Timeline for Polytope with indegree-increasing property.
Current License: CC BY-SA 3.0
8 events
when toggle format | what | by | license | comment | |
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Mar 9, 2015 at 14:25 | comment | added | Yunhyung Cho | @Joseph O'Rourke : Hao Chen pointed out what I exactly want to say. Sorry for the confusion. | |
Mar 9, 2015 at 6:28 | comment | added | Hao Chen | I would rephrase the question as follows: can we always find a convex polytope $P'$ such that $P'$ is combinatorially equivalent to $P$ and $P'$ is index increasing w.r.t. a given vector (after all, we can fix the vector wlog) | |
Mar 8, 2015 at 23:38 | comment | added | Joseph O'Rourke | @TheMaskedAvenger: Wikipedia says, a "simple polytope is a $d$-dimensional polytope each of whose vertices are adjacent to exactly $d$ edges (also $d$ facets)." | |
Mar 8, 2015 at 23:16 | comment | added | The Masked Avenger | For the geometric incognoscenti among us, are there brief descriptions which would convey an appropriate meaning of "simple" for this context? My guess is no holes. | |
Mar 8, 2015 at 23:09 | comment | added | Joseph O'Rourke | @HaoChen: A deformation should be "preserving ... convexity," a stipulation I do not understand. | |
Mar 8, 2015 at 17:25 | comment | added | Joseph O'Rourke | @HaoChen: Ah, I was interpreting "simple" as in "simple polyhedron," rather than "simple polytope": all vertices degree $d$. Thanks for the clarifications. | |
Mar 8, 2015 at 16:52 | comment | added | Hao Chen | He said "simple convex polytope". I think "deformation" means any combinatorial equivalent realization. | |
Mar 8, 2015 at 15:44 | history | answered | Joseph O'Rourke | CC BY-SA 3.0 |