On the super edge-magic deficiency of some graphs

Abstract

A graph G is called super edge-magic if there exists a bijection $f:V(G)\cup E(G)⟶\{1,2,\cdots ,|V(G)|+|E(G)|\}$, where $f(V(G))=\{1,2,\cdots ,|V(G)|\}$, such that $f(u)+f(uv)+f(v)$ is a constant for every edge $uv\in E(G)$. Such a case, f is called a super edge magic labeling of G. A bipartite graph G with partite sets A and B is called consecutively super edge-magic if there exists a super edge-magic labeling f with the property that $f(A)=\{1,2,\cdots ,|A|\}$ and $f(B)=\{|A|+1,|A|+2,\cdots ,|V(G)|\}$. The super edge-magic deficiency of a graph G, denoted by ${\mu }_s(G)$, is either the minimum nonnegative integer n such that $G\cup n{K}_1$ is super edge-magic or +∞ if there exists no such n. The consecutively super edge-magic deficiency of a bipartite graph G, denoted by ${\mu }_c(G)$, is either the minimum nonnegative integer n such that $G\cup n{K}_1$ is consecutively super edge-magic or +∞ if there exists no such n. In this paper, we study the super edge-magic deficiency of some graphs. We investigate the (consecutively) super edge-magic deficiency of forests with two components. We also investigate the super edge-magic deficiency of a 2-regular graph $2C_3\cup C_n$ and join product of $K_{1,n}\cup P_m$ with an isolated vertex.

(Consecutively) super edge-magic graph; (Consecutively) super edge-magic deficiency; Forest with two components; 2-regular graph; Join product graph

1. Introduction

Let G be a finite and simple graph having vertex set $V(G)$ and edge set $E(G)$, where $p=|V(G)|$ and $q=|E(G)|$. A labeling of G is a bijection $f:V(G)\cup E(G)⟶\{1,2,\cdots ,p+q\}$. Weight of an edge xy under the labeling f is $w_f(xy)=f(x)+f(y)+f(xy)$. The labeling f is called an edge-magic labeling if $w_f(xy)=k$ for any $xy\in E(G)$. In such a case, G is called an edge-magic graph and the constant k is called the magic constant of the labeling f. An edge-magic labeling f of G with additional property that $f(V(G))=\{1,2,\cdots ,p\}$ is called a super edge-magic (SEM) labeling. Thus, a SEM graph is a graph that admits a SEM labeling. The concept of an edge-magic labeling was introduced by Kotzig and Rosa in 1970 [1]. Meanwhile, the terminology of a SEM labeling was introduced by Enomoto et al. in 1998 [2]. In 2001, Muntaner-Batle [3] introduced the concept of special SEM labeling of a bipartite graph. A special SEM labeling of a bipartite graph G with partite sets A and B is a SEM labeling f of G with the property that $f(A)=\{1,2,\cdots ,|A|\}$ and $f(B)=\{|A|+1,|A|+2,\cdots ,p\}$. In 2007, Oshima [4] called such a labeling consecutively SEM labeling.

The following lemma proved by Figueroa-Centeno et al. [5] provides sufficient and necessary conditions for a SEM graph.

Lemma 1.1

[5] A graph G is SEM if and only if there exists a bijection $f:V(G)\to \{1,2,\cdots ,p\}$ such that the set of all edge-sums $S=\{f(x)+f(y):xy\in E(G)\}$ consists of q consecutive integers. In this case, f extends to be a SEM labeling of G with magic constant $k=p+q+min(S)$.

The next lemma proved by Enomoto et al. [2] gives sufficient condition for non-existence of a SEM labeling of a graph.

Lemma 1.2

[2] If G is a SEM graph then $q\le 2p-3$.

Moreover, Kotzig and Rosa [1] also proved that for every graph G there exists a nonnegative integer n such that $G\cup n{K}_1$ is an edge-magic graph. This fact leads them to introduce the concept of edge-magic deficiency of a graph. The edge-magic deficiency of a graph G, $\mu (G)$, is defined as the minimum nonnegative integer n such that $G\cup n{K}_1$ is an edge-magic graph. This concept motivated Figueroa-Centeno et al. [6] to introduce the concept of super edge-magic deficiency of a graph. The super edge-magic deficiency (SEMD) of a graph G, ${\mu }_s(G)$, is defined as either the minimum nonnegative n such that $G\cup n{K}_1$ is a SEM graph or +∞ if there exists no such n. In 2016, Ichishima et al. [7] give the notion of consecutively super edge-magic deficiency of a bipartite graph. The consecutively super edge-magic deficiency (consecutively SEMD) of a bipartite graph G, ${\mu }_c(G)$, is defined to be either the smallest nonnegative integer n with the property that $G\cup n{K}_1$ is consecutively SEM or +∞ if there exists no such n.

Many researchers have investigated the SEMD of some classes of graphs. The complete results on this subject can be seen in a dynamic survey of graph labeling by Gallian [8]. In this paper, we study the (consecutively) SEMD of forests with two components, where its components are non isomorphic combs, isomorphic a subdivision of $K_{1,3}$ and $K_{1,4}$, and union of a comb and a subdivision of $K_{1,3}$ or $K_{1,4}$. Moreover, we find SEMD of 2-regular graph $2C_3\cup C_n$. By applying Ichishima's et al. [9] and Cichacz's et al. [10] methods to this result, we prove that some 2-regular graphs with a large order have zero SEMD, and then we also obtain the SEMD of union of cycles and paths. In addition, we study the SEMD of graph $(K_{1,n}\cup P_m)+K_1$ for any integer $n\ge 1$ and $m\ge 3$.

2. The SEMD of forests with two components

Figueroa-Centeno et al. [6] proved that the SEMD of all forests is finite. Next, the same authors [11] investigated the SEMD of some classes of forests with two components, such as $P_m\cup K_{1,n},K_{1,n}\cup K_{1,m}$, and $P_m\cup P_n$. Based on these results, they gave the following conjecture.

Conjecture 2.1

[11] If F is a forest with two components then ${\mu }_s(F)\le 1$.

Inspired by Conjecture 2.1, many researchers have investigated the SEMD of some forests with two components. Baig et al. [12] found the SEMD of union of combs and stars. Javed et al. [13] studied the SEMD of forests with two components consisting of combs, generalized combs, and stars. In particular, they found the SEMD of two isomorphic combs. Imran and Mukhtar [14] showed that forests with two components consisting of stars and subdivision of stars have zero SEMD. Krisnawati et al. [15] investigated SEMD of two non isomorphic of a subdivision of $K_{1,3}$ and $K_{1,4}$.

In this section, we study SEMD of disjoint union of two non isomorphic combs. We also continue Krisnawati's et al. [15] work to find SEMD of disjoint union of two isomorphic of a subdivision of $K_{1,3}$ and $K_{1,4}$. We prove that these graphs have zero SEMD. Some of these graphs also have zero consecutively SEMD. Additionally, we investigate the (consecutively) SEMD of union of a comb and a subdivision of $K_{1,3}$ or $K_{1,4}$.

A comb, $C{b}_n$ with $n\ge 1$, is a tree consisting of path $P_{n+1}$, whose vertices are $u_0,u_1,u_2,\cdots ,u_n$, together with edges $\{u_i{v}_i:1\le i\le n\}$. We investigate zero (consecutively) SEMD of two non isomorphic combs in the following theorem.

Theorem 2.2

Let $G_{n,m}=C{b}_n\cup C{b}_m$. For any $n,m\ge 1$ and $n\ne m$, ${\mu }_s(G_{n,m})={\mu }_c(G_{n,m})=0$.

Proof

Let $G_{n,m}$ be a graph with vertex set and edge set as follows.

$$\begin{gathered} V(G_{n,m})=\{u_i:0\le i\le n\}\cup \{v_i:1\le i\le n\}\cup \{w_j:0\le j\le m\}\cup \{x_j:1\le j\le m\}\text{and} \\ E(G_{n,m})=\{u_i{u}_{i+1}:0\le i\le n-1\}\cup \{u_i{v}_i:1\le i\le n\}\cup \{w_j{w}_{j+1}:0\le j\le m-1\}\cup \{w_j{x}_j:1\le j\le m\}. \end{gathered}$$

Thus, $G_{n,m}$ is a graph of order $2(n+m+1)$ and size $2(n+m)$.

For any $n,m\ge 1$ and $n\ne m$, define a labeling $f:V(G_{n,m})⟶\{1,2,\cdots ,2(n+m+1)\}$. Without loss of generality, we assume $n<m$. We consider the proof based on the value of n.

