Double Total Domination in Circulant Graphs 1

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1 Applied Mathematical Sciences, Vol. 12, 2018, no. 32, HIKARI Ltd, Double Total Domination in Circulant Graphs 1 Qin Zhang and Chengye Zhao 2 College of Science China Jiliang University Hangzhou, , China Copyright c 2018 Qin Zhang and Chengye Zhao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract A subset S of V (G) is a double total dominating set of G if for every vertex v V (G), N(v) S 2. We prove a necessary and sufficient condition for circulant graphs with an efficient double total dominating set and get the double total domination number of C n (1, k) for k = 2, 3, 4, 5. Mathematics Subject Classification: 05C35 Keywords: Circulant graphs; Double total dominating set; Efficient double total dominating set 1 Introduction We only consider the finite and undirected graphs without loops or multiple edges. Let G = (V (G), E(G)) be a graph with the order of vertex set V (G) and the order of edge set E(G). The open neighborhood and the closed neighborhood of a vertex v V are denoted by N(v) = {u V (G) : vu E(G)} and N[v] = N(v) {v}, respectively. For a vertex set S V (G), N(S) = N(v), 1 The research is supported by the Natural Science Fundation of Zhejiang province (No.LY14F020040) and Natural Science Foundation of China (No ). 2 Corresponding author v S

2 1624 Qin Zhang and Chengye Zhao and N[S] = v S N[V ]. A set S V (G) is a total domination set if and only if N(S) = V (G). The total domination number γ t (G) is the minimum cardinality of minimal total dominating sets. Henning etc.[2] defined a subset S of V (G) is a k-tuple total dominating set of G if for every vertex v V (G), N(v) S k. The k-tuple total domination number of G, denoted by γ t k (G), is the minimum cardinality of a k-tuple total domination set of G. A k-tuple total domination set where k = 2 is called a double total domination set. Henning etc.[3] also had discussed in double total domination in graphs. Zhao etc.[7] had given some results in double total domination in Generalized Petersen graphs. The k-tuple total domination had been studied by several authors(see [4, 5] and elsewhere). A subset S of V (G) is an efficient double total dominating set of G if for each vertex of G is dominated by exactly two vertices in S, i.e. N(v) S = 2. The circulant graph C n (1, k) is defined by the vertex set V (C n (1, k)) = {v i : 0 i n 1} and the edge set E(C n (1, k)) = {v i v i+1, v i v i+k : 0 i n 1, subscripts module n}. Rad[6] studied domination number and total domination number of circulant graph C n (1, 3). Obradovic ect.[1] studied efficient domination in circulant graphs. Here we study double total domination in circulant graphs and show its exact values or bounds. 2 The Exact Value of γ t 2 (C n (1, k)) Let S be a double total dominating set of C n (1, k) and let v s be the dominated times of the vertex v by S. The black vertices in figures indicate double total dominating set of circulant graphs. Lemma 2.1 If C n (1, k) has an efficient double total dominating set, then γ t 2 (C n (1, k)) = n/2, when k is an even number and 2 n, or k is an odd number(k 3) and 4 n. Proof. Since C n (1, k) is a 4-regular graph with n vertices, if C n (1, k) has an efficient double total dominating set, then according to the definition of efficient double total dominating set, 2n = 4 S, thus, γ t 2 (C n (1, k)) = S = n/2. Because S = n/2, whether k is odd or even, to make S an integer, then 2 n. when k is an odd number, without loss of generality we can assume k = 3, v 0 S. According to the definition of efficient double total dominating set, we consider four cases as below: Case 1 v n 3 S, v n 1 S, v 1 / S, v 3 / S v 0 S, since v 1 is dominated exactly twice by S, we consider three cases as below: Case 1.1 v n 2 S, v 2 / S, v 4 / S (see Figure 1(1))

