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I am tasked with designing a flight assistant and one of the problems at hand is finding an optimum route that visits every airport and returns to the airport of origin. In this problem I've taken the airports as nodes and the flights as edges. Since the task is only to visit every airport it doesn't matter if I visit any airport more than once as long as I've visited them all. The graph is directed since a flight need not return to the airport it came from.

Since vertices may be repeated this wouldn't be an actual TSP implementation.

The answer I've found to this online is to make the graph a complete one where if airport A and B weren't connected, the edge connecting them will have the weight of the shortest path from A to B. Solve the TSP there and then in that cycle replace edges that don't actually exist with the actual path they represent.

Problem is that this still isn't enough because since the problem isn't a metric TSP, even if it isn't necessary to repeat vertices the optimum path may require you to repeat them.

For example, take the graph with nodes A, B and C with edges (A,B),(B,A) of weight 1, (A,C),(C,A) also with weight 1 and (B,C),(C,B) with weight 10. In this problem you wouldn't need to repeat vertices but the shortest path would be A-B-A-C-A of weight 4 where A-B-C-A, which doesn't repeat vertices, would have weight 12.

Thanks in advance for the help.

azurox
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  • I think your approach still works, if you replace existing edges with their shortest-path dummy edges if the shortest-path edge is cheaper. So in your example you would replace (B,C) and (C,B) with edges of weight 2 since you would never take (B,C) or (C,B), you'd take (B,A)-(A,C) or (C,A)-(A,B). (By the way I think you meant the second pair of edges to be (A,C),(C,A) not (A,C),(B,C).) – LarrySnyder610 Nov 08 '17 at 01:16
  • I agree that probably should work, I'll try implementing that (you're also right about the edges being wrong). Thank you very much. – azurox Nov 08 '17 at 11:33

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