Joining the $2^k$ points of $\{0,1\}^k$ with the shortest tree

Question

Let $k$ be a given positive integer, and then consider the unit hypercube $\{0, 1\}^k \subset \mathbb{R}^k$ (i.e., a $k$-dimensional "cube" in the well-known Euclidean space).
We need to join all the $2^k$ vertices of the stated hypercube by using a Euclidean Steiner tree, which is a spanning tree of the original vertices plus zero or more new vertices (Steiner points) such that we minimise the total Euclidean length of the edges in the tree. Denote the total length of these optimal trees as $l_{min}(k)$, for any integer $k$ (e.g., $l_{min}(1) = 1$, $l_{min}(2) = 1 + \sqrt{3}$, $l_{min}(3) \approx 6.196$, and so forth).

I wonder if solving the problem for every $k$ would be doable. I can only point out that an upper bound is trivially given by $2^k-1$ and that the trick to join $2^{k-1}$ pairs of opposite vertices does not improve the mentioned trivial solution as $k$ goes above $3$.

The figure below constructively shows that $l_{min}(3) \leq 4 \cdot \sqrt{2} + 1$.

P.S. For any $k + 1$ we can also recycle our best tree, double it and then spend a unit segment to join the two subtrees, so $l_{min}(k+1) \le 1 + 2 l_{min}(k)$.

P.P.S. The user Noam D. Elkies suggested in a comment of the present thread that $l_{min}(k) < l_{upper}(k)$ holds for every $k>2$ (since my value of $l_{upper}(2)$ is certainly bigger than the optimal solution of the square Steiner tree problem and thus $l_{min}(3) < 4 \cdot \sqrt{2} + 1$ follows). Noam was definitely right since in this paper the solution $l_{upper}(2) = 1 + \sqrt{3}$ is given (see Figure 2 from the Reference above).

Answer 1

I suspect that the optimum, for a cube of side length $2$, is $2^k \sqrt{3} - 2 \sqrt{3}+2$. Note that the optimum if we use edges of the cube is $2 (2^k-1)$, so this is better by a factor of roughly $\sqrt{3}/2$. (Side comment: For trees in the plane, there is a famous conjecture of Gilbert and Pollack (1968) that adding extra vertices can never decrease the length by a factor better than $\sqrt{3}/2\approx 0.866$. However, in $k$-dimensional space, better improvements are possible; Gilbert and Chung (1976) give examples with ratio approaching $\tfrac{\sqrt{3}}{4-\sqrt{2}} \approx 0.669$. It is peculiar that this particular $k$-dimensional problem seems to have the same ratio as the conjectured worst case of the $2$-dimensional problem.)

The form of the solution: I'll write $\vec{e}_1$,..., $\vec{e}_k$ for the standard basis of $\mathbb{R}^k$. The vertices of my cube are at $x_1 \vec{e}_1 + \cdots + x_k \vec{e}_k$ for $x_1$, ..., $x_k \in \{ \pm 1 \}$.

I will add an extra $2$-valent vertex denoted $v( )$ at the center $(0,0,\ldots, 0)$ of the cube. It will be at the midpoint of an edge, so we can erase it at the end, but it makes the notation more symmetric.

Combinatorially, my tree is the perfect binary rooted tree with $1+2+\cdots + 2^{k-1}+2^k$ vertices. A general vertex will be denoted $v(x_1, x_2, \ldots, x_j)$ where $0 \leq j \leq k$ and each $x_i$ is $\pm 1$. The root is $v()$ at position $(0,0,\ldots,0)$; the $2^k$ leaves are the vertices of the cube, with $v(x_1, x_2, \ldots, x_k)$ at $(x_1, x_2, \ldots, x_k) = x_1 \vec{e}_1 + \cdots + x_k \vec{e}_k$. In general, the children of $v(x_1, x_2, \cdots, x_j)$ are $v(x_1, x_2, \cdots, x_j, 1)$ and $v(x_1, x_2, \cdots, x_j, (-1))$.

