Timeline for Detecting coplanarity by given pairwise distances
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:48 | history | edited | CommunityBot | replaced http://cs.stackexchange.com/ with https://cs.stackexchange.com/ | |
| Oct 21, 2014 at 23:02 | vote | accept | padawan | ||
| Jul 29, 2014 at 21:38 | history | edited | Craig Gidney | CC BY-SA 3.0 | edited body |
| Jul 29, 2014 at 20:59 | history | bounty awarded | CommunityBot | ||
| Jul 22, 2014 at 18:27 | history | edited | Craig Gidney | CC BY-SA 3.0 | added 197 characters in body |
| Jul 22, 2014 at 16:34 | comment | added | Craig Gidney | @AmitKumarGupta If you have a K_4 then put one of them at (0,0,0), another at (x,0,0), the next at (y,z,0), and then the last at (a,b,c) with c > 0. If you can't solve for a unique embedding, up to translation/rotation/mirroring, then I don't think there's a well defined solution. You'd get things like rigid groups of points connected by a lever that you can rotate around without violating the distance constraints, and I don't think the planes can survive that loss of information. | |
| Jul 22, 2014 at 16:27 | comment | added | Amit Kumar Gupta | How do you compute the plane's normal, or the point's delta to some point on the plane? It's easy when you know the coordinates of the points. How do you do that when you don't know the coordinates? | |
| Jul 22, 2014 at 16:18 | comment | added | Craig Gidney | @AmitKumarGupta To be a bit more precise, the question is not about how to discover which points are on a plane. It's about how to pick a small set of planes covering the points, with the additional wrinkle that you only get some of the interpoint distances instead of the points themselves. | |
| Jul 22, 2014 at 16:15 | comment | added | Craig Gidney | @AmitKumar Oh, I see that you meant how do you test if a point is on a plane. Basically, dot product the the plane's normal against the point's delta to a point on the plane. When the dot product is zero (or near zero if you're approximating), the point is on the plane. You do this for each not-disqualified point. | |
| Jul 22, 2014 at 16:11 | comment | added | Craig Gidney | @AmitKumar Any arbitrary triplet of points is co-planar, because they define a plane. A corner case is when they're degenerate (two of the points are the same, or the three points fall on a line), in which case they're still co-planar but many planes can cover them. | |
| Jul 22, 2014 at 15:37 | comment | added | Amit Kumar Gupta | Your very first sentence of the first algorithm basically says "pick 3 points, and disqualify all the points which are coplanar with those 3." But how do you determine which points are coplanar? That's the whole point of this question. Assuming that you don't know the coordinates of the points, and only know some of their pairwise distances, how do you detect coplanarity? | |
| Jul 22, 2014 at 15:24 | history | edited | Craig Gidney | CC BY-SA 3.0 | added 271 characters in body |
| Jul 22, 2014 at 15:13 | history | answered | Craig Gidney | CC BY-SA 3.0 |