guhbkji

3 ms·

1. 1.▲
   by guhbkji 2y ago

   You are conflating. Integer factorization is unsolved and it’s decision problem is in NP. IF’s decision problem’s complexity “may be in NP” because the question of whether P equalling NP is unknown. Meaning IF is NP, but may well be P if P=
2. 2.▲
   by guhbkji 2y ago

   > not even in NP This is incorrect. Integer factorization is NP-intermediate. Very much “in NP”. https://en.m.wikipedia.org/wiki/NP-intermediate Also, saying factorization lacks “complexity” because sieves exist mi
3. 3.▲
   by guhbkji 2y ago

   > if someone proves that P = NP > we’ll just have to reprioritize and update the list of algorithms we teach to undergrads, issue performance-enhancement updates to some software libraries, and patch any security vulnerabilities. Wow,