Can decision problems be NP-complete?

Yes, NP is commonly only defined for decision problems In computational complexity theory, NP (nondeterministic polynomial time) is a complexity class used to classify decision problems.

What is a decision problem NP?

NP is the set of decision problems for which the problem instances, where the answer is “yes”, have proofs verifiable in polynomial time by a deterministic Turing machine, or alternatively the set of problems that can be solved in polynomial time by a nondeterministic Turing machine.

What is P NPC and NPH problem?

NP Hard Problems (NPH) In other words if a NPH problem is non-deterministic polynomial time solvable, it is a NPC problem. Example of a NPH problem that is not NPC is Halting Problem (halting problem is undecidable and all undecidable problem is guaranteed not to be in NP and hence not NPC also).

How do you prove a decision is NP-complete?

We say X is NP-complete if: X ∈ NP • for all Y ∈ NP, Y ≤P X. If these hold, then X can be used to solve every problem in NP. Therefore, X is definitely at least as hard as every problem in NP.

Is Travelling salesman NP-complete?

Traveling Salesman Optimization(TSP-OPT) is a NP-hard problem and Traveling Salesman Search(TSP) is NP-complete. However, TSP-OPT can be reduced to TSP since if TSP can be solved in polynomial time, then so can TSP-OPT(1). I thought for A to be reduced to B, B has to be as hard if not harder than A.

What is NP-hard and NP-complete problems?

A problem is NP-hard if all problems in NP are polynomial time reducible to it, even though it may not be in NP itself. If a polynomial time algorithm exists for any of these problems, all problems in NP would be polynomial time solvable. These problems are called NP-complete.

Which of the following is NP-complete problem?

Explanation: Hamiltonian circuit, bin packing, partition problems are NP complete problems.

What is P NP NP-complete?

NP is set of decision problems that can be solved by a Non-deterministic Turing Machine in Polynomial time. P is subset of NP (any problem that can be solved by a deterministic machine in polynomial time can also be solved by a non-deterministic machine in polynomial time).

Is independent set NP-complete?

The independent set decision problem is NP-complete, and hence it is not believed that there is an efficient algorithm for solving it. The maximum independent set problem is NP-hard and it is also hard to approximate.