Case 1. n is odd.

$$f(a)=\{\begin{array}{cc}i+2,\hfill & \text{if}a=u_i\text{and}i\text{is even},\hfill \\ n+m+i+1,\hfill & \text{if}a=u_i\text{and}i\text{is odd},\hfill \\ i+2,\hfill & \text{if}a=v_i\text{and}i\text{is odd},\hfill \\ n+m+i+1,\hfill & \text{i}a=v_i\text{and}i\text{is even},\hfill \\ 2n+m+j+2,\hfill & \text{if}a=w_j\text{and}j\text{is even},\hfill \\ n+j+2,\hfill & \text{if}a=w_j\text{and}j\text{is odd},1\le j\le n,\hfill \\ n+j+1,\hfill & \text{if}a=w_j\text{and}j\text{is odd},n+2\le j\le m,\hfill \\ 2n+m+j+2,\hfill & \text{if}a=x_j\text{and}j\text{is odd},\hfill \\ n+j+2,\hfill & \text{if}a=x_j\text{and}j\text{is even},2\le j\le n-1,\hfill \\ 1,\hfill & \text{if}a=x_{n+1},\hfill \\ n+j+1,\hfill & \text{if}a=x_j\text{and}j\text{is even},n+3\le j\le m.\hfill \end{array}$$

Case 2. n is even.

$$f(a)=\{\begin{array}{cc}n+m+i+2,\hfill & \text{if}a=u_i\text{and}i\text{is even},\hfill \\ i+1,\hfill & \text{if}a=u_i\text{and}i\text{is odd},\hfill \\ n+m+i+2,\hfill & \text{if}a=v_i\text{and}i\text{is odd},\hfill \\ i+1,\hfill & \text{if}a=v_i\text{and}i\text{is even},\hfill \\ n+j+2,\hfill & \text{if}a=w_j\text{and}j\text{is even},0\le j\le n\hfill \\ n+j+1,\hfill & \text{if}a=w_j\text{and}j\text{is even},n+2\le j\le m,\hfill \\ 2n+m+j+2,\hfill & \text{if}a=w_j\text{and}j\text{is odd},\hfill \\ n+j+2,\hfill & \text{if}a=x_j\text{and}j\text{is odd},1\le j\le n-1\hfill \\ 1,\hfill & \text{if}a=x_{n+1},\hfill \\ n+j+1,\hfill & \text{if}a=x_j\text{and}j\text{is odd},n+3\le j\le m,\hfill \\ 2n+m+j+2,\hfill & \text{if}a=x_j\text{and}j\text{is even}.\hfill \end{array}$$

It is not hard to verify that the set of all edge-sums generated by f is $S=\{n+m+4,n+m+5,\cdots ,3(n+m+1)\}$. By Lemma 1.1, f extends to a SEM labeling of $G_{n,m}$ with magic constant $k=5(n+m)+6$. Hence, for any $n,m\ge 1$ and $n\ne m$, ${\mu }_s(G_{n,m})=0$.

Next, let A and B be partite sets of $G_{n,m}$.

For odd n,

$$\begin{gathered} A=\{u_i:i\equiv 0(\text{mod}2)\}\cup \{v_i:i\equiv 1(\text{mod}2)\}\cup \{w_j:j\equiv 1(\text{mod}2)\}\cup \{x_j:j\equiv 0(\text{mod}2)\}\text{and} \\ B=\{u_i:i\equiv 1(\text{mod}2)\}\cup \{v_i:i\equiv 0(\text{mod}2)\}\cup \{w_j:j\equiv 0(\text{mod}2)\}\cup \{x_j:j\equiv 1(\text{mod}2)\}. \end{gathered}$$

For even n,

$$\begin{gathered} A=\{u_i:i\equiv 1(\text{mod}2)\}\cup \{v_i:i\equiv 0(\text{mod}2)\}\cup \{w_j:j\equiv 0(\text{mod}2)\}\cup \{x_j:j\equiv 1(\text{mod}2)\}\text{and} \\ B=\{u_i:i\equiv 0(\text{mod}2)\}\cup \{v_i:i\equiv 1(\text{mod}2)\}\cup \{w_j:j\equiv 1(\text{mod}2)\}\cup \{x_j:j\equiv 0(\text{mod}2)\}. \end{gathered}$$

Since $f(A)=\{1,2,\cdots ,n+m+1\}$ and $f(B)=\{n+m+2,n+m+3,\cdots ,2(n+m+1)\}$, so $G_{n,m}$ is a consecutively SEM graph. Thus, for any $n,m\ge 1$ and $n\ne m$, ${\mu }_c(G_{n,m})=0$. □

As an illustration of the proof of Theorem 2.2, see the Fig. 1.

Figure 1.

The (consecutively) SEM labeling of Cb₃ ∪ Cb₆ and Cb₄ ∪ Cb₆.

A subdivision of a star $K_{1,r}$, denoted by $T(n_1,n_2,\cdots ,n_r)$, where $n_i\ge 1$, $1\le i\le r$, is a graph obtained by inserting $n_i-1$ vertices to each edge of the star $K_{1,r}$. Now, we study (consecutively) SEMD of disjoint union of two isomorphic of subdivision of $K_{1,3}$ and $K_{1,4}$ in the following theorems.

Theorem 2.3

Let $G_n=2T(n,n-1,n-1)$. For any even $n\ge 2$, ${\mu }_s(G_n)={\mu }_c(G_n)=0$ and for any odd $n\ge 3$, ${\mu }_s(G_n)\le 1$.

Proof

Let $G_n$ be a graph with

$$\begin{gathered} V(G_n)=\{y,y_{1,j},y_{2,l},y_{3,l}:1\le j\le n,1\le l\le n-1\}\cup \{z,z_{1,j},z_{2,l},z_{3,l}:1\le j\le n,1\le l\le n-1\}\text{and} \\ E(G_n)=\{y{y}_{i,1},z{z}_{i,1}:1\le i\le 3\}\cup \{y_{1,j}{y}_{1,j+1},z_{i,j}{z}_{i,j+1}:1\le j\le n-1\}\cup \{y_{2,l}{y}_{2,l+1},y_{3,l}{y}_{3,l+1},z_{2,l}{z}_{2,l+1},z_{3,l}{z}_{3,l+1}:1\le l\le n-2\}. \end{gathered}$$

Thus, $G_n$ is a graph of order $6n-2$ and size $6n-4$.

Let $n\ge 2$ be an even integer. Define a labeling $f:V(G_n)⟶\{1,2,\cdots ,6n-2\}$ as follows.

$$f(y)=\frac{1}{2}(7n),f(z)=\frac{1}{2}(11n-4).$$

$$f(a)=\{\begin{array}{cc}\frac{1}{2}(n-j+1),\hfill & \text{if}a=y_{1,j}\text{and}j\text{is odd},\hfill \\ \frac{1}{2}(7n-j),\hfill & \text{if}a=y_{1,j}\text{and}j\ne n\text{is even},\hfill \\ 6n-2,\hfill & \text{if}a=y_{1,n},\hfill \\ \frac{1}{2}(n+l+1),\hfill & \text{if}a=y_{2,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(7n+l),\hfill & \text{if}a=y_{2,l}\text{and}l\text{is even},\hfill \\ \frac{1}{2}(3n-l+1),\hfill & \text{if}a=y_{3,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(9n-l-2),\hfill & \text{if}a=y_{3,l}\text{and}l\text{is even},\hfill \\ \frac{1}{2}(3n+j+1),\hfill & \text{if}a=z_{1,j}\text{and}j\text{is odd},\hfill \\ \frac{1}{2}(9n+j-4),\hfill & \text{if}a=z_{1,j}\text{and}j\text{is even},\hfill \\ \frac{1}{2}(5n-l+1),\hfill & \text{if}a=z_{2,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(11n-l-4),\hfill & \text{if}a=z_{2,l}\text{and}l\text{is even},\hfill \\ \frac{1}{2}(5n+l+1),\hfill & \text{if}a=z_{3,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(11n+l-4),\hfill & \text{if}a=z_{3,l}\text{and}l\text{is even}.\hfill \end{array}$$

The set of all edge-sums generated by f is $S=\{3n+2,3n+3,\cdots ,9n-3\}$. By Lemma 1.1, f extends to a SEM labeling of $G_n$ with magic constant $k=15n-4$. Hence, for any even $n\ge 2$, ${\mu }_s(G_n)=0$.

Moreover, let A and B be partite sets of $G_n$, where

$$\begin{gathered} A=\{y_{1,j},y_{2,l},y_{3,l}:j,l\equiv 1(\text{mod}2)\}\cup \{z_{1,j},z_{2,l},z_{3,l}:j,l\equiv 1(\text{mod}2)\}\text{and} \\ B=\{y,y_{1,j},y_{2,l},y_{3,l}:j,l\equiv 0(\text{mod}2)\}\cup \{z,z_{1,j},z_{2,l},z_{3,l}:j,l\equiv 0(\text{mod}2)\}. \end{gathered}$$

It is easy to see that $f(A)=\{1,2,\cdots ,3n\}$ and $f(B)=\{3n+1,3n+2,\cdots ,6n-2\}$. So, $G_n$ is a consecutively SEM graph. Thus, ${\mu }_c(G_n)=0$ for any even $n\ge 2$.

Now, to show ${\mu }_s(G_n)\le 1$ for any odd $n\ge 3$, let us consider the graph $H_n=G_n\cup K_1$, where $V(H_n)=V(G_n)\cup \{w\}$ and $E(H_n)=E(G_n)$. Define a labeling $g:V(H_n)⟶\{1,2,\cdots ,6n-1\}$ as follows.

$$g(y)=\frac{1}{2}(7n+1),g(z)=6n-2,g(w)=\frac{1}{2}(11n-3).$$

$$g(a)=\{\begin{array}{cc}\frac{1}{2}(n-j+2),\hfill & \text{if}a=y_{1,j}\text{and}j\text{is odd},\hfill \\ \frac{1}{2}(7n-j+1),\hfill & \text{if}a=y_{1,j}\text{and}j\text{is even},\hfill \\ \frac{1}{2}(n+l+2),\hfill & \text{if}a=y_{2,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(7n+l+1),\hfill & \text{if}a=y_{2,l}\text{and}l\ne n-1\text{is even},\hfill \\ 6n-1,\hfill & \text{if}a=y_{2,n-1},\hfill \\ \frac{1}{2}(3n-l),\hfill & \text{if}a=y_{3,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(9n-l-1),\hfill & \text{if}a=y_{3,l}\text{and}l\text{is even},\hfill \\ \frac{1}{2}(4n-j+1),\hfill & \text{if}a=z_{1,j}\text{and}j\text{is odd},\hfill \\ \frac{1}{2}(10n-j-2),\hfill & \text{if}a=z_{1,j}\text{and}j\text{is even},\hfill \\ \frac{1}{2}(4n+l+1),\hfill & \text{if}a=z_{2,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(10n+l-4),\hfill & \text{if}a=z_{2,l}\text{and}l\text{is even},\hfill \\ 3n+l-3,\hfill & \text{if}a=z_{3,l},l\text{is odd},\text{and}n=5,\hfill \\ 3n-1,\hfill & \text{if}a=z_{3,1}\text{and}n\ne 5,\hfill \\ \frac{1}{2}(12n+l-11),\hfill & \text{if}a=z_{3,l},l=3,5,\text{and}n\ne 5,\hfill \\ \frac{1}{2}(11n+l-8),\hfill & \text{if}a=z_{3,l},l\ge 7\text{is odd},\hfill \\ 27,\hfill & \text{if}a=z_{3,2}\text{and}n=5,\hfill \\ 14,\hfill & \text{if}a=z_{3,4}\text{and}n=5,\hfill \\ 3n,\hfill & \text{if}a=z_{3,2}\text{and}n\ne 5.\hfill \end{array}$$

To label the rest of vertices, we consider the following cases.

