Does NP-complete mean NP?
A problem is called NP (nondeterministic polynomial) if its solution can be guessed and verified in polynomial time; nondeterministic means that no particular rule is followed to make the guess. If a problem is NP and all other NP problems are polynomial-time reducible to it, the problem is NP-complete.
What is meant by NP-hard and NP-complete?
A problem is NP-hard if all problems in NP are polynomial time reducible to it, even though it may not be in NP itself. If a polynomial time algorithm exists for any of these problems, all problems in NP would be polynomial time solvable. These problems are called NP-complete.
Is NP and NP-complete the same?
NP-complete problems are those problems that are both NP-Hard and in the complexity class NP. Therefore, to show that any given problem is NP-complete, you need to show that the problem is both in NP and that it is NP-hard.
Is scheduling NP-complete?
As a consequence, the general preemptive scheduling problem is also NP-complete. These results are tantamount to showing that the scheduling problems mentioned are intractable.
What makes a problem NP-complete?
NP-complete problems are a set of problems to each of which any other NP-problem can be reduced in polynomial time, and whose solution may still be verified in polynomial time. That is, any NP problem can be transformed into any of the NP-complete problems.
Is scheduling NP-hard or NP-complete?
Sometimes people use terms ”NP-hard” and ”NP-complete” as if they are synonyms (usually to mean ”hard”), but this is not quite correct: a problem can easily be NP-hard without being NP-complete. So, the scheduling, where the output is the optimal set of schedule, would be NP-hard, but not NP-complete.
What is NP-hard problem in algorithm?
A problem is NP-hard if an algorithm for solving it can be translated into one for solving any NP-problem (nondeterministic polynomial time) problem. NP-hard therefore means “at least as hard as any NP-problem,” although it might, in fact, be harder.
How do you prove NP-completeness?
We can solve Y in polynomial time: reduce it to X. Therefore, every problem in NP has a polytime algorithm and P = NP. then X is NP-complete. In other words, we can prove a new problem is NP-complete by reducing some other NP-complete problem to it.
Why is NP-complete useful?
NP-complete languages are significant because all NP-complete languages are thought of having similar hardness, in that process solving one implies that others are solved as well. If some NP-complete languages are proven to be in P, then all of NPs are proven to be in P.