# On the (Consecutively) Super Edge-Magic Deficiency of Subdivision of Double Stars

Authors: Vira Hari Krisnawati; Anak Agung Gede Ngurah; Noor Hidayat; Abdul Rouf Alghofari
Journal: Journal of Mathematics
 (2020)

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## Abstract

LetG be a finite, simple, and undirected graph with vertex set VG and edge set EG. A super edge-magic labeling of G is a bijection f:VG∪EG⟶1,2,…,VG+EG such that fVG=1,2,…,VG and fu+fuv+fv is a constant for every edge uv∈EG. The super edge-magic labeling f of G is called consecutively super edge-magic if G is a bipartite graph with partite sets A and B such that fA=1,2,…,A and fB=A+1,A+2,…,VG. A graph that admits (consecutively) super edge-magic labeling is called a (consecutively) super edge-magic graph. The super edge-magic deficiency of G, denoted by μsG, is either the minimum nonnegative integer n such that G∪nK1 is super edge-magic or +∞ if there exists no such n. The consecutively super edge-magic deficiency of a graph G is defined by a similar way. In this paper, we investigate the (consecutively) super edge-magic deficiency of subdivision of double stars. We show that, some of them have zero (consecutively) super edge-magic deficiency.


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## Body

## 1. Introduction

Besides having an interesting theoretical study, graph labelings can be applied in various fields of computer science such as coding theory, cryptography, circuit design, database management system, design of algorithms, and communication networks [1–5].All graphs considered here are finite, simple, and undirected graphs. We denote the vertex and edge sets of a graphG by VG and EG, respectively, where p=VG and q=EG. For most terminology and notation in graph theory used in this paper, we follow Chartrand et al. [6].Kotzig and Rosa [7] in 1970 introduced the concept of an edge-magic graph. A graph G is called edge-magic if there is a bijection f:VG∪EG⟶1,2,…,p+q such that fx+fxy+fy=k is a constant for every edge xy∈EG. In such a case, f is called an edge-magic labeling of G and k is called the magic constant of f. Meanwhile, Enomoto et al. [8] in 1998 introduced the terminology of a super edge-magic graph. An edge-magic graph G with an edge-magic labeling f is called super edge-magic (SEM) if fVG=1,2,…,p. In this case, f is called a SEM labeling of G. The following lemma, proved by Figueroa-Centeno et al. [9], provides sufficient and necessary conditions for a SEM graph.Lemma 1 (see [9]).
A graphG is SEM if and only if there exists a bijection f:VG⟶1,2,…,p such that the set of all edge-sums S=fx+fy:xy∈EG consists of q consecutive integers. In this case, f extends to be a SEM labeling of G with magic constant k=p+q+minS.Enomoto et al. [8] provided a sufficient condition for nonexistence of a SEM labeling of a graph.Lemma 2 (see [8]).
IfG is a SEM graph, then q≤2p−3.They also proposed the following conjecture.Conjecture 1 (see [8]).
