Kinds of Matchings Extending to Hamiltonian Cycles in Hypercube Networks

Abstract

The hypercube $Q_n$ is a well-known and efficient interconnection network. Ruskey and Savage posed the following question: does every matching in a hypercube $Q_n$ for $n\ge 2$ extend to a Hamiltonian cycle? Fink addressed this by proving that every perfect matching extends to a Hamiltonian cycle in $Q_n$, thereby resolving Kreweras’ conjecture. Ruskey and Savage’s problem is still open and has been proven only for small matchings. An edge of $Q_n$ is an i-edge when the binary representations of its endpoints differ at the ith coordinate. In this paper, we consider $Q_n$ for $n\ge 3$ and show that any matching consisting of edges of at most six types, which does not cover every pair of vertices at a distance of 3, extends to a Hamiltonian cycle.

1. Introduction

Matchings and cycles are basic structures in graph theory. These structures in hypercubes are a class of graphs that have been extensively explored in computer science and mathematics. Let the set $\{1,2,\dots ,n\}$ be denoted as $\left[n\right]$. The n-dimensional hypercube $Q_n$, where the vertex set consists of binary strings of length n, i.e., $V\left(Q_n\right)=\{u:u=u^1,u^2\dots u^n\mathrm{and}{u}^i\in \{0,1\}\mathrm{for}\mathrm{all}i\in \left[n\right]\}$. Two vertices are adjacent if and only if their corresponding binary strings differ at exactly one position.

The hypercube $Q_n$ is a well-known and efficient interconnection network. For any $n\ge 2$, $Q_n$ is known to be Hamiltonian. This statement dates back to 1872 [1]. Since then, there has been considerable interest in studying Hamiltonian cycles in hypercubes, and this topic has received considerable attention [2,3,4,5,6]. A hypercube graph is bipartite. Alaeiyan [7] presented the bipartite Ala graph, denoted as $\mathrm{Ala}(m,G,k)$, where m and k are positive integers, and investigated various properties of this graph family, focused on Hamiltonian cycles and Eulerian properties.

If no two edges in a graph (G) share a vertex, the set of edges is called a matching. If a matching contains every vertex of G, it is considered as a perfect matching. If every vertex in a graph G appears exactly once in a cycle, the cycle is called a Hamiltonian cycle.

The following problem was asked by Ruskey and Savage [8]: for $n\ge 2$, does any matching in $Q_n$ extend to a Hamiltonian cycle of $Q_n$? Kreweras [9] conjectured that for $n\ge 2$, any perfect matching in $Q_n$ can be extended to a Hamiltonian cycle of $Q_n$. This conjecture was proved by Fink [10,11] and strengthened for $K\left(Q_n\right)$. $K\left(Q_n\right)$ is a complete graph of the vertices of hypercube $Q_n$.

Theorem 1

([10,11]). For $n\ge 2$ and every perfect matching P of $K\left(Q_n\right)$, there exists a perfect matching R of $Q_n$ such that $P\cup R$ extends to a Hamiltonian cycle in $K\left(Q_n\right)$.

Gregor [12] strengthens the Fink result [10] and finds that when we split the hypercube into subcubes of non-zero dimensions, every perfect matching of the hypercube extends to a Hamiltonian cycle if and only if the matching interconnects these subcubes. Regarding perfect matchings, Dimitrov et al. [13] showed that for a given perfect matching P in $Q_n$, a Hamiltonian cycle that does not contain any edge from P exists if and only if the graph $Q_n-P$ is connected.

In the study of Hamiltonian cycles in hypercubes, both long edges and short edges play crucial roles. A long edge in a hypercube ($Q_n$) typically refers to an edge connecting two vertices that differ in more than one coordinate; in other words, their Hamming distance is greater than 1. In contrast, a short edge connects vertices that differ in exactly one coordinate, which corresponds to the standard edges of the hypercube. Moreover, Fink [10] established that the following result holds for $n\in \{2,3,4\}$. Later, Wang and Zhao [14] extended the proof to the 5-cube.

Theorem 2

([10,14]). For $n\in \{2,3,4,5\}$, any matching in $Q_n$ extends to a Hamiltonian cycle of $Q_n$.

Since in $Q_n$, not every matching is perfect or extendable to a perfect matching, Limaye and Sarvate, in [15], showed that matching in $Q_n$ is k-extendable for $k<n$; also, any matching is extendable to a perfect matching of size n if and only if it does not cover the neighborhood of an uncovered vertex. Moreover, Vandenbussche and West [16] proved that the extendability of matching M (with vertex set U) is k-suitable if the graph $G-U$ contains no deficient set less than k. For $f_k\left(n\right)$ as the largest value of r, any k-suitable matching with at most $f_k\left(n\right)=k(n-k)+\left({\displaystyle \genfrac{}{}{0pt}{}{k-1}{2}}\right)$ for $k\le n-3$ is extendable to a perfect matching. It is important to note that a positive solution for a matching that extends to a perfect matching is the induced matching [16]. However, this result does not completely solve the problem, as hypercubes of a matching extending to the maximal matching by inclusion (i.e., adding more edges of M) may still not be perfect. Thus, the kind of matching we are focusing on is not extendable to perfect matching.

Dvořák [5] showed that for all $n\ge 2$, any set of at most $2n-3$ edges in the hypercube $Q_n$, where the edges form vertex-disjoint paths, can be extended to a Hamiltonian cycle of $Q_n$. As a corollary, this implies that matching in $Q_n$ with no more than $2n-3$ edges can be extended to a Hamiltonian cycle. The authors in [17] proved that for $n\ge 4$, every matching of at most $3n-10$ edges in $Q_n$ extended to a Hamiltonian cycle of $Q_n$. Dvořák and Fink [18] showed that matchings of a size bounded by an $O\left(n^2\right)$ function in $K\left(Q_n\right)$ for $n\ge 5$, extend to a Hamiltonian cycle, using the edges of $Q_n$.

Fink proved that every perfect matching in $Q_n$ extends to a Hamiltonian cycle. The broader question, posed by Ruskey and Savage, of whether every matching in $Q_n$ extends to a Hamiltonian cycle remains unresolved for general matchings. A solution is only known for some particular cases, including perfect matchings, linear matchings, and matchings of a quadratic size. In this paper, we consider $Q_n$ for $n\ge 3$ and show that any matching consisting of edges of at most six types, which does not cover every pair of vertices at a distance of 3, extends to a Hamiltonian cycle. Section 2 contains preliminaries and lemmas, while Section 3 presents the extension of the given conditional matching to a Hamiltonian cycle.

2. Preliminaries and Lemmas

Terminology and notation used in this paper are defined as follows: terms that remain undefined can be found in [19]. The vertex and edge sets of graph G are denoted by $V\left(G\right)$ and $E\left(G\right)$, respectively. Given a set $S\subseteq E\left(G\right)$, graph $G-S$ is obtained by removing all edges of S from G. Let H and $H^{\prime }$ be two subgraphs of G. We use $H+H^{\prime }$ to denote the graph formed by combining vertex set $V\left(H\right)\cup V\left(H^{\prime }\right)$ and edge set $E\left(H\right)\cup E\left(H^{\prime }\right)$. For a set $S\subseteq E\left(G\right)$, $H+S$ denotes the graph with the vertex set $V\left(H\right)\cup V\left(S\right)$ and edge set $E\left(H\right)\cup S$, where $V\left(S\right)$ denotes the set of vertices incident to the edges in S.

The distance between vertices u and v is the number of edges in the shortest path connecting u and v in G, denoted by $d_G(u,v)$, with the subscript omitted when the context is clear. For every $i\in [n-1]$, a path between two vertices, a and b, represented as $P_{ab}$, is a set of unique vertices $a=v_1,v_2,\dots ,v_n=b$ and edges $v_i{v}_{i+1}$. A path $P_{aa}$ is composed of only a single vertex a if $a=b$, and $n=1$. When two paths $P_{ab}$ and $P_{cd}$ do not share any vertex and vertices b, c are adjacent, the combined path $P_{ab}+P_{cd}$ represents a new path from a to d, created by joining $P_{ab}$ and $P_{cd}$. A path P is considered a subpath of another path $P^{\prime }$ if P is a subgraph of $P^{\prime }$.

The n-dimensional hypercube $Q_n$ has a vertex set V and edge set E. The expression $\chi \left(u\right)={\sum }_{i=1}^n{u}^i$(mod 2) provides the parity of the vertex u. There are equal numbers of vertices $2^{n-1}$ with parities of 1 and 0. These vertices are often labeled black and white, respectively. Because $Q_n$ is bipartite, the vertices of the hypercube $Q_n$ can be separated into two sets according to their parity. Hence, $\chi \left(x\right)\ne \chi \left(y\right)$ if and only if $d(x,y)$ is odd.

An edge is a j-edge in the hypercube $Q_n$ for each $j\in \left[n\right]$, implying that the endpoints of the edge differ only in the j-th coordinate. The set of all such j-edges in $Q_n$ is denoted as $E_j$. Consequently, the complete set of edges in $Q_n$ expressed as the union of all j-edges: $E\left(Q_n\right)={\bigcup }_{j=1}^n{E}_j$. Based on the vertex sets $\{u\in V\left(Q_n\right):u^j=0\}$ and $\{u\in V\left(Q_n\right):u^j=1\}$, respectively. We consider the subcubes $Q_{n-j}^0$ and $Q_{n-j}^1$, which are isomorphic subcubes of $Q_n$ produced by deleting the j-edges. This relationship can be written as $Q_n-E_j=Q_{n-j}^0+Q_{n-j}^1$. Additionally, let $Q_m^{ϵL}$ and $Q_m^{ϵR}$ represent the subgraphs of $Q_{n-j}$, where $ϵ\in \{0,1\}$. Both $Q_m^{ϵL}$ and $Q_m^{ϵR}$ are isomorphic, with L and R indicating the left and right orientations.

Wang and Zhang obtained the following results [20]: matching M in $Q_3-\{u,v\}$ extends to a spanning 2-path, and the matching in $Q_n-u$ extends to the Hamiltonian path. Similarly, in this section, the results show a matching extension to the Hamiltonian path and spanning 2 paths in small cubes.

Lemma 3

([20]). Let u, v, x, y be pairwise distinct vertices in $Q_3$ such that $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(y\right)$, and let M be matching in $Q_3-\{u,v\}$ with $d(u,x)=d(v,y)=1$. Then, there exists a spanning 2-path $P_{ux}+P_{vy}$ in $Q_3$ passing through M.

Lemma 4

([20]). Let $u,v\in V\left(Q_n\right)$ for $n\in \{3,4\}$, such that $\chi \left(u\right)\ne \chi \left(v\right)$ and M is a matching in $Q_n-u$. Then, $Q_n$ contains a Hamiltonian path passing through M, connecting u and v.

