaction in uranium), how does this affect the world politics???"
Turns out, it affected the global politics tremendously.
Anthropic also claims a disproof of the ETH, posted here:
https://news.ycombinator.com/item?id=50255159
I’m shaking, so much has happened in the past few years, I think it’s inevitable we will find the efficient algorithm for P=NP. Not sure if I’m ready to live in that world yet.
NP looking a little more P than usual
No, this doesn't touch NP ?= P
ETH (which was commonly thought to be true) implies P≠NP.
sure, this proof itself doesn't settle the P/NP problem, but we've lost one of our core beliefs.
I think that statement is too strong -- if ETH was a core belief then its implication P≠NP must also have been a core belief ... but I don't think that those who assumed ETH held it that strongly. As you say, it was commonly thought to be true ... as was P≠NP ... and the latter still is. But is not a "core belief" -- I think virtually everyone believes that they could be wrong, whichever side they're on.
At this point, I have no expectations. The future is alien and strange, so time to talk the silicon runes...
I wonder how long it'll pay for frontier labs to use compute this way?
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?
the presumption is that these results are a side-effect of reinforcement learning to make the models smarter.
Uh no, reinforcemnt learning doesn't magically pop out math proofs. OpenAI has a ton of mathematicians on staff and is spending millions of dollars in compute to pursue these solutions.
Dropping ultra hard problems on those things and giving them reward when lean accepts the proof sound like an obvious RL environment tbh
That's not necessarily incompatible. The raw output could come from RL rollouts where the model learnt from the process of solving these problems.
It would be interesting to see if this new counting method has any implications to counting 2SAT or parity-counting of 2SAT. Since problem posed in XORSAT and 2SAT clauses is NP-complete, this could have implications about P vs NP.
I am long out of this game, but the complexity class they show still has an n in the exponent, even if divided by log(n).
I have no intuition whatsoever on how that algorithm scales. Can anyone enlighten me?
Can someone ELI5 ETH for those of us who slept through theoretical CS classes? Asking for a friend...
I like this https://www.reddit.com/r/explainlikeimfive/comments/33fmrq/e...
2n should read as 2^n
Unless OP already solved p=np...
that threw me, thanks
> the number of operations needed to solve satisfiability will always be exponential in the size of the problem, no matter what algorithm is chosen
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".
My 2nd grader once brought home a homework sheet with some simple problems involving combinations of coins and I was like "this is unsolvable!"
There's a set of problems that are (it is conjectured) easy to verify but hard to solve. One of them, called 3SAT, looks for values (a, b, c, ...) such that an expression like ((a or b or c) and (b or not-c or e) and ...) evaluates to true. And it turns out that all of these supposedly hard problems can be converted to 3SAT, so if you find a solution to one you find a solution to all.
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.
The exponential time hypothesis is essentially that NP-complete problems' best solutions are all exponential.
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)
> 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 linear time.
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.
Sorry, you're correct. I meant "polynomial" rather than linear.
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.
Ok, I read your edited comment. Looks a lot better know.
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.)
the language around complexity domains is really confusing.
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.
This isn't quite true, problems in NP can still require exponential time to solve.
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.
thanks for the correction. Corrected.
But even if every problem in NP is sub-exponential (which, as tfpgh notes, this result doesn't prove), how does that make it "more likely" that every such problem is solvable in a polynomial # of steps? I feel like it's a bit like claiming that, because Alex Honnold has shown that a human can free solo El Capitan in less than 4 hours, that makes it "more likely" that I could. To me, his success does not budge the latter probability at all.
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.
3SAT can't be solved in sub-exponential time
Fwiw never thought it was a well-named conjecture
*Can be solved
This proves that EXP != NP
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
I'm stating the conjecture as requested
I think you misunderstood, I believe they were requesting a layperson-friendly explanation of the conjecture and not a statement of it.
How is "This proves that EXP != NP", which is false, a layperson-friendly explanation of the actual proof that 3SAT is sub-exponential?
Note https://news.ycombinator.com/item?id=50253213
Whenever someone asks for an ELI5, there's always got to be somebody who further confuses the issue.
Which is a separate issue than the one you commented originally
This seems huge, like not quite p=np but in the same league. Am I missing something ?
[deleted, originally wrote the hypothesis being true would imply NP=EXPTIME which is wrong, since there can be problems in EXPTIME which aren't in NP.
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.]
> Instead we now know that NP is a strict subset of EXP (there are problems in NP that can be solved in subexponential time).
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.
I mean the result that k-SAT cannot be solved in 2^{o(n)} clearly would imply P!=NP, but how does ETH imply NP = EXP? And how does \neg ETH imply NP != EXP?
