The Italian domination numbers of some products of directed cycles
aa r X i v : . [ c s . D M ] A ug THE ITALIAN DOMINATION NUMBERS OF SOMEPRODUCTS OF DIRECTED CYCLES
KIJUNG KIM
Abstract.
An Italian dominating function on a digraph D with vertexset V ( D ) is defined as a function f : V ( D ) → { , , } such that everyvertex v ∈ V ( D ) with f ( v ) = 0 has at least two in-neighbors assigned 1under f or one in-neighbor w with f ( w ) = 2. In this paper, we determinethe exact values of the Italian domination numbers of some products ofdirected cycles. Key words:
Italian dominating function; Italian domination number;cartesian product; strong product Introduction and preliminaries
Let D = ( V, A ) be a finite simple digraph with vertex set V = V ( D )and arc set A = A ( D ). An arc joining v to w is denoted by v → w . The maximum out-degree and maximum in-degree of D are denoted by ∆ + ( D )and ∆ − ( D ), respectively.Let D = ( V , A ) and D = ( V , A ) be two digraphs. The cartesianproduct of D and D is the digraph D (cid:3) D with vertex set V × V and fortwo vertices ( x , x ) and ( y , y ),( x , x ) → ( y , y )if one of the following holds: (i) x = y and x → y ; (ii) x → y and x = y .The strong product of D and D is the digraph D ⊗ D with vertex set V × V and for two vertices ( x , x ) and ( y , y ),( x , x ) → ( y , y )if one of the following holds: (i) x → y and x → y ; (ii) x = y and x → y ; (iii) x → y and x = y .An Italian dominating function (IDF) on a digraph D is defined as afunction f : V ( D ) → { , , } such that every vertex v ∈ V ( D ) with f ( v ) = 0has at least two in-neighbors assigned 1 under f or one in-neighbor w with f ( w ) = 2. In other words, we say that a vertex v for which f ( v ) ∈ { , } dominates itself, while a vertex v with f ( v ) = 0 is dominated by f if ithas at least two in-neighbors assigned 1 under f or one in-neighbor w with f ( w ) = 2. An Italian dominating function f : V ( D ) → { , , } gives apartition { V , V , V } of V ( D ), where V i := { x ∈ V ( D ) | f ( x ) = i } . The weight of an Italian dominating function f is the value ω ( f ) = f ( V ( D )) = Date : August 25, 2020.2010
Mathematics Subject Classification. u ∈ V ( D ) f ( u ). The Italian domination number of a digraph D , denotedby γ I ( D ), is the minimum taken over the weights of all Italian dominatingfunctions on D . A γ I ( D ) -function is an Italian dominating function on D with weight γ I ( D ).The Italian dominating functions in graphs and digraphs have studiedin [1, 2, 3, 4, 8, 9]. The authors of [1] introduce the concept of Italiandomination and present bounds relating the Italian domination number tosome other domination parameters. The authors of [2] characterize the trees T for which γ ( T ) + 1 = γ I ( T ) and also characterize the trees T for which γ I ( T ) = 2 γ ( T ). After that, there are some studies on the cartesian productsof undirected cycles or undirected paths in [3, 5, 6, 7]. Recently, the author of[8] initiated the study of the Italian domination number in digraphs. In thispaper, we