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I'm looking for a solution of symmetric Traveling Salesman Problem in which we take into account shapes of cities. For example, in the graph from geeksforgeeks.org, the shortest path is 1-2-4-3-1 TSP sample graph

We can imagine that cities aren't points but e.g. horizontal lines (because in a case of choose another shapes the problem will be much harder to solve). In above case we can make two points from one point e.g. point 1 became points: 1' and 1'' and the weight (connected to point 1) in the graph will change.

Are there well-known algorithms to solve TSP in case that if I visit city 1' I must visit city 1'' (and symmetrically)?

trojek
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  • The tour will include both 1' and 1" in any case !? –  Dec 20 '16 at 10:16
  • Yes, we can assume that you must visit both points but you can start from 1' or 1''. – trojek Dec 20 '16 at 10:32
  • You missed my point. A tour always visits all points (and the starting point is immaterial). –  Dec 20 '16 at 10:34
  • More precise when you visit 1' the next point MUST be 1'', of course you can visit firstly 1'' and then 1'. – trojek Dec 20 '16 at 10:37
  • Please rephrase "(because in a case of choose another shapes the problem will be much harder to solve)". Unless your distances are asymmetrical, I think your problem can be reduced to a standard TSP in which you replace the distance from 1' and 1'' to other cities by the respective minimal distance. The distances between N' and N'' are a constant to the total length and do not change regardless of the order in which you travel through the cities. – Andreas Mar 06 '17 at 21:46

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