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points by demibabs 3 days ago | 0 comments

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.

[–] demibabs 3 days ago

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.)

[–] edanm 3 days ago

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.

[–] isaacfrond 2 days ago

Right. To be clear: the halting problem is not in the class NP. Any problem in NP can trivially be reduced to the halting problem and so the latter is NP-hard.

[–] triMichael 3 days ago

Thanks for the correction! I updated my answer to fix that :)

[–] bluepoint 1 day ago

I don’t know what you mean by negating the subexponential time. If the reduction is polynomial the time is still dominated by the 3SAT. Btw wikipedia says that the proposed algorithms correspond to 2^(n (log log n)^2/log n) which is subexponential but not in a practical way.

Edit: Ok, I a guess I am wrong if the reduction is n^ log n/(log log n)^2 we are back to exponential.

[–] bobmarleybiceps 3 days ago

oh that actually helps a lot. I was tying myself in knots trying to think it through.

[–] sourdecor 3 days ago

"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.

[–] gbacon 2 days ago

The class NP-complete contains the hardest problems in NP. If you pick any problem in NP-complete, all problems in NP can be reduced to it (meaning that a solution to the problem you chose can be used to solve everything in the rest of NP) in polynomial time or “efficiently.” This applies to all of NP, including every other NP-complete problem, so in a sense, NP-complete problems are the same problem dressed in different clothing.

So the reduction to 3SAT could happen in principle, but it isn’t a necessary step.

[–] imadr 3 days ago

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

[–] demibabs 3 days ago

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.

[–] nightshade1511 3 days ago

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.

[–] raegis 3 days ago

Hey, this is a common error. (2^(n^0.9))^2 is not 2^(n^1.8) but 2^(2 * n^0.9).

[–] raegis 2 days ago

Apologies, I'm incorrect. I misread your math.

[–] Aunche 3 days ago

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.

[–] stbede 3 days ago

(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

[–] raegis 3 days ago

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.

[–] Dylan16807 3 days ago

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

[–] raegis 2 days ago

I was just correcting the statement as written. I see they meant to say something else. You are correct about my other comment--I was in a misdirected, math-correcting rage.

[–] jballanc 3 days ago

...that's...not how exponents work

[–] Dylan16807 3 days ago

The result is right, they wrote the parentheses in the wrong spot, don't be so shocked.

[–] caughtinthought 3 days ago

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

[–] demibabs 3 days ago

Gotcha, thanks

[–] csheehan10 3 days ago

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.

[–] imadr 3 days ago

Misunderstood your question sorry, I have no idea!

[–] vote_farisa_202 3 days ago

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.

[–] calf 3 days ago

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.

[–] shoo 3 days ago

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)