investigate the Italian domination numbers of cartesian productsand strong products of directed cycles.The following results are useful to our study. Proposition 1.1 ([8]) . Let D be a digraph of order n . Then γ I ( D ) ≥⌈ n + ( D ) ⌉ . Proposition 1.2 ([8]) . Let D be a digraph of order n . Then γ I ( D ) ≤ n and γ I ( D ) = n if and only if ∆ + ( D ) , ∆ − ( D ) ≤ . Proposition 1.3 ([8]) . If D is a directed path or a directed cycle of order n , then γ I ( D ) = n . The Italian domination numbers of some products ofdirected cycles
In this section, we determine the exact values of the Italian dominationnumbers of some products of directed cycles.First, we consider the cartesian product of directed cycles. We denotethe vertex set of a directed cycle C m by { , , . . . , m } , and assume that i → i + 1 is an arc of C m . For every vertex ( i, j ) ∈ V ( C m (cid:3) C n ), the firstand second components are considered modulo m and n , respectively. Foreach 1 ≤ k ≤ n , we denote by C km the subdigraph of C m (cid:3) C n induced bythe set { ( j, k ) | ≤ j ≤ m } . Note that C km is isomorphic to C m . Let f bea γ I ( C m (cid:3) C n )-function and set a k = P x ∈ V ( C km ) f ( x ). Then γ I ( C m (cid:3) C n ) = P nk =1 a k . It is easy to see that C m (cid:3) C n is isomorphic to C n (cid:3) C m . So, γ I ( C m (cid:3) C n ) = γ I ( C n (cid:3) C m ). Theorem 2.1. If m = 2 r and n = 2 s for some positive integers r, s , then γ I ( C m (cid:3) C n ) = mn .Proof. Define f : V ( C m (cid:3) C n ) → { , , } by f ((2 i − , j − f ((2 i, j )) = 1for each 1 ≤ i ≤ r and 1 ≤ j ≤ s , and f (( x, y )) = 0otherwise. It is easy to see that f is an IDF of C m (cid:3) C n with weight mn andso γ I ( C m (cid:3) C n ) ≤ mn . Since ∆ + ( D ) = 2, it follows from Proposition 1.1that γ I ( C m (cid:3) C n ) ≥ mn . Thus, we have γ I ( C m (cid:3) C n ) = mn . (cid:3) heorem 2.2. For an odd integer n ≥ , γ I ( C (cid:3) C n ) = n + 1 .Proof. Define f : V ( C (cid:3) C n ) → { , , } by f ((1 , j − ≤ j ≤ n +12 , f ((2 , j )) = 1for each 1 ≤ j ≤ n − , f ((2 , n )) = 1and f (( x, y )) = 0otherwise. It is easy to see that f is an IDF of C (cid:3) C n with weight n + 1and so γ I ( C (cid:3) C n ) ≤ n + 1.Now we claim that γ I ( C (cid:3) C n ) ≥ n + 1. Suppose to the contrary that γ I ( C (cid:3) C n ) ≤ n . Let f be a γ I ( C (cid:3) C n )-function. If a k = 0 for some k ,assume without loss of generality k = 3, then f ((1 , f ((2 , ,
3) and (2 , f ((1 , f ((2 , g : V ( C (cid:3) C n ) → { , , } by g ((1 , g ((2 , g ((2 , , g ((2 , g (( x, y )) = f (( x, y ))otherwise. Then g is an IDF of C (cid:3) C n with weight less than ω ( f ), which isa contradiction. Thus, a k ≥ k . By assumption, a k = 1 for each k . Without loss of generality, we assume that f ((1 , , f ((2 , a = 1 and f ((2 , f ((2 , f ((1 , i )) = 1 foreach 1 ≤ i ≤ n − , f ((2 , i − ≤ i ≤ n +12 and f (( x, y )) = 0otherwise. But, the vertex (1 ,
1) is not dominated, a contradiction. Thuswe have γ I ( C (cid:3) C n ) ≥ n + 1. This completes the proof. (cid:3) Theorem 2.3.
