Anti-periodic dynamics on high-order inertial Hopfield neural networks involving time-varying delays

1.   Introduction

Let $G = (V, E)$ be a finite undirected simple graph and $\overrightarrow{G} = (V(\overrightarrow{G}), E(\overrightarrow{G}))$ be a finite directed graph. If $x, y\in V(\overrightarrow{G})$ and there exists an arrow from $x$ into $y$, then $x$ is called a head of $y$ and $y$ is called a tail of $x$. Let $N^+(x)$ and $N^-(x)$ denote the set of the head and the tail of $x$, respectively. The neighborhood $N(x)$ of a vertex $x\in V(G)$ consists of the vertices adjacent to $x$.

Let $G_{1}, G_2$ be two graphs, and let $\overrightarrow{G_1}, \overrightarrow{G_2}$ be two directed graphs. The Cartesian product $G_1\square G_2$ is the graph with vertex set $V(G_1)\times V(G_2)$, and two vertices $(g_1, h_1)$ and $(g_2, h_2)$ are adjacent in $G_1\square G_2$ if and only if $g_1 = g_2$ and $h_1$ is adjacent to $h_2$ in $G_2$, or $h_1 = h_2$ and $g_1$ is adjacent to $g_2$ in $G_1$. Similarly, the Cartesian product $\overrightarrow{G_1}\square \overrightarrow{G_2}$ is the directed graph with vertex set $V(\overrightarrow{G_1})\times V(\overrightarrow{G_2})$ and there is a directed arrow from $(g_1, h_1)$ to $(g_2, h_2)$ in $\overrightarrow{G_1}\square \overrightarrow{G_2}$ if and only if $g_1 = g_2$ and $\overrightarrow{h_1h_2}\in E(\overrightarrow{G_2})$, or $h_1 = h_2$ and $\overrightarrow{g_1g_2}\in E(\overrightarrow{G_1})$. Let $C_n$ and $\overrightarrow{C_n}$ denote the cycle of length $n$ and directed cycle of length $n\geq 3$, respectively.

The concept of group distance magic labeling of undirected graphs was introduced by Froncek in ^[1]. Let $\Gamma$ be an abelian group and let $\varphi:V\longrightarrow \Gamma$ be a mapping. We call $w(v): = \sum_{x\in N(v)}\varphi(x)$ the weight of $v$ with respect to $\varphi$ for any $v\in V$. Then, $\varphi$ is called a $\Gamma$-distance magic labeling if and only if $\varphi$ is bijective and the weight is independent of the choice of $v$. The unique weight is called the magic constant of the magic labeling. A graph $G$ is called $\Gamma$-distance magic if there exists a $\Gamma$-distance magic labeling of $G$.

Froncek ^[1] showed that ${C_m\square C_n}$ is $\mathbb{Z}_{mn}$-distance magic if and only if $2\mid mn$ for any integers $m, n\geq 3$, and he asked for the full characterization full characterization of abelian groups $\Gamma$ such that $C_m\square C_n$ is $\Gamma$-distance magic. Cichacz et al. ^[2] answered the above question for the case that $m = n$. Recently, we generalized the idea and determined all abelian groups $\Gamma$ such that $C_m\square C_n$ admits a $\Gamma$-distance magic labeling for any $m, n\geq 3$ in ^[3]. The result is stated as follows.

Theorem 1.1 (^[3]). Let $m, n > 2$ be integers and let $\Gamma$ be an abelian group of order $mn$. Then, $C_m\square C_n$ admits a $\Gamma$-distance magic labeling if and only if $2\mid mn$ and $\frac{l}{2d_1}\mid\exp(\Gamma)$, where $l = \mathrm{lcm}[m, n]$, $d = \gcd(m, n)$ and

This type of labeling was generalized to the directed graph $\overrightarrow{G} = (V(\overrightarrow{G}), E(\overrightarrow{G}))$ in ^[4]. Suppose $\varphi:V(\overrightarrow{G})\longrightarrow\Gamma$ is a map. For any $v\in V(\overrightarrow{G})$, we call $w(v): = \sum_{y\in N^+(v)}\varphi(y)-\sum_{y\in N^-(v)}\varphi(y)$ the weight of $v$ with respect to $\varphi.$ Then, $\varphi$ is called a $\Gamma$-distance magic labeling of $\overrightarrow{G}$ if and only if $\varphi$ is bijective and there exists $c\in\Gamma$ such that $w(v) = c$ for any $v\in V(\overrightarrow{G})$. The constant $c$ is called the magic constant of the labeling $\varphi.$ It was shown that $\overrightarrow{C_m}\square \overrightarrow{C_n}$ is $\mathbb{Z}_{mn}$-distance magic in ^[4].

