algorithm - Why do we say that NP complete problems are NP? -


i have gone through links regarding topic still confused why consider np complete np. can verify in polynomial time np complete problems np, have np problems can solved in polynomial time np complete problems can't solved in polynomial time doesn't contradict property of calling np complete problems np?

there's set p of decision problems can solved deterministic turing machine in polynomial time.

then there's set np of decision problems can solved non-deterministic turing machine in polynomial time, i.e. solution can verified in polynomial time given witness string.

a deterministic turing machine can simulate non-deterministic one, know there exponential-time algorithm solve np problems. don't know whether don't in fact have p = np.

an np-complete problem np problem at least hard other np problem. example, sat np-complete because can encode non-deterministic turing machine , solving sat means simulating machine. can show np-completeness of problem decision problem demonstrating known np-complete problem b can reduced in polynomial time. means if can solved in polynomial time, b can solved in polynomial time too, in sense @ least hard b.

but have np problems can solved in polynomial time well

exactly, because p subset of np.

np complete problems can't solved in polynomial time

we don't know sure.

doesn't contradict property of calling np complete problems np

not @ all. yes there problems in np know in p. doesn't mean there no problems in np not in p. of course don't know latter. case every np-complete problem in p, in case p = np.


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