THE RAINBOW DOMINATION NUMBER OF A DIGRAPH

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1 Kragujevac Journal of Mathematics Volume 37() (013), Pages THE RAINBOW DOMINATION NUMBER OF A DIGRAPH J. AMJADI 1, A. BAHREMANDPOUR 1, S. M. SHEIKHOLESLAMI 1, AND L. VOLKMANN Abstract. Let D =(V,A) be a finite and simple digraph. A II-rainbow dominating function (RDF) of a digraph D is a function f from the vertex set V to the set of all subsets of the set {1, } such that for any vertex v V with f(v) =; the condition S un (v) f(u) ={1, } is fulfilled, where N (v) is the set of in-neighbors of v. The weight of a RDF f is the value!(f) = P vv f(v). The -rainbow domination number of a digraph D, denoted by r (D), is the minimum weight of a RDF of D. In this paper we initiate the study of rainbow domination in digraphs and we present some sharp bounds for r(d). 1. Introduction Let D be a finite simple digraph with vertex set V (D) =V and arc set A(D) =A. A digraph without directed cycles of length is an oriented graph. The order n = n(d) of a digraph D is the number of its vertices. We write d + D (v) fortheoutdegreeof avertexv and d D (v) foritsindegree. Theminimum and maximum indegree and minimum and maximum outdegree of D are denoted by = (D), = (D), + = + (D) and + = + (D), respectively. If uv is an arc of D, then we also write u! v, and we say that v is an out-neighbor of u and u is an in-neighbor of v. For avertexv of a digraph D, we denote the set of in-neighbors and out-neighbors of v by N (v) =N D (v) andn + (v) =N + D (v), respectively. Let N [v] =N (v) [ {v} and N + [v] =N + (v) [ {v}. ForS V (D), we define N + [S] = S vs N + [v]. If X V (D), then D[X] is the subdigraph induced by X. If X V (D)andv V (D), then E(X, v) is the set of arcs from X to v. Consult [, 7] for the notation and terminology which are not defined here. For a real-valued function f : V (D)! R the weight of f is w(f) = P vv f(v), and for S V, we define f(s) =P vs f(v), so w(f) =f(v ). A vertex v dominates all vertices in N + [v]. A subset S of vertices of D is a dominating set if S dominates V (D). The domination number (D) istheminimum Key words and phrases. Rainbow dominating function, Rainbow domination number, Digraph. 010 Mathematics Subject Classification. Primary: 05C69. Secondary: 05C0. Received: July 5,

2 58 J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN cardinality of a dominating set of D. The domination number of digraphs was introduced by Chartrand, Harary and Yue [1] as the out-domination number and has been studied by several authors (see, for example [4, 8]). A Roman dominating function (RDF) on a digraph D is a function f : V! {0, 1, } satisfying the condition that every vertex v for which f(v) = 0 has an in-neighbor u for which f(u) =. The weight of an RDF f is the value!(f) = P vv f(v). The Roman domination number of a digraph D, denoted by R (D), equals the minimum weight of an RDF on D. A R (D)-function is a Roman dominating function of D with weight R(D). The Roman domination number in digraphs was introduced by Kamaraj and Jakkammal [3] and has been studied in [5]. A Roman dominating function f : V! {0, 1, } can be represented by the ordered partition (V 0,V 1,V )(or(v f 0,V f 1,V f )toreferf) ofv, where V i = {v V f(v) =i}. For a positive integer k, ak-rainbow dominating function (krdf) of a digraph D is afunctionf from the vertex set V (D) tothesetofallsubsetsoftheset{1,,...,k} such that for any vertex v V (D) withf(v) =; the condition S un (v) f(u) = {1,,...,k} is fulfilled. The weight of a krdf f is the value!