The Italian bondage and reinforcement numbers of digraphs
aa r X i v : . [ c s . D M ] A ug THE ITALIAN BONDAGE AND REINFORCEMENT NUMBERSOF DIGRAPHS
KIJUNG KIM
Abstract. An Italian dominating function on a digraph D with vertex set V ( D ) is defined as a function f : V ( D ) → { , , } such that every vertex v ∈ V ( D ) with f ( v ) = 0 has at least two in-neighbors assigned 1 under f or onein-neighbor w with f ( w ) = 2. The weight of an Italian dominating function f isthe value ω ( f ) = f ( V ( D )) = P u ∈ V ( D ) f ( u ). The Italian domination number of a digraph D , denoted by γ I ( D ), is the minimum taken over the weights of allItalian dominating functions on D . The Italian bondage number of a digraph D , denoted by b I ( D ), is the minimum number of arcs of A ( D ) whose removalin D results in a digraph D ′ with γ I ( D ′ ) > γ I ( D ). The Italian reinforcementnumber of a digraph D , denoted by r I ( D ), is the minimum number of extraarcs whose addition to D results in a digraph D ′ with γ I ( D ′ ) < γ I ( D ). In thispaper, we initiate the study of Italian bondage and reinforcement numbers indigraphs and present some bounds for b I ( D ) and r I ( D ). We also determinethe Italian bondage and reinforcement numbers of some classes of digraphs. Key words:
Italian domination number, Italian bondage number, Italian rein-forcement number Introduction
Let D = ( V, A ) be a finite simple digraph with vertex set V = V ( D ) and arc set A = A ( D ). The order n ( D ) of a digraph is the size of V ( D ). For an arc uv ∈ A ( D ),we say that v is an out-neighbor of u and u is an in-neighbor of v . We denote theset of in-neighbors and out-neighbors of v by N − D ( v ) and N + D ( v ), respectively. Wewrite deg − D ( v ) and deg + D ( v ) for the size of N − D ( v ) and N + D ( v ), respectively. Let N − D [ v ] = N − D ( v ) ∪ { v } and N + D [ v ] = N + D ( v ) ∪ { v } . For s subset S of V ( D ), we define N + ( S ) = S v ∈ S N + D ( v ) and N + [ S ] = S v ∈ S N + D [ v ]. The maximum out-degree and maximum in-degree of a digraph D are denoted by ∆ + ( D ) and ∆ − ( D ), respectively.For a digraph D , a subset S of V ( D ) is a dominating set if S v ∈ S N + D [ v ] = V ( D ).The domination number γ ( D ) is the minimum cardinality of a dominating set of D . The concept of the domination number of a digraph was introduced in [2].The bondage number b ( D ) of a digraph D is the minimum number of arcs of A ( D )whose removal in D results in a digraph D ′ with γ ( D ′ ) > γ ( D ). The concept of thebondage number of a digraph was proposed in [1]. The reinforcement number r ( D )of a digraph D is the minimum number of extra arcs whose addition to D resultsin a digraph D ′ with γ ( D ′ ) < γ ( D ). The concept of the reinforcement number of adigraph was introduced in [6].Among the variations of domination, so called Italian domination of graphs isintroduced in [3]. The authors of [3] present bounds relating the Italian dominationnumber to some other domination parameters. The authors of [5] characterize thetrees T for which γ ( T ) + 1 = γ I ( T ) and also characterize the trees T for which Date : August 25, 2020.2010
