On (super) edge-magic deficiency of some classes of graphs

Anak Agung Gede Ngurah^∗a , Rinovia Simanjuntakb,c, Edy Tri Baskorob,c

Jalan Terusan Raya Dieng 62–64 Malang, Indonesia

aag.ngurah@unmer.ac.id, {rino, ebaskoro}@itb.ac.id

A graph G of order p and size q is called edge-magic total if there exists a bijection ϕ from V (G)∪ E(G) to the set {1, 2, . . . , p+q} such that ϕ(s) +ϕ(st) +ϕ(t) is a constant for every edge st in E(G). An edge-magic total graph with ϕ(V (G)) = {1, 2, . . . , p} is called super edge-magic total. Furthermore, the edge-magic deficiency of a graph G is the smallest integer n ≥ 0 such that G ∪ nK¹ is edge-magic total. The super edge-magic deficiency of a graph G is either the smallest integer n ≥ 0 such that G ∪ nK¹ is super edge-magic total or +∞ if there exists no such integer n. In this paper, we study the (super) edge-magic deficiency of join product graphs and 2-regular graphs.

Keywords: (super) edge-magic graph, (super) edge magic deficiency, join product graph, 2-regular graph

Mathematics Subject Classification : 05C78

DOI: 10.5614/ejgta.2025.13.1.4

1. Introduction

Let G be a finite and simple graph with vertex set V (G) and edge set E(G) such that p = |V (G)| and q = |E(G)|. An edge-magic total labeling (EMTL) of a graph G is a bijection ϕ : V (G) ∪ E(G) → {1, 2, · · · , p + q} such that ϕ(s) + ϕ(st) + ϕ(t) is a constant k, called the magic

Received: 9 July 2019, Revised: 27 January 2025, Accepted: 2 February 2025.

^aDepartment of Civil Engineering, Universitas Merdeka Malang

^bCombinatorial Mathematics Research Group, Faculty of Mathematics and Natural Sciences,

Institut Teknologi Bandung, Indonesia ^cCenter for Reseach Collaboration on Graph Theory and Combinatorics, Indonesia

^∗corresponding author

constant of ϕ, for every st ∈ E(G). An EMTL ϕ of G is called a super edge-magic total labeling (SEMTL) if ϕ(V (G)) = {1, 2, · · · , p}. A graph that admits a (S)EMTL is called (super) edgemagic total ((S)EMT). The notions of an EMTL and an EMT graph were introduced in [12] while the concepts of a SEMTL and a SEMT graph were introduced in [3].

The next lemma provides necessary and sufficient conditions for a graph to be a SEMT graph.

Lemma 1.1. [4] A graph G is SEMT if and only if there exists a bijection ϕ : V (G) → {1, 2, · · · , p} such that the set S = {ϕ(s) + ϕ(t) : st ∈ E(G)} is a set of q consecutive integers.

In [12], Kotzig and Rosa also introduced the notion of edge-magic deficiency (EMD) of a graph. The EMD of a graph G, µ(G), is the minimum integer n ≥ 0 such that G ∪ nK¹ is an EMT graph. Kotzig and Rosa proved that every graph has finite EMD. Figueroa-Centeno et al. [5] introduced the notion of super edge-magic deficiency (SEMD) of a graph. The SEMD of a graph G, µs(G), is defined as either the minimum integer n ≥ 0 such that G ∪ nK¹ is a SEMT graph or +∞ if there exists no such integer n. Thus, a (S)EMT graph is a graph with zero (S)EMD. Unlike the EMD, not all graphs have finite SEMD. Lemma 1.2 provides necessary conditions for a graph to have infinite SEMD.

Lemma 1.2. [5] If G is a graph with q ≡ 2 (mod 4) edges and every vertex of G has even degrees, then µs(G) = +∞.

Several papers dealing with (S)EMD of graphs for instants see [1, 14, 15, 16, 17, 18]. The latest developments in these labelings can be found in [7]. Here, we study the (S)EMD of join product graphs and 2-regular graphs.

2. SEMD of Join Product Graphs

The join product of two graphs G and H, G + H, is a graph having vertex set V (G) ∪ V (H) and edge set E(G) ∪ E(H) ∪ {st : s ∈ V (G), t ∈ V (H)}. If H is an isolated vertex, then it is denoted by G+K1. To present the results on SEMD of join product graphs, we need the following lemma. First, we define a star Sⁿ as a graph with n + 1 vertices where one vertex has degree n and n vertices have degree one.

Lemma 2.1. Let G be a graph with p ≥ 8 vertices and q = 2p − 3 edges. If µs(G) = 0, then 2C3, C³ ∪ S³ or 2S³ are subgraphs of G.

Proof. Let ϕ be a SEMTL of G. Since q = 2p − 3, then {ϕ(s) + ϕ(t) : st ∈ E(G)} is a unique set S = {3, 4, 5, . . . 2p − 1}. To get 3, 4, and 5 in S, vertices with labels 1, 2, and 3 should form a cycle C³ or vertices with labels 1, 2, 3, and 4 should form a star S3, where vertices with labels 2, 3, and 4 are adjacent to the vertex with label 1. In a similar way, vertices with labels p, p − 1, and p − 2 should form a cycle C³ or vertices with labels p, p − 1, p − 2 and p − 3 should form a star S3. Since p ≥ 8, so 1, 2, 3, 4, p − 3, p − 2, p − 1 and p are distinct integers. Thus, we have desire results.

Lemma 2 is not true in reverse. Let's look at graph G in Figure 1 as an example. This graph has p = 8 vertices, q = 2p−3 = 13 edges, and having subgraphs isomorphic to 2C3, C3∪S3, and 2S3. Suppose µs(G) = 0 with a SEMTL ϕ. Then, {ϕ(u) + ϕ(v) : uv ∈ E(G)} = {3, 4, 5, . . . , 15}. So,

\[5\phi(z) + \phi(v_1) + \phi(v_2) + \phi(v_3) + 2\phi(u) = \sum_{i=3}^{15} i - 2\sum_{i=1}^{8} i = 45.\]

It can be verified that we do not get a SEMTL of G for any solutions of this equation. Thus, µs(G) ≥ 1. By this fact and the labeling of G ∪ K¹ in Figure 1, we conclude that µs(G) = 1.

Figure 1. A grap G with 8 vertices and 13 edges having subgraphs 2C3, C³ ∪ S3, and 2S³ and a vertex labeling of G ∪ K1.

