Research Article On the Domination Number of Cartesian Product of Two Directed Cycles

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1 Applied Mathematics Volume 013, Article ID , 7 pages Research Article On the Domination Number of Cartesian Product of Two Directed Cycles Zehui Shao, 1, Enqiang Zhu, 3 and Fangnian Lang 1, 1 School of Information Science and Technology, Chengdu University, Chengdu , China Key Laboratory of Pattern Recognition and Intelligent Information Processing, Institutions of Higher Education of Sichuan Province, Chengdu , China 3 School of Electronic Engineering and Computer Science, Peking University, Beijing , China Correspondence should be addressed to Fangnian Lang; fnlang56@163com Received 18 September 013; Revised 11 October 013; Accepted 11 October 013 Academic Editor: Carla Roque Copyright 013 Zehui Shao et al 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 Denote by γ(g) the domination number of a digraph G and C m C n the Cartesian product of C m and C n, the directed cycles of length m, n In 010, Liu et al determined the exact values of γ(c m C n ) for m =, 3, 4, 5, 6 In 013, Mollard determined the exact values of γ(c m C n ) for m=3k+ In this paper, we give lower and upper bounds of γ(c m C n ) with m=for different cases In particular, (k + 1)n/ γ(c C n ) (k + 1)n/ + k Based on the established result, the exact values of γ(c m C n ) are determined for m=7and 10 by the combination of the dynamic algorithm, and an upper bound for γ(c 13 C n ) is provided 1 Introduction All of the digraphs considered in this paper are finite and simple, that is, without multiple edges or loops For a digraph G = (V, E) and a vertex V V, N + G (V) and N G (V) denote the set of out-neighbors and in-neighbors of V Given two vertices u and V in G, wesayu dominates V if u = V or uv ELet N + G [V] = N+ G (V) V} AvertexV dominates all vertices in N + G [V] AsetD Vis a dominating set of G if D dominates V(G) The domination number of G, denoted by γ(g),isthe minimum cardinality of a dominating set of G The Cartesian product of graphs G and H is the graph G H with the vertex set G H,and(g, h)(g,h ) E(G H) if either gg E(G) and h = h or hh E(H) and g = g The Cartesian product is commutative and associative, having the one-vertex graph as a unit The subgraph of G Hinduced by V(G) h, whereh V(H), is isomorphic to G, calledag-layer (over h) and denoted by G h For more information on the Cartesian product of graphs see [1] We use a 0-1 matrix pattern A with m rows and n columns to represent a dominating set D of C m C n,where(i, j) D if and only if the value at the entry (i, j) of A equals 1 For example, the pattern in Figure represents the dominating set (the set of black circles) shown in Figure 1 LetP be a pattern; we denote by c(p) the column number of PLetA and B be patterns; then the pattern AB denotes the concatenation of patterns A and B, anda k denotes the concatenation of patterns A for k times Let [n] denote the set 0,1,,n 1}Foranintegern 3, the directed cycle of length n is the graph C n whose vertices are 0,1,,n 1andwhoseedgesarethepairsi, i+1,wherethe arithmetic is done modulo nwedenotev(c n ) by [n]inthis paper, when we consider the Cartesian product graph C m C n of two cycles C m and C n,theedgesetofc m is (i, i + 1) : i [m]} and C n is (i + 1, i) : i [n]} Throughoutthepaper, when considering the vertex (i, j) in the graph C m C n,we use the arithmetic operations of the index i over modulo m and j modulo n The dominating set problem requires determining the domination number of a given graph It has natural applications in numerous facility location problems In such problems, the vertices of a graph correspond to locations, adjacency represents some notion of accessability, and the goal is to find a subset of locations accessible from all other locations at which to install fire stations, bus stops, post

