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Mar 23, 2020 at 0:36 history edited Delaram Nematollahi CC BY-SA 4.0
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Mar 17, 2020 at 17:40 vote accept Delaram Nematollahi
Mar 17, 2020 at 15:00 history tweeted twitter.com/StackMma/status/1239929611460509698
Mar 17, 2020 at 7:42 answer added Henrik Schumacher timeline score: 5
Mar 17, 2020 at 6:59 comment added Henrik Schumacher Another issue (related to that) is that the final matrix seems to be a diagonal matrix. So there is also no need to compute the off-diagonal entries.
Mar 17, 2020 at 6:59 comment added Henrik Schumacher Yes, it is certainly possible to speed this up considerably. Basically 99.9 percent of the numbers in the Table you Total over are zeroes. There should be a simple logic to figure out whether the entries have to be computed in the first place. To point out what I mean: constructions like Sum[f[i] g[j] KroneckerDelta[i, j], {i, 1, n}, {j, 1, n}] are popular with physicists as they allow simple paper calculation, but super inefficient on a compute because they require $\Theta(n^2)$. Instead, Sum[f[i] g[i], {i, 1, n}] leads to the same result in $\Theta(n)$.
Mar 17, 2020 at 1:36 history edited MikeY CC BY-SA 4.0
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Mar 17, 2020 at 0:16 history asked Delaram Nematollahi CC BY-SA 4.0