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LPisGood • today at 2:36 AM • 1 reply • view on HN

> To be clear though, it does not imply that NP is subexponential

Is there a subset of NP problems for which a subexponential algorithm would witness the entire NP class being subexponential?


Replies

eli_gottlieb • today at 3:08 AM

~~Yeah, that's NP-complete problems.~~

EDIT: DISREGARD THAT I WAS MISTAKEN

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