this post was submitted on 25 Sep 2026
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[–] LodeMike@lemmy.today 1 points 4 weeks ago (1 children)

Factoring is at most NP-complete because we have a polynomial time verification for it.

[–] a_non_monotonic_function@lemmy.world 0 points 4 weeks ago (1 children)

Factoring is not NP-complete because it is not a member of NP-hard.

You are correct that verifiability is in there, but you have to also be able to do the reduction.

There are plenty of problems out there where we lack polynomial time solutions, while the problem also lacks the expressiveness required to reduce back and forth.

[–] LodeMike@lemmy.today 1 points 4 weeks ago* (last edited 4 weeks ago) (1 children)

Factoring is not in any category because we don't have a proof for it. As it stands, the algorithms we have require exponential time in the worst case, but solutions can be verified in polynomial time, which is NP-complete. I said "at most" for a reason.

Also, every algorithm is verifiable. If there's no better way to do it then re-running the algorithm and that algorithm is exponential, then it's NP-Hard.

Factoring is in a category. Everything computable exists somewhere. We know factoring to be in NP.

What I was disagreeing with was the "at least" and "at most" characterization of NP completeness. It is a set, not a boundary. The actual diagram of the complexity zoo is much more complicated than concentric circles.

And for verifiability, I was not referring to it as an existential sort of thing. I was simply saying that I agreed with you in that particular facet, but it isn't sufficient to describe NP completeness. You also need NP-hardness.