Timeline for Sparse block matrices all wrong
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 5, 2018 at 8:12 | history | tweeted | twitter.com/StackMma/status/1003912373097390080 | ||
| Jun 5, 2018 at 7:40 | answer | added | kglr | timeline score: 3 | |
| Jun 5, 2018 at 7:23 | comment | added | Henrik Schumacher | Let us continue this discussion in chat. | |
| Jun 5, 2018 at 7:22 | comment | added | მამუკა ჯიბლაძე | @HenrikSchumacher Yes, what confused me is the very existence of ArrayFlatten - I thought it was precisely for such cases. While in fact it seems to become arrayflattened from the beginning... | |
| Jun 5, 2018 at 7:18 | comment | added | Henrik Schumacher | Ah, this last command actually gave me a hint on what you were thinking how it should work. | |
| Jun 5, 2018 at 7:08 | vote | accept | მამუკა ჯიბლაძე | ||
| Jun 5, 2018 at 6:46 | review | Close votes | |||
| Jun 5, 2018 at 6:56 | |||||
| Jun 5, 2018 at 6:44 | answer | added | Henrik Schumacher | timeline score: 9 | |
| Jun 5, 2018 at 6:34 | comment | added | მამუკა ჯიბლაძე | Tried this too. Now (with the new example that I replaced the old one with) the first two rows are ok; however the first rows of other blocks still become chopped off by some reason, and these blocks are shifted up. And I still believe that dimensions of sa must be {n, n}, it is ArrayFlatten[sa] that must have dimensions {Total[dim], Total[dim]}, not sa itself, no? | |
| Jun 5, 2018 at 6:30 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 4.0 | more revealing example |
| Jun 5, 2018 at 6:30 | comment | added | Henrik Schumacher | Then you should use Total[Most[dim]] + 1. | |
| Jun 5, 2018 at 6:28 | comment | added | მამუკა ჯიბლაძე | Maybe actually I will replace the example with this one, it is more illustrative... | |
| Jun 5, 2018 at 6:27 | comment | added | მამუკა ჯიბლაძე | @HenrikSchumacher Now it gives correct answer. However with, say, dim = {2, 3, 4} it is all wrong again. As for your explanation: it should be an $n\times n$ array of matrices, should not it? | |
| Jun 5, 2018 at 6:24 | comment | added | Henrik Schumacher | Then try Band[{1, n (n - 1)/2 + 1}]. The second index of the band of the last matrix should the sum of the dimensions of the others +1 | |
| Jun 5, 2018 at 6:21 | comment | added | მამუკა ჯიბლაძე | @HenrikSchumacher Tried. Wrong again (in a different way) - among other things, it is not square anymore. Besides, I don't see why should it be placed there... | |
| Jun 5, 2018 at 6:14 | comment | added | Henrik Schumacher | Shouldn't the last matrix be placed on Band[{1, Quotient[n (n + 1), 2]}]? | |
| Jun 5, 2018 at 6:12 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 4.0 | added 57 characters in body |
| Jun 5, 2018 at 6:02 | history | asked | მამუკა ჯიბლაძე | CC BY-SA 4.0 |