Many people study the group magic labeling of directed graph, which included the directed antiprism, lexicographic product of graphs, Cartesian product of graphs, and complete multipartite graphs, see ^[5,6,7]. In this paper, we focus on the group distance magic labeling on the Cartesian product of two directed cycles $\overrightarrow{C_m}\square \overrightarrow {C_n}$. We give a necessary and sufficient condition for that $\overrightarrow{C_m}\square \overrightarrow {C_n}$ admits a group distance magic labeling. This is another version of Theorem 1.1.

The present paper is organized as follows. In Section 2, we obtain some necessary conditions for a group magic labeling of $\overrightarrow{C_m}\square \overrightarrow{C_n}$ to be $\Gamma$-distance magic. We present and prove the main result in Section 3.

2.   Necessity

In the remainder of this paper, we fix two integers $m, n\geq 3$ and let $l = \mathrm{lcm}[m, n]$ and $d = \gcd(m, n)$. Let $\Gamma$ denote an abelian group of order $mn.$ For convenience, we identity the vertices set of $\overrightarrow{C_m}\square \overrightarrow {C_n}$ as the set of the equivalent classes on $\mathbb{Z}\times\mathbb{Z}$ defined by

There is a directed edge from $(i, j)_{\mathcal{R}}$ to $(i^{\prime}, j^{\prime})_{\mathcal{R}}$ if and only if either $i\equiv i'\pmod{m}$ and $j'-j \equiv 1 \pmod{n}$, or $j\equiv j'\pmod{m}$ and $i'-i \equiv 1 \pmod{m}$. Let $D^s = \{(i, j)_{\mathcal{R}}\in V(\overrightarrow{C_m}\square \overrightarrow {C_n}):j-i\equiv s\pmod{d}\}$ for $s = 0, 1, \ldots, d-1.$ We have a partition of $V(\overrightarrow{C_m}\square \overrightarrow {C_n}) = \cup_{s = 0}^{d-1}D^s$, and we call $D^0, D^1, \ldots, D^{d-1}$ the diagonals of $\overrightarrow{C_m}\square \overrightarrow {C_n}$.

We need the following basic fact, see ^[8].

Lemma 2.1. Let $n_i, a_i$ be integers, $i = 1, 2$. The following system of congruence equations

is solvable if and only if $\gcd(n_1, n_2)\mid a_1-a_2.$

Lemma 2.2. Let $k$ be the minimal positive integer such that the following congruence equation

has a solution $t_0$. Then,

Proof. If $2\nmid d$, then $k = d$ and

Without loss of generality, we may assume that $2\nmid n$. So, $\gcd(\frac{t_0}{d}, \frac{m}{d}) = \gcd(-2, \frac{m}{d}) = \gcd(2, l)$ and $\gcd(\frac{t_0}{d}, \frac{n}{d}) = \gcd(2, \frac{n}{d}) = 1$. Hence,

If $d\equiv 2\pmod{4}$, then $k = \frac{d}{2}$. So, $\gcd(t_0, l) = d\gcd(\frac{t_0}{d}, \frac{mn}{d^2}) = d$.