2 Applied Mathematics Proof For a dominating set D of C C n, we define a functionf as follows: Figure 1: A dominating set of C 7 C Figure : Pattern of the dominating set shown in Figure 1 offices, or similar facilities [] Dominating sets have also been applied in coding theory [] and social networks [3] For more information on the history and applications of the dominating set problem, see [4] In 1990, Faudree and Schelp [5] initially discussed the domination number of the Cartesian product of two undirected graphs The domination number for Cartesian product oftwographshasattractedlotsofattention,andthereare many works on undirected graphs (see, eg, [6 1]) Recently, there are also some works on the domination number of Cartesian product of two directed graphs, in particular, the Cartesian product of directed paths and cycles [13 15] In [13, 15], Liu et al determined the exact values of γ(c m C n ) for m =, 3, 4, 5, 6 andshowedthefollowing Theorem 1 γ(c m C n )=mn/3 if m 0(mod 3) and n 0(mod 3) Mollard [14] determined exact values of γ(c m C n ) for m = 3k + This paper investigates properties on the domination of Cartesian product of C m and C n with m = 3k + 1 Wegivelowerandupperboundsofγ(C m C n ) with m=for different cases In particular, (k + 1)n/ γ(c C n ) (k + 1)n/ + k Based on the established result, the exact values of γ(c m C n ) are determined for m=7 and 10 by the combination of the dynamic algorithm, and an upper bound for γ(c 13 C n ) is provided Main Results 1 Bounds on C C n Proposition (Mollard [14]) Let S be a dominating set of C m C n ;thenforalli [n] considered modulo n one has S C i 1 m + S Ci m m Lemma 3 Let n 3, k 1 Then there exists a minimum dominating set D of C C n such that C i D kfor every i V(C n ) where f (D) = n 1 t=0 x t, (1) x t = 0, if D Ct k+1, D Ct, if D Ct k () We now assume that D is a minimum dominating set of C C n and suppose to the contrary that C i+1 D < kfor some i such that f(d) is maximized Then there exists a set T = (j, i+1), (j+1, i+1), (j+, i+1)} such that u Dfor any u TThus,(j, i), (j + 1, i)} DByProposition,wehave x t =0if 0<x t+1 kthusd = D\(j+1,i)} (j+1,i+1)} is a dominating set with D = D such that f(d )>f(d), a contradiction with the fact that f(d) is maximized Lemma 4 Let n 3, k 1 Then there exists a minimum dominating set D of C C n such that (C i Ci+1 ) D k + 1 for every i V(C n ) Proof By Lemma 3, it is sufficient to consider a minimum dominating set D of C C n such that C i D k for every i V(C n )If C i D = k,togetherwith Proposition, wehavethat C i 1 D k+1,which completes the proof By Lemmas 3 and 4, thereisanobviouslowerboundin terms of γ(c C n ) as follows Lemma 5 Let k 1and n 3be two integers; then (k + 1) n γ(c C n ) (3) Lemma 6 For two integers k 1and n 3,thenforany l [k],onehas (1) γ(c C n ) (k+1)n/ +l if n 6l(mod 6k+), () γ(c C n ) (k+1)n/ +k lif n + 6l (mod 6k + ), (3) γ(c C n ) (k+1)n/ +k if n 4+6l(mod 6k+ ), (4) γ(c C n ) (k + 1)n/ + l + 1 if n 3 + 6l (mod 6k + ), (5) γ(c C n ) (k+1)n/ +k lif n 5 + 6l (mod 6k + ), (6) γ(c C n ) (k+1)n/ +k if n 7+6l(mod 6k+ ), (7) γ(c C n ) (k+1)n/ +k for n 1(mod 6k+) Proof For any i [n],let (3q + i,i):0 q k}, A i = (3q + + i,i):0 q k 1}, if n is even, if n is odd (4)

