The Exponential Time Hypothesis Is False [pdf] (github.com)
svat 6 hours ago
- 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.
slashtrends 6 hours ago
devy 5 hours ago
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.
hathawsh 5 hours ago
murderfs 3 hours ago
Maxion 2 hours ago
treebeard901 2 hours ago
skissane 5 hours ago
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.
fpoling 3 hours ago
Dylan16807 an hour ago
mattm 4 hours ago
sampo 4 hours ago
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)
steezeburger 4 hours ago
bananaflag 3 hours ago
Animats 5 hours ago
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.
svat 4 hours ago
> 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.
Animats 4 hours ago
FlyingCanoe 4 hours ago
It also not probabilistic, so linear programing is not rely a good example
vlovich123 3 hours ago
zahlman an hour ago
> 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.
lukasgelbmann 6 minutes ago
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.
dingdongditchme an hour ago
kscarlet 38 minutes ago
vanderZwan 35 minutes ago
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.
cochne 9 hours ago
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.
demibabs 9 hours ago
imadr 8 hours ago
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
demibabs 8 hours ago
imadr 8 hours ago
caughtinthought 8 hours ago
So 3SAT+poly reduction ends up being exponential
demibabs 8 hours ago
nightshade1511 8 hours ago
raegis 6 hours ago
Aunche 8 hours ago
jballanc 7 hours ago
Dylan16807 2 hours ago
stbede 6 hours ago
but
2^(sqrt(n)^2) = 2^n
raegis 6 hours ago
Dylan16807 2 hours ago
This also means your other comment here is wrong to suggest any error: https://news.ycombinator.com/item?id=50254142
csheehan10 5 hours ago
In this case they have c = 1-0.5(\log \log n)/(\log n) which is approaching 1 for large n.
vote_farisa_202 8 hours ago
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.
triMichael 8 hours ago
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.
demibabs 8 hours ago
(By the way, I think you mean NP-complete, not NP-hard. The Halting Problem is NP-hard.)
triMichael 8 hours ago
edanm 2 hours ago
bobmarleybiceps 7 hours ago
sourdecor 6 hours ago
calf 8 hours ago
shoo 7 hours ago
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)
marshray 8 hours ago
Wouldn't that also include functions that can be evaluated in hardware using a fixed number of gates? E.g., symmetric cryptography?
pcbouman 4 hours ago
ReptileMan 19 minutes ago
LPisGood 6 hours ago
Is there a subset of NP problems for which a subexponential algorithm would witness the entire NP class being subexponential?
eli_gottlieb 5 hours ago
EDIT: DISREGARD THAT I WAS MISTAKEN
LPisGood 5 hours ago
csheehan10 5 hours ago
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.
zodiac 4 hours ago
criemen 9 hours ago
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.
infamouscow 8 hours ago
kingstnap 8 hours ago
bsubs 8 hours ago
ashkankiani 8 hours ago
yyyk 8 hours ago
** Yes, I know that's proven to be impossible.
MichaelDickens 7 hours ago
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.
yyyk 7 hours ago
adrianN 5 hours ago
adgjlsfhk1 4 hours ago
kite42 8 hours ago
gradus_ad 7 hours ago
zahlman an hour ago
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.
sebmellen 7 hours ago
chr15m 7 hours ago
angry_octet 31 minutes ago
strommec 9 hours ago
> 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?
rokob 9 hours ago
dnautics 9 hours ago
Xirdus 8 hours ago
programjames 8 hours ago
caughtinthought 9 hours ago
margalabargala 7 hours ago
jibalt 7 hours ago
What is relevant is that the logic goes the other way: SETH implies quadratic, !SETH implies nothing.
varyherb 6 hours ago
refulgentis 7 hours ago
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.