• (i)
  For $n\equiv 3(\mathrm{mod}4)$ and $n>5$, set

  $$g(z_{3,l})=\{\begin{array}{cc}\frac{1}{2}(6n-l-2)\hfill & \text{for}l=4r,1\le r\le \frac{n-3}{4},\hfill \\ \frac{1}{2}(6n-l+2)\hfill & \text{for}l=4r+2,1\le r\le \frac{n-3}{4}.\hfill \end{array}$$
• (ii)
  For $n\equiv 1(\mathrm{mod}4)$ and $n>5$, set

  $$g(z_{3,l})=\{\begin{array}{cc}3n-l+2\hfill & \text{for}l=2r+2,1\le r\le \frac{n-1}{4},\hfill \\ \frac{1}{2}(7n-2l+1)\hfill & \text{for}l=\frac{1}{2}(n+4r+3),1\le r\le \frac{n-5}{4}.\hfill \end{array}$$

The set of all edge-sums generated by g is $S=\{3n+2,3n+3,\cdots ,9n-3\}$. By Lemma 1.1, g extends to a SEM labeling of $H_n$ with magic constant $k=15n-3$. Hence, for any odd $n\ge 3$, ${\mu }_s(G_n)\le 1$. □

Theorem 2.4

Let $G_n=2T(n,n-1,n-1,n)$. For any $n\ge 2$, ${\mu }_s(G_n)=0$.

Proof

The graph $G_n$ is a graph of order $8n-2$ and size $8n-4$. Let

$$\begin{gathered} V(G_n)=\{y,y_{1,j},y_{2,l},y_{3,l},y_{4,j}:1\le j\le n,1\le l\le n-1\}\cup \{z,z_{1,j},z_{2,l},z_{3,l},z_{4,j}:1\le j\le n,1\le l\le n-1\}\text{and} \\ E(G_n)=\{y{y}_{i,1},z{z}_{i,1}:1\le i\le 4\}\cup \{y_{i,j}{y}_{i,j+1},z_{i,j}{z}_{i,j+1}:i=1,4;1\le j\le n-1\}\cup \{y_{i,l}{y}_{i,l+1},z_{i,l}{z}_{i,l+1}:i=2,3;1\le l\le n-2\}. \end{gathered}$$

Define a labeling $f:V(G_n)⟶\{1,2,\cdots ,8n-2\}$ as follows.

$$f(y)=5n+1,$$

$$f(a)=\{\begin{array}{cc}\frac{1}{2}(j+1),\hfill & \text{if}a=y_{1,j}\text{and}j\text{is odd},\hfill \\ \frac{1}{2}(8n+j+2),\hfill & \text{if}a=y_{1,j}\text{and}j\text{is even},\hfill \\ \frac{1}{2}(2n-l+1),\hfill & \text{if}a=y_{2,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(10n-l+2),\hfill & \text{if}a=y_{2,l}\text{and}l\text{is even},\hfill \\ \frac{1}{2}(2n+l+1),\hfill & \text{if}a=y_{3,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(10n+l+2),\hfill & \text{if}a=y_{3,l}\text{and}l\text{is even},\hfill \\ \frac{1}{2}(4n-j+1),\hfill & \text{if}a=y_{4,j}\text{and}j\text{is odd},\hfill \\ \frac{1}{2}(12n-j+2),\hfill & \text{if}a=y_{4,j}\text{and}j\text{is even}.\hfill \end{array}$$

To label the vertices of the second component of $G_n$, we consider two following cases.

Case 1. n is even.

For $n=4$, set $f(z)=27$ and

$$\begin{array}{cccc}f(z_{1,1})=10,\hfill & f(z_{1,2})=26,\hfill & f(z_{1,3})=9,\hfill & f(z_{1,4})=25,\hfill \\ f(z_{2,1})=11,\hfill & f(z_{2,2})=28,\hfill & f(z_{2,3})=12,\hfill & \hfill \\ f(z_{3,1})=14,\hfill & f(z_{3,2})=29,\hfill & f(z_{3,3})=13,\hfill & \hfill \\ f(z_{4,1})=17,\hfill & f(z_{4,2})=16,\hfill & f(z_{4,3})=30,\hfill & f(z_{4,4})=15.\hfill \end{array}$$

For $n\ne 4$, set $f(z)=7n$ and

$$f(a)=\{\begin{array}{cc}\frac{1}{2}(4n+j+1),\hfill & \text{if}a=z_{1,j}\text{and}j\text{is odd},\hfill \\ \frac{1}{2}(12n+j),\hfill & \text{if}a=z_{1,j}\text{and}j\text{is even},\hfill \\ \frac{1}{2}(6n-l+1),\hfill & \text{if}a=z_{2,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(14n-l),\hfill & \text{if}a=z_{2,l}\text{and}l\text{is even},\hfill \\ \frac{1}{2}(6n+l+1),\hfill & \text{if}a=z_{3,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(14n+l),\hfill & \text{if}a=z_{3,l}\text{and}l\text{is even},\hfill \\ 4n,\hfill & \text{if}a=z_{4,1},\hfill \\ \frac{1}{2}(16n+j-9),\hfill & \text{if}a=z_{4,j}\text{and}j=3,5,\hfill \\ \frac{1}{2}(15n+j-7),\hfill & \text{if}a=z_{4,j}\text{and}j\ge 7\text{is odd},\hfill \\ 4n+1,\hfill & \text{if}a=z_{4,2}.\hfill \end{array}$$

For even $j\ge 4$, label $z_{4,j}$ as follows.

• (i)
  For $n\equiv 2(\mathrm{mod}4)$,

  $$f(z_{4,j})=\{\begin{array}{cc}\frac{1}{2}(8n-j)\hfill & \text{for}j=4r,1\le r\le \frac{n-2}{4},\hfill \\ \frac{1}{2}(8n-j+4)\hfill & \text{for}j=4r+2,1\le r\le \frac{n-2}{4}.\hfill \end{array}$$
• (ii)
  For $n\equiv 0(\mathrm{mod}4)$,

  $$f(z_{4,j})=\{\begin{array}{cc}4n-j+3\hfill & \text{for}j=2r+2,1\le r\le \frac{n}{4},\hfill \\ \frac{1}{2}(9n-2j+4)\hfill & \text{for}j=\frac{1}{2}(n+4r+4),1\le r\le \frac{n-4}{4}.\hfill \end{array}$$

Case 2. n is odd.

Set $f(z)=7n$ and

$$f(a)=\{\begin{array}{cc}\frac{1}{2}(4n+j+1),\hfill & \text{if}a=z_{1,j}\text{and}j\text{is odd},\hfill \\ \frac{1}{2}(12n+j),\hfill & \text{if}a=z_{1,j}\text{and}j\text{is even},\hfill \\ \frac{1}{2}(6n-l+1),\hfill & \text{if}a=z_{2,l}\text{and}l\text{is odd},\hfill \\ \frac{1}{2}(14n-l),\hfill & \text{if}a=z_{2,l}\text{and}l\text{is even},\hfill \\ 4n+1,\hfill & \text{if}a=z_{3,1},\hfill \\ \frac{1}{2}(16n-l-1),\hfill & \text{if}a=z_{3,l}\text{and}l\ge 3\text{is odd},\hfill \\ \frac{1}{2}(8n-l+2),\hfill & \text{if}a=z_{3,l}\text{and}l\ge 3\text{is even},\hfill \\ \frac{1}{2}(6n+j+1),\hfill & \text{if}a=z_{4,j}\text{and}j\text{is odd},\hfill \\ \frac{1}{2}(14n+j),\hfill & \text{if}a=z_{4,j}\text{and}j\text{is even}.\hfill \end{array}$$

For all cases, the set of all edge-sums generated by the labeling f is $S=\{4n+3,4n+4,\cdots ,12n-2\}$. By Lemma 1.1, f extends to a SEM labeling of $G_n$ with magic constant $k=20n-3$. Therefore, for any $n\ge 2$, ${\mu }_s(G_n)=0$. □

To clarify the proof of Theorem 2.3, Theorem 2.4, see Fig. 2.

Figure 2.

The (consecutively) SEM labeling of 2T(6,5,5) and 2T(7,6,6,7).

Next, we give the upper bound of (consecutively) SEMD of $C{b}_{n-1}\cup T(n,n-1,n-1)$ and $C{b}_{n-1}\cup T(n,n-1,n-1,n)$.

Theorem 2.5

Let $G_n=C{b}_{n-1}\cup T(n,n-1,n-1)$. For any $n\ge 2$, ${\mu }_s(G_n)={\mu }_c(G_n)\le \lfloor \frac{n-2}{2}\rfloor$.

Proof

Let $G_n$ be a graph having vertex and edge sets as follows.

$$\begin{gathered} V(G_n)=\{u_i:0\le i\le n-1\}\cup \{v_i:1\le i\le n-1\}\cup \{y,y_{1,r},y_{2,s},y_{3,s}:1\le r\le n,1\le s\le n-1\}\text{and} \\ E(G_n)=\{u_i{u}_{i+1}:0\le i\le n-2\}\cup \{u_i{v}_i:1\le i\le n-1\}\cup \{y{y}_{j,1},y_{1,r}{y}_{1,r+1},y_{2,s}{y}_{2,s+1},y_{3,s}{y}_{3,s+1}:1\le j\le 3,1\le r\le n-1,1\le s\le n-2\}. \end{gathered}$$

For any $n\ge 2$, let $H_n=G_n\cup \lfloor \frac{n-2}{2}\rfloor K_1$ be a graph with

$$V(H_n)=V(G_n)\cup \{z_t:1\le t\le \lfloor \frac{n-2}{2}\rfloor \}\text{and}E(H_n)=E(G_n).$$

Thus, $H_n$ has $\lfloor \frac{11n-6}{2}\rfloor$ vertices and $5n-4$ edges.