(f) = P vv f(v). The k-rainbow domination number of a digraph D, denoted by rk (D), is the minimum weight of a krdf of D. A rk (D)-function is a k-rainbow dominating function of D with weight rk(d). Note that r1(d) istheclassicaldominationnum- ber (D). A -rainbow dominating function (briefly, rainbow dominating function) f : V! P({1, }) canberepresentedbytheorderedpartition(v 0,V 1,V,V 1, )(or (V f 0,V f 1,V f,v1,) f toreferf) ofv, where V 0 = {v V f(v) =;}, V 1 = {v V f(v) ={1}}, V = {v V f(v) ={}} and V 1, = {v V f(v) ={1, }}. In this representation, its weight is!(f) = V 1 + V + V 1,. Since V 1 [ V [ V 1, is adominatingsetwhenf is a RDF, and since assigning {1, } to the vertices of a dominating set yields an RDF, we have (1.1) (D) apple r (D) apple (D). Our purpose in this paper is to establish some bounds for the rainbow domination number of a digraph. For S V (D), the S-out-private neighbors of a vertex v of S are the vertices of N + [v]\n + [S {v}]. The vertex v is its own out-private neighbor if v 6 N + [S {v}]. The other out-private neighbors are external, i.e., belong to V S. We make use of the following result in this paper. Proposition 1.1. [3] Let f =(V 0,V 1,V ) be any R(D)-function of a digraph D. Then (a) If w V 1, then N D (w) \ V = ;. (b) Let H = D[V 0 [ V ]. Then each vertex v V with N (v) \ V 6= ;, has at least two out-private neighbors relative to V in the digraph H. Proposition 1.. Let k 1beaninteger.IfD is a digraph of order n, then min{k, n} apple rk (D) apple n.

3 THE RAINBOW DOMINATION NUMBER OF A DIGRAPH 59 In particular, rn(d) =n. Proof. Let f be a rk (D)-function. If there exists a vertex v such that f(v) =;, then the definition yields to f(n (v)) = {1,,...,k} and thus k apple rk (D). If f(v) is nonempty for all vertices v V (D), then n apple rk (D), and the first inequality is proved. Next consider the function g, defined by g(v) ={1} for each v V (D). Then g is a k-rainbow dominating function of weight n, and so rk (D) apple n. Proposition 1.3. Let k 1beaninteger.IfD is a digraph of order n, then rk(d) apple n + (D)+k 1. Proof. Let v be a vertex of maximum outdegree + (D). Define f : V (D)! P({1,,...,k}) byf(v) ={1,,...,k}, f(x) =; if x N + (v) andf(x) ={1} otherwise. It is easy to see that f is a k-rainbow dominating function of D and thus rk(d) apple n + (D)+k 1. Let k 1beaninteger,andletD be a digraph of order n k and maximum outdegree + (D) =n 1. Since n k, Proposition 1. leads to rk (D) k. Hence it follows from Proposition 1.3 that and therefore k apple rk (D) apple n + (D)+k 1=k rk (D) =k. This example shows that Proposition 1.3 is sharp.. Bounds on the rainbow domination number of digraphs Theorem.1. For a digraph D, 3 R (D) apple r (D) apple R (D). Proof. The upper bound is immediate by definition. To prove the lower bound, let f be a r(d)-function and let X i = {v V (D) i f(v)} for i = 1,. We may assume that X 1 apple X. Then X 1 apple ( X 1 + X )/ = r (D)/. Define g : V (D)! {0, 1, } by g(u) =0iff(u) =;, g(u) =1whenf(u) ={} and g(u) =if1f(u). Obviously, g is a Roman dominating function on D with!(g) apple X 1 + X apple 3 r (D) andtheresultfollows. Lemma.1. Let k 0beanintegerandletD be a digraph of order n. (i) If R(D) = r (D) +k, then there exists a subset U of V (D) such that N + [U] = n R(D)+ U. In particular, there exists a subset U of V (D) such that N + [U] = n r(d)+ U k. (ii) For each U V (D), R(D) apple n ( N + [U] U ). In particular, if there exists a subset U of V (D) suchthat N + [U] = n r(d) + U k, then R(D) apple r (D)+k. (iii) If U is a subset of V (D)suchthat N + [U ] U is maximum, then R (D) = n ( N + [U ] U ).

