If lower bound of a problem is exponential then is it NP?

Question

Assuming that we have a problem $p$ and we showed that the lower bound for solving $p$ is $\mathcal{\Omega}(2^n)$.

• can lower bound $\mathcal{\Omega}(2^n)$ implies the problem in $NP$?

Answer 1

No. For example, the halting problem has an $\Omega(2^n)$ lower bound, but it is not in NP (since it is not computable).

The nondeterministic time hierarchy theorem shows that any NEXP-complete problem is another example (with $2^n$ potentially replaced by a smaller exponential function $c^{n^\epsilon}$).

NP is an upper bound on the complexity of a problem.

Answer 2

No. First, as Yuval points out, the problem could be much harder than the lower bound that you've proven.

Second, even if the problem takes time $\Theta(2^n)$ to solve, we don't know how this relates to $\mathbf{NP}$. It's possible that $\mathbf{P}=\mathbf{NP}$, in which case any problem in $\mathrm{TIME}[\Omega(2^n)]$ is certainly not in $\mathbf{NP}$ by the time hierarchy theorem. But even if $\mathbf{P}\neq\mathbf{NP}$, it's possible that the problem requires exponential space so isn't in $\mathbf{NP}$.

The best algorithms we know for $\mathbf{NP}$-complete problems take exponential time but you shouldn't assume that "in $\mathbf{NP}$" means "takes exponential time" or vice-versa.