Total Roman domination on the digraphs

Xinhong Zhang; Xin Song; Ruijuan Li

doi:10.1515/math-2022-0575

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Total Roman domination on the digraphs

Xinhong Zhang , Xin Song and Ruijuan Li

Published/Copyright: April 1, 2023

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From the journal Open Mathematics Volume 21 Issue 1

Abstract

Let D = ( V , A ) be a simple digraph with vertex set V , arc set A , and no isolated vertex. A total Roman dominating function (TRDF) of D is a function h : V → { 0 , 1 , 2 } , which satisfies that each vertex x ∈ V with h ( x ) = 0 has an in-neighbour y ∈ V with h ( y ) = 2 , and that the subdigraph of D induced by the set { x ∈ V : h ( x ) ≥ 1 } has no isolated vertex. The weight of a TRDF h is ω ( h ) = ∑ x ∈ V h ( x ) . The total Roman domination number γ t R ( D ) of D is the minimum weight of all TRDFs of D . The concept of TRDF on a graph G was introduced by Liu and Chang [Roman domination on strongly chordal graphs, J. Comb. Optim. 26 (2013), no. 3, 608–619]. In 2019, Hao et al. [Total Roman domination in digraphs, Quaest. Math. 44 (2021), no. 3, 351–368] generalized the concept to digraph and characterized the digraphs of order n ≥ 2 with γ t R ( D ) = 2 and the digraphs of order n ≥ 3 with γ t R ( D ) = 3 . In this article, we completely characterize the digraphs of order n ≥ k with γ t R ( D ) = k for all integers k ≥ 4 , which generalizes the results mentioned above.

Keywords: total Roman dominating function; total Roman domination number; digraph

MSC 2010: 05C20; 05C69

1 Introduction

In recent years, domination theory in digraphs has inspired widespread interest. Some variations have fallen within the scope of research (see [1–9]).

For the notation and terminology in this article, see [10]. Let D = ( V , A ) be a simple digraph with vertex set V and arc set A . In this article, if not otherwise specified, we assume that digraph D has no isolated vertex. Let x ∈ V . The out-neighbourhood, N + ( x ) , of x is { y : ( x , y ∈ A ) } , and N + [ x ] = N + ( x ) ∪ { x } is called the closed out-neighbourhood. Similarly, the in-neighbourhood N − ( x ) and the closed in-neighbourhood N − [ x ] of x can be defined. Furthermore, the set N ( x ) = N + ( x ) ∪ N − ( x ) is called the neighbourhood of x . We call the vertices in N ( x ) the neighbours of x . For a vertex subset W ⊆ V , the subdigraph induced by W is denoted by D [ W ] .

A total Roman dominating function (TRDF) in a digraph D is a function h : V → { 0 , 1 , 2 } that satisfies the following conditions:

each vertex x ∈ V with h ( x ) = 0 has an in-neighbour y ∈ V with h ( y ) = 2 ;
the subdigraph of D induced by the set { x ∈ V : h ( x ) ≥ 1 } has no isolated vertex.

For the sake of simplicity, let V i = { x ∈ V : h ( x ) = i } for i = 0 , 1 , 2 and we also write h = ( V 0 , V 1 , V 2 ) . For a vertex subset W ⊆ V , we define h ( W ) = ∑ v ∈ W h ( v ) and ω ( h ) = h ( V ) . The total Roman domination number of D is

γ t R ( D ) = min { ω ( h ) : h is a TRDF of D } .

A TRDF h is called a γ t R ( D ) -function if ω ( h ) = γ t R ( D ) . For more results about this, see [11–15].

In 2019, Hao et al. [15] characterized the digraphs of order n ≥ 2 with γ t R ( D ) = 2 and the digraphs of order n ≥ 3 with γ t R ( D ) = 3 . In this article, we generalize the results mentioned above to all positive integers k ≥ 4 and characterize the digraphs of order n ≥ k with γ t R ( D ) = k . In Section 2, we discuss the case k = 4 , and in Section 3, we proceed for k ≥ 5 .

2 The digraphs of order n ≥ 4 with γ t R ( D ) = 4

In this section, we characterize the digraphs D of order n ≥ 4 with γ t R ( D ) = 4 . To show the main result, we need to use the following propositions.

Proposition 2.1

[15] For any digraph D of order n ≥ 2 with no isolated vertex, γ t R ( D ) = 2 if and only if n = 2 .

Proposition 2.2

[15] For any digraph D of order n ≥ 3 with no isolated vertex, γ t R ( D ) = 3 if and only if one of the following hold:

n = 3 ;
n ≥ 4 and there exist two vertices u and v of D such that V ( D ) \ { u , v } ⊆ N + ( v ) and D [ { u , v } ] is connected in D.

