[hexomancer]

How?

edit: The wikipedia mentions that the TSP problem is NP-hard and explicitly says that the decision version of this problem is NP-complete. My assumption is that if the optimization version was proven to be NP-hard there would be no need to explicitly mention the decision version.

I can think of a way to use the decision version to find a solution to the optimization version but i feel like it must be flawed:

First we do a binary search on `L` (the length of the tour- input to the decision version) so we can find the optimal L within a factor of epsilon (maybe this epsilon is the flaw? But I don't think that is the case.) Now we pick an edge and increase its weight to infinity. Now we do the binary search again on the new graph. If the value of the optimal solution has changed it means that the edge must be in the optimal optimization solution. By doing the same process on all the edges we can find the optimal solution.

As I said there must be a flaw in the above algorithm but I can't find it.

[mihaild]

First, use binary search to find answer. Then, remove edges one-by-one if removing this edge will not destroy all remaining path with shortest length, until your graph is reduced to a single cycle.

[hexomancer]

Thanks for the answer, I have edited my post and I have basically the same solution. (Only instead of removing the edges I increase their weight to infinity because I think the TSP problem is defined only on complete graphs) But as I have mentioned in my comment I feel like this solution is flawed.

[zmonx]

You can without loss of generality assume that the weights are integers (if they are rational, you can multiply all weights so that they become integral).

Hence, you can use bisection to compute the actual optimum, not involving epsilon at all.

[mihaild]

Epsilon isn't a problem, as TSP is NP-complete even for integer weights. Your solution needs some modification for case where we have multiple optimal cycles (as you will find edges that are included in at least one cycle).

I don't think there is anything wrong with optimization been reducible to decision - it's quite common method both in theory and in practice.

[hexomancer]

I didn't say there is anything wrong with reducing the optimization to decision in general. But I feel this specific application to the TSP problem may be flawed. (for the reasons i mentioned before. i.e wikipedia being very specific about the decision version) For another resource see: https://www.ibm.com/developerworks/community/blogs/jfp/entry...

note: I am not implying that the above source is reputable. But it does hint that the solution to this problem probably is not this trivial.

[mihaild]

Decision version of TSP is NP-complete. Optimization version is Cook-reducible to decision version of TSP. And it problem is Cook-reducible to P-problem, then the problem is itself in P. (note that it's not true if you replace P with NP, for example)

[hexomancer]

Do you have a reference to a reputable source that proves the optimization version is cook-reducable to decision version?

[mihaild]

I don't. (it's always hard to find reference for easy problems)

[PrunJuice]

One minor foible. Don't select edges that are part of the minimum path, instead remove edges that are not part of the minimum path until only the minimum path remains. This strategy works even on graphs with multiple minimum paths.

[hexomancer]

I don't get why having multiple optimal paths would falsify my algorithm. Can you provide an explicit counter-example?

Here is why I think the algorithm is correct:

At each step the edge that we are considering is either contained in all the optimal solutions or only some of them. If the edge is contained in all the solutions, increasing its weight to infinity would change the optimal solution and we pick that edge in our solution. Otherwise (if the edge is contained in only some of the solutions) increasing the weight would not change the solution because there is another optimal solution that does not contain that edge so we do not pick that edge.

So we can prove this theorem: At every step of the algorithm if an edge is picked, it is contained in all the optimal solutions.

So the algorithm does not pick any extra edges. Now we have to prove that it includes all the necessary edges. But that is easy because each time that we choose not to including an edge, we are sure that there is an optimal solution in the remaining graph so we are never left with a graph with no optimal solution.

I think you assumed that I meant we change the edge weights from infinity back to their original value at each step but that is not what I meant.

[PrunJuice]

> If the value of the optimal solution has changed it means that the edge must be in the optimal optimization solution.

I interpreted this as meaning you were selecting the falsifying removals' edges, instead of removing the non-falsifying ones. You've got it.