In [19], Ngurah and Simanjuntak proved the following lemma. They also showed that the lemma is attainable by some classes of trees and forests.

Lemma 2.2. [19] Let G be a graph without cycle and isolated vertices. If the SEMD of G + K¹ is zero, then G is a tree or a forest.

We now relax the condition of the Lemma 2.2 as in Lemma 2.3. The proof of Lemma 2.3 is identical to the proof of Lemma 2.2.

Lemma 2.3. Let G be a graph without isolated vertices. If µs(G + K1) = 0, then G is a tree, a forest, a union of cycles and trees, or a union of unicyclic graphs and trees.

The corona product, G ⊙ H, of two graphs G and H is defined as the graph formed by taking one copy of G and |V (G)| copies of H, then connecting the ith vertex of G to every vertex in the ith copy of H. If H is an isolated vertex, then it is denoted by G ⊙ K1. The next results show that Lemma 2.3 is attainable.

Theorem 2.1. a) µs([Cⁿ ∪ Pm] + K1) = 0 if and only if n = 3 and 2 ≤ m ≤ 4. b) µs([(C³ ⊙ K1) ∪ Pm] + K1) = 0 if and only if 2 ≤ m ≤ 4.

Proof. a) First, let Fn,m = [Cⁿ ∪ Pm] + K¹ be a graph with V (Fn,m) = {xⁱ : 1 ≤ i ≤ n} ∪ {y^j : 1 ≤ j ≤ m} ∪ {z} and E(Fn,m) = {xixi+1 : 1 ≤ i ≤ n − 1} ∪ {xnx1} ∪ {yjyj+1 : 1 ≤ j ≤ m − 1} ∪ {zxⁱ , zy^j : 1 ≤ i ≤ n, 1 ≤ j ≤ m}. Thus, Fn,m has p = n + m + 1 vertices and q = 2n + 2m − 1 = 2(n + m + 1) − 3 edges.

Next, we show that, for m ≥ 5, µs(F3,m) > 0. For m ≥ 5, suppose that µs(F3,m) = 0 with a SEMTL ϕ. Then, S = {ϕ(u) + ϕ(v) : uv ∈ E(F3,m)} = {3, 4, 5, . . . , 2m + 7}. Notice that F3,m has no subgraphs isomorphic to 2S3. By Lemma 2.1, we should consider the following cases.

Case 1. Vertices with labels 1, 2, 3 and vertices with label p = m + 4, p - 1, p - 2 form a \(2C_3\). If \(\{\phi(x_1), \phi(x_2), \phi(x_3)\} = \{1, 2, 3\}\) and \(\phi(z) \in \{p, p - 1, p - 2\}\) then, we fail to get 6 in S. If \(\{\phi(x_1), \phi(x_2), \phi(x_3)\} = \{p, p - 1, p - 2\}\) and \(\phi(z) \in \{1, 2, 3\}\) then, we fail to get 2p - 4 in S.

Case 2. Vertices with labels 1, 2, 3 form a \(C_3\) and vertices with labels p, p-1, p-2, p-3 form a \(S_3\).

In this case, it is easy to check that we fail to get 6 in S.

Case 3. Vertices with labels p, p-1, p-2, form a \(C_3\) and vertices with label 1, 2, 3, 4 form a \(S_3\).

In this case, we fail to get 2p-4 in S. Thus, for \(m \ge 5\), \(\mu_s(F_{3,m}) > 0\).

Now, we show that, for any \(n \geq 4\) and \(m \geq 2\), \(\mu_s(F_{n,m}) > 0\). Since, for \(n \geq 4\) and m > 2, \(F_{n,m}\) is a graph with \(|V(F_{n,m})| \geq 8\) and has no subgraphs isomorphic to \(2C_3\), \(C_3 \cup S_3\) and \(2S_3\), by Lemma 2.1, \(\mu_s(F_{n,m}) > 0\). Thus, the remaining case is to show that \(\mu_s(F_{4,2}) > 0\). Let \(\mu_s(F_{4,2}) = 0\) and let \(\phi\) be a SEMTL of \(F_{4,2}\). Then, \(4\phi(z) + \sum_{i=1}^4 \phi(x_i) = 32\). It is simple to confirm that none of the equation's solutions result in a SEMTL of \(F_{4,2}\). Thus, \(\mu_s(F_{4,2}) > 0\).

Finally, we show that, for \(2 \le m \le 4\), \(\mu_s(F_{3,m}) = 0\). Define a vertex labeling \(\phi\) as follows. \(\{\phi(x_1), \phi(x_2), \phi(x_3)\} = \{1, 2, 3\}\), \(\phi(z) = 5\), for m = 2, set \(\{\phi(y_1), \phi(y_2)\}\) to \(\{4, 6\}\), for m = 3, set \(\{\phi(y_1), \phi(y_2), \phi(y_3)\}\) to \(\{4, 6, 7\}\), and for m = 4, set \(\{\phi(y_1), \phi(y_2), \phi(y_3), \phi(y_4)\}\) to \(\{4, 6, 8, 7\}\).

b) Let \(H_m = [(C_3 \odot K_1) \cup P_m] + K_1\) be a graph with \(V(H_m) = \{u_i, v_i : 1 \le i \le 3\} \cup \{w_j : 1 \le j \le m\} \cup \{z\}\) and \(E(H_m) = \{u_i u_{i+1} : 1 \le i \le 2\} \cup \{u_3 u_1\} \cup \{u_i v_i : 1 \le i \le 3\} \cup \{w_j w_{j+1} : 1 \le j \le m-1\} \cup \{zu_i, zv_i, zw_j : 1 \le i \le 3, 1 \le j \le m\}\). Thus \(H_m\) has p = m+7 vertices and q = 2m+11 = 2(m+7)-3 edges.

First, we prove that, for \(2 \le m \le 4\), \(\mu_s([(C_3 \odot K_1) \cup P_m] + K_1) = 0\). Define a vertex labeling \(\phi\) as follows. \((\phi(u_1), \phi(u_2), \phi(u_3)) = (1, 2, 3)\), \((\phi(v_1), \phi(v_2), \phi(v_3)) = (5, 6, 4)\), \(\phi(z) = 8\), for m = 2, set \(\{\phi(w_1), \phi(w_2)\}\) to \(\{7, 9\}\), for m = 3, set \((\phi(w_1), \phi(w_2), \phi(w_3))\) to (7, 9, 10), and for m = 4, set \((\phi(w_1), \phi(w_2), \phi(w_3), \phi(w_4))\) to (7, 9, 11, 10). It is simple to confirm that \(\phi\) extends to a SEMTL of \([(C_3 \odot K_1) \cup P_m] + K_1\) for \(2 \le m \le 4\).