For $n=2$, label $(u_0,u_1,v_1)$ and $(y,y_{1,1},y_{1,2},y_{2,1},y_{3,1})$ with $(6,1,7)$ and $(8,4,5,2,3)$, respectively. These vertex labelings can be extended to a SEM labeling of $H_2$ with magic constant $k=21$.

For $n\ge 3$, define a labeling $f:V(H_n)⟶\{1,2,\cdots ,\lfloor \frac{11n-6}{2}\rfloor \}$ as follows.

$$f(u_i)=\{\begin{array}{cc}3n+i-1\hfill & \text{for even }i,\hfill \\ i\hfill & \text{for odd }i.\hfill \end{array}f(v_i)=\{\begin{array}{cc}3n+i-1\hfill & \text{for odd }i,\hfill \\ i\hfill & \text{for even }i.\hfill \end{array}$$

$$f(a)=\{\begin{array}{cc}\frac{1}{2}(2n+r-1),\hfill & \text{if}a=y_{1,r}\text{and}r\text{is odd},\hfill \\ \frac{1}{2}(4n-s-1),\hfill & \text{if}a=y_{2,s}\text{and}s\text{is odd},\hfill \\ \frac{1}{2}(4n+s-1).\hfill & \text{if}a=y_{3,s}\text{and}s\text{is odd}.\hfill \end{array}$$

Case 1. n is even.

$f(y)=5n-1$, $f(z_1)=4n-1$, $f(z_t)=\frac{1}{2}(5n+2t-2)$ for $t=2,3,\cdots ,\lfloor \frac{n-2}{2}\rfloor$, and

$$f(b)=\{\begin{array}{cc}\frac{1}{2}(8n+r-2),\hfill & \text{if}b=y_{1,r}\text{and}r\text{is even},\hfill \\ \frac{1}{2}(10n-s-2),\hfill & \text{if}b=y_{2,s}\text{and}s\text{is even}\hfill \\ \frac{1}{2}(10n+s-2),\hfill & \text{if}b=y_{3,s}\text{and}s\ne n-2\text{is even},\hfill \\ \frac{1}{2}(5n),\hfill & \text{if}b=y_{3,n-2}.\hfill \end{array}$$

Case 2. n is odd.

$f(y)=5n-2$, $f(z_t)=\frac{1}{2}(5n+2t-1)$ for $t=1,2,\cdots ,\lfloor \frac{n-2}{2}\rfloor$, and

$$f(b)=\{\begin{array}{cc}\frac{1}{2}(8n+r-4),\hfill & \text{if}b=y_{1,r}\text{and}r\text{is even}\hfill \\ \frac{1}{2}(10n-s-4),\hfill & \text{if}b=y_{2,s}\text{and}s\text{is even}\hfill \\ \frac{1}{2}(10n+s-4),\hfill & \text{if}b=y_{3,s}\text{and}s\ne n-1\text{is even},\hfill \\ \frac{1}{2}(5n-1),\hfill & \text{if}b=y_{3,n-1}.\hfill \end{array}$$

For all cases, it can be verified that the set of all edge-sums generated by the labeling f is $S=\{3n,3n+1,\cdots ,8n-5\}$. By Lemma 1.1, f extends to a SEM labeling of $H_n$ with magic constant $k=\lfloor \frac{27n-14}{2}\rfloor$. Hence, ${\mu }_s(G_n)\le \lfloor \frac{n-2}{2}\rfloor$ for any $n\ge 2$.

Furthermore, let A and B be partite sets of $H_n$ where

$$\begin{gathered} A=\{u_i:i\equiv 1(\text{mod}2)\}\cup \{v_i:i\equiv 0(\text{mod}2)\}\cup \{y_{1,r},y_{2,s},y_{3,s}:r,s\equiv 1(\text{mod}2)\}\text{and} \\ B=\{u_i:i\equiv 0(\text{mod}2)\}\cup \{v_i:i\equiv 1(\text{mod}2)\}\cup \{y,y_{1,r},y_{2,s},y_{3,s}:r,s\equiv 0(\text{mod}2)\}\cup \{z_t:1\le t\le \lfloor \frac{n-2}{2}\rfloor \}. \end{gathered}$$

Since for any $n\ge 2$, $f(A)=\{1,2,\cdots ,\lfloor \frac{5n-2}{2}\rfloor \}$ and $f(B)=\{\lfloor \frac{5n}{2}\rfloor ,\lfloor \frac{5n+2}{2}\rfloor ,\cdots ,\lfloor \frac{11n-6}{2}\rfloor \}$, then $H_n$ is a consecutively SEM graph. Thus, ${\mu }_c(G_n)\le \lfloor \frac{n-2}{2}\rfloor$. □

Theorem 2.6

Let $G_n=C{b}_{n-1}\cup T(n,n-1,n-1,n)$. For any $n\ge 2$,

$${\mu }_s(G_n)\le \{\begin{array}{cc}n-1,\hfill & \mathit{\text{if}}n\mathit{\text{is even}},\hfill \\ n-2,\hfill & \mathit{\text{if}}n\mathit{\text{is odd}}.\hfill \end{array}$$

Proof

Define the vertex and edge sets of $G_n$ as follows.

$$\begin{gathered} V(G_n)=\{u_i:0\le i\le n-1\}\cup \{v_i:1\le i\le n-1\}\cup \{y,y_{1,r},y_{2,s},y_{3,s},y_{4,r}:1\le r\le n,1\le s\le n-1\}\text{and} \\ E(G_n)=\{u_i{u}_{i+1}:0\le i\le n-2\}\cup \{u_i{v}_i:1\le i\le n-1\}\cup \{y{y}_{j,1},y_{1,r}{y}_{1,r+1},y_{2,s}{y}_{2,s+1},y_{3,s}{y}_{3,s+1},y_{4,r}{y}_{4,r+1}:1\le j\le 4,1\le r\le n-1,1\le s\le n-2\}. \end{gathered}$$

For any even $n\ge 2$, let $H_n=G_n\cup (n-1)K_1$ be a graph with

$$V(H_n)=V(G_n)\cup \{z_t:1\le t\le n-1\}\text{and}E(H_n)=E(G_n).$$

Next, define a labeling $f:V(H_n)⟶\{1,2,\cdots ,7n-3\}$ as follows.

$$f(u_i)=\{\begin{array}{cc}4n+i\hfill & \text{for even }i,\hfill \\ i\hfill & \text{for odd }i.\hfill \end{array}f(v_i)=\{\begin{array}{cc}4n+i\hfill & \text{for odd }i,\hfill \\ i\hfill & \text{for even }i.\hfill \end{array}$$

$$f(z_t)=3n+t\text{for}t=1,2,\cdots ,n-1.$$

For $n=4$, set $f(y)=22$ and

$$\begin{array}{cccc}f(y_{1,1})=5,\hfill & f(y_{1,2})=21,\hfill & f(y_{1,3})=4,\hfill & f(y_{1,4})=20,\hfill \\ f(y_{2,1})=6,\hfill & f(y_{2,2})=23,\hfill & f(y_{2,3})=7,\hfill & \hfill \\ f(y_{3,1})=9,\hfill & f(y_{3,2})=24,\hfill & f(y_{3,3})=8,\hfill & \hfill \\ f(y_{4,1})=12,\hfill & f(y_{4,2})=11,\hfill & f(y_{4,3})=28,\hfill & f(y_{4,4})=10.\hfill \end{array}$$

For $n\ne 4$, set $f(y)=6n-1$ and

$$f(a)=\{\begin{array}{cc}\frac{1}{2}(2n+r-1),\hfill & \text{if}a=y_{1,r}\text{and}r\text{is odd},\hfill \\ \frac{1}{2}(10n+r-2),\hfill & \text{if}a=y_{1,r}\text{and}r\text{is even},\hfill \\ \frac{1}{2}(4n-s-1),\hfill & \text{if}a=y_{2,s}\text{and}s\text{is odd},\hfill \\ \frac{1}{2}(12n-s-2),\hfill & \text{if}a=y_{2,s}\text{and}s\text{is even},\hfill \\ \frac{1}{2}(4n+s-1),\hfill & \text{if}a=y_{3,s}\text{and}s\text{is odd},\hfill \\ \frac{1}{2}(12n+s-2),\hfill & \text{if}a=y_{3,s}\text{and}s\text{is even},\hfill \\ 3n-1,\hfill & \text{if}a=y_{4,1},\hfill \\ \frac{1}{2}(14n+r-11),\hfill & \text{if}a=y_{4,r}\text{and}r=3,5,\hfill \\ \frac{1}{2}(13n+r-9),\hfill & \text{if}a=y_{4,r}\text{and}r\ge 7\text{is odd},\hfill \\ 3n,\hfill & \text{if}a=y_{4,2}.\hfill \end{array}$$

For even $r\ge 4$, label $y_{4,r}$ as follows.

