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Comment by fpoling

10 hours ago

And vise versa, a proof that P!=NP do not exclude an algorithm that is fast enough for practical n.

Maybe, but that feels significantly more exotic and unlikely to me. You have to basically negate the natural cost scaling of brute force in the thousand to million n range but lose that speed benefit later.

  • Well, if someone found a quasipolynomial time algorithm – O(n^log(n)) – for an NP-complete problem, that would technically be compatible with P!=NP, but might have quite significant practical implications.