If $4\mid d$, then $k = \frac{d}{4}$ and $\gcd(t_0, l) = \frac{d}{2}\gcd(\frac{2t_0}{d}, \frac{2mn}{d^2}) = \frac{d}{2}$. This finishes the proof. □

Suppose $m = \prod_{i = 1}^rp_i^{t_i}$ and $n = \prod_{i = 1}^rp_i^{s_i}$ are the prime factorizations of $m$ and $n$. Define $f(m, n): = \prod_{i = 1}^rp_i^{u_i}$, where

The following lemma gives many information about the labeling of the diagonals and the magic constant for a group magic labeling of $\overrightarrow{C_m}\square \overrightarrow{C_n}$. Recall that the order of an element $x$ of a finite abelian group, which is denoted by $o(x)$, is equal to the minimal positive integer $k$ such that $kx = 0.$

Lemma 2.3. Suppose that $\varphi:V(\overrightarrow{C_m}\square \overrightarrow{C_n})\longrightarrow\Gamma$ is a $\Gamma$-distance magic labeling with magic constant $c$. Let $x_{ij} = \varphi((i, j)_{\mathcal{R}})$ and $a_{ij} = x_{ij}-x_{i+1, j+1}$ for any integer $i, j$. Let

Then,

(1) $d^{\prime}$ is a period of the sequence $\{a_{i+k, j+k}\}_{k = 0}^{\infty}$;

(2) $\{x_{i+kd^{\prime}, j+kd^{\prime}}\}_{k = 0}^{\infty}$ is an arithmetic sequence of common difference $h_{ij}$, whose order is equal to $\frac{l}{d^{\prime}}$;

(3) If $(i, j)_{\mathcal{R}}$ and $(i^{\prime}, j^{\prime})_{\mathcal{R}}$ are in the same diagonal of $\overrightarrow{C_m}\square \overrightarrow{C_n}$, then $h_{ij} = h_{i^{\prime}, j^{\prime}}$;

(4) The set of the labeling on the diagonal which contained $(i, j)_{\mathcal{R}}$ is a union of $d^{\prime}$ cosets of the cyclic group generated by $h_{ij}$;

(5) If $2\nmid l$, then $o(c) = l$;

(6) If $2\mid l$ and $4\nmid d$, then $f(\frac{l}{2}, \frac{d^{\prime}}{2})\mid o(c)\mid\frac{l}{2}$.

Proof. (1) By hypothesis, we have

for any $i, j.$ Replacing $i, j$ by $i+1, j-1$ respectively in the above equation, we obtain $a_{i-2, j+1}+a_{i-1, j} = c$, and thus $a_{i-1, j} = a_{i+1, j-2}$. It follows that $a_{ij} = a_{i-2k, j+2k}$ for any integers $i, j, k$. By Lemma 2.2, there exist integers $k$ and $t_0$ such that

and $\gcd(t_0, l) = d^{\prime}$. We obtain $a_{ij} = a_{i-2k, j+2k} = a_{i+t_0, j+t_0}$. Since both $l$ and $t_0$ are periods of $\{a_{i+k, j+k}\}_{k = 0}^{\infty}$, it follows that $d^{\prime} = \gcd(t_0, l)$ is also a period of $\{a_{i+k, j+k}\}_{k = 0}^{\infty}$.

(2) Now, for any $i, j, k$, we have

Thus, $\{x_{i+kd^{\prime}, j+kd^{\prime}}\}_{k = 0}^{\frac{l}{d^{\prime}}-1}$ is an arithmetic sequence of common difference $h_{ij} = -\sum_{r = 0}^{d^{\prime}-1}a_{i+r, j+r}$. Since $\varphi$ is a bijection, $x_{i+kd^{\prime}, j+kd^{\prime}} = x_{ij}+kh_{ij} = x_{ij}$ if and only if $l\mid kd^{\prime}$. It follows that $\frac{l}{d^{\prime}}h_{ij} = 0$ and $rh_{ij}\neq 0$ for any positive integer $r < \frac{l}{d^{\prime}}$. So, $o(h_{ij}) = \frac{l}{d^{\prime}}$.

(3) Assume that $i^{\prime}\equiv i+t\pmod{m}$ and $j^{\prime}\equiv j+t\pmod{n}$ for some integers $t$. By (1), the sum of any consecutive $d^{\prime}$ terms in the sequence $\{a_{i+k, j+k}\}_{k = 0}^{\infty}$ are equal. Combining this with the discussion in (2), we have

(4) By (1) and (2), we have

This proves (4).

