Why is TSP NP-hard while the Hamiltonian path NP-complete?

Question

Why aren't these 2 problems, namely TSP and Hamiltonian path problem, both NP-complete?

They seem identical.

Answer 1

For a problem X to be NP-complete, it has to satisfy:

1. X is in NP, given a solution to X, the solution can be verified in polynomial time.
2. X is in NP-hard, that is, every NP problem is reduceable to it in polynomial time (you can do this through a reduction from a known NP-hard problem (e.g. Hamiltonian Path)).

There are two versions of the The Travelling Salesman Problem (TSP):

1. The optimization version (probably the one you are looking at), namely, find the optimum solution to the TSP. This is not a decision problem, and hence cannot be in NP, but it is however in NP-hard which can be proven via a Hamiltonian Path reduction. Therefore this isn't an NP complete problem.
2. The decision version - given an integer K is there a path through every vertex in the graph of length < K? This is a decision (yes/no) problem, and a solution can be verified in polynomial time (just traverse the path and see if it touches every vertex) and so it is in NP, but it is also in NP-hard (by an identical proof as above). Since it satisfies both requirements for NP-completeness, it is an NP-complete problem.

Answer 2

The definitions of NP-hardness and NP-completeness are related but different. Specifically, a problem is NP-hard if every problem in NP reduces to it in polynomial time, and a problem is NP-complete if it's both NP-hard and itself in NP.

The class NP consists of decision problems, problems that have a yes/no answer. As a result, TSP cannot be in NP because the expected answer is a number rather than yes or no. Therefore, TSP can be NP-hard, but it can't be NP-complete.

On the other hand, the Hamiltonian path problem asks for a yes/no answer, and it happens to be in NP. Therefore, since it's NP-hard as well, it's NP-complete.

Now, you can take TSP and convert it to a decision problem by changing the question from "what's the cheapest path?" to "is there a path that costs X or less?," and that latter formulation is in NP and also happens to be NP-complete.