Every tree is SEM.A SEM labeling can be considered as a particular case of ana,d-super edge antimagic labeling for d=0. An a,d-super edge antimagic labeling of a graph G is a bijection f:VG∪EG⟶1,2,…,p+q such that fx+fxy+fy:xy∈EG=a,a+d,a+2d,…,a+q−1d, where a>0 and d≥0, and fVG=1,2,…,p. The recent results of this labeling can be seen in [10, 11]. In [11], Liu et al. computed the bounds of a of super edge-antimagic labelings of subdivided caterpillars. They also presented a partial support of Conjecture 1 by proving that some classes of subdivided caterpillars are SEM.As same as SEM graph, Muntaner-Batle [12] in 2001 introduced the concept of a special SEM bipartite graph. In 2007, Oshima [13] called such a graph as a consecutively SEM graph. A bipartite graph G with partite sets A and B is called consecutively SEM if there exists a SEM labeling f of G with the property that fA=1,2,…,A and fB=A+1,A+2,…,p.Kotzig and Rosa [7] proved that, for every graph G there exists a nonnegative integer n such that G∪nK1 is an edge-magic graph. This fact encourages the concept of edge-magic deficiency of a graph. The edge-magic deficiency of a graph G, μG, is defined as the minimum nonnegative integer n such that G∪nK1 is an edge-magic graph. This concept motivated Figueroa-Centeno et al. [14] to introduce the concept of super edge-magic deficiency of a graph. The super edge-magic deficiency (SEMD) of a graph G, μsG, is defined as either the minimum nonnegative n such that G∪nK1 is a SEM graph or +∞ if there exists no such n. Moreover, Ichishima et al. [15] defined a similar notion for consecutively SEM labeling. The consecutively SEMD of a graph G, μcG, is defined to be either the smallest nonnegative integer n with the property that G∪nK1 is consecutively SEM or +∞ if there exists no such n. Hence, a (consecutively) SEM graph is a graph has zero (consecutively) SEMD. As an immediate consequence of these definitions, μG≤μsG≤μcG holds for every graph G.Motivated by Conjecture1, many researchers have investigated the SEM labeling of some families of trees. The SEM labeling of subdivision of stars K1,3 was studied by Ngurah et al. [16]. Hussain et al. [17] studied the SEM labeling of banana trees. Ahmad et al. [18] investigated the existence of SEM labeling of subdivision of banana trees. Javaid et al. [19] found SEM labeling of subdivision of stars K1,4 and w-trees. In [20], Ali et al. investigated the SEM labeling on w-trees. Next, Ali et al. [21] studied the SEM labeling of subdivision of stars K1,n for n≥5. However, Conjecture 1 is still open. Meanwhile, in [15], Ichishima et al. presented some results on consecutively SEMD of forets with two components, where its components are (non) isomorphic stars and union of paths and stars. In this paper, we study the (consecutively) SEMD of subdivision of double stars. We find the upper bound of (consecutively) SEMD of particular subdivision of double stars. We also prove that subdivision of double stars with a large order has zero (consecutively) SEMD.


## 2. The Results

A double starDSm,n is a tree obtained from two disjoint stars K1,m and K1,n, by joining the center vertices through an edge. For ri,tj≥1, 1≤i≤m, 1≤j≤n, and s≥0, a subdivision of a double starDTr1,r2,…,rm;s;t1,t2,…,tn is a graph obtained from a double star DSm,n by inserting ri−1 vertices to each edge of K1,m, s vertices to the edge which link the centre vertex of two stars, and tj−1 vertices to each edge of K1,n. Thus, DTr1,r2,…,rm;s;t1,t2,…,tn≅Gm;s;n is a graph of order 2+s+∑i=1mri+∑j=1ntj.Let vertex and edge sets ofGm;s;n be defined as follows:(1)VGm;s;n=x,xi,a:1≤i≤m,1≤a≤ri∪yl:1≤l≤s∪z,zj,b:1≤j≤n,1≤b≤tj,EGm;s;n=xxi,1,xi,axi,a+1:1≤i≤m,1≤a≤ri−1∪xy1∪ylyl+1:1≤l≤s−1∪ysz∪zzj,1,zj,bzj,b+1:1≤j≤n,1≤b≤tj−1.In this paper, we assume thats≥0 is odd. We denote the partite sets of Gm;s;n as follows:(2)A=xi,a:1≤i≤m,a≡1mod2∪z,zj,b:1≤j≤n,b≡0mod2,fors=0,xi,a:1≤i≤m,a≡1mod2∪yl:l≡1mod2∪zj,b:1≤j≤n,b≡1mod2,for odds≥1,B=x,xi,a:1≤i≤m,a≡0mod2∪zj,b:1≤j≤n,b≡1mod2,fors=0,x,xi,a:1≤i≤m,a≡0mod2∪yl:l≡0mod2∪z,zj,b:1≤j≤n,b≡0mod2,for odds≥1.First, we investigate the upper bound of (consecutively) SEMD ofGm;0;3.Theorem 1.
Letγ≥2 be an even integer. If r1=r2=t1=γ, t2=t3=γ−1, and ri=2i−3γ+1, 3≤i≤m, then μsGm;0;3=μcGm;0;3≤2m−3−1γ+1+1 for any m≥3.Proof.