3 Applied Mathematics 3 Itcanbeseenthatifi 0(mod ), A i =k+1;otherwise, A i = klets = n 1 i=0 A i;then S = (k + 1)n/ and there are at most k vertices, (1,0),(4,0),(7,0),,(3k,0) of C C n of which each is not dominated by SDenotedby B=(1,n 1),(4,n 1),,(3k,n 1)}, we consider the following cases (1) n 6l(mod 6k + ) Inthiscase,itiseasytoseethat A n 1 B =k l By adding at most l vertices to S,wegeta dominating set of γ(c C n ) as required () n +6l(mod 6k + ) Since A n 1 B = l,wewill obtain a desired dominating set of γ(c C n ) by adding at most k lvertices to S (3) n 4+6l(mod 6k+)Since A n 1 B =Ø, we have to add at most k vertices so that a dominating set of γ(c C n ) is constructed based on S (4) n 3+6l(mod 6k+)Sincen 1is an even and A n 1 B = k l, wehavetoaddatmostl vertices to S so that a dominating set of γ(c C n ) will be formed (5) n 5+6l(mod 6k+)Since A n 1 B = l,wewill obtain a dominating set of γ(c C n ) by adding at most k l vertices to S (6) n 7+6l(mod 6k+)LetA n 3 = (1 + 3l + 3k, n 3),(+3l,n 3),(5+3l,n 3),,(+3l+3(k 1),n 3)} when + 3l > + 3l + 3k (mod 3k + 1); otherwise, let A n 3 = (1+3l, n 3), (5+3l, n 3), (8+3l, n 3),, (+3l+3k, n 3)} when +3l < +3l+3k (mod );leta n =Cn \(1, n ),(4,n ),(7,n ),,(3k,n ),(1+3l+3k,n )}when +3l > +3l+3k (mod ); otherwise, let A n =Cn \ (1,n ),(4,n ),(7,n ),,(3k,n ),(1+3l,n )} when + 3l < + 3l + 3k (mod );leta n 1 = (1, n 1), (4, n 1),, (3k, n 1)}, A i =A i for i [n 4] In this way, if we let D= n 1 i=0 A i,thend is a dominating set of C C n In addition, A n 3 =k+1, A n =k, and A n 1 =kso,d = (((k + 1)n)/) + ((k 1)/) = ((k + 1)n)/ + k (7) n 1(mod 6k+) LetD= n i=0 A i (1+3q,n 1),(3+ 3q,n 1) : 0 q k 1}ThenD is a dominating set of C C n with cardinality (k + 1)n/ + k The result of Lemmas 5 and 6 show the following Corollary 7 Let k 1and n 3be two integers; then (k+1) n (k + 1) n γ(c C n ) +k (5) 3 On the Domination of C 7 C n Lemma 8 Let k 1and n 3be two integers Suppose D is a minimum dominating set of C C n LetS i =D C i and a i = S i for i=0,1,,n 1 (a) If there is an integer i such that a i +a i+1 = k + 1 (considered modulo n), then one of a i,a i+1 } is k and the other is k+1 (b) If there is an integer i such that a i =k+1, a i+1 =k, a i+ =k+1,anda i+3 =k, then for any distinct vertices (j 1,i+1),(j,i+1)in S i+1 and (j 3,i+3),(j 4,i+3)in S i+3, j 1 j 0(mod 3), j 3 j 4 0(mod 3);foranydistinct vertex (j 1,i),(j,i)in S i and (j 3,i+),(j 4,i+)in S i+, j 1 j 0(mod 3) or j 1 j =1and j 3 j 4 0(mod 3) or j 3 j 4 = 1,wherej 1 < j, j 3 < j 4, j 1 <j, j 3 <j 4,andtheyaretakenmodulo3k + 1 Proof It follows immediately that (a) holds by Lemmas 3 and 4 In addition, since a i =k+1, a i+1 =k, a i+ =k+1,and a i+3 =k, each of vertices of C i+1 and Ci+3 is dominated by only one vertex; also, only one vertex of C i+ is dominated by two vertices Now we will first show that it is impossible that there are three consecutive vertices in S i Otherwise,suppose that (s, i), (s + 1, i), and(s +, i) (modulo ) are three consecutive vertices We can deduce that vertices (s+1,i+) and (s+, i+3) are in S i+ and S i+3,respectively,so(s+1, i+3) is dominated by both (s+1, i+) and (s+, i+3), a contradiction Suppose that there are (j 1,i+1), (j,i+1)(or (j 3,i+3), (j 4,i+3))inS i+1 (or S i+3 )suchthatj 1 j 0(mod 3) (or j 3 j 4 0(mod 3))Thenitiseasytoshowthatthereexist three consecutive vertices S i or three consecutive vertices (s 1,i+),(s 1 +1,i+),and(s 1 +,i+) S i+ (modulo ), thus (s 1,i+1),(s 1 +1,i+1),and(s 1 +,i+1)(modulo ) canonlybedominatedbys i,and(s 1 +4,i+1) S i+1 since it is impossible that there are four consecutive vertices in S i+3 by a i+3 =k So, it is an easy