If I am understanding this correctly (and I am still reading this!) and if this is true, this means NP is sub-exponential. Sub-exponential does not mean P=NP. But yes, this is huge if true.
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.
It doesn't imply NP is sub-exponential, only that 3SAT is. Conversions to other problems can still end up exponential.
No. NP is always poly time reduceable to 3SAT
See this from the paper: "Note that runtime bounds of the form 2^o(n) do not transfer under general polynomial-time reductions (2^n^o(1) would be needed for this), so our result does not imply that general NP problems admit subexponential time algorithms."
OTOH, it would be surprising if this sort of technique didn't apply to other NP problems.
It's not clear what you're trying to say or the relevance. The fact remains that "NP is always poly time reduceable to 3SAT" and "3SAT is sub-exponential" does not imply that NP is sub-exponential.
right, but anything linearly reducable to kSAT is trivially sub-exponential, and a discovery like this means that linear reductions are a whole lot more important than they used to be.
Also for NP problems that can't linear reduce to kSAT just having the knowledge that SAT is sub-exponential (and the tools), makes it a lot more likely that we see a whole lot more problems become exponential.
Yeah that would be interesting, like could there be different NP languages being sub-exponential but for entirely different reasons.
practically, a poly-time reduction can increase teh size of hte problem instance. So maybe you transform a 3-color instance (graph on n vertices) into a n^2 sized SAT instance. You can then solve this in time 2^{n^2 / sqrt{log{n^2}}}, but this is not sub-exponential.
Your "No" is wrong. As he said, "Conversions to other problems can still end up exponential", so (NP is poly time reduceable to 3SAT + 3SAT is sub-exponential) does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
Subexponential(poly(n)) may be exponential in n
Did you reduce in the wrong direction or am I misunderstanding what you're trying to say?
This defeats a few hypotheses that were widely relied on.
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).
Assuming P != NP (and that this NP != EXP proof holds up), what's the next most "natural" candidate for how fast NP must grow? This paper appears to show Omega(2^(n/sqrt(log(n))) with a couple additional fudge factors which would be a very disappointing answer.
This is not a proof of NP ≠ EXP. NP ≠ EXP would mean that there is a problem that can be solved in exponential time, but not in non-deterministic polynomial time. That’s completely different from what they have shown.
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
Even if all NP problems could be solved in sub-exponential time, it would not imply that NP ≠ EXP. It seems that many people here are confused about what EXP means – it’s the set of all problems that can be solved in at most exponential time, including those in P.
Definitely puts some weight on the P=NP side of the scale!
My (very amateur) interpretation is that this weakens confidence in P!=NP. Would like to hear from some experts in this area, if this is so or not.
Scott Aaronson comments on his blog "Given this news, I’ll up my personal probability of P=NP from 3% to 5%." (https://scottaaronson.blog/?p=10239#comment-2042428)
I’m not as smart as Scott but for me it’s more like 10% -> 20%. Do we have a prediction market for this?
It seems highly irrational to bet that your evaluation is more likely than Aaronson's.
Predictions markets are terrible mechanisms for this, because they require imposing an arbitrary time limit at which point the prediction will pay out. IOW, even if you religiously believed P=NP, but you think it would take at least two years to prove it, you would vote against the prediction if the deadline were one year.
Why must prediction markets have a predefined end date?
Without it, capital would be trapped indefinitely.
This is not necessarily true, you could have contracts rather than money down
How would it work otherwise...?
Anybody can exit their contract by selling their contract to an interested party
This proves that EXP != NP, which was a stronger claim than P != NP.
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.
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
There's a chain of classes P, NP(aka Sigma_1), Sigma_2, .., Sigma_n, ... PSPACE, EXPTIME. What we know for sure is that P < EXPTIME, others are unknown (well, maybe this paper adds something to this, time will tell).
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.
It basically just says that NP is not the same as EXP
This is not a proof that NP != EXP. NP is poly time reduceable to 3SAT + 3SAT is sub-exponential does not imply that NP is sub-exponential -- the cost of the reduction may be enough to push the conjunction from sub-exponential to exponential.
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
Subexponential time still can mean much larger than polynomial time. There's a huge space between the two.
It’s not that it isn’t quite p=np; it actually isn’t p=np.
Still, an impressive accomplishment.
no. It's huge.
Double-checking, is this more like the recent DFT improvement where even the word-RAM model of computation doesn't justify the fraction of a log factor they eliminated because of an assumption of arbitrary-precision arithmetic, or has ETH actually fallen with the Lean proof being formulated appropriately? Do we know yet?
