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Good4bootheeyesterday at 3:10 PM4 repliesview on HN

Isn't it a bit Catch 22 anyway? If someone finds a algorithm to reduce some NP task X to class P, then that just means X wasn't a true NP task and P!=NP is still undecided?


Replies

layer8yesterday at 3:22 PM

If it’s an NP-complete [0] problem like SAT, as many NP problems are, then we are done, because all NP problems can be reduced to it (in polynomial time).

[0] https://en.wikipedia.org/wiki/P_versus_NP_problem#NP-complet...

Tyr42yesterday at 3:18 PM

You can prove something is in NP by providing a (polynomial) reduction from a known NP hard task, and vice versa. All the known NP problems (Knapsack, SAT, etc) are mutually reducable in this way, so solving one lets you solve the others. So if X was shown to be NP, then given a polynomial time solution to X, you can stack the polynomial time reduction from X to SAT to solve SAT in polynomial time too.

SetTheoristyesterday at 3:16 PM

AIUI if you have an (polynomial-time) algorithm to reduce some NP-complete task to P then you have indeed shown that P=NP.

subsistence234yesterday at 10:30 PM

NP doesn't mean "we don't know a polynomial time algorithm for it", it means "a proposed answer can be verified as correct in polynomial time"