TITLE: Triangulating hypercubes
QUESTION [8 upvotes]: Motivation: I'm working on a computational problem at the moment, and have some very good routines for natively working with simplicial complexes and calculating homology, but the structures I'm dealing with arise naturally as cubical complexes.
Problem: Is there an efficient way to triangulate the n-cube, i.e. calculate a (relatively) small list of n-simplices on the same vertices as the cube, and which define a simplicial complex spanning the cube?
I've done some reference-chasing and there seems to be no decently-sharp estimate (as an upper or lower bound) for the asymptotic complexity of the problem, although the best upper-bounds I'm aware of (for the size of the smallest solution-set) seem to indicate something exponentially smaller than factorial (see Haiman, 91). This paper also exhibits a lower bound, given below
$\frac{2^n\,n!}{(n+1)^{{}^{(n+1)/2}}}$
Orden and Santos improved the upper bound somewhat, by reducing the base of the exponential.

REPLY [6 votes]: A slight improvement on the lower bound was recently obtained by Alexey Glazyrin (http://arxiv.org/abs/0910.4200, http://www.sciencedirect.com/science/article/pii/S0012365X12003974). Basically, it changes the $2^n$ factor in Nick's initial post to an $e^n$. Smith's hyperbolic volume method gave an $\sqrt{6}^n$.
But, of course, as Greg points out, the real open question is whether the simplicity of the cube grows as the $c^n n!/n^{n/2}$ of the lower bounds or as the $c^n n!$ that can be achieved by actual constructions (or something in between).