Timeline for How to optimize for the best fit nonuniform-scale-rotation to a given 3×3 matrix?
Current License: CC BY-SA 4.0
13 events
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| Nov 24, 2022 at 9:40 | answer | added | fedja | timeline score: 1 | |
| Nov 24, 2022 at 4:40 | comment | added | fedja | OK, after some experimenting I came to the conclusion that the approach that works best is usual 2D rotations (you can solve 2D exactly in one step) occasionally shaken by one step of your procedure (shaking is necessary or you can get stuck at the maximum with respect to every coordinate 2D rotation that isn't a global maximum. The stable examples of that are rare: about 1 in 1000, but they exist). I'll try to post it as a separate answer. | |
| Nov 19, 2022 at 0:23 | answer | added | Alec Jacobson | timeline score: 0 | |
| Nov 17, 2022 at 16:53 | answer | added | fedja | timeline score: 3 | |
| Nov 17, 2022 at 1:06 | history | edited | Alec Jacobson | CC BY-SA 4.0 |
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| Nov 16, 2022 at 21:09 | comment | added | fedja | I'll post when I have some free time (which, probably, meand late evening today) :-) | |
| Nov 16, 2022 at 21:08 | comment | added | fedja | Then I have some good news for you: 1) Your algorithm is guaranteed to work unless you are doing one stupid thing that you shouldn't be doing, i.e., sticking to rotations ($\det=1$) rather than rotations+reflections (full orthogonal group) in solving Procrustes. 2) There might be a quicker algorithm though it is not theoretically guaranteed (it is an old MO problem whether it always works). Practically it merely means that you can run them in parallel and just choose the better iteration at each step to feed into both to have the best of two speeds at the expense of doubling the best time). | |
| Nov 16, 2022 at 16:52 | comment | added | Alec Jacobson | err, that wouldn't make it unique since you could still pick which negative entry in $s$ to keep. I guess it's just that we can always strip $s$ down to a single negative entry. | |
| Nov 16, 2022 at 16:30 | comment | added | Alec Jacobson | Yes. Negative entries in $s$ are OK. (I believe that means non-uniqueness, e.g, could multiply any $s$,$R$ by $diag([1,-1,-1])$, so I guess(?) to make it unique we could say that $s$ is allowed to have at most one negative entry). | |
| Nov 16, 2022 at 15:39 | comment | added | fedja | Just to make sure that we are on the same page: you allow negative entries in $s$, don't you? | |
| Nov 15, 2022 at 9:24 | comment | added | Rodrigo de Azevedo | Interesting to see that you are still interested in Procrustes-like problems 5 years later. | |
| Nov 15, 2022 at 9:23 | history | edited | Rodrigo de Azevedo |
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| Nov 14, 2022 at 20:29 | history | asked | Alec Jacobson | CC BY-SA 4.0 |