Timeline for When does there exist a convex polyhedron with given edge lengths?
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9 events
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| Apr 19, 2016 at 15:21 | comment | added | Gerhard Paseman | I thought so too, then I tried placing edges together. So far the idea that seems to be most inclusive is to take three edges at a time of not too dissimilar lengths and "tent" them over an existing triangle. If the partition piece sizes don't grow too fast, this seems to work, but one still has to figure out which partitions those are. If one goes the route of starting with faces with large numbers of sides, one hits a large number of constraints pretty quickly. Gerhard "Building Such Polyhedra Is Hard" Paseman, 2016.04.19. | |
| Apr 19, 2016 at 11:49 | comment | added | Moritz Firsching | @StefanKohl I just wanted to make it more plausible that the fraction of all "edgy" partitions, i.e. partitions which are a length list of some polytope, might not tend to zero. The prisms allow are not enough, but they do give a positive fraction of all partitions of the special form described above. The tetrahedron already seems to give a positive fraction of all partitions into $6$ parts. | |
| Apr 19, 2016 at 10:34 | comment | added | Stefan Kohl♦ | @MoritzFirsching: For polygons you are obviously right. -- But your conclusion for polyhedra does not seem to be correct: since we have $\lim_{n \rightarrow \infty} p(n)/p(3n) = 0$, it seems your construction of prisms does only work for a fraction of all partitions which tends to $0$. -- Or did I misunderstand what you mean? | |
| Apr 19, 2016 at 7:47 | comment | added | Moritz Firsching | I would expect that the number of partitions, that correspond to a convex polygon is asymptotically the same (up to a constant) as the number of all partitions, is this true? Then if you only consider prisms over convex polygons you already get a positive fraction of all partitions of $3n$ for the form $(p)+(p)+1+\dots+1$, where p is a partition of $n$ that corresponds to a convex polygon. @GerhardPaseman: I am not convinced that the fraction should be tending to $0$. | |
| Apr 17, 2016 at 0:20 | answer | added | Moritz Firsching | timeline score: 7 | |
| Apr 16, 2016 at 20:31 | comment | added | j.c. | A related question covering the case of simplicial polyhedra math.stackexchange.com/questions/574323/… | |
| Apr 16, 2016 at 20:21 | comment | added | Moritz Firsching | to start simple: for the tetrahedron there is the Cayley-Menger determinant in addition to having all the face obeying the triangle inequality (which is not sufficient) | |
| Apr 16, 2016 at 20:17 | comment | added | Gerhard Paseman | Note that you need two polygons to share the longest edge, so an edge of length n/3 or longer will be a deal breaker, and I imagine n/4 will present some challenges. This suggests to me a fraction tending to 0 for the probability of a partition to be a length list of feasible polytopes. For d dimensions, I think n/d is a similar obstacle, but am less sure. Gerhard "Fractions, Now There's A Chance" Paseman, 2016.04.16. | |
| Apr 16, 2016 at 19:54 | history | asked | Stefan Kohl♦ | CC BY-SA 3.0 |