4 60 J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN Proof. (i) Let f =(V 0,V 1,V )bea R (D)-function. By Proposition 1.1 (a), N + [V ]= V 0 [ V. Since R (D) = V 1 + V,wehave N + [V ] = V 0 + V =(n V 1 V )+ V = n ( R (D) V ). Hence V is the desired subset of V (D). (ii) Let U be a subset of V (D). Clearly, f =(N + [U] U, V (D) N + [U],U)isan RDF of D of weight n ( N + [U] U ) andhence R (D) apple n ( N + [U] U ). (iii) It follows from (i) and the choice of U that when R (D) = r (D)+k. By (ii), we obtain N + [U ] U n r(d) k = n R(D), R(D) apple n ( N + [U ] U ) apple n (n R(D)) = R (D) and the proof is complete. Theorem.. Let k 0beanintegerandletD be a digraph of order n. Then R(D) = r (D)+k if and only if (i) there exists no subset U of V (D) suchthat n r(d)+ U k +1apple N + [U] apple n r(d)+ U and (ii) there exists a subset U of V (D) suchthat N + [U] = n r(d)+ U k. Proof. The proof is by induction on k. Let k =0. If R(D) = r (D), then (i) is trivial and (ii) follows from Lemma.1 (i). Now assume that (i) and (ii) hold. It follows from Lemma.1 (ii) that R(D) apple r (D). By Theorem.1, we have R(D) = r (D), as desired. Therefore we may assume that k 1andthetheorem is true for each m apple k 1. Let R(D) = r (D) +k. By Lemma.1 (i), it su ces to show that (i) holds. Assume to the contrary that there exists an integer 0 apple i apple k 1 and a subset U V (D) suchthatn r(d)+ U i apple N + [U]. By choosing (U, i) sothatiis as small as possible and by the inductive hypothesis we have R (D) apple r (D)+i< r(d)+k which is a contradiction. This proves the only part of the theorem. Conversely, assume that (i) and (ii) hold. By Lemma.1 (ii) we need to show that R(D) r(d) +k. Suppose i = R (D) r(d). If 0 apple i apple k 1, then by the inductive hypothesis there exists a set U V (D) suchthatn r(d)+ U i = N + [U] which contradicts (i). Hence R (D) = r (D)+k and the proof is complete. Theorem.3. Let D be a digraph of order n. Then n r(d). + + (D)

5 THE RAINBOW DOMINATION NUMBER OF A DIGRAPH 61 Proof. Let f =(V 0,V 1,V,V 1, )bea r (D)-function. Then r (D) = V 1 + V + V 1, and n = V 0 + V 1 + V + V 1,. Since each vertex of V 0 has at least one in-neighbor in V 1, or at least one in-neighbor in V and at least one in-neighbor in V 1, we deduce that V 0 apple + ( V + V 1, )and V 0 apple + ( V 1 + V 1, ). Hence, we conclude that as desired. ( + +) r (D) =( =( + +)( V 1 + V + V 1, ) + +)( V 1 + V )+4 V 1, + + V 1, V 1 + V + V 0 +4 V 1, =n + V 1, n, The proof of Theorem.3 shows that r(d) n (D) if there exists a r (D)-function f such that V f 1, 6= ;. Let G be a graph and r (G) its -rainbow domination number. If G is of order n and maximum degre,then Theorem.3 implies immediately the known bound r (G) dn/( +)e, given by Sheikhoelslami and Volkmann [6]. Next we characterize the digraphs D with r(d) =. Proposition.1. Let D be a digraph of order n. Then r (D) =ifandonly if n =orn 3and + (D) =n 1orthereexisttwodi erent vertices u and v such that V (D) {u, v} N + (u) andv (D) {u, v} N + (v). Proof. If n =orn 3and + (D) =n 1 or there exist two di erent vertices u and v such that V (D) {u, v} N + (u) andv (D) {u, v} N + (v), then it is easy to see that r(d) =. Conversely, assume