Theorem 2.3

Let D = ( V , A ) be a digraph of order n ≥ 4 with no isolated vertex. Then, γ t R ( D ) = 4 if and only if one of the following is true:

n = 4 and D does not contain a vertex subset X such that ∣ X ∣ = 2 , V \ X ⊆ N + ( x ) for a vertex x ∈ X , and D [ X ] has no isolated vertex;
n ≥ 5 , Δ + ( D ) = n − 2 ,
1. D [ V \ N + ( v ) ] is not connected for any vertex v with maximum out-degree of D, and
2. D contains a vertex subset X such that ∣ X ∣ = 3 , V \ X ⊆ N + ( x ) for a vertex x ∈ X , and D [ X ] has no isolated vertex;
n ≥ 5 , Δ + ( D ) ≤ n − 3 ,
1. D contains a vertex subset X such that ∣ X ∣ = 3 , V \ X ⊆ N + ( x ) for a vertex x ∈ X , and D [ X ] has no isolated vertex, or
2. D contains a vertex subset X such that ∣ X ∣ = 2 , V \ X ⊆ N + ( X ) , and D [ X ] has no isolated vertex.

Proof

( ⇒ ) Assume that γ t R ( D ) = 4 . Let h = ( V 0 , V 1 , V 2 ) be a γ t R ( D ) -function.

We claim that Δ + ( D ) ≤ n − 2 . Suppose not. Then, Δ + ( D ) = n − 1 , and there is a vertex v ∈ V with d + ( v ) = n − 1 . Let u be an out-neighbour of v . Define a function g 0 : V → { 0 , 1 , 2 } such that g 0 ( v ) = 2 , g 0 ( u ) = 1 , and g 0 ( w ) = 0 otherwise. This results in a TRDF with weight ω ( g 0 ) = 3 < 4 = γ t R ( D ) , a contradiction. Hence, Δ + ( D ) ≤ n − 2 .

Since γ t R ( D ) = ∣ V 1 ∣ + 2 ∣ V 2 ∣ = 4 and Δ + ( D ) ≤ n − 2 , we distinguish two cases: either Δ + ( D ) = n − 2 or Δ + ( D ) ≤ n − 3 .

Case 1: Δ + ( D ) = n − 2 .

Subcase 1.1: ∣ V 1 ∣ = 4 and ∣ V 2 ∣ = 0 .

Since ∣ V 2 ∣ = 0 , we have ∣ V 0 ∣ = 0 , and then ∣ V ∣ = ∣ V 1 ∣ = 4 . For n = 4 , it is easy to see that the condition 2.3 ( 1 ) is true by Proposition 2.2 ( 2 ) .

Subcase 1.2: ∣ V 1 ∣ = 2 and ∣ V 2 ∣ = 1 .

First, we show that the condition (2.a) is true. Suppose to the contrary that there is a vertex v ∈ V with d + ( v ) = Δ + ( D ) such that D [ V \ N + ( v ) ] is connected. Then, ( u , v ) ∈ A for the vertex u ∈ V \ N + [ v ] , as shown in Figure 1(1). Define a function g 1 : V → { 0 , 1 , 2 } such that g 1 ( v ) = 2 , g 1 ( u ) = 1 and g 1 ( x ) = 0 for each vertex x ∈ N + ( v ) . Then g 1 is a TRDF with weight ω ( g 1 ) = 3 < 4 = γ t R ( D ) , a contradiction. Therefore, the condition (2.a) holds.

Furthermore, let V 2 = { v 0 } and X = V 1 ∪ V 2 . Then, we have V \ X = V 0 ⊆ N + ( v 0 ) and D [ X ] has no isolated vertex by the definition of TRDF. This implies that the condition (2.b) holds, see Figure 1(2).

Subcase 1.3: ∣ V 1 ∣ = 0 and ∣ V 2 ∣ = 2 .

Let V 2 = { v 0 , v 1 } , V 1 = ∅ . Since h is a γ t R ( D ) -function, there exists a vertex with the maximum out-degree Δ + ( D ) in V 2 . Without loss of generality, assume that d + ( v 0 ) = Δ + ( D ) .

If v 1 ∉ N + ( v 0 ) (as shown in Figure 1(3)), then define a function g 2 : V → { 0 , 1 , 2 } such that g 2 ( v 0 ) = 2 , g 2 ( v 1 ) = 1 , and g 2 ( x ) = 0 otherwise. Then, g 2 is a TRDF with weight ω ( g 2 ) = 3 < 4 = γ t R ( D ) , a contradiction. Thus, v 1 ∈ N + ( v 0 ) . Let v 2 ∈ V \ N + [ v 0 ] (see Figure 1(4)). It is not difficult to see that f = ( V 0 f , V 1 f , V 2 f ) = ( V \ { v 0 , v 1 , v 2 } , { v 1 , v 2 } , { v 0 } ) is a γ t R ( D ) -function with ∣ V 1 f ∣ = 2 and ∣ V 2 f ∣ = 1 . Then, by the proof of Subcase 1.2, conditions (2.a) and (2.b) hold.

