Can NP-hard problems be solved in polynomial-time?

If P ≠ NP, then NP-hard problems could not be solved in polynomial time. Some NP-hard optimization problems can be polynomial-time approximated up to some constant approximation ratio (in particular, those in APX) or even up to any approximation ratio (those in PTAS or FPTAS).

What is NP completeness problem?

NP-complete problem, any of a class of computational problems for which no efficient solution algorithm has been found. Many significant computer-science problems belong to this class—e.g., the traveling salesman problem, satisfiability problems, and graph-covering problems.

Which is the class of decision problems?

Polynomial time and space NP = the class of all decision problems solvable by nondeterministic polynomial-time algorithms. CoNP the class of all complements of problems that are in NP. PSPACE= the class of all problems solvable by deterministic polynomial-space algorithms.

Which of the problem Cannot be solved by backtracking method?

Which of the problems cannot be solved by backtracking method? Explanation: N-queen problem, subset sum problem, Hamiltonian circuit problems can be solved by backtracking method whereas travelling salesman problem is solved by Branch and bound method.

Is there a polynomial time algorithm for NP complete problems?

No polynomial time algorithm has yet been discovered for any NP complete problem, nor has anybody yet been able to prove that no polynomial-time algorithm exist for any of them. The interesting part is, if any one of the NP complete problems can be solved in polynomial time, then all of them can be solved.

How do you prove that a problem is NP-complete?

From the definition of NP-complete, it appears impossible to prove that a problem L is NP-Complete. By definition, it requires us to that show every problem in NP is polynomial time reducible to L. Fortunately, there is an alternate way to prove it.

Is the clique decision problem NP-hard or NP-complete?

Now, let the Clique Decision Problem by C. To prove that C is NP-Hard, we take an already known NP-Hard problem, say S, and reduce it to C for a particular instance. If this reduction can be done in polynomial time, then C is also an NP-Hard problem. The Boolean Satisfiability Problem (S) is an NP-Complete problem as proved by the Cook’s theorem.

What is the hardest problem in NP?

NP-complete problems are the hardest problems in NP set. A decision problem L is NP-complete if: 1) L is in NP (Any given solution for NP-complete problems can be verified quickly, but there is no efficient known solution). 2) Every problem in NP is reducible to L in polynomial time (Reduction is defined below).