LetG≅Gm;0;3∪2m−3−1γ+1+1K1 be a graph with the vertex set VG=VGm;0;3∪wh:1≤h≤2m−3−1γ+1+1 and edge set EG=EGm;0;3.
Define a labelingf:VG⟶1,2,…,p, where p=2m−3·3+3γ+2m−3·3−1, as follows:(3)fx=122m−2+5γ+2m−2,fz=122m−1+7γ+2m−1−2,for 1≤h≤2m−3−1γ+1+1,(4)fwh=2m−2+4γ+h+2m−2−2,fu=12γ−a+1,ifu=x1,aandais odd,122m−2+5γ−a+2m−2,ifu=x1,aanda≠γis even,2m−3·3+3γ+2m−3·3−1,ifu=x1,γ,12γ+a+1ifu=x2,aandais odd,122m−2+5γ+a+2m−2,ifu=x2,aandais even,122i−2+1γ+1−a,ifu=xi,a,3≤i≤mandais odd,122m−2+2i−2+5γ−a+2m−2+2i−2,ifu=xi,a,3≤i≤mandais even,122m−2+1γ+1+b,ifu=z1,bandbis odd,122m−1+5γ+b+2m−1−2,ifu=z1,bandbis even,122m−2+3γ−b+2m−2+1,ifu=z2,bandbis odd,122m−1+7γ−b+2m−1−2,ifu=z2,bandbis even,122m−2+3γ+b+2m−2+1,ifu=z3,bandbis odd,122m−1+7γ+b+2m−1−2,ifu=z3,bandbis even.
It can be checked that the set of all edge-sums isS=c,c+1,…,c+q−1, where c=2m−3+2γ+1 and q=2m−2+4γ+2m−2−2. Hence, by Lemma 1, f extends to a SEM labeling of G with magic constant k=2m−2·3+9γ+2m−2·3−1. Therefore, for any m≥3, μsGm;0;3≤2m−3−1γ+1+1.
Furthermore, letA′ and B′ be partite sets of G, where A′=A and B′=B∪wh:1≤h≤2m−3−1γ+1+1. Since fA′=1,2,…,d and fB′=d+1,d+2,…,p where d=2m−3+2γ+2m−3, G is a consecutively SEM graph. Hence, μcGm;0;3≤2m−3−1γ+1+1, for any m≥3.
As an illustration of the proof of Theorem1, see Figure 1.Figure 1
The (consecutively) SEM labeling ofDT4,4,5,10;0;4,3,3∪6K1.Theorem 2.
Letγ≥3 be an odd integer. If r1=r2=t1=t2=t3=γ and ri=2i−3γ, 3≤i≤m, then μsGm;0;3=μcGm;0;3≤2m−3−1γ+2 for any m≥3.Proof.
LetH≅Gm;0;3∪2m−3−1γ+2K1 be a graph with the vertex set VH=VGm;0;3∪wh:1≤h≤2m−3−1γ+2 and edge set EH=EGm;0;3.
Define a labelingf:VH⟶1,2,…,p, where p=2m−3·3+3γ+4, as follows:
The label of isolated vertices is(5)fw1=2m−3+2γ+4,fw2=122m−1+5γ+7,fwh=2m−2+4γ+h+1,for3≤h≤2m−3−1γ+2,and the label of the remaining vertices is as follows:(6)fx=122m−2+5γ+7,fz=122m−1+7γ+7,fu=12γ−a+2,ifu=x1,aandais odd,122m−2+5γ−a+7,ifu=x1,aanda≠γ−1is even,2m−3·3+3γ+4,ifu=x1,γ−1,12γ+a+2,ifu=x2,aandais odd,122m−2+5γ+a+7,ifu=x2,aandais even,122i−2+1γ−a+4,ifu=xi,a,3≤i≤mandais odd,122m−2+2i−2+5γ−a+7,ifu=xi,a,3≤i≤mandais even,122m−2+1γ+b+4,ifu=z1,bandbis odd,122m−1+5γ+b+7,ifu=z1,bandbis even,122m−2+3γ−b+6,ifu=z2,bandbis odd,122m−1+7γ−b+7,ifu=z2,bandbis even,122m−2+3γ+b+6,ifu=z3,bandbis odd,122m−1+7γ+b+7,ifu=z3,bandbis even.