Mathematics Subject Classification. I ( T ) = 2 γ ( T ). After that, there are many studies on Italian domination of graphsin [7, 8, 9, 11, 13]. Recently, the author of [14] initiated the study of the Italiandomination number in digraphs. Related results was given in [12, 15]. Our aim inthis paper is to initiate the study of Italian bondage and reinforcement numbersfor digraphs.An Italian dominating function (IDF) on a digraph D with vertex set V ( D )is defined as a function f : V ( D ) → { , , } such that every vertex v ∈ V ( D )with f ( v ) = 0 has 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 ) → { , , } givesan ordered partition ( V , V , V ) (or ( V f , V f , V f ) to refer to f ) of V ( D ), where V i := { x ∈ V ( D ) | f ( x ) = i } . The weight of an Italian dominating function f isthe value ω ( f ) = f ( V ( D )) = P u ∈ V ( D ) f ( u ). The Italian domination number of adigraph D , denoted by γ I ( D ), is the minimum taken over the weights of all Italiandominating functions on D . A γ I ( D ) -function is an Italian dominating function on D with weight γ I ( D ).The Italian bondage number of a digraph D , denoted by b I ( D ), is the minimumnumber of arcs of A ( D ) whose removal in D results in a digraph D ′ with γ I ( D ′ ) >γ I ( D ).The Italian reinforcement number of a digraph D , denoted by r I ( D ), is theminimum number of extra arcs whose addition to D results in a digraph D ′ with γ I ( D ′ ) < γ I ( D ). The Italian reinforcement number of a digraph D is defined to be0 if γ I ( D ) ≤
2. A subset R of A ( D ) is called an Italian reinforcement set (IRS) of D if γ I ( D + R ) < γ I ( D ). An r I ( D )-set is an IRS of D with size r I ( D ).This paper is organized as follows. In Section 2, we prepare basic results onthe Italian domination number. In Section 3, we give some bounds of the Italianbondage number and determine the exact values of Italian bondage numbers of someclasses of digraphs. In Section 4, we characterize all digraphs D with r I ( D ) = 1.We give some bounds of the Italian reinforcement number and also determine theexact values of Italian reinforcement numbers of compositions of digraphs.2. The Italian domination numbers
In this paper, we make use of the following results.
Observation 2.1.
For a digraph D , γ I ( D ) ≤ n − ∆ + ( D ) + 1 .Proof. Let D be a digraph, and let v be a vertex with deg + D ( v ) = ∆ + ( D ). Definea function f : V ( D ) → { , , } by f ( v ) = 2, f ( x ) = 0 if x ∈ N + ( v ), and f ( x ) = 1otherwise. It is easy to see that f is an IDF of D . (cid:3) The following result is the exact value of Italian domination number of a com-plete bipartite graph (see [3] for the definition of Italian dominating function anddomination number on a graph).
Lemma 2.2 ([4]) . For a complete bipartite graph K m,n with ≤ m ≤ n and n ≥ , γ I ( K m,n ) = if m ≤ ; if m = 3 ; if m ≥ . Theorem 2.3 ([14]) . Let D be a digraph of order n . Then γ I ( D ) ≥ ⌈ n + ( D ) ⌉ . Theorem 2.4.
Let D be a digraph of order n ≥ . Then γ I ( D ) = 2 if and only if ∆ + ( D ) = n − or there exist two distinct vertices u and v such that V ( D ) \{ u, v } ⊆ N + D ( u ) and V ( D ) \ { u, v } ⊆ N + D ( v ) . roof. If ∆ + ( D ) = n − u and v such that V ( D ) \ { u, v } ⊆ N + D ( u ) and V ( D ) \ { u, v } ⊆ N + D ( v ), then it is easy to see that γ I ( D ) = 2.Assume that γ I ( D ) = 2. Let ( V , V , V ) be a γ I ( D )-function. Then γ I ( D ) =2 = | V | + 2 | V | and | V | ≤
1. If | V | = 1, then | V | = 0 and hence ∆ + ( D ) = n − | V | = 0, then | V | = 2 and, by the definition of IDF, there exist two distinctvertices u and v such that V ( D ) \ { u, v } ⊆ N + D ( u ) and V ( D ) \ { u, v } ⊆ N + D ( v ) (cid:3) Theorem 2.5 ([14]) . Let D be a digraph of order n ≥ . Then γ I ( D ) < n if andonly if ∆ + ( D ) ≥ or ∆ − ( D ) ≥ . Corollary 2.6. If D is a directed path or cycle of order n , then γ I ( D ) = n . The Italian bondage numbers
Bounds of the Italian bondage numbers.
