Are NP languages decidable?
All languages in P and NP are decidable. Any undecidable language, such as the halting problem, cannot be in NP. The nondeterministic time hierarchy theorem can be used to show that there are languages in NEXP that are not in NP.
Is NP-complete problem and NP-hard decidable or computable?
Since NP problems are computable, every NP problem is polynomial-time reducible to H, so H is NP-hard.
Is every problem in P decidable?
No, because that isn’t even a computational problem.
Are all problems in NP NP-complete?
A problem is called NP (nondeterministic polynomial) if its solution can be guessed and verified in polynomial time; nondeterministic means that no particular rule is followed to make the guess. If a problem is NP and all other NP problems are polynomial-time reducible to it, the problem is NP-complete.
Are all languages decidable?
All finite languages are regular. Some infinite languages are regular. Only infinite languages can be undecidable. To be undecidable, there must be individual strings in the language that cause the TM to fail.
Is every context-free language decidable?
1. (a) True, since every regular language is context-free, every context-free language is decidable, and every decidable language is Turing-recognizable.
Is every NP-hard language is in NP?
The answer to this question depends on the complexity assumptions. If P=NP, then every nontrivial language L is NP-hard. [A language is said to be nontrivial if it contains at least one yes instance and at least one no instance.]
Can an Undecidable language be NP-complete?
An NP-hard is a problem that is at least as hard as any NP-complete problem. Therefore an undecidable problem can be NP-hard.
Is every language in P also in NP?
Clearly, any languages in P is also in NP, since NTM can linearly simulate TM. It is possible that for every language in NP, there always exists a polynomial TM that decides it, but we are not smart enough yet to discover those polynomial algorithms for certain hard problems.
Are all NP languages in NP-complete?
The definition of nondeterministic polynomial-time computable is almost the same, except that the turing machine is allowed to be nondeterministic. A language L is NP-hard if it is harder than every NP problem. A language is NP-complete if it is both NP-hard and NP.
Which problem is not NP-complete?
Which of the following problems is not NP complete? Explanation: Hamiltonian circuit, bin packing, partition problems are NP complete problems. Halting problem is an undecidable problem.
How do you know if a language is decidable?
A language is decidable if and only if it and its complement are recognizable. Proof. If a language is decidable, then its complement is decidable (by closure under complementation). Either w ∈ L, or w ∈ L .
Can a p or NP language be decidable?
Yes, any language in P or NP is decidable. First, 2 definitions: NP is the class of languages that can be recognized by a nondeterministic Turing machine (TM) in polynomial time (http://en.wikipedia.org/wiki/NP_…).
Are there any undecidable problems in the NP Set?
Clearly there aren’t any undecidable problems in NP. However, according to Wikipedia: NP is the set of all decision problems for which the instances where the answer is “yes” have [.. proofs that are] verifiable in polynomial time by a deterministic Turing machine.
Which is an equivalent definition of the problem NP?
An equivalent definition of NP is that it consists of all problems that are decidable (not just verifiable) in polynomial time by a non-deterministic Turing machine.
Is there an undecidable problem in a NTM?
NTMs are known to be no more powerful than TMs in the sense that the set of problems decidable by NTMs is identical to the set of problems decidable by TMs, so clearly by this definition there can be no undecidable problems in NP.