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This is probably NP-hard. Suppose you have points A, B, C, D, E. Let's try a simple deterministic algorithm. Suppose you calculate the distance from point B to line A-C and it's smaller than your threshold (1 meter). So you delete B. Then you try the same for C to line A-D, but it's bigger and D for C-E, which is also bigger. But it turns out that the optimal solution is A, B, E, because point C and D are close to the line B-E, yet on opposite sides. If you delete 1 point, you cannot be sure that it should be a point that you should keep, unless you try every single possible solution (which can be <code>n^n</code> in size, so on <code>n=80</code> that's more than the minimum number of atoms in the known universe). Next step: try a brute force or branch and bound algorithm. Doesn't scale, doesn't work for real-world size. You can safely skip this step :) Next step: First do a determinstic algorithm and improve upon that with a metaheuristic algorithm (tabu search, simulated annealing, genetic algorithms). In java there are a couple of open source implementations, such as <a href="http://www.jboss.org/drools/drools-planner" rel="nofollow">Drools Planner</a>. All in all, you 'll probably have a workable solution (although not optimal) with the first simple deterministic algorithm, because you only have 1 constraint. A far cousin of this problem is probably the Traveling Salesman Problem variant in which the salesman cannot visit all cities but has to select a few.
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1. POSimplification / optimization of GPS track
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1. USGeoffrey De Smet
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1. COIf we strive to find an really optimal solution (with a given target number of points that should be left), then yes, it's probably NP-hard, but ''good enough'' solution is fine for me. Basically, it boils down to finding a good enough approximation of a line in a 3D space of ''(x, y, t)'', composed using straight segments.
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