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A slight improvement on the lower bound was recently obtained by Alexey Glazyrin (Lower bounds for the simplexity of the $n$-cube: https://arxiv.org/abs/0910.4200, https://doi.org/10.1016/j.disc.2012. …

Finding the exact number of unimodular triangulations of a cube in higher dimension is, I would say, out of reach. In fact, I would say the number is doubly exponential in n (that is, exponential in t …

Here is a lower bound of $2^{\Omega(2^n)}$ for the number of unimodular triangulations. Let me start with the triangulation described by Włodzimierz Holsztyński, which is indeed quite classical and I …

Let me point out that this also admits a purely combinatorial proof. Consider your triangulated $S^n$ as the boundary of a triangulated ball $B^{n+1}$ (for example, but not necessarily, cone your tria …

The answer is no, as follows from the following Lemma of Klee and Walkup: Lemma: If P is a d-polytope with n facets and we perform a "wedge" over any facet F we get a (d+1)-polytope P' with n+1 facet …

Let me point out a very significant difference between dimensions 2 and 3. In dimension 2 ANY triangulation (Delaunay or not, random or not) has average degree strictly smaller than six (and going to …

The answer is YES in the following stronger form: Lemma 1: Let $d>n$ and let $p\in dP\cap \mathbb{Z}^n$. Then there is an $i\in\{1,\dots, n\}$ and points $p'\in iP\cap \mathbb{Z}^n$, $p_1,\dots, p_{d …

The conditions you pose on $P_0$ imply that it is a reflexive polytope. (That is, a lattice polytope with the origin in its interior and such that its polar dual is also a lattice polytope). There are …

A positive answer to this question has just appeared in the arXiv: Tiling with arbitrary tiles; Vytautas Gruslys, Imre Leader, Ta Sheng Tan; http://arxiv.org/abs/1505.03697

Every simplicial polytope is rational, hence there are infinitely many rational polytopes in any fixed dimension. In contrast, there are finitely many 0/1 polytopes, since they cannot have more than $ …

Unless you have some additional structure on the points $A_1$,...,$A_n$ your problem is as hard (and as easy) as computing the convex hull of $2n$ points in $\mathbb{R}^n$, and triangulating this conv …

I think the answer is yes and the following is a sketch of proof: The second barycentric subdivision of a $\Delta$-complex is a triangulation. The f-vector of a $\Delta$-complex and of its barycentr …

The answer is "never" (except in the obvious case $d=2$, $n\ge 4$). $C(n,d)$ is neighborly, meaning that every $d/2$ or less vertices define a simplex. In particular, for $d\ge 4$ its graph is comple …