The DFT result isn't RAM specific. It holds up in multi-tape turing machines. (it is, however, galactic).
That doesn't sound right. Previous results proved that with bounded values you couldn't beat O(nlogn), among other things. The OpenAI result explicitly used arbitrary-precision complex arithmetic. Yes, it's galactic [0], but either it's wrong or the previous results were wrong, no?
[0] Very much not something I care much about in these sorts of conversations. Even if it weren't galactic, a random logn factor here and there usually loses to cache efficiency and whatnot. Their speedups were less than that, and I'd already assumed we were in the realm of applied math rather than real engineering.
> but either it's wrong or the previous results were wrong, no?
Not quite. This result uses polylog(n) precision which isn't bounded, so it escapes the previous bounds. This doesn't noticably affect the runtime because not all of the math needs this extra precision (in this case you need a couple computations in higher precision but less than n operations), so the extra precision can be payed for even in a TM/circuit model.
Are you thinking of the int multiplication result instead? The DFT result uses a computation model with unlimited precision reals.
no. The DFT result as published uses unlimited precision, but the comparison with the int multiplication shows that the unlimited precision just makes the paper shorter/easier rather than being an exploit in the computation model.
No practical implications, we dont use 10^87 variables (a point where the proposed new method surpasses the already existing sat solvers), however a good theoritical point.
This construction splits the space into 'small' models with a quadratic lookup cost and 'large' with sub-exponential cost. The small one is by enumerating all SAT problems with their answers, then looking up the answer. The large one is something in polynomials I don't understand. An answer from openai about where the cutoff is was:
I'm not inclined to consider "wrote down all the answers ahead of time, so that's free" as a quadratic time solution to SAT. Nor very convinced by the other half.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.
I still think P != NP, I would assume 2^{n/log(n)} is an elegant optimal boundary.
If I can "count the satisfying assignments of a k-CNF in subexponential time and polynomial space", could I just count the the solutions for assignments of 0 and 1 to x_1, then proceed with whichever is greater (or nonzero), counting the solutions for x_2=0, and so on.
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?
What about OpenAI allowed ETH to be determined to be false that couldn't be done by humans? Was there something in particular that OpenAI's AI did that help bridge the gap to a solution?
What about a particular mathematician solving a problem that couldn't be solved by other mathematicians?
(For those unfamiliar with the form: https://en.wikipedia.org/wiki/Double_dactyl )
We are living in such a weird, amazing, and miraculous time. It’s completely surreal.
Perhaps this will be the push we need to finally, seriously consider the possibility that our formalisms of “efficient” and “efficiently verifiable” algorithms correspond to nothing resembling reality, and P \in NP is actually ill-posed.
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.)
> Perhaps this will be the push we need to finally, seriously consider the possibility that our formalisms of “efficient” and “efficiently verifiable” algorithms correspond to nothing resembling reality,
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?
"Ill-posed" here means that the formalisms we've chosen to represent "efficient" and "efficiently verifiable" algorithms could turn out to not correspond to anything humans would consider meaningful ("we posed the problem poorly"). For most people, a solution to an NP-complete problem that's barely-subexponential would be clear evidence of these formalisms being ill-posed.
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).
> "Ill-posed" here means that the formalisms we've chosen to represent "efficient" and "efficiently verifiable" algorithms could turn out to not correspond to anything humans would consider meaningful ("we posed the problem poorly").
OK, that's a reasonable worry and one that practitioners share. Thanks for explaining. Though I don't think ill-posed is a good term for this, the technical meaning of the term points in a different direction.
So far, as far as I can tell this new result gives much of any extra credence to that worry.
It is completely acceptable in math to use "ill-posed" to indicate that something is insufficiently specified, improperly formulated, or ambiguous. Math people do this all the time.
I remember the olden days when LLMs couldn't count.
Any implications on what could be achieved with this information, for a non academic?
There is nothing immediately useful or practical from this.
> Given this news, I’ll up my personal probability of P=NP from 3% to 5%.
— Scott Aaronson
I'm not the kind of programmer that even understands P=NP, so I'll either be way off base, or be blessed with an outsider perspective.
Considering that no human will be able to sift through these in no less than a team-decade, to verify that there's no mistake, yet still that will be massively less compute than the one necessary to produce the 'solution', isn't the empirical finding contrary to the theoretical finding being suggested? Namely that in practice we are generating these 'proofs' in a massive amount of compute time, but it will take tiny amounts of time to verify them.