It can be verified that the set of all edge-sums isS=c,c+1,…,c+q−1, where c=2m−3+2γ+7 and q=2m−2+4γ+1. Hence, by Lemma 1, f extends to a SEM labeling of H with magic constant k=2m−2·3+9γ+12. Therefore, for any m≥3, μsGm;0;3≤2m−3−1γ+2.
Moreover, letA′ and B′ be partite sets of H, where A′=A and B′=B∪wh:1≤h≤2m−3−1γ+2. Since fA′=1,2,…,d and fB′=d+1,d+2,…,p, where d=2m−3+2γ+3, H is a consecutively SEM graph. Hence, μcGm;0;3≤2m−3−1γ+2 for any m≥3.Based on Theorems1 and 2, we propose the following open problem.Open Problem 1.
Find an upper bound of (consecutively) SEMD ofGm;0;3 for the remaining cases. Furthermore, find a better upper bound of (consecutively) SEMD of Gm;0;3 for the same cases as in Theorems 1 and 2.
Next, we show that the graphsG2;s;n have zero (consecutively) SEMD.Theorem 3.
Letγ≥1 be an odd integer. If r1=r2=s=t1=t2=γ and tj=2j−3γ+1, 3≤j≤n, then μsG2;s;n=μcG2;s;n=0 for every n≥1 and odd s≥1.Proof.
Based on the sufficient conditions,G2;s;n has order(7)p=4γ+2,forn=1,2n−2+4γ+2n−2+1,forn≥2.
We consider the proof into two cases depending on the value ofn.(i)
Case 1:n=1.Define a labelingf:VG2;s;1⟶1,2,…,p as follows:(8)fx=125γ+5,fz=4γ+2,fu=12γ−a+2,ifu=x1,aandais odd,125γ−a+5,ifu=x1,aandais even,12γ+a+2,ifu=x2,aandais odd,125γ+a+5,ifu=x2,aandais even,123γ−l+4,ifu=ylandlis odd,127γ−l+5,ifu=ylandlis even,124γ−b+5,ifu=z1,bandbis odd,128γ−b+4,ifu=z1,bandbis even.It can be examined that the set of all edge sums isS=2γ+4,2γ+5,…,6γ+4. Hence, by Lemma 1, f extends to a SEM labeling of G2;s;1 with magic constant k=10γ+7. Therefore, for every odd s≥1, μsG2;s;1=0.(ii)
Case 2:n≥2.Define a labelingf:VG2;s;1⟶1,2,…,p as follows:(9)fx=122n−2+5γ+1,fz=2n−3+4γ+2n−3+2,fu=12γ−a+2,ifu=x1,aandais odd,122n−2+5γ+1−a,ifu=x1,aandais even,12γ+a+2,ifu=x2,aandais odd,122n−2+5γ+1+a,ifu=x2,aandais even,123γ−l+4,ifu=ylandlis odd,122n−2+7γ−l+2n−2+5,ifu=ylandlis even,124γ−b+5,ifu=z1,bandbis odd,122n−2+8γ−b+2n−2+4,ifu=z1,bandbis even,124γ+b+5,ifu=z2,bandbis odd,122n−2+8γ+b+2n−2+4,ifu=z2,bandbis even,122j−2+4γ−b+2j−2+5,ifu=zj,b,3≤j≤nandbis odd,122n−2+2j−2+8γ−b+2n−2+2j−2+4,ifu=zj,b,3≤j≤nandbis even.
We can examine that the set of all edge-sums isS=c,c+1,…,c+p−2, where c=2n−3+2γ+2n−3+4. Hence, by Lemma 1, f extends to a SEM labeling of G2;s;n with magic constant k=2n−3·5+10γ+2n−3·5+5. Therefore, for every n≥2 and odd s≥1, μsG2;s;n=0.
