It is known that the general traveling salesman problem is NP-hard. Even when the distances follow the triangle inequality. But let's take the problem very literally. There are actual cities (points) on a 2-d euclidean space and a tour must be devised to visit all of them in the smallest total distance. We are given the $x$ and $y$ coordinates of each city and the distances between them are the euclidean distances. Is this simplified problem also NP-hard?

  • $\begingroup$ The TSP is defined on graphs. I don't think that the planarity property makes the problem easier, just by obviousness: no textbook reports a "map" version as being tractable. Also, most graphs can have a planar embedding. $\endgroup$
    – user16034
    Jan 9, 2023 at 9:45

1 Answer 1


This is called the Geometric Traveling Salesman problem. It was proved to be NP-complete by Garey, Graham and Johnson in 1976 and Papadimitriou in 1977.

Source: Computers and Intractability (Garey and Johnson)

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    $\begingroup$ Does that include the case where the map is planar (no crossing lines)? If so, are there any non-NP-complete special cases? $\endgroup$ Jan 9, 2023 at 3:57
  • $\begingroup$ My guess is that it is still $\mathsf{NP}$-complete, yes. $\endgroup$
    – Nathaniel
    Jan 9, 2023 at 5:59
  • $\begingroup$ If you restrict it enough then it won’t be NP complete anymore. For example with the restriction “all distances must be the same “. $\endgroup$
    – gnasher729
    Jan 9, 2023 at 22:06

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