that r(d) =. Letf =(V 0,V 1,V,V 1, )bea r (D)-function. Clearly, = r (D) = V 1 + V + V 1, and thus V 1, apple 1. If V 1, =1,then V 1 = V =0andhence + (D) =n 1. If V 1, =0,then V 1, V apple. If V 1 =0 or V =0,thenwededucethatn =. Intheremainingcasethat V 1 = V =1, we assume that V 1 = {u} and V = {v}. The definition of the -rainbow dominating function implies that each vertex of V (D) {u, v} has u and v as an in-neighbor. Consequently, V (D) {u, v} N + (u) andv (D) {u, v} N + (v). Proposition.. Let D be a digraph of order n. Then + (D) or (D). r (D) <nif and only if Proof. Let f =(V 0,V 1,V,V 1, )bea r (D)-function of D. The hypothesis V 0 + V 1 + V + V 1, = n> r (D) = V 1 + V + V 1, implies V 0 V 1, +1. If some vertex w V 0 has no in-neighbor in V 1,, then N (w)\v 1 1and N (w)\v 1

6 6 J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN which implies that (D) d (w). So we may assume each vertex w V 0 has at least one in-neighbor in V 1,. Then we have X uv 1, d + D (u) V 0 V 1, +1. If we suppose on the contrary that + (D) apple 1, then we arrive at the contradiction X V 1, d + D (u) V 1, +1. uv 1, If + (D), then Proposition 1.3 implies that r(d) apple n + (D)+1<n. If (D), then assume that u is a vertex with in-degree (D) andletv, w N (u) betwodistinctvertices.then({u},v(d) {u, v}, {v}, ;) isarainbowdominating function of D implying that r(d) <n, and the proof is complete. Corollary.1. If D is a directed path or directed cycle of order n, then Next we characterize the digraphs which attain the lower bound in (1.1). r (D) =n. Proposition.3. Let D be a digraph on n vertices. Then (D) = r (D) ifand only if D has a (D)-set S that partitions into two nonempty subsets S 1 and S such that N + (S 1 )=V(D) (S 1 [ S )andn + (S )=V(D) (S 1 [ S ). Proof. Assume that (D) = r (D) andletf =(V 0,V 1,V,V 1, )bea r -function of D. If (D) = r (D) =n, then clearly D is empty and the result is immediate. Let (D) = r (D) <n. Then the assumption implies that we have equality in (D) apple V 1 + V + V 1, apple V 1 + V + V 1, = r (D). This implies that V 1, =0 and hence we deduce that each vertex in V 0 has at least one in-neighbor in V 1 and one in-neighbor in V. Therefore, V (D) (V 1 [ V ) N + (V 1 )andv(d) (V 1 [ V ) N + (V ). If there is an arc (a, b) ind[v 1 [ V ], then obviously (V 1 [ V ) {b} is a dominating set of D which is a contradiction. Hence V 1 [ V is dominating set of D with V (D) (V 1 [ V )=N + (V 1 )andv(d) (V 1 [ V )=N + (V ). Conversely, assume that D has a minimum dominating set S that partitions into two nonempty subsets S 1 and S such that N + (S 1 )=V(D) (S 1 [ S )andn + (S )= V (D) (S 1 [ S ). It is straightforward to verify that the function (V (D) (S 1 [ S ),S 1,S, ;)isarainbowdominatingfunctionofd of weight (D)andhence (D) = r(d). This completes the proof. 3. Cartesian product of directed cycles Let D 1 =(V 1,A 1 )andd =(V,A )betwodigraphswhichhavedisjointvertex sets V 1 and V and disjoint arc sets A 1 and A, respectively. The Cartesian product D 1 D is the digraph with vertex set V 1 V and for any two vertices (x 1,x )and (y 1,y )ofd 1 D,(x 1,x )(y 1,y ) A(D 1 D )ifoneofthefollowingholds: (i) x 1 = y 1 and x y A(D ); (ii) x 1 y 1 A(D 1 )andx = y.

