Timeline for What is the number of equitriangulations of the n-cube?
Current License: CC BY-SA 3.0
26 events
| when toggle format | what | by | license | comment | |
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| Dec 15, 2014 at 8:58 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
A thank you note
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| Dec 15, 2014 at 8:51 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
A thank you note
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| Dec 15, 2014 at 7:55 | comment | added | Daniel Soltész | I'll just leave this here: meta.mathoverflow.net/questions/1987/politeness-on-mo-metamo | |
| Dec 15, 2014 at 7:49 | comment | added | Pietro Majer | In other words, for any vertex V of the cube, following TMA's construction, we have 8 triangulations made by simplices that share that vertex V. So there are 64 such "conic" triangulations, but this way we count more times these triangulations that admit more common vertices, and these are indeed possible exactly for 2 opposite vertices of a diagonal. If we fix a diagonal of the cube there is exactly one nice triangulation all of whose simplices share that diagonal. So I see 60 conic nice triangulations of a cube. | |
| Dec 15, 2014 at 7:26 | comment | added | Włodzimierz Holsztyński | @TheMaskedAvenger, I am sorry for not associating TMA with you on this occasion (without @ in front of TMA it was hard for me, but not anymore!). Thank you David for your help :-) | |
| Dec 15, 2014 at 7:24 | history | edited | David Eppstein | CC BY-SA 3.0 |
edited body
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| Dec 15, 2014 at 7:21 | comment | added | The Masked Avenger | It seems challenging to communicate this idea effectively to you. | |
| Dec 15, 2014 at 7:19 | comment | added | Włodzimierz Holsztyński | @TheMaskedAvenger and David and Pietro, and whoever would like to provide a 3-dim counterexample to bound 4--you may now fill up the stars at my the bottom of my Answer by a 5th nice triangulation. (Sorry, David, we were writing at the same time). | |
| Dec 15, 2014 at 7:18 | comment | added | David Eppstein | Look, it's really simple. It's just the pulling triangulation from one vertex, after you triangulate the three square faces opposite that vertex. Each of those square faces has two different diagonals that can be used to triangulate it before you pull. (And "TMA" = initials of The Masked Avenger.) | |
| Dec 15, 2014 at 7:16 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
Counterexample(?)
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| Dec 15, 2014 at 7:03 | comment | added | Włodzimierz Holsztyński | @DavidEppstein -- David, I still don't understand (TMA? what you connect to what? etc). Perhaps a $3$-dim picture would help. Or the best of all, just one triangulation of $3$-cube different from the 4 given by me in the answer. It'd take $3\cdot 4=12$ coordinates (like a $12$-vector) per simplex, and six simplexes, for a total of $72$ coordinates. Symmetries should cut down this number at least by half, I am sure. | |
| Dec 15, 2014 at 6:39 | comment | added | Włodzimierz Holsztyński | @AndréHenriques -- so your opinion about what is or is not polite/rude makes you to down-vote on mathematics. Actually, it was me who was patient and perfectly polite. (This of course makes me wonder about you but never mind). MO can be so wonderful :-) Go on others like André. down-vote to your heart desire. | |
| Dec 15, 2014 at 6:11 | comment | added | André Henriques | @Włodzimierz Holsztyński: you get a downvote from me. TheMasked avenger has been very patiently trying to engage into a discussion, and you have been extremely rude and unwilling to listen to him. | |
| Dec 15, 2014 at 6:10 | comment | added | David Eppstein | Wlod: here's my description of TMA's construction. Connect one vertex to the three opposite faces of a cube, forming three (skewed) square pyramids. Each pyramid can be divided into two orthoschemes (the convex hull of a right-angled-path) or it can be divided into an equilateral-triangle pyramid (the convex hull of a right-angled $K_{1,3}$) and another nice tetrahedron (congruent to the tet with vertices 000 111 110 101). So there are indeed two distinct possibilities per pyramid, eight per pyramid-decomposition, and 48 total among triangulations of this type. | |
| Dec 15, 2014 at 5:51 | comment | added | Włodzimierz Holsztyński | @TheMaskedAvenger -- again, no need to ask. Again, please, simply write down your (full) Answer. | |
| Dec 15, 2014 at 5:49 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
Construction
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| Dec 15, 2014 at 5:41 | comment | added | The Masked Avenger | Ok. Let a and b range over {0,1} and consider the five points (0,0,0) and (1,a,b). This describes the vertices whose convex hull is a pyramid with square base, and by omitting one of the last four points, contains four different simplices. By considering (a,b,1) and (a,1,b) as well, one gets 8 distinct partitions of the unit cube. Which of these are not triangulations? | |
| Dec 15, 2014 at 4:52 | comment | added | Włodzimierz Holsztyński | @TheMaskedAvenger -- your construction must be quite symmetric. This would simplify the job of describing your 8 (?) different nice triangulations. Say, there might be just 4 to describe and 4 more which are symmetric (I don't know, I don't know your construction). | |
| Dec 15, 2014 at 4:48 | comment | added | Włodzimierz Holsztyński | @TheMaskedAvenger -- no need to ask. Go, please, beyond a vague description. Simply provide enough of details. I do not know your 8 constructions (if they truly provide triangulations), and I don't know whether or not some of them coincide, give you fewer than your 8. Why don't you write your own adequate answer. | |
| Dec 15, 2014 at 4:28 | comment | added | The Masked Avenger | Suppose I divide the cube into three congruent right angled pyramids sharing a common vertex. Each of these pyramids can be decomposed into two simplices in at least two ways. I thus have 8 different decompositions of the cube into simplices, 64 if I vary the common apex. Which of these decompositions is not a triangulation? | |
| Dec 14, 2014 at 22:33 | comment | added | Pietro Majer | (sorry, actually I meant partitions in simplices) | |
| Dec 14, 2014 at 17:17 | comment | added | Pietro Majer | For instance, among the $T(n)$ nice triangulations of the $n$-cube, exactly $T(n-1)^n$ are made by simplices all having $(0,0,\dots,0)$ as a common vertex. | |
| Dec 14, 2014 at 15:41 | comment | added | Pietro Majer | I think there are much more… | |
| Dec 14, 2014 at 9:56 | history | undeleted | Włodzimierz Holsztyński | ||
| Dec 14, 2014 at 9:41 | history | deleted | Włodzimierz Holsztyński | via Vote | |
| Dec 14, 2014 at 9:28 | history | answered | Włodzimierz Holsztyński | CC BY-SA 3.0 |