▲ | redleader55 6 days ago | |||||||||||||||||||
This [1] diagram from Wikipedia represents you are saying in a visual way. NP-hard and NP-complete are defined as basically an equivalence class modulo an algorithm in P which transforms from one problem to the other. In more human terms both NP-hard and NP-complete are reducible with a polynomial algorithm to another problem from their respective class, making them at least as hard. The difference between the two classes, again in human terms, is that NP-complete problems are decision problems "Does X have property Y?", while NP-hard problems can include more types - search problems, optimization, etc, making the class NP-hard strictly larger than NP-complete in set terms. The statement in the article means that NP-hard problems require solving a NP-complete problem plus a P problem - hence being at least as hard. [1] https://en.m.wikipedia.org/wiki/File:P_np_np-complete_np-har... | ||||||||||||||||||||
▲ | edanm 5 days ago | parent | next [-] | |||||||||||||||||||
I don't believe this is accurate. The difference between NP-Hard and NP-Complete is that NP-Complete problems are both NP-Hard, and in NP. As opposed to NP-Hard problems that are not NP-Complete, meaning they are not themselves in NP. | ||||||||||||||||||||
▲ | zahlman 5 days ago | parent | prev [-] | |||||||||||||||||||
If NP-hard isn't even a subset of NP, why is it called "NP-hard"? | ||||||||||||||||||||
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