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points by lynndotpy 3 days ago | 0 comments

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.

[–] cochne 3 days ago

It doesn't imply NP is sub-exponential, only that 3SAT is. Conversions to other problems can still end up exponential.

[–] wbl 3 days ago

No. NP is always poly time reduceable to 3SAT

[–] cochne 3 days ago

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."

[–] adgjlsfhk1 3 days ago

OTOH, it would be surprising if this sort of technique didn't apply to other NP problems.

[–] jibalt 3 days ago

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.

[–] adgjlsfhk1 3 days ago

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.

[–] calf 3 days ago

Yeah that would be interesting, like could there be different NP languages being sub-exponential but for entirely different reasons.

[–] mswphd 3 days ago

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.

[–] jibalt 3 days ago

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.

[–] LPisGood 3 days ago

Subexponential(poly(n)) may be exponential in n

[–] dataflow 3 days ago

Did you reduce in the wrong direction or am I misunderstanding what you're trying to say?

[–] lynndotpy 2 days ago

Yeah, I was wrong. Page two too, I commented too soon.

NP is not yet proven to be subexponential but still, ETH disproven is huge