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Research article

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

  • Received: 17 April 2020 Accepted: 12 June 2020 Published: 23 June 2020
  • MSC : 34C25, 34K13, 34K25

  • Taking into accounting time-varying delays and anti-periodic environments, this paper deals with the global convergence dynamics on a class of anti-periodic high-order inertial Hopfield neural networks. First of all, with the help of Lyapunov function method, we prove that the global solutions are exponentially attractive to each other. Secondly, by using analytical techniques in uniform convergence functions sequence, the existence of the anti-periodic solution and its global exponential stability are established. Finally, two examples are arranged to illustrate the effectiveness and feasibility of the obtained results.

    Citation: Qian Cao, Xiaojin Guo. Anti-periodic dynamics on high-order inertial Hopfield neural networks involving time-varying delays[J]. AIMS Mathematics, 2020, 5(6): 5402-5421. doi: 10.3934/math.2020347

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  • Taking into accounting time-varying delays and anti-periodic environments, this paper deals with the global convergence dynamics on a class of anti-periodic high-order inertial Hopfield neural networks. First of all, with the help of Lyapunov function method, we prove that the global solutions are exponentially attractive to each other. Secondly, by using analytical techniques in uniform convergence functions sequence, the existence of the anti-periodic solution and its global exponential stability are established. Finally, two examples are arranged to illustrate the effectiveness and feasibility of the obtained results.


    Let G=(V,E) be a finite undirected simple graph and G=(V(G),E(G)) be a finite directed graph. If x,yV(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 xV(G) consists of the vertices adjacent to x.

    Let G1,G2 be two graphs, and let G1,G2 be two directed graphs. The Cartesian product G1G2 is the graph with vertex set V(G1)×V(G2), and two vertices (g1,h1) and (g2,h2) are adjacent in G1G2 if and only if g1=g2 and h1 is adjacent to h2 in G2, or h1=h2 and g1 is adjacent to g2 in G1. Similarly, the Cartesian product G1G2 is the directed graph with vertex set V(G1)×V(G2) and there is a directed arrow from (g1,h1) to (g2,h2) in G1G2 if and only if g1=g2 and h1h2E(G2), or h1=h2 and g1g2E(G1). Let Cn and Cn denote the cycle of length n and directed cycle of length n3, respectively.

    The concept of group distance magic labeling of undirected graphs was introduced by Froncek in [1]. Let Γ be an abelian group and let φ:VΓ be a mapping. We call w(v):=xN(v)φ(x) the weight of v with respect to φ for any vV. Then, φ is called a Γ-distance magic labeling if and only if φ 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 Γ-distance magic if there exists a Γ-distance magic labeling of G.

    Froncek [1] showed that CmCn is Zmn-distance magic if and only if 2mn for any integers m,n3, and he asked for the full characterization full characterization of abelian groups Γ such that CmCn is Γ-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 Γ such that CmCn admits a Γ-distance magic labeling for any m,n3 in [3]. The result is stated as follows.

    Theorem 1.1 ([3]). Let m,n>2 be integers and let Γ be an abelian group of order mn. Then, CmCn admits a Γ-distance magic labeling if and only if 2mn and l2d1exp(Γ), where l=lcm[m,n], d=gcd(m,n) and

    d1={d,2dd2,2d,4dd4,4d.

    This type of labeling was generalized to the directed graph G=(V(G),E(G)) in [4]. Suppose φ:V(G)Γ is a map. For any vV(G), we call w(v):=yN+(v)φ(y)yN(v)φ(y) the weight of v with respect to φ. Then, φ is called a Γ-distance magic labeling of G if and only if φ is bijective and there exists cΓ such that w(v)=c for any vV(G). The constant c is called the magic constant of the labeling φ. It was shown that CmCn is Zmn-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 CmCn. We give a necessary and sufficient condition for that CmCn 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 CmCn to be Γ-distance magic. We present and prove the main result in Section 3.

    In the remainder of this paper, we fix two integers m,n3 and let l=lcm[m,n] and d=gcd(m,n). Let Γ denote an abelian group of order mn. For convenience, we identity the vertices set of CmCn as the set of the equivalent classes on Z×Z defined by

    (x1,y1)R(x2,y2)x1y1(modm),x2y2(modn).