(5) Since $2\nmid l$, we have $d^{\prime} = d$. Recall that $a_{i-1, j+1} = a_{i-1-2k, j+1+2k}$ for any $i, j, k$. Setting $k = \frac{d-1}{2}$, one has

Let $t$ satisfy that

Then, $d$ divides $t$, and we obtain that $a_{i-1, j+1} = a_{i-d, j+d} = a_{i+t, j+t} = a_{ij}$ by (1). It follows that $2a_{ij} = a_{i-1, j+1}+a_{ij} = c.$ Therefore, we deduce that $2a_{ij} = 2a_{i^{\prime}j^{\prime}} = c$ for any $i, j, i^{\prime}, j^{\prime}$. Since $2\nmid mn = |\Gamma|$, one has $a_{ij} = a_{i^{\prime}, j^{\prime}} = a$. Thus, the sequence $\{x_{i+k, j+k}\}_{k = 0}^{l-1}$ is an arithmetic sequence of common difference $-a$ with pairwise distinct terms. So, the order of $-a$ is exactly $l$, and $o(c) = o(2a) = \frac{o(a)}{\gcd(2, o(a))} = o(a) = l.$

(6) We have $d^{\prime}\equiv2\pmod{4}$ in this case. So,

Let $t$ satisfy that

Since $2\mid l$ and $\frac{d^{\prime}}{2}$ is odd, at least one of $m, n$ is even and $t$ is odd. There exists $r\in\mathbb{Z}$ such that $t = \frac{d^{\prime}}{2}(2r+1)$. So, we have $a_{i-1, j+1} = a_{i+t, j+t} = a_{i+\frac{d^{\prime}}{2}, j+\frac{d^{\prime}}{2}}$ by (1). Hence,

Hence,

Combining with (2), we obtain $\frac{l}{2}c = \frac{l}{d^{\prime}}\cdot \frac{d^{\prime}}{2}c = \frac{l}{d^{\prime}}(-h_{ij}) = 0$, and thus $o(c)\mid\frac{l}{2}$. Suppose $\frac{l}{2} = \prod_{i = 1}^rp_i^{t_i}$, $\frac{d^{\prime}}{2} = \prod_{i = 1}^rp_i^{s_i}$, and $o(c) = \prod_{i = 1}^rp_i^{u_i}$ are the prime factorizations of $\frac{l}{2}$, $\frac{d^{\prime}}{2}$, and $o(c)$. Since $o(\frac{d^{\prime}}{2}c) = \frac{o(c)}{\gcd(o(c), \frac{d^{\prime}}{2})} = \frac{l}{d^{\prime}} = \prod_{i = 1}^rp_i^{t_i-s_i}$, we obtain $u_i-\min\{s_i, u_i\} = t_i-s_i$. It implies that $u_i = t_i$ if $t_i > s_i.$ Therefore, $f(\frac{l}{2}, \frac{d^{\prime}}{2})\mid o(c)$. □

3.   Main result

In this section, we present and prove the main result of this paper. We begin with two useful lemmas.

Lemma 3.1 (^[3]). Let $B$ be a subgroup of an abelian group $\Gamma$ such that $\frac{|\Gamma|}{|B|} = 2k.$ Then, there exist $x_i, y_i, c\in \Gamma$, $i = 0, 1, \ldots, k-1$ such that $x_i+y_i = c$ and

We need the following lemma in the construction of magic labeling in our main result.

Lemma 3.2. Let $H$ be an abelian group. Suppose $|H| = l$ is even and $d$ is an odd divisor of $l$. Suppose $c\in H$ and $f(\frac{l}{2}, d)\mid o(c)\mid \frac{l}{2}$. Then we can arrange the elements in $H$ as a sequence $z_0, z_1, \ldots, z_{l-1}$ such that

for all $i$, where the subscripts are calculated modulo $l$.