The underlying graph G [ D ] of adigraph D is the graph obtained by replacing each arc uv by an edge uv . Note that G [ D ] has two parallel edges uv when D contains the arc uv and vu . A digraph D is connected if the underlying graph G [ D ] is connected. For a graph G , we denote thedegree of v ∈ V ( G ) by deg G ( v ). In particular, ∆( G ) means the maximum degreein G . Theorem 3.1. If D is a digraph, and xyz a path of length in G [ D ] such that yx, yz ∈ A ( D ) , then b I ( D ) ≤ deg G [ D ] ( x ) + deg − D ( y ) + deg G [ D ] ( z ) − | N − ( x ) ∩ N − ( y ) ∩ N − ( z ) | . Moreover, if x and z are adjacent in G [ D ] , then b I ( D ) ≤ deg G [ D ] ( x ) + deg − D ( y ) + deg G [ D ] ( z ) − − | N − ( x ) ∩ N − ( y ) ∩ N − ( z ) | . Proof.
Let B be the set of all arcs incident with x or z and all arcs terminating at y with the exception of all arcs from N − ( x ) ∩ N − ( z ) to y . Then | B | ≤ deg G [ D ] ( x ) + deg − D ( y ) + deg G [ D ] ( z ) − | N − ( x ) ∩ N − ( y ) ∩ N − ( z ) | and | B | ≤ deg G [ D ] ( x ) + deg − D ( y ) + deg G [ D ] ( z ) − − | N − ( x ) ∩ N − ( y ) ∩ N − ( z ) | when x and z are adjacent.Let D ′ = D − B . In D ′ , x and z are isolated, and all in-neighbors of y in D ′ ,if any, lie in N − ( x ) ∩ N − ( z ). Let f = ( V , V , V ) be a γ I ( D ′ )-function. Then f ( x ) = f ( z ) = 1. If f ( y ) = 2, then( V ∪ { x, z } , V \ { x, z } , V )is an IDF of D with weight less than ω ( f ). If f ( y ) = 1, then( V ∪ { x, z } , V \ { x, y, z } , V ∪ { y } )is an IDF of D with weight less than ω ( f ). However, if f ( y ) = 0, then thereexists w ∈ N − ( x ) ∩ N − ( y ) ∩ N − ( z ) such that f ( w ) = 2 or there exist w , w ∈ N − ( x ) ∩ N − ( y ) ∩ N − ( z ) such that f ( w ) = f ( w ) = 1. Since w, w and w arein-neighbors of x and z in D ,( V ∪ { x, z } , V \ { x, z } , V )is an IDF of D with weight less than ω ( f ). This completes the proof. (cid:3) Theorem 3.2.
Let D be a digraph of order n ≥ . If G [ D ] is connected, then b I ( D ) ≤ ( γ I ( D ) − G [ D ]) . roof. We proceed by induction on γ I ( D ). Assume that γ I ( D ) = 2. For a vertex u ∈ V ∪ V , let B u be the set of arcs incident with u . Since γ I ( D − u ) ≥ n ≥
3, we have γ I ( D − B u ) = γ I ( D − u ) + 1 ≥ . This implies that b I ( D ) ≤ | B u | for u ∈ V ∪ V . Thus, b I ( D ) ≤ ∆( G [ D ]).Assume that the result is true for every digraph with the Italian dominationnumber k ≥
3. Let D be a digraph with γ I ( D ) = k + 1. Suppose to the contrarythat b I ( D ) > ( γ I ( D ) − G [ D ]). Let u be an arbitrary vertex of D , and let B u be the set of arcs incident with u . Then we have γ I ( D ) = γ I ( D − B u ). Let f be a γ I ( D − B u )-function. Then f ( u ) = 1 and the function f restricted to D − u is also a γ I ( D − u )-function. This implies that γ I ( D − u ) = γ I ( D ) −
1. So, b I ( D ) ≤ b I ( D − u ) + deg G [ D ] ( u ). By the induction hypothesis, we have b I ( D ) ≤ b I ( D − u ) + deg G [ D ] ( u ) ≤ ( γ I ( D − u ) − G [ D − u ]) + deg G [ D ] ( u ) ≤ ( γ I ( D − u ) − G [ D ]) + ∆( G [ D ])= γ I ( D − u )∆( G [ D ])= ( γ I ( D ) − G [ D ]) . This is a contradiction. (cid:3)
The Italian bondage numbers of some classes of digraphs.