7 THE RAINBOW DOMINATION NUMBER OF A DIGRAPH 63 We denote the vertices of a directed cycle C n by the integers {1,,...,n} considered modulo n. Note that N + ((i, j)) = {(i, j +1), (i +1,j)} for any vertex (i, j) V (C m C n ), the first and second digit are considered modulo m and n, respectively. For any k {1,,...,n}, we will denote by Cm k the subdigraph of C m C n induced by the vertices {(j, k) j {1,,...,m}}. Note that Cm k is isomorphic to C m. Let f be a r(c m C n )-function and set a k = P xv (Cm k ) f(x) for any k {1,,...,n}. Then r(c m C n )= P n k=1 a k. It is easy to see that C m C n = Cn C m for any directed cycles of length m, n, and so r (C m C n )= r (C n C m ). n if n is even Theorem 3.1. For n, r(c C n )= n +1 if n is odd. Proof. First let n be even. Define f : V (C C n )! P({1, }) byf((1, i 1)) = {1} for 1 apple i apple n, f((, i)) = {} for 1 apple i apple n and f(x) =; otherwise. Obviously, f is ardfofc C n of weight n and hence r (C C n ) apple n. On the other hand, since + (C C n )=,itfollowsfromtheorem.3that r(c C n )=n. Now let n be odd. We claim that r(c C n ) n+1. Assume to the contrary that r(c C n ) apple n. Let f be a r (C C n )-function and set a k = P xv (C k ) f(x) for any k {1,,...,n}. If a k =0forsomek, say k =3,thenf((1, 3)) = f((, 3)) = ; and to dominate the vertices (1, 3) and (, 3) we must have f((1, )) = {1, } and f((, )) = {1, }, respectively. Then the function g : V (C C n )! P({1, }) by g((1, )) = {1},g((, )) = ;,g((, 1)) = g((, 3)) = {} and g(x) =f(x) forx V (C C n ) {(1, ), (, ), (, 1), (, 3)} is a RDF of C C n of weight less than!(f) which is a contradiction. Thus a k 1foreachk. By assumption a k =1foreachk. We may assume, without loss of generality, that f((1, )) = {1}. To dominate (, ), we must have f((, 1)) = {}. Since a 3 =1andf((, )) = ;, todominate((1, 3)), we must have f((, 3)) = {}. Repeating this process we obtain f((1, i)) = {1} for 1 apple i apple n 1 and f((, i 1)) = {} for 1 apple i apple n+1 and f(x) =; otherwise. But then the vertex (1, 1) is not dominated, a contradiction. Thus r(c C n ) n +1. Define g : V (C C n )! P({1, }) byg((1, 1)) = {1},g((1, i)) = {1} for 1 apple i apple n 1 and g((, i 1)) = {} for 1 apple i apple n+1 and g(x) =; otherwise. Clearly, g is a RDF of C C n of weight n +1andhence r (C C n )=n +1. Theorem 3.. For n, r(c 3 C n )=n. Proof. First we prove r (C 3 C n ) apple n. Define g : V (C 3 C n )! P({1, })asfollows: If n 0(mod3),theng((1, 3i+1)) = g((, 3i+)) = g((3, 3i+3)) = {1},g((1, 3i+ 3)) = g((, 3i +1))=g((3, 3i +))={} for 0 apple i apple n 1andg(x) =; otherwise, 3 if n 1(mod3),theng((3,n))) = {1},g((,n)) = {},g((1, 3i +1))=g((, 3i + )) = g((3, 3i +3))={1},g((1, 3i +3))=g((, 3i +1))=g((3, 3i +))={} for 0 apple i apple n 1 1andg(x) =; otherwise, 3 if n (mod3),theng((1,n)) = g((1,n 1)) = g((3,n)) = {1},g((,n 1)) = {},g((1, 3i+1)) = g((, 3i+)) = g((3, 3i+3)) = {1},g((1, 3i+3)) = g((, 3i+1)) =

