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 $C_3$?


Given a unit vector $u \in \mathbb R^d$, the "heights" of vertices of the $n$-cube where $u$ is regarded as the vertical direction are the sums of subsets of the entries of $u$. Thus the minimum separation is the minimum difference between the sums of two distinct subsets of these entries. If you take $$u = [1,2,\ldots,2^{d-1}]/\sqrt{1^2 + 2^2 + \ldots (2^{d-1})^2} = \sqrt{\dfrac{3}{4^d-1}}[1,2,\ldots,2^{d-1}]$$ you get uniform separation of $\sqrt{3/(4^d-1)}$.

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  • $\begingroup$ Beautiful! $\mbox{}$ $\endgroup$ – Joseph O'Rourke Jan 5 '15 at 2:01
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    $\begingroup$ Is it clear that uniform separation gives the optimal separation? (if you change the vector $u$, you might in principle destroy the uniformity, but obtain separation over a (reasonably substantially) larger range: if $u=[1,1,1,\ldots,1]$ then the image of the diagonal is of length $\sqrt d$, whereas here it is $O(1)$). This gives an upper bound for the minimal separation of $\sqrt d/2^d$, whereas here you have a lower bound of approximately $\sqrt 3/2^d$. $\endgroup$ – Anthony Quas Jan 5 '15 at 5:02
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    $\begingroup$ @AnthonyQuas: Good point! I can verify that for $d=2,3$, the max separation is achieved with uniform separation. $\endgroup$ – Joseph O'Rourke Jan 5 '15 at 11:30
  • $\begingroup$ (Initially I accepted this lucid answer, but as Anthony points out, it doesn't necessarily solve the original question [not that Robert ever claimed it did].) $\endgroup$ – Joseph O'Rourke Jan 6 '15 at 12:37

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