Theorem 5

([5]). Let $n\ge 2$, and let $F\subseteq E\left(Q_n\right)$ such that $\left|F\right|\le 2n-3$. If F is a linear forest, then there exists a Hamiltonian cycle in $Q_n$ that contains every edge in F.

Lemma 6

([20,21]). Let x and y be vertices in $Q_n$ for $n\in \{3,4\}$ such that $\chi \left(x\right)=\chi \left(y\right)$ and M matching in $Q_4-x$. Then there exists a spanning 2-path $P_{xy}+P_{uv}$ that passes through M, where u and v are two distinct vertices such that $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)$.

Lemma 7

([21]). Let $u,x,y$ be three distinct vertices in $Q_3$ such that $\chi \left(u\right)\ne \chi \left(x\right)=\chi \left(y\right)$. Suppose M is a matching in $Q_3-u$, then there exists a spanning $\left(\right\{u,v\},\{x,y\left\}\right)$-path $P_{u,*}+P_{v,*}$, where $v\in V\left(Q_3\right)$ is a vertex satisfying $\chi \left(v\right)=\chi \left(u\right)$.

Lemma 8

([21]). Let $u,v,x,y$ be four pairwise distinct vertices in the three-dimensional hypercube $Q_3$, such that $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(y\right)$. Suppose that M is a matching in the subgraph $Q_3-\{u,v\}$. Then, there exists a spanning $\left(\right\{u,v\},\{x,y\left\}\right)$-path $P_{u,*}+P_{v,*}$ in $Q_3$ that contains all edges of M.

Lemma 9

([21]). If M is a matching in $Q_3$, then there exist two distinct Hamiltonian paths in $Q_3$ starting at vertex u, each ending at a different vertex, such that both paths include all edges of M.

Lemma 10.

Let u, v, x, and y be pairwise distinct vertices in $Q_4$ such that $\chi \left(x\right)=\chi \left(y\right)\ne \chi \left(u\right)=\chi \left(v\right)$. If $M\le 5$ is a matching in $Q_4-\{u,v\}$, then there is a spanning 2-path $P_{ux}+P_{vy}$ in $Q_4$ containing M such that $d(u,x)=d(v,y)=1$.

Proof. 

Choose $j\in \left[4\right]$ such that $|M\cap E_j|$ is as small as possible. As M is a given matching, and thus $\left|M\right|\le 5$, it follows that $|M\cap E_j|\le 1$. If M contains no edge from $E_j$, we may assume without loss of generality that $j=4$. In the case where $|M\cap E_j|=1$, M must include exactly one edge from each $E_i$ for all $i\in \left[4\right]$. Thus, there are four possible choices for j, and we can select one such that the corresponding edge in $M\cap E_j$ is not incident with either x or y. Next, $Q_4$ is decomposed into two subcubes $Q_3^0$ and $Q_3^1$ along the edge set $E_4$. Using the vertex-transitivity of $Q_4$, we may assume that vertex $u\in V\left(Q_3^0\right)$.

Case 1. $M\cap E_j=\varnothing$.

Subcase 1.1. $v\in V\left(Q_3^1\right)$.

Given that $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(y\right)$ and that both u and v are uncovered by M, the lemma’s statement is symmetric for pairs $\{u,v\}$ and $\{x,y\}$. Therefore, we only need to consider the following two cases.

Subcase 1.1.1. $x\in V\left(Q_3^0\right)$, $y\in V\left(Q_3^1\right)$ (or $y\in V\left(Q_3^0\right)$, $x\in V\left(Q_3^1\right)$).

Because $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(y\right)$ and matching M is in $Q_4-\{u,v\}$, Lemma 4 ensures that there exists a Hamiltonian path $P_{ux}^0$ in $Q_3^0$ and another path $P_{vy}^1$ in $Q_3^1$, such that $P_{ux}^0$ and $P_{vy}^1$ pass through matchings $M_0$ and $M_1$, respectively. Thus, the required spanning path is $P_{ux}^0+P_{vy}^1$.

Subcase 1.1.2. $\{x,y\}\subseteq V\left(Q_3^0\right)$ (or $\{x,y\}\subseteq V\left(Q_3^1\right)$).

Since $u\notin V\left(M\right)$ and $\chi \left(u\right)\ne \chi \left(x\right)=\chi \left(y\right)$, by Lemma 7, there exists a spanning 2-path $P_{u{s}_0}+P_{xy}$ in $Q_3^0$ passing through $M_0$, where $s_0\in V\left(Q_3^0\right)$ satisfies $\chi \left(s_0\right)=\chi \left(u\right)$. Furthermore, since $v\notin V\left(M\right)$ and $\chi \left(v\right)\ne \chi \left(s_1\right)$, by Lemma 4, there exists a Hamiltonian path $P_{v{s}_1}^1$ in $Q_3^1$ that contains $M_1$. Consequently, the required spanning path is $P_{ux}^0+P_{s_{0}y}^0+P_{v{s}_1}^1+s_0{s}_1$.

Subcase 1.2. $v\in V\left(Q_3^0\right)$

Subcase 1.2.1. $x\in V\left(Q_3^0\right)$, $y\in V\left(Q_3^1\right)$ (or $y\in V\left(Q_3^0\right)$, $x\in V\left(Q_3^1\right)$)

By Lemma 9, there exists a Hamiltonian path $P_{s_{1}y}^1$ in $Q_3^1$ that includes the matching $M_1$ for some $i\in \{1,2\}$, where $s_1^1$ and $s_1^2$ are two distinct vertices in $Q_3^1$ satisfying $\chi \left(s_1^1\right)=\chi \left(s_1^2\right)\ne \chi \left(y\right)$. Choose an index i such that $s_0^i\ne x$; without loss of generality, let $i=1$. Because M is a matching in $Q_3^0-\{u,v\}$ and $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(s_0^1\right)$, by Lemma 3, there exists a spanning 2-path $P_{ux}+P_{v{s}_0^1}$ in $Q_3^0$ containing $M_0$. Thus, the required spanning 2-path is $P_{ux}^0+P_{v{s}_0^1}^0+P_{s_1^{1}y}^1+\left\{s_0{s}_1\right\}$.

Subcase 1.2.2: $\{x,y\}\subseteq V\left(Q_3^1\right)$

Because $\chi \left(x\right)=\chi \left(y\right)$, Lemma 6 implies the existence of a spanning 2-path $P_{x{s}_1}^1+P_{y{t}_1}^1$ in $Q_3^1$ containing $M_1$, where $s_1$ and $t_1$ are distinct vertices in $Q_3^1$ satisfying $\chi \left(s_1\right)=\chi \left(t_1\right)\ne \chi \left(x\right)$. Because M is a matching in $Q_3^0-\{u,v\}$ and $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(s_0\right)=\chi \left(t_0\right)$, Lemma 3 provides a spanning 2-path $P_{uv}^0+P_{s_0{t}_0}^0$ in $Q_3^0$ that contains $M_0$. Therefore, the resulting spanning 2-path is $P_{ux}^0+P_{vy}^0+P_{x{s}_1}^1+P_{y{t}_1}^1+\{s_0{s}_1,t_0{t}_1\}$.

Subcase 1.2.3. $\{x,y\}\subseteq V\left(Q_3^0\right)$.

If $|M_1|=4$, then $|M_0|=1$. According to Theorem 2, there exists a Hamiltonian cycle $C_1$ in $Q_3^1$ that passes through $M_1$. Since $M_1$ is a perfect matching and $|E\left(C_1\right)\setminus M_1|-2\ge 1$, we can choose an edge $s_1{t}_1\in E\left(C_1\right)\setminus M_1$ such that $\{s_0,t_0\}\cap \{u,v\}=\varnothing$. Given $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(y\right)$, Lemma 3 ensures the existence of a spanning 2-path is $P_{ux}+P_{vy}$ in $Q_3^0$ that includes the edge $s_0{t}_0$. Thus, the spanning path in $Q_4$ is $P_{ux}^0+P_{vy}^0+C_1+\{s_0{s}_1,t_0{t}_1\}-\{s_0{t}_0,s_1{t}_1\}$.

If $|M_1|\le 3$, we claim that at most one edge $e\in E\left(Q_3^1\right)\setminus M_1$ exists such that $M_1\cup \left\{e\right\}$ is not part of any Hamiltonian cycle in $Q_3^1$. For $|M_1|\le 2$, any $e\in E\left(Q_3^1\right)\setminus M_1$ ensures $M_1\cup \left\{e\right\}$ is a linear forest with at most three edges. By Theorem 5, such a forest is always included in a Hamiltonian cycle. For $|M_1|=3$, there are exactly three non-isomorphic size-3 matchings in $Q_3$, labeled $P_1,P_2$, and $P_3$ (see Figure 1). Checking each possible pair $(M_1,e)$ confirms this claim (see Figure 1).

Using Lemma 3 again, since $M_0\subseteq Q_3^0\setminus \{u,v\}$, there exists a spanning 2-path $P_{uv}+P_{xy}$ in $Q_3^0$ that contains $M_0$. Because the path contains more than $|M_0|+|M_1|+1$ edges, there is an edge $s_0{t}_0\in E(P_{ux}^0+P_{vy}^0)\setminus M_0$ such that $s_1{t}_1\notin M_1$ and $M_1+\left\{s_1{t}_1\right\}$ lies in some Hamiltonian cycle $C_1$ of $Q_3^1$. Therefore, the required spanning path in $Q_4$ is $P_{ux}^0+P_{vy}^0+C_1+\{s_0{s}_1,t_0{t}_1\}-\{s_0{t}_0,s_1{t}_1\}$.

Case 2. $|M\cap E_4|=1$. Let $M\cap E_4=\left\{w_0{w}_1\right\}$. Note that $\{w_0,w_1\}\cap \{x,y\}=\varnothing$.

Subcase 2.1. $v\in V\left(Q_3^1\right)$.

Subcase 2.1.1. $x\in V\left(Q_3^0\right),y\in V\left(Q_3^1\right)$ (or $x\in V\left(Q_3^1\right),y\in V\left(Q_3^0\right)$).

Now $\chi \left(w_0\right)=\chi \left(u\right)$ or $\chi \left(w_1\right)=\chi \left(v\right)$. Without loss of generality, we assume $\chi \left(w_0\right)=\chi \left(u\right)$. Since $v\notin V\left(M_1\right)$, by Lemma 4, there exists a Hamiltonian path $P_{vy}^1$ of $Q_3^1$ passing through $M_1$. Let $s_1$ be a neighbor of $w_1$ such that $s_0\ne x$. If $s_1\in V\left(P_{vy}^1\right)$, then because $M_0$ is a matching in $Q_3^0-\{u,w_0\}$, and $\chi \left(w_0\right)=\chi \left(u\right)\ne \chi \left(x\right)=\chi \left(s_0\right)$, by Lemma 3 there exists a spanning 2-path $P_{ux}+P_{w_0{s}_0}$ of $Q_3^0$ passing through $M_0$. Hence the desired spanning 2-path $P_{ux}+P_{vy}$ of $Q_4$ is $P_{ux}^0+P_{w_0{s}_0}^0+P_{vy}^1+\{s_0{s}_1,w_0{w}_1\}-\left\{s_1{w}_1\right\}$.

