Timeline for What is the number of equitriangulations of the n-cube?
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| when toggle format | what | by | license | comment | |
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| Dec 15, 2014 at 8:27 | comment | added | Pietro Majer | So I guess there is a lower bound of the form $$A(n)\ge 2^nB(n-1)-2^{n-1}C(n-2)$$ where A(n) B(n) C(n) count the number of (nice, pulled) triangulations, respectively on the n-cube, on a bouquet of n+1 n-cubes, on a complex of (n+1)(n+2) n-cubes, as described above. | |
| Dec 15, 2014 at 8:26 | history | edited | David Eppstein | CC BY-SA 3.0 |
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| Dec 15, 2014 at 8:21 | comment | added | Pietro Majer | Also, we can fix an "axial" triangulation (= all simplices have two opposite vertices V, V' in common, that is a diagonal of the cube as a common edge) fixing a n-2 dimensional triangulation on the union of all those n-2 dimensional faces of the cube, that do not have neither V nor V' as vertex. These are n(n-1) (just one half of the total number 2n(n-1) of n-2 dimensional faces of $I^n$). | |
| Dec 15, 2014 at 8:16 | comment | added | Pietro Majer | So in dimension n we can fix a pulled partition from a vertex V (= by simplices all having V as a vertex) exactly by defining a triangulation of the bouquet of n facets of the cube around the vertex opposite to V. | |
| S Dec 15, 2014 at 7:51 | history | answered | David Eppstein | CC BY-SA 3.0 | |
| S Dec 15, 2014 at 7:51 | history | made wiki | Post Made Community Wiki by David Eppstein |