• (i)
  For $n\equiv 2(\mathrm{mod}4)$,

  $$f(y_{4,r})=\{\begin{array}{cc}\frac{1}{2}(6n-r-2)\hfill & \text{for}r=4l,1\le l\le \frac{n-2}{4},\hfill \\ \frac{1}{2}(6n-r+2)\hfill & \text{for}r=4l+2,1\le l\le \frac{n-2}{4}.\hfill \end{array}$$
• (ii)
  For $n\equiv 0(\mathrm{mod}4)$,

  $$f(y_{4,r})=\{\begin{array}{cc}3n-r+2\hfill & \text{for}r=2l+2,1\le l\le \frac{n}{4},\hfill \\ \frac{1}{2}(7n-2r+2)\hfill & \text{for}r=\frac{1}{2}(n+4l+4),1\le l\le \frac{n-4}{4}.\hfill \end{array}$$

It can be checked that the set of all edge-sums is $S=\{4n+1,4n+2,\cdots ,10n-4\}$. Hence by Lemma 1.1, f extends to a SEM labeling of $H_n$ with magic constant $k=17n-6$. Therefore, ${\mu }_s(G_n)\le n-1$ for any even $n\ge 2$.

Now, for any odd $n\ge 3$, let $H_n^{\prime }=G_n\cup (n-2)K_1$ be a graph with

$$V(H_n^{\prime })=V(G_n)\cup \{z_t:1\le t\le n-2\}\text{and}E(H_n^{\prime })=E(G_n).$$

Next, define a labeling $g:V(H_n^{\prime })⟶\{1,2,\cdots ,7n-4\}$ as follows.

$$g(u_i)=\{\begin{array}{cc}4n+i-2\hfill & \text{for even }i,\hfill \\ i\hfill & \text{for odd }i.\hfill \end{array}g(v_i)=\{\begin{array}{cc}4n+i-2\hfill & \text{for odd }i,\hfill \\ i\hfill & \text{for even }i.\hfill \end{array}$$

$$g(y)=6n-3\text{and}g(z_t)=3n+t-1\text{for}t=1,2,\cdots ,n-2.$$

$$g(a)=\{\begin{array}{cc}\frac{1}{2}(2n+r-1),\hfill & \text{if}a=y_{1,r}\text{and}r\text{is odd},\hfill \\ \frac{1}{2}(10n+r-6),\hfill & \text{if}a=y_{1,r}\text{and}r\text{is even},\hfill \\ \frac{1}{2}(4n-s-1),\hfill & \text{if}a=y_{2,s}\text{and}s\text{is odd},\hfill \\ \frac{1}{2}(12n-s-6),\hfill & \text{if}a=y_{2,s}\text{and}s\text{is even},\hfill \\ \frac{1}{2}(4n+s-1),\hfill & \text{if}a=y_{3,s}\text{and}s\text{is odd},\hfill \\ \frac{1}{2}(12n+s-6),\hfill & \text{if}a=y_{3,s}\text{and}s\text{is even},\hfill \\ 3n-1,\hfill & \text{if}a=y_{4,1},\hfill \\ \frac{1}{2}(14n-r-5),\hfill & \text{if}a=y_{4,r}\text{and}r\ge 3\text{is odd},\hfill \\ \frac{1}{2}(6n-r-2),\hfill & \text{if}a=y_{4,r}\text{and}r\text{is even}.\hfill \end{array}$$

Under the labeling g, the set of all edge-sums is $S=\{4n-1,4n,\cdots ,10n-6\}$. Hence by Lemma 1.1, g extends to a SEM labeling of $H_n^{\prime }$ with magic constant $k=17n-9$. Therefore, ${\mu }_s(G_n)\le n-2$ for any odd n. □

Fig. 3 shows an illustration of the proof of Theorem 2.5, Theorem 2.6.

Figure 3.

The (consecutively) SEM labeling of Cb₇ ∪ T(8,7,7)∪3K₁ and the SEM labeling of Cb₆ ∪ T(7,6,6,7)∪5K₁.

3. The SEMD of $2C_3\cup C_n$ and related graphs

Enomoto et al. [2] stated that a cycle $C_n$ is SEM if and only if n is odd. Figueroa-Centeno et al. [6] investigated the SEMD of the cycle $C_n$. They also proved the following result.

Theorem 3.1

[6] If G is a graph of size $q\equiv 2(\mathrm{mod}4)$ such that every vertex in $V(G)$ has even degree then ${\mu }_s(G)=+\infty$.

In 2005, Figueroa-Centeno et al. [11] investigated the SEMD of $2C_n$, $3C_n$, $4C_n$, and conjectured that

$${\mu }_s(m{C}_n)=\{\begin{array}{ccc}0,\hfill & \text{if}\hfill & mn\equiv 1(\text{mod}2),\hfill \\ 1,\hfill & \text{if}\hfill & mn\equiv 0(\text{mod}4),\hfill \\ +\infty ,\hfill & \text{if}\hfill & mn\equiv 2(\text{mod}4).\hfill \end{array}$$

In 2009, Holden et al. [16] showed that $C_4\cup (2t-1)C_3$ for any integer $t\ge 3$, $C_5\cup (2t)C_3$ for any integer $t\ge 3$, and $C_7\cup (2t)C_3$ for any integer $t\ge 1$ have a strong vertex-magic labeling, which is equivalent to a SEM labeling. Based on their results, they proposed the following conjecture.

Conjecture 3.2

[16] A 2-regular graph of odd order is SEM if and only if it is not one of $C_3\cup C_4$, $3C_3\cup C_4$, or $2C_3\cup C_5$.

Motivated by Conjecture 3.2, Figueroa-Centeno et al. [17] showed that some 2-regular graphs with two components are SEM. They proved that $C_3\cup C_n$ is SEM if and only if $n\ge 4$ is even, $C_4\cup C_n$ is SEM if and only if $n\ge 5$ is odd, $C_5\cup C_n$ is SEM if and only if $n\ge 4$ is even, and $C_m\cup C_n$ is SEM for any even $m\ge 4$ and odd $n\ge \frac{1}{2}(m+4)$. Ichishima and Oshima [18] determined the SEMD of $C_m\cup C_n$ for even m and n, and for arbitrary n when $m=3,4,5$ and 7. In this section, we investigate the SEMD of $2C_3\cup C_n$.

Theorem 3.3

The SEMD of a 2-regular graph $2C_3\cup C_n$ is given by

$${\mu }_s(2C_3\cup C_n)=\{\begin{array}{ccc}0,\hfill & \mathit{\text{if}}\hfill & n\equiv 1,3(\mathrm{mod}4)\mathit{\text{and}}n\ne 5,\hfill \\ 1,\hfill & \mathit{\text{if}}\hfill & n\equiv 2(\mathrm{mod}4),\hfill \\ +\infty ,\hfill & \mathit{\text{if}}\hfill & n\equiv 0(\mathrm{mod}4).\hfill \end{array}$$

and ${\mu }_s(2C_3\cup C_5)\le 2$.

Proof

First, let $2C_3\cup C_n$ be a graph with vertex and edge sets

$$\begin{gathered} V(2C_3\cup C_n)=\{u_1,u_2,u_3\}\cup \{v_1,v_2,v_3\}\cup \{w_j:1\le j\le n\}\text{and } \\ E(2C_3\cup C_n)=\{u_1{u}_2,u_2{u}_3,u_3{u}_1\}\cup \{v_1{v}_2,v_2{v}_3,v_3{v}_1\}\cup \{w_j{w}_{j+1},w_n{w}_1:1\le j\le n-1\}. \end{gathered}$$

Next, we show that ${\mu }_s(2C_3\cup C_n)=0$ for $n\equiv 1(\mathrm{mod}4)$. Let $n=4r+1$ where $r\ge 2$. Define a labeling $f:V(2C_3\cup C_n)⟶\{1,2,\cdots ,4r+7\}$ as follows.

$$\begin{array}{ccc}f(u_1)=2r,\hfill & f(u_2)=2r+2,\hfill & f(u_3)=2r+3,\hfill \\ f(v_1)=3r+4,\hfill & f(v_2)=3r+5,\hfill & f(v_3)=3r+6,\hfill \end{array}$$

$$f(w_j)=\{\begin{array}{cc}i\hfill & \text{for }j=2i-1\text{ and }1\le i\le r-1,\hfill \\ 2r+i+3\hfill & \text{for }j=2i\text{ and }1\le i\le r-1,\hfill \\ 3r+3\hfill & \text{for }j=2r-1.\hfill \end{array}$$

To label $w_j$, for $2r\le j\le 4r+1$, we consider the following cases.

Case 1. $r\ge 2$ is even.

$$f(w_j)=\{\begin{array}{cc}r+2i-1\hfill & \text{for }j=2r+4i-4\text{ and }1\le i\le \frac{r}{2},\hfill \\ 3r+2i+6\hfill & \text{for }j=2r+4i-3\text{ and }1\le i\le \frac{1}{2}(r-2),\hfill \\ r+2i-2\hfill & \text{for }j=2r+4i-2\text{ and }1\le i\le \frac{1}{2}(r-2),\hfill \\ 3r+2i+5\hfill & \text{for }j=2r+4i-1\text{ and }1\le i\le \frac{1}{2}(r-2),\hfill \\ 4r-i+8\hfill & \text{for }j=4r+2i-5\text{ and }1\le i\le 3,\hfill \\ 2r-3i+4\hfill & \text{for }j=4r+2i-4\text{ and }1\le i\le 2.\hfill \end{array}$$

Case 2. $r\ge 3$ is odd.

$$f(w_j)=\{\begin{array}{cc}r+2i-1\hfill & \text{for }j=2r+4i-4\text{ and }1\le i\le \frac{1}{2}(r-1),\hfill \\ r+2i-2\hfill & \text{for }j=2r+4i-2\text{ and }1\le i\le \frac{1}{2}(r+1),\hfill \\ 3r+2i+7\hfill & \text{for }j=2r+4i-1\text{ and }1\le i\le \frac{1}{2}(r-1),\hfill \\ 3r+2i+6\hfill & \text{for }j=2r+4i+1\text{ and }1\le i\le \frac{1}{2}(r-3),\hfill \\ 4r-2i+9\hfill & \text{for }j=4r+2i-3\text{ and }1\le i\le 2,\hfill \\ 3r+7\hfill & \text{for }j=2r+1\hfill \\ 2r+1\hfill & \text{for }j=4r-2.\hfill \end{array}$$

It can be checked that for all cases, the set of all edge-sums is $S=\{2r+5,2r+6,\cdots ,6r+11\}$. By Lemma 1.1, f extends to a SEM labeling of $2C_3\cup C_n$ with magic constant $k=10r+19$. Hence, ${\mu }_s(2C_3\cup C_n)=0$ for $n\equiv 1(\mathrm{mod}4)$.