Subcase 2.1.2. $\{x,y\}\subseteq V\left(Q_3^1\right)$ (or $\{x,y\}\subseteq V\left(Q_3^0\right)$).

If $\chi \left(w_0\right)\ne \chi \left(u\right)$, then $\chi \left(v\right)=\chi \left(w_1\right)$. Because $M_1$ is a matching in $Q_3^1-\{v,w_1\}$ and $\chi \left(v\right)=\chi \left(w_1\right)\ne \chi \left(x\right)=\chi \left(y\right)$, by Lemma 8 there exists a spanning 2-path $P_{vy}+P_{w_{1}x}$ in $Q_3^1$ through $M_1$. Since $M_0$ is a matching in $Q_3^0-u$ and $\chi \left(u\right)\ne \chi \left(w_0\right)$, by Lemma 4, there exists a Hamiltonian path $P_{u{w}_0}^0$ in $Q_3^0$ through $M_0$. Hence the desired spanning path is $P_{u{w}_0}^0+P_{vy}+P_{w_{1}x}+\left\{w_0{w}_1\right\}$.

If $\chi \left(w_0\right)=\chi \left(u\right)$, then since $v\notin V\left(M_1\right)$ and $\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(y\right)$, by Lemma 7, there exists a spanning 2-path $P_{vy}^1+P_{x{s}_1}^1$ in $Q_3^1$ through $M_1$ with $p\left(s_1\right)=p\left(v\right)$. If $w_1\in V\left(P_{vy}^1\right)$, let $t_1$ be the neighbor of $w_1$ on $P_{vy}^1$. Because $\chi \left(u\right)=\chi \left(w_0\right)\ne \chi \left(s_0\right)=\chi \left(t_0\right)$ and $M_0$ is a matching in $Q_3^0-\{u,w_0\}$, by Lemma 8, there exists a spanning 2-path $P_{u{s}_0}^0+P_{t_0{w}_0}^0$ of $Q_3^0$ through $M_0$. The resulting spanning path is $P_{u{s}_0}^0+P_{t_0{w}_0}^0+P_{vy}^1+P_{s_{1}x}^1+\{w_0{w}_1,t_0{t}_1,s_0{s}_1\}-\left\{t_1{w}_1\right\}.$

Subcase 2.2. $v\in V\left(Q_3^0\right)$ and $p\left(u\right)\ne p\left(w_0\right)$.

Subcase 2.2.1. $x\in V\left(Q_3^0\right),y\in V\left(Q_3^1\right)$ (or $y\in V\left(Q_3^0\right),x\in V\left(Q_3^1\right)$).

Because $M_1$ is a matching in $Q_3^1-w_1$ and $\chi \left(y\right)\ne \chi \left(w_1\right)$, by Lemma 4, there exists a Hamiltonian path $P_{w_{1}y}^1$ in $Q_3^1$ passing through $M_1$. Because $M_0$ is a matching in $Q_3^0-\{u,v\}$ and $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(w_0\right)$, by Lemma 8, there exists a spanning $P_{ux}+P_{v{w}_0}$ in $Q_3^0$. The desired path is $P_{ux}^0+P_{v{w}_0}^0+P_{w_{1}y}^1+\left\{w_0{w}_1\right\}$.

Subcase 2.2.2. $\{x,y\}\subseteq V\left(Q_3^0\right)$.

Because $M_0$ is a matching in $Q_3^0-\{u,v\}$ and $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(y\right)$, by Lemma 8, there exists a spanning $P_{ux}^0+P_{vy}^0$. Let $s_0$ be a neighbor of $w_0$ in this path. Because $p\left(w_1\right)\ne p\left(s_1\right)$ and $M_1$ is a matching in $Q_3^1-w_1$, by Lemma 4, there exists a Hamiltonian path $P_{w_1{s}_1}^1$ through $M_1$. The desired path is $P_{ux}^0+P_{vy}^0+P_{w_1{s}_1}^1+\{w_0{w}_1,s_0{s}_1\}-\left\{w_0{s}_0\right\}$.

Subcase 2.2.3. $\{x,y\}\subseteq V\left(Q_3^1\right)$.

Because $\chi \left(w_1\right)\ne \chi \left(x\right)=\chi \left(y\right)$ and $M_1$ is a matching in $Q_3^1-w_1$, by Lemma 7, there exists a spanning 2-path $P_{x{w}_1}^1+P_{y{s}_1}^1$ in $Q_3^1$ with $\chi \left(s_1\right)=\chi \left(w_1\right)$. Since $M_0$ is a matching in $Q_3^0-\{u,v\}$, by Lemma 8, there exists a spanning $P_{u{w}_0}^0+P_{v{s}_0}^0$ in $Q_3^0$. The complete path is $P_{x{w}_1}^1+P_{y{s}_1}^1+P_{u{w}_0}^0+P_{v{s}_0}^0+\{w_0{w}_1,s_0{s}_1\}$.

Subcase 2.3. $v\in V\left(Q_3^0\right)$ and $\chi \left(u\right)=\chi \left(w_0\right)$.

We divide $Q_4$ into subcubes $Q_3^0$ and $Q_3^1$ by $E_3$ such that $|M\cap E_3|\ne 2$. Let $M\cap E_3=\left\{w_0^{\prime }{w}_1^{\prime }\right\}$, where $w_0^{\prime }\in V\left(Q_3^0\right)$. From the beginning of this Lemma, $\{w_0^{\prime },w_1^{\prime }\}\cap \{x,y\}=\varnothing$.

Subcase 2.3.1: $v\in V\left(Q_3^{1,3}\right)$

In this case, the vertices $u,v,x,y,w_0^{\prime },w_1^{\prime }$ correspond to the roles played by $u,v,x,y,w_0,w_1$ in Subcase 2.1. Given that $\{w_0^{\prime },w_1^{\prime }\}\cap \{x,y\}=\varnothing$ and, similarly, $\{w_0,w_1\}\cap \{x,y\}=\varnothing$, the configuration is equivalent to that considered in Subcase 2.1. Thus, the result directly follows from the conclusion established in the subcase.

Subcase 2.3.2: $v\in V\left(Q_3^{0,3}\right)$ and $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(w_0^{\prime }\right)$

Here, we interpret the vertices $u,v,x,y,w_0^{\prime },w_1^{\prime }$ in the same manner as in Subcase 2.2. The conditions $\{w_0^{\prime },w_1^{\prime }\}\cap \{x,y\}=\varnothing$ and $\{w_0,w_1\}\cap \{x,y\}=\varnothing$ ensure that the structure is consistent with that in Subcase 2.2. Hence, the corresponding conclusion applies to this case.

Subcase 2.3.3: $v\in V\left(Q_3^{0,3}\right)$ and $\chi \left(u\right)=\chi \left(v\right)=\chi \left(w_0^{\prime }\right)$

Since $u,v\in V\left(Q_3^{0,4}\right)\cap V\left(Q_3^{0,3}\right)$ and given that $\chi \left(u\right)=\chi \left(v\right)=\chi \left(w_0^{\prime }\right)$, $w_0\in V\left(Q_3^{0,4}\right)$ and $\chi \left(w_0\right)=\chi \left(u\right)$. Now consider $M\setminus \{w_0{w}_1,w_0^{\prime }{w}_1^{\prime }\}$. The remaining edges must be in $E_1\cup E_2$. Given that $|M\cap E_1|=2$, and $|M\cap E_2|=1$, there are two possible matchings for M.

We proceed by analyzing two cases depending on the placement of the vertices x and y. If either the edge from $M\cap E_1$ or from $M\cap E_2$ is not adjacent to x or y, assume without loss of generality that the edge in $M\cap E_1$ is not incident to $\{x,y\}$. In this case, $Q_4$ is decomposed along $E_1$, resulting in $u\in V\left(Q_3^{0,1}\right)$ and $v\in V\left(Q_3^{1,1}\right)$. The configuration of $u,v,x,y$ are now the same as that of Subcase 2.1, and we may apply the same reasoning. In the alternative situation where both edges in $M\cap E_1$ and $M\cap E_2$ are incident to $\{x,y\}$, note that $\chi \left(u\right)=\chi \left(v\right)\ne \chi \left(x\right)=\chi \left(y\right)$. Consequently, a spanning 2-path $P_{ux}+P_{vy}$ exists in $Q_4$ that containing M (see Figure 2). □

The following lemma extends the given kind of matching M to a Hamiltonian path $P_{uv}$ in hypercube $Q_5$, ensuring that vertex u is not included as an endpoint of any edge of M.

Lemma 11.

If $u,v\in Q_5$ and a matching M does not cover every two vertices at a distance of 3, in $Q_5-u$ such that $\chi \left(u\right)\ne \chi \left(v\right)$. Then there exists a Hamiltonian path in which $P_{uv}$ contains M.

Proof. 

Consider a matching M in $Q_5-u$ such that M does not cover every pair of vertices at a distance of three. Let $j\in \left[5\right]$ denote the edges with endpoints that differ in the j-th position, and choose a direction j such that $|M\cap E_j|$ is as small as possible. Since $|M\cap E_j|\le 1$, we will discuss all cases. By removing the set of edges $E_j$, we split $Q_5$ into the subcubes $Q_4^0$ and $Q_4^1$. By symmetry, we may assume that vertex $u\in Q_4^0$. Define $M_{ϵ}=M\cap E\left(Q_4^{ϵ}\right)$ for each $ϵ\in \{0,1\}$ and, by symmetry, assume $|M_0|\le |M_1|$. It is important to note that each vertex $x_{ϵ}\in V\left(Q_4^{ϵ}\right)$ has a unique neighbor, denoted $x_{1-ϵ}$, in $Q_4^{1-ϵ}$, where $ϵ\in \{0,1\}$.

Case 1. $M\cap E_j=\varnothing$.

Subcase. 1.1. $u\in V\left(Q_4^0\right)$ and $v\in V\left(Q_4^1\right)$ or ($v\in V\left(Q_4^0\right)$ and $u\in V\left(Q_4^1\right)$).