Is it very useful to talk about computing in purely mathematical terms and not in empirical terms? There's very obvious truths that we are seeing here, and it seems like those that go through some sort of rigorous mathematical training kind of fail to see it. Similar to how 70% of physicians or something believe in the Many Worlds Interpretation or that time isn't real, a sort of algebraic psychosis where one is removed from a human reality that we have access to in order to attempt getting close to an 'objective reality' that is far removed from the human experience.
To some extent we are computing entities, so I don't even think that we can conceive of computing as decoupled from ourselves[1]. I know the observer effect is very often mis-cited by laymen, but I'll be damned if some computer scientists like Godel don't sound like they were up to something similar to this.
Also, here's another outsider perspective. This is obviously slop, I'm not sure how but it's obviously nerd-sniping the CS majors that understand this, because it's obviously very hard to find the flaw, but it gives epicycles energy. It's also sufficiently complex that under a bayesian analysis it is more likely that the AI has reached the limit of its processing power until it becomes confused enough to believe it has reached something significant, kind of like how I'm doing now.
[1] xkcd://1163
> Considering that no human will be able to sift through these in no less than a team-decade
How do you figure that? For ETH, OpenAI is engaging with top experts in this area. It's unlikely that they'll need 10 man-years combined to go through the results:
> This work will be verified and shepherded by a group led by Russell Impagliazzo (UC San Diego), Ramamohan Paturi (UC San Diego), and Ryan Williams (MIT). All three are experts in algorithm design and complexity theory.
> Impagliazzo and Paturi, along with Francis Zane, are the original proposers of the Exponential Time Hypothesis, and have authored a number of influential papers on the topic over the last 30 years.
> Williams has refuted strong versions of the ETH, used ETH for hardness results, and connected algorithms and lower bounds.
I think I misread something, the pdf is like 80 pages long, and there’s a list of like 200 pdfs, each around that length, I thought they were all related to this proof, but they are unrelated.
[flagged]
math major here, I can't read the paper for shit
Not a math major here. But is this paper way too long? So long that it raises the possibility that OpenAI's staff mathematicians passed through AI's raw output as-is?
That problem seems independent of the 'stochastic parrot' accusation?
A parrot might squeak the theory of everything and it won't matter to any of us cause nobody speaks parrot.
We can record a lot of parrot speech and train LLMs against it and eventually build a translator.
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.
Yes we can just train llms against it and it's all solved. Just a few more trillion dollars and a bunch of deforestation for datacenters bro.
Ironically yeah.. if it can solve more complex problems across domains like biology, econ, finance then all the deforestation is worth it
It is just that we never considered how advanced stochastic parrots could be so powerful when dealing with basically the whole library of acquired human knowledge and guided by experts to generate.
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.
Is this really as interpetable math is fundamentally. Like wtf is this accessibility? who is this for? Do top mathematicians read this like a chapter book? Absolutely ridiculous they have all this compute and creative resources and cant make something more intuitive to help people understand it more natively
It's a math paper written for mathematicians.
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).
Even mathematicians take a long time to read math books and articles, especially if they go into the details. They don't read math like you would read a novel.
> They don't read math like you would read a novel.
siting on a recliner making an occasional sensible chuckle while flipping through the pages?
It's sometimes okay to look at something, go "this isn't for me", and move on. You don't need to have an opinion on everything.
Same to you bud
Sometimes hard things are hard.
Actually, no - this is what controllability is all about. Humans have to understand it.
Humans can understand it. Not all of them, because to understanding it, there's a lot of other things you need to understand before.
> Absolutely ridiculous they have all this compute and creative resources and cant make something more intuitive to help people understand it more natively
OpenAI has that too - open their website and ask.
And get some hallucinated slop infotainment?
I agree with you insofar that the primary motivator for these press releases is public relations and investor confidence, so it should be made as minimally hard to understand as possible just by sheer virtue that it exposes oAI to more cash thrown at them.
I totally disagree with you on the precept that all difficult to understand things need be needlessly dumbed-down/watered-down until palatable for a mass (average) audience. Requiring 'math50' before 'trig101' isn't gate-keeping in any moral sense, it's an effort to save everyones' time involved when the effort is futile.
Similarly I don't demand that haute couture fashion companies explain the style to me until 'I get it' ; I just take it for granted that i'm not going to be able to understand the entirety of the world in one lifetime.
wtf......
for context (ai convo):
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.
Humans in this thread have already explained that concept with much more straightforward prose.
From the HN comment guidelines:
> Don't post generated text or AI-edited text. HN is for conversation between humans.
Disclosure doesn't address the objection (not that it wouldn't be obvious anyway). If you learned something from an AI convo, tell us what you learned, in your own words, not what the LLM said.