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So the Euclidean TSP decision problem is NP-complete (see http://dx.doi.org/10.1016/0304-3975(77)90012-3 ) so the TSP optimization problem should be NP-hard.

On the other hand there is a PTAS for the Euclidean TSP (see http://dx.doi.org/10.1145/290179.290180 ).

Wikipedia says that a PTAS is not possible for strongly NP-hard Problems.

So is this a contradiction, or is the Euclidean TSP "only" weakly NP-hard? Did I miss something else?

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  • $\begingroup$ see also Why we can't have FPTAS for strong NP complete problems $\endgroup$ – vzn Apr 30 '15 at 18:15
  • $\begingroup$ Why would it be a contradiction? "Strongly NP-hard" is not the same concept as "NP-hard". $\endgroup$ – D.W.♦ Apr 30 '15 at 21:11
  • $\begingroup$ @D.W. Of course it is not. I just wanted to know where the failure in my understanding was. Maybe the question is not optimally formulated. $\endgroup$ – surt91 May 1 '15 at 15:39
  • $\begingroup$ You could have PTAS for Strongly NP-hard Problem!!! But you cannot have FPTAS for Strongly NP-hard problem unless P=NP $\endgroup$ – user777 Mar 1 '18 at 10:09
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You're confusing a polynomial time approximation scheme (PTAS) and a fully polynomial time approximation scheme (FPTAS). Euclidean TSP has a PTAS, but it is not an FPTAS because the polynomial increases in degree as 1 / ε decreases. Only an FPTAS is disallowed for strongly NP-hard problems, assuming P $\neq$ NP.

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  • $\begingroup$ feel something is missing here. has euclidean TSP been proven strongly NP hard? $\endgroup$ – vzn Apr 30 '15 at 18:17
  • $\begingroup$ @vzn [Papadimitriou 1977] show that Euclidean TSP is NP-complete. The reduction from Exact Cover by 3-Sets problem (this problem is known to be as another way of 3-Dimentional Matching problem which has reduction from 3-SAT, all these problems are known to be NP-hard in the strong sense "hard for even small number" so this implies that Euclidean TSP is NP-hard in the strong sense). [sciencedirect.com/science/article/pii/0304397577900123/… $\endgroup$ – user777 Mar 1 '18 at 10:06

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