Because $M_1$ is a matching in $Q_4^1$, by Theorem 2, a Hamiltonian cycle $C_1$ exists in $Q_4^1$ containing the edges of $M_1$. Given that $v\in V\left(Q_4^1\right)$ and $u\in V\left(Q_4^0\right)$, there exists a neighbor $s_1$ of v such that $\chi \left(u\right)\ne \chi \left(s_0\right)$ and $v{s}_1\notin M$. We have $\chi \left(u\right)\ne \chi \left(s_0\right)$ and $u\notin V\left(M_0\right)$ with $\chi \left(v\right)\ne \chi \left(u\right)$ and $\chi \left(s_1\right)\ne \chi \left(s_0\right)$. By Lemma 4, there exists a Hamiltonian path $P_{u{s}_0}$ that includes the edges of $M_0$ in $Q_4^0$. Thus, the required Hamiltonian path is $P_{uv}:=P_{u{s}_0}+C_1+\left\{s_0{s}_1\right\}-\left\{v{s}_1\right\}$ in $Q_5$ containing M.

Subcase. 1.2. $u,v\in V\left(Q_4^0\right)$

We observe that the matching M in $Q_5-u$ satisfies $M\cap E_j=\varnothing$, to investigate the case when $v,u\in V\left(Q_4^0\right)$. Since $u\notin V\left(M_0\right)$ and $\chi \left(u\right)\ne \chi \left(v\right)$. We have, $M_0$ is a matching in $Q_4^0-u$. By Lemma 4, there is a Hamiltonian path $P_{uv}$ in $Q_4^0$ that contains the edges of $M_0$. Because $M_1$ is not perfect, there exists an edge $s_0{t}_0\in E\left(P_{uv}\right)\setminus M_0$ in $P_{uv}$, and $s_1\notin M_1$. Similarly, by Lemma 4, the edges of $M_1$ extend to a Hamiltonian path $P_{s_1{t}_1}$ in $Q_4^1-s_1$. Consequently, $P_{uv}+P_{s_1{t}_1}+\{s_0{s}_1,t_0{t}_1\}-\left\{s_0{t}_0\right\}$ is the desired Hamiltonian path in $Q_5$ containing M.

Case 2. If $|M\cap E_j|=1$.

Subcase. 2.1. $u\in V\left(Q_4^0\right)$ and $v\in V\left(Q_4^1\right)$ or ($v\in V\left(Q_4^0\right)$ and $u\in V\left(Q_4^1\right)$).

Since $u\in V\left(Q_4^0\right)$ and $v\in V\left(Q_4^1\right)$, let $M\cap E_j=w_0{w}_1$. If $\chi \left(w_0\right)\ne \chi \left(u\right)$, this implies $\chi \left(w_1\right)\ne \chi \left(v\right)$ such that $\chi \left(v\right)\ne \chi \left(u\right)$ and $\chi \left(w_1\right)\ne \chi \left(w_0\right)$. Given that $u\notin V\left(M_0\right)$ and $w_1\notin V\left(M_1\right)$, by Lemma 4, Hamiltonian paths $P_{u{w}_0}$ in $Q_4^0$ and $P_{w_{1}v}$ in $Q_4^1$ pass through $M_0$ and $M_1$, respectively. Thus, the edges of $P_{u{w}_0}+P_{w_{1}v}+\left\{w_0{w}_1\right\}$ construct the desired Hamiltonian path in $Q_5$ passing through M.

If $\chi \left(w_0\right)=\chi \left(u\right)$ and $\chi \left(w_1\right)=\chi \left(v\right)$, $M_1$ is a matching in $Q_4^1-w_1$ and $|M_0|\le |M_1|$. Choose $s_1$ and $t_1$ are chosen as distinct vertices in $Q_4^1$ such that $\chi \left(s_1\right)=\chi \left(t_1\right)\ne \chi \left(w_1\right)$. According to Lemma 6, there exists a spanning 2-path $P_{w_{1}v}+P_{s_1{t}_1}$ in $Q_4^1$ that includes the edges of $M_1$, Therefore, we have $\chi \left(u\right)=\chi \left(w_0\right)$ and $\chi \left(s_0\right)=\chi \left(t_0\right)$. Because $s_0\ne t_0$, obtain $d(u,s_0)=d(w_0,t_0)=1$ in the subcube $Q_4^0$. Now, $M_0$ is matching in $Q_4^0-\{u,w_0\}$, by Lemma 10, and there exists a spanning 2-path $P_{u{s}_0}+P_{w_0{t}_0}$ in $Q_4^0$ passing through $M_0$. Therefore, the desired Hamiltonian path in $Q_5$ is $P_{u{s}_0}+P_{w_0{t}_0}+P_{s_1{t}_1}+P_{w_{1}v}+\{s_0{s}_1,t_0{t}_1,w_0{w}_1\}$ containing M.

Subcase. 2.2. $u,v\in V\left(Q_4^0\right)$

Since $M\cap E_j=\left\{w_0{w}_1\right\}$ and $w_0\in V\left(Q_4^0\right)$, according to Lemma 4, there exists a Hamiltonian path $P_{uv}$ in $Q_4^0$ containing $M_0$. Let $r_0$ be $w_0$ neighbor of $P_{uv}$. We have $w_0{r}_0\notin M$, and $w_0{w}_1\in M$. We have $w_1\notin V\left(M_1\right)$ and $\chi \left(w_1\right)\ne \chi \left(r_1\right)$. Again by Lemma 4, there exists a Hamiltonian path $P_{w_1{r}_1}$ in $Q_4^1$ that contains the edges of $M_1$. Therefore, the required Hamiltonian path in $Q_5$ is $P_{uv}+P_{w_1{r}_1}+\{w_0{w}_1,r_0{r}_1\}-\left\{w_0{r}_0\right\}$.

Case 3. $|M\cap E_j|=2$.

Suppose $M\cap E_5=\{w_0{w}_1,z_0{z}_1\}$, in this case, we can split $Q_5$ into subcubes $Q_4^0$ and $Q_4^1$ such that vertices $u,v$ both are in same subcube; by symmetry, we can consider $u,v\in V\left(Q_4^1\right)$ such that $|M\cap E_j|=2$. In this study, we considered two cases.

If $\chi \left(w_{ϵ}\right)\ne \chi \left(z_{ϵ}\right)$ for $ϵ\in [0,1]$, by Lemma 4, there exists a Hamiltonian path $P_{w_0{z}_0}$ in $Q_4^0-w_0$ that passes through $M_0$. Let u and $w_1$ be the vertices of the subgraph $Q_4^1$. Assume that $M_1$ is matching in $Q_4^1-u$; Lemma 6 states that there is a spanning 2-path $P_{u{w}_1}+P_{v{z}_1}$ that passes through $M_1$, satisfying $\chi \left(w_1\right)=\chi \left(u\right)\ne \chi \left(v\right)$. Therefore, the required path in $Q_5$ is $P_{w_0{z}_0}+P_{u{w}_1}+P_{v{z}_1}+\{w_0{w}_1,z_0{z}_1\}$.

If $\chi \left(w_{ϵ}\right)=\chi \left(z_{ϵ}\right)$, since $u,v\in V\left(Q_4^1\right)$, then by Lemma 4, there exists a Hamiltonian path $P_{uv}$ in $Q_4^1$ passing through $M_1$. As vertices $w_1,z_1$ are on path $P_{uv}$. Suppose $s_1$ is a neighbor of $z_1$ and $t_1$ is a neighbor of $w_1$ on $P_{uv}$ such that $\chi \left(w_1\right)=\chi \left(z_1\right)$. Let $z_0$, $w_0$, $t_0$, $s_0$ be pairwise distinct vertices in $Q_4^0$. Suppose $\chi \left(w_0\right)=\chi \left(z_0\right)$ and $\chi \left(s_0\right)=\chi \left(t_0\right)$ such that $d(w_0,t_0)=d(z_0,s_0)=1$. Then by Lemma 10, there exists a spanning 2-path $P_{w_0{t}_0}+P_{z_0{s}_0}$ in $Q_4^0$ passing through $M_0$. Therefore, $P_{uv}+P_{w_0{t}_0}+P_{z_0{s}_0}+\{w_0{w}_1,t_0{t}_1,z_0{z}_1,s_0{s}_1\}-\{s_1{z}_1,t_1{w}_1\}$ is the Hamiltonian path in $Q_5$ passing through M. □

3. Matchings in at Most Six Directions

In this section, we discuss matchings in the hypercube $Q_n$ that do not cover every pair of vertices at a distance of three and consist of edges in at most six directions. These matchings are not perfect due to the specified constraints, and their size is less than $2^{n-2}$ edges in $Q_n$. The following theorems demonstrate that any matching satisfying these conditions can be extended to a Hamiltonian cycle in $Q_n$.

Theorem 12.

Every matching M in the hypercube $Q_6$ such that M does not cover every two vertices at a distance of three extends to a Hamiltonian cycle.

Proof. 

Suppose M is matching in the hypercube $Q_6$ with M matchings that do not cover every two vertices at a distance of three. Choose a direction $j\in \left[6\right]$, and split $Q_6$ into subcubes $Q_5^0$ and $Q_5^1$ such that $|M\cap E_j|$ is as small as possible. For $j=6$ such that $|M\cap E_6|\le 3$, divide $Q_6-E_6$ into subcubes $Q_5^{ϵ}$ and their respective matchings $M_{ϵ}$ where $ϵ\in \{0,1\}$, in these subcubes. Assume by symmetry $|M_0|\le |M_1|$. When $|M\cap E_j|=1$, we denote $M\cap E_j=\left\{u_0{u}_1\right\}$ and $M\cap E_j=2$, and also denote $M\cap E_j=\{u_0{u}_1,v_0{v}_1\}$ when $M\cap E_j=3$, then $M\cap E_j=\{u_0{u}_1,v_0{v}_1,w_0{w}_1\}$. Note that every vertex $x_{ϵ}\in V\left(Q_5^{ϵ}\right)$ has a distinct adjacent vertex, $x_{1-ϵ}$, within $Q_5^{1-ϵ}$, where $ϵ\in \{0,1\}$. The following four main cases assist M into the Hamiltonian cycle:

Case 1. $M\cap E_j=\varnothing$.

Because $|M_0|\le |M_1|$, assume $|M_1|\le 15$, it indicates that $M_1$ is not a perfect matching. Assume the size of $M_0\cup M_1$ satisfies $|M_0\cup M_1|\le 20$. Consequently, Theorem 2 extends $M_1$ to a Hamiltonian cycle $C_1$ in $Q_5^1$. For a vertex $v_1$ on $C_1$, there is a neighboring vertex $u_1$ on $C_1$ such that $u_1{v}_1\notin M_1$ and $v_0\notin M_0$. Given that $\chi \left(u_0\right)\ne \chi \left(v_0\right)$ such that $M_0$ is matching in $Q_5^0-v_0$. According to Lemma 11, there exists a Hamiltonian path $P_{u_0{v}_0}$ in $Q_5^0$ passing through $M_0$. Thus, $C_1+P_{u_0{v}_0}+\{u_0{u}_1,v_0{v}_1\}-u_1{v}_1$ constructs a Hamiltonian cycle in $Q_6$ containing M, as depicted in Figure 3a.