I'll abbreviate $v(\overbrace{1,1,\ldots,1}^j)$ to $v(1^j)$. If I figure out the coordiantes of the $v(1^j)$, then the other vertices at level $j$ are obtained by switching the sign of the coordinates in all possible ways. Let $\vec{u}_j$ be the unit vector in direction $v(1^j) - v(1^{j-1})$ and let $r_j$ be the length of the edge $(v(1^{j-1}), v(1^j))$. So $v(1^j)$ is in position $r_1 \vec{u}_1 + \cdots + r_j \vec{u}_j$. The rest of this post is computing the values of the $\vec{u}_j$ and $r_j$.

Computing the $\vec{u}_j$ Look at the vertex $v(1^j)$ for $1 \leq j \leq k$. Its neighbors are $v(1^{j-1})$, $v(1^{j+1})$ and $v(1^j (-1))$. These four vertices lie in an affine two-plane $P$. By symmetry, the vertices $v(1^{j-1})$ and $v(1^j)$ are on $\vec{e}_{j+1}^{\perp}$ and $v(1^{j+1})$ and $v(1^j(-1))$ are reflections of each other over $e_{j+1}^{\perp}$. Thus, $(\vec{u}_j, \vec{e}_{j+1})$ is an orthonormal basis for $P$. We want $\vec{u}_{j+1}$ and $-\vec{u}_j$ to make a $120^{\circ}$ angle, so we should have $$\vec{u}_{j+1} = (\cos 60^{\circ}) \vec{u}_j + (\sin 60^{\circ}) \vec{e}_{j+1} = \frac{1}{2} \vec{u}_j +\frac{\sqrt{3}}{2} \vec{e}_{j+1} \tag{1}$$ Using $(1)$ recursively with the base case $\vec{u}_1 = \vec{e}_1$, we get $$\vec{u}_j = \frac{1}{2^{j-1}} \vec{e}_1 + \frac{\sqrt{3}}{2^{j-1}} \vec{e}_2 + \cdots + \frac{\sqrt{3}}{4} \vec{e}_{j-1} + \frac{\sqrt{3}}{2} \vec{e}_j. \tag{2}$$

Computing the $r_j$: We want to have $$r_1 \vec{u}_1 + \cdots + r_k \vec{u}_k = (1,1,\ldots,1) \tag{3}.$$ Extracting the $j$-th component of $(3)$, and using $(2)$, we get $$\begin{array}{rcll} r_1+\frac{1}{2} r_2 + \cdots + \frac{1}{2^{k-1}} r_k &=& 1 & \\ \frac{\sqrt{3}}{2} r_j + \frac{\sqrt{3}}{4} r_{j+1} + \cdots + \frac{\sqrt{3}}{2^{k-j+1}} r_k &=& 1 & \text{for}\ j \geq 2 \\ \end{array} \tag{4}$$ Solving $(4)$ recursively (starting at $r_k$ and working back to $r_1$), I get $$r_j = \begin{cases} 1-1/\sqrt{3} & j=1 \\ 1/\sqrt{3} & 2 \leq j \leq k-1 \\ 2/\sqrt{3} & j=k. \end{cases} \tag{5}.$$

The total length of the tree is $$ \begin{align} 2 r_1 + 4 r_2 + \cdots + 2^k r_k &= 2(1-1/\sqrt{3}) + (4+8+\cdots+2^{k-1})/\sqrt{3} + 2^k \cdot 2/\sqrt{3} \\ &= 2^k \sqrt{3} - 2 \sqrt{3}+2. \end{align}$$

In case this formula is useful to anyone, for $1 \leq j \leq k-1$, we have $$v(1^j) = {\Big(}1-\frac{1}{2^{j-1} \sqrt{3}},\ 1-\frac{1}{2^{j-1}}, 1-\frac{1}{2^{j-2}},\ \ldots,\ 1-\frac{1}{4},\ 1-\frac{1}{2},\ 0,0,\ldots,0 {\Big)}.$$

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As mentioned above, an optimal Steiner tree always has all interior vertices trivalent and all angles equal to $120^{\circ}$. See, for example, the first paragraph of Section 6.1 in

Hwang, Frank K.; Richards, Dana S.; Winter, Pawel, The Steiner tree problem, Annals of Discrete Mathematics. 53. Amsterdam: North-Holland. xi, 339 p. (1992). ZBL0774.05001.

So the remaining problem is to guess which trivalent tree with $2^k$ labeled leaves to use. This one seems like the best choice, but I have no idea how to prove that it can't be improved on.