If a problem X is in C and hard for C, then X is said to be complete for C. This means that X is the hardest problem in C. (Since many problems could be equally hard, one might say that X is one of the hardest problems in C.) Thus the class of NP-complete problems contains the most difficult problems in NP, in the sense that they are the ones most likely not to be in P. Because the problem P = NP is not solved, being able to reduce a known NP-complete problem, Π2, to another problem, Π1, would indicate that there is no known polynomial-time solution for Π1. This is because a polynomial-time solution to Π1 would yield a polynomial-time solution to Π2. Similarly, because all NP problems can be reduced to the set, finding an NP-complete problem that can be solved in polynomial time would mean that P = NP.