Case 2: Δ + ( D ) ≤ n − 3 .

Subcase 2.1: ∣ V 1 ∣ = 4 and ∣ V 2 ∣ = 0 .

Since ∣ V 2 ∣ = 0 , we have ∣ V 0 ∣ = 0 , and then ∣ V ∣ = ∣ V 1 ∣ = 4 . For n = 4 , it is easy to see that the condition (1) is true by Proposition 2.2 ( 2 ) .

Subcase 2.2: ∣ V 1 ∣ = 2 and ∣ V 2 ∣ = 1 .

In the same manner as the proof of the condition (2.b), it can be obtained that D contains a set X of order 3 such that V \ X ⊆ N + ( x ) for a vertex x ∈ X and D [ X ] has no isolated vertex. That is, condition (3.a) holds.

Subcase 2.3: ∣ V 1 ∣ = 0 and ∣ V 2 ∣ = 2 .

Let X = V 2 . Since V 1 = ∅ and h is a γ t R ( D ) -function, we have V \ X = V 0 ⊆ N + ( X ) and D [ X ] has no isolated vertex. Therefore, condition (3.b) holds.

( ⇐ ) To show the sufficiency, assume that one of the three conditions (1), (2), and (3) holds in the statement of the theorem.

If ( 1 ) holds, we obtain γ t R ( D ) ≥ 4 by Propositions 2.1 and 2.2(2). Furthermore, define a function g 3 : V → { 0 , 1 , 2 } such that g 3 ( v ) = 1 for every vertex v ∈ V . Then γ t R ( D ) ≤ ω ( g 3 ) = 4 .

If ( 2 ) holds, since D [ V \ N + ( v ) ] is not connected for each vertex v with maximum out-degree Δ + ( D ) = n − 2 by (2.a), there do not exist two vertices u and v of D such that V \ { u , v } ⊆ N + ( v ) and D [ { u , v } ] is connected in D . Then, γ t R ( D ) ≥ 4 follows trivially from Propositions 2.1 and 2.2. On the other hand, by condition (2.b), we have the function g 4 = ( V \ X , X \ { x } , { x } ) as a TRDF on D with weight ω ( g 4 ) = 4 , and so γ t R ( D ) ≤ 4 .

If ( 3 ) holds, we first have Δ + ( D ) ≤ n − 3 , then there do not exist two vertices u and v of D such that V \ { u , v } ⊆ N + ( v ) . Furthermore, we obtain γ t R ( D ) ≥ 4 by Propositions 2.1 and 2.2. On the other hand, if (3.a) holds, then define the function g 5 = ( V \ X , X \ { x } , { x } ) as a TRDF on D with weight ω ( g 5 ) = 4 , and so γ t R ( D ) ≤ 4 . If (3.b) holds, then the function g 5 = ( V \ X , ∅ , X ) is a TRDF on D with weight ω ( g 5 ) = 4 , and so γ t R ( D ) ≤ 4 .

Consequently, we have γ t R ( D ) = 4 .□

$Figure 1 (1) and (2): The illustrations of Subcase 1.2, where the edge with no direction means that both directions are possible. (3) and (4): The illustrations of Subcase 1.3.$

Figure 1

(1) and (2): The illustrations of Subcase 1.2, where the edge with no direction means that both directions are possible. (3) and (4): The illustrations of Subcase 1.3.

3 The digraphs of order n ≥ k with γ t R ( D ) = k for any positive integer k ≥ 5

Definition 3.1

Let t ≥ 2 be a positive integer and D = ( V , A ) a digraph. Then, D has an ( X , W , t ) -structure if there exists a subset X ⊆ V such that for a subset W ⊂ V with 0 ≤ ∣ W ∣ ≤ ⌊ t 2 ⌋ the following hold:

W ⊆ X , V \ X ⊆ N + ( W ) , and D [ X ] have no isolated vertex if 1 ≤ ∣ W ∣ ≤ ⌊ t 2 ⌋ , which includes two cases (Figure 2);
X = V and D [ X ] = D has no isolated vertex if ∣ W ∣ = 0 .

$Figure 2 (1): N + ( W ) ∩ X ≠ ∅ {N}^{+}\left(W)\cap X\ne \varnothing ; (2): N + ( W ) ∩ X = ∅ {N}^{+}\left(W)\cap X=\varnothing .$

Figure 2

(1): N + ( W ) ∩ X ≠ ∅ ; (2): N + ( W ) ∩ X = ∅ .

Theorem 3.2

Let D = ( V , A ) be a digraph of order n ≥ 4 with no isolated vertex. Then, γ t R ( D ) ≤ n . Furthermore, the equality holds if and only if there does not exist an ( X , W , n − 1 ) -structure with ∣ X ∣ ≤ n − 1 − ∣ W ∣ .

Proof

Define a function h : V → { 0 , 1 , 2 } such that h ( v ) = 1 for each vertex v ∈ V . Then