Timeline for Choosing directed subgraph in a triangulation
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 19, 2016 at 18:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| S May 20, 2016 at 17:51 | history | suggested | user 1 | CC BY-SA 3.0 |
some $$ added
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| May 20, 2016 at 17:36 | review | Suggested edits | |||
| S May 20, 2016 at 17:51 | |||||
| May 20, 2016 at 16:44 | answer | added | Elena Yudovina | timeline score: 1 | |
| May 20, 2016 at 6:54 | comment | added | Walkiria | Triangulation which I consider is on the plane. I saw that another question about spanning tree, but what if we need just spanning forest or spanning forest with some cycles which are not connected? | |
| May 19, 2016 at 14:02 | comment | added | Elena Yudovina | It might depend on what you're triangulating. If the triangulation is of a surface of genus $g > 1$, the answer is "no" by a counting argument. Let $F$ be the number of triangles (faces) of your triangulation, then the number of edges is $E = 3/2 F$ and the number of vertices is $E-F+(2-2g) < F/2$. The number of edges in $H$ must be bounded above by the number of vertices and below by $F/2$, which is a problem. For the remaining cases, you might find this question useful: mathoverflow.net/questions/112661/… | |
| May 16, 2016 at 12:46 | review | First posts | |||
| May 16, 2016 at 13:45 | |||||
| May 16, 2016 at 12:39 | history | asked | Walkiria | CC BY-SA 3.0 |