3 Double total domination in circulant graphs 1625 v n 1 S, v 1 / S, v 3 / S, since v 2 is dominated exactly twice by S, thus v 5 S. v 0 S, v 2 / S, v 4 / S, since v 3 is dominated exactly twice by S, thus v 6 S. v 1 / S, v 3 / S, v 5 S, since v 4 is dominated exactly twice by S, v 7 S. v 2 / S, v 4 / S, v 6 S, since v 5 is dominated exactly twice by S, Case 1.2 v n 2 / S, v 2 S, v 4 / S (see Figure 1(2)) v n 1 S, v 1 / S, v 3 / S, since v 2 is dominated exactly twice by S, thus v 5 S. v 0 S, v 2 S, v 4 / S, since v 3 is dominated exactly twice by S, thus v 6 / S. v 1 / S, v 3 / S, v 5 S, since v 4 is dominated exactly twice by S, v 7 S. v 2 S, v 4 / S, v 6 / S, since v 5 is dominated exactly twice by S, Case 1.3 v n 2 / S, v 2 / S, v 4 S (see Figure 1(3)) v n 1 S, v 1 / S, v 3 / S, since v 2 is dominated exactly twice by S, thus v 5 S. v 0 S, v 2 / S, v 4 S, since v 3 is dominated exactly twice by S, thus v 6 / S. v 1 / S, v 3 / S, v 5 S, since v 4 is dominated exactly twice by S, v 7 S. v 2 / S, v 4 S, v 6 / S, since v 5 is dominated exactly twice by S, (1) v n 2 S, v 2 / S, v 4 / S (2) v n 2 / S, v 2 S, v 4 / S (3) v n 2 / S, v 2 / S, v 4 S Figure 1: The Case 1 for proof of Lemma 2.1 Case 2 v n 3 / S, v n 1 S, v 1 S, v 3 / S v 0 S, since v 1 is dominated exactly twice by S, we consider three cases as below: Case 2.1 v n 2 S, v 2 / S, v 4 / S (see Figure 2(1)) v n 1 S, v 1 S, v 3 / S, since v 2 is dominated exactly twice by S, thus v 5 / S. v 0 S, v 2 / S, v 4 / S, since v 3 is dominated exactly twice by S, thus v 6 S. v 1 S, v 3 / S, v 5 / S, since v 4 is dominated exactly twice by

4 1626 Qin Zhang and Chengye Zhao S, v 7 S. v 2 / S, v 4 / S, v 6 S, since v 5 is dominated exactly twice by S, Case 2.2 v n 2 / S, v 2 S, v 4 / S (see Figure 2(2)) v n 1 S, v 1 S, v 3 / S, since v 2 is dominated exactly twice by S, thus v 5 / S. v 0 S, v 2 S, v 4 / S, since v 3 is dominated exactly twice by S, thus v 6 / S. v 1 S, v 3 / S, v 5 / S, since v 4 is dominated exactly twice by S, v 7 S. v 2 S, v 4 / S, v 6 / S, since v 5 is dominated exactly twice by S, Case 2.3 v n 2 / S, v 2 / S, v 4 S (see Figure 2(3)) v n 1 S, v 1 S, v 3 / S, since v 2 is dominated exactly twice by S, thus v 5 / S. v 0 S, v 2 / S, v 4 S, since v 3 is dominated exactly twice by S, thus v 6 / S. v 1 S, v 3 / S, v 5 / S, since v 4 is dominated exactly twice by S, v 7 S. v 2 / S, v 4 S, v 6 / S, since v 5 is dominated exactly twice by S, (1) v n 2 S, v 2 / S, v 4 / S (2) v n 2 / S, v 2 S, v 4 / S (3) v n 2 / S, v 2 / S, v 4 S Figure 2: The Case 2 for proof of Lemma 2.1 Case 3 v n 3 S, v n 1 / S, v 1 / S, v 3 S v 0 S, since v 1 is dominated exactly twice by S, we consider three cases as below: Case 3.1 v n 2 S, v 2 / S, v 4 / S (see Figure 3(1)) v n 1 / S, v 1 / S, v 3 S, since v 2 is dominated exactly twice by S, thus v 5 S. v 0 S, v 2 / S, v 4 / S, since v 3 is dominated exactly twice by S, thus v 6 S. v 1 / S, v 3 S, v 5 S, since v 4 is dominated exactly twice by S, v 7 / S. v 2 / S, v 4 / S, v 6 S, since v 5 is dominated exactly twice by S,

