For a triangulation $T$ of an $n$-dimensional cube, whose vertices are the $2^n$ original vertices, let $d(T)$ be the largest Hamming-distance of two vertices that are in the same simplex.
How small can $d(T)$ be as a function of $n$?
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Assume that the vertices of the cube form the set $\{-1,1\}^n$. Consider a simplex containing $0$, with vertices $a_1,a_2,\dots,a_{n+1}$. It cannot happen that $\langle a_1,a_i\rangle>0$ always, otherwise $0$ is separated from the simplex by a hyperplane orthogonal to $a_1$. Hence $\langle a_1,a_i\rangle\leq 0$ for some $i$, which means that $d(T)\geq n/2$.
[ADDED] We may choose some of the $a_i$ so that $0$ will be in the relative interior of their convex hull. Then we will get the strict inequality, getting the lower bound of $(n+1)/2$.
This seems to be close to optimal. If there exists a Hadamard matrix of order $n+1$, we can choose a simplex containing $0$ with Hamming distances $(n+1)/2$ (take a Hadamard matrix with the first row and column consisting of $-1$s, remove those row and column --- you will get the set of vertices, along with the all-$-1$s vertex). I hope that, starting from this, we can construct a good triangulation, although I do not have an explicit construction.
