alt Hacker News[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.]
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luc4 • yesterday at 11:20 PM
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?
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> 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.