For an integer n ≥ , γ I ( C (cid:3) C n ) = 2 n .Proof. When n = 3 r for some positive integer r , define f : V ( C (cid:3) C n ) →{ , , } by f ((1 , j + 1)) = f ((2 , j + 1)) = 1for each 0 ≤ j ≤ n − f ((2 , j + 2)) = f ((3 , j + 2)) = 1for each 0 ≤ j ≤ n − f ((1 , j + 3)) = f ((3 , j + 3)) = 1for each 0 ≤ j ≤ n − f (( x, y )) = 0otherwise.When n = 3 r + 1 for some positive integer r , define f : V ( C (cid:3) C n ) →{ , , } by f ((2 , n )) = f ((3 , n )) = 1 nd f (( x, y )) = f (( x, y ))otherwise.When n = 3 r + 2 for some positive integer r , define f : V ( C (cid:3) C n ) →{ , , } by f ((1 , n − f ((2 , n − f ((1 , n )) = f ((3 , n )) = 1and f (( x, y )) = f (( x, y ))otherwise. It is easy to see that f i ( i = 0 , ,
2) is an IDF of C (cid:3) C n withweight 2 n and so γ I ( C (cid:3) C n ) ≤ n .Now we prove that γ I ( C (cid:3) C n ) ≥ n . Let f be a γ I ( C (cid:3) C n )-function. Claim 1. a k ≥ ≤ k ≤ n .Proof. Suppose to the contrary that a k = 0 for some k , say k = n . Todominate (1 , n ), (2 , n ) and (3 , n ), we must have f ((1 , n − f ((2 , n − f ((3 , n − g defined by g ((1 , n − g ((2 , n − g ((3 , n − ,g ((1 , n )) = g ((2 , n )) = 1and g (( x, y )) = f (( x, y ))otherwise, is an IDF of C (cid:3) C n with weight less than ω ( f ). This is ancontradiction. (cid:3) We choose a γ I ( C (cid:3) C n )-function h so that the size of M h := { k | a k = 1 } is as small as possible. Claim 2. | M h | = 0.Proof. Suppose to the contrary that | M h | ≥
1. Without loss of generality,assume that a n = 1 and h ((1 , n )) = 1. To dominate (2 , n ) and (3 , n ), wemust have h ((2 , n − h ((3 , n − n = 3, then clearly a ≥ γ I ( C (cid:3) C ) ≥
6. However, when n = 3, the previously definedfunction f is a γ I ( C (cid:3) C )-function such that ω ( f ) = 6 and | M f | = 0.This contradicts the choice of h . From now on, assume n ≥
4. We divideour consideration into the following two cases.Case 1. a n − = 1.By the same argument as above, we have a n − ≥
3. So a n − + a n − + a n − + a n ≥
8. If n = 4, then the previously defined function f induces acontradiction. Suppose n ≥
5. Since a n − ≥ h (( i, n − i ∈ { , , } . Without loss of generality, we may assume h ((1 , n − t : V ( C (cid:3) C n ) → { , , } by t ((1 , n − t ((2 , n − t ((1 , n − t ((2 , n )) = 0 ,t ((2 , n − t ((1 , n − t ((2 , n − t ((1 , n )) = 1 ,t ((3 , n − t ((3 , n − t ((3 , n − t ((3 , n )) = 1and t (( x, y )) = h (( x, y )) therwise. Then it is easy to see that t is an IDF of C (cid:3) C n such that | M t | < | M h | . This contradicts the choice of h .Case 2. a n − ≥ a n − + a n − + a n ≥
6. Since a n − ≥ h (( i, n − i ∈ { , , } . Without loss of generality, we may assume h ((1 , n − t : V ( C (cid:3) C n ) → { , , } by t ((1 , n − t ((2 , n − t ((3 , n )) = 0 ,t ((2 , n − t ((1 , n − t ((1 , n )) = 1 ,t ((3 , n − t ((3 , n − t ((2 , n )) = 1and t (( x, y )) = h (( x, y ))otherwise. Then it is easy to see that t is an IDF of C (cid:3) C n such that | M t | < | M h | . This contradicts the choice of h . (cid:3) By Claims 1 and 2, we have γ I ( C (cid:3) C n ) ≥ n . This completes the proof. (cid:3) Next, we consider the strong product of directed cycles. We denote thevertex set of a directed cycle C m by { , , . . . , m } , and assume that i → i + 1 is an arc of C m . For every vertex ( i, j ) ∈ V ( C m ⊗ C n ), the first andsecond components are considered modulo m and n , respectively. For each1 ≤ k ≤ n , we denote by C km the subdigraph of C m ⊗ C n induced by theset { ( j, k ) | ≤ j ≤ m } . Note that C km is isomorphic to C m . Let f be a γ I ( C m ⊗ C n )-function and set a k = P x ∈ V ( C km ) f ( x ). Then γ I ( C m ⊗ C n ) = P nk =1 a k . Lemma 2.4.
