2017年5月22日 · This means that any complete problem for a class (e.g. PSPACE) which contains NP is also NP-hard. In order to get a problem which is NP-hard but not NP-complete, it suffices to find a computational class which (a) has complete problems, (b) provably contains NP, and (c) is provably different from NP. So:

The current hot topic has been the use of PCP Theorem to prove various NP-Hardness results for approximation versions of NP-Hard problems. So, non-determinism is still part of the definition of NP and is historically important, but the current status is that it is not mandatory.

A problem in P is in NP by definition, but the converse may not be the case; probably the most important open question in computer science is whether classes P and NP are the same, that is P=NP. NP-complete is a family of NP problems for which you know that if one of them had a polynomial solution then everyone of them has.

2019年9月2日 · A problem is said to be NP-hard if everything in NP can be transformed in polynomial time into it, and a problem is NP-complete if it is both in NP and NP-hard. The NP-complete problems represent the hardest problems in NP. If any NP-complete problem has a polynomial time algorithm, all problems in NP do.".

Such a problem is NP-hard and in NP. How do we know if a problem is hardest in NP, and no harder problem exists. I understand that let's assume that somehow magically we know that a problem L is hardest in NP and then we can find out more hardest problems H if we can reduce H to L and vice versa.But my question is how does it all begin?

2016年6月15日 · NP-hard if all problems in NP polynomially reduce to it; and NP-complete if all problems in NP polynomially transform to it. What is the difference between polynomial reduction and polynomial transformation?

NP is the set of all problems that can be checked for correctness if you guess an answer. NP-hard is the set of all problems that are at least as hard as the hardest NP problems. NP-complete is the set of all problems are NP-hard problems that are also NP. Have I covered anything? Are there any important aspects that I still need to know?

2011年5月17日 · I don't know if you would consider it "real life", but exact inference in Bayesian networks is NP-Hard; it's a straightforward reduction from 3-SAT (I guess one could argue that Bayesian inference is used in the playing of many games).

2019年11月12日 · $\begingroup$ It might be good to add that while the decision variant of MILP lies in NP, the proof of this is not easy (at the least, proving MILP is NP-hard is a lot easier). Although an optimal point would work as a certificate, the tricky part is to show that we can always find an optimal point that can be encoded in polynomial size.

2015年7月8日 · The problem 3-COLOURABILITY is NP-hard because there is a polynomial time reduction from 3-SAT to 3-COLOURABILITY and there is a reduction from SAT to 3-SAT. It is proven that if you can solve SAT in polynomial time, you can solve any NP problem in polynomial time (Cook's theorem).