For a graph G , the associated digraph G ∗ is the digraph obtained from G by replacing each edgeof G by two oppositely oriented arcs. Note that γ I ( G ) = γ I ( G ∗ ) for any graph G . Theorem 3.3.
Let K ∗ n be the complete digraph of order n ≥ . Then b I ( K ∗ n ) = n .Proof. Note that γ I ( K ∗ n ) = 2. Let B be an arc set of K ∗ n . Define D := K ∗ n − B . If D contain a vertex x such that deg + D ( x ) = n −
1, then it follows from Observation2.1 that γ I ( D ) = 2. This implies that b I ( K ∗ n ) ≥ n .Let { x , x , . . . , x n } be the vertex set of K ∗ n , and let B := { x x , x x , . . . , x n x } be the arc set of a directed cycle in K ∗ n . Define D := K ∗ n − B . Then one can observethat there do not exist two distinct vertices u and v in D such that V ( D ) \ { u, v } ⊆ N + D ( u ) and V ( D ) \ { u, v } ⊆ N + D ( v ). It follows from Theorem 2.4 that γ I ( D ) ≥ (cid:3) The following result follows from the definition of associated digraph and Lemma2.2. For a complete bipartite digraph K ∗ m,n with 1 ≤ m ≤ n ,(1) γ I ( K ∗ m,n ) = m ≤ m = 3;4 if m ≥ Theorem 3.4.
Let K ∗ m,n be the complete bipartite digraph such that ≤ m < n .Then b I ( K ∗ m,n ) = if m ≤ ; if m = 3 ; m + 2 if m ≥ .Proof. We denote K ∗ m,n by D . Let X = { x , x , . . . , x m } and Y = { y , y , . . . , y n } be the partite sets of D . The result is clear for m ≤ m = 3. It follows from (1) that γ I ( D ) = 3. If we remove two arcsterminating at some vertex y j ∈ Y , then the Italian domination number of resultingdigraph increases. So, b I ( D ) ≤
2. For any arc e of A ( D ), there exist two vertices x i and x j such that N + D − e ( x i ) = n and N + D − e ( x j ) = n . Thus, we have b I ( D ) = 2.Assume that m ≥
4. It follows from (1) that γ I ( D ) = 4. Let B = { x i y | ≤ i ≤ m } ∪ { y x , y x } . It is easy to see that γ I ( D − B ) ≥
5. So, b I ( D ) ≤ m + 2. ext, we show that b I ( D ) ≥ m + 2. Let B ′ be a subset of A ( D ) such that | B ′ | = m + 1, and let D ′ = D − B ′ . Then D ′ has at least n − D and D ′ . Let E = { v ∈ V ( D ) | d + D ( v ) = d + D ′ ( v ) } . If E ∩ X = ∅ 6 = E ∩ Y , then clearly γ I ( D ′ ) = 4. Henceforth, we assume that E ∩ X = ∅ or E ∩ Y = ∅ . Without loss of generality, assume that E ∩ X = ∅ . Then E ⊆ Y and B ′ contains one outgoing arc for each x