Timeline for Efficient algorithm for solving a convex quadratic program [duplicate]
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 18, 2018 at 22:39 | history | closed |
Suvrit Neil Strickland András Bátkai Stefan Kohl♦ CommunityBot |
Duplicate of Complexity for solving linear equations? | |
| S Apr 17, 2018 at 13:54 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor edits
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| Apr 17, 2018 at 13:12 | review | Suggested edits | |||
| S Apr 17, 2018 at 13:54 | |||||
| Apr 16, 2018 at 18:18 | comment | added | Cristóbal Guzmán | See my answer here mathoverflow.net/a/244126/39129 | |
| Apr 16, 2018 at 17:52 | comment | added | user35593 | thats actually the image of A. If A has full rank you can set x=Ay and minimize over y | |
| Apr 16, 2018 at 17:49 | comment | added | O. Richard | @user35593 Could you elaborate more on finding the base of the orthogonal complement? | |
| Apr 16, 2018 at 17:16 | review | Close votes | |||
| Apr 18, 2018 at 22:43 | |||||
| Apr 16, 2018 at 17:11 | comment | added | user35593 | If b is not orthogonal to the kernel of the transpose of A the quadratic expression can get arbitrarily small. On the other hand if b is orthogonal to the kernel you can choose a base of the orthogonal complement and solve the corresponding linear equation. | |
| Apr 16, 2018 at 16:58 | vote | accept | O. Richard | ||
| Apr 16, 2018 at 16:30 | answer | added | Federico Poloni | timeline score: 1 | |
| Apr 16, 2018 at 16:22 | history | edited | O. Richard | CC BY-SA 3.0 |
added 34 characters in body; edited title
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| Apr 16, 2018 at 16:10 | history | asked | O. Richard | CC BY-SA 3.0 |