5 Double total domination in circulant graphs 1627 Case 3.2 v n 2 / S, v 2 S, v 4 / S (see Figure 3(2)) v n 1 / S, v 1 / S, v 3 S, since v 2 is dominated exactly twice by S, thus v 5 S. v 0 S, v 2 S, v 4 / S, since v 3 is dominated exactly twice by S, thus v 6 / S. v 1 / S, v 3 S, v 5 S, since v 4 is dominated exactly twice by S, v 7 / S. v 2 S, v 4 / S, v 6 / S, since v 5 is dominated exactly twice by S, Case 3.3 v n 2 / S, v 2 / S, v 4 S (see Figure 3(3)) v n 1 / S, v 1 / S, v 3 S, since v 2 is dominated exactly twice by S, thus v 5 S. v 0 S, v 2 / S, v 4 S, since v 3 is dominated exactly twice by S, thus v 6 / S. v 1 / S, v 3 S, v 5 S, since v 4 is dominated exactly twice by S, v 7 / S. v 2 / S, v 4 S, v 6 / S, since v 5 is dominated exactly twice by S, (1) v n 2 S, v 2 / S, v 4 / S (2) v n 2 / S, v 2 S, v 4 / S (3) v n 2 / S, v 2 / S, v 4 S Figure 3: The Case 3 for proof of Lemma 2.1 Case 4 v n 3 S, v n 1 / S, v 1 S, v 3 / S v 0 S, since v 1 is dominated exactly twice by S, we consider three cases as below: Case 4.1 v n 2 S, v 2 / S, v 4 / S (see Figure 4(1)) v n 1 / S, v 1 S, v 3 / S, since v 2 is dominated exactly twice by S, thus v 5 S. v 0 S, v 2 / S, v 4 / S, since v 3 is dominated exactly twice by S, thus v 6 S. v 1 S, v 3 / S, v 5 S, since v 4 is dominated exactly twice by S, v 7 / S. v 2 / S, v 4 / S, v 6 S, since v 5 is dominated exactly twice by S, Case 4.2 v n 2 / S, v 2 S, v 4 / S (see Figure 4(2)) v n 1 / S, v 1 S, v 3 / S, since v 2 is dominated exactly twice by S, thus v 5 S. v 0 S, v 2 S, v 4 / S, since v 3 is dominated exactly twice by S,

6 1628 Qin Zhang and Chengye Zhao thus v 6 / S. v 1 S, v 3 / S, v 5 S, since v 4 is dominated exactly twice by S, v 7 / S. v 2 S, v 4 / S, v 6 / S, since v 5 is dominated exactly twice by S, Case 4.3 v n 2 / S, v 2 / S, v 4 S (see Figure 4(3)) v n 1 / S, v 1 S, v 3 / S, since v 2 is dominated exactly twice by S, thus v 5 S. v 0 S, v 2 / S, v 4 S, since v 3 is dominated exactly twice by S, thus v 6 / S. v 1 S, v 3 / S, v 5 S, since v 4 is dominated exactly twice by S, v 7 / S. v 2 / S, v 4 S, v 6 / S, since v 5 is dominated exactly twice by S, this case, which is 4. Thus 4 n. (1) v n 2 S, v 2 / S, v 4 / S (2) v n 2 / S, v 2 S, v 4 / S (3) v n 2 / S, v 2 / S, v 4 S Figure 4: The Case 4 for proof of Lemma 2.1 By Cases 1-4, when k is an odd number, if C n (1, k) has an efficient double total dominating set, then 4 n. Lemma 2.2 If k is an even number and 2 n, or k is an odd number(k 3) and 4 n, then γ t 2 (C n (1, k)) = n/2 and therefore C n (1, k) has an efficient double total dominating set. Proof. If k is an even number and 2 n, we construct an efficient double total dominating set S = {v i : i 0(mod2), 0 i n 2}. The neighborhood of v i are v i+1 / S, v i 1 / S, v i+k S, v i k S. The neighborhood of v i+1 are v i+2 S, v i S, v i+1+k / S, v i+1 k / S. So S = γ t 2 (C n (1, k)) = n/2 and therefore S is an efficient double total dominating set. If k is an odd number and 4 n, we construct an efficient double total dominating set S = {v i, v i+1 : i 0(mod4), 0 i n 4}. When k 3(mod4). The neighborhood of v i are v i+1 S, v i 1 / S, v i+k / S, v i k S. The neighborhood of v i+1 are v i+2 / S, v i S, v i+1+k S, v i+1 k / S. The neighborhood of v i+2 are v i+3 / S, v i+1 S, v i+2+k S,