Next, suppose that, for any \(m \geq 5\), \(\mu_s(H_m) = 0\). Then, there exists a SEMTL \(\phi\) of \(H_m\) such that \(S = \{\phi(u) + \phi(v) : uv \in E(H_m)\} = \{3, 4, 5, \dots, 2p-1\}\), in which we note that 2p-1=2m+13. By Lemma 2.1, we should consider three cases.

Case 1. Vertices with labels 1, 2, 3 and vertices with labels p, p-1, p-2 form a \(2C_3\). If \((\phi(u_1), \phi(u_2), \phi(u_3)) = (1, 2, 3)\) and \(\phi(z) \in \{p, p-1, p-2\}\) then, to get 6, 7, and 8 in S, set \((\phi(v_1), \phi(v_2), \phi(v_3))\) to (5, 6, 4) or (6, 4, 5). Thus, we fail to get 9 in S. If \((\phi(u_1), \phi(u_2), \phi(u_3)) = (p, p-1, p-2)\) and \(\phi(z) \in \{1, 2, 3\}\) then, to get 2p-4, 2p-5, and 2p-6, set \((\phi(v_1), \phi(v_2), \phi(v_3))\) to (p-5, p-3, p-4) or (p-4, p-5, p-3). In this case, it is not possible to get 2p-7 in S.

Case 2. Vertices with labels 1, 2, 3 form a \(C_3\) and vertices with labels p, p-1, p-2, p-3 form a \(S_3\).

If \((\phi(u_1), \phi(u_2), \phi(u_3)) = (1, 2, 3)\) and \(\phi(z) \in \{p, p-1, p-2, p-3\}\) then, by a similar argument as in the Case 1, we fail to get 9 in S. If \(\phi(z) \in \{1, 2, 3\}\) and \((\phi(u_1), \phi(u_2), \phi(u_3), \phi(v_1)) = (p, p-2, p-3, p-1)\), then we fail to get 2p-4 in S. If \(\phi(z) \in \{1, 2, 3\}\) and \((\phi(u_1), \phi(u_2), \phi(u_3), \phi(v_1)) = (p, p-1, p-3, p-2)\), then we fail to get either 2p-5 or 2p-6 in S.

Case 3. Vertices with labels p, p-1, p-2 form a \(C_3\) and vertices with labels 1, 2, 3, 4 form a \(S_3\).

By a similar argument as in the Case 2, this case also do not lead to a SEMTL of \(H_m\).

Case 4. Vertices with labels 1, 2, 3, 4 and vertices with labels p, p-1, p-2, p-3 form a \(2S_3\). If \(\phi(z) \in \{1, 2, 3, 4\}\) and \((\phi(u_1), \phi(u_2), \phi(u_3), \phi(v_1)) = (p, p-2, p-3, p-1)\), then we fail to get 2p-4 in S. If \(\phi(z) \in \{1, 2, 3, 4\}\) and \((\phi(u_1), \phi(u_2), \phi(u_3), \phi(v_1)) = (p, p-1, p-3, p-2)\), then we fail to get either 2p-5 or 2p-6 in S. If \((\phi(u_1), \phi(u_2), \phi(u_3), \phi(v_1)) = (1, 3, 4, 2)\) and \(\phi(z) \in \{p, p-1, p-2, p-3\}\) then, it is not possible to get 6 in S. If \((\phi(u_1), \phi(u_2), \phi(u_3), \phi(v_1)) = (1, 2, 4, 3)\) and \(\phi(z) \in \{p, p-1, p-2, p-3\}\) then, it is not possible to get either 7 or 8 in S. Thus, \(\mu_s(H_m) > 0\), if \(m \ge 5\). This complete the proof of part b).

3. (S)EMD of 2-Regular Graphs

In [8], Holden et al. proved that \(C_5 \cup (2t)C_3\), \(C_4 \cup (2t-1)C_3\), and \(C_7 \cup (2t)C_3\) are strong vertex-magic total (SVMT) graphs. It is easy to verify that a (S)VMT 2-regular graph is equivalent to a (S)EMT 2-regular graph. Based on these results, they posed the Conjecture 1.

Conjecture 1. [8]. Let G be a 2-regular graph of odd order. The graph G is SVMT if and only if G is not one of \(C_4 \cup C_3\), \(C_4 \cup 3C_3\) or \(C_5 \cup 2C_3\).

Cichacz et al. [2] provide a partial solution to Conjecture 1, introducing a method for generating (S)VMT labelings of 2-regular graphs.

Theorem 3.1. [2]. Let \(k \geq 1\) be an integer and \(F = \bigcup_{i=1}^k C_{n_i}\) be a 2-regular graph. If F is a (S)VMT graph, then \(G = \bigcup_{i=1}^k C_{mn_i}\) is a (S)VMT graph for every odd \(m \geq 3\).

In [6], Figueroa-Centeno et al. provided the following result.

Theorem 3.2. [6]. If a 3-colorable graph G is (S)EMT, then mG is (S)EMT for any odd integer m.

As mentioning in [7], Ichishima and Oshima [11] investigate the SEMD of 2-regular graphs \(C_m \cup C_n\) for m = 3, 4, 5, 7 and any n. Krisnawati et al. [13] investigate the SEMD of a 2-regular graph with three components, namely \(2C_3 \cup C_n\). In the next theorem, we study the SEMD of a 2-regular graph \(2C_4 \cup C_n\).

Theorem 3.3. Let \(n \ge 4\) be a positive integer. Then

\[\mu_s(2C_4 \cup C_n) = \begin{cases} 0, & \text{if } n \equiv 1 \pmod{2}, \\ 1, & \text{if } n = 4 \text{ and } n \equiv 8, 12 \pmod{16}, \\ +\infty, & \text{if } n \equiv 2 \pmod{4}, \end{cases}\]

and \(\mu_s(2C_4 \cup C_n) \le 2\), if \(4 < n \equiv 0, 4 \pmod{16}\).