i ∈ X . Since | B ′ | = m + 1 < m , B ′ contains exactly one outgoing arc for some x i ∈ X . Without loss of generality,assume that i = 1 and x y ∈ B ′ . If E = Y , then( V ( D ′ ) \ { x , y } , ∅ , { x , y } )is an IDF of D ′ with weight 4. Let E ⊂ Y . We may assume that E ⊆ { y , y , . . . , y n − } .Thus, B ′ contains one outgoing arc from y n , say y n x m . Since | B ′ | = m + 1, B ′ contains exactly one outgoing arc for each x i ∈ X and one outgoing arc from y n .If x i y j ∈ B ′ for some 1 ≤ i ≤ m and j < n , then( V ( D ′ ) \ { x i , y j } , ∅ , { x i , y j } )is an IDF of D ′ with weight 4. Thus, we assume that x i y n ∈ B ′ for each 1 ≤ i ≤ m .But, ( V ( D ′ ) \ { x m , y n } , ∅ , { x m , y n } )is an IDF of D ′ with weight 4. Thus, we have b I ( D ) ≥ m + 2. (cid:3) The Italian reinforcement numbers
Digraphs with r I ( D ) = 1 .Lemma 4.1. Let D be a digraph with γ I ( D ) ≥ . Let F be an r I ( D ) -set, and let g be a γ I ( D ) -function of D + F . Then the following hold: (i) For each arc v v ∈ F , g ( v ) = 0 and g ( v ) = 0 . (ii) γ I ( D + F ) = γ I ( D ) − .Proof. If there exists an arc v v ∈ F such that either g ( v i ) ≥ i ∈ { , } or g ( v ) = g ( v ) = 0, then g is also an IDF of D +( F \{ v v } ), and hence F \{ v v } is an IRS of D , which contradicts the definition of F . Thus, (i) holds.By the definition of F , we have γ I ( D + F ) ≤ γ I ( D ) −
1. Suppose that γ I ( D + F ) ≤ γ I ( D ) −
2. Let v v ∈ F . By (i), g ( v ) = 0 and g ( v ) = 0. Then the function g ′ : V ( D + ( F \ { v v } )) → { , , } with g ′ ( x ) = (cid:26) x = v ; g ( x ) otherwiseis an IDF of D + ( F \ { v v } ) such that ω ( g ′ ) = ω ( g ) + 1 ≤ γ I ( D ) −
1. This impliesthat F \ { v v } is an IRS of D , which contradicts the definition of F . Thus, (ii)holds. (cid:3) Lemma 4.2.
Let D be a digraph of order n ≥ , ∆ + ( D ) ≥ and γ I ( D ) = n . Then r I ( D ) = 1 .Proof. It follows from Theorem 2.5 that ∆ + ( D ) = 1. Since P v ∈ V ( D ) deg + ( v ) = P v ∈ V ( D ) deg − ( v ), we have ∆ − ( D ) ≥
1. It also follows from Theorem 2.5 that∆ − ( D ) = 1. Thus, D is disjoint union of directed paths, cycles or isolated vertices.Let uv ∈ A ( D ) and w ∈ V ( D ) \ { u, v } . It is easy to see that( { v, w } , V ( D ) \ { u, v, w } , { u } )is an IDF of D + uw with weight n −