Case 2. $|M\cap E_6|=1$.

We assume that $M\cap E_6=\left\{u_0{u}_1\right\}$. Because $M_1$ is matching in $Q_5^1$, by Theorem 2, we can extend $M_1$ to a Hamiltonian cycle $C_1$. Let $v_1$ be a neighbor of $u_1$ in $Q_5^1$ such that $u_1{v}_1\notin M_1$, there exists $u_0{v}_0\notin M_0$ in $Q_5^0$. Consequently, $\chi \left(u_0\right)\ne \chi \left(v_0\right)$ and $\chi \left(u_1\right)\ne \chi \left(v_1\right)$, implying that $M_0$ is matching in $Q_5^0-u_0$. By applying Lemma 11, there exists a Hamiltonian path $P_{u_0{v}_0}$ containing $M_0$ in $Q_5^0$. Therefore, $P_{u_0{v}_0}+C_1+\{u_0{u}_1,v_0{v}_1\}-u_1{v}_1$ constitutes a Hamiltonian cycle containing M within hypercube $Q_6$, as illustrated in Figure 3b.

Case 3. $|M\cap E_6|=2$.

Since $M\cap E_6=\{u_1{u}_0,v_1{v}_0\}$, where $u_{ϵ}$ and $v_{ϵ}$ belong to $V\left(Q_5^{ϵ}\right)$ for $ϵ\in \{0,1\}$. If $u_1,v_1$ are adjacent on a cycle such that $\chi \left(u_1\right)\ne \chi \left(v_1\right)$, then the proof is similar to that of Case 2. Therefore, for further analysis, assume that $u_1,v_1$ are non-adjacent and $\chi \left(u_0\right)=\chi \left(v_0\right)$, which means $\chi \left(u_1\right)=\chi \left(v_1\right)$, two vertices with the same color match. Given that $|M_5^0|\le |M_5^1|$, it follows that $|M_5^1|\le 15$. Thus, we conclude that $\left|M\right|\le 20$, because ${\sum }_{i\in \left[5\right]}|M\cap E_i|=|M_0|+|M_1|\le 18$.

Select a position $k\in \left[5\right]$ to minimize $|M\cap E_k|$, such that $|M\cap E_k|\le 3$. Without loss of generality, we choose $k=5$. Split $Q_5^{ϵ}$ into their subcubes $Q_4^{ϵL}$ and $Q_4^{ϵR}$ at $k=5$ for $ϵ\in \{0,1\}$. Therefore, $Q_5^{ϵ}-E_k=Q_4^{ϵL}+Q_4^{ϵR}$. Now, define $M_{ϵ\mu }=M_{ϵ}\cap E\left(Q_4^{ϵ\mu }\right)$ for every $\mu \in \{L,R\}$, as shown in Figure 4. In $V\left(Q_4^{ϵL}\right)$, every vertex $a_{ϵL}$ is uniquely connected to a neighboring vertex $a_{ϵR}$ in $Q_4^{ϵR}$. Similarly, in $V\left(Q_4^{ϵR}\right)$, every vertex $b_{ϵR}$ is uniquely connected to a neighboring vertex $b_{ϵL}$ in $Q_4^{ϵL}$.

By symmetry, we may assume that $|M_0\cap E_5|\le |M_1\cap E_5|$. Moreover, it is clear that $|M_0\cap E_5|\le 1$, since the total number of edges satisfies $|M\cap E_5|=|M_1\cap E_5|+|M_0\cap E_5|\le 3$. Assuming vertex $u_0\in V\left(Q_5^0\right)$, without loss of generality, consider $u_0\in V\left(Q_4^{0L}\right)$, thus $u_1\in V\left(Q_4^{1L}\right)$. We then consider two distinct subcases.

Subcase 3.1. $v_1\in V\left(Q_4^{1R}\right)$ and $v_0\in V\left(Q_4^{0R}\right)$.

Subcase 3.1.1. $M_0\cap E_5=\varnothing$.

According to Theorem 2, a Hamiltonian cycle $C_1$ exists in $Q_5^1$ passing through $M_1$. There exists a vertex $x_1$ adjacent to $u_1$ on $C_1$ that belongs to $V\left(Q_4^{1L}\right)$, and $y_1$ is a neighbor of $v_1$ on $C_1$ in $V\left(Q_4^{1R}\right)$. Given $\{v_0{v}_1,u_0{u}_1\}\subseteq M$, we have $\{v_1{y}_1,u_1{x}_1\}\cap M=\varnothing$. Since $\chi \left(x_0\right)\ne \chi \left(u_0\right)$ and $M_{0L}$ is matching in $Q_4^{0L}-u_0$, by Lemma 4, there exists a Hamiltonian path $P_{u_0{x}_0}$ in $Q_4^{0L}$ containing $M_{0L}$. Similarly, there exists a Hamiltonian path $P_{v_0{y}_0}$ in $Q_4^{0R}$ containing $M_{0R}$. Consequently, combining $P_{v_0{y}_0}+P_{u_0{x}_0}+C_1+\{x_0{x}_1,u_0{u}_1,y_0{y}_1,v_0{v}_1\}-\{v_1{y}_1,u_1{x}_1\}$ constructs a Hamiltonian cycle in $Q_6$ containing M, as shown in Figure 5a.

Subcase 3.1.2. $|M_0\cap E_5|=1$.

Let $M_0\cap E_5=\left\{s_{0L}{s}_{0R}\right\}$, where $s_{0\mu }\in V\left(Q_4^{0\mu }\right)$ for $\mu \in \{L,R\}$. Given that $\chi \left(u_0\right)=\chi \left(v_0\right)$ and $\chi \left(s_{0L}\right)\ne \chi \left(s_{0R}\right)$, without loss of generality, assume $\chi \left(u_0\right)=\chi \left(v_0\right)=\chi \left(s_{0L}\right)\ne \chi \left(s_{0R}\right)$.

First, we establish the existence of a Hamiltonian cycle $C_1$ extending $M_1$, such that $v_1$ on $C_1$ has two consecutive vertices that both belong to $V\left(Q_4^{1R}\right)$. If $|M_1\cap E_5|=2$ such that $M_1\cap E_5=\{a_{1L}{a}_{1R},b_{1L}{b}_{1R}\}$. If $\chi \left(a_{1\mu }\right)=\chi \left(b_{1\mu }\right)$, then by Theorem 2, there exists a Hamiltonian cycle $C_1$, similar to Subcase 3.1.1. Suppose $\chi \left(a_{1\mu }\right)\ne \chi \left(b_{1\mu }\right)$ where $a_{1\mu }\in V\left(Q_4^{1\mu }\right)$ given that $|M_1\cap E_5|=1$ such that $M_1\cap E_5=\left\{a_{1L}{a}_{1R}\right\}$. Because $a_{1R}$ has four neighbors in $Q_4^{1R}$, we select $b_{1R}$, a neighbor of $a_{1R}$, such that $v_1\ne b_{1R}$. Consequently, it follows that $\chi \left(a_{1\mu }\right)\ne \chi \left(b_{1\mu }\right)$ for every $\mu \in \{L,R\}$.

In the situation described above, there are matchings $M_{1L}$ and $M_{1R}$ in $Q_4^{1L}-a_{1L}$ and $Q_4^{1R}-a_{1R}$, respectively. According to Lemma 4, there exist Hamiltonian paths $P_{b_{1L}{a}_{1L}}$ and $P_{b_{1R}{a}_{1R}}$ that are in $Q_4^{1L}$ and $Q_4^{1R}$ passing through $M_{1L}$ and $M_{1R}$, respectively. We assume that $C_1=P_{a_{1R}{b}_{1R}}+P_{a_{1L}{b}_{1L}}+\{b_{1L}{b}_{1R},a_{1L}{a}_{1R}\}$. Since $\{v_1{v}_0,a_{1L}{a}_{1R},b_{1L}{b}_{1R}\}\subseteq M$, this implies $y_1\notin \{a_{1R},b_{1R}\}$. Thus, both neighboring vertices of $y_1$ on $C_1$ belong to $V\left(Q_4^{1R}\right)$, and there is a Hamiltonian cycle $C_1$ in $Q_5^1$ containing $M_1$.

Choose a neighbor $x_1$ of $u_1$ on $C_1$ such that $x_1\in V\left(Q_4^{1L}\right)$, and a neighbor $y_1$ of $v_1$ on $C_1$ such that $y_0\ne s_{0R}$. By Lemma 4 there exists a Hamiltonian path $P_{v_0{y}_0}$ in $Q_4^{0R}$ passing through $M_{0R}$. Since $M_{0R}$ is a matching in $Q_4^{0R}-v_0$. We choose a neighbor $t_{0R}$ of $s_{0R}$ on $P_{v_0{y}_0}$ such that $t_{0L}\ne x_0$, since $s_{0R}\notin \{v_0,y_0\}$. The vertices $u_0$, $x_0$, $s_{0L}$, and $t_{0L}$ are all distinct vertices in $Q_4^{0L}$, such that $\chi \left(u_0\right)=\chi \left(s_{0L}\right)$, $\chi \left(x_0\right)=\chi \left(t_{0L}\right)$, and $d(u_0,x_0)=d(s_{0L},t_{0L})=1$. Since $M_{0L}$ is a matching in $Q_4^{0L}-\{u_0,s_{0L}\}$, and $|M_{0L}|\le |M_{0R}|$, by Lemma 10, there is a spanning 2-path $P_{u_0{x}_0}+P_{s_{0L}{t}_{0L}}$ in $Q_4^{0L}$ passing through $M_{0L}$. Consequently, $P_{u_0{x}_0}+C_1+P_{s_{0L}{t}_{0L}}+P_{v_0{y}_0}+\{x_0{x}_1,u_0{u}_1,v_0{v}_1,y_0{y}_1,s_{0L}{s}_{0R},t_{0L}{t}_{0R}\}-\{v_1{y}_1,u_1{x}_1,s_{0R}{t}_{0R}\}$ constructs a Hamiltonian cycle in $Q_6$ passing through M, as shown in Figure 5b.

Subcase 3.2. $v_1\in V\left(Q_4^{1L}\right)$ and $v_0\in V\left(Q_4^{0L}\right)$.

