Optimal cycles enclosing all the nodes of a $k$-dimensional hypercube

TL;DR

For any $k$ above one, it is constructively proved that it is possible to visit once and only once all the aforementioned nodes of a hypercube, H(2,k), with a cycle having only $3 \cdot 2^{k-2}$ links.

Abstract

We solve the general problem of visiting all the $2^k$ nodes of a $k$-dimensional hypercube by using a polygonal chain that has minimum link-length, and we show that this optimal value is given by $h(2,k):=3 \cdot 2^{k-2}$ if and only if $k \in \mathbb{N}-\{0,1\}$. Furthermore, for any $k$ above one, we constructively prove that it is possible to visit once and only once all the aforementioned nodes, $H(2,k):=\{\{0,1\} \times \{0,1\} \times \dots \times \{0,1\}\} \subset \mathbb{R}^k$, with a cycle (i.e., a closed path) having only $3 \cdot 2^{k-2}$ links.

Figures (5)

• Figure 1: The minimum-link perfect covering cycle $\mathcal{\bar{C}}(h(2,2)):=\left(\frac{1}{2},\frac{3}{2}\right)$-$(-1,0)$-$(2,0)$-$\left(\frac{1}{2},\frac{3}{2}\right)$ joins all the nodes of $H(2,3)$ (picture realized with GeoGebra Geogebra:32).
• Figure 2: The minimum-link closed polygonal chain $\mathcal{P}(6):=\left(\frac{1}{2},\frac{1}{2},2 \right)$-$\left(-\frac{1}{2},-\frac{1}{2},0 \right)$-$(\frac{3}{2},\frac{3}{2},0)$-$\left(\frac{1}{2},\frac{1}{2},2 \right)$-$(-\frac{1}{2},\frac{3}{2},0)$-$(\frac{3}{2},-\frac{1}{2},0)$-$\left(\frac{1}{2},\frac{1}{2},2 \right)$ visits all the nodes of $H(2,3)$ once and only once (picture realized with GeoGebra Geogebra:32).
• Figure 3: The minimum-link closed polygonal chain $\mathcal{P}(12):=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2} \right)$-$\left(-1,-1,-1, 0 \right)$-$(2,2,2,0)$-$\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2} \right)$-$(-1,-1,2,0)$-$(2,2,-1,0)$- $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2} \right)$-$(-1,2,-1,0)$-$(2,-1,2,0)$-$\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2} \right)$-$(-1,2,2,0)$-$(2,-1,-1,0)$-$\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2} \right)$ joins all the nodes of $H(2,4)$ (picture realized with GeoGebra Geogebra:32).
• Figure 4: The minimum-link perfect covering cycle $\mathcal{\bar{C}}(h(2,3)):=\left(\frac{1}{2},\frac{1}{2},2 \right)$-$\left(-\frac{1}{2},-\frac{1}{2},0 \right)$-$\left(\frac{4}{3},\frac{4}{3},0 \right)$-$\left(\frac{1}{2},\frac{1}{2},\frac{5}{2} \right)$-$\left(-\frac{1}{3},\frac{4}{3},0 \right)$-$\left(\frac{3}{2},-\frac{1}{2},0 \right)$-$\left(\frac{1}{2},\frac{1}{2},2 \right)$ joins all the nodes of $H(2,3)$ (picture realized with GeoGebra Geogebra:32).
• Figure 5: The minimum-link perfect covering cycle $\mathcal{\bar{C}}(h(2,4))\space:=\space\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2} \right)$-$\left(-1,-1,-1,0 \right)$-$\left(\frac{3}{2},\frac{3}{2},\frac{3}{2},0 \right)$-$\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},2 \right)$-$\left(-\frac{1}{2},-\frac{1}{2},\frac{3}{2},0 \right)$-$\left(\frac{4}{3},-\frac{1}{3},-\frac{1}{3},0 \right)$- $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{5}{2} \right)$-$\left(-\frac{1}{3},\frac{4}{3},-\frac{1}{3},0 \right)$-$\left(\frac{5}{4},-\frac{1}{4},\frac{5}{4},0 \right)$-$\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},3 \right)$-$\left(-\frac{1}{4},\frac{5}{4},\frac{5}{4},0 \right)$-$\left(2,-1,-1,0 \right)$-$\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{3}{2} \right)$ visits all the nodes of $H(2,4)$ once and only once (picture realized with GeoGebra Geogebra:32).

Theorems & Definitions (11)

• Definition 1.1
• Definition 1.2
• Definition 1.3
• Lemma 2.1
• proof
• Theorem 2.2
• proof
• Theorem 2.3
• proof
• Corollary 2.4