Proof. Let $C = < c >$ and $B = < dc > = < d_1c >$ be the cyclic group generated by $c$ and $dc$, where $d_1 = \gcd(o(c), d)$. We first compute $|B| = o(dc)$. Let $l = 2^{m_0}\cdot\prod_{i = 1}^kp_i^{m_i}$ and $d = \prod_{i = 1}^kp_i^{n_i}$ be the prime factorization of $l$ and $d$, where $p_i\geq 3$, $m_i > n_i$ for $i\in\{1, 2, \ldots, t\}$, and $m_i = n_i$ for $i\in\{t+1, t+2, \ldots, k\}$. Then, $f(\frac{l}{2}, d) = 2^{m_0-1}\prod_{i = 1}^tp_i^{m_i}$. Since $f(\frac{l}{2}, d)\mid o(c)\mid \frac{l}{2}$, we may assume that $o(c) = 2^{m_0-1}\prod_{i = 1}^kp_i^{s_i}$, where $s_i = m_i$ for $i\in\{1, 2, \ldots, t\}$ and $s_i\leq m_i$ for $i > t$. It follows that

We obtain $\frac{|H|}{|C|} = \frac{2d}{d_1}$. By Lemma 3.1, there exists $x_0, x_1, \ldots, x_{\frac{2d}{d_1}-1}, b\in H$ such that $H = \cup_{i = 0}^{\frac{2d}{d_1}-1}(x_i+C)$, and $x_i+x_{i+\frac{d}{d_1}} = b$ for each $i$ calculated modulo $\frac{2d}{d_1}$. Combining this with that $C = \cup_{j = 0}^{d_1-1}(jc+B)$, we have

We first claim that there exist integers $t_0, t_1$ such that such that $t_0+t_1 = \frac{d}{d_1}$ and $\gcd(t_0, d_1) = \gcd(t_1, d_1) = 1.$

Let $d_1 = \prod_{i = 1}^rp_i^{e_i}$ be the prime factorization of $d_1$. Suppose $\frac{d}{d_1}\equiv y_i\pmod{p_i^{e_i}}$ where $0\leq y_i\leq p_i^{e_i}-1$. Let

By the Chinese Remainder Theorem, there exist $t_0, q$ such that $t_0\equiv u_i\pmod{p_i}$ and $q\equiv v_i\pmod{p_i}$. Since $t_0+q\equiv y_i\equiv \frac{d}{d_1}\pmod{p_i^{e_i}}$, we have $d_1\mid \frac{d}{d_1}-t_0-q$, and there exists an integer $k$ such that $kd_1 = \frac{d}{d_1}-t_0-q$. Let $t_1 = q+kd_1$. Then, $\frac{d}{d_1} = t_0+t_1$ and $\gcd(t_0, p_i) = \gcd(t_1, p_i) = 1$.

Now we define a sequence $z_i$ for any integer $i$ as follows. Write

Then, by the division algorithm, we set

Recall that the subscript of $x_i^{\prime}$ is calculated modulo $\frac{2d}{d_1}$. We see that $i\mapsto z_i$ is well-defined and $l$ is a period of $\{z_i\}_{i\in\mathbb{Z}}$. For $i$ given in the form (3.1), we have

Hence,

Combining (3.2) and (3.3), we obtain that

for any $i.$ Therefore,

Replacing $z_i$ by $-z_i$, we may assume that $z_i-z_{i+1}+z_{i+d}-z_{i+d+1} = c.$

It remains to show that $H = \{z_i:0\leq i < l\}$.

We claim that:

(i) $\{z_{i+2jd}:0\leq j < \frac{l}{2d}\} = z_i+B$ is a coset of $B$ for $i = 0, 1, \ldots, 2d-1$;

(ii) $X_i = \{z_{i+j\frac{2d}{d_1}}:0\leq j < \frac{ld_1}{2d}\} = x_i+C$ is a coset of $C$ for $i = 0, 1, \ldots, \frac{2d}{d_1}-1$.

Indeed, we have $z_{i+2jd}-z_i = jdc$ by (3.2) for any $j$. It follows that $\{z_{i+2jd}:0\leq j < \frac{l}{2d}\} = \{z_i+jdc:0\leq j < \frac{l}{2d}\} = z_i+B.$ This proves (i).

For $j_1\in\{0, 1, \ldots, \frac{ld_1}{2d}-1\}$, write $j_1 = rd_1+j,$ where $0\leq r < \frac{l}{2d}$, $0\leq j < d_1$. By (i) of the claim, we have

The proof of (ii) is divided into two cases.