According to Theorem 2, a Hamiltonian cycle $C_1$ exists in $Q_5^1$ passing through $M_1$. Let $x_1$ be a neighbor of $u_1$ on $C_1$ such that $x_1\in V\left(Q_4^{1L}\right)$, and let $y_1$ be a neighbor of $v_1$ on $C_1$ such that $y_1\in V\left(Q_4^{1L}\right)$. Thus, $\chi \left(v_1\right)=\chi \left(u_1\right)$ and $\chi \left(y_1\right)=\chi \left(x_1\right)$. Moreover, we have $d(u_1,x_1)=d(v_1,y_1)=1$, and $y_1\ne x_1$. These properties also apply to the corresponding vertices $u_0,x_0,v_0,y_0$. By symmetry, we may assume $|M_{0L}|\le |M_{0R}|$, and using Lemma 10, there exists a spanning 2-path $P_{u_0{x}_0}+P_{v_0{y}_0}$ in $Q_4^{0L}$ that passes through $M_{0L}$. We now consider two subcases.

Subcase 3.2.1. $M_0\cap E_5=\varnothing$.

Consider vertex $s_{0L}$ on the vertices of path $P_{u_0{x}_0}$ or $P_{v_0{y}_0}$, where $s_{0\mu }\in V\left(Q_4^{0\mu }\right)$. Let $s_{0L}$ be in the vertices of path $P_{v_0{y}_0}$ and have a neighbor on $P_{v_0{y}_0}$. We select a neighbor $t_{0L}$ of $s_{0L}\in V\left(Q_4^{0L}\right)$ for every $\mu \in \{L,R\}$ such that $\chi \left(s_{0\mu }\right)\ne \chi \left(t_{0\mu }\right)$. Because $M_{0R}$ is a matching in $Q_4^{0R}-s_{0R}$, by Lemma 4, there exists a Hamiltonian path $P_{s_{0R}{t}_{0R}}$ in $Q_4^{0R}$ containing $M_{0R}$. Then, $P_{v_0{y}_0}+P_{u_0{x}_0}+P_{s_{0R}{t}_{0R}}+C_1+\{u_0{u}_1,v_0{v}_1,x_0{x}_1,y_0{y}_1,s_{0R}{s}_{0L},t_{0R}{t}_{0L}\}-\{v_1{y}_1,u_1{x}_1,s_{0L}{t}_{0L}\}$ is a Hamiltonian cycle in $Q_6$ passing through M, as shown in Figure 6.

Subcase 3.2.2. $|M_0\cap E_5|=1$.

Let $M_0\cap E_5=\left\{s_{0L}{s}_{0R}\right\}$, where $s_{0\mu }\in V\left(Q_4^{0\mu }\right)$. Given that the vertex $s_{0L}$ is either on $P_{u_0{x}_0}$ or $P_{v_0{y}_0}$, we may assume $s_{0L}\in P_{v_0{y}_0}$ and has a neighbor on $P_{v_0{y}_0}$. Select a neighbor $t_{0R}$ of $s_{0L}\in V\left(Q_4^{0L}\right)$, for every $\mu \in \{L,R\}$ and holds that $\chi \left(s_{0\mu }\right)\ne \chi \left(t_{0\mu }\right)$. Because $M_{0R}$ is a matching in $Q_4^{0R}-s_{0R}$, by Lemma 4, there is a Hamiltonian path $P_{s_{0R}{t}_{0R}}$ in $Q_4^{0R}$ containing $M_{0R}$. Thus, $C_1+P_{v_0{y}_0}+P_{u_0{x}_0}+P_{s_{0R}{t}_{0R}}+\{u_0{u}_1,v_0{v}_1,x_0{x}_1,y_0{y}_1,s_{0R}{s}_{0L},t_{0R}{t}_{0L}\}-\{v_1{y}_1,u_1{x}_1,s_{0L}{t}_{0L}\}$ constructs a Hamiltonian cycle in $Q_6$ containing M. (The presentation of subcase 3.2.1 and subcase 3.2.2 is similar to that depicted in Figure 6).

Case 4. $|M\cap E_6|=3$.

Let $M\cap E_6=\{u_0{u}_1,v_0{v}_1,w_0{w}_1\}$, where $u_{ϵ},v_{ϵ},w_{ϵ}\in V\left(Q_5^{ϵ}\right)$. Given that $|M\cap E_i|=3$ for $i\in \left[6\right]$, divide $Q_6$ at position $i=6$ and assume that $|M_5^0|\le |M_5^1|$, and it follows that $|M_5^1|\le 13$. Thus, $\left|M\right|\le 20$, since ${\sum }_{i\in \left[5\right]}|M\cap E_i|=|M_0|+|M_1|\le 17$. Hence, we distinguish this case in two ways. If two vertices from $\{u_{ϵ},v_{ϵ},w_{ϵ}\}$ are in one partite set while the remaining vertex is in the other partite set such that $\chi \left(u_{ϵ}\right)=\chi \left(v_{ϵ}\right)\ne \chi \left(w_{ϵ}\right)$. If all three vertices $u_{ϵ},v_{ϵ},w_{ϵ}$ are in the same partite set then, $\chi \left(u_{ϵ}\right)=\chi \left(v_{ϵ}\right)=\chi \left(w_{ϵ}\right)$.

Subcase 4.1. If two vertices are in one partite set.

Because $\chi \left(u_{ϵ}\right)=\chi \left(v_{ϵ}\right)\ne \chi \left(w_{ϵ}\right)$. Splitting $Q_5^{ϵ}$ into 4-cube $Q_4^{ϵL}$ and $Q_4^{ϵR}$ at position k such that $u_{ϵ}\in V\left(Q_4^{ϵL}\right)$ and $v_{ϵ}\in V\left(Q_4^{ϵR}\right)$. Without loss of generality, let $k=5$. Since $\chi \left(u_{ϵ}\right)=\chi \left(v_{ϵ}\right)\ne \chi \left(w_{ϵ}\right)$, by symmetry, we may suppose $w_{ϵ}\in V\left(Q_4^{ϵL}\right)$. Given that $|M_0\cap E_5|+|M_1\cap E_5|=|M\cap E_5|=3$, similarly, assume $|M_0\cap E_5|\le 1$. Let $M_{ϵ\mu }=M_{ϵ}\cap E\left(Q_4^{ϵ\mu }\right)$ for every $\mu \in \{R,L\}$.

Subcase 4.1.1. $M_0\cap E_5=\varnothing$.

Given that $\chi \left(u_1\right)\ne \chi \left(w_1\right)$ and $M_1$ is a matching in $Q_5^1-u_1$, by Lemma 11, there exists a Hamiltonian path $P_{u_1{w}_1}$ in $Q_5^1$ containing $M_1$. Since $v_1$ has only one adjacent vertex in $Q_4^{1L}$, we choose a neighbor $y_1$ of $v_1$ on $P_{u_1{w}_1}$ such that $y_1\in V\left(Q_4^{1R}\right)$. Given that $y_0\in V\left(Q_4^{0R}\right)$ and $\chi \left(u_0\right)=\chi \left(v_0\right)\ne \chi \left(w_0\right)=\chi \left(y_0\right)$, note that $M_{0L}$ is a matching in $Q_4^{0L}-u_0$; while $M_{0R}$ is matching in $Q_4^{0R}-v_0$, by Lemma 4, there exists a Hamiltonian paths $P_{u_0{w}_0}$ in $Q_4^{0L}$ and $P_{v_0{y}_0}$ in $Q_4^{0R}$, containing $M_{0L}$ and $M_{0R}$, respectively. Consequently, $P_{u_1{w}_1}+P_{u_0{w}_0}+P_{v_0{y}_0}+\{u_0{u}_1,w_0{w}_1,v_0{v}_1,y_0{y}_1\}-\left\{v_1{y}_1\right\}$ constitutes a Hamiltonian cycle in $Q_6$ that contains M, illustrated in Figure 7a.

Subcase 3.1.2. $|M_0\cap E_5|=1$ and $|M_1\cap E_5|=2$.

Let $M_0\cap E_5=\left\{s_{0L}{s}_{0R}\right\}$ and $M_1\cap E_5=\{a_{1L}{a}_{1R},b_{1L}{b}_{1R}\}$, where $s_{0\mu }\in V\left(Q_4^{0\mu }\right)$ and $a_{1\mu },b_{1\mu }\in V\left(Q_4^{1\mu }\right)$. If $\chi \left(a_{1L}\right)\ne \chi \left(b_{1L}\right)$ and $\chi \left(u_1\right)\ne \chi \left(w_1\right)$, without loss of generality, we may assume $\chi \left(u_1\right)=\chi \left(b_{1L}\right)$ and $\chi \left(w_1\right)=\chi \left(a_{1L}\right)$. Consequently, $\chi \left(v_1\right)=\chi \left(a_{1R}\right)\ne \chi \left(b_{1R}\right)$.

We claim that $Q_5^1$ contains a Hamiltonian cycle $C_1$ containing $M_1$, with both neighbors of $v_1$ on $C_1$ also belonging to $V\left(Q_4^{1R}\right)$. Given that $|M_1\cap E_5|=2$, it follows that $|M\cap E_5|=3$. Consequently, $|M\cap E_i|=3$ for every $i\in \left[5\right]$. Let $M_1\cap E_5=\{a_{1L}{a}_{1R},b_{1L}{b}_{1R}\}$, where $a_{1\mu },b_{1\mu }\in V\left(Q_4^{1\mu }\right)$. Therefore, $\chi \left(a_{1\mu }\right)\ne \chi \left(b_{1\mu }\right)$ for every $\mu \in \{L,R\}$.

If $|M_1\cap E_5|=1$, then denote $M_1\cap E_5=\left\{a_{1L}{a}_{1R}\right\}$, where $a_{1\mu }\in V\left(Q_4^{1\mu }\right)$. We choose a neighbor $b_{1R}$ of $a_{1R}$ in $Q_4^{1R}$ such that $b_{1R}\ne v_1$, since $a_{1R}$ has four neighbors in $Q_4^{1R}$. Thus, for every $\mu \in \{L,R\}$, $\chi \left(a_{1\mu }\right)\ne \chi \left(b_{1\mu }\right)$.

For the above two cases, since $M_{1\mu }$ is matching in $Q_4^{1\mu }-a_{1\mu }$, by Lemma 4, the Hamiltonian paths $P_{a_{1\mu }{b}_{1\mu }}$ in $Q_4^{1\mu }$ passing through $M_{1\mu }$ and $u_1{w}_1$ for each $\mu \in \{L,R\}$. Let $C_1=P_{a_{1L}{b}_{1L}}+P_{a_{1R}{b}_{1R}}+\{a_{1L}{a}_{1R},b_{1L}{b}_{1R}\}$. In the first case, since $\{u_0{u}_1,v_0{v}_1,a_{1L}{a}_{1R},b_{1L}{b}_{1R}\}\subseteq M$, we deduce that $v_1\notin \{a_{1R},b_{1R}\}$. In the second case, because $\{u_0{u}_1,v_0{v}_1,a_{1L}{a}_{1R}\}\subseteq M$, we find $v_1\ne a_{1R}$; consequently, $v_1\notin \{a_{1R},b_{1R}\}$. Since $u_1{w}_1$ is an edge in $Q_4^{1L}$, by Theorem 2, construct a Hamiltonian cycle $C_1$ in $Q_5^1$ that includes $M_1$ and $u_1{w}_1$, ensuring that the two neighbors of $v_1$ on $C_1$ are both vertices in $V\left(Q_4^{1R}\right)$.