task to prove that either (s 1 +3,i+ ) isnotdominatedbyd or (s 1 +4,i+)isdominatedby two vertices, a contradiction By using the same approach as above, we also can prove the cases of vertices in S i and S i+ Lemma 9 Let k 1and n 3IfD is a minimum dominating set of C C n such that (a i,a i+1,a i+,a i+3,,a i+l ) = (k + 1,k,k+1,k,,k+1),whereS i =D C i and a i = S i for some i, l,then(j, i+1) S i+1 ifandonlyif(j+t 1,i+t 1) S i+t 1 for any j 0,1,,3k}and t 1,,,l} Proof By Lemma 8 and the symmetrical structure of C m C n, we assume without loss of generality S i+l 1 = (3q,i+l 1) : 0 q k 1} Since each vertex in (3q 1, i + l 1) : 0 q k 1} (,i+l 1)is not dominated by C i+l 1,wehave S i+l = (3q 1, i+l ) : 0 q k 1} (, i+l )By observing that each vertex in S i+l needs to be dominated, we have the desired set S i+l 3 Repeating this process until S i+1 is determined, we have the desired result Theorem 10 Let n 7Then γ(c 7 C n ) 5n, n 0 (mod 14), = 5n + 1, n 3, 4, 6, 8, 10, 11 (mod 14), 5n +, n 1,, 5, 7, 9, 1, 13 (mod 14) (6)

4 4 Applied Mathematics Figure 3: A dominating set of C 7 C 4 Proof Consider the following Case 1 n 0(mod 14) The lower bound follows from Lemma 5, andtheupper bound follows from Lemma 6 with l=0in Case 1 Case n 3, 4, 6, 8, 10, 11 (mod 14) By using Lemmas 3 and 4, weobtainthatγ(c 7 C n ) = 5n/ if and only if n 0(mod 14) Therefore, the desired lower bound is established Upper bound can be obtained by applying Lemma 6withk =inthefollowingcases: (i) n 3(mod 14), l=0in case 4; (ii) n 6(mod 14), l=1in case 1; (iii) n 8(mod 14), l=1in case ; (iv) n 11(mod 14), l=1in case 5 We give the upper bound for the case n 4, 10 (mod 14) by constructions Let n = 14k + 4 The pattern depicted in Figure is a dominating set of C 7 C 18 with 46 vertices Moreover, the leftmost 14 columns induce a dominating set of C 7 C 14 with 35 vertices By repeating the leftmost 14 columns for k times, wegetadominatingsetofc 7 C 14k+4 with 35k + 11 vertices Therefore, γ(c 7 C 14k+4 ) 35k + 11, andthus,γ(c 7 C n ) 5n/ + 1 Let n = 14k + 10 The pattern depicted in Figure 3 is a dominating set of C 7 C 4 with 61 vertices Moreover, the leftmost 14 columns induce a dominating set of C 7 C 14 with 35 vertices By repeating the leftmost 14 columns for k times, wegetadominatingsetofc 7 C 14k+10 with 35k + 6 vertices Therefore, γ(c 7 C 14k+10 ) 35k+6,andthus,γ(C 7 C n ) 5n/ + 1 Case 3 n 1,, 5, 7, 9, 1, 13 (mod 14) The upper bound follows from Lemma 6 Wethenshow the lower bound Let n 1(mod 14) If there exists a minimum dominating set of C C n with cardinality 5n/ +1, then by Lemmas 3 and 4, wehavesuchadominatingsetd with (a 0,a 1,a,a 3,,a n,a n 1 )=(3,,3,,,,3),where S i =D C i 7 and a i = S i We without loss of generality assume S n = (0, n ), (3, n )}ThenS 1 = (1, 1), (4, 1)} and so S 0 = (0, 0), (3, 0), (6, 0)} We now consider S n, S n 1,and S 0 with S n and S 0 knownthentherearefourverticesin S n 1 not dominated by S n S 0, a contradiction Therefore, γ(c 7 C n ) 5n/ + if n 1(mod 14) Byusingthe same approach, we can obtain the desired lower bound if n, 5, 7, 9, 1, 13 (mod 14) and the proof is omitted Figure 4: Two vertices of D 7 4 On the Domination of C 10 C n We use the dynamic algorithm to provide the result on γ(c m C n ) for m=10and 13 We first describe the concept needed to describe our computer checking The idea is introduced in [16] in a more general framework and extended in [17], but for our purpose the following description will be sufficient For a graph G and D V(G), we define a function f: V(G) 0, 1} such that f(v) =0if V Dand f(v) =1 if V