If $0\leq i < \frac{d}{d_1}$, by (3.2), we obtain that

Hence, $\{z_{i+j\frac{2d}{d_1}}:0\leq j < d_1\} = \{x_i+(i+jt_0)c:0\leq j < d_1\}$ and

If $\frac{d}{d_1}\leq i < \frac{2d}{d_1}$, then $i+j\frac{2d}{d_1} = (2j+1)\frac{d}{d_1}+(i-\frac{d}{d_1})$. By (3.2) again, we deduce that

In this case, we have $\{z_{i+j\frac{2d}{d_1}}:0\leq j < d_1\} = \{x_i+jt_1c:0\leq j < d_1\}$ and

Note that both $\{i+jt_0:0\leq j < d_1-1\}$ and $\{jt_1:0\leq j < d_1\}$ are complete systems of residues modulo $d_1$ since $\gcd(t_0, d_1) = \gcd(t_1, d_1) = 1$. Recall that $C = < c >$ and $B = < d_1c >$. So,

Coming with (3.4), (3.5), and (3.6), we obtain that

for $i = 0, 1, \ldots, \frac{2d}{d_1}-1.$ Finally, $\{z_i:0\leq i < l\} = \cup_{i = 0}^{\frac{2d}{d_1}-1}X_i = \cup_{i = 0}^{\frac{2d}{d_1}-1}(x_i+C) = H.$ The proof is finished. □

Theorem 3.3. $\overrightarrow{C_m}\square \overrightarrow {C_n}$ is $\Gamma$-distance magic if and only if $l_1\mid \exp(\Gamma)$, where

Proof. Let

It is straightforward to verify that

So, the necessity follows from $(2), (5)$, and (6) of Lemma 2.3.

Let $H$ be a subgroup of $\Gamma$ with order $l$, and let $V = V(\overrightarrow{C_m}\square \overrightarrow {C_n})$. We have the following claim.

Claim 1: Suppose there exists a labeling $\psi:V\longrightarrow H$ with the following properties.

(i) The restriction $\psi|_{D^s}$ is a bijection for $s = 0, 1, \ldots, d-1$;

(ii) There exists $c\in H$ such that the weight of any vertex with respect to $\psi$ is $c$.

Then, there exists a $\Gamma$-distance magic labeling of $\overrightarrow{C_m}\square \overrightarrow {C_n}$ with magic constant $c.$

Proof of the Claim 1: Since $\frac{|\Gamma|}{|H|} = d$, there exist $b_0, b_1, \ldots, b_{d-1}\in \Gamma$ such that $\Gamma = \cup_{i = 0}^{d-1}(b_i+H)$. We define $\varphi:V\longrightarrow \Gamma$ by

Then, $\varphi$ is a bijection by (i), and

Claim 2: We have $D^s = \{(i, i+s)_{\mathcal{R}}:0\leq i < l\}$. Let $v = (i, i+s)_{\mathcal{R}}\in D^s$. Then,

(1) If $1\leq s\leq d-2$, then $N^+(v) = \{(i-1, i+s)_{\mathcal{R}}, (i, i+s-1)_{\mathcal{R}}\}$;

(2) If $s = 0$, then $N^+(v) = \{(i-1, i)_{\mathcal{R}}, (i+k_0, i+k_0+d-1)_{\mathcal{R}}\}$;

(3) If $s = d-1$, then $N^+(v) = \{(i-k_0-1, i-k_0-1)_{\mathcal{R}}, (i, i+d-2)_{\mathcal{R}}\}$;

where $m\mid k_0$ and $k_0\equiv -d\pmod{n}$.

Proof of the Claim 2: (1) is obvious. If $s = 0$, then $N^+(v) = \{(i-1, i)_{\mathcal{R}}, (i, i-1)_{\mathcal{R}}\}$. By the choice of $k_0$, we see that

So, $(i, i-1)_{\mathcal{R}} = (i+k_0, i+k_0+d-1)_{\mathcal{R}}$. The proof of (3) is similar.

We divide the proof into three cases.