7 Double total domination in circulant graphs 1629 v i+2 k / S. The neighborhood of v i+3 are v i+4 S, v i+2 / S, v i+3+k / S, v i+3 k S. When k 1(mod4). The neighborhood of v i are v i+1 S, v i 1 / S, v i+k S, v i k / S. The neighborhood of v i+1 are v i+2 / S, v i S, v i+1+k / S, v i+1 k S. The neighborhood of v i+2 are v i+3 / S, v i+1 S, v i+2+k / S, v i+2 k S. The neighborhood of v i+3 are v i+4 S, v i+2 / S, v i+3+k / S, v i+3 k S. So S = γ t 2 (C n (1, k)) = n/2 and therefore S is an efficient double total dominating set. On the basis of the lemma 2.1 and lemma 2.2, we can obtain the following theorem. Theorem 2.1 A circulant graph C n (1, k) has an efficient double total dominating set if and only if n 0(mod2) and k is even or n 0(mod4) and k is odd. Theorem 2.2 γ t 2 (C n (1, k)) n/2. Proof. Let S is a double total dominating set. The dominated times of every vertex of C n (1, k) by S is at least two, then v S 2n. Because v V C n (1, k) are 4-regular graphs, 4 S v S 2n, then S n/2. Thus, γ t 2 (C n (1, k)) n/2. v V Theorem 2.3 γ t 2 (C n (1, 2)) = n/2. Proof. Let S 0 = {v i : i 0(mod2), 0 i n 1}(see Figure 5). Note that S 0 is a double total dominating set of C n (1, 2) and S 0 = n/2. Thus, γ t 2 (C n (1, 2)) n/2. From Theorem 2.2, we have γ t 2 (C n (1, 2)) = n/2. vn 8 vn 7 vn 6 vn 5 vn 4 (1) n 0(mod2) vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 v9 (2) n 1(mod2) Figure 5: The proof of Theorem 2.3 { n/2, n 0, 1, 3 (mod 4), Theorem 2.4 γ t 2 (C n (1, 3)) = n/2 + 1, n 2 (mod 4). Proof. Let S 0 = {v i, v i+1 : i 0 (mod 4), 0 i n 4}. We consider four cases as below: Case 1 n 0 (mod 4). Note that S 0 (see Figure 6(1)) is a double total dominating set of C n (1, 3) and S 0 = n/2. Thus, γ t 2 (C n (1, 3)) n/2. From Theorem 2.2, γ t 2 (C n (1, 3)) = n/2 = n/2. Case 2 n 1 (mod 4). Let S = S 0 {v n 1 }(see Figure 6(2)). Note that S is a double total dominating set of C n (1, 3) and S = S = 2 n/4 + 1 = 2 (4m + 1)/4 + 1 = 2m + 1 = n/2. Thus, γ t 2 (C n (1, 3)) n/2. From Theorem 2.2, γ t 2 (C n (1, 3)) = n/2.

