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by cochne10 hours ago |
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upvoteby gewetensleegte41 minutes ago|
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> 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.

No, subexponential is not the same as polynomial time. There's sub-exponential functions that grow faster than polynomial.

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upvoteby marshray9 hours ago|
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> 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?

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upvoteby pcbouman5 hours ago|
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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.
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upvoteby ReptileMan1 hours ago|
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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.
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upvoteby 2 hours ago|
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upvoteby LPisGood7 hours ago|
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> 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?

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upvoteby eli_gottlieb6 hours ago|
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~~Yeah, that's NP-complete problems.~~

EDIT: DISREGARD THAT I WAS MISTAKEN

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upvoteby csheehan106 hours ago|
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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.

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upvoteby LPisGood6 hours ago|
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That is unfortunately not the case
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upvoteby demibabs10 hours ago|
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Can you explain more?
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upvoteby triMichael10 hours ago|
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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.

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upvoteby demibabs9 hours ago|
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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.)

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upvoteby edanm3 hours ago|
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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.
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upvoteby triMichael9 hours ago|
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Thanks for the correction! I updated my answer to fix that :)
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upvoteby sourdecor8 hours ago|
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"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.
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upvoteby bobmarleybiceps9 hours ago|
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oh that actually helps a lot. I was tying myself in knots trying to think it through.
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upvoteby imadr10 hours ago|
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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

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upvoteby demibabs10 hours ago|
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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.
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upvoteby nightshade15119 hours ago|
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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.
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upvoteby raegis7 hours ago|
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Hey, this is a common error. (2^(n^0.9))^2 is not 2^(n^1.8) but 2^(2 * n^0.9).
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upvoteby Aunche9 hours ago|
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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.
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upvoteby stbede7 hours ago|
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(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

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upvoteby raegis7 hours ago|
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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.
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upvoteby Dylan168073 hours ago|
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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

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upvoteby jballanc8 hours ago|
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...that's...not how exponents work
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upvoteby Dylan168073 hours ago|
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The result is right, they wrote the parentheses in the wrong spot, don't be so shocked.
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upvoteby caughtinthought9 hours ago|
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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

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upvoteby demibabs9 hours ago|
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Gotcha, thanks
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upvoteby csheehan106 hours ago|
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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.

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upvoteby imadr9 hours ago|
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Misunderstood your question sorry, I have no idea!
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upvoteby 9 hours ago|
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upvoteby vote_farisa_20210 hours ago|
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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.

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upvoteby calf9 hours ago|
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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.
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upvoteby shoo8 hours ago|
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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)

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