Let $G_2^4 :=\{0,1\} \times \{0,1\} \times \{0,1\} \times \{0,1\}$ be a set of $2^4$ points in $\mathbb{R}^4$.
Which is the minimum number of straight lines connected at their endpoints (i.e., the shortest polygonal chain) needed to join all the vertices of $G_2^4$ ?
Now, let us call this optimum value $h(2,k)$, where $k$ is the number of dimensions we are taking into account.
We know that $h(2,2)=3$ and $h(2,3)=6$ (e.g., $(1,0,0)$-$(0,0,0)$-$(2,2,2)$-$(\frac{1}{2},-1,\frac{1}{2})$-$(-\frac{1}{2},1,\frac{3}{2})$-$(1,1,0)$-$(1,1,0)$ is a valid solution - for details see GENERAL UNCROSSING COVERING PATHS INSIDE THE AXIS-ALIGNED BOUNDING BOX) so that $h(2,4) \leq 2 \cdot h(2,3) + 1 = 13$.
Is this value optimal? Could we answer the same question for any $k > 3$?
I am persuaded that $h(2,4) \leq 12$, but the problem is completely open.
Happy to announce that we have fully solved the general case of the above, formally and constructively proving the existence of optimal covering cycles, for any given choice of $k \in \mathbb{N}-\{0,1\}$ for the set $G_2^k :=\{0,1\} \times \{0,1\} \times \cdots \times \{0,1\}$, with exactly $3 \cdot 2^{k-2}$ links.
Since no covering trail for the same set of nodes (i.e., the $2^k$ vertices of a $k$-cube, for any $k \geq 2$) can have less than $3 \cdot 2^{k-2}$ links, those cycles are also minimum-link covering trails.
Moreover, it is possible to touch once (and only once) all the Steiner points of any of the minimum-link covering cycles mentioned above, with the only exception of the first/last one (this is mandatory in order to close the covering path and get a cycle).
Here is the link of our preprint that has just been announced on the arXiv, any comments are welcome: https://arxiv.org/abs/2212.11216