    There is a directed edge from (i,j)R to (i,j)R if and only if either ii(modm) and jj1(modn), or jj(modm) and ii1(modm). Let Ds={(i,j)RV(CmCn):jis(modd)} for s=0,1,,d1. We have a partition of V(CmCn)=d1s=0Ds, and we call D0,D1,,Dd1 the diagonals of CmCn.

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

    Lemma 2.1. Let ni,ai be integers, i=1,2. The following system of congruence equations

    {xa1(modn1)xa2(modn2),

    is solvable if and only if gcd(n1,n2)a1a2.

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

    {t2k(modm)t2k(modn)

    has a solution t0. Then,

    gcd(t0,l)={dgcd(2,l),2dd,2d,4dd2,4d.

    Proof. If 2d, then k=d and

    {t02d(modm)t02d(modn).

    Without loss of generality, we may assume that 2n. So, gcd(t0d,md)=gcd(2,md)=gcd(2,l) and gcd(t0d,nd)=gcd(2,nd)=1. Hence,

    gcd(t0,l)=dgcd(t0d,mnd2)=dgcd(2,l).

    If d2(mod4), then k=d2. So, gcd(t0,l)=dgcd(t0d,mnd2)=d.

    If 4d, then k=d4 and gcd(t0,l)=d2gcd(2t0d,2mnd2)=d2. This finishes the proof.

    Suppose m=ri=1ptii and n=ri=1psii are the prime factorizations of m and n. Define f(m,n):=ri=1puii, where

    ui={max{ti,si},siti0,si=ti.

    The following lemma gives many information about the labeling of the diagonals and the magic constant for a group magic labeling of CmCn. 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 φ:V(CmCn)Γ is a Γ-distance magic labeling with magic constant c. Let xij=φ((i,j)R) and aij=xijxi+1,j+1 for any integer i,j. Let

    d={dgcd(2,l),2dd,2d,4dd2,4d.

    Then,

    (1) d is a period of the sequence {ai+k,j+k}k=0;

    (2) {xi+kd,j+kd}k=0 is an arithmetic sequence of common difference hij, whose order is equal to ld;

    (3) If (i,j)R and (i,j)R are in the same diagonal of CmCn, then hij=hi,j;

    (4) The set of the labeling on the diagonal which contained (i,j)R is a union of d cosets of the cyclic group generated by hij;

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

    (6) If 2l and 4d, then f(l2,d2)o(c)l2.

    Proof. (1) By hypothesis, we have

    w((i,j)R)=xi1,jxi,j+1+xi,j1xi+1,j=ai1,j+ai,j1=c,

    for any i,j. Replacing i,j by i+1,j1 respectively in the above equation, we obtain ai2,j+1+ai1,j=c, and thus ai1,j=ai+1,j2. It follows that aij=ai2k,j+2k for any integers i,j,k. By Lemma 2.2, there exist integers k and t0 such that

    {t02k(modm)t02k(modn),

    and gcd(t0,l)=d. We obtain aij=ai2k,j+2k=ai+t0,j+t0. Since both l and t0 are periods of {ai+k,j+k}k=0, it follows that d=gcd(t0,l) is also a period of {ai+k,j+k}k=0.

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

    xi+kd,j+kdxi+(k+1)d,j+(k+1)d=d1r=0(xi+kd+r,j+kd+rxi+kd+r+1,j+kd+r+1)=d1r=0ai+kd+r,j+kd+r=d1r=0ai+r,j+r=xijxi+d,j+d.

    Thus, {xi+kd,j+kd}ld1k=0 is an arithmetic sequence of common difference hij=d1r=0ai+r,j+r. Since φ is a bijection, xi+kd,j+kd=xij+khij=xij if and only if lkd. It follows that ldhij=0 and rhij0 for any positive integer r<ld. So, o(hij)=ld.

    (3) Assume that ii+t(modm) and jj+t(modn) for some integers t. By (1), the sum of any consecutive d terms in the sequence {ai+k,j+k}k=0 are equal. Combining this with the discussion in (2), we have

    hi,j=d1r=0ai+t+r,j+t+r=d1r=0ai+r,j+r=hij.

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

    {xi+t,j+t:0tl1}=d1r=0{xi+r+kd,j+r+kd:0k<ld}=d1r=0{xi+r,j+r+khij:0k<ld}=d1r=0(xi+r,j+r+<hij>).