8 1630 Qin Zhang and Chengye Zhao Case 3 n 3 (mod 4). Let S = S 0 {v n 3, v n 2 }(see Figure 6(3)). Note that S is a double total dominating set of C n (1, 3) and S = S = 2 n/4 +2 = 2 (4m+3)/4 +2 = 2m+2 = n/2. Thus, γ t 2 (C n (1, 3)) n/2. From Theorem 2.2, γ t 2 (C n (1, 3)) = n/2. Case 4 n 2 (mod 4). By Lemma 2.1, C n (1, 3) is not efficient, so in this case γ t 2 (C n (1, 3)) n/ Let S = S 0 {v n 2, v n 1 }(see Figure 6(4)). Note that S is a double total dominating set of C n (1, 3) and S = S = 2 n/4 + 2 = 2 (4m + 2)/4 + 2 = 2m + 2 = n/ Thus, C n (1, 3) = n/ By Cases 1-4, the proof is completed. vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (1) n 0 (mod 4) vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (2) n 1(mod4) vn 11 vn 10 vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (3) n 3(mod4) vn 10 vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (4) n 2 (mod 4) Figure 6: The proof of Theorem 2.4 Theorem 2.5 γ t 2 (C n (1, 4)) = n/2. Proof. We consider two cases as below: Case 1 n 0 (mod 2). Let S = {v i : i 0(mod2), 0 i n 2}(see Figure 7(1)). Note that S is a double total dominating set of C n (1, 4) and S = n/2. Thus, γ t 2 (C n (1, 4)) n/2. From Theorem 2.2, γ t 2 (C n (1, 4)) = n/2 = n/2. Case 2 n 1 (mod 2). Let S = {v 0, v 1, v 6, v 7, v 8 } {v i : i 0(mod2), 10 i n 1} (see Figure 7(2)). Note that S is a double total dominating set of C n (1, 4) and S = 5 + (n 9)/2 = (n + 1)/2 = (2m + 2)/2 = m + 1 = n/2. Thus, γ t 2 (C n (1, 4)) n/2. From Theorem 2.2, γ t 2 (C n (1, 4)) = n/2. By Cases 1-2, the proof is completed. { n/2, n 0, 1, 3 (mod 4), Theorem 2.6 γ t 2 (C n (1, 5)) = n/2 + 1, n 2 (mod 4).

9 Double total domination in circulant graphs 1631 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (1) n 0 (mod 2) vn 4 v9 v10 v11 v12 (2) n 1 (mod 2) Figure 7: The proof of Theorem 2.5 Proof. We consider four cases as below: Case 1 n 0 (mod 4). Let S = {v i, v i+1 : i 0(mod4), 0 i n 4}(see Figure 8(1)). Note that S is a double total dominating set of C n (1, 5) and S = n/2. Thus, γ t 2 (C n (1, 5)) n/2. From Theorem 2.2, γ t 2 (C n (1, 5)) = n/2 = n/2. Case 2 n 1 (mod 4). We consider two cases as below: Case 2.1 n 1 (mod 8). Let S = {v i, v i+1, v i+2, v i+3 : i 0(mod8), 0 i n 9} {v n 1 }(see Figure 8(2)). Note that S is a double total dominating set of C n (1, 5) and S = 4 n/8 + 1 = 4 (8m + 1)/8 + 1 = 4m + 1 = n/2. Thus, γ t 2 (C n (1, 5)) n/2. From Theorem 2.2, γ t 2 (C n (1, 5)) = n/2. Case 2.2 n 5 (mod 8). Let S = {v i, v i+1, v i+3, v i+6 : i 1(mod8), 1 i n 12} {v n 4, v n 3, v n 1 }(see Figure 8(3)). Note that S is a double total dominating set of C n (1, 5) and S = 4 n/8 + 3 = 4 (8m + 5)/8 + 3 = 4m + 3 = n/2. Thus, γ t 2 (C n (1, 5)) n/2. From Theorem 2.2, γ t 2 (C n (1, 5)) = n/2. Case 3 n 3 (mod 4). We consider two cases as below: Case 3.1 n 3 (mod 8). Let S = {v 4, v 5, v 6, v 7 } {v i, v i+3, v i+5, v i+6 : i 1(mod8), 9 i n 10} {v n 3, v n 2 }(see Figure 8(4)). Note that S is a double total dominating set of C n (1, 5) and S = 4 n/8 + 2 = 4 (8m + 3)/8 + 2 = 4m + 2 = n/2. Thus, γ t 2 (C n (1, 5)) n/2. From Theorem 2.2, γ t 2 (C n (1, 5)) = n/2. Case 3.2 n 7 (mod 8). Let S = {v i, v i+1, v i+2, v i+3 : i 0(mod8), 0 i n 15} {v n 7, v n 6, v n 5, v n 4 }(see Figure 8(5)). Note that S is a double total dominating set of C n (1, 5) and S = 4 n/8 + 4 = 4 (8m + 7)/8 + 4 = 4m + 4 = n/2. Thus, γ t 2 (C n (1, 5)) n/2. From Theorem 2.2, γ t 2 (C n (1, 5)) = n/2. Case 4 n 2 (mod 4). By Lemma 2.1, C n (1, 5) is not efficient, so in this case γ t 2 (C n (1, 5)) n/ We consider two cases as below: Case 4.1 n 2 (mod 8). Let S = {v i, v i+1, v i+2, v i+3 : i 0(mod8), 0 i n 10} {v n 2, v n 1 }(see Figure 8(6)). Note that S is a double total dominating set of C n (1, 5) and S = 4 n/8 + 2 = 4 (8m + 2)/8 + 2 = 4m + 2 = n/ Thus, C n (1, 5) = n/ Case 4.2 n 6 (mod 8). Let S = {v i, v i+1, v i+2, v i+3 : i 0(mod8), 0 i

