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Oct 12, 2017 at 22:42 comment added whatsisname @FSMaxB: The travelling salesman problem is one of the most studied challenges in computer science. You need to review the scientific literature.
Oct 12, 2017 at 18:36 comment added FSMaxB @Philipp: Good enough.
Oct 12, 2017 at 18:19 comment added Philipp Do you need the provable best solution or would a "good enough" solution be sufficient?
Oct 12, 2017 at 16:45 comment added FSMaxB @whatsisname: I understand that I was very vague. In the end all I can do is to try different ways to handle the data and analyze the result. But I can't do that if I don't find any way to finish the computation in let's say a few hours at max. Think of the dataset as arrays of 32bit integers on a regular x86_64 CPU (with 4-32 cores).
Oct 12, 2017 at 16:41 vote accept FSMaxB
Oct 12, 2017 at 15:27 comment added whatsisname What algorithms have you found, especially in academic journals? Also you haven't defined "too slow", as big-O complexity and real world speed are unrelated when looking at a single dataset size. There are algorithms out there that can process datasets of several millions of points in "reasonable" timeframes, but you haven't defined what reasonable is.
Oct 12, 2017 at 15:12 answer added TheCatWhisperer timeline score: 0
Oct 12, 2017 at 14:54 history edited FSMaxB CC BY-SA 3.0
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Oct 12, 2017 at 14:45 comment added FSMaxB Preferably better than randomly choosing two points and swapping them if it improves the total distance. (which would be one approach that is O(1), but maybe not really desirable)
Oct 12, 2017 at 14:43 comment added FSMaxB @BobDalgleish I know that I can not calculate the distance between every pair of points, because that would be O(n²). I'm asking for a heuristic that produces a lower distance for the entire roundtrip than just randomly choosing points, but with a time complexity of less than O(n²).
Oct 12, 2017 at 13:29 comment added BobDalgleish You are asking if you can compute the hamming distance matrix between each pair of points in less than O(n^2). The matrix will n rows and n columns. You have to fill half the entries in the matrix. The best you can do isn ^ 2 / 2 - n, which is O(n^2).
Oct 12, 2017 at 8:31 review First posts
Oct 12, 2017 at 8:46
Oct 12, 2017 at 8:28 history asked FSMaxB CC BY-SA 3.0