Proof. First, let \(H = 2C_4 \cup C_n\) be a graph with \(V(H) = \{u_i, v_i : 1 \le i \le 4\} \cup \{w_i : 1 \le i \le n\}\) and \(E(H) = \{u_i u_{i+1}, v_i v_{i+1} : 1 \le i \le 3\} \cup \{u_4 u_1, v_4 v_1\} \cup \{w_i w_{i+1} : 1 \le i \le n-1\} \cup \{w_n w_1\}.\)

Next, we show that the graph H is SEMT if and only if n is odd. If H is a SEMT graph then its magic constant is \(\frac{1}{2}(5n+43)\). So, n should be an odd integer. Next, for odd \(n \geq 5\) define \(\phi: V(H) \longrightarrow \{1, 2, \ldots, n+8\}\) as follows. \((\phi(u_1), \phi(u_2), \phi(u_3), \phi(u_4)) = (1, \frac{1}{2}(n+1))\)

9), 2, \(\frac{1}{2}(n+13)\)), \((\phi(v_1), \phi(v_2), \phi(v_3), \phi(v_4)) = (3, \frac{1}{2}(n+15), 5, \frac{1}{2}(n+17))\). When n=7, label \((\phi(w_1), \phi(w_2), \phi(w_3), \phi(w_4), \phi(w_5), \phi(w_6), \phi(w_7)\) with (6, 7, 13, 9, 14, 4, 15). When \(n \geq 9\), define \(\phi(w_1) = 4, \phi(w_2) = \frac{1}{2}(n+11), \phi(w_3) = n+7, \phi(w_4) = \frac{1}{2}(n+7),\) and label the remaining vertices as follows.

Case 1. \(n \equiv 1 \pmod{4}\).

\[\phi(w_i) = \begin{cases} \frac{1}{2}(2n+21-i), & \text{if } i = 5, 9, 13, \dots, n, \\ \frac{1}{2}(n+9-i), & \text{if } i = 6, 10, 14, \dots, n-3, \\ \frac{1}{2}(2n+17-i), & \text{if } i = 7, 11, 15, \dots, n-2, \\ \frac{1}{2}(n+13-i), & \text{if } i = 8, 12, 16, \dots, n-1. \end{cases}\]

Case 2. \(7 < n \equiv 3 \pmod{4}\).

\[\phi(w_i) = \begin{cases} n+8, & \text{if } i=5, \\ n+6, & \text{if } i=7, \\ 6, & \text{if } i=n-1, \\ \frac{1}{2}(n+9-i), & \text{if } i=6,10,14,\ldots,n-5, \\ \frac{1}{2}(n+13-i), & \text{if } i=8,12,16,\ldots,n-3, \\ \frac{1}{2}(2n+17-i), & \text{if } i=9,13,17,\ldots,n-2, \\ \frac{1}{4}(2n+21-i), & \text{if } i=11,15,19,\ldots,n. \end{cases}\]

It can be checked that \(\{\phi(x) + \phi(y) : xy \in E(H)\} = \{\frac{1}{2}(n+11), \frac{1}{2}(n+13), \dots, \frac{1}{2}(3n+25)\}.\)By Lemma 1.1, H is a SEMT graph.

Now, we prove that, for n=4 and \(n\equiv 8,12 \pmod{16}\), \(\mu_s(H)=1\). If n=4, \(\mu_s(3C_4)=1\)have been proved in [10] (see Corollary 3.6). Next, for \(n \equiv 8,12 \pmod{16}\), consider the graph \(H \cup K_1\) and define \(\phi: V(H \cup K_1) \longrightarrow \{1, 2, \dots, n+9\}\) as follows. \((\phi(u_1), \phi(u_2), \phi(u_3), \phi(u_4)) =\)\((1, \frac{1}{2}(n+10), 2, \frac{1}{2}(n+14)), (\phi(v_1), \phi(v_2), \phi(v_3), \phi(v_4)) = (4, \frac{1}{2}(n+12), 5, \frac{1}{2}(n+16)), \phi(w_1) = 3, \phi(w_2) = \frac{1}{2}(n+24), \phi(w_i) = \frac{1}{2}(i+9) \text{ for } i = 3, 5, 7, \dots, n-1, \phi(w_n) = \frac{1}{2}(n+22), \text{ and}\)\(\phi(K_1) = \frac{1}{4}(3n + 28).\)

Case 3. \(n \equiv 8 \pmod{16}\).

\[\phi(w_i) = \begin{cases} \frac{1}{2}(n+20), & \text{if } i = 4, \\ \frac{1}{2}(n+22+i), & \text{if } i = 6, 14, 22, \dots, \frac{n}{2} - 6, \\ & \cup i = 8, 16, 24, \dots, \frac{n}{2} - 4, \\ & \cup i = 10, 18, 26, \dots, \frac{n}{2} - 2, \\ \frac{1}{2}(n+6+i), & \text{if } i = 12, 20, 28, \dots, \frac{n}{2}, \\ \frac{1}{2}(n+20+i), & \text{if } i = \frac{n}{2} + 2, \frac{n}{2} + 4, \frac{n}{2} + 6, \dots, n-2. \end{cases}\]

Case 4. \(n \equiv 12 \pmod{16}\).

\[\phi(w_i) = \begin{cases} \frac{1}{2}(n+26), & \text{if } i = 4, \\ \frac{1}{2}(n+18), & \text{if } i = 6, \\ \frac{1}{2}(n+22+i), & \text{if } i = 8, 16, 24, \dots, \frac{n}{2} - 6, \\ & \cup i = 10, 18, 26, \dots, \frac{n}{2} - 4, \\ & \cup i = 12, 20, 28, \dots, \frac{n}{2} - 2, \\ \frac{1}{2}(n+6+i), & \text{if } i = 14, 22, 30, \dots, \frac{n}{2}, \\ \frac{1}{2}(n+20+i), & \text{if } i = \frac{n}{2} + 2, \frac{n}{2} + 4, \frac{n}{2} + 6, \dots, n-2. \end{cases}\]

It is simple to confirm that, by Lemma 1.1, \(H \cup K_1\) a SEMT graph. Hence, \(\mu_s(H) = 1\) for n = 4 and \(n \equiv 8, 12 \pmod{16}\).

