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To reformulate my comments as an answer, and to expand a bit: We don't know whether NPNP = NP — it's a notoriously open problem in complexity theory, though as with P versus NP we suspect that they a …

A Turing machine, it must be remembered, is a kind of flowchart. So is the structure of a computer program generally. So turning "a flowchart" into a formal answer to the problem should be fairly easy …

For a language $L' \in \mathsf{coNP}$, there are certificates for all NO instances, which is to say inputs $x \notin L'$, which could be efficiently verified and used as proof that $x \notin L$. Suppo …

The class BQP (see also Complexity Zoo: BQP) was defined by E. Bernstein and U. Vazirani. Quantum complexity theory. SIAM Journal on Computing 26, pp. 1411-1473 (1997). The class QMA (see also Co …

The crux is that you have to consider the decision problem: Travelling Salesman Problem (Decision Version). Given a weighted graph G and a target cost C, is there a Hamiltonian cycle in G whose we …

By the nondeterministic version of the time-hierarchy theorem, we have $\mathsf{NP} \subsetneq \mathsf{NEXP}$, where $\mathsf{NEXP}$ is the class of problems solvable in non-deterministic exponential- …

By the nondeterministic time hierarchy, you could show that the problem is $\mathsf {NEXP}$-hard; as $\mathsf {NP} \ne \mathsf {NEXP}$, it is impossible to reduce the problem in polynomial time to any …

Here's another way of looking at the point that Shaull makes with respect to "deciders". A problem is in NP if and only if there is an algorithm $V: \{0,1\}^n \times \{0,1\}^{\mathrm{poly}(n)} \to \{ …

Short answer. If we formulate an appropriate decision problem version of the Discrete Logarithm problem, we can show that it belongs to the intersection of the complexity classes NP, coNP, and BQP. …

Because you don't know ahead of time whether or not any given input is a 'yes' instance — that is, whether there exists any accepting path — it makes sense for the sake of uniformity to bound the run- …