Case 1: If $2\nmid l$, then $l\mid\exp(\Gamma)$. Let $h\in \Gamma$ with order $l$, and let $H$ be the cyclic subgroup of $\Gamma$ generated by $h$. We label the $(i, i+s)_{\mathcal{R}}\mapsto x_{i, i+s}: = ih$ for $(i, i+s)_{\mathcal{R}}\in D^s$. It is straightforward to check that this labeling satisfies the conditions in Claim 1.

Case 2: $4\mid d$. Since $\frac{2l}{d}\mid \exp(\Gamma)$, there exists an element $h\in\Gamma$ with order $\frac{2l}{d}$. Let $H$ be a subgroup of $\Gamma$ such that $|H| = l$ and $h\in H$. Suppose $H = \cup_{j = 0}^{\frac{d}{2}-1}(z_j+ < h > )$. We first label $D^{2s}$ by

where $0\leq s < \frac{d}{2}$, $i+s = q\frac{d}{2}+r, 0\leq r < \frac{d}{2}.$

For $(i, j)_{\mathcal{R}}\in D^{2s+1}$, we label $(i, j)_{\mathcal{R}}$ by $x_{ij}: = x_{i, j-1},$ which is a translation of the labeling on $D^{2s}$.

It is straightforward to verify that

Now we compute the weight of $v = (i, i+2s+1)\in D^{2s+1}$ with respect to the labeling. Suppose $i+s = q\frac{d}{2}+r, 0\leq r < \frac{d}{2}.$

If $s < \frac{d-2}{2}$, then $(i-1, i+2s+1)_{\mathcal{R}}, (i, i+2s+2)_{\mathcal{R}}\in D^{2s+2}$ and $(i, i+2s)_{\mathcal{R}}, (i+1, i+2s+1)_{\mathcal{R}}\in D^{2s}$. Then,

If $s = \frac{d-2}{2}$, then $N^+(v) = \{(i-k_0-1, i-k_0-1)_{\mathcal{R}}, (i, i+d-2)_{\mathcal{R}}\}$ by (3) of Claim 2, where $m\mid k_0$ and $k_0\equiv -d\pmod{n}$. So, $N^-(v) = \{(i-k_0, i-k_0)_{\mathcal{R}}, (i+1, i+d-1)_{\mathcal{R}}\}$. In this case, we have

Hence,

Case 3: $2\mid l$ and $4\nmid d.$ Note that $l_1 = f(\frac{l}{2}, \frac{d^{\prime}}{2})\mid \exp(\Gamma)$ and $\frac{d^{\prime}}{2}$ is odd. Fix an element $c\in \Gamma$ such that $f(\frac{l}{2}, \frac{d^{\prime}}{2})\mid o(c)\mid \frac{l}{2}$. There exists a subgroup $H$ of $\Gamma$ with order $l$ containing $c$. By Lemma 3.2, we can order the elements in $H$ as a sequences $z_0, z_1, \ldots, z_{l-1}$ such that

where the subscript is calculated modulo $l$.

It is easy to deduce that

from (3.7).

We divide the remains of the proof of Case 3 into two subcases. Without loss of generality we may assume that $2\mid m.$

Subcase 3.1: $2\mid d$ and $d = d^{\prime}$. We label $D^{s}$ by

It is clear that the labeling satisfies $(i)$ of Claim 1. Note that the labeling on $D^{s}$ for odd $s$ is a translation of the labeling of $D^s$ for even $s$. So, we only need to compute the weight of the vertices on $D^{2s+1}$.

Let $v = (i, i+2s+1)_{\mathcal{R}}\in D^{2s+1}.$ If $s\leq \frac{d-3}{2}$, then

by Eq (3.7) and the definition of the labeling.

If $s = \frac{d-3}{2}$ and $v = (i, i+d-1)_{\mathcal{R}}$, then $N^+(v) = \{(i-k_0-1, i-k_0-1)_{\mathcal{R}}, (i, i+d-2)_{\mathcal{R}}\}$ by (3) of Claim 2.

