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graph concept – Clustering of vertices in an $n$-dimensional dice

You know {that a} sq. has a diagonal of sqrt(2), a dice has a diagonal of sqrt(3), and many others. i.e. an nD hypercube has a diagonal of sqrt(n).

If you additional align your hypercube alongside this diagonal, the vertices will fall in precisely n+1 equally spaced layers: the sq. is vertex atop diagonal (orthogonal to the chosen axis) atop contrary vertex, the dice is vertex atop triangle atop twin triangle atop contrary vertex, and many others.

Thus your requested for minimal distance merely is counting the respective layers to the place the second vertex is in, when the primary vertex is taken to breathe the top-most.

— rk

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