I'm afraid that the title is not clear. The problem at hand is on the maximum number of vertices of a convex hull generated by the following $2^n$ points: \begin{align} \pm\boldsymbol{v}_1\pm\boldsymbol{v}_2\ldots\pm\boldsymbol{v}_n \end{align} where $\boldsymbol{v}_k \in \mathbb{R}^3$. If $n\leq 3$, linearly independent $\{\boldsymbol{v}_k\}$ makes these $2^n$ points be vertices. I guess when $n$ becomes larger only a few of them can be vertices and the others are inside the convex hull (almost certain when considering the two-dimensional analogy). My questions are:
- Is there any better upper bound on the number of vertices?
- If the bound is exact, how can we construct such $\{\boldsymbol{v}_k\}$?
Any comments would be appreciated. Thank you.