    This proves (4).

    (5) Since 2l, we have d=d. Recall that ai1,j+1=ai12k,j+1+2k for any i,j,k. Setting k=d12, one has

    ai1,j+1=aid,j+d.

    Let t satisfy that

    {td(modm)td(modn).

    Then, d divides t, and we obtain that ai1,j+1=aid,j+d=ai+t,j+t=aij by (1). It follows that 2aij=ai1,j+1+aij=c. Therefore, we deduce that 2aij=2aij=c for any i,j,i,j. Since 2mn=|Γ|, one has aij=ai,j=a. Thus, the sequence {xi+k,j+k}l1k=0 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)=o(a)gcd(2,o(a))=o(a)=l.

    (6) We have d2(mod4) in this case. So,

    ai1,j+1=ai12d24,j+1+2d24=aid2,j+d2.

    Let t satisfy that

    {td2(modm)td2(modn).

    Since 2l and d2 is odd, at least one of m,n is even and t is odd. There exists rZ such that t=d2(2r+1). So, we have ai1,j+1=ai+t,j+t=ai+d2,j+d2 by (1). Hence,

    aij+ai+d2,j+d2=aij+ai1,j+1=w(i,j+1)=c.

    Hence,

    xijxi+d,j+d=d1k=0(xi+k,j+kxi+k+1,j+k+1)=d1k=0ai+k,j+k=d21k=0(ai+k,j+k+ai+k+d2,j+k+d2)=d2c.

    Combining with (2), we obtain l2c=ldd2c=ld(hij)=0, and thus o(c)l2. Suppose l2=ri=1ptii, d2=ri=1psii, and o(c)=ri=1puii are the prime factorizations of l2, d2, and o(c). Since o(d2c)=o(c)gcd(o(c),d2)=ld=ri=1ptisii, we obtain uimin{si,ui}=tisi. It implies that ui=ti if ti>si. Therefore, f(l2,d2)o(c).

    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 Γ such that |Γ||B|=2k. Then, there exist xi,yi,cΓ, i=0,1,,k1 such that xi+yi=c and

    Γ=(k1i=0(xi+B))(k1i=0(yi+B)).

    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 cH and f(l2,d)o(c)l2. Then we can arrange the elements in H as a sequence z0,z1,,zl1 such that

    zizi+1+zi+dzi+d+1=c

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

    Proof. Let C=<c> and B=<dc>=<d1c> be the cyclic group generated by c and dc, where d1=gcd(o(c),d). We first compute |B|=o(dc). Let l=2m0ki=1pmii and d=ki=1pnii be the prime factorization of l and d, where pi3, mi>ni for i{1,2,,t}, and mi=ni for i{t+1,t+2,,k}. Then, f(l2,d)=2m01ti=1pmii. Since f(l2,d)o(c)l2, we may assume that o(c)=2m01ki=1psii, where si=mi for i{1,2,,t} and simi for i>t. It follows that

    o(dc)=o(c)gcd(o(c),d)=2m01ki=1psiiki=1pmin{si,ni}i=2m01ki=1psimin{si,ni}i=2m01ti=1pminii=l2d.

    We obtain |H||C|=2dd1. By Lemma 3.1, there exists x0,x1,,x2dd11,bH such that H=2dd11i=0(xi+C), and xi+xi+dd1=b for each i calculated modulo 2dd1. Combining this with that C=d11j=0(jc+B), we have

    H=2dd11i=0(xi+C)=2dd11i=0d11j=0(xi+jc+B).

    We first claim that there exist integers t0,t1 such that such that t0+t1=dd1 and gcd(t0,d1)=gcd(t1,d1)=1.

    Let d1=ri=1peii be the prime factorization of d1. Suppose dd1yi(modpeii) where 0yipeii1. Let

    (ui,vi)={(1,yi1),piyi1(2,yi2),piyi1.

    By the Chinese Remainder Theorem, there exist t0,q such that t0ui(modpi) and qvi(modpi). Since t0+qyidd1(modpeii), we have d1dd1t0q, and there exists an integer k such that kd1=dd1t0q. Let t1=q+kd1. Then, dd1=t0+t1 and gcd(t0,pi)=gcd(t1,pi)=1.