Furthermore, since for anyn≥1, fA=1,2,…,d and fB=d+1,d+2,…,p, where(10)d=2γ+2,ifn=1,125γ+5,ifn=2,2n−3+2γ+2n−3+2,ifn≥3,then G2;s;n is a consecutively SEM graph. Hence, μcG2;s;n=0 for every n≥1 and odd s≥1.Figure2 shows the vertex labelings of the graph G2;5;1 and G2;3;4, given in the proof of Theorem 3.Figure 2
The (consecutively) SEM labelings ofDT5,5;5;5 and DT3,3;3;3,3,4,8.
(a)(b)Theorem 4.
Letγ≥2 be an even integer. If r1=t1=γ, r2=s=t2=γ−1, t3=γ+2, and tj=2j−4γ+4, 4≤j≤n, then μsG2;s;n=μcG2;s;n=0 for any n≥2 and odd s≥1.Proof.
Based on the properties of the theorem, the order ofG2;s;n is(11)p=5γ−1,forn=2,2n−3+5γ+2n−1−3,forn≥3.
Thus, there are two following cases to prove the theorem depending on the value ofn.(i)
Case 1:n=2.Define a labelingf:VG2;s;n⟶1,2,…,p as follows:(12)fx=127γ,fz=5γ−1,fu=12a+1,ifu=x1,aandais odd,125γ+a,ifu=x1,aandais even,122γ+a+1,ifu=x2,aandais odd,127γ+a,ifu=x2,aandais even,122γ−l+1,ifu=ylandlis odd,127γ−l,ifu=ylandlis even,123γ+b+1,ifu=z1,bandbis odd,128γ+b−2,ifu=z1,bandbis even.125γ−b+1,ifu=z2,bandbis odd,1210γ−b−2,ifu=z2,bandbis even.It can be checked that the set of all edge-sums isS=1/25γ+4,1/25γ+6,…,6γ+4. Hence, by Lemma 1, f extends to a SEM labeling of G2;s;2 with magic constant k=1/225γ−2. Therefore, for any odd s≥1, μsG2;s;2=0.(ii)
Case 2:n≥3.Define a labelingf:VG2;s;2⟶1,2,…,p as follows:(13)fx=122n−3+7γ+2n−1−2,fz=122n−3+10γ+2n−1−4,fu=12a+1,ifu=x1,aandais odd,122n−3+5γ+a+2n−1−2,ifu=x1,aandais even,122γ+a+1,ifu=x2,aandais odd,122n−3+7γ+a+2n−1−2,ifu=x2,aandais even,122γ−l+1,ifu=ylandlis odd,122n−3+7γ−l+2n−1−2,ifu=ylandlis even,123γ+b+1,ifu=z1,bandbis odd,122n−3+8γ+b+2n−1−4,ifu=z1,bandbis even,125γ−b+1,ifu=z2,bandbis odd,122n−3+10γ−b+2n−1−4,ifu=z2,bandbis even,125γ+b+1,ifu=z3,bandbis odd,122n−3+10γ+b+2n−1−4,ifu=z3,bandbis even,122j−3+5γ−b+2j−1−1,ifu=zj,b,4≤j≤nandbis odd,122n−3+2j−3+10γ−b+2n−1+2j−1−4,ifu=zj,b,4≤j≤nandbis even.
We can check that the set of all edge-sums isS=c,c+1,…,c+p−2, where c=1/22n−3+5γ+2n−1+2. Hence, by Lemma 1, f extends to a SEM labeling of G2;s;n with magic constant k=2p+c−1. Therefore, for any n≥3 and odd s≥1, μsG2;s;n=0.
Moreover, since for anyn≥2, fA=1,2,…,d, and fB=d+1,d+2,…,p, where(14)d=125γ,ifn=2,3γ+1,ifn=3,122n−3+5γ+2n−1−2,ifn≥4,then G2;s;n is a consecutively SEM graph. Hence, μcG2;s;n=0 for any n≥2 and odd s≥1.The open problem related to Theorems3 and 4 is as follows.Open Problem 2.
Decide if the graphsG2;s;n have zero (consecutively) SEMD for the remaining cases.
