Timeline for Sorting a symmetric matrix alphanumerically
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2023 at 14:58 | comment | added | IntroductionToProbability | Thanks @Edmund, I was wondering about this and I'm glad OP got a solution | |
| Mar 27, 2023 at 12:59 | comment | added | Edmund | @IntroductionToProbability Your solution does not have the same eigenvalues as the starting matrix. The "staying together" means the rows can change position and the numbers in a row change position within the row. Numbers should not leave their row, only be rearranged within it. | |
| Nov 17, 2022 at 13:18 | comment | added | Hauke Reddmann | Edited. I hope it is clear now. | |
| Nov 17, 2022 at 13:15 | comment | added | IntroductionToProbability | I'm sorry I can't help further without some examples or a definition of the vectors "staying together" | |
| Nov 17, 2022 at 13:09 | comment | added | Hauke Reddmann | Here is an idea for an almost working poly-alg: 1.Sort all row vectors of mat element-wise. 2.Lump them together to a matrix mat2 and sort the cols by vector order alphanumeric. 3.Remember the permutation of step 2 (easy: patch 1...n to the end of the vectors before step 2 and extract perm afterwards). 4.Apply perm to both row and col of mat simultaneously: mat3=mat[perm[i],perm[j]]. Done. Unfortunately, with identical vectors occuring the algorithm still might fail. | |
| Nov 17, 2022 at 13:06 | comment | added | IntroductionToProbability | @HaukeReddmann it is still not clear to me what you mean by stay together. Can you edit your question to include this constraint as well as give a concrete example? | |
| Nov 17, 2022 at 13:02 | comment | added | Hauke Reddmann | Bad wording...The numbers of the vector stay "together". Obviously, this can't work with just sorting all numbers of the matrix - it's just a lucky accident that (0123),(0145),(0246) and (0356) stay together after sorting. OK, I reword it: if a matrix contains the vector (a1,a2,...) before, it must also contain this vector after sorting (in any order of the vector elements). | |
| Nov 16, 2022 at 23:12 | comment | added | IntroductionToProbability | What do you mean by “the vectors stay in place” | |
| Nov 16, 2022 at 14:18 | history | edited | IntroductionToProbability | CC BY-SA 4.0 | improved formatting and made code more obvious |
| Nov 16, 2022 at 14:01 | comment | added | IntroductionToProbability | If the second example is {{0, 1, 2, 4}, {1, 0, 3, 5}, {2, 3, 0, 1}, {4, 5, 1, 0}} then it sorts to {{0, 1, 1, 2}, {1, 0, 3, 4}, {1, 3, 0, 5}, {2, 4, 5, 0}}. I will edit my answer to make the algorithm clearer | |
| Nov 16, 2022 at 12:00 | comment | added | Hauke Reddmann | This is definitely correct (as a result), since the vectors stay in place, but how does it work exactly? And, even more relevant, does it work for the critical second example? | |
| Nov 15, 2022 at 14:53 | history | answered | IntroductionToProbability | CC BY-SA 4.0 |