DWecallsuchafunctionthe-coloringofG Iffor f(v) =0there exists a vertex u N G (V) such that f(v) =1, then we say V is dominated We can see that this notation is equivalent to the dominating set and we will use it to describe the dynamic algorithm for this problem Let C=u i u i+1 :i [n]}beadirected cycle in a weighted digraph G Wesay thatthesum oftheweightsofalltheedges in C to be the weight of C Let u be a -coloring of C m P ; u is said to be an (a, b)- vertexifthenumberofonesinthefirstcopyofc m equals a and the second equals b (see, eg, the left side of the vertex in Figure 4 is a (3, )-vertex) An (a, b)-vertexissaidtobe redundant if it is dominated and any removal of one vertex in the first column destroys the domination Let T=(a i,b i ):1 i t} We define a weighted digraph D T m (when no confusion can arise, the set T is omitted and the digraph is denoted by D m ) as follows The vertices of D T m consist of all the nonredundant - colorings of C m P Letu=u 1 u be a vertex of D T m Thenu 1 and u represent the -colorings of C m P restricted to the firstandsecondcopiesofc m Letu and V be two vertices of D T m Then uv denotes the -coloring of C m P 3 obtained by applying u and V 1 to the consecutive copies of C m Foreach vertex u of D T m,wemakeanarcfromu to V in DT m if and only if the following conditions are fulfilled: (i) u equals V 1, (ii) each of the vertices of C 1 m C m is dominated (by C0 m C 1 m ), (iii) for each (a, b)-vertex of D T m, (a, b) T Finally, the weight of uv is the number of ones in C 1 m of V

5 Applied Mathematics 5 Let T = (, 3), (3, )} Figure 4 shows two vertices of D T 7 We can see that the -coloring ( ) of the second copy of C 7 of the first vertex equals the first copy of C 7 of the second vertex Moreover, the weight of the second copy of C 7 of the first vertex is It follows that A(D T 7 ) has an arc from the first (left) vertex to the second (right) with weight Lemma 11 Let m, n 3ThenD m possesses a directed cycle of length n with weight k if and only if C m C n contains a minimal dominating set S with S = k Proof If S is a minimal dominating set of C m C n,thens restricted to C i m Ci+1 m is a -coloring C of C m P SinceS is minimal, we have that C is nonredundant If D m possesses a directed cycle of length n, weconcatenatethesecondcyclec m of each vertex of the cycle and then obtain a dominating set S of C m C n Since each vertex in the cycle isnonredundant, we have thats is minimal We define a weighted adjacency matrix A(D m ) of D m as follows Let V(D m )=V 1, V,,V K }Then A(D m ) i,j = w(v iv j ), V i V j D m, +, V i V j D m, where w(v i V j ) is the weight of the edge V i V j Let A = (a) p q (p rows, q columns) and B = (b) q r (q rows, r columns) be two weighted adjacency matrices We define A multiplied by B as follows a 1,1 a 1, a 1,q b 1,1 b 1, b 1,r a,1 a, a,p b,1 b, b,r ( ) ( ) a p,1 a p, a p,q b q,1 b q, b q,r c 1,1 c 1, c 1,r c,1 c, c,r =( ), c p,1 c p, c p,r where c s,t = min 1 i q a s,i +b i,t } for s 1,,,p}and t 1,,,r} We denote by A n the nth power of A; wehavethe following result Theorem 1 Let m, n 3 and A(D m ) be the weighted adjacency matrix of D m with K vertices Then one has (7) (8) γ(c m C n )=min ((A (D m )) n ) i,i :1 i K} (9) Lemma 13 Let n 10, γ(c 10 C n ) 7n/ +,andd a dominating set with cardinality 7n/ + (a) If n is odd, then D (C i 10 Ci+1 10 ) 8 for any i V(C n) (b) If n is even, then D (C i 10 Ci+1 10 ) 9 for any i V(C n) butatmostoneintegeri 0 such that D (C i 0 10 C i ) = 9 Proof (a) Suppose that there exists i 0 [n 1]such that D (C i 0 ) 9Wecandeducethat 10 C i D j [n 1], j