1. Thus, we have r I ( D ) = 1. (cid:3) Theorem 4.3.
Let D be a digraph with γ I ( D ) ≥ . Then r I ( D ) = 1 if and onlyif there exist a γ I ( D ) -function f = ( V , V , V ) of D and a vertex v ∈ V satisfyingone of the following conditions: i) f ( N − ( v )) = 1 and f ( N − ( x ) \ { v } ) ≥ for each x ∈ N + ( v ) ∩ V . (ii) f ( N − ( v )) = 0 , f ( N − ( x ) \ { v } ) ≥ for each x ∈ N + ( v ) , and V = ∅ .Proof. First, assume that (i) holds. Then it follows from f ( N − ( v )) = 1 that thereexists u ∈ V ∩ N − ( v ). Since γ I ( D ) ≥
3, there exists w ∈ ( V ∪ V ) \ { v, u } . Since uv ∈ A ( D ) and f ( N − ( x ) \ { v } ) ≥ x ∈ N + ( v ) ∩ V , ( V ∪ { v } , V \ { v } , V )is an IDF of D + wv with weight γ I ( D ) −
1. Thus, we have r I ( D ) = 1.Next, assume that (ii) holds. Let w ∈ V . Then it follows from f ( N − ( v )) = 0that wv A ( D ). Since f ( N − ( x ) \ { v } ) ≥ x ∈ N + ( v ), ( V ∪ { v } , V \{ v } , V ) is an IDF of D + wv with weight γ I ( D ) −
1. Thus, we have r I ( D ) = 1.Conversely, assume that r I ( D ) = 1, and let uv be an arc of D with γ I ( D + uv ) <γ I ( D ). Let g be a γ I ( D + uv )-function. Then g ( u ) = 0 and g ( v ) = 0 by Lemma4.1(i). The function f : V ( D ) → { , , } with f ( x ) = (cid:26) x = v ; g ( x ) otherwiseis an IDF of D . It follows from Lemma 4.1(ii) that f is a γ I ( D )-function.Suppose that f ( N − ( v )) ≥
2. Then g ( N − ( v )) ≥
2. So, g is an IDF of D .This means that γ I ( D ) ≤ ω ( g ) = γ I ( D + uv ), a contradiction. Thus, we have f ( N − ( v )) ≤ f ( N − ( x ) \ { v } ) = h ( N − ( x ) \ { v } ) ≥ x ∈ N + ( v ) ∩ V f , since g is a γ I ( D + uv )-function with g ( v ) = 0. If f ( N − ( v )) = 1, then (i) holds. Nowassume that f ( N − ( v )) = 0. Then we have h ( u ) = f ( u ) = 2, since g ( v ) = 0 and u is an in-neighbor of v in D + uv . As V f = ∅ , (ii) holds. (cid:3) Bounds of the Italian reinforcement numbers.Theorem 4.4.
If D is a digraph of order n with γ I ( D ) ≥ , then r I ( D ) ≤ n − ∆ + ( D ) − γ I ( D ) + 2 . Proof.
Since γ I ( D ) ≥
3, it follows from Theorem 2.4 that ∆ + ( D ) ≤ n −
2. Let u be a vertex with deg + D ( u ) = ∆ + ( D ) and let R := { uv | v ∈ V ( D ) \ N + [ u ] } . Then( V ( D ) \ { u } , ∅ , { u } ) is an IDF of D + R . Thus, r I ( D ) ≤ n − ∆ + ( D ) − . There exist r I ( D ) − v , v , . . . , v r I ( D ) − in V ( D ) \ N + [ u ].Let D ′ be a digraph obtained from D by adding r I ( D ) − uv i . Then, bythe definition of r I ( D ) and Observation 2.1, γ I ( D ) = γ I ( D ′ ) ≤ n − ∆ + ( D ′ ) + 1 . Since ∆ + ( D ′ ) = ∆ + ( D ) + r I ( D ) −
1, we have r I ( D ) ≤ n − ∆ + ( D ) − γ I ( D ) + 2. (cid:3) Theorem 4.5. If D is a digraph such that γ I ( D ) = 3 and γ ( D ) = 2 , then r ( D ) ≤ r I ( D ) + 1 .Proof. Let R be a r I ( D )-set. Then γ I ( D + R ) = 2. If r ∈ R , then clearly r I ( D +( R \ { r } )) = 1.By Theorem 4.3, there exist a γ I ( D + ( R \ { r } ))-function f = ( V , V , V ) of D + ( R \ { r } ) and a vertex v ∈ V satisfying one of the following conditions:(i) f ( N − ( v )) = 1 and f ( N − ( x ) \ { v } ) ≥ x ∈ N + ( v ) ∩ V .