Timeline for Generalizing the internal angle of a graph in $\mathbb{E}^2$ to $\mathbb{S}^2$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 16, 2012 at 7:34 | vote | accept | Samuel Reid | ||
| May 16, 2012 at 5:28 | answer | added | Gerry Myerson | timeline score: 1 | |
| May 16, 2012 at 1:27 | history | edited | Samuel Reid | CC BY-SA 3.0 |
Added comments to address the comments.
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| May 16, 2012 at 1:22 | comment | added | Samuel Reid | @Will Jagy: According to Pach and Agarwal, there is no $b_{6}$, maybe this has to do with the restriction which is forgot to mention which is that the edge length of the graph must be 1 and no two vertices of the graph (connected by an edge or not) can be closer than 1 unit distance away. For my case I am looking at in $\mathbb{S}^2$, the graph is not required to be in only one hemisphere, but I know that it does not partition $\mathbb{S}^2$ into equal size cells. The edges of the arcs are great circles of edge length $\frac{pi}{3}$, by $s=r \theta$. | |
| May 16, 2012 at 0:56 | comment | added | Will Jagy | Why is there no $b_6$? I can understand being bounded by a closed polygon, essentially if we demand no isolated vertices and none of valence 1. Is a graph required to be in a hemisphere or some such? Are the edges arcs of great circles? | |
| May 16, 2012 at 0:50 | comment | added | Samuel Reid | @Joseph O'Rourke: Thank you for the suggestion, I have added more context. Let me know if it is still unclear. | |
| May 16, 2012 at 0:49 | history | edited | Samuel Reid | CC BY-SA 3.0 |
added 920 characters in body
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| May 16, 2012 at 0:15 | comment | added | Joseph O'Rourke | @Samuel: I don't have the P&A book with me, and just from the words you quoted I cannot figure out what it means. Perhaps others may have the same difficulty. What does it mean for a "polygon to bound a graph at a vertex"? | |
| May 16, 2012 at 0:03 | history | asked | Samuel Reid | CC BY-SA 3.0 |