Traveling salesman problem with a lot of time, a variation; using incremental method and the second order Delaunay triangulation(Open 22/Aug/2010 : The 0th Revision Sunday, 22-Aug-2010 22:48:17 JST)
In this page, we can find the shortest route passing through the points you clicked.
Please click black screen. When you click more than 4 points, the screen shows the shortest route to pass through the points. But if the points are a lot (about more than 10), it takes a lot of time to obtain the solution, sorry.
In this version 2, I delete some combinations for quick computation. The condition of combinations are that two line segments cross each other.
In this version 3, further, comparing the shortest paths from Delaunay triangulation. In some cases, two paths have difference.
In this version 4, further, FURTHER, for more quick response, some technic are used.
1 Set Delaunay shortest path.
2. Compute Exact shortest paths. But I don't compute when lengths are longer than Delaunay path.
In this version 5, further, some revision I tried. At first, using incremental method for initial solution. By comparing initial solution, obtaining earlier in Delaunay step. But still very slow...
In this variation, using the second order Delaunay triangulation in place of the (first order )Delaunay triangulation. In case of first order triangulation, some path for TSP is not exact shortest path. But second order triangulation has more edges than first order, so I believe the probability that we can obtain exact TSP path is higher, but no guarantee.
idela2tsp.java
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