Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with polyhedra
Search options not deleted
user 2233
19
votes
Does every convex polyhedron have a combinatorially isomorphic counterpart whose faces all h...
The answer to the third question is no. This is a rather counter-intuitive discovery of Micha Perles from the sixties. See this paper of Ziegler, for a simpler construction and other pertinent infor …
- 32.1k
1
vote
Testing whether two vertices are neighbours
You should first check that $v_i$ and $v_j$ are both vertices of $X$. Given the description $X = \{ x \in \mathbb{R}^k \, : \, A x \leq b \}$, this is easy.
Suppose that $A$ is an $m \times n$ mat …
- 32.1k
12
votes
Visibility of vertices in polyhedra
One can attempt to do the same thing in 3 dimensions, but somewhat surprisingly there exist polyhedra that cannot be decomposed into tetrahedra (without adding additional vertices). … See here, where they show that the problem of deciding if a 3-dimensional polyhedra can be decomposed into tetrahedra is NP-complete. The references in that paper might be helpful. …
- 32.1k