algorithmics | Solutions for some algorithmic problems | Learning library

I think you need something like a descision tree. It should have a root node and split off at every possible descision. It will be a tree graph ADT, something like this.

The decision problem is NP-complete because you can both have a polynomial time verifier for the solution, as well as the fact that the hamiltonian cycle problem is reducible to TSP_DECIDE in polynomial time.

However, the optimization problem is strictly NP-hard, because even though TSP_OPTIMIZE is reducible from the hamiltonian (HAM) cycle problem in polynomial time, you don't have a poly time verifier for a claimed hamiltonian cycle C, whether it is the shortest or not, because you simply have to enumerate all possibilities (which consumes the factorial order space & time).

What the given reference define is, bottleneck TSP

The Bottleneck traveling salesman problem (bottleneck TSP) is a problem in discrete or combinatorial optimization. The problem is to find the Hamiltonian cycle in a weighted graph which minimizes the weight of the most weighty edge of the cycle.

The problem is known to be NP-hard. The decision problem version of this, "for a given length x is there a Hamiltonian cycle in a graph G with no edge longer than x?", is NP-complete. NP-completeness follows immediately by a reduction from the problem of finding a Hamiltonian cycle.

This problem can be solved by performing a binary search or sequential search for the smallest x such that the subgraph of edges of weight at most x has a Hamiltonian cycle. This method leads to solutions whose running time is only a logarithmic factor larger than the time to find a Hamiltonian cycle.

The mistake is to say that the TSP is NP complete. Truth is that TSP is NP hard. Let me explain a bit:

The TSP is a problem defined by a set of cities and the distances between each city pair. The problem is to find a circuit that goes through each city once and that ends where it starts. This in itself isn't difficult. What makes the problem interesting is to find the shortest circuit among all those that are possible.

Solving this problem is quite simple. One merely need to compute the length of all possible circuits, then keep the shortest one. Issue is that the number of such circuits grows very quickly with the number of cities. If there are n cities then this number is factorial of n-1 = (n-1)(n-2)...3.2.

A problem is NP if one can easily (in polynomial time) check that a proposed solution is indeed a solution.

Here is the trick.

In order to check that a proposed tour is a solution of the TSP we need to check two things, namely

1. That each city is is visited only once
2. That there is no shorter tour than the one we are checking

We didn't check the second condition! The second condition is what makes the problem difficult to solve. As of today, no one has found a way to check condition 2 in polynomial time. It means that the TSP isn't in NP, as far as we know.

Therefore, TSP isn't NP complete as far as we know. We can only say that TSP is NP hard.

When they write that TSP is NP complete, they mean that the following decision problem (yes/no question) is NP complete:

TSP_DECISION : Given a number L, a set of cities, and distance between all city pairs, is there a tour visiting each city exactly once of length less than L?

This problem is indeed NP complete, as it is easy (polynomial time) to check that a given tour leads to a yes answer to TSPDECISION.