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 u N (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 ω(f) = v V f(v). The -rainbow domination number of a digraph D, denoted by γ 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 γ 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) for the outdegree of a vertex v and d D (v) for its indegree. The minimum and maximum indegree and minimum and maximum outdegree of D are denoted by δ = δ (D), = (D), δ + = δ + (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 a vertex v of a digraph D, we denote the set of in-neighbors and out-neighbors of v by N (v) = N D (v) and N + (v) = N + D (v), respectively. Let N [v] = N (v) {v} and N + [v] = N + (v) {v}. For S V (D), we define N + [S] = v S N + [v]. If X V (D), then D[X] is the subdigraph induced by X. If X V (D) and v 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) = v V f(v), and for S V, we define f(s) = v S 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 γ(d) is the minimum 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 ω(f) = v V f(v). The Roman domination number of a digraph D, denoted by γ R (D), equals the minimum weight of an RDF on D. A γ R (D)-function is a Roman dominating function of D with weight γ 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 ) to refer f) of V, where V i = {v V f(v) = i}. For a positive integer k, a k-rainbow dominating function (krdf) of a digraph D is a function f from the vertex set V (D) to the set of all subsets of the set {1,,..., k} such that for any vertex v V (D) with f(v) = the condition u N (v) f(u) = {1,,..., k} is fulfilled. The weight of a krdf f is the value ω(f) = v V f(v). The k-rainbow domination number of a digraph D, denoted by γ rk (D), is the minimum weight of a krdf of D. A γ rk (D)-function is a k-rainbow dominating function of D with weight γ rk (D). Note that γ r1 (D) is the classical domination number γ(d). A -rainbow dominating function (briefly, rainbow dominating function) f : V P({1, }) can be represented by the ordered partition (V 0, V 1, V, V 1, ) (or (V f 0, V f 1, V f, V f 1,) to refer f) of V, 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 ω(f) = V 1 + V + V 1,. Since V 1 V V 1, is a dominating set when f is a RDF, and since assigning {1, } to the vertices of a dominating set yields an RDF, we have (1.1) γ(d) γ r (D) γ(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 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 γ 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, has at least two out-private neighbors relative to V in the digraph H. Proposition 1.. Let k 1 be an integer. If D is a digraph of order n, then min{k, n} γ rk (D) n.

3 THE RAINBOW DOMINATION NUMBER OF A DIGRAPH 59 In particular, γ rn (D) = n. Proof. Let f be a γ 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 γ rk (D). If f(v) is nonempty for all vertices v V (D), then n γ 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 γ rk (D) n. Proposition 1.3. Let k 1 be an integer. If D is a digraph of order n, then γ rk (D) n + (D) + k 1. Proof. Let v be a vertex of maximum outdegree + (D). Define f : V (D) P({1,,..., k}) by f(v) = {1,,..., k}, f(x) = if x N + (v) and f(x) = {1} otherwise. It is easy to see that f is a k-rainbow dominating function of D and thus γ rk (D) n + (D) + k 1. Let k 1 be an integer, and let D be a digraph of order n k and maximum outdegree + (D) = n 1. Since n k, Proposition 1. leads to γ rk (D) k. Hence it follows from Proposition 1.3 that k γ rk (D) n + (D) + k 1 = k and therefore γ 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 γ R(D) γ r (D) γ R (D). Proof. The upper bound is immediate by definition. To prove the lower bound, let f be a γ r (D)-function and let X i = {v V (D) i f(v)} for i = 1,. We may assume that X 1 X. Then X 1 ( X 1 + X )/ = γ r (D)/. Define g : V (D) {0, 1, } by g(u) = 0 if f(u) =, g(u) = 1 when f(u) = {} and g(u) = if 1 f(u). Obviously, g is a Roman dominating function on D with ω(g) X 1 + X 3γ r(d) and the result follows. Lemma.1. Let k 0 be an integer and let D be a digraph of order n. (i) If γ R (D) = γ r (D) + k, then there exists a subset U of V (D) such that N + [U] = n γ R (D) + U. In particular, there exists a subset U of V (D) such that N + [U] = n