Joker_vD 3 hours ago
Turns out, it affected the global politics tremendously.
bananaflag 2 hours ago
qgin 9 hours ago
jibalt 7 hours ago
mathgladiator 6 hours ago
pxx 4 hours ago
sure, this proof itself doesn't settle the P/NP problem, but we've lost one of our core beliefs.
jibalt 3 hours ago
softwaredoug 8 hours ago
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?
adgjlsfhk1 8 hours ago
paxy 6 hours ago
Philpax 6 hours ago
baq 3 hours ago
choeger 34 minutes ago
I have no intuition whatsoever on how that algorithm scales. Can anyone enlighten me?
mlmonkey 9 hours ago
caughtinthought 9 hours ago
Fwiw never thought it was a well-named conjecture
pcstl 9 hours ago
This proves that EXP != NP
caughtinthought 9 hours ago
pcstl 8 hours ago
caughtinthought 8 hours ago
moishe_syzslak 8 hours ago
jibalt 6 hours ago
jibalt 6 hours ago
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
ehershey 9 hours ago
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.gberger 9 hours ago
newsy-combi 9 hours ago
pvab3 7 hours ago
nullc 8 hours ago
A common complexity theory misunderstanding is that statements like this applies to all or even most instances-- and it leads to objectively bad engineering of the form "oh look this is actually 3sat, it's unsolvable".
(Great fodder for a midwit meme: idiot and genius say "my heuristics go brrr!"-- midwit says "No no no! NP problems can't be efficiently solved!")
Any specific instance may be efficiently solvable (and perhaps all instances coming from your particular problem set -- a subset of all possible 3sats).
So I would add to that paragraph "at least for some 3sat instances".
marshray 7 hours ago
pcstl 9 hours ago
NP-complete problems are essentially those where you need to make some kind of choice over a superlinearly-growing set of options where each option can be solved in polynomial time. Like, for example, finding a route between cities on a map under a certain total distance. You might need to go over all choices of routes.
The ETH essentially assumes that the best solution (in terms of complexity, i.e. how it grows) you can get for these kinds of problems is exponential. OpenAI just proved that this is false, you can have solutions that are below exponential complexity (but still above polynomial - if anyone finds a polynomial solution then we'd have proven that P=NP)
eru 8 hours ago
Seems like a weird way to describe NP, and I'm not even sure it's true?
NP is the class of decision problems where if the answer is YES and someone slips you a hint, you can solve them quickly (or quickly figure out that the hint was useless and report that). Where quickly means in polynomial time.
There's nothing about linear time in the classic definition.
pcstl 8 hours ago
NP = nondeterministically polynomial. i.e. if you could evaluate all choices in parallel (or just "know" the right choice), then you'd get a polynomial runtime. That's what I was trying to informally hint at, but I ended up using the wrong term. I'll edit my original post.
That's equivalent to your description - the "hint" is knowing which of the options to choose.
The reason why superlinearity comes into play is that if you had a linearly growing amount of solutions to test then you'd just have a sum of polynomials, which is itself polynomial.
eru 8 hours ago
One nitpick: I think your set of options needs to grow a lot faster than just superlinearly, I suspect it needs to grew superpolynomially?
However otherwise what you describe is essentially true of all problems in NP, not just the NP hard problems:
Sorting is (trivially) in NP, and you can say that to sort you need to pick one out of n! many permutations (which certainly grows superlinearly and even superpolynomially).
NP hard problems have the distinction that if you solve one of them efficiently, you can solve all of them efficiently, because you can reduce them to each other. (Basically because NP hard problems allow you to sort-of describe a computer, and reducing them to each other is equivalent to writing an interpreter or compiler.)
SyzygyRhythm 9 hours ago
The Exponential Time Hypothesis in effect says that 3SAT can only be solved by brute force--trying every possible combination until you find the right one. If your 3SAT expression has 100 variables, then this will take about 2^100 tries (exponential time).
