Timeline for $n$-dimensional Voronoi diagram
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 15, 2010 at 20:07 | comment | added | Michael Hardy | Thank you. I've put in an interlibrary loan request for Brown's paper. | |
| Dec 15, 2010 at 18:33 | comment | added | Joseph O'Rourke | @Michael: Whether the asymmetries are "messy" may be a matter of taste :-). The top of the paraboloid-lifted hull records furthest-point diagram adjacencies, while the bottom of the hull records closest-point adjacencies. For the sphere viewpoint, see Kevin Q. Brown: Voronoi Diagrams from Convex Hulls. Inf. Process. Lett. 9(5): 223-228 (1979). | |
| Dec 15, 2010 at 17:48 | comment | added | Michael Hardy | Is the sphere version in the literature somewhere? I can imagine it having a nice proof, but I'd have to think about the specifics. | |
| Dec 15, 2010 at 17:47 | comment | added | Michael Hardy | I like the "sphere" version of the geometric proposition involved, since it lacks these messy asymmetries. Take a finite set of points on a sphere. Form their convex hull. Then the edges of that polygon are the Delaunay triangulation and the dual is the Voronoi tesselation. No need to talk about "lower" and "upper". | |
| Dec 15, 2010 at 17:45 | comment | added | Michael Hardy | Oh! You meant when the set of points whose Delaunay triangulation you're trying to find it bounded. I was picturing an unbounded set of points. | |
| Dec 15, 2010 at 17:42 | comment | added | Michael Hardy | Doesn't the whole convex hull lie above its boundary in this case? Thus the whole boundary is the lower boundary; the convex hull is not bounded above. | |
| Dec 15, 2010 at 16:00 | comment | added | Joseph O'Rourke | @Michael: Yes, that's it; thanks for clarifying. But it is only the lower hull that projects to the Delaunay triangulation. | |
| Dec 15, 2010 at 15:41 | history | answered | Michael Hardy | CC BY-SA 2.5 |