Because $u_1{w}_1$ is an edge in $Q_4^{1L}$, a Hamiltonian cycle $C_1$ passes through this edge. Select a neighbor $y_1$ of $v_1$ on $C_1$ such that $y_0\ne s_{0R}$. Given that $M_{0R}$ is a matching in $Q_4^{0R}-v_0$, according to Lemma 4, there exists a Hamiltonian path $P_{v_0{y}_0}$ in $Q_4^{0R}$ containing $M_{0R}$. Since $s_{0R}\notin \{v_0,y_0\}$, we select a neighbor $t_{0R}$ of $s_{0R}$ on $P_{v_0{y}_0}$ such that $t_{0L}\ne w_0$. Now, $u_0$, $w_0$, $s_{0L}$, $t_{0L}$ are pairwise distinct vertices in $Q_4^{0L}$, such that $\chi \left(u_0\right)=\chi \left(s_{0L}\right)$ and $\chi \left(w_0\right)=\chi \left(t_{0L}\right)$ with $d(u_0,w_0)=d(s_{0L},t_{0L})=1$. Since $M_{0L}$ is a matching in $Q_4^{0L}-\{u_0,s_{0L}\}$ and $|M_{0L}|\le |M_{0R}|$, by Lemma 10, there exists a spanning 2-path $P_{u_0{w}_0}+P_{s_{0L}{t}_{0L}}$ in $Q_4^{0L}$ containing $M_{0L}$. Hence, $P_{u_0{w}_0}+C_1+P_{s_{0L}{t}_{0L}}+P_{v_0{y}_0}+\{w_0{w}_1,u_0{u}_1,v_0{v}_1,y_0{y}_1,s_{0L}{s}_{0R},t_{0L}{t}_{0R}\}-\{v_1{y}_1,u_1{w}_1,s_{0R}{t}_{0R}\}$ constitutes a Hamiltonian cycle in $Q_6$ containing M. (Note that the structure is analogous to subcase 3.1 in Subcase 3.1.2; readers can refer to the diagram (b) in Figure 5 by replacing $x_{ϵ}$ by $w_{ϵ}$ for more detail.)

Now, if $\chi \left(a_{1L}\right)=\chi \left(b_{1L}\right)$ and $\chi \left(u_1\right)\ne \chi \left(w_1\right)$, without loss of generality, we may assume $\chi \left(u_1\right)=\chi \left(b_{1L}\right)$ and $\chi \left(w_1\right)=\chi \left(a_{1L}\right)$. Consequently, $\chi \left(v_1\right)=\chi \left(a_{1R}\right)=\chi \left(b_{1R}\right)$.

We know that vertices $w_1,u_1\in V\left(Q_4^{1L}\right)$, since $\chi \left(u_1\right)\ne \chi \left(w_1\right)$. Consequently, $d(w_1,u_1)=1$. According to Lemma 11, there exists a Hamiltonian path $P_{u_1{w}_1}$ in $Q_5^1$ containing $M_1$, since $M_1$ is a matching in $Q_5^1-u_1$. We choose a neighbor $y_1$ of $v_1$ on $P_{u_1{w}_1}$ such that $y_1\in V\left(Q_4^{1R}\right)$ since $v_1$ has just one neighbor in $Q_4^{1R}$. By Lemma 4, there exists a Hamiltonian path $P_{v_0{y}_0}$ in $Q_4^{0R}$ containing $M_{0R}$, as $M_{0R}$ is a matching in $Q_4^{0R}-v_0$. Since $s_{0R}\notin \{v_0,y_0\}$, we choose a neighbor $t_{0R}$ of $s_{0R}$ on $P_{v_0{y}_0}$ such that $t_{0L}\ne w_0$. Now, $u_0$, $w_0$, $s_{0L}$, $t_{0L}$ are pairwise distinct vertices in $Q_4^{0L}$, with $\chi \left(u_0\right)=\chi \left(s_{0L}\right)$, $\chi \left(w_0\right)=\chi \left(t_{0L}\right)$, and $d(u_0,w_0)=d(s_{0L},t_{0L})=1$. Given that $M_{0L}$ is a matching in $Q_4^{0L}-\{u_0,s_{0L}\}$ and $|M_{0L}|\le |M_{0R}|$, by Lemma 10, there exists a spanning 2-path $P_{u_0{w}_0}+P_{s_{0L}{t}_{0L}}$ in $Q_4^{0L}$ containing $M_{0L}$. Hence, $P_{u_0{w}_0}+P_{u_1{w}_1}+P_{s_{0L}{t}_{0L}}+P_{v_0{y}_0}+\{w_0{w}_1,u_0{u}_1,v_0{v}_1,y_0{y}_1,s_{0L}{s}_{0R},t_{0L}{t}_{0R}\}-\{v_1{y}_1,s_{0R}{t}_{0R}\}$ constitutes a Hamiltonian cycle in $Q_6$ containing M, shown in Figure 7b.

Subcase 4.2. If $\chi \left(u_{ϵ}\right)=\chi \left(v_{ϵ}\right)=\chi \left(w_{ϵ}\right)$.

The vertices $u_{ϵ},v_{ϵ},w_{ϵ}$ are all in the same partite set. The 5-cube $Q_5^{ϵ}$ can be divided into two 4-cubes, $Q_4^{ϵL}$ and $Q_4^{ϵR}$, at position k where $u_{ϵ}\in V\left(Q_4^{ϵL}\right)$ and $v_{ϵ}\in V\left(Q_4^{ϵR}\right)$. Without loss of generality, suppose that $k=5$, and given that $\chi \left(u_{ϵ}\right)=\chi \left(v_{ϵ}\right)=\chi \left(w_{ϵ}\right)$, by symmetry, we may consider that $w_{ϵ}\in V\left(Q_4^{ϵL}\right)$. Considering $|M_0\cap E_5|+|M_1\cap E_5|=|M\cap E_5|=3$, by symmetry, we may assume $|M_0\cap E_5|\le 1$. Define $M_{ϵ\mu }=M_{ϵ}\cap E\left(Q_4^{ϵ\mu }\right)$ for each $\mu \in \{L,R\}$.

Subcase 4.2.1. $M_0\cap E_5=\varnothing$, since $\chi \left(u_1\right)=\chi \left(w_1\right)$.

Let $x_1$ be the unique vertex in $Q_4^{1L}$, such that $d(x_1,w_1)=4$. It follows that $d(x_1,u_1)=1$. Given that $M_1$ is a matching in $Q_5^1-u_1$ by Lemma 11, there is a Hamiltonian path $P_{u_1{x}_1}$ in $Q_5^1$ passing through $M_1$. We choose a neighbor $z_1$ of $w_1$ on $P_{u_1{x}_1}$ such that $z_1\in V\left(Q_4^{1L}\right)$, as $w_1$ has only one neighbor in $Q_4^{1L}$. Since $d(x_1,w_1)=4$, it follows that $z_1\ne x_1$. Suppose $M_{0L}$ is matching in $Q_4^{0L}-u_0$, satisfying $\chi \left(w_0\right)=\chi \left(x_0\right)\ne \chi \left(u_0\right)$ such that $|M_{0R}|\le |M_{0L}|$. By Lemma 6, there exists a spanning 2-path $P_{u_0{z}_0}+P_{x_0{w}_0}$ in $Q_4^{0L}$ that containing $M_{0L}$. We choose a neighbor $y_1$ of $v_1$ on $P_{u_1{x}_1}$ such that $y_1\in V\left(Q_4^{1R}\right)$ and $y_0\in V\left(Q_4^{0R}\right)$, since $v_1$ has only one neighbor in $Q_4^{1R}$. By Lemma 4, there exists a Hamiltonian path $P_{v_0{y}_0}$ in $Q_4^{0R}$ containing $M_{0R}$, since $M_{0R}$ is a matching in $Q_4^{0R}-v_0$. Thus, $P_{u_1{x}_1}+P_{u_0{z}_0}+P_{x_0{w}_0}+P_{v_0{y}_0}+\{u_0{u}_1,x_0{x}_1,w_0{w}_1,v_0{v}_1,z_0{z}_1,y_0{y}_1\}-\{v_1{y}_1,w_1{z}_1\}$ is a Hamiltonian cycle in $Q_6$ passing through M, as depicted in Figure 8a.

Subcase 4.2.2. $|M_0\cap E_5|=1$ and $|M_1\cap E_5|=2$.

Let $M_0\cap E_5=\left\{s_{0L}{s}_{0R}\right\}$ and $M_1\cap E_5=\{a_{1L}{a}_{1R},b_{1L}{b}_{1R}\}$, where $s_{0\mu }\in V\left(Q_4^{0\mu }\right)$ and $a_{1\mu },b_{1\mu }\in V\left(Q_4^{1\mu }\right)$. If $\chi \left(a_{1L}\right)\ne \chi \left(b_{1L}\right)$ and $\chi \left(u_1\right)=\chi \left(w_1\right)$, without loss of generality, we may assume $\chi \left(u_1\right)=\chi \left(b_{1L}\right)$ and $\chi \left(w_1\right)=\chi \left(a_{1L}\right)$. Consequently, $\chi \left(v_1\right)=\chi \left(a_{1R}\right)\ne \chi \left(b_{1R}\right)$. We claim that $M_1$ can be extended to a Hamiltonian cycle $C_1$ in $Q_5^1$. Since $|M_1\cap E_5|=2$, we deduce that $|M\cap E_5|=3$. This implies $|M\cap E_i|=3$ for every $i\in \left[5\right]$. Therefore, $\chi \left(a_{1\mu }\right)\ne \chi \left(b_{1\mu }\right)$ for each $\mu \in \{L,R\}$.

$|M_1\cap E_5|=1$, as stated in subcase 4.1. Let $M_1\cap E_5=\{a_{1L},a_{1R}\}$, where $a_{1\mu }\in V\left(Q_4^{1\mu }\right)$. Since $a_{1R}$ has four neighbors in $Q_4^{1R}$. Therefore, we may choose a neighbor $b_{1R}$ of $a_{1R}$ in $Q_4^{1R}$ such that $b_{1R}\ne v_1$ and $\chi \left(a_{1\mu }\right)\ne \chi \left(b_{1\mu }\right)$ for every $\mu \in \{L,R\}$.