8 64 J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN g((3, 3i +))={} for 0 apple i apple n 3 1andg(x) =; otherwise. It is easy to see that in each case, g is a RDF of C 3 C n of weight n and hence r (C 3 C n ) apple n. Now we show that r(c 3 C n ) n. First we prove that for any r (C 3 C n )- function f, P xv (Cm) f(x) 1foreachk. Let to the contrary that P k xv (Cm) f(x) = k 0 for some k, say k = n. Then we must have f((1,n 1)) = f((,n 1)) = f((3,n 1)) = {1, }. Then the function f 1 defined by f 1 ((1,n 1)) = f 1 ((,n 1)) = f 1 ((3,n 1)) = {1}, f 1 ((1,n)) = f 1 ((,n)) = {} and f 1 (x) =f(x) otherwise,isa RDF of G of weight less than!(f), a contradiction. P Let g be a r(c 3 C n )-function such that the size of the set S = {k a k = xv (C g(x) =1and1appleiapplen} is minimum. We claim that S =0implyingthat m) k r(c 3 C n )=!(g) n. Assume to the contrary that S 1. Suppose, without loss of generality, that a n =1andthatg((1,n)) = {1} and g((,n)) = g((3,n)) = ;. To dominate (,n)and(3,n) we must have g((,n 1)) and g((3,n 1)) = {1, }, respectively. If n =3,thenclearlya 1 andso r (C 3 C n ) n. Let n 4. Consider two cases. Case 1. a n =1. As above we have a n 3 3. Thus a n 3 + a n + a n 1 + a n 8. If n =4,weare done. Suppose n 5. Since a n 4 1, 1 [ 3 i=1g((i, n 4)) or [ 3 i=1g((i, n 4)). Let 1 [ 3 i=1g((i, n 4)) (the case [ 3 i=1g((i, n 4)) is similar). If 1 g((3,n 4)), then define h : V (C 3 C n )! P({1, }) byh((1,n 3)) = h((1,n 1)) = h((,n 3)) = h((3,n 1)) = {},h((1,n)) = h((,n)) = h((,n )) = h((3,n )) = {1},h((1,n )) = h((,n 1)) = h((3,n 3)) = h((3,n)) = ; and h(x) =g(x) otherwise. If 1 g((,n 4)), then define h : V (C 3 C n )! P({1, }) byh((1,n 3)) = h((,n 1)) = h((3,n 1)) = h((3,n 3)) = {},h((1,n)) = h((1,n )) = h((,n )) = h((3,n)) = {1},h((,n )) = h((1,n 1)) = h((,n 3)) = h((,n)) = ; and h(x) =g(x) otherwise. If 1 g((1,n 4)), then define h : V (C 3 C n )! P({1, }) byh((,n 3)) = h((,n 1)) = h((3,n 3)) = h((3,n 1)) = {},h((1,n)) = h((1,n )) = h((3,n )) = h((3,n)) = {1},h((,n )) = h((1,n 1)) = h((1,n 3)) = h((,n)) = ; and h(x) =g(x) otherwise. Clearly, h is a RDF of C 3 C n for which {k P xv (C h(x) =1and1appleiapple m) k n} < {k P xv (C g(x) =1and1appleiapplen} which contradicts the choice of g. m) k Case. a n. Then a n +a n 1 +a n 6. Since a n 3 1, 1 [ 3 i=1g((i, n 3)) or [ 3 i=1g((i, n 3)). Assume 1 [ 3 i=1g((i, n 3)) (the case [ 3 i=1g((i, n 3)) is similar). If 1 g((3,n 3)), then define h : V (C 3 C n )! P({1, }) byh((1,n )) = h((,n )) = h((,n 1)) = {},h((1,n)) = h((3,n)) = h((3,n 1)) = {1},h((1,n 1)) = h((,n)) = h((3,n )) = ; and h(x) =g(x) otherwise.