Lastly, we show that subdivision of double starsGm;s;n has zero (consecutively) SEMD, as the following theorems.Theorem 5.
Letγ≥1 be an odd integer. If r1=r2=s=γ, ri=2i−2γ, t1=t2=s+∑i=3mri, and tj=2j−3t1+1, 3≤j≤n, then μsGm;s;n=μcGm;s;n=0 for arbitrary m≥3, n≥1, and odd s≥1.Proof.
(i) Case 1:n=1.
Define a labelingf:VGm;s;1⟶1,2,…,p, where p=2mγ+2 as follows:(15)fx=122m+1γ+5,fz=2mγ+2,fu=12γ−a+2,ifu=x1,aandais odd,122m+1γ−a+5,ifu=x1,aandais even,12γ+a+2,ifu=x2,aandais odd,122m+1γ+a+5,ifu=x2,aandais even,122i−1+1γ−a+4,ifu=xi,a,3≤i≤mandais odd,122m+2i−1+1γ−a+5,ifu=xi,a,3≤i≤mandais even,123γ−l+4,ifu=ylandlis odd,122m+3γ−l+5,ifu=ylandlis even,122mγ−b+5,ifu=z1,bandbis odd,2mγ−b+3,ifu=z1,bandbis even.(ii)
Case 2:n≥2.
Define a labelingf:VGm;s;n⟶1,2,…,p, where p=2n−22m−1−1+2mγ+2n−2+1, as follows: (16)fx=122n−22m−1−1+2m+1γ+2n−2+5,fz=122n−22m−1−1+2m+1γ+2n−2+4,fu=12γ−a+2,ifu=x1,aandais odd,122n−22m−1−1+2m+1γ−a+2n−2+5,ifu=x1,aandais even,12γ+a+2,ifu=x2,aandais odd,122n−22m−1−1+2m+1γ+a+2n−2+5,ifu=x2,aandais even,122i−1+1γ−a+4,ifu=xi,a,3≤i≤mandais odd,122n−22m−1−1+2m+2i−1+1γ−a+2n−2+5,ifu=xi,a,3≤i≤mandais even,123γ−l+4,ifu=ylandlis odd,122n−22m−1−1+2m+3γ−l+2n−2+5,ifu=ylandlis odd,122mγ−b+5,ifu=z1,bandbis odd,122n−22m−1−1+2m+1γ−b+2n−2+4,ifu=z1,bandbis even.122mγ+b+5,ifu=z2,bandbis odd,122n−22m−1−1+2m+1γ+b+2n−2+4,ifu=z2,bandbis even,and for 3≤j≤n,(17)fzj,b=122m2j−3+1−2j−2γ−b+2j−2+5,ifbis odd,122n−22m−1−1+2m2j−3+2−2j−2γ−b+2n−2+2j−2+4,ifbis even.
It can be verified that, forn≥1, the set of all edge-sums is S=c,c+1,…,c+p−2, where(18)c=2m−1γ+4,forn=1,2n−32m−1−1+2m−1γ+2n−3+4,forn≥2.
Hence, by Lemma1, f extends to a SEM labeling of Gm;s;n with magic constants k=2p+c−1. Therefore, for arbitrary m≥3, n≥1, and odd s≥1, μsGm;s;n=0.
Moreover, since for anyn≥1, fA=1,2,…,d and fB=d+1,d+2,…,p, where(19)d=2m−1γ+2,ifn=1,122m−1·3−1γ+5,ifn=2,2m−12n−3+1−2n−3γ+2n−3+2,ifn≥3,then Gm;s;n is a consecutively SEM graph. Hence, μcGm;s;n=0 for arbitrary m≥3, n≥1, and odd s≥1.Theorem 6.
Letγ≡2mod4. If r1=γ, r2=s=γ−1, ri=2i−3γ, 3≤i≤m, t1=1/22m−2+1γ, t2=t1−1, t3=t2+2, and tj=2j−4t3+2, 4≤j≤n, then μsGm;s;n=μcGm;s;n=0 for any m≥3, n≥2, and s≡1mod4.Proof.