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

    i=2qd+kdd1+s,0k<2d1,0s<dd1. (3.1)

    Then, by the division algorithm, we set

    zi:={xi+(qd+s+kt0)c,2k,0k<d1xi+(qd+s+(kd1)t0)c,2k,d1k<2d1xi+(qd+kt1)c,2k,0k<d1xi+(qd+(kd1)t1)c,2k,d1k<2d1. (3.2)

    Recall that the subscript of xi is calculated modulo 2dd1. We see that izi is well-defined and l is a period of {zi}iZ. For i given in the form (3.1), we have

    i+d={2qd+(k+d1)dd1+s,0k<d12(q+1)d+(kd1)dd1+s,d1k<2d1.

    Hence,

    zi+d={xi+d+(qd+kt1)c,2k,0k<d1xi+d+((q+1)d+(kd1)t1)c,2k,d1k<2d1xi+d+(qd+s+kt0)c,2k,0k<d1xi+d+((q+1)d+s+(kd1)t0)c,2k,d1k<2d1. (3.3)

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

    zi+zi+d=b+(2qd+kdd1+s)c=b+ic,

    for any i. Therefore,

    zizi+1+zi+dzi+d+1=c.

    Replacing zi by zi, we may assume that zizi+1+zi+dzi+d+1=c.

    It remains to show that H={zi:0i<l}.

    We claim that:

    (i) {zi+2jd:0j<l2d}=zi+B is a coset of B for i=0,1,,2d1;

    (ii) Xi={zi+j2dd1:0j<ld12d}=xi+C is a coset of C for i=0,1,,2dd11.

    Indeed, we have zi+2jdzi=jdc by (3.2) for any j. It follows that {zi+2jd:0j<l2d}={zi+jdc:0j<l2d}=zi+B. This proves (i).

    For j1{0,1,,ld12d1}, write j1=rd1+j, where 0r<l2d, 0j<d1. By (i) of the claim, we have

    {zi+j12dd1:0j1<ld12d}={zi+(rd1+j)2dd1:0r<l2d,0j<d1}=d11j=0{zi+j2dd1+2dr:0r<l2d}=d11j=0(zi+j2dd1+B).

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

    If 0i<dd1, by (3.2), we obtain that

    zi+j2dd1={xi+(i+2jt0)c,0j<d1+12xi+(i+(2jd1)t0)c,d1+12j<d1.

    Hence, {zi+j2dd1:0j<d1}={xi+(i+jt0)c:0j<d1} and

    Xi=d11j=0(zi+j2dd1+B)=d11j=0(xi+(i+jt0)c+B). (3.4)

    If dd1i<2dd1, then i+j2dd1=(2j+1)dd1+(idd1). By (3.2) again, we deduce that

    zi+j2dd1={xi+(2j+1)t1c,0j<d112xi+(2j+1d1)t1c,d112j<d1.

    In this case, we have {zi+j2dd1:0j<d1}={xi+jt1c:0j<d1} and

    Xi=d11j=0(zi+j2dd1+B)=d11j=0(xi+jt1c+B). (3.5)

    Note that both {i+jt0:0j<d11} and {jt1:0j<d1} are complete systems of residues modulo d1 since gcd(t0,d1)=gcd(t1,d1)=1. Recall that C=<c> and B=<d1c>. So,

    C=d11j=0(jc+B)=d11j=0((i+jt0)c+B)=d11j=0(jt1c+B). (3.6)

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

    Xi=d11j=0(xi+jc+B)=xi+C

    for i=0,1,,2dd11. Finally, {zi:0i<l}=2dd11i=0Xi=2dd11i=0(xi+C)=H. The proof is finished.

    Theorem 3.3. CmCn is Γ-distance magic if and only if l1exp(Γ), where

    l1={2ld,4d12f(l,d),2l,4d,2ldf(l,d),ld2(mod4)l,2l.

    Proof. Let

    d={dgcd(2,l),2dd,d2(mod4)d2,4d.

    It is straightforward to verify that

    f(l2,d2)={12f(l,d),2l,4d,2ldf(l,d),ld2(mod4).

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

    Let H be a subgroup of Γ with order l, and let V=V(CmCn). We have the following claim.

    Claim 1: Suppose there exists a labeling ψ:VH with the following properties.