9 THE RAINBOW DOMINATION NUMBER OF A DIGRAPH 65 If 1 g((,n 3)), then define h : V (C 3 C n )! P({1, }) byh((1,n )) = h((,n 1)) = h((3,n )) = {},h((1,n)) = h((3,n)) = h((3,n 1)) = {1},h((1,n 1)) = h((,n )) = h((,n)) = ; and h(x) =g(x) otherwise. If 1 g((1,n 3)), then define h : V (C 3 C n )! P({1, }) byh((,n )) = h((3,n )) = h((3,n 1)) = {},h((1,n)) = h((1,n 1)) = h((,n)) = {1},h((3,n)) = h((1,n )) = h((,n 1)) = ; and h(x) =g(x) otherwise. Clearly, h is a RDF of C 3 C n for which {k P xv (Cm k ) h(x) =1and1appleiapple n} < {k P xv (Cm k ) g(x) =1and1apple i apple n} which is a contradiction again. Therefore, {k a k = P xv (C g(x) = 1 and 1 apple i apple n} = 0 and hence m) k r(c 3 C n )=!(g) n. Thus r(c 3 C n )=n and the proof is complete. Proposition 3.1. If m = r and n = s for some positive integers r, s, then r(c m C n )= mn. Proof. Define f : V (C m C n )! P({1, }) byf((i 1, j 1)) = {1} for each 1 apple i apple r and 1 apple j apple s, f((i, j)) = {} for each 1 apple i apple r and 1 apple j apple s and f(x) =; otherwise. It is easy to see that f is a RDF of C m C n with weight mn and so r (C m C n ) apple mn. Now the results follows from Theorem Strong product of directed cycles Let D 1 =(V 1,A 1 )andd =(V,A )betwodigraphswhichhavedisjointvertex sets V 1 and V and disjoint arc sets A 1 and A, respectively. The strong product D 1 D is the digraph with vertex set V 1 V and for any two vertices (x 1,x )and (y 1,y )ofd 1 D,(x 1,x )(y 1,y ) A(D 1 D )ifoneofthefollowingholds: (i) x 1 y 1 A(D 1 )andx y A(D ); (ii) x 1 = y 1 and x y A(D ); (iii) x 1 y 1 A(D 1 )andx = y. We denote the vertices of a directed cycle C n by the integers {1,,...,n} considered modulo n. There is an arc xy from x to y in C n if and only if y = x +1(modn) and N + ((i, j)) = {(i, j +1), (i +1,j), (i +1,j +1)} for any vertex (i, j) V (C m C n ), the first and second digit are considered modulo m and n, respectively. We use the notation defined in Section 3. Lemma 4.1. For positive integers m, n, r(c m C n ) d mn e. Proof. Observe that the vertices of Cm k are dominated by vertices of Cm k 1 or Cm k, k =1,,...,n. Especially, the vertices of Cm 1 are dominated by Cm 1 and Cm. n We show that P n k=1 a k d mne. In order to this, we show that a k + a k+1 m for each k =1,,...,n, where a n+1 = a 1. First let a k+1 =0. Thentorainbowlydominate (i, k +1) for each 1apple i apple m, we must have f((i 1,k)) + f((i, k)). Then a k = P m i=1 ( f((i 1,k)) + f((i, k)) ) m and hence a k + a k+1 m. If a k+1 = t,

10 66 J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN then it is not hard to see that a k m t and so a k + a k+1 m. Therefore, nx nx r (C m C n )= a k = (a k + a k+1 ) nm k=1 where a n+1 = a 1. This implies that r(c m C n ) d mn e. Proposition 4.1. If m =r and n =sfor some positive integers r, s, then C n )=d mne. k=1 r (C m Proof. Define f : V (D)! P({1, }) byf((i 1, j 1)) = {1, } for each 1 apple i apple r and 1 apple j apple s, andf(x) =; otherwise. It is easy to see that f is a RDF of C m C n with weight mn and so r (C m C n ) apple mn. Now the results follows from Lemma 4.1. Proposition 4.. If m =4r and n =s +1 for some positive integers r, s, then r(c m C n )=d mn e. Proof. Let W 1 = {(4i +1, 1) 0 apple i apple r 1}, W = {(4i +3, 1) 0 apple i apple r 1}, Z 1 1 = {(4i +, j) 0 apple i apple r Z 1 = {(4i +4, j) 0 apple i apple r Z 1 = {(4i +4, j +1) 0 apple i apple r Z = {(4i +, j +1) 0 apple i apple r 1and1apple j apple s}, 1and1apple j apple s}, 1and1apple j apple s}, 1and1apple j apple s}. Define f : V (C m C n )! P({1, }) byf(x) ={1} for x W 1 [ Z1 1 [ Z, 1 f(x) ={} for x W [ Z1 [ Z,andf(x) =; otherwise. It is easy to see that f is a RDF of C m C n with weight mn and so r (C m C n ) apple mn. It follows from Lemma 4.1 that r(c m C n )=d mne. Proposition 4.3. If m =4r +andn =s +1forsomepositiveintegersr, s, then d mneapple r(c m C n ) appled mne +1. Proof. Let C 1 = {(4r +, s +1)} [ {(4i +4, s +1) 0 apple i apple r 1}, C = {(4r +1, s +1)} [ {(4i +, s +1) 0 apple i apple r 1}, C 1, = {(4r +1, j +1) 0 apple j apple s 1}, Z 1 1 = {(4i +1, j +1) 0 apple i apple r 1and0apple j apple s 1}, Z 1 = {(4i +1, j) 0 apple i apple r Z 1 = {(4i +3, j) 0 apple i apple r 1and1apple j apple s}, 1and1apple j apple s}, Z = {(4i +3, j +1) 0 apple i apple r 1and0apple j apple s 1}.

