NP problemleri var mı, P de değil, NP Complete değil mi?

$\mathsf{NP}$ ( $\mathsf{P}$ değil) Complete değil, bilinen herhangi bir sorun var mı? Benim anladığım şu ki, durumun olduğu yerde şu anda bilinen bir sorun yok, ancak bir olasılık olarak göz ardı edilmedi. $\mathsf{NP}$

Bir sorun varsa $\mathsf{NP}$ (ve $\mathsf{P}$ ), ancak $\mathsf{NP\text{-}complete}$ , bu durum sorun örnekleri arasında mevcut bir izomorfik bir sonucu olacaktır $\mathsf{NP\text{-}complete}$ grubu? Bu durumda, nasıl bunu bilemez $\mathsf{NP}$ sorun 'sert' şu anda olduğu gibi tespit olandan değil $\mathsf{NP\text{-}complete}$ ayarladınız mı?

NP tamamlanmayan (P de değil) NP'de bilinen bir problem var mı? Benim anladığım şu ki, durumun olduğu yerde şu anda bilinen bir sorun yok, ancak bir olasılık olarak göz ardı edilmedi.

Hayır, bu bilinmemektedir (önemsiz dillerin haricinde ve Σ * birçok kimse azalmalar dikkate alındığında, bu iki nedeni birçok kimse indirimleri tanımının tam değildir, genellikle bu iki göz ardı edilir). Bir varlığı K p için tam değildir sorun , N , P , birçok-on polinom zaman azalmalar ima wrt bu P ≠ N P (yaygın olarak inanılan rağmen) bilinmemektedir. İki sınıfları daha sonra farklı ise biz orada sorunlar olduğunu biliyoruz N P bunun için tam değildir, herhangi bir sorunla almak P .$\emptyset$$\Sigma^*$$\mathsf{NP}$$\mathsf{NP}$$\mathsf{P}\neq\mathsf{NP}$$\mathsf{NP}$$\mathsf{P}$

NP (P değil) fakat NP Tamamlanmamış bir sorun varsa, bu, bu sorunun örnekleri ile NP Tamamlandı arasında var olan hiçbir izomorfizmin sonucu olmaz mıydı?

İki karmaşıklık sınıfları ile daha sonra farklı ise Ladner teoremi problemler vardır yani arasında olan, aramaddesini P ve N , P - C o m p l e t e .$\mathsf{NP}$$\mathsf{P}$$\mathsf{NP\text{-}complete}$

Bu durumda, NP sorununun şu an NP Complete seti olarak tanımladığımızdan daha zor olmadığını nasıl bilebiliriz?

Bunlar yine polinom zaman indirgenebilir olan problemleri de daha zor olamaz bu yüzden , N , P - C o m p l e t e sorunları.$\mathsf{NP\text{-}complete}$$\mathsf{NP\text{-}complete}$

As @Kaveh stated, this question is only interesting if we assume $P \neq NP$; the rest of my answer takes this as an assumption, and mostly provides links to further wet your appetite. Under that assumption, by Ladner's theorem we know that there are problems that are neither in $P$ nor $NPC$; these problems are called $NP$-intermediate or $NPI$. Interestingly enough, Ladner's theorem can be generalized to many other complexity classes to produce similar intermediate problems. Further, the theorem also implies, that there is an infinite hierarchy of intermediate problems that are not poly-time reducible to each other in $NPI$.

Unfortunately, even with the assumption $P \neq NP$ it is very difficult to find natural problems that would be provably $NPI$ (of course you have the artificial problems coming from the proof of Ladner's theorem). Thus, even assuming $P \neq NP$ at this time we can only believe some problems to be $NPI$ but not prove it. We come to such beliefs when we have reasonable evidence to believe that an $NP$ problem is not in $NPC$ and/or not in $P$; or just when it has been studied for a long time and avoided fitting into either class. There is a pretty comprehensive list of such problems in this answer. It includes such all-time favorites as factoring, discrete log, and graph-isomorphism.

Interestingly, some of these problems (notable: factoring and discrete log) have polynomial time solutions on quantum computers (i.e. they are in $BQP$). Some other problems (such as graph-isomorphism) are not known to be in $BQP$, and there is ongoing research to resolve the question. On the other hand, it is suspected that $NPC \not\subseteq BQP$, thus people don't believe we will have an efficient quantum algorithm for SAT (although we can get a quadratic speed up); it is an interesting question to worry about what sort of structure $NPI$ problems need in order to be in $BQP$.

No NP-complete problems are known to be in P. If there is a polynomial-time algorithm for any NP-complete problem, then P = NP, because any problem in NP has a polynomial-time reduction to each NP-complete problem. (That's actually how "NP-complete" is defined.) And obviously, if every NP-complete problem lies outside of P, this means that P ≠ NP. We're not really sure why it's hard to show it one way or the other; if we knew the answer to that question, we'd probably know a lot more about P and NP. We have a few proof techniques that we know don't work (relativization and natural proofs, for example), but don't have a principled explanation as to why this problem is hard.

If there are problems in NP which are not in P, then there is actually an infinite hierarchy of problems in NP between those in P and those which are NP complete: this is a result called Ladner's theorem.

Hope this helps!

There are some problems which are NP, but no one knows they are NP-Complete or $P$, like graph isomorphism¹. But as I know there isn't special complexity class for such a problems, may be I'm wrong, though.

May be it's $P$, e.g before AKS algorithm no one knows primality testing is $P$ or NPC.

Also there are some problems which are NPC but not in strong sense or weakly NP-Complete, like 2-Partition problem, means, if the input numbers are in polynomial order of input size, this problems can be solved in $P$ (or there is a pseudo polynomial time algorithm for them).

¹ Similar problem: sub graph isomorphism is NP-Complete in strong sense.