What they found was an algorithm faster than brute force. 3SAT is still potentially one of these hard problems, but finding anything faster than brute force is a major advance.
dnautics 8 hours ago
P means "this problem can be solved in polynomial # of steps relative to the problem size"
NP means "we don't know if the problem can be solved in polynomial # of steps relative to problem size, but we know we can check if the result is correct in a polynomial # of steps relative to the problem size"
!ETH: it was widely suspected that the flagship problem in NP, 3SAT, requires an exponential # of steps relative to problem size. We now know that 3SAT requires AT MOST something more than polynomial but strictly less than exponential to solve.
"P = NP" means "if you can check it in polynomial time, there's some way to solve it in polynomial time". "P != NP" means "there are some problems checkable in polynomial time for which there is no polynomial way to obtain the result"
Today's new proof makes "P = NP" "feel more likely", because we know that the flagship problem in NP is not exponential.
tfpgh 8 hours ago
Solving an arbitrary NP problem looks like first reducing that problem to SAT and then solving the equivalent SAT problem. We know (via the Cook–Levin theorem) that any NP problem can be reduced to SAT in polynomial time, and we now know that solving SAT is subexponential in terms of the size of the SAT problem, but it's possible for the conversion to SAT to grow the problem enough that it's exponential in terms of the original problem.
dnautics 8 hours ago
jibalt 4 hours ago
It seems to me that the logic only goes the other way ... if ETH had been proven then it would follow that P != NP. With !ETH we really don't know any more about P ?= NP then we did before. Reasoning that P = NP is "more likely" strikes me as a form of a fallacy of affirming the consequent.
lisper 9 hours ago
adgjlsfhk1 9 hours ago
krackers 9 hours ago
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.]
luc4 9 hours ago
eru 9 hours ago
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.
odyssey7 9 hours ago
Still, an impressive accomplishment.
lynndotpy 9 hours ago
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.
cochne 9 hours ago
wbl 9 hours ago
cochne 9 hours ago
adgjlsfhk1 9 hours ago
jibalt 6 hours ago
adgjlsfhk1 5 hours ago
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.
calf 4 hours ago
dataflow 8 hours ago
mswphd 8 hours ago
jibalt 6 hours ago
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
LPisGood 6 hours ago
jaggederest 9 hours ago
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).
thatoneengineer 9 hours ago
zeroonetwothree 9 hours ago
jibalt 6 hours ago
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
kakapo5672 9 hours ago
caughtinthought 9 hours ago
jibalt 6 hours ago
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
pcstl 9 hours ago
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.
jibalt 6 hours ago
Note https://news.ycombinator.com/item?id=50253213 which explains it better.
mudita 9 hours ago
zeroonetwothree 9 hours ago
kibwen 7 hours ago
internetter 6 hours ago
jibalt 6 hours ago
howunfortunate 5 hours ago
Tadpole9181 5 hours ago
internetter 4 hours ago
jibalt 6 hours ago
IsTom 9 hours ago
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.
amelius 9 hours ago
js8 3 hours ago
bozdemir 2 hours ago
airtnp 3 hours ago
JonChesterfield 8 hours ago
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.
hansvm 5 hours ago
adgjlsfhk1 4 hours ago
hansvm 3 hours ago
[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.
adgjlsfhk1 3 hours ago
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.
lg5689 2 hours ago
adgjlsfhk1 2 hours ago
marshray 8 hours ago
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?
reenorap 4 hours ago
hardbass 13 minutes ago
rnbnnnn 4 hours ago
Higgledy-piggledy,
Exp. Time Hypothesis:
Runtime complexity
Soars to the sky.
Subexponentially,
Tensor contractions have
Tallied solutions for
OpenAI.zahlman 43 minutes ago
nemo44x 8 hours ago
antics 9 hours ago
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.)
eru 8 hours ago
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?
antics 8 hours ago
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).
eru 7 hours ago
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.
antics 7 hours ago
colordrops an hour ago
sevenzero 2 hours ago
Kina 2 hours ago
cubefox 2 hours ago
— Scott Aaronson
TZubiri 6 hours ago
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
mattm 4 hours ago
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.
TZubiri 4 hours ago