(ii) f ( N − ( v )) = 0, f ( N − ( x ) \ { v } ) ≥ x ∈ N + ( v ), and V = ∅ .Suppose that (i) holds. Since γ I ( D +( R \{ r } )) = 3, it follows from f ( N − ( v )) = 1that there exists u ∈ V such that u N − ( v ). Let w ∈ V ∩ N − ( v ). Since f ( N − ( x ) \ { v } ) ≥ x ∈ N + ( v ) ∩ V , we have ux, wx ∈ A ( D + ( R \ { r } )for each x ∈ N + ( v ) ∩ V . Since γ I ( D + ( R \ { r } )) = 3, we have u, w ∈ N − ( x ) for ach x ∈ V \ N + ( v ). Thus, { u } is a dominating set of D + (( R \ { r } ) ∪ { uv, uw } ).This implies that r ( D ) ≤ r I ( D ) + 1.Suppose that (ii) holds. Let V = { u } . Then we have V ⊆ N + ( u ). Thus, { u } isa dominating set of D + (( R \ { r } ) ∪ { uv } ). This implies that r ( D ) ≤ r I ( D ). (cid:3) The Italian reinforcement numbers of compositions of digraphs.
Fortwo digraphs G and H , two kinds of joins G → H and G ↔ H were defined in [6].The digraph G → H consists of G and H with extra arcs from each vertex of G toevery vertex of H . The digraph G ↔ H can be obtained from G → H by addingarcs from each vertex of H to every vertex of G . Theorem 4.6.
Let G and H be two digraphs such that ∆ + ( G ) ≥ and ∆ + ( H ) ≥ .Then (i) γ I ( G → H ) = γ I ( G ) , (ii) r I ( G → H ) = r I ( G ) ,Proof. (i) Let f be a γ I ( G )-function. Then it follows from the definition of IDFthat f is extended to an IDF of G → H by assigning 0 to every vertex of H . Thus, γ I ( G → H ) ≤ γ I ( G ). On the other hand, if g = ( V , V , V ) is a γ I ( G → H )-function, then clearly g | G := ( V ∩ V ( G ) , V ∩ V ( G ) , V ∩ V ( G )) is an IDF of G .Thus, γ I ( G ) ≤ γ I ( G → H ).(ii) If γ I ( G ) = 2, then it follows from (i) that γ I ( G → H ) = 2. So, r I ( G → H ) = r I ( G ). From now on, we assume γ I ( G ) ≥
3. Let R be a r I ( G )-set. Then γ I (( G → H ) + R ) = γ I (( G + R ) → H ) = γ I ( G + R ) < γ I ( G ) = γ I ( G → H ) . Thus, r I ( G → H ) ≤ r I ( G ).Now we claim that r I ( G ) ≤ r I ( G → H ). Let R be a r I ( G → H )-set. Supposethat R is a subset of R such that two ends of arcs in R lie in V ( G ). Let f = ( V f , V f , V f ) be a γ I (( G → H ) + R )-function, and let g = f | G . We divide ourconsideration into the following two cases. Case 1. g is an IDF of G + R .Then we have γ I (( G → H ) + R ) = ω ( f ) ≥ ω ( g ) ≥ γ I ( G + R )= γ I (( G + R ) → H )= γ I (( G → H ) + R ) ≥ γ I (( G → H ) + R ) . Since R ⊆ R and R is a r I ( G → H )-set, we have R = R . So, γ I ( G + R ) ≤ ω ( g ) = γ I (( G → H ) + R ) < γ I ( G → H ) = γ I ( G ). Thus, r I ( G ) ≤ | R | = | R | = r I ( G → H ). Case 2. g is not an IDF of G + R .Then some vertex u ∈ V f ∩ V ( G ) has an in-neighbor w ∈ V ( H ) such that wu ∈ R . Fix v ∈ V ( G ), and let R = { vu | u ∈ N } , where N = { u ∈ V f ∩ V ( G ) | u does not dominated by the vertices of G under f } . Then clearly | R ∪ R | ≤| R | . It is easy to see that the function h : V ( G ) → { , , } defined by h ( v ) = max { f ( v ) , max { f ( N − ( u ) ∩ V ( H )) | u ∈ N }} and h ( x ) = f ( x ) otherwise, is an IDF f G + ( R ∪ R ) with weight at most ω ( f ). Now we have γ I ( G + ( R ∪ R )) ≤ ω ( h ) ≤ ω ( f )= γ I (( G → H ) + R ) < γ I ( G → H )= γ I ( G ) . Thus, r I ( G ) ≤ | R ∪ R | ≤ | R | = r I ( G → H ). (cid:3) The corona G −→◦ H of two digraphs G and H is formed from one copy of G and n ( G ) copies of H by joining v i to every vertex of H i , where v i is the i th vertex of G and H i is the i th copy of H . Theorem 4.7.
