, 2014

We describe a recursive procedure for constructing the desired Steiner tree. The case $n = 1$ is handled by a single line segment.

Assume a Steiner tree of length $1 + (2^{n-1} - 1)\sqrt{3}$ associated with the vertex set of a unit $n$-cube has been constructed such that each vertex of the $n$-cube is the endpoint of just one segment, and the length of this segment is greater than $\sqrt{3}/6$. Now consider a unit $(n+1)$-cube. Select a pair of opposite $n$-faces and consider the $n$-cube whose vertices are the midpoints of the edges joining the corresponding vertices of these faces. Start with the assumed Steiner tree associated with the vertex set of this $n$-cube. For each vertex $v$ of this $n$-cube, delete a segment of length $\sqrt{3}/6$ from the segment terminating at $v$, then add the segments joining the new endpoint to the two vertices of the $(n+1)$-cube that are endpoints of the edge containing $v$. By the Pythagorean theorem, these segments each have length $1/\sqrt{3}$. The net effect of these changes is to add $2^n(2/\sqrt{3} - \sqrt{3}/6) = 2^{n-1}\sqrt{3}$ to the length of the tree. Thus the resulting tree has length $1 + (2^n - 1)\sqrt{3}$. It is clearly a Steiner tree associated with the vertex set of the $(n+1)$-cube and has the additional properties required for the induction.