What is you understanding on NP and co-NP problems?

Complementary Problems While an NP problem asks whether a given instance is a yes-instance, its complement asks whether an instance is a no-instance, which means the complement is in co-NP. Any yes-instance for the original NP problem becomes a no-instance for its complement, and vice versa.

How and when a NP-hard problem can also be NP-complete?

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.

What is the relation between P and NP?

NP is set of 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 deterministic machine in polynomial time can also be solved by non-deterministic machine in polynomial time) but P≠NP.

Which of the following problems are in co-NP?

So co-NP contains problems such as non-3-colourability, Boolean unsatisfiability and so on.

Is NP-complete in NP?

A problem p in NP is NP-complete if every other problem in NP can be transformed (or reduced) into p in polynomial time. It is not known whether every problem in NP can be quickly solved—this is called the P versus NP problem.

What is the relationship between the classes of NP complete and NP?

A problem is said to be NP-hard if everything in NP can be transformed in polynomial time into it even though it may not be in NP. Conversely, a problem is NP-complete if it is both in NP and NP-hard. The NP-complete problems represent the hardest problems in NP.

What happens when P will be equal to NP?

If P equals NP, every NP problem would contain a hidden shortcut, allowing computers to quickly find perfect solutions to them. But if P does not equal NP, then no such shortcuts exist, and computers’ problem-solving powers will remain fundamentally and permanently limited.

Is there any NP-complete problem that is Turing-reducible to Q?

(D) Incorrect because there is no NP-complete problem that is polynomial time Turing-reducible to Q. Let X be a problem that belongs to the class NP. Then which one of the following is TRUE? If X can be solved deterministically in polynomial time, then P = NP. If X is NP-hard, then it is NP-complete.

What is the difference between NP and NP-complete problems?

(1) The problem of determining whether there exists a cycle in an undirected graph is in P. (2) The problem of determining whether there exists a cycle in an undirected graph is in NP. (3) If a problem A is NP-Complete, there exists a non-deterministic polynomial time algorithm to solve A.

Can all NP-complete problems be solved in polynomial time?

The answer is B (no NP-Complete problem can be solved in polynomial time). Because, if one NP-Complete problem can be solved in polynomial time, then all NP problems can solved in polynomial time. If that is the case, then NP and P set become same which contradicts the given condition.

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).