Next, we show that, or \(4 < n \equiv 0, 4 \pmod{16}, \mu_s(H) \leq 2\). Let us consider the graph \(H \cup 2K_1\) and define \(\phi: V(H \cup 2K_1) \longrightarrow \{1, 2, \dots, n+10\}\) as follows. \((\phi(u_1), \phi(u_2), \phi(u_3), \phi(u_4)) = (1, \frac{1}{2}(n+10), 2, \frac{1}{2}(n+14), (\phi(v_1), \phi(v_2), \phi(v_3), \phi(v_4)) = (4, \frac{1}{2}(n+12), 5, \frac{1}{2}(n+16)), \phi(w_1) = 3, \phi(w_2) = \frac{1}{2}(n+24), \phi(w_4) = \frac{1}{2}(n+26), \phi(w_6) = \frac{1}{2}(n+18), \phi(w_i) = \frac{1}{2}(i+9), i = 3, 5, 7, \dots, n-1,\) and \(\phi(w_n) = \frac{1}{2}(n+22)\).

Case 1. \(n \equiv 0 \pmod{16}\).

\[\phi(\{2K_1\}) = \{\frac{1}{4}(3n+40), n+7\}\]

and

\[\phi(w_i) = \begin{cases} \frac{1}{2}(n+22+i), & \text{if } i = 8, 16, 24, \dots, \frac{n}{2} - 8, \\ & \cup i = 10, 18, 26, \dots, \frac{n}{2} - 6, \\ & \cup i = 12, 20, 28, \dots, \frac{n}{2} - 4, \end{cases}\] \[\text{[rumus tidak dapat ditampilkan dengan baik — lihat PDF asli]}\]

Case 2. \(4 < n \equiv 4 \pmod{16}\).

\[\phi(\{2K_1\}) = \{\frac{1}{4}(3n+36), n+8\}\]

and

\[f(w_i) = \begin{cases} \frac{1}{2}(n+22+i), & \text{if } i = 8, 16, 24, \dots, \frac{n}{2} - 2, \\ & \cup i = 10, 18, 26, \dots, \frac{n}{2} - 8, \\ & \cup i = 12, 20, 28, \dots, \frac{n}{2} - 6, \\ \frac{1}{4}(3n+48), & \text{if } i = \frac{n}{2}, \\ \frac{1}{4}(3n+52), & \text{if } i = \frac{n}{2} + 2, \\ \frac{1}{2}(n+6+i), & \text{if } i = 14, 22, 30, \dots, \frac{n}{2} + 4, \end{cases}\]

The labeling of \(w_i\) when i is even and \(i \ge \frac{n}{2} + 6\) is determined by the following subcases. Subcase 2.1. \(4 < n \equiv 4 \pmod{48}\).

\[\phi(w_i) = \begin{cases} \frac{1}{2}(n+24+i), & \text{if } i = \frac{n}{2}+6, \frac{n}{2}+12, \frac{n}{2}+18, \dots, n-14, \\ & \cup i = \frac{n}{2}+8, \frac{n}{2}+14, \frac{n}{2}+20, \dots, n-12, \\ n+10, & \text{if } i = n-8, \\ n+7, & \text{if } i = n-6, \\ n+9, & \text{if } i = n-2, \\ \frac{1}{2}(n+12+i), & \text{if } i = \frac{n}{2}+10, \frac{n}{2}+16, \frac{n}{2}+22, \dots, n-4. \end{cases}\]

Subcase 2.2. \(n \equiv 20 \pmod{48}\).

When n=20, define \(\phi(w_{16})=26\) and \(\phi(w_{18})=29\). When \(n\geq 68\), label the remaining vertices by the following formula.

\[\phi(w_i) = \begin{cases} \frac{1}{2}(n+24+i), & \text{if } i = \frac{n}{2}+6, \frac{n}{2}+12, \frac{n}{2}+18, \dots, n-16, \\ & \cup i = \frac{n}{2}+8, \frac{n}{2}+14, \frac{n}{2}+20, \dots, n-14, \\ n+9, & \text{if } i = n-10, \\ n+6, & \text{if } i = n-8, \\ n+10, & \text{if } i = n-4, \\ n+7, & \text{if } i = n-2, \\ \frac{1}{2}(n+12+i), & \text{if } i = \frac{n}{2}+10, \frac{n}{2}+16, \frac{n}{2}+22, \dots, n-6. \end{cases}\]

Subcase 2.3. Subcase \(n \equiv 36 \pmod{48}\).

\[\phi(w_i) = \begin{cases} \frac{1}{2}(n+24+i), & \text{if } i = \frac{n}{2}+6, \frac{n}{2}+12, \frac{n}{2}+18, \dots, n-6, \\ & \cup i = \frac{n}{2}+8, \frac{n}{2}+14, \frac{n}{2}+20, \dots, n-4, \\ \frac{1}{2}(n+12+i), & \text{if } i = \frac{n}{2}+10, \frac{n}{2}+16, \frac{n}{2}+22, \dots, n-2. \end{cases}\]

By Lemma 1.1, \(H \cup 2K_1\) is a SEMT graph. Thus, \(\mu_s(H) \le 2\) for \(n \equiv 0, 4 \pmod{16}\). Finally, as a direct consequence of Lemma 1.2, \(\mu_s(H) = +\infty\), if \(n \equiv 2 \pmod{4}\).

The following corollary is a direct consequence of applying Theorem 3.2 to the Theorem 3.3.

Corollary 3.1. Let \(l \equiv 2 \pmod{4}\), \(m \equiv 1 \pmod{2}\) and \(n \geq 4\) be positive integers. Then

\[\mu_s(lC_4 \cup mC_n) = \begin{cases} 0, & \text{if } n \equiv 1 \text{ (mod 2)}, \\ +\infty, & \text{if } n \equiv 2 \text{ (mod 4)}, \end{cases}\]

\(\mu_s(lC_4 \cup mC_n) \le m\), if n = 4 and \(n \equiv 8, 12 \pmod{16}\), and \(\mu_s(lC_4 \cup mC_n) \le 2m\), if \(4 < n \equiv 0, 4 \pmod{16}\).

The following result is obtained by applying Theorems 3.1 and 3.2 to the facts that \(\mu_s(2C_4 \cup C_n) = 0\) for \(n \equiv 1 \pmod 2\) and \(\mu_s(2C_4 \cup C_n) = +\infty\) for \(n \equiv 2 \pmod 4\).

Corollary 3.2. Let \(l \equiv 2 \pmod{4}\), \(m \equiv 1 \pmod{2}\), \(n \geq 4\) and r be positive integers. Then

\[\mu_s(lC_{4(2r+1)} \cup mC_{n(2r+1)}) = \begin{cases} 0, & \text{if } n \equiv 1 \text{ (mod 2)}, \\ +\infty, & \text{if } n \equiv 2 \text{ (mod 4)}, \end{cases}\]

Unresolved issues related to these results are detailed as follows.