(i) Case 1:n=2.
Define a labelingf:VGm;s;2⟶1,2,…,p, where p=2m−1+3γ−1, as follows:(20)fx=122m−1+5γ,fz=2m−1+3γ−1,fu=12a+1,ifu=x1,aandais odd,122m−1+3γ+a,ifu=x1,aandais even,122γ+a+1,ifu=x2,aandais odd,122m−1+5γ+a,ifu=x2,aandais even,122i−2+2γ−a+1,ifu=xi,a,3≤i≤mandais odd,122m−1+2i−2+5γ−a,ifu=xi,a,3≤i≤mandais even,122γ−l+1,ifu=ylandlis odd,122m−1+5γ−l,ifu=ylandlis even,122m−2+2γ+b+1,ifu=z1,bandbis odd,122m−2·3+5γ+b−2,ifu=z1,bandbis even,122m−1+3γ−b+1,ifu=z2,bandbis odd,122m+6γ−b−2,ifu=z2,bandbis even.(ii)
Case 2:n≥3.
Define a labelingf:VGm;s;n⟶1,2,…,p, where p=2n−42m−2+1+2m−1+3γ+2n−3·3−3, as follows:(21)fx=122n−42m−2+1+2m−1+5γ+2n−3·3−2,fz=122n−42m−2+1+2m+6γ+2n−3·3−4,fu=12a+1,ifu=x1,aandais odd,122n−42m−2+1+2m−1+3γ+a+2n−3·3−2,ifu=x1,aandais even,122γ+a+1,ifu=x2,aandais odd,122n−42m−2+1+2m−1+5γ+a+2n−3·3−2,ifu=x2,aandais even,122i−2+2γ−a+1,ifu=xi,a,3≤i≤m,andais odd,122n−42m−2+1+2m−1+2i−2+5γ−a+2n−3·3−2,ifu=xi,a,3≤i≤mandais even,122γ−l+1,ifu=ylandlis odd,122n−42m−2+1+2m−1+5γ−l+2n−3·3−2,ifu=ylandlis even,122m−2+2γ+b+1,ifu=z1,bandbis odd,122n−42m−2+1+2m−2⋅3+5γ+b+2n−3·3−4,ifu=z1,bandbis even,122m−1+3γ−b+1,ifu=z2,bandbis odd,122n−42m−2+1+2m+6γ−b+2n−3·3−4,ifu=z2,bandbis even,122m−1+3γ+b+1,ifu=z3,bandbis odd,122n−42m−2+1+2m+6γ+b+2n−3·3−4,ifu=z3,bandbis even,and for 4≤j≤n,(22)fzj,b=122m−22j−4+2+2j−4+3γ−b+2j−3·3−1,ifbis odd,122n−42m−2+1+2m−22j−4+4+2j−4+6γ−b+2n−3·3+2j−3·3−4,ifbis even.
We can examine that, forn≥2, the set of all edge-sums is S=c,c+1,…,c+p−2, where(23)c=122m−1+3γ+4,forn=2,122n−42m−2+1+2m−1+3γ+2n−3·3+2,forn≥3.
Hence, by Lemma1, f extends to a SEM labeling of Gm;s;n with magic constants k=2p+c−1. Therefore, for any m≥3, n≥2, and s≡1mod4, μsGm;s;n=0.
Furthermore, since for anyn≥2, fA=1,2,…,d, and fB=d+1,d+2,…,p, where(24)d=122m−1+3γ,ifn=2,142m−2·5+7γ+2,ifn=3,122m−22n−4+2+2n−4+3γ+2n−3·3−2,ifn≥4,then Gm;s;n is a consecutively SEM graph. Hence, μcGm;s;n=0 for any m≥3, n≥2, and s≡1mod4.
To clarify the proof of Theorem6, see Figure 3.Figure 3
The (consecutively) SEM labelings ofDT2,1,2,4,8;1;9,8 and DT2,1,2,4,8;1;9,8,10,12.
(a)(b)Theorem 7.
