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I am not talking about NP-indeterminate class because those problems have to be shown to not exist either in P or NP-complete class and existence of such problems proves P!=NP. I am interested to know if we are half-way there i.e. problems that have been proven to be not NP-complete but are in NP but we don't know if they are in P as well.

What about such problems whose deterministic polynomial solution was found only recently? Was primality testing such a problem?

I would appreciate if the answers are given for someone like me who only has a high level understanding of computational complexity theory.

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    $\begingroup$ Integer Factorization is suspected to be not NP-Hard (and not P as well) - but not proof yet. $\endgroup$ – amit Oct 17 '12 at 19:41
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Any problem that is provably not $\mathsf{NP}$ complete (w.r.t. polynomial time reductions) would mean that the class of problems reducible to it are distinct from $\mathsf{NP}$. In other words if you show some problem is in $\mathsf{NP}$ but is not $\mathsf{NP}$ complete would imply that $\mathsf{P}$ is not equal to $\mathsf{NP}$.

In short, there is no problem in $\mathsf{NP}$ that we know it is not $\mathsf{NP}$-complete.

Regarding primality and AKS algorithm for it:

Note that the according to the current state of knowledge, it is possible that primality is NP-complete. Existence of a deterministic polynomial time algorithm for a problem does not rule out the possibility of the problem being $\mathsf{NP}$-complete.

You are probably interested in the list of problems that are conjectured to be between $\mathsf{P}$ and $\mathsf{NP}$ (known as $\mathsf{NP}$-intermediate) in which case you can check the answers to Problems Between P and NPC.

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  • $\begingroup$ This is not right since integer factorization is in NP but is suspected to be outside of NP complete. $\endgroup$ – argentage Oct 17 '12 at 20:16
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    $\begingroup$ @airza, what I have written is correct. Suspected is not equal to proven. The OP is asking if there is any problem that is proven not to be NP-complete and there is no such problem and will not be as long as we have not proven P is not equal to NP. $\endgroup$ – Kaveh Oct 17 '12 at 20:19
  • $\begingroup$ @ Kaveh Thanks for the answer. Could you please explain yourself here: "Existence of a deterministic polynomial time algorithm for a problem does not rule out the possibility of the problem being 𝖭𝖯-complete". I thought that if a deterministic polynomial time algorithm exists for a problem then it is in P. $\endgroup$ – Shimano Oct 18 '12 at 7:52
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    $\begingroup$ Yes, that's correct, but it's mot known that P and NP-complete are disjoint sets. For example, if P=NP then all (non-trivial) problems in P are NP-complete. $\endgroup$ – Kaveh Oct 18 '12 at 11:45

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