Talk:Traveling Salesman Problem is NP-Complete

Is the optimization problem really NP-Complete? It is easy to understand it's NP-Hard, but the given proof doesn't seem to hold for it to be NP. In plain: How do you check whether or not a given path is the shortest of all possible paths in the graph? You can easily calculate the lenght of this path, true, but checking if it is optimal would require solving the original optimization problem.


 * The short answer is yes, they are both NP-complete. The section with the title "The Function Version of the Problem is Reducible to the Decision Version of the Problem" is intended to show that the function problem is NP.  The basic idea is that one can calculate the length of the shortest Hamilton cycle by using the decision problem to establish an upper and lower bound and then any Hamilton cycle with that length is a minimal cycle.  Doing this does require solving about $ln(sn)$ instances of the decision problem to find the tight bound, and another $O(n^2)$ instances to find a path with that length, but in the end it is still just a polynomial increase to the length of the calculation.  I am sorry if this was not clear on the page proper.  Is there a way the article can be rewritten to make this more clear? --Emichael 11:06, 21 September 2011 (CDT)

According to this page [] For now TSP-OPTIMIZE is in NP-Hard but not (necessarily) in NP, so its not in NP-Complete. TSP-DECIDE is both in NP-Hard as well as NP, and is therefore NP-Complete.Slacka (talk) 13:50, 10 November 2012 (UTC)