Open Problem 1. Determine the exact value of \(\mu_s(2C_4 \cup C_n)\) when \(n \equiv 0, 4\) (mod 16). Further, for \(l \equiv 2 \pmod{4}\) and \(m \equiv 1 \pmod{2}\), find the upper bound of the \(\mu_s(lC_{4(2r+1)} \cup mC_{n(2r+1)})\) when \(n \equiv 0 \pmod{4}\).

In [2], Chicacz et al. prove that \(2C_n\) is VMT for \(n \in \{4,6,8,10\}\) and posed a conjecture that, for even value of n, the graph \(2C_n\) is VMT. We able to give support of the validity of this conjecture by proving \(2C_n\) is EMT for \(n \in \{5,7,9,11,12,13,14\}\), see Figure 2. Theorem 3.4 is a direct consequence of applying Theorems 3.1 and 3.2 to these results.

Figure 2. An EMTL of 2Cⁿ for n ∈ {5, 7, 9, 11, 12, 13, 14}.

Theorem 3.4. Let m ≡ 2(mod 4), r be positive integers and n ∈ {5, 7, 9, 11, 12, 13, 14}. Then mCn(2r+1) has zero EMD.

Based on Theorem 3.4, the following conjecture is proposed.

Conjecture 2. For any integers m ≡ 2 (mod 4), n ≥ 3, and r ≥ 1, the 2-regular graph mCn(2r+1) has zero EMD.

A graph H ∪ kK¹ is called pseudo super edge-magic (PSEM) if there exists a bijection ϕ : V (H ∪kK1) → {1, 2, . . . , |V (H)|+k} such that {ϕ(s)+ϕ(t) : st ∈ E(H)}∪ {2ϕ(u) : u ∈ kK1} is a set of |E(H)| + k consecutive integers. In such a case ϕ is called a pseudo super edge-magic labeling (PSEML) of H ∪ kK1. These concepts was introduced in [9]. It is easy to verify that if H is a 2-regular graph and H ∪ kK¹ is a PSEM graph, then |V (H)| + k should be an odd integer. We discover that Theorem 3.1 also valid for a PSEM graph H ∪ kK1. As an example, see Figure 3. This fact is not mentioned by the authors in [2].

Theorem 3.5. For any integer n ≥ 4, the following graphs are PSEM.

• a) Cⁿ ∪ K1, if n ≡ 0 (mod 2).
• b) Cⁿ ∪ 2K1, if n ≡ 1 (mod 2).
• c) C⁴ ∪ Cⁿ ∪ K1, if n ≡ 0 (mod 4).
• d) C⁴ ∪ Cⁿ ∪ K1, if n ∈ {6, 10, 14, 18, 22, 26}.
• e) 2C⁴ ∪ Cⁿ ∪ K1, if n ≡ 8, 12 (mod 16).

Proof. a) In [5], Ichishima et al. proved that \(\mu_s(C_n) = 1\), if \(n \equiv 0 \pmod 4\). The labeling defined to prove this result actually is a PSEML of \(C_n \cup K_1\). Next, for \(n \equiv 2 \pmod 4\), let \(V(C_n \cup K_1) = \{u_i : 1 \le i \le n\} \cup \{v\}\) and \(E(C_n \cup K_1) = \{u_i u_{i+1} : 1 \le i \le n-1\} \cup \{u_n u_1\}\). Define a vertex labeling \(\phi : V(C_n \cup K_1) \to \{1, 2, \dots, n+1\}\) as follows. When n = 6, 10 and 14, set \((\phi(u_1), \phi(u_2), \dots, \phi(u_6); \phi(v))\) to (1, 7, 2, 5, 6, 4; 3), \((\phi(u_1), \phi(u_2), \dots, \phi(u_{10}); \phi(v))\) to (1, 8, 2, 9, 3, 11, 5, 10, 7, 6; 4), and \((\phi(u_1), \phi(u_2), \dots, \phi(u_{14}); \phi(v))\) to (1, 10, 2, 12, 3, 13, 4, 9, 14, 6, 15, 7, 11, 8, 1; 5), respectively. When \(n \ge 18\), define

\[\phi(v) = \frac{1}{4}(n+6)\]

and

\[\phi(u_i) = \begin{cases} \frac{1}{2}(1+i), & \text{if } i = 1, 3, 5, \dots, \frac{n}{2}, \\ \frac{1}{4}(n+10), & \text{if } i = \frac{2}{2} + 2, \\ \frac{1}{2}(n+4+i), & \text{if } i = 2, 4, 6, \dots, \frac{1}{2}(n-2). \end{cases}\]

To label the remaining vertices, we consider the following two cases.

Case 1. \(n \equiv 2 \pmod{8}\).

\[\phi(u_i) = \begin{cases} \frac{1}{2}(5+i), & \text{if } i = \frac{n}{2} + 4, \frac{n}{2} + 8, \frac{n}{2} + 12, \dots, n-1, \\ \frac{1}{2}(1+i), & \text{if } i = \frac{n}{2} + 6, \frac{n}{2} + 10, \frac{n}{2} + 14, \dots, n-3, \\ \frac{1}{2}(n+6+i), & \text{if } i = \frac{n}{2} + 1, \frac{n}{2} + 5, \frac{n}{2} + 9, \dots, n-4, \\ \frac{1}{2}(n+2+i), & \text{if } i = \frac{n}{2} + 3, \frac{n}{2} + 7, \frac{n}{2} + 11, \dots, n-2, \\ \frac{1}{2}(n+2), & \text{if } i = n. \end{cases}\]

Case 2. \(n \equiv 6 \pmod{8}\).

\[\phi(u_i) = \begin{cases} \frac{1}{2}(5+i), & \text{if } i = \frac{n}{2}+4, \frac{n}{2}+8, \frac{n}{2}+12, \dots, n-7, \\ \frac{1}{2}(1+i), & \text{if } i = \frac{n}{2}+6, \frac{n}{2}+10, \frac{n}{2}+14, \dots, n-5, \\ \frac{n}{2}, & \text{if } i = n-3, \\ \frac{1}{2}(n+4), & \text{if } i = n-1, \\ \frac{1}{2}(n+6+i), & \text{if } i = \frac{n}{2}+1, \frac{n}{2}+5, \frac{n}{2}+9, \dots, n-10, \\ \frac{1}{2}(n+2+i), & \text{if } i = \frac{n}{2}+3, \frac{n}{2}+7, \frac{n}{2}+11\dots, n-8, \\ n-1, & \text{if } i = n-6, \\ n+1, & \text{if } i = n-4, \\ n, & \text{if } i = n-2, \\ \frac{1}{2}(n+2), & \text{if } i = n. \end{cases}\]

