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Jun 2, 2021 at 13:12 comment added Demetri Pananos "Machine languish" Oh that's good, I should use that.
Jun 2, 2021 at 12:56 comment added PierreSimonDeLaplace I was thinking about including some randomness and iterate over the matrix in a different order few times, with changing order of the rows, the i variable in the algorithm. It will not find the best solution, but maybe good enough. It would be more machine languagish way approach, but I'm not sure about it.
Jun 2, 2021 at 12:36 comment added Demetri Pananos That is a problem, but I’ve already noted that by identifying the approach as greedy.
Jun 2, 2021 at 12:26 comment added PierreSimonDeLaplace I can see one problem with this approach. If one has selected in the first iteration a column with the lowest value of some distance it doesn't mean that this column with comparison to another row could not give better results in the total, if we sum up all the norms. For example if A=[10 10 1] and B=[30 20 1] one can see that the best pairs would be $\{\{1,1\},\{10,20\},\{10,30\}\}$, but the algorithm will return $\{\{10,1\},\{10,20\},\{1,30\}\}$.
Jun 2, 2021 at 3:01 history edited Demetri Pananos CC BY-SA 4.0
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Jun 2, 2021 at 2:51 history edited Demetri Pananos CC BY-SA 4.0
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Jun 2, 2021 at 2:13 history answered Demetri Pananos CC BY-SA 4.0