All NP complete problems are NP hard problems when solved in a different way.

But, all NP hard problems are not NP complete.

Ex:

1. Traveling salesman problem. It is both NP hard and NP complete. We can find that whether the solution is correct or not in the given period of time. In this way, it is NP complete.

But, to find the shortest path, i.e. optimization of Traveling Salesman problem is NP hard. If there will be changing costs, then every time when the salesperson returns to the source node, then he will be having different shortest path. In this way, it is hard to solve. It cannot be solved in the polynomial time. In this way, it is NP hard problem.

2. Halting problem.

3. Sum of subset problem.

A problem is 'in NP' if there exists a polynomial time complexity algorithm which runs on a Non-Deterministic Turing Machine that solves it.

A problem is 'NP Hard' if all problems in NP can be reduced to it in polynomial time, or equivalently if there is a polynomial-time reduction of any other NP Hard problem to it.

A problem is NP Complete if it is both in NP and NP hard.

No. Verifying if I'm lying is NP-Complete.

- a problem in NP means that it can be solved in polynomial time with a non-deterministic turing machine

- a problem that is NP-hard means that all problems in NP are "easier" than this problem

- a problem that is NP-complete means that it is in NP and it is NP-hard

example - Hamiltonian path in a graph:

The problem is: given a graph as input, an algorithm must say whether there is a hamiltonian path in it or not.

in NP: here is an algorithm that works in polynomial time on a non-deterministic turing machine:

guess a path in the graph. Check that it is really a hamiltonian path.

NP-hard: we use reduction from a problem that is NP-comlete (SAT for example). Given an input for the other problem we construct a graph for the hamiltonian-path problem. The graph should have a path iff the original problem should return "true".

Therefore, if there is an algorithm that executes in polynomial time, we solve all the problems in NP in polynomial time.j

P is the class of problems that can be solved in polynomial time. That is, the size of the input affects the length of the computation multiplicatively.

NP is the class of problems in which the effect of input size on the length of the computation is exponential or factorial. In addition, for a problem to be in this class, a proposed or candidate solution must be checkable in polynomial time.

The usual example here has to do with multiplication and factoring. You can take two very long prime numbers and quickly multiply them. So multiplication is in P. Given the result of that multiplication, the task of finding its prime factors is not easy. That is, there is no known algorithm that can solve the factoring problem (given very large numbers) in polynomial time.

Within the NP class is a subclass consisting of the hardest problems in NP. A problem belonging to this class is called NP-complete. This means that, if a solution can be found to this problem (examples include the travelling salesman problem and the trunk-packing problem), then that solution can be transformed into a solution for all NP problems.