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There are n points in the 2D plane. A robot wants to visit all of them but can only move horizontally or vertically. How should it visit all of them so that the total distance it covers is minimal?

SD.
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This is the Travelling Salesman Problem where the distance between each pair of points is |y2-y1|+|x2-x1| (called Rectilinear distance or Manhattan distance). It's NP-hard which basically means that there is no known efficient solution.

Methods to solve it on Wikipedia.

The simplest algorithm is a naive brute force search, where you calculate the distance for every possible permutation of the points and find the minimum. This has a running time of O(n!). This will work for up to about 10 points, but it will very quickly become too slow for larger numbers of points.

Mark Byers
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  • Are you sure the distance between pair of points is |y2-y1|+|x2-x1|? I'm quite sure its': sqrt((y2-y1)^2+(x2-x1)^2) – Waleed Amjad Dec 05 '09 at 22:47
  • That said, if you want a pretty good solution but not necessarily the absolute optimal one, that wikipedia page also has some good suggestions. – UncleO Dec 05 '09 at 22:49
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    @Sbm007 Apparently, the robot can't move on a diagonal, so Mark is correct. – UncleO Dec 05 '09 at 22:50
  • I voted for your answer, but I'd like to see a little better definition of NP-hard. There's much more to it than what you posted, obviously. – San Jacinto Dec 05 '09 at 23:02
  • @San Jacinto: OK, I linked 'NP-hard' to Wikipedia. It's described there better than I could do myself. – Mark Byers Dec 05 '09 at 23:05