Next, we prove that ${\mu }_s(2C_3\cup C_n)=0$ for $n\equiv 3(\mathrm{mod}4)$. Let $n=4r+3$, $r\ge 0$. For $r=0$, label $(u_1,u_2,u_3),(v_1,v_2,v_3)$, and $(w_1,w_2,w_3)$ with $(1,5,9),(2,6,7)$, and $(3,4,8)$, respectively. For $r\ge 1$, define a labeling $f:V(2C_3\cup C_n)⟶\{1,2,\cdots ,4r+9\}$ as follows.

$$\begin{array}{ccc}f(u_1)=2r+2,\hfill & f(u_2)=2r+3,\hfill & f(u_3)=2r+4,\hfill \\ f(v_1)=3r+5,\hfill & f(v_2)=3r+6,\hfill & f(v_3)=3r+7,\hfill \end{array}$$

$$f(w_j)=\{\begin{array}{cc}i\hfill & \text{for }j=2i-1\text{and}1\le i\le 2r+1,\hfill \\ 2r+i+5\hfill & \text{for }j=2i\text{and}1\le i\le r-1,\hfill \\ 2r+i+8\hfill & \text{for }j=2i\text{and}r\le i\le 2r+1,\hfill \\ 2r+5\hfill & \text{for }j=n=4r+3.\hfill \end{array}$$

It is not hard to verify that the set of all edge-sums is $S=\{2r+6,2r+7,\cdots ,6r+14\}$. Hence by Lemma 1.1, f extends to a SEM labeling of $2C_3\cup C_n$ with magic constant $k=10r+24$. Therefore, ${\mu }_s(2C_3\cup C_n)=0$ for $n\equiv 3(\mathrm{mod}4)$.

Now, let $G=2C_3\cup C_n\cup K_1$ for $n=4r+2$, $r\ge 1$. Define a labeling $f:V(G)⟶\{1,2,\cdots ,4r+9\}$ as follows.

$$f(u_1)=2r+4,f(u_2)=2r+5,f(u_3)=4r+8,f(z)=3r+7,$$

where z is the vertex of $K_1$. For the other vertices, we consider the following cases.

Case 1. For any odd $r\ge 1$, set

$$f(v_1)=r+2,f(v_2)=r+4,f(v_3)=3r+8,\text{and}$$

$$f(w_j)=\{\begin{array}{cc}i\hfill & \text{for }j=2i-1\text{ and }1\le i\le 2,\hfill \\ 2r+7\hfill & \text{for }j=2,\hfill \\ 4r+9\hfill & \text{for }j=4,\hfill \\ 2r-2i+4\hfill & \text{for }j=4i+1\text{ and }1\le i\le r,\hfill \\ 4r-2i+8\hfill & \text{for }j=4i+2\text{ and }1\le i\le \frac{1}{2}(r-1),\hfill \\ 2r-2i+5\hfill & \text{for }j=4i+3\text{ and }1\le i\le \frac{1}{2}(r-1),\hfill \\ 4r-2i+9\hfill & \text{for }j=4i+4\text{ and }1\le i\le \frac{1}{2}(r-1),\hfill \\ 3r-2i+7\hfill & \text{for }j=2r+4i\text{ and }1\le i\le \frac{1}{2}(r+1),\hfill \\ 3r-2i+8\hfill & \text{for }j=2r+4i+2\text{ and }1\le i\le \frac{1}{2}(r-1),\hfill \\ r-2i+2\hfill & \text{for }j=2r+4i+4\text{ and }1\le i\le \frac{1}{2}(r-1).\hfill \end{array}$$

Case 2. For any even $r\ge 2$, set

$$f(v_1)=r+1,f(v_2)=r+5,f(v_3)=3r+9,\text{and}$$

$$f(w_j)=\{\begin{array}{cc}i\hfill & \text{for }j=2i-1\text{ and }1\le i\le \frac{1}{2}(r+2),\hfill \\ 2r+2i+5\hfill & \text{for }j=4i-2\text{ and }1\le i\le \frac{r}{2},\hfill \\ 2r+2i+6\hfill & \text{for }j=4i\text{ and }1\le i\le r,\hfill \\ 3r+2i+9\hfill & \text{for }j=2r+4i-2\text{ and }1\le i\le \frac{r}{2},\hfill \\ 2r+2i-2\hfill & \text{for }j=4r+2i-5\text{ and }1\le i\le 2,\hfill \\ r+3\hfill & \text{for }j=4r+1,\hfill \\ 2r+6\hfill & \text{for }j=n=4r+2.\hfill \end{array}$$

To label the rest of vertices, we consider two subcases.

• •
  Subcase 2.1. For $r\equiv 2(\mathrm{mod}4)$, set

  $$f(w_j)=\{\begin{array}{cc}\frac{1}{2}(3r-2)\hfill & \text{for }j=r+3,\hfill \\ \frac{1}{2}(r+2i+2)\hfill & \text{for }j=r+2i+3\text{ and }1\le i\le \frac{1}{2}(r-4),\hfill \\ r+2i+5\hfill & \text{for }j=2r+4i-3\text{ and }1\le i\le \frac{1}{4}(r-2),\hfill \\ r+2i-2\hfill & \text{for }j=2r+4i-1\text{ and }1\le i\le \frac{1}{4}(r-2),\hfill \\ \frac{1}{2}(3r+4i-2)\hfill & \text{for }j=3r+4i-5\text{ and }1\le i\le \frac{1}{4}(r+2),\hfill \\ \frac{1}{2}(3r+4i+8)\hfill & \text{for }j=3r+4i-3\text{ and }1\le i\le \frac{1}{4}(r-2).\hfill \end{array}$$
• •
  Subcase 2.2. For $r\equiv 0(\mathrm{mod}4)$, set

  $$f(w_j)=\{\begin{array}{cc}\frac{1}{2}(r+2i+4)\hfill & \text{for }j=r+2i+1\text{ and }1\le i\le \frac{1}{2}(r-4),\hfill \\ r+2i\hfill & \text{for }j=2r+4i-5\text{ and }1\le i\le \frac{r}{4},\hfill \\ r+2i+5\hfill & \text{for }j=2r+4i-3\text{ and }1\le i\le \frac{1}{4}(r-4),\hfill \\ \frac{1}{2}(r+4)\hfill & \text{for }j=3r-3,\hfill \\ \frac{1}{2}(3r+4i+6)\hfill & \text{for }j=3r+4i-5\text{ and }1\le i\le \frac{r}{4},\hfill \\ \frac{1}{2}(3r+4i)\hfill & \text{for }j=3r+4i-3\text{ and }1\le i\le \frac{r}{4}.\hfill \end{array}$$

It can be checked that under the labeling f, the set of all edge-sums is $S=\{2r+6,2r+7,\cdots ,6r+13\}$. By Lemma 1.1, f extends to a SEM labeling of G with magic constant $k=10r+23$. Thus, ${\mu }_s(2C_3\cup C_n)=1$ for $n\equiv 2(\mathrm{mod}4)$.

To show that ${\mu }_s(2C_3\cup C_5)\le 2$, label $\{u_1,u_2,u_3\}\cup \{v_1,v_2,v_3\}\cup \{w_1,w_2,w_3,w_4,w_5\}\cup \{2K_1\}$ with $\{3,7,9\}\cup \{5,6,12\}\cup \{1,8,11,2,13\}\cup \{4,10\}$, respectively. Finally, ${\mu }_s(2C_3\cup C_n)=+\infty$ for $n\equiv 0(\mathrm{mod}4)$ is a direct consequence of Theorem 3.1. □

In 2013, Ichishima et al. [9] introduced the concept of a pseudo SEM graph. A graph G with isolated vertices is called pseudo SEM if there exists a bijection $f:V(G)⟶\{1,2,...,|V(G)|\}$ such that the set $\{f(u)+f(v):uv\in E(G)\}\cup \{2f(u):deg(u)=0\}$ consists of $|E(G)|+|\{u\in V(G):deg(u)=0\}|$ consecutive integers. Such a function is called a pseudo SEM labeling. They also proved the following results.

Theorem 3.4

[9] Let m be an odd integer. If $G=({\cup }_{i=1}^{k_1}{C}_{n_i})\cup k_2{K}_1$ is any pseudo SEM 2-regular graph then $H=({\cup }_{i=1}^{k_1}(m,n_i)C_{[m,n_i]})\cup k_2{C}_m$ is a SEM 2-regular graph, where $(m,n_i)$ and $[m,n_i]$ are the greatest common divisor and the least common multiple of integers m and $n_i$, respectively.

Corollary 3.5

[9] Let m be an odd integer such that $(m,n_i)=1$ and $(m,n_i)=n_i$, $1\le i\le k_1$. If $G=({\cup }_{i=1}^{k_1}{C}_{n_i})\cup k_2{K}_1$ is any pseudo super edge-magic graph then $H=({\cup }_{i=1}^{k_1}{C}_{m{n}_i})\cup k_2{C}_m$ is a SEM graph.

By applying Theorem 3.4 and Corollary 3.5 to a part of Theorem 3.3, we have the following Lemma.