By Lemma 4, there exist the Hamiltonian paths $P_{a_{1\mu }{b}_{1\mu }}$ in $Q_4^{1\mu }$ passing through $M_{1\mu }$ for every $\mu \in \{L,R\}$ for these two cases, since $M_{1\mu }$ is a matching in $Q_4^{1\mu }-a_{1\mu }$. Hence, $C_1=P_{a_{1L}{b}_{1L}}+P_{a_{1R}{b}_{1R}}+\{a_{1L}{a}_{1R},b_{1L}{b}_{1R}\}$. In the first case, since $\{u_0{u}_1,v_0{v}_1,a_{1L}{a}_{1R},b_{1L}{b}_{1R}\}\subseteq M$, we get $v_1\notin \{a_{1R},b_{1R}\}$. In the second case, because $\{u_0{u}_1,v_0{v}_1,a_{1L}{a}_{1R}\}\subseteq M$, it follows that $v_1\ne a_{1R}$, and hence, $v_1\notin \{a_{1R},b_{1R}\}$. Therefore, $C_1$ is a Hamiltonian cycle in $Q_5^1$ passing through $M_1$ such that both neighbors of $v_1$ on $C_1$ are in $V\left(Q_4^{1R}\right)$.

Let $x_1$ be neighbor of $u_1$ such that vertices $u_1,x_1\in V\left(Q_4^{1L}\right)$ on $C_1$, and choose a neighbor $z_1$ of $w_1$ on $C_1$ such that $w_0\ne s_{0L}$. Suppose vertices $u_0$ and $z_0$ are in the graph $Q_4^{0L}$, where $\chi \left(u_0\right)=\chi \left(z_0\right)$. Suppose $M_{0L}$ is matching in $Q_4^{0L}-u_0$, satisfying $\chi \left(w_0\right)=\chi \left(x_0\right)\ne \chi \left(u_0\right)$; by Lemma 6, there is a spanning 2-path $P_{u_0{z}_0}+P_{x_0{w}_0}$ in $Q_4^{0L}$ passing through $M_{0L}$. Assume $s_{0L}$ is a vertex on $P_{w_0{x}_0}$ such that $\chi \left(s_{0L}\right)\ne \chi \left(s_{0R}\right)$. Let vertices $s_{0R}$ and $v_0$ be in the subgraph $Q_4^{0R}$, where $\chi \left(s_{0R}\right)=\chi \left(v_0\right)$. Now, $s_{0R}$, $v_0$, $t_{0R}$, $y_0$ are pairwise distinct vertices in $Q_4^{0R}$, with $\chi \left(s_{0R}\right)=\chi \left(v_0\right)$, $\chi \left(t_{0R}\right)=\chi \left(y_0\right)$, and $d(s_{0R},t_{0R})=d(v_0,y_0)=1$. Since $M_{0R}$ is a matching in $Q_4^{0R}-\{s_{0R},v_0\}$, and we have $|M_{0R}|\le |M_{0L}|$, by Lemma 10, there exists a spanning 2-path $P_{s_{0R}{t}_{0R}}+P_{v_0{y}_0}$ in $Q_4^{0R}$ passing through $M_{0R}$. Hence, $C_1+P_{s_{0R}{t}_{0R}}+P_{u_0{z}_0}+P_{v_0{y}_0}+P_{w_0{x}_0}+\{s_{0L}{s}_{0R},t_{0L}{t}_{0R},x_0{x}_1,y_0{y}_1,u_0{u}_1,v_0{v}_1,s_{0R}{s}_{0L},t_{0R}{t}_{0L}\}-\{v_1{y}_1,u_1{x}_1,w_1{z}_1,s_{0L}{t}_{0L}\}$ is a Hamiltonian cycle containing M in $Q_6$, as shown in Figure 8b.

Now, if $\chi \left(a_{1L}\right)=\chi \left(b_{1L}\right)$ and $\chi \left(u_1\right)=\chi \left(w_1\right)$, without loss of generality that we may assume $\chi \left(u_1\right)=\chi \left(b_{1L}\right)$ and $\chi \left(w_1\right)=\chi \left(a_{1L}\right)$. Consequently, $\chi \left(v_1\right)=\chi \left(a_{1R}\right)=\chi \left(b_{1R}\right)$.

Because $\chi \left(u_1\right)=\chi \left(w_1\right)$ such that $w_1,u_1\in V\left(Q_4^{1L}\right)$, according to Lemma 11, there exists a Hamiltonian path $P_{u_1{x}_1}$ in $Q_5^1$ containing $M_1$, since $M_1$ is a matching in $Q_5^1-u_1$. Let $x_1$ be a neighbor of $u_1$ such that vertices $u_1,x_1\in V\left(Q_4^{1L}\right)$ on $C_1$, and choose a neighbor $z_1$ of $w_1$ on $C_1$. Suppose vertices $u_0$ and $z_0$ are in the graph $Q_4^{0L}$, where $\chi \left(u_0\right)=\chi \left(z_0\right)$. Suppose $M_{0L}$ is matching in $Q_4^{0L}-u_0$, satisfying $\chi \left(w_0\right)=\chi \left(x_0\right)\ne \chi \left(u_0\right)$, then by Lemma 6, there exists a spanning 2-path $P_{u_0{z}_0}+P_{x_0{w}_0}$ in $Q_4^{0L}$ passing through $M_{0L}$. Suppose $s_{0L}$ is a vertex on $P_{w_0{x}_0}$ such that $\chi \left(s_{0L}\right)\ne \chi \left(s_{0R}\right)$. Let vertices $s_{0R}$ and $v_0$ be in the subgraph $Q_4^{0R}$, where $\chi \left(s_{0R}\right)=\chi \left(v_0\right)$. Now, $s_{0R}$, $v_0$, $t_{0R}$, $y_0$ are pairwise distinct vertices in $Q_4^{0R}$, such that $\chi \left(s_{0R}\right)=\chi \left(v_0\right)$, $\chi \left(t_{0R}\right)=\chi \left(y_0\right)$ and $d(s_{0R},t_{0R})=d(v_0,y_0)=1$. Because $M_{0R}$ is a matching in $Q_4^{0R}-\{s_{0R},v_0\}$ and we have $|M_{0R}|\le |M_{0L}|$, according to Lemma 10, there exists a spanning 2-path $P_{s_{0R}{t}_{0R}}+P_{v_0{y}_0}$ in $Q_4^{0R}$ containing $M_{0R}$. Hence, $P_{u_1{x}_1}+P_{s_{0R}{t}_{0R}}+P_{u_0{z}_0}+P_{v_0{y}_0}+P_{w_0{x}_0}+\{s_{0L}{s}_{0R},t_{0L}{t}_{0R},x_0{x}_1,y_0{y}_1,u_0{u}_1,v_0{v}_1,s_{0R}{s}_{0L},t_{0R}{t}_{0L}\}-\{v_1{y}_1,w_1{z}_1,s_{0L}{t}_{0L}\}$ is a Hamiltonian cycle in $Q_6$ containing M, as shown in Figure 9. □

Next, we aim to generalize this result to the n-dimensional hypercube, denoted by $Q_n$, specifying the matchings as described above. In particular, we will consider matchings that contain edges at most in six directions and that avoid covering all pairs of vertices at a distance of 3 and can extend to a Hamiltonian cycle in $Q_n$.

Theorem 13.

For $n\ge 3$, let M be a matching in the hypercube $Q_n$ such that $\left|\right\{i\in \left[n\right]:M\cap E_i\ne \varnothing \left\}\right|\le 6$ and does not cover every two vertices at a distance of 3. Then, there exists a Hamiltonian cycle in $Q_n$ containing M.

Proof. 

Given that in the hypercube $Q_n$, for $n\ge 3$, any matching consisting of edges at most in six directions and does not cover every pair of vertices at a distance of 3, we prove the theorem by induction on n.

First, we address the base case. For $n\in \{2,3,4,5\}$, where M is any matching in $Q_n$, the result directly follows Theorem 2. When $n=6$, then Theorem 12 proves the result. Now assume $n\ge 7$ and with the given restrictions on matching M; thus, M is not perfect. The matching M is at most in six directions and does not cover a pair of vertices at a distance of 3. Without loss of generality, we may assume that these directions are among $\{1,2,3,4,5,6\}$. Therefore, $M\subseteq E_1\cup E_2\cup E_3\cup E_4\cup E_5\cup E_6$.

Select a coordinate position $j\in \{1,\dots ,n\}$ such that one of the following holds:

• $|M\cap E_j|=2$ and the edges in $M\cap E_j$ are adjacent;
• $|M\cap E_j|=1$ with $M\cap E_j=\left\{u_0{u}_1\right\}$, and there exists a neighbor $v_{ϵ}\notin M$;
• $M\cap E_j=\varnothing$, and there exists an edge $u_{ϵ}{v}_{ϵ}\notin M$ for each $ϵ\in \{0,1\}$.

Split $Q_n$ into its two $(n-1)$-dimensional subcubes $Q_{n-1}^0$ and $Q_{n-1}^1$ along the j-th dimension. This is because the matching M is restricted and the size is at most $2^{n-2}$. Without loss of generality, $|M_0|\le 2^{(n-1)-2}$ and $|M_1|\le 2^{(n-1)-2}$ according to the given restrictions on matching M. By the induction hypothesis, there exists a Hamiltonian cycle $C_0$ in the subcube $Q_{n-1}^0$ which contains all the edges of $M_0$. Choose an edge $u_0{v}_0\in C_0$ such that the corresponding vertices $u_1,v_1\in V\left(Q_{n-1}^1\right)$ satisfy $u_1,v_1\notin V\left(M_1\right)$. Then, $M_1\cup \left\{u_1{v}_1\right\}$ is a matching in $Q_{n-1}^1$; by the induction hypothesis, there exists a Hamiltonian cycle $C_1$ in $Q_{n-1}^1$ that passes through all edges in $M_1\cup \left\{u_1{v}_1\right\}$. Thus, the required Hamiltonian cycle in $Q_n$ is $C:=C_0+\{u_0{u}_1,v_0{v}_1\}+C_1-\{u_0{v}_0,u_1{v}_1\}$, which passes through all edges of M in $Q_n$. This is a required proof. □

4. Conclusions

In this study, we examined the extension of matchings to Hamiltonian cycles in the hypercube $Q_n$ for $n\ge 3$. Building on previous work that resolved the case for perfect matchings, we focused on matchings with structural constraints, specifically for those that avoid covering pairs of vertices at a distance of three and consist of edges at most in six distinct directions. We proved that such matchings are always extendable to a Hamiltonian cycles. This result contributes a new partial solution to the open problem posed by Ruskey and Savage, advancing our understanding of the Hamiltonian extensions in hypercubes.