It can be checked that \(\{\phi(u_i) + \phi(u_{i+1}) : 1 \le i \le n-1\} \cup \{\phi(u_n) + \phi(u_1)\} \cup \{2\phi(v)\}\) is \(\{\frac{1}{2}(n+4), \frac{1}{2}(n+6), \dots \frac{1}{2}(3n+4)\}\). Hence, \(C_n \cup K_1\) is a PSEM graph.

b) For odd n, let \(V(C_n \cup 2K_1) = \{u_i : 1 \le i \le n\} \cup \{v, w\}\) and \(E(C_n \cup 2K_1) = \{u_i u_{i+1} : 1 \le i \le n-1\} \cup \{u_n u_1\}\). Define a vertex labeling \(\phi\) of \(C_n \cup 2K_1\) as follows.

Case 1. \(n \equiv 1 \pmod{4}\).

\[f({v, w}) = {\frac{1}{2}(n+1), \frac{1}{4}(3n+5)}\]

and

\[\phi(u_i) = \begin{cases} \frac{1}{2}(1+i), & \text{if } i = 1, 3, 5, \dots, n-2, \\ \frac{1}{2}(n+3), & \text{if } i = n, \\ \frac{1}{2}(n+3+i), & \text{if } i = 2, 4, 6, \dots, \frac{1}{2}(n-5), \\ \frac{1}{2}(n+5+i), & \text{if } i = \frac{1}{2}(n-1), \frac{1}{2}(n+3), \frac{1}{2}(n+7), \dots, n-1. \end{cases}\]

Case 2. \(n \equiv 3 \pmod{4}\).

When n=7 and 11, define \((\phi(u_1),\phi(u_2),\ldots,\phi(u_7);\phi(v),\phi(w))=(1,7,2,9,4,8,6;3,5)\) and \((\phi(u_1),\phi(u_2),\ldots,\phi(u_{11});\phi(v),\phi(w))=(1,9,2,11,3,13,5,10,7,12,8;4,6)\), respectively. When \(n\geq 15\), define

\[\phi(\{v, w\}) = \{\frac{1}{2}(n+3), \frac{1}{4}(n+5)\}\]

and

\[\phi(u_i) = \begin{cases} \frac{1}{2}(1+i), & \text{if } i = 1, 3, 5, \dots, \frac{1}{2}(n-1), \\ \frac{1}{2}(n+5+i), & \text{if } i = 2, 4, 6, \dots, \frac{1}{2}(n-3), \\ \frac{1}{4}(n+9), & \text{if } i = \frac{1}{2}(n+3), \end{cases}\]

Subcase 2.1. \(n \equiv 3 \pmod{8}\).

\[\phi(u_i) = \begin{cases} \frac{1}{2}(n+7+i), & \text{if } i = \frac{1}{2}(n+1), \frac{1}{2}(n+9), \dots, n-9, \\ \frac{1}{2}(n+3+i), & \text{if } i = \frac{1}{2}(n+5), \frac{1}{2}(n+13), \dots, n-7, \\ n, & \text{if } i = n-5, \\ n+2, & \text{if } i = n-3, \\ n+1, & \text{if } i = n-1, \\ \frac{1}{2}(5+i), & \text{if } i = \frac{1}{2}(n+7), \frac{1}{2}(n+15), \dots, n-6, \\ \frac{1}{2}(1+i), & \text{if } i = \frac{1}{2}(n+11), \frac{1}{2}(n+19), \dots, n-4, \\ \frac{1}{2}(n+1), & \text{if } i = n-2, \\ \frac{1}{2}(n+5), & \text{if } i = n. \end{cases}\]

Subcase 2.1. \(n \equiv 7 \pmod{8}\).

\[\phi(x) = \begin{cases} \frac{1}{2}(n+7+i), & \text{if } i = \frac{1}{2}(n+1), \frac{1}{2}(n+9), \dots, n-3, \\ \frac{1}{2}(n+3+i), & \text{if } i = \frac{1}{2}(n+5), \frac{1}{2}(n+13), \dots, n-1, \\ \frac{1}{2}(5+i), & \text{if } i = \frac{1}{2}(n+7), \frac{1}{2}(n+15), \dots, n, \\ \frac{1}{2}(1+i), & \text{if } i = \frac{1}{2}(n+11), \frac{1}{2}(n+19), \dots, n-2. \end{cases}\]

It is a routine procedure to verify that \(\phi\) is a PSEML of G.

c) Let us define \(G_n \cong C_4 \cup C_n \cup K_1\), \(n \equiv 0 \pmod{4}\), as a graph having

\[V(G_n) = \{u_i : 1 \le i \le 4\} \cup \{v_i : 1 \le i \le n\} \cup \{K_1\}\]

and

\[E(G_n) = \{u_i u_{i+1} : 1 \le i \le 3\} \cup \{u_4 u_1\} \cup \{v_i v_{i+1} : 1 \le i \le n-1\} \cup \{v_n v_1\}.\]

Next, define \(\phi:V(G_n)\longrightarrow \{1,2,\ldots,n+5\}\) as follows.

Case 1. n = 4.

\[(\phi(u_1),\phi(u_2),\phi(u_3),\phi(u_4))=(1,5,3,6),\ (\phi(v_1),\phi(v_2),\phi(v_3),\phi(v_4))=(2,8,4,9),\ \phi(K_1)=7.\]

Case 2. \(n \ge 8\).

\[\phi(x) = \begin{cases} (1, \frac{1}{2}(n+6), 2, \frac{1}{2}(n+10)), & \text{if } x = (u_1, u_2, u_3, u_4), \\ \frac{1}{2}(i+7), & \text{if } x = v_i \text{ for } i = 1, 5, 9, \dots, n-3, \\ \frac{1}{2}(i+3), & \text{if } x = v_i \text{ for } i = 3, 7, 11, \dots, n-1, \\ \frac{1}{2}(n+8), & \text{if } x = v_n, \\ \frac{1}{4}(3n+16), & \text{if } x = K_1. \end{cases}\]

To label the remaining vertices, we consider the following two subcases.