Lemma 3.6

For any odd m and $n\equiv 2(\mathrm{mod}4)$, the following SEMD of some graphs are hold.

$$\begin{array}{cccc}\hfill \mathit{\text{(i)}}& {\mu }_s(2C_{3m}\cup (m,n)C_{[m,n]}\cup C_m)=0.\hfill & \hfill \mathit{\text{(iii)}}& {\mu }_s(2C_{3m}\cup C_{5m}\cup 2C_m)=0.\hfill \\ \hfill \mathit{\text{(ii)}}& {\mu }_s((n+6)P_m\cup C_m)=0.\hfill & \hfill \mathit{\text{(iv)}}& {\mu }_s(12P_m\cup C_m)=0.\hfill \end{array}$$

Proof

It is easy to verify that the labeling f in the proof of Theorem 3.3 is a pseudo SEM labeling of $2C_3\cup C_n\cup K_1$ for $n\equiv 2(\mathrm{mod}4)$ and $2C_3\cup C_5\cup 2K_1$. By applying Theorem 3.4 and Corollary 3.5 to these 2-regular pseudo SEM graphs, we obtain (i) and (iii). By removing $n+6$ edges of the SEM 2-regular graph $2C_{3m}\cup (m,n)C_{[m,n]}\cup C_m$ with the smallest edge-sums $\{a,a+1,\cdots ,a+n+5\}$, where $a=\frac{1}{2}[(n+10)+(m-1)(n+7)]$, we get (ii). In a similar way, we obtain the result (iv) by removing 12 edges of the SEM 2-regular graph $2C_{3m}\cup C_{5m}\cup 2C_m$ with the smallest edge-sums $\{b,b+1,\cdots ,b+11\}$, where $b=\frac{1}{2}(13m+3)$. □

As an illustration of the proof of Lemma 3.6 (i) and (ii), see Fig. 4.

Figure 4.

(a) The pseudo SEM labeling of 2C₃ ∪ C₁₀ ∪ K₁; (b) The SEM labeling of 2C₉ ∪ C₃₀ ∪ C₃, which is obtained by applying Corollary 3.5 to Fig. 4(a) for m = 3; (c) The SEM labeling of 16P₃ ∪ C₃, which is obtained by removing 16 edges with the smallest edge-sums {27,28,⋯,42} of the graph in Fig. 4(b).

Cichacz et al. [10] in 2017 introduced a new method to expand the classes of known (strong) vertex-magic labeling of 2-regular graphs as the following theorem.

Theorem 3.7

[10] For $1\le i\le t$, let $n_i\ge 3$ be an integer and $G={\cup }_{i=1}^t{C}_{n_i}$. If G is (strong) vertex-magic then $H={\cup }_{i=1}^t{C}_{n_i(2r+1)}$ is (strong) vertex-magic for every integer r.

Ngurah [19] discovered that the method introduced in the proof of Theorem 3.7 is also valid for pseudo SEM graphs. By applying this fact to our results, we have the following lemma.

Lemma 3.8

For any odd m and $n\equiv 2(\mathrm{mod}4)$, the following SEMD of some graphs are satisfied.

$$\begin{array}{cccc}\hfill \mathit{\text{(i)}}& {\mu }_s(2C_{3m}\cup C_{nm}\cup C_m)=0.\hfill & \hfill \mathit{\text{(v)}}& {\mu }_s(2C_{3m}\cup C_{5m}\cup C_m)\le m.\hfill \\ \hfill \mathit{\text{(ii)}}& {\mu }_s(m{P}_3\cup C_{3m}\cup C_{nm}\cup C_m)=0.\hfill & \hfill \mathit{\text{(vi)}}& {\mu }_s(2C_{3m}\cup m{P}_5\cup C_m)\le m.\hfill \\ \hfill \mathit{\text{(iii)}}& {\mu }_s(m{P}_3\cup C_{3m}\cup m{P}_n\cup C_m)=0.\hfill & \hfill \mathit{\text{(vii)}}& {\mu }_s(m{P}_3\cup C_{3m}\cup m{P}_5\cup C_m)\le m.\hfill \\ \hfill \mathit{\text{(iv)}}& {\mu }_s(2C_{3m}\cup C_{5m}\cup 2C_m)=0.\hfill & \hfill \mathit{\text{(viii)}}& {\mu }_s(2m{P}_3\cup m{P}_5\cup C_m)\le m.\hfill \end{array}$$

Proof

The result (i) and (iv) are obtained by applying Theorem 3.7 to the fact that $2C_3\cup C_n\cup K_1$ for $n\equiv 2(\mathrm{mod}4)$ and $2C_3\cup C_5\cup 2K_1$ are pseudo SEM, respectively. We gain (ii) and (iii) from (i) by removing edges with the edge-sums $\{a,a+1,\cdots ,a+m-1\}$ and $\{a,a+1,\cdots ,a+2m-1\}$, where $a=\frac{1}{2}(nm+7m+3)$, respectively. Moreover, we obtain the result (v), (vi), (vii), and (viii) from (iv) by deleting edges with the edge-sums $\{b,b+1,\cdots ,b+m-1\}$, $\{b,b+1,\cdots ,b+2m-1\}$, $\{b,b+1,\cdots ,b+3m-1\}$, and $\{b,b+1,\cdots ,b+4m-1\}$, where $b=\frac{1}{2}(13m+3)$, respectively. □

Fig. 5 shows the construction given in the proof of Lemma 3.8 (i), (ii), and (iii).

Figure 5.

(a) The SEM labeling of 2C₉ ∪ C₃₀ ∪ C₃, which is obtained by applying Theorem 3.7 to Fig. 4(a) for r = 1; (b) The SEM labeling of 3P₃ ∪ C₉ ∪ C₃₀ ∪ C₃, which is obtained by removing 3 edges with the smallest edge-sums {27,28,29} in Fig. 5(a); (c) The SEM labeling of 3P₃ ∪ C₉ ∪ 3P₁₀ ∪ C₃, which is obtained by removing 3 edges with the smallest edge-sums {30,31,32} in Fig. 5(b).

4. The SEMD of join product of union of a star and a path with an isolated vertex

To present our results, we need the concept of dual labeling. A dual labeling $f^{\prime }$ of a SEM labeling f is defined as $f^{\prime }(x)=p+1-f(x)$ for all $x\in V(G)$ and $f^{\prime }(xy)=2p+q+1-f(xy)$ for all $xy\in E(G)$. Baskoro et al. [20] proved that the dual of a SEM labeling is also a SEM labeling.

Ngurah and Simanjuntak [21] studied the SEMD of join products of a path, a star, and a cycle, respectively, with isolated vertices. Generally, they showed that the join product of a SEM graph with isolated vertices has finite SEMD. In [22], the same authors investigated the SEMD of join product of a graph G which has certain properties with an isolated vertex. They gave a necessary condition for $G+K_1$ to have zero SEMD as the following lemma.

Lemma 4.1

[22] Let G be a graph with no cycle and minimum degree one. If ${\mu }_s(G+K_1)=0$ then G is a tree or a forest.

They showed that Lemma 4.1 is attainable. In particular, they proved that the join product of some forests with an isolated vertex has zero SEMD, such as ${\mu }_s([P_n\cup P_2]+K_1)=0$ if and only if $3\le n\le 5$ and ${\mu }_s([K_{1,n}\cup P_2]+K_1)=0$ if and only if $n=2$. In this section, we study the SEMD of join product graph $(K_{1,n}\cup P_m)+K_1$ for any integer $n\ge 1$ and $m\ge 3$.

For any integer $n\ge 1$ and $m\ge 3$, let $G_{n,m}=(K_{1,n}\cup P_m)+K_1$ be a graph having vertex and edge sets:

$$\begin{gathered} V(G_{n,m})=\{c,z\}\cup \{x_i:1\le i\le n\}\cup \{y_j:1\le j\le m\}\text{and } \\ E(G_{n,m})=\{cz\}\cup \{c{x}_i,z{x}_i:1\le i\le n\}\cup \{z{y}_j:1\le j\le m\}\cup \{y_j{y}_{j+1}:1\le l\le m-1\}. \end{gathered}$$

Thus, the graph $G_{n,m}$ has $n+m+2$ vertices and $2n+2m$ edges.

Now, we give necessary and sufficient conditions of $G_{n,m}$ to have zero SEMD for $m=3,4,5$.

Theorem 4.2

${\mu }_s(G_{n,3})=0$ if and only if $n=1,2,3$; and for $m=4,5$, ${\mu }_s(G_{n,m})=0$ if and only if $n=1,2$.

Proof

Firstly, we show that for $n=1,2,3$, ${\mu }_s(G_{n,3})=0$. For $n=1,2,3$, label $(c,z),(x_1,\cdots ,x_n),(y_1,y_2,y_3)$ with $(3,2),(1),(5,4,6)$; $(2,3),(1,4),(6,5,7)$; and $(1,4),(2,3,7),(5,8,6)$, respectively. These vertex labelings can be extended to a SEM labeling of $G_{n,3}$ for $n=1,2,3$.

Next, we prove that ${\mu }_s(G_{n,3})>0$ for every $n\ge 4$. Suppose that ${\mu }_s(G_{n,3})=0$ for every $n\ge 4$. Then by Lemma 1.1, there exists a bijection $f:V(G_{n,3})⟶\{1,2,\cdots ,n+5\}$ such that $S=\{f(x)+f(y):xy\in E(G_{n,3})\}$ is a set of $2n+6$ consecutive integers. Since $G_{n,3}$ has $n+5$ vertices and $2n+6$ edges, there are two possibilities of S, namely $\{3,4,\cdots ,2n+8\}$ or $\{4,5,\cdots ,2n+9\}$ and they are dual to each other. Thus, we can consider $S=\{3,4,\cdots ,2n+8\}$. Based on the degree of all vertices of $G_{n,3}$, the sum of all elements in S contains $n+4$ times of label of z, $n+1$ times of label of c, three times of label of $y_2$ and two times of label of the remaining vertices. Hence,

$$(n+4)f(z)+(n+1)f(c)+3f(y_2)+2[f(y_1)+f(y_3)+\sum _{i=1}^{n}f(x_i)]=\sum _{s\in S}s=(n+4)(2n+9)-3$$

or

$$(n+2)f(z)+(n-1)f(c)+f(y_2)=n^2+6n+3.$$

Furthermore, to get edge-sums 3, 4, and 5 in S, then the vertices of labels 1, 2 and 3 are adjacent to each other or the vertex of label 1 is adjacent to the vertices of labels 2, 3, and 4. By using the above equation and this fact, we have two following cases.

