TITLE: Tilting the $d$-cube to vertically separate its vertices
QUESTION [18 upvotes]: Let $C_d$ be a unit edge-length cube in $d$ dimensions.
I would like to orient it ("tilt" it) so that the vertical (last) coordinates
of its $2^d$ vertices are maximally separated, in the sense
that the minimum vertical distance between any two vertices is maximized
over all orientations.
For $C_2$ (in standard orientation, edges parallel to
Cartesian axes), tilting $\arctan \frac{1}{2} \approx 26.6^\circ$ separates
the vertices by $\delta=1/\sqrt{5} \approx 0.447$:

 
 
 
 
 


For $C_3$ (in standard orientation), I believe that rotating the vector $(0,0,1)$ to lie on
the vector $(\frac{1}{4},\frac{1}{2},1)$ results in a vertex separation of $\delta=1/\sqrt{21} \approx 0.218$:

 
 
 
 
 


My question is:

Q. What is the generalization to $C_d$ for $d>3$? What is the largest
  vertex separation $\delta$ achievable?
  Can one always achieve a uniform vertex separation
  (the same $\delta$ between each vertically adjacent pair),
  as in $C_2$ and $C_3$?

REPLY [6 votes]: Robert Israel's answer is best possible.
As pointed out there, the problem is equivalent to finding a vector in $\mathbb{R}^d$ with minimal separation $1$ among sums of subsets of entries, and with the least possible length ($\ell_2$ norm). The vector Robert constructed, after rescaling to make the separation $1$, is $v_d=[2^0,2^1,\dots,2^{d-1}]$. Clearly this has the smallest possible $\ell_1$ norm.
It turns out that it also has the smallest possible $\ell_2$ norm, but for $d>3$ not the smallest $\ell_\infty$ norm.
Proof. It is a straightforward consequence of these 2 facts:

Lemma. The following equality holds in $\mathbb{Z}[x_1,\dots x_d]$:

$$\sum_{S,T\subseteq I_d}(x_S-x_T)^2=C_d \sum_{i=1}^d x_i^2$$
where $I_d=\{1,2,\dots d\}$, $\displaystyle x_S=\sum_{i\in S} x_i$ and the positive integer $C_d$ depends only on $d$.

if $x=v_d$ then $\{x_S\}=\{0,1,2,\dots 2^d-1\}$ clearly has the tightest possible packing of values (among vectors with separation $1$, and uniquely up to permutations) and therefore must minimize the left hand side in the lemma.

Proof of Lemma.
There are 16 equally frequent cases for the occurrence of $x_ix_j$ in $(x_S-x_T)^2$:
$$\begin{array}{@{}l|l|l|c|c@{}}
\mbox{case} & \in S & \in T & x_S-x_T & x_ix_j \text{ coef in }\\
            &     &      &     & (x_S-x_T)^2\\ \hline
\mbox{1}  & \_  & \_  & \cdots           & 0 \\
\mbox{2}  & i   & \_  & x_i\cdots        & 0 \\
\mbox{3}  & j   & \_  & x_j\cdots        & 0 \\
\mbox{4}  & i,j & \_  & x_i+x_j\cdots    & 2 \\
\mbox{5}  & \_  & i   & -x_i\cdots       & 0 \\
\mbox{6}  & i   & i   & \cdots           & 0 \\
\mbox{7}  & j   & i   & x_j-x_i\cdots    & -2 \\
\mbox{8}  & i,j & i   & x_j\cdots        & 0 \\
\mbox{9}  & \_  & j   & -x_j\cdots       & 0 \\
\mbox{10} & i   & j   & x_i-x_j\cdots    & -2 \\
\mbox{11} & j   & j   & \cdots           & 0 \\
\mbox{12} & i,j & j   & x_i\cdots        & 0 \\
\mbox{13} & \_  & i,j & -x_i-x_j\cdots   & 2 \\
\mbox{14} & i   & i,j & -x_j\cdots       & 0 \\
\mbox{15} & j   & i,j & -x_i\cdots       & 0 \\
\mbox{16} & i,j & i,j & \cdots           & 0 \end{array}$$
the total coefficient of $x_ix_j$ over all the cases is 0. Therefore $\sum_{S,T\subseteq I_d}(x_S-x_T)^2$ only contains square terms and by symmetry must then be a multiple of $\sum_{i=1}^d x_i^2 \quad\blacksquare$
Regarding the fact that $v_d$ above is not minimal $\ell_\infty$-wise, counterexamples are provided via the Atkinson-Negro-Santoro sequence:
$d=4: [3,5,6,7]$
$d=5: [6,9,11,12,13]$
$d=6: [11,17,20,22,23,24]$
etc.