Timeline for Maximal number of visible vertices
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Nov 13, 2020 at 9:12 | answer | added | M. Winter | timeline score: 2 | |
| Nov 13, 2020 at 8:57 | comment | added | M. Winter | What are you taking the maximum over? All orientations and translations of a fixed polyhedron $P$? All polyhedra $P$ with $N$ facets? | |
| Nov 13, 2020 at 8:35 | comment | added | Ilya Bogdanov | @FedorPetrov As F.C. mentions, you may start with a tetrahedron all of whose vertices are visible, and on each stage truncate a top vertex creating a new face and increasing the number of (visible) vertices by 2. (A top vertex here is a vertex such that all its three faces are visible.) So the answer is $2N-4$. | |
| Nov 13, 2020 at 8:30 | comment | added | Fedor Petrov | @IlyaBogdanov so why is it optimal? So far I see the bounds like $N\le f(N) \le 2N-4$. | |
| Nov 12, 2020 at 10:14 | comment | added | Ilya Bogdanov | F.C. is right; this is clearly optimal, as a convex polytope with $N$ facets cannot have more than $2N-4$ vertices, by Euler's formula (and relation $e\geq 3v/2$). | |
| Nov 12, 2020 at 9:13 | comment | added | F. C. | Oh, I see that I had somehow missed the point. But then one can truncate the top vertex and iterates this kind of truncation, adding each time 2 vertices and one face. | |
| Nov 12, 2020 at 8:49 | history | edited | YCor |
edited tags
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| Nov 12, 2020 at 8:47 | comment | added | YCor | @F.C. This shows that $f(N)\ge N$. | |
| Nov 12, 2020 at 7:46 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
Made the title more informative
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| Nov 12, 2020 at 7:42 | comment | added | F. C. | Maybe the pyramid over a regular polygon would be a good example. | |
| Nov 12, 2020 at 7:18 | history | asked | Dmitry Maximov | CC BY-SA 4.0 |