11 THE RAINBOW DOMINATION NUMBER OF A DIGRAPH 67 Define f : V (C m C n )! P({1, }) byf(x) ={1, } for x C 1,, f(x) ={1} for x C 1 [ Z1 1 [ Z, 1 f(x) ={} for x C [ Z1 [ Z,andf(x) =; otherwise. It is easy to see that f is a RDF of C m C n with weight mn +1 and so r(c m C n ) apple mn +1. It follows from Lemma 4.1 that d mneapple r(c m C n ) appled mne +1. Lemma 4.. If m =4r +3andn =4s +3forsomenon-negativeintegersr, s, then r(c m C n )=d mn e. Proof. Let R 1 1 = {(4i +1, 4j +1) 0 apple i apple r and 0 apple j apple s}, R 1 = {(4i +1, 4j +3) 0 apple i apple r and 0 apple j apple s}, R 1 = {(4i +, 4j +) 0 apple i apple r and 0 apple j apple s}, R = {(4i +, 4j +4) 0 apple i apple r and 0 apple j apple s 1}, R 1 3 = {(4i +3, 4j +3) 0 apple i apple r and 0 apple j apple s}, R 3 = {(4i +3, 4j +1) 0 apple i apple r and 0 apple j apple s}, R 1 4 = {(4i +4, 4j +4) 0 apple i apple r 1and0apple j apple s 1}, R 4 = {(4i +4, 4j +) 0 apple i apple r 1and0apple j apple s}. Define f : V (C m C n )! P({1, }) byf(x) ={1} for x R1 1 [ R 1 [ R3 1 [ R4, 1 f(x) ={} for x R1 [ R [ R3 [ R4,andf(x) =; otherwise. It is easy to see that f is a RDF of C m C n with weight b mnc +1 and so r(c m C n ) appled mn e. By Lemma 4.1 we obtain r (C m C n )=d mne. ce of Azarbai- Acknowledgment: This work has been supported by the Research O jan Shahid Madani University. References [1] G. Chartrand, F. Harary and B.Q. Yue, On the out-domination and in-domination numbers of a digraph, Discrete Mathematics 197/198 (1999), [] T. W. Haynes, S. T. Hedetniemi and P. J. Slater, Fundamentals of Domination in graphs, Marcel Dekker, Inc. New York, [3] M. Kamaraj and V. Hemalatha, Directed Roman domination in digraphs, International Journal of Combinatorial Graph Theory and Applications 4 (011), [4] C. Lee, Domination in digraphs, Journal of the Korean Mathematical Society 35 (1998), [5] S. M. Sheikholeslami and L. Volkmann, The Roman domination number of a digraph, Acta Universitatis Apulensis. Mathematics. Informatics 7 (011), [6] S. M. Sheikholeslami and L. Volkmann, The k-rainbow domatic number of a graph, Discussiones Mathematicae. Graph Theory 3 (01),

12 68 J. AMJADI, A. BAHREMANDPOUR, S. M. SHEIKHOLESLAMI, AND L. VOLKMANN [7] D. B. West, Introduction to Graph Theory, Prentice-Hall, Inc, 000. [8] X. D. Zhang, J. Liu, X. Chen and J. Meng, On domination number of Cartesian product of directed cycles, Information Processing Letters 111 (010), Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, I.R. Iran address: j-amjadi@azaruniv.edu address: s.m.sheikholeslami@azaruniv.edu Lehrstuhl II für Mathematik, RWTH Aachen University, 5056 Aachen, Germany address: volkm@math.rwth-aachen.de