    (i) The restriction ψ|Ds is a bijection for s=0,1,,d1;

    (ii) There exists cH such that the weight of any vertex with respect to ψ is c.

    Then, there exists a Γ-distance magic labeling of CmCn with magic constant c.

    Proof of the Claim 1: Since |Γ||H|=d, there exist b0,b1,,bd1Γ such that Γ=d1i=0(bi+H). We define φ:VΓ by

    φ((i,j)R):=ψ((i,j)R)+bs,(i,j)RDs.

    Then, φ is a bijection by (i), and

    φ((i1,j)R)φ((i,j+1)R)+φ((i,j1)R)φ((i+1,j)R)=ψ((i1,j)R)ψ((i,j+1)R)+ψ((i,j1)R)ψ((i+1,j)R)=c.

    Claim 2: We have Ds={(i,i+s)R:0i<l}. Let v=(i,i+s)RDs. Then,

    (1) If 1sd2, then N+(v)={(i1,i+s)R,(i,i+s1)R};

    (2) If s=0, then N+(v)={(i1,i)R,(i+k0,i+k0+d1)R};

    (3) If s=d1, then N+(v)={(ik01,ik01)R,(i,i+d2)R};

    where mk0 and k0d(modn).

    Proof of the Claim 2: (1) is obvious. If s=0, then N+(v)={(i1,i)R,(i,i1)R}. By the choice of k0, we see that

    {ii+k0(modm)i1i+k0+d1d(modn).

    So, (i,i1)R=(i+k0,i+k0+d1)R. The proof of (3) is similar.

    We divide the proof into three cases.

    Case 1: If 2l, then lexp(Γ). Let hΓ with order l, and let H be the cyclic subgroup of Γ generated by h. We label the (i,i+s)Rxi,i+s:=ih for (i,i+s)RDs. It is straightforward to check that this labeling satisfies the conditions in Claim 1.

    Case 2: 4d. Since 2ldexp(Γ), there exists an element hΓ with order 2ld. Let H be a subgroup of Γ such that |H|=l and hH. Suppose H=d21j=0(zj+<h>). We first label D2s by

    (i,i+2s)Rxi,i+2s:=(1)s(zr+qh),

    where 0s<d2, i+s=qd2+r,0r<d2.

    For (i,j)RD2s+1, we label (i,j)R by xij:=xi,j1, which is a translation of the labeling on D2s.

    It is straightforward to verify that

    xi,i+2sxi+1,i+2s+1={(1)s(zrzr+1),0r<d21(1)s(zd21z0h),r=d21.

    Now we compute the weight of v=(i,i+2s+1)D2s+1 with respect to the labeling. Suppose i+s=qd2+r,0r<d2.

    If s<d22, then (i1,i+2s+1)R,(i,i+2s+2)RD2s+2 and (i,i+2s)R,(i+1,i+2s+1)RD2s. Then,

    w(v)=xi1,i+2s+1xi,i+2s+2+xi,i+2sxi+1,i+2s+1={(1)s+1(zrzr+1)+(1)s(zrzr+1),0r<d21(1)s+1(zd21z0h)+(1)s(zd21z0h),r=d21=0.

    If s=d22, then N+(v)={(ik01,ik01)R,(i,i+d2)R} by (3) of Claim 2, where mk0 and k0d(modn). So, N(v)={(ik0,ik0)R,(i+1,i+d1)R}. In this case, we have

    xik01,ik01xik0,ik0={zrzr+1,0r<d21zd21z0h,r=d21.

    Hence,

    w(v)=xik01,ik01xik0,ik0+xi,i+d2xi+1,i+d1=0.

    Case 3: 2l and 4d. Note that l1=f(l2,d2)exp(Γ) and d2 is odd. Fix an element cΓ such that f(l2,d2)o(c)l2. There exists a subgroup H of Γ with order l containing c. By Lemma 3.2, we can order the elements in H as a sequences z0,z1,,zl1 such that

    zizi+1+zi+d2zi+d2+1=c, (3.7)

    where the subscript is calculated modulo l.

    It is easy to deduce that

    zizi+1+zi+(2k+1)d2zi+(2k+1)d2+1=c (3.8)

    from (3.7).