For positive integers m, n ≥ , γ I ( C m ⊗ C n ) ≥ ⌈ mn ⌉ .Proof. Note that the vertices of C km are dominated by vertices of C k − m or C km . It suffices to verify that P nk =1 a k ≥ ⌈ mn ⌉ . In order to do, we claim a k + a k +1 ≥ m for each k . First of all, we assume that a k +1 = 0. Then todominate ( i, k + 1) for each 1 ≤ i ≤ m , we must have f (( i − , k )) + f (( i, k )) ≥ . Then 2 a k = P mi =1 ( f (( i − , k )) + f (( i, k ))) ≥ m and hence a k + a k +1 ≥ m .If a k +1 = t >
0, then there will be at least m − t vertices in V that willbe dominated only by vertices of C km . This fact induces a k ≥ m − t and so a k + a k +1 ≥ m . Therefore, we have2 γ I ( C m ⊗ C n ) = 2 n X k =1 a k = n X k =1 ( a k + a k +1 ) ≥ nm. This completes the proof. (cid:3)
Theorem 2.5.
For positive integers m, n ≥ , γ I ( C m ⊗ C n ) = ⌈ mn ⌉ .Proof. We divide our consideration into the following two cases.Case 1. m or n is even. ince C m ⊗ C n is isomorphic to C n ⊗ C m , we may assume that n = 2 s forsome positive integers s .Define f : V ( C m ⊗ C n ) → { , , } by f (( i, j − ≤ i ≤ m and 1 ≤ j ≤ s , and f (( x, y )) = 0otherwise. It is easy to see that f is an IDF of C m ⊗ C n with weight mn and so γ I ( C m ⊗ C n ) ≤ ⌈ mn ⌉ . Thus, it follows from Lemma 2.4 that γ I ( C m ⊗ C n ) = ⌈ mn ⌉ .Case 2. m = 2 r + 1 , n = 2 s + 1 for some positive integers r, s .Define f : V ( C m ⊗ C n ) → { , , } by f ((2 i + 1 , j + 1)) = 1for each 0 ≤ i ≤ r and 0 ≤ j ≤ s , f ((2 i, j )) = 1for each 1 ≤ i ≤ r and 1 ≤ j ≤ s and f (( x, y )) = 0otherwise. It is easy to see that f is an IDF of C m ⊗ C n with weight( r + 1)( s + 1) + rs and so γ I ( C m ⊗ C n ) ≤ ⌈ mn ⌉ . Thus, it follows fromLemma 2.4 that γ I ( C m ⊗ C n ) = ⌈ mn ⌉ . (cid:3) Conclusions
In this paper, we determined the exact values of γ I ( C (cid:3) C l ), γ I ( C (cid:3) C l )and γ I ( C m (cid:3) C n ) for an integer l and even integers m, n . The other cases arestill open. We conclude by giving a conjecture. Conjecture 3.1.
For an odd integer n , γ I ( C (cid:3) C n ) = 2 n + 2 . References [1] M. Chellali, T. W. Haynes, S.T. Hedetniemi, A. A. McRae, Roman { } -domination, Discrete Appl. Math.
204 (2016) 22–28.[2] M. A. Henning, W. F. Klostermeyer, Italian domination in trees,
Discrete Appl. Math.
217 (2017) 557–564.[3] H. Gao, P. Wang, E. Liu, Y. Yang, More results on Italian domination in C n (cid:3) C m , Mathematics P ( n, Mathematics C n (cid:3) P m . IEEE Access { } -domination number of Cartesian products ofcycles, J. Comb. Optim.
35 (2018) 75-85.[7] Z. Stepien, A. Szymaszkiewicz, L. Szymaszkiewicz, M. Zwierzchowski, 2-Rainbowdomination number of C n (cid:3) C . Discret. Appl. Math.
107 (2014) 113-116.[8] L. Volkmann, Italian domination in digraphs,
J. Combin. Math. Combin. Comput. ,to appear.[9] L. Volkmann, The Italian domatic number of a digraph,
Commun. Comb. Optim. epartment of Mathematics, Pusan National University, Busan 46241, Re-public of Korea E-mail address : [email protected]@pusan.ac.kr