γ r (D) + U k. (ii) For each U V (D), γ R (D) n ( N + [U] U ). In particular, if there exists a subset U of V (D) such that N + [U] = n γ r (D) + U k, then γ R (D) γ r (D) + k. (iii) If U is a subset of V (D) such that N + [U ] U is maximum, then γ 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 ) be a γ R (D)-function. By Proposition 1.1 (a), N + [V ] = V 0 V. Since γ R (D) = V 1 + V, we have N + [V ] = V 0 + V = (n V 1 V ) + V = n (γ 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) is an RDF of D of weight n ( N + [U] U ) and hence γ R (D) n ( N + [U] U ). (iii) It follows from (i) and the choice of U that when γ R (D) = γ r (D) + k. By (ii), we obtain N + [U ] U n γ r (D) k = n γ R (D), γ R (D) n ( N + [U ] U ) n (n γ R (D)) = γ R (D) and the proof is complete. Theorem.. Let k 0 be an integer and let D be a digraph of order n. Then γ R (D) = γ r (D) + k if and only if (i) there exists no subset U of V (D) such that n γ r (D) + U k + 1 N + [U] n γ r (D) + U and (ii) there exists a subset U of V (D) such that N + [U] = n γ r (D) + U k. Proof. The proof is by induction on k. Let k = 0. If γ R (D) = γ 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 γ R (D) γ r (D). By Theorem.1, we have γ R (D) = γ r (D), as desired. Therefore we may assume that k 1 and the theorem is true for each m k 1. Let γ R (D) = γ r (D) + k. By Lemma.1 (i), it suffices to show that (i) holds. Assume to the contrary that there exists an integer 0 i k 1 and a subset U V (D) such that n γ r (D) + U i N + [U]. By choosing (U, i) so that i is as small as possible and by the inductive hypothesis we have γ R (D) γ r (D) + i < γ 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 γ R (D) γ r (D) + k. Suppose i = γ R (D) γ r (D). If 0 i k 1, then by the inductive hypothesis there exists a set U V (D) such that n γ r (D) + U i = N + [U] which contradicts (i). Hence γ R (D) = γ r (D) + k and the proof is complete. Theorem.3. Let D be a digraph of order n. Then n γ r (D). + + (D)

5 THE RAINBOW DOMINATION NUMBER OF A DIGRAPH 61 Proof. Let f = (V 0, V 1, V, V 1, ) be a γ r (D)-function. Then γ 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 + ( V + V 1, ) and V 0 + ( V 1 + V 1, ). Hence, we conclude that as desired. ( + + )γ r (D) = ( + + )( V 1 + V + V 1, ) = ( + + )( V 1 + V ) + 4 V 1, + + V 1, V 1 + V + V V 1, = n + V 1, n, The proof of Theorem.3 shows that γ r (D) n (D) if there exists a γ r (D)-function f such that V f 1,. Let G be a graph and γ r (G) its -rainbow domination number. If G is of order n and maximum degre, then Theorem.3 implies immediately the known bound γ r (G) n/( + ), given by Sheikhoelslami and Volkmann [6]. Next we characterize the digraphs D with γ r (D) =. Proposition.1. Let D be a digraph of order n. Then γ r (D) = if and only if n = or n 3 and + (D) = n 1 or there exist two different vertices u and v such that V (D) {u, v} N + (u) and V (D) {u, v} N + (v). Proof. If n = or n 3 and + (D) = n 1 or there exist two different vertices u and v such that V (D) {u, v} N + (u) and V (D) {u, v} N + (v), then it is easy to see that γ r (D) =. Conversely, assume that γ r (D) =. Let f = (V 0, V 1, V, V 1, ) be a γ r (D)-function. Clearly, = γ r (D) = V 1 + V + V 1, and thus V 1, 1. If V 1, = 1, then V 1 = V = 0 and hence + (D) = n 1. If V 1, = 0, then V 1, V. If V 1 = 0 or V = 0, then we deduce that n =. In the remaining case that 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) and V (D) {u, v} N + (v). Proposition.. Let D be a digraph of order n. Then γ r (D) < n if and only if + (D) or (D). Proof. Let f = (V 0, V 1, V, V 1, ) be a γ r (D)-function of D. The hypothesis V 0 + V 1 + V + V 1, = n > γ 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 1 and 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 u V 1, d + D (u) V 0 V 1, + 1. If we suppose on the contrary that + (D) 1, then we arrive at the contradiction V 1, u V 1, d + D (u) V 1, + 1. If + (D), then Proposition 1.3 implies that γ r (D) n + (D) + 1 < n. If (D), then assume that u is a vertex with in-degree (D) and let v, w N (u) be two distinct vertices. Then ({u}, V (D) {u, v}, {v}, ) is a rainbow dominating function of D implying that γ r (D) < n, and the proof is complete. Corollary.1. If D is a directed path or directed cycle of