Timeline for Does there exist an algorithm that decomposes a matrix into a minimal number of elementary matrices for $F_{2}$?
Current License: CC BY-SA 3.0
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Apr 12, 2022 at 6:53 | comment | added | Mariano Suárez-Álvarez | @JosephVanName, in doi:10.1016/j.aam.2006.08.008 there is an algorithm that improves the n^2 of Gaussian elimination to n^2/\log 2n, and this is asymtotically optimal according to doi:10.48550/ARXIV.1406.5826 | |
S Dec 5, 2017 at 19:38 | history | bounty ended | CommunityBot | ||
S Dec 5, 2017 at 19:38 | history | notice removed | CommunityBot | ||
S Nov 27, 2017 at 17:56 | history | bounty started | Rodrigo de Azevedo | ||
S Nov 27, 2017 at 17:56 | history | notice added | Rodrigo de Azevedo | Canonical answer required | |
S Nov 27, 2017 at 15:12 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
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Nov 27, 2017 at 14:08 | review | Suggested edits | |||
S Nov 27, 2017 at 15:12 | |||||
Nov 27, 2017 at 7:19 | comment | added | Joseph Van Name | Rodrigo de Azevedo. I am to a large extent interested in the linear transformations where it is known that $r\ll n^{2}$ but where Gaussian elimination will take about $n^{2}$ steps. My uneducated guess is that the $r$ produced Gaussian elimination could be improved a little bit for random matrices though. | |
Nov 26, 2017 at 18:31 | comment | added | Rodrigo de Azevedo | Does Gaussian elimination produce too large an $r$? | |
Nov 26, 2017 at 14:21 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
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Nov 26, 2017 at 14:18 | comment | added | Joseph Van Name | I did define CNOT gate $(x,y)\mapsto(x,x\oplus y)$ in the question. en.wikipedia.org/wiki/Controlled_NOT_gate . | |
Nov 26, 2017 at 14:07 | comment | added | Turbo | please define $CNOT$ gate. | |
Nov 26, 2017 at 13:15 | comment | added | Joseph Van Name | Yes. $\oplus$ is the standard notation for XOR which is addition modulo 2. | |
Nov 26, 2017 at 13:08 | history | rollback | Joseph Van Name |
Rollback to Revision 3 - I reverted unwarranted old edits. An XOR gate deletes information. The CNOT gate does not.
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Nov 26, 2017 at 12:42 | history | edited | Turbo | CC BY-SA 3.0 |
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Nov 26, 2017 at 11:36 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
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Nov 25, 2017 at 17:36 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
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Nov 25, 2017 at 17:31 | history | asked | Joseph Van Name | CC BY-SA 3.0 |