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This isn't correct. What you may be remembering is that some (not all) NP complete problems have limits on how accurately they can be approximated (unless P = N
by kalkin 3y ago
This isn't correct. What you may be remembering is that some (not all) NP complete problems have limits on how accurately they can be approximated (unless P = NP). But approximation algorithms for NP complete problems form a whole subfield of CS.
- moyix 3y agoThe theorem that proves this is the PCP Theorem, in case anyone wants to read more about it: https://en.wikipedia.org/wiki/PCP_theorem#PCP_and_hardness_of_approximation https://en.wikipedia.org/wiki/PCP_theorem#PCP_and_hardness_o...