Timeline for Drawing Zonotopes from Adjacency Matrix
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:56 | history | edited | CommunityBot | replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/ | |
| Apr 13, 2017 at 12:48 | history | edited | CommunityBot | replaced http://cs.stackexchange.com/ with https://cs.stackexchange.com/ | |
| Aug 4, 2013 at 1:47 | vote | accept | Alex R. | ||
| Aug 3, 2013 at 21:48 | answer | added | cormullion | timeline score: 6 | |
| Aug 3, 2013 at 21:44 | comment | added | cormullion | I would like this to be possible, but I'm not convinced. After all, it's easy to make a weighted graph that can't be drawn accurately in 2 dimensions (Graph[{1 <-> 2, 2 <-> 3, 3 <-> 1}, EdgeLabels -> "EdgeWeight", EdgeWeight -> {1, 42, -8}]) presumably Mathematica puts topology before accuracy? | |
| Aug 3, 2013 at 20:50 | history | tweeted | twitter.com/#!/StackMma/status/363763817198534656 | ||
| Aug 3, 2013 at 20:25 | comment | added | Alex R. | @Daniel Lichtblau: I have the $(x,y)$ coordinates of all points for the geometric graph which allows me to identify the boundary and more or less where the regions go. It's not clear however how to efficiently determine the angles in the rhombi of the zonotope that corresponds to the geometric graph. Right now I'm just hoping Mathematica has a built in function that can arrange and scale everything properly. In other words I'm hoping that fixing edge lengths to be equal will force it to draw a zonotope | |
| Aug 3, 2013 at 19:57 | comment | added | Daniel Lichtblau | If you have the (x,y) coordinates already, why not use those in locating your graph vertices? | |
| Aug 3, 2013 at 18:39 | review | First posts | |||
| Aug 3, 2013 at 19:04 | |||||
| Aug 3, 2013 at 18:30 | history | edited | Alex R. | CC BY-SA 3.0 | added 37 characters in body |
| Aug 3, 2013 at 18:23 | history | asked | Alex R. | CC BY-SA 3.0 |