Timeline for How to plot planar graphs as such?
Current License: CC BY-SA 3.0
21 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 6, 2018 at 21:15 | answer | added | Szabolcs | timeline score: 4 | |
| Jan 28, 2013 at 16:34 | history | edited | Szabolcs | CC BY-SA 3.0 | added 139 characters in body |
| Jan 28, 2013 at 0:16 | vote | accept | Szabolcs | ||
| Jan 27, 2013 at 21:01 | answer | added | Sjoerd C. de Vries | timeline score: 19 | |
| Mar 2, 2012 at 14:44 | answer | added | Heike | timeline score: 17 | |
| Feb 16, 2012 at 10:42 | comment | added | Szabolcs | MathGroup version here: groups.google.com/d/topic/comp.soft-sys.math.mathematica/… | |
| Feb 16, 2012 at 0:48 | history | edited | J. M.'s missing motivation | edited tags | |
| Feb 15, 2012 at 17:46 | history | edited | Szabolcs | CC BY-SA 3.0 | added 198 characters in body |
| Feb 15, 2012 at 17:44 | comment | added | Szabolcs | @Daniel I found that function, but it requires a set of points (coordinate pairs) as input and it simply shows a Delaunay triangulation of these points. It does not seem to take a graph as input without explicit point coordinates. I only have the graph (e.g. as an adjacency list), I can test using PlanarQ that it is indeed a planar graph, and now I would like to show it as a planar graph (i.e. without intersecting edges) | |
| Feb 15, 2012 at 17:14 | comment | added | Daniel Lichtblau | Have a look at Help > Documentation Center > ComputationalGeometry/ref/PlanarGraphPlot | |
| Feb 15, 2012 at 15:43 | comment | added | Szabolcs | @kguler Even the graphs returned by GraphData don't contain vertex position information, so it doesn't solve the visualization problem. Testing planarity is not a problem, PlanarQ does that. Also, lookup tables with graphs are really problematic because they require a lot of isomorphism testing (which can be slow---and version 8.0.4 IsomorphicGraphQ is still buggy unfortunately) | |
| Feb 15, 2012 at 15:37 | comment | added | Szabolcs | This paper, posted by @halirutan in chat, seems to suggest this is not a trivial thing to do: math.uni-hamburg.de/home/schacht/2011/untangle.pdf | |
| Feb 15, 2012 at 15:14 | comment | added | kglr | GraphData["Classes"] returns 163 classes of which oneis "Planar" and GraphData["Planar"] returns 2923 graphs. Perhaps, this list could serve as some sort of look-up table? | |
| Feb 15, 2012 at 15:02 | comment | added | rm -rf♦ | @kguler I tried that and other options... it gives the correct plot for the first example and for a few in the second bigger list of graphs, but not all... | |
| Feb 15, 2012 at 14:59 | history | tweeted | twitter.com/#!/StackMma/status/169798113807175682 | ||
| Feb 15, 2012 at 14:53 | comment | added | J. M.'s missing motivation | On the other hand, GraphData["TetrahedralGraph"] is drawn such that it is obviously planar... | |
| Feb 15, 2012 at 14:48 | comment | added | kglr | Did you try selecting RadialDrawing in the right-click context menu under Graph Layout. For this example it gives the graph on the left. | |
| Feb 15, 2012 at 14:47 | comment | added | Szabolcs | @Mr.Wizard The full documentation of Combinatorica` is not included. It has to be bought separately. | |
| Feb 15, 2012 at 14:46 | history | edited | Szabolcs | CC BY-SA 3.0 | edited title |
| Feb 15, 2012 at 14:35 | comment | added | Mr.Wizard | What do you mean: the documentation is not included with Mathematica? | |
| Feb 15, 2012 at 14:13 | history | asked | Szabolcs | CC BY-SA 3.0 |