10 1632 Qin Zhang and Chengye Zhao n 14} {v n 6, v n 5, v n 4, v n 3 }(see Figure 8(7)). Note that S is a double total dominating set of C n (1, 5) and S = 4 n/8 + 4 = 4 (8m + 6)/8 + 4 = 4m + 4 = n/ Thus, C n (1, 5) = n/ By Cases 1-4, the proof is completed. vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (1) n 0 (mod 4) vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (2) n 1 (mod 8) vn 13 vn 12 vn 11 vn 10 vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (3) n 5 (mod 8) vn 11 vn 10 vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 v9 v10 v11 v12 v13 v14 v15 (4) n 3 (mod 8) vn 15 vn 14 vn 13 vn 12 vn 11 vn 10 vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (5) n 7 (mod 8) vn 10 vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (6) n 2 (mod 8) vn 14 vn 13 vn 12 vn 11 vn 10 vn 9 vn 8 vn 7 vn 6 vn 5 vn 4 vn 3 vn 2 vn 1 v0 v1 v2 v3 v4 v5 v6 v7 (7) n 6 (mod 8) Figure 8: The proof of Theorem Further Research By the proof of double total domination number of C n (1, 2), C n (1, 3), C n (1, 4) and C n (1, 5). For any positive integer m, we conjecture : Conjecture 3.1 γ t 2 (C n (1, 2m)) = n/2.

11 Double total domination in circulant graphs 1633 Conjecture 3.2 γ t 2 (C n (1, 2m + 1)) = References { n/2, n 0, 1, 3 (mod 4), n/2 + 1, n 2 (mod 4). [1] N. Obradovic, J. Peters and G. Ruzic, Efficient domination in circulant graphs with two chord lengths, Information Processing Letters, 102 (2007), [2] M.A. Henning and A.P. Kazemi, k-tuple total domination in graphs, Discrete Applied Mathematics, 158 (2010), [3] M.A. Henning and A. Yeo, Strong transversals in hypergraphs and double total domination in graphs, SIAM Journal on Discrete Mathematics, 24 (2010), no. 4, [4] D. Pradhan, Algorithmic aspects of k-tuple total domination in graphs, Information Processing Letters, 112 (2012), [5] J.K. Lan and G.J. Chang, On the algorithmic complexity of k-tuple total domination, Discrete Applied Mathematics, 174 (2014), [6] N.J. Rad, Domination in Circulant graphs, Analele Stiintifice Ale Universitatii Ovidius Constanta Seria Matematica, 17 (2014), [7] Chengye Zhao and Linlin Wei, Double total domination on generalized Petersen graphs, Applied Mathematical Sciences, 11 (2017), no. 19, Received: November 27, 2018; Published: December 18, 2018