Letγ≡0mod4 be a positive integer. If r1=γ, r2=s=γ−1, ri=2i−3γ, 3≤i≤m, t1=t3=1/22m−2+1γ, t2=t1−2, and tj=2j−4t3+2, 4≤j≤n, then μsGm;s;n=μcGm;s;n=0 for every m≥3, n≥2, and s≡3mod4.Proof.
(i) Case 1:n=2.
Define a labelingf:VGm;s;2⟶1,2,…,p, where p=2m−1+3γ−2, as follows:(25)fx=122m−1+5γ−2,fz=2m−1+3γ−2,fu=12a+1,ifu=x1,aandais odd,122m−1+3γ+a−2,ifu=x1,aandais even,122γ+a+1,ifu=x2,aandais odd,122m−1+5γ+a−2,ifu=x2,aandais even,122i−2+2γ−a+1,ifu=xi,a,3≤i≤mandais odd,122m−1+2i−2+5γ−a−2,ifu=xi,a,3≤i≤mandais even,122γ−l+1,ifu=ylandlis odd,122m−1+5γ−l−2,ifu=ylandlis even,122m−2+2γ+b+1,ifu=z1,bandbis odd,122m−2·3+5γ+b−4,ifu=z1,bandbis even,122m−1+3γ−b−1,ifu=z2,bandbis odd,122m+6γ−b−4,ifu=z2,bandbis even.(ii)
Case 2:n≥3.
Define a labelingf:VGm;s;n⟶1,2,…,p, where p=2n−42m−2+1+2m−1+3γ+2n−2−4, as follows:(26)fx=122n−42m−2+1+2m−1+5γ+2n−2−4,fz=2n−52m−2+1+2m−1+3γ+2n−3−3,fu=12a+1,ifu=x1,aandais odd,122n−42m−2+1+2m−1+3γ+a+2n−2−4,ifu=x1,aandais even,122γ+a+1,ifu=x2,aandais odd,122n−42m−2+1+2m−1+5γ+a+2n−2−4,ifu=x2,aandais even,122i−2+2γ−a+1,ifu=xi,a,3≤i≤mandais odd,122n−42m−2+1+2m−1+2i−2+5γ−a+2n−2−4,ifu=xi,a,3≤i≤mandais even,122γ−l+1,ifu=ylandlis odd,122n−42m−2+1+2m−1+5γ−l+2n−2−4,ifu=ylandlis even,122m−2+2γ+b+1,ifu=z1,bandbis odd,122n−42m−2+1+2m−2⋅3+5γ+b+2n−2−6,ifu=z1,bandbis even,122m−1+3γ−b−1,ifu=z2,bandbis odd,122n−42m−2+1+2m+6γ−b+2n−2−6,ifu=z2,bandbis even,122m−1+3γ+b−1,ifu=z3,bandbis odd,122n−42m−2+1+2m+6γ+b+2n−2−6,ifu=z3,bandbis even,and for 4≤j≤n,(27)fzj,b=122m−22j−4+2+2j−4+3γ−b+2j−2−3,ifbis odd,122n−42m−2+1+2m−22j−4+4+2j−4+6γ−b+2n−2+2j−2−6,ifbis even.
It can be verified that, forn≥2, the set of all edge-sums is S=c,c+1,…,c+p−2, where(28)c=122m−1+3γ+2,forn=2,122n−42m−2+1+2m−1+3γ+2n−2,forn≥3.
Hence, by Lemma1, f extends to a SEM labeling of Gm;s;n with magic constants k=2p+c−1. Therefore, for every m≥3, n≥2, and s≡3mod4, μsGm;s;2=0.
Moreover, since for anyn≥2, fA=1,2,…,d and fB=d+1,d+2,…,p, where(29)d=122m−1+3γ−2,ifn=2,142m−2·5+7γ−4,ifn=3,122m−22n−4+2+2n−4+3γ+2n−2−4,ifn≥4,then Gm;s;n is a consecutively SEM graph. Hence, μcGm;s;n=0 for every m≥3, n≥2, and s≡3mod4.According to Theorems5–7, we present the following open problems.Open Problems 3.
Find (consecutively) SEMD of the graphsGm;s;n for the remaining cases.



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