Let G and H be two digraphs with n ( H ) ≥ . Then (i) γ I ( G −→◦ H ) = 2 n ( G ) , (ii) r I ( G −→◦ H ) = if n ( G ) = 1 ; n ( H ) if G is the empty digraph and n ( G ) ≥ ; n ( H ) − otherwise.Proof. (i) If n ( G ) = 1, then clearly γ I ( G −→◦ H ) = 2. Assume that n ( G ) ≥
2. It iseasy to see that ( V ( G −→◦ H ) \ V ( G ) , ∅ , V ( G )) is an IDF of G −→◦ H . So, γ I ( G −→◦ H ) ≤ n ( G ).Let f be a γ I ( G −→◦ H )-function. To dominate the vertices of H i ,we must have P x ∈ V ( H i ) ∪{ v i } f ( x ) ≥
2. Since a single vertex of G does not dominate vertices indifferent copies of H , we have γ I ( G −→◦ H ) ≥ n ( G ).(ii) If n ( G ) = 1, then clearly r I ( G −→◦ H ) = 0. Assume that n ( G ) ≥
2. We divideour consideration into the following two cases.
Case 1. A ( G ) = ∅ .Let R = { v u | u ∈ V ( H n ( G ) ) } . Then it is easy to see that( n ( G ) [ i =1 V ( H i ) , { v n ( G ) } , V ( G ) \ { v n ( G ) } )is an IDF of ( G −→◦ H ) + R with weight 2 n ( G ) −
1. Thus, r I ( G −→◦ H ) ≤ n ( H ).Let F be a r I ( G −→◦ H )-set. By Lemma 4.1(ii), γ I (( G −→◦ H ) + F ) = γ I ( G −→◦ H ) − U i = { v i } ∪ V ( H i ) for 1 ≤ i ≤ n ( G ), and let f = ( V , V , V ) be a r I (( G −→◦ H ) + F )-function. Then P x ∈ U i f ( x ) ≤ i , say i = n ( G ). To dominate thevertices in U n ( G ) , F must contain at least n ( H ) arcs which go from some vertices in( V ∪ V ) ∩ ( S n ( G ) − i =1 U i ) to vertices in U n ( G ) . Thus, | F | ≥ n ( H ) and so r I ( G −→◦ H ) ≥ n ( H ). Case 2. A ( G ) = ∅ .Without loss of generality, we assume that v v n ( G ) ∈ A ( G ). Let V ( H n ( G ) ) = { w , . . . , w n ( H ) } , and let R = { v w j | w j ∈ V ( H n ( G ) ) \ { w }} . Then( n ( G ) [ i =1 V ( H i ) ∪ { v n ( G ) } , { w } , { v , . . . , v n ( G ) − } )is an IDF of ( G −→◦ H ) + R with weight 2 n ( G ) −
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Department of Mathematics, Pusan National University, Busan 46241, Republic ofKorea
E-mail address : [email protected]@pusan.ac.kr