Timeline for Polytope with indegree-increasing property.
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 9, 2015 at 15:42 | comment | added | Yunhyung Cho | Of course. I just want to have an example. Thank you so much! | |
| Mar 9, 2015 at 14:56 | comment | added | Hao Chen | @YunhyungCho I believe that "most" polytope can not be realized with such a plane. I don't have a characterization for those that can. | |
| Mar 9, 2015 at 14:52 | comment | added | Yunhyung Cho | @ Hao Chen : Yes. I definitely agree with you that there must exist such plane. I should keep trying to construct a counter-example, because we allow to deform a polytope even after taking several truncations. | |
| Mar 9, 2015 at 14:41 | history | edited | Hao Chen | CC BY-SA 3.0 |
edited body
|
| Mar 9, 2015 at 14:29 | comment | added | Hao Chen | @YunhyungCho I just adjusted my answer to include extreme cases as simplex. | |
| Mar 9, 2015 at 14:28 | history | edited | Hao Chen | CC BY-SA 3.0 |
added 57 characters in body
|
| Mar 9, 2015 at 14:18 | comment | added | Yunhyung Cho | Nice approach! But how do you know that every face of $\mathcal{P}$ contains a vertex of indegree 1 and a vertex of indegree 2 ? For example, for 3-dimensional simplex case, every facet is a triangle so that there must be a facet which contains vertices of indegree 0,2 (or 4),and 6. | |
| Mar 9, 2015 at 12:20 | history | edited | Hao Chen | CC BY-SA 3.0 |
added 81 characters in body
|
| Mar 9, 2015 at 12:06 | history | edited | Hao Chen | CC BY-SA 3.0 |
added 7 characters in body
|
| Mar 9, 2015 at 11:58 | history | edited | Hao Chen | CC BY-SA 3.0 |
added 22 characters in body
|
| Mar 9, 2015 at 11:52 | history | answered | Hao Chen | CC BY-SA 3.0 |