Case 2.1 \(n \equiv 0 \pmod{8}\).

\[\phi(x) = \begin{cases} \frac{1}{2}(n+12+i), & \text{if } x = v_i \text{ for } i = 2, 6, 10, \dots, \frac{n}{2} - 6, \\ \frac{1}{2}(n+8+i), & \text{if } x = v_i \text{ for } i = 4, 8, 12, \dots, \frac{n}{2} - 4, \\ \frac{1}{2}(n+14+i), & \text{if } x = v_i \text{ for } i = \frac{n}{2} - 2, \frac{n}{2} + 2, \frac{n}{2} + 6, \dots, n - 6, \\ \frac{1}{2}(n+10+i), & \text{if } x = v_i \text{ for } i = \frac{n}{2}, \frac{n}{2} + 4, \frac{n}{2} + 8, \dots, n - 4, \\ n+5, & \text{if } x = v_{n-2}. \end{cases}\]

Case 2.2 \(n \equiv 4 \pmod{8}\).

\[\phi(x) = \begin{cases} \frac{1}{2}(n+12), & \text{if } x = v_i \text{ for } i = 2, \\ \frac{1}{2}(n+12+i), & \text{if } x = v_i \text{ for } i = 4, 8, 12, \dots, \frac{n}{2} - 6, \\ \frac{1}{2}(n+8+i), & \text{if } x = v_i \text{ for } i = 6, 10, 14, \dots, \frac{n}{2} - 4, \\ \frac{1}{2}(n+14+i), & \text{if } x = v_i \text{ for } i = \frac{n}{2} - 2, \frac{n}{2} + 2, \frac{n}{2} + 6, \dots, n - 4, \\ \frac{1}{2}(n+10+i), & \text{if } x = v_i \text{ for } i = \frac{n}{2}, \frac{n}{2} + 4, \frac{n}{2} + 8, \dots, n - 2. \end{cases}\]

It can be checked that \(\phi\) is a PSEML of G.

d) For \(n \in \{6, 10, 14, 18, 22, 26\}\), let \(H_n = C_4 \cup C_n \cup K_1\), \(V(H_n) = \{s_i : 1 \le i \le 4\} \cup \{t_i : 1 \le i \le n\} \cup \{w\}\) and \(E(H_n) = \{s_i s_{i+1} : 1 \le i \le 3\} \cup \{s_4 s_1\} \cup \{t_i t_{i+1} : 1 \le i \le n-1\} \cup \{t_n t_1\} \cup \{ww\}\). It is simple to confirm that \(\phi : V(H_n) \to \{1, 2, \dots, n+5\}\) is a PSEML of \(H_n\).

Case 1. n = 6. \((\phi(s_1), \phi(s_2), \phi(s_3), \phi(s_4)) = (1, 11, 2, 7), (\phi(t_1), \phi(t_2), \dots, \phi(t_6)) = (3, 8, 9, 6, 10, 4),\) and \(\phi(w) = 5\).

Case 2. n=10. \((\phi(s_1),\phi(s_2),\phi(s_3),\phi(s_4))=(1,15,2,9), (\phi(t_1),\phi(t_2),\dots,\phi(t_{10}))=(4,10,3,12,11,8,14,7,13,5),\) and \(\phi(w)=6.\)

Case 3. n = 14. \((\phi(s_1), \phi(s_2), \phi(s_3), \phi(s_4)) = (1, 19, 2, 11), (\phi(t_1), \phi(t_2), \dots, \phi(t_{14})) = (4, 12, 3, 15, 14, 9, 18, 10, 16, 8, 17, 5, 6, 13),\) and \(\phi(w) = 7\).

Case 4. n = 18. \((\phi(s_1), \phi(s_2), \phi(s_3), \phi(s_4)) = (1, 23, 2, 13), (\phi(t_1), \phi(t_2), \dots, \phi(t_{18})) = (4, 14, 3, 20, 10, 17, 18, 11, 21, 12, 22, 9, 19, 7, 6, 16, 5, 15), and <math>\phi(w) = 8\).

Case 5. n = 22. \((\phi(s_1), \phi(s_2), \phi(s_3), \phi(s_4)) = (1, 27, 2, 15), (\phi(t_1), \phi(t_2), \dots, \phi(t_{22})) = (4, 16, 3, 24, 12, 21, 20, 11, 23, 14, 26, 13, 25, 10, 22, 8, 7, 19, 6, 18, 5, 17), and <math>f(w) = 9\).

Case 6. n = 26. \((\phi(s_1), \phi(s_2), \phi(s_3), \phi(s_4)) = (1, 31, 2, 17), (\phi(t_1), \phi(t_2), \dots, \phi(t_{26})) = (4, 18, 3, 28, 13, 30, 16, 29, 15, 27, 12, 26, 14, 23, 24, 11, 25, 9, 8, 22, 7, 21, 6, 20, 5, 19), and <math>\phi(w) = 10\).

e) The labeling \(\phi\) in the proof of Theorem 3.3 is a PSEML of \(2C_4 \cup C_n \cup K_1\) for \(n \equiv 8, 12 \pmod{16}\).

Corollary 3.3 is a direct consequence of applying first Theorem 3.1 and then Theorem 3.2 to the graphs in Theorem 3.5.

Figure 3. (a). A PSEML of C⁵ ∪ 2K1. (b). A SEMTL of C²⁵ ∪ 2C⁵ which is obtained by applying Theorem 3.1 to C⁵ ∪ 2K1.

Corollary 3.3. Let m, r ≥ 1, l ≥ 2, and n ≥ 4 be positive integers such that m is odd and l ≡ 2 (mod 4). Then all the following graphs have zero SEMD.

• a) m[Cn(2r+1) ∪ C2r+1], if n ≡ 0 (mod 2).
• b) m[Cn(2r+1) ∪ 2C2r+1], if n ≡ 1 (mod 2).
• c) m[C4(2r+1) ∪ Cn(2r+1) ∪ C2r+1], if n ≡ 0 (mod 4).
• d) m[C4(2r+1) ∪ Cn(2r+1) ∪ C2r+1], if n ∈ {6, 10, 14, 18, 22, 26}.
• e) lC4(2r+1) ∪ m[Cn(2r+1) ∪ C2r+1], if n ≡ 8, 12 (mod 16).

Acknowledgement

The authors express their gratitude to the referee for his/her insightful comments. The first author has been supported by "Hibah Kompetensi 2018", 019/SP2H/K2/KM/2018, from the Directorate General of Higher Education, Indonesia.

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