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Oct 22, 2014 at 13:18 history edited padawan CC BY-SA 3.0
deleted 1244 characters in body
Oct 22, 2014 at 12:37 history edited padawan CC BY-SA 3.0
An extra restriction is added.
Oct 21, 2014 at 23:02 vote accept padawan
S Jul 29, 2014 at 20:59 history bounty ended CommunityBot
S Jul 29, 2014 at 20:59 history notice removed CommunityBot
Jul 23, 2014 at 0:50 answer added Craig Gidney timeline score: 3
Jul 22, 2014 at 21:57 comment added vzn you mentioned sensors in your comments. think this would be significantly improved by stating the bkg/ motivation/ application of the problem which seems to be wrt sensor data etc... the rewritten version sounds more like a machine learning or curve(-like) fitting problem to me with incomplete data.
S Jul 22, 2014 at 17:16 history suggested Amit Kumar Gupta CC BY-SA 3.0
make question clear and precise
Jul 22, 2014 at 16:58 answer added D.W. timeline score: 3
Jul 22, 2014 at 16:07 comment added padawan @AmitKumarGupta Points on the lines of intersections are uneffective to the nature of the problem. It might belong to multiple planes but giving the answer "this point is coplanar with those ones" is enough for a solution. We can assume that $k$ planes cover all the points.
Jul 22, 2014 at 16:05 comment added Amit Kumar Gupta It's also not clear if you're okay with some of the points belonging to multiple planes.
Jul 22, 2014 at 16:04 comment added Amit Kumar Gupta In general, the $k$ planes might not be unique. For instance if the vertices are vertices of a cube, and $k = 2$, then the top and bottom plane work, as do the front and back, as do the left and right. You were talking about the $k$ planes, but presumably you're okay with finding any $k$ planes that still do the trick of covering all points in $V$?
Jul 22, 2014 at 16:02 review Suggested edits
S Jul 22, 2014 at 17:16
Jul 22, 2014 at 15:24 comment added Craig Gidney (It's much better now.)
Jul 22, 2014 at 15:13 answer added Craig Gidney timeline score: 3
Jul 22, 2014 at 14:54 answer added vzn timeline score: -1
Jul 22, 2014 at 14:30 history edited padawan CC BY-SA 3.0
added 721 characters in body
Jul 22, 2014 at 14:23 comment added Craig Gidney Why isn't the graph complete? So we're not actually given the distances between every pair? Is the graph sparse? Is it given as an adjacency list, an adjacency matrix, or something else? Give an example.
Jul 22, 2014 at 14:20 comment added padawan @Strilanc input is an edge weighted graph. The vertices are points in 3D and edge weights are distances between points. The graph is not complete.
Jul 22, 2014 at 14:18 comment added Craig Gidney It's not clear what you're asking. Do we get the points or not? What do you mean exactly by "given set of points and pairwise distances (not the coordinates)"? Give an example.
Jul 22, 2014 at 8:03 comment added padawan @vzn That is a fine idea but what if there are more than one coplanar sensor groups?
Jul 22, 2014 at 4:37 comment added vzn idea: think this generalizes. put all the points in a matrix and find its rank. if its 2, they are all coplanar. this might go over better on Mathematics...?
Jul 22, 2014 at 3:10 comment added vzn a ref that the Cayley-Menger determinant is sufficient for that purpose would be helpful if you know one. what do you mean "all pairwise distances between each 4 nodes"? isnt there as many pairwise distances as edges with pts taken as vertices on a graph? is the algorithm you have in mind looking at coplanarity of all 4-point choices? more on coplanar
S Jul 21, 2014 at 19:10 history bounty started padawan
S Jul 21, 2014 at 19:10 history notice added padawan Draw attention
Jul 18, 2014 at 6:08 history tweeted twitter.com/#!/StackCompSci/status/490015283009323008
Jul 17, 2014 at 22:54 history asked padawan CC BY-SA 3.0