Note that $(i, i+d-2)_{\mathcal{R}}\mapsto z_{i+\frac{d-2}{2}(\frac{d}{2}+1)}$ and $(i-k_0-1, i-k_0-1)_{\mathcal{R}}\mapsto z_{i-k_0-1}$. Recalling that $2\mid m$ and $m\mid k_0$, we obtain

and thus $\frac{d-2}{2}(\frac{d}{2}+1)+k_0+1$ is an odd multiple of $\frac{d}{2}$. By Eq (3.8), we have

Subcase 3.2: $2\nmid d$ and $d^{\prime} = 2d$. We label $D^{s}$ in the following way. Set

By the same discussion as in subcase 3.1, the weight of each vertex with respect to the labeling in $D^{j}$ is also $c$, for $j = 1, 2, \ldots, d-2$. We only write the details for $j = 0, d-1.$

Let $v_0 = (i, i)\in D^{0}$ and $v_1 = (i, i+d-1)\in D^{d-1}$. By Claim 2, $N^+(v) = \{(i-1, i)_{\mathcal{R}}, (i+k_0, i+k_0+d-1)_{\mathcal{R}}\}$ and $N^+(v) = \{(i-k_0-1, i-k_0-1)_{\mathcal{R}}, (i, i+d-2)_{\mathcal{R}}\}$, where $m\mid k_0$ and $k_0\equiv -d\pmod{n}$. By the construction of the labeling, we have

and

By a simple computation, we see that

and

Since $2\mid m$ and $2\mid d$, both $n_0$ and $n_1$ are odd multiples of $d = \frac{d^{\prime}}{2}$. We get

from (3.8).

For $v = (i, i)\in D^{0}$, we have $(i-1, i)\mapsto z_{i-1+\frac{d+1}{2}(d+1)}$ and $(i, i-1) = (i-k_0-1, i-k_0+d-2)\mapsto z_{i-k_0-1+\frac{d-1}{2}(d+1)}$. Again, we observe that

is an odd multiple of $d$. Hence, the weight of $v$ is also $c$ by Claim 2.

Finally, we construct the labeling satisfied Claim 1 in Cases 1–3. So, $\overrightarrow{C_m}\square \overrightarrow {C_n}$ is $\Gamma$-distance magic. The proof is complete. □

4.   Final remark

It is natural to study the same problem on the Cartesian product of several cycles. But, it seems hard to obtain a full characterization for the existence of the group distance magic labeling product on Cartesian product of several cycles. However, we can prove an analogous result to Theorem 2.1 in ^[2], which describes a relation between the labeling of two graphs $G_1, G_2$ and the labeling of their Cartesian product $G_1\square G_2$. This is a sufficient condition.

We say the a directed graph $\overrightarrow{G}$ is locally regular if $|N^+(x)| = |N^-(x)|$ for any $x\in V(\overrightarrow{G})$.

Theorem 4.1. Let $\overrightarrow{G_i}$ be directed graphs and let $\Gamma_i$ be abelian groups, $i = 1, 2$. Suppose both $\overrightarrow{G_1}$ and $\overrightarrow{G_2}$ are locally regular and $G_i$ is $\Gamma_i$-distance magic. Then, $\overrightarrow{G_1}\square \overrightarrow{G_2}$ is $\Gamma_1\oplus \Gamma_2$-distance magic.

Proof. Let $\varphi_i:V(\overrightarrow{G_i})\longrightarrow \Gamma_i$ be the $\Gamma_i$-distance magic labeling with magic constant $c_i$, $i = 1, 2$. For convenience, we identity $\Gamma_1$ as the subgroup $\{(x, 0):x\in \Gamma_1\}$ and identity $\Gamma_2$ as the subgroup $\{(0, y):y\in \Gamma_2\}$ of $\Gamma_1\oplus \Gamma_2$. Define the labeling $\varphi:V(\overrightarrow{G_1}\square \overrightarrow{G_2})\longrightarrow\Gamma_1\oplus\Gamma_2$, as:

It is clear that $\varphi$ is bijective. For any $v = (x, y)\in V(\overrightarrow{G_1}\square \overrightarrow{G_2})$, we have

Hence, $\overrightarrow{G_1}\square \overrightarrow{G_2}$ admits a $\Gamma_1\oplus \Gamma_2$-distance magic labeling. □

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work was supported by the Guangxi Natural Science Foundation (2025GXNSFAA069810) and the National Natural Science Foundation of China (12161059).

Conflict of interest

The authors declare there is no conflicts of interest.