i 0,i 0 +1} V(C j 10 ) 7 (n ) 1 (10) Thus, by the pigeonhole principle it can be seen that there exist some i [n 1]such that D (C i 10 Ci+1 10 ) < 7,which contradicts with Lemma 4 (b) The proof is similar to (a) Theorem 14 Let n 10Then γ(c 10 C n ) 7n, n 0 (mod 0), 7n + 1, n 3, 6, 14, 17 (mod 0), = 7n +, n 4, 7, 8, 9, 10, 11, 1, 13, 16 (mod 0), 7n + 3, n 1,, 5, 15, 18, 19 (mod 0) (11) Proof When n 0, 3, 6, 14, 17 (mod 0), theresultfollows from Lemmas 6, 8, and 9 Otherwise, let T = (3, 4), (4, 3), (3, 5), (5, 3), (4, 4)} Weimplementedthealgorithm as described above, and the digraph D T 10 was created and it has 4905 vertices By Lemma 13, it can be seen that there may exist at most one (x, y)-vertex such that x+y = 9in the auxiliary graph D 10 We consider each possible auxiliary graph D 10, whose vertex set is the union of that of D T 10 and one (x, y)-vertex with x+y = 9NotethatD 10 has only 4906 vertices; the matrix multiplication can be done easily By using matrix multiplication, we found a formula as stated in Theorem 14 (Such parameters a and b exist; see [17]) If γ(c 10 C n ) (for some n) isgreaterthan 7n/ +, then γ(c 10 C n )= 7n/ +3 Remark 15 Note that when Lemmas 4 and 13 are applied, we can operate matrix multiplication on the graph D T 1 10 whose rownumberis4906itcanbeclearlyseenthat D (C i 10 C i+1 10 ) 9 for any i V(C n) with the condition of Lemma 13, and it is natural to consider the graph D T 10,where T = T (3, 6), (4, 5), (5, 4), (6, 3)}However,D T 10 has 4845 vertices, whose vertex number is much more than that of D 10 Therefore, Lemmas 4 and 13 play the key role in computing the exact values of γ(c 10 C n ),andthishelpstomakemany hard instances become easy to solve

6 6 Applied Mathematics Table 1: Constructions for upper bounds of C 13 C n r (c, l) (1, 0) (7, 0) (, 0) (4, 0) A k B (5, 0) (1, 1) A k C (, 1) r (c, l) (4, 1) A k D (5, 1) (1, ) (6, 1) (, ) (4, ) A k E (5, ) r (c, l) (1, 3) A k F (, 3) (4, 3) A k G (5, 3) (1, 4) (6, 3) 5 Bounds on the Domination Number of C 13 C n We will conclude this work with upper bounds for the domination number of C 13 C n Theorem 16 Let n 13Then γ(c 13 C n ) 9n, n 0 (mod 6), n + 1, n 3, 6, 0, 3 (mod 6), 9n +, n 9, 10, 1, 14, 16, 17 (mod 6), 9n + 3, n 4, 7, 8, 11, 13, 15, 18, 19, (mod 6), 9n + 4, n 1,, 5, 1, 4, 5 (mod 6) (1) Proof We show the desired weight for all cases in Table 1, where r n(mod 6) and (c, l) stands for case (c) with l in Lemma 6 For example, the result of the case n 0(mod 6) follows from case 1 with l=0in Lemma 6 Also, patterns A G with 13 rows are presented (see Figures 5 and 6) By repeating the pattern of A for k times and concatenating H B,C,D,E,F,G},we will obtain a pattern A k H of C 13 C n,wheren = 6k+c(H)Fromtheabove,itcan be seen that all the desired upper bounds are established A B C D Figure 5: AB: dominating set of C 13 C 30 ; AC: dominating set of C 13 C 33 ; AD: dominating set of C 13 C E F G Figure 6: AE: dominating set of C 13 C 4 ; AF: dominating set of C 13 C 45 ; AG: dominating set of C 13 C 48 Disclosure The authors confirm that they have given due consideration to the protection of intellectual property associated with this work and that there are no impediments to publication, including the timing of publication, with respect to intellectual property In so doing, the authors confirm that they have followed the regulations of their institutions concerning intellectual property The authors understand that the corresponding author is the sole contact for the editorial process (including editorial manager and direct communications with