Case 1. There is a triangle with the vertices of labels 1, 2 and 3. Since every triangle in $G_{n,3}$ share a common vertex z, hence $f(z)\in \{1,2,3\}$ and $f(c),f(y_2)\in \{1,2,3\}-\{f(z)\}$. We can check that the equation has no solution for $f(z)\in \{1,2\}$. If $f(z)=3$ then $(n-1)f(c)+f(y_2)=n^2+3n-3$ and its solution is $f(y_2)=1$ and $f(c)=n+4$. However, the solution does not lead to a SEM labeling of $G_{n,3}$ for any $n\ge 4$.

Case 2. The vertex of label 1 is adjacent to the vertices of label 2, 3, and 4. Since the vertex of label 1 must have minimum degree 3, the possibilities of its vertex are z, c or $y_2$.

• (i)If $f(z)=1$ then $(n-1)f(c)+f(y_2)=n^2+5n+3$, where $f(c),f(y_2)\in \{2,3,\cdots ,n+5\}$. It is easy to verify that this equation has no solution.
• (ii)If $f(c)=1$ then $(n+2)f(z)+f(y_2)=n^2+5n+4$. It can be checked that this equation has no solution if $f(z)\in \{2,3,\cdots ,n+1\}\cup \{n+3,n+4,n+5\}$. If $f(z)=n+2$ then $f(y_2)=n$. However, this solution does not lead to a SEM labeling of $G_{n,3}$ for any $n\ge 4$.
• (iii)If $f(y_2)=1$ then $(n+2)f(z)+(n-1)f(c)=n^2+6n+2$. Since $y_2$ has degree three, which is adjacent to z, thus $f(z)\in \{2,3,4\}$. If $f(z)=2$ then $f(c)>n+5=p$. If $f(z)=3$ then either $f(z)+f(y_1)=f(y_2)+f(y_3)=5$ or $f(z)+f(y_3)=f(y_1)+f(y_2)=5$. If $f(z)=4$ then $f(c)=n+3-\frac{3}{n-1}$, thus $n=4$. However, this solution cannot be extended to a SEM labeling of $G_{4,3}$.

Hence, ${\mu }_s(G_{n,3})>0$ for every $n\ge 4$. Therefore, ${\mu }_s(G_{n,3})=0$ if and only if $n=1,2,3$.

Now, we show that for $m=4,5$, ${\mu }_s(G_{n,m})=0$ if and only if $n=1,2$. For $n=1,2$ and $m=4$, label $(c,z),(x_1,\cdots ,x_n),(y_1,y_2,y_3,y_4)$ with $(3,2),(1),(4,6,5,7)$ and $(2,3),(1,4),(5,7,6,8)$, respectively. Moreover, for $n=1,2$ and $m=5$, label $(c,z),(x_1,\cdots ,x_n),(y_1,y_2,y_3,y_4,y_5)$ with $(2,1),(3),(4,7,5,8,6)$ and $(3,2),(1,4),(5,8,6,9,7)$, respectively. It can be checked that these labelings extend to a SEM labeling of $G_{n,4}$ and $G_{n,5}$ for $n=1,2$. Conversely, suppose that ${\mu }_s(G_{n,4})={\mu }_s(G_{n,5})=0$ for $n\ge 3$. Then for $m=4$, we have

$$(n+3)f(z)+(n-1)f(c)+f(y_2)+f(y_3)=n^2+8n+10$$

and for $m=5$,

$$(n+4)f(z)+(n-1)f(c)+f(y_2)+f(y_3)+f(y_4)=n^2+10n+19.$$

By a similar argument as in the proof of $G_{n,3}$, we can show that ${\mu }_s(G_{n,4})>0$ and ${\mu }_s(G_{n,5})>0$ for every $n\ge 3$. Hence, for $m=4,5$, ${\mu }_s(G_{n,m})=0$ if and only if $n=1,2$. □

Since $G_{n,3}$, $G_{n,4}$, and $G_{n,5}$ do not SEM for almost all of n, thus we try to find its SEMD. Our result is as follows.

Theorem 4.3

For any integer $m,n\ge 3$, the SEMD of $G_{n,m}$ is given by:

• (i)${\mu }_s(G_{n,m})=1$ if $m=3$ and $n=4,5$; $m=4$ and $n=3,4,5,6$; and $m=5$ and $n=3,4,5$.
• (ii)${\mu }_s(G_{n,m})\le n-1$ if $m\ge 3$ and $n=r\lfloor \frac{m-1}{2}\rfloor$, where r is any positive integer.

Proof

• (i)
  As a consequence of Theorem 4.2, we obtain that ${\mu }_s(G_{n,3})\ge 1$ for $n\ge 4$, ${\mu }_s(G_{n,4})\ge 1$ for $n\ge 3$, and ${\mu }_s(G_{n,5})\ge 1$ for $n\ge 3$. Hence, the remaining case is to show that ${\mu }_s(G_{n,3})\le 1$ for $n=4,5$; ${\mu }_s(G_{n,4})\le 1$ for $n=3,4,5,6$; and ${\mu }_s(G_{n,5})\le 1$ for $n=3,4,5$. Thus, let us consider the graph $G_{n,m}\cup K_1$ and suppose w is the vertex of $K_1$.

  • –
    For $n=4,5$ and $m=3$, label $(c,z),(x_1,x_2,\cdots ,x_n),(y_1,y_2,y_3),(w)$ with $(1,5),(2,3,4,9),(7,6,10),(8)$ and $(1,6),(2,3,4,5,11),(7,8,10),(9)$, respectively.
  • –
    For $n=3,4,5,6$ and $m=4$, label $(c,z),(x_1,x_2,\cdots ,x_n),(y_1,y_2,y_3,y_4),(w)$ with $(1,4),(2,3,7),(6,10,5,8),(9)$; $(1,5),(2,3,4,9),(6,11,7,8),(10)$; $(1,6),(2,3,4,5,11),(8,12,7,9),(10)$; and $(1,7),(2,3,4,5,6,13),(8,10,12,9),(11)$, respectively.
  • –
    For $n=3,4,5$ and $m=5$, label $(c,z),(x_1,x_2,\cdots ,x_n),(y_1,y_2,\cdots ,y_5),(w)$ with $(1,4),(2,3,7),(5,8,10,6,11),(9)$; $(1,5),(2,3,4,9),(6,10,8,12,7),(11)$; and $(1,6),(2,3,4,5,11),(7,9,13,8,12),(10)$, respectively.

  It is not hard to check that these labelings extend to a SEM labeling of $G_{n,3}\cup K_1$ for $n=4,5$; $G_{n,4}\cup K_1$ for $n=3,4,5,6$; and $G_{n,5}\cup K_1$ for $n=3,4,5$.
• (ii)
  Let $H=G_{n,m}\cup (n-1)K_1$, where $m\ge 3$ and $n=r\lfloor \frac{m-1}{2}\rfloor$. Define a labeling $f:V(H)⟶\{1,2,\cdots ,2n+m+1\}$ as follows.

  $$f(c)=m+1,f(z)=\lfloor \frac{3m+1}{2}\rfloor ,$$

  $$f(x_i)=\{\begin{array}{cc}\frac{(2l+1)m-2l+5}{2}+i-((l-1)(\frac{m-1}{2})+1),\hfill & \text{ if }m\text{is odd,}\hfill \\ \frac{(2l+1)m-4l+6}{2}+i-((l-1)(\frac{m-2}{2})+1),\hfill & \text{ if }m\text{ is even,}\hfill \end{array}$$

  where $(l-1)\lfloor \frac{m-1}{2}\rfloor +1\le i\le l\lfloor \frac{m-1}{2}\rfloor$, $1\le l\le r$. For $1\le j\le m$,

  $$f(y_j)=\{\begin{array}{cc}l\hfill & \text{for}j=2l-1\text{and}1\le l\le \lfloor \frac{m+1}{2}\rfloor ,\hfill \\ \lfloor \frac{m+1}{2}\rfloor +l\hfill & \text{for}j=2l\text{and}1\le l\le \lfloor \frac{m}{2}\rfloor .\hfill \end{array}$$

  The remaining labels are used to label isolated vertices of H. It can be verified that the set of all edge-sums is a consecutive integer $S=\{a,a+1,\cdots ,a+2n+2m-1\}$, where $a=\lfloor \frac{m+5}{2}\rfloor$. By Lemma 1.1, f extends to a SEM labeling of H with magic constant $k=4n+3m+a+1$. Therefore, for any $m\ge 3$ and $n=r\lfloor \frac{m-1}{2}\rfloor$, $r\in Z^{+}$, ${\mu }_s(G_{n,m})\le n-1$. □

5. Conclusion

In this paper, we study the (consecutively) SEMD of some graphs. We find the exact value or upper bound of the (consecutively) SEMD of forests with two components. We also find the exact value of the SEMD of a 2-regular graph $2C_3\cup C_n$ for almost all of n. By using this and previous known results, we obtain the exact value or upper bound of the SEMD of some 2-regular graphs and union of cycles and paths. Moreover, we provide the necessary and sufficient conditions of $(K_{1,n}\cup P_m)+K_1$ to gain zero SEMD and the upper bound of SEMD of this graph for the remaining cases.

Declarations

Author contribution statement

V.H. Krisnawati: Conceived and designed the analysis; Analyzed and interpreted the data; Wrote the paper.

A.A.G. Ngurah: Analyzed and interpreted the data; Contributed analysis tools or data; Wrote the paper.

N. Hidayat, A.R. Alghofari: Conceived and designed the analysis; Contributed analysis tools or data.

Funding statement

This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

Declaration of interests statement

The authors declare no conflict of interest.

Additional information

No additional information is available for this paper.