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

    Subcase 3.1: 2d and d=d. We label Ds by

    (i,i+s)R{zi+s2(d+1),2szi+(s+d2)(d+1),2s,0il1.

    It is clear that the labeling satisfies (i) of Claim 1. Note that the labeling on Ds for odd s is a translation of the labeling of Ds for even s. So, we only need to compute the weight of the vertices on D2s+1.

    Let v=(i,i+2s+1)RD2s+1. If sd32, then

    w(v)=zi+s(d2+1)zi+s(d2+1)+1+zi+(s+1)d2+szi+(s+1)d2+s+1=c

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

    If s=d32 and v=(i,i+d1)R, then N+(v)={(ik01,ik01)R,(i,i+d2)R} by (3) of Claim 2.

    Note that (i,i+d2)Rzi+d22(d2+1) and (ik01,ik01)Rzik01. Recalling that 2m and mk0, we obtain

    d22(d2+1)+k0+1d22(modm),

    and thus d22(d2+1)+k0+1 is an odd multiple of d2. By Eq (3.8), we have

    w(v)=zi+d22(d2+1)zi+d22(d2+1)+1+zik01zik0=c.

    Subcase 3.2: 2d and d=2d. We label Ds in the following way. Set

    (i,i+s)R{zi+s2(d+1),2szi+(s+d2)(d+1),2s,0il1.

    By the same discussion as in subcase 3.1, the weight of each vertex with respect to the labeling in Dj is also c, for j=1,2,,d2. We only write the details for j=0,d1.

    Let v0=(i,i)D0 and v1=(i,i+d1)Dd1. By Claim 2, N+(v)={(i1,i)R,(i+k0,i+k0+d1)R} and N+(v)={(ik01,ik01)R,(i,i+d2)R}, where mk0 and k0d(modn). By the construction of the labeling, we have

    w(v0)=zi1+(d+1)22zi+(d+1)22+zi+k0+(d1)(d+1)2zi+k0+1+(d1)(d+1)2,

    and

    w(v1)=zik01zik0+zi+(d1)(d+1)zi+1+(d1)(d+1).

    By a simple computation, we see that

    n0=(d+1)221k0(d1)(d+1)2d(modm),

    and

    n1=(d1)(d+1)+k0+1d2(modm).

    Since 2m and 2d, both n0 and n1 are odd multiples of d=d2. We get

    w(v0)=w(v1)=c

    from (3.8).

    For v=(i,i)D0, we have (i1,i)zi1+d+12(d+1) and (i,i1)=(ik01,ik0+d2)zik01+d12(d+1). Again, we observe that

    (i1+d+12(d+1))(ik01+d12(d+1))=k0+d+1

    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, CmCn is Γ-distance magic. The proof is complete.

    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 G1,G2 and the labeling of their Cartesian product G1G2. This is a sufficient condition.

    We say the a directed graph G is locally regular if |N+(x)|=|N(x)| for any xV(G).

    Theorem 4.1. Let Gi be directed graphs and let Γi be abelian groups, i=1,2. Suppose both G1 and G2 are locally regular and Gi is Γi-distance magic. Then, G1G2 is Γ1Γ2-distance magic.

    Proof. Let φi:V(Gi)Γi be the Γi-distance magic labeling with magic constant ci, i=1,2. For convenience, we identity Γ1 as the subgroup {(x,0):xΓ1} and identity Γ2 as the subgroup {(0,y):yΓ2} of Γ1Γ2. Define the labeling φ:V(G1G2)Γ1Γ2, as:

    φ((x,y))=φ1(x)+φ2(y).

    It is clear that φ is bijective. For any v=(x,y)V(G1G2), we have

    w(v)=wN+(v)φ(w)wN(v)φ(w)=[zN+(x)(φ1(z)+φ2(y))+zN+(y)(φ1(x)+φ2(z))][zN(x)(φ1(z)+φ2(y))+zN(y)(φ1(x)+φ2(z))]=[zN+(x)φ1(z)zN(x)φ1(z)]+[zN+(y)φ1(z)zN(y)φ1(z)]+(|N+(y)||N(y)|)φ1(x)+(|N+(x)||N(x)|)φ2(y)=c1+c2.

    Hence, G1G2 admits a Γ1Γ2-distance magic labeling.

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

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

    The authors declare there is no conflicts of interest.



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