order n, then γ r (D) = n. Next we characterize the digraphs which attain the lower bound in (1.1). Proposition.3. Let D be a digraph on n vertices. Then γ(d) = γ r (D) if and only if D has a γ(d)-set S that partitions into two nonempty subsets S 1 and S such that N + (S 1 ) = V (D) (S 1 S ) and N + (S ) = V (D) (S 1 S ). Proof. Assume that γ(d) = γ r (D) and let f = (V 0, V 1, V, V 1, ) be a γ r -function of D. If γ(d) = γ r (D) = n, then clearly D is empty and the result is immediate. Let γ(d) = γ r (D) < n. Then the assumption implies that we have equality in γ(d) V 1 + V + V 1, V 1 + V + V 1, = γ 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 ) and V (D) (V 1 V ) N + (V ). If there is an arc (a, b) in D[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 ) and V (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 ) and N + (S ) = V (D) (S 1 S ). It is straightforward to verify that the function (V (D) (S 1 S ), S 1, S, ) is a rainbow dominating function of D of weight γ(d) and hence γ(d) = γ r (D). This completes the proof. 3. Cartesian product of directed cycles Let D 1 = (V 1, A 1 ) and D = (V, A ) be two digraphs which have disjoint vertex 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 ) of D 1 D, (x 1, x )(y 1, y ) A(D 1 D ) if one of the following holds: (i) x 1 = y 1 and x y A(D ); (ii) x 1 y 1 A(D 1 ) and x = 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 γ r (C m C n )-function and set a k = x V (Cm k ) f(x) for any k {1,,..., n}. Then γ r (C m C n ) = 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 γ r (C m C n ) = γ r (C n C m ). { n if n is even Theorem 3.1. For n, γ r (C C n ) = n + 1 if n is odd. Proof. First let n be even. Define f : V (C C n ) P({1, }) by f((1, i 1)) = {1} for 1 i n, f((, i)) = {} for 1 i n and f(x) = otherwise. Obviously, f is a RDF of C C n of weight n and hence γ r (C C n ) n. On the other hand, since + (C C n ) =, it follows from Theorem.3 that γ r (C C n ) = n. Now let n be odd. We claim that γ r (C C n ) n+1. Assume to the contrary that γ r (C C n ) n. Let f be a γ r (C C n )-function and set a k = x V (C k ) f(x) for any k {1,,..., n}. If a k = 0 for some k, say k = 3, then f((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) for x V (C C n ) {(1, ), (, ), (, 1), (, 3)} is a RDF of C C n of weight less than ω(f) which is a contradiction. Thus a k 1 for each k. By assumption a k = 1 for each k. We may assume, without loss of generality, that f((1, )) = {1}. To dominate (, ), we must have f((, 1)) = {}. Since a 3 = 1 and f((, )) =, to dominate ((1, 3)), we must have f((, 3)) = {}. Repeating this process we obtain f((1, i)) = {1} for 1 i n 1 and f((, i 1)) = {} for 1 i n+1 and f(x) = otherwise. But then the vertex (1, 1) is not dominated, a contradiction. Thus γ r (C C n ) n + 1. Define g : V (C C n ) P({1, }) by g((1, 1)) = {1}, g((1, i)) = {1} for 1 i n 1 and g((, i 1)) = {} for 1 i n+1 and g(x) = otherwise. Clearly, g is a RDF of C C n of weight n + 1 and hence γ r (C C n ) = n + 1. Theorem 3.. For n, γ r (C 3 C n ) = n. Proof. First we prove γ r (C 3 C n ) n. Define g : V (C 3 C n ) P({1, }) as follows: If n 0 (mod 3), then g((1, 3i+1)) = g((, 3i+)) = g((3, 3i+3)) = {1}, g((1, 3i+ 3)) = g((, 3i + 1)) = g((3, 3i + )) = {} for 0 i n 1 and g(x) = otherwise, 3 if n 1 (mod 3), then g((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 i n 1 1 and g(x) = otherwise, 3 if n (mod 3), then g((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 i n 3 1 and g(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 γ r (C 3 C n ) n. Now we show that γ r (C 3 C n ) n. First we prove that for any γ r (C 3 C n )- function f, x V (Cm) f(x) 1 for each k. Let to the contrary that k x V (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, is a RDF of G of weight less than ω(f), a contradiction. Let g be a γ r (C 3 C n )-function such that the size of the set S = {k a k = x V (C k m) g(x) = 1 and 1 i n} is minimum. We claim that S = 0 implying that γ r (C 3 C n ) = ω(g) n. Assume to the contrary that S 1. Suppose, without loss of generality, that a n = 1 and that g((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, then clearly a 1 and so γ 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, we are 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, }) by