the office) He/she is responsible for communicating with the other authors about progress, submissions of revisions, and final approval of proofs The authors confirm that they have provided current, correct addresses (fnlang56@163com and zshao@cdueducn) which is accessible by the corresponding author and which has been configured to accept from Conflict of Interests The authors wish to confirm that there is no known conflict of interests associated with this paper and there has been no significant financial support for this work that could have influenced its outcome The authors confirm that the paper has been read and approved by all named authors and that there are no other persons who satisfied the criteria for authorship but are not listed The authors further confirm that the order of authors listed in the paper has been approved by all of them

7 Applied Mathematics 7 Acknowledgments The authors would like to acknowledge the referees for their valuable comments and suggestions This work was supported by the National Natural Science Foundation of China under Grants nos and References [1] R Hammack, W Imrich, and S Klavžar, Handbook of Product Graphs, Discrete Mathematics and Its Applications, CRC Press, Boca Raton, Fla, USA, nd edition, 011 [] I Gorodezky, Domination in kneser graphs [PhD thesis], University of Waterloo, Ontario, Canada, 007 [3] L L Kelleher and M B Cozzens, Dominating sets in social network graphs, Mathematical Social Sciences,vol16,no3,pp 67 79, 1988 [4] T W Haynes, S T Hedetniemi, and P J Slater, Fundamentals of Domination ingraphs,vol08ofmonographs and Textbooks in Pure and Applied Mathematics, Marcel Dekker, New York, NY, USA, 1998 [5] R J Faudree and R H Schelp, The domination number for the product of graphs, Congressus Numerantium,vol79,pp9 33, 1990 [6] M El-Zahar, S Khamis, and Kh Nazzal, On the domination number of the cartesian product of the cycle of length n and any graph, Discrete Applied Mathematics, vol155,no4,pp515 5, 007 [7] M El-Zahar and C M Pareek, Domination number of products of graphs, Ars Combinatoria, vol 31, pp 3 7, 1991 [8] T Y Chang and W E Clark, The domination numbers of the 5 nand 6 ngrid graphs, Graph Theory,vol17,no 1, pp , 1993 [9] S Gravier and M Mollard, On domination numbers of cartesian product of paths, Discrete Applied Mathematics, vol 80, no -3, pp 47 50, 1997 [10] B L Hartnell and D F Rall, On dominating the cartesian product of a graph and K, Discussiones Mathematicae Graph Theory,vol4,no3,pp389 40,004 [11] M S Jacobson and L F Kinch, On the domination number of products of graphs I, Ars Combinatoria, vol 18, pp 33 44, 1984 [1] S Klavžar and N Seifter, Dominating cartesian products of cycles, Discrete Applied Mathematics,vol59,no,pp19 136, 1995 [13] J Liu, X Zhang, X Chen, and J Meng, On domination number of cartesian product of directed cycles, Information Processing Letters,vol110,no5,pp ,010 [14] M Mollard, On the domination of cartesian product of directed cycles: results for certain equivalence classes of lengths, Discussiones Mathematicae Graph Theory, vol33,no,pp , 013 [15] X Zhang, J Liu, X Chen, and J Meng, Domination number of cartesian products of directed cycles, Information Processing Letters, vol 111, no 1, pp 36 39, 010 [16] S Klavžar and J Žerovnik, Algebraic approach to fasciagraphs and rotagraphs, Discrete Applied Mathematics, vol68,no1-, pp , 1996 [17] P Pavlič and J Žerovnik, Roman domination number of the cartesian products of paths and cycles, Electronic Combinatorics,vol19,no3,p19,01

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