h((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, }) by h((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, }) by h((, 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 x V (C h(x) = 1 and 1 i m) k n} < {k x V (C g(x) = 1 and 1 i n} 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, }) by h((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, }) by h((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, }) by h((, 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 x V (Cm k ) h(x) = 1 and 1 i n} < {k x V (Cm k ) g(x) = 1 and 1 i n} which is a contradiction again. Therefore, {k a k = x V (C g(x) = 1 and 1 i n} = 0 and hence m) k γ r (C 3 C n ) = ω(g) n. Thus γ 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 γ r (C m C n ) = mn. Proof. Define f : V (C m C n ) P({1, }) by f((i 1, j 1)) = {1} for each 1 i r and 1 j s, f((i, j)) = {} for each 1 i r and 1 j 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 γ r (C m C n ) mn. Now the results follows from Theorem Strong product of directed cycles Let D 1 = (V 1, A 1 ) and D = (V, A ) be two digraphs which have disjoint vertex 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 ) of D 1 D, (x 1, x )(y 1, y ) A(D 1 D ) if one of the following holds: (i) x 1 y 1 A(D 1 ) and x y A(D ); (ii) x 1 = y 1 and x y A(D ); (iii) x 1 y 1 A(D 1 ) and x = 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 (mod n) 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, γ r (C m C n ) mn. 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 n k=1 a k mn. 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. Then to rainbowly dominate (i, k + 1) for each 1 i m, we must have f((i 1, k)) + f((i, k)). Then a k = 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, n n γ r (C m C n ) = a k = (a k + a k+1 ) nm k=1 where a n+1 = a 1. This implies that γ r (C m C n ) mn. Proposition 4.1. If m = r and n = s for some positive integers r, s, then γ r (C m C n ) = mn. Proof. Define f : V (D) P({1, }) by f((i 1, j 1)) = {1, } for each 1 i r and 1 j 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 γ r (C m C n ) 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 γ r (C m C n ) = mn. Proof. Let k=1 W 1 = {(4i + 1, 1) 0 i r 1}, W = {(4i + 3, 1) 0 i r 1}, Z 1 1 = {(4i +, j) 0 i r 1 and 1 j s}, Z 1 = {(4i + 4, j) 0 i r 1 and 1 j s}, Z 1 = {(4i + 4, j + 1) 0 i r 1 and 1 j s}, Z = {(4i +, j + 1) 0 i r 1 and 1 j s}. Define f : V (C m C n ) P({1, }) by f(x) = {1} for x W 1 Z1 1 Z, 1 f(x) = {} for x W Z1 Z, and f(x) = otherwise. It is easy to see that f is a RDF of C m C n with weight mn and so γ r (C m C n ) mn. It follows from Lemma 4.1 that γ r (C m C n ) = mn. Proposition 4.3. If m = 4r + and n = s + 1 for some positive integers r, s, then mn γ r(c m C n ) mn + 1. Proof. Let C 1 = {(4r +, s + 1)} {(4i + 4, s + 1) 0 i r 1}, C = {(4r + 1, s + 1)} {(4i +, s + 1) 0 i r 1}, C 1, = {(4r + 1, j + 1) 0 j s 1}, Z 1 1 = {(4i + 1, j + 1) 0 i r 1 and 0 j s 1}, Z 1 = {(4i + 1, j) 0 i r 1 and 1 j s}, Z 1 = {(4i + 3, j) 0 i r 1 and 1 j s}, Z = {(4i + 3, j + 1) 0 i r 1 and 0 j s 1}.

11 THE RAINBOW DOMINATION NUMBER OF A DIGRAPH 67 Define f : V (C m C n ) P({1, }) by f(x) = {1, } for x C 1,, f(x) = {1} for x C 1 Z1 1 Z, 1 f(x) = {} for x C Z1 Z, and f(x) = otherwise. It is easy to see that f is a RDF of C m C n with weight mn +1 and so γ r(c m C n ) mn +1. It follows from Lemma 4.1 that mn γ r(c m C n ) mn + 1. Lemma 4.. If m = 4r + 3 and n = 4s + 3 for some non-negative integers r, s, then γ r (C m C n ) = mn. Proof. Let R 1 1 = {(4i + 1, 4j + 1) 0 i r and 0 j s}, R 1 = {(4i + 1, 4j + 3) 0 i r and 0 j s}, R 1 = {(4i +, 4j + ) 0 i r and 0 j s}, R = {(4i +, 4j + 4) 0 i r and 0 j s 1}, R 1 3 = {(4i + 3, 4j + 3) 0 i r and 0 j s}, R 3 = {(4i + 3, 4j + 1) 0 i r and 0 j s}, R 1 4 = {(4i + 4, 4j + 4) 0 i r 1 and 0 j s 1}, R 4 = {(4i + 4, 4j + ) 0 i r 1 and 0 j s}. Define f : V (C m C n ) P({1, }) by f(x) = {1} for x R1 1 R 1 R3 1 R4, 1 f(x) = {} for x R1 R R3 R4, and f(x) = otherwise. It is easy to see that f is a RDF of C m C n with weight mn + 1 and so γ r(c m C n ) mn. By Lemma 4.1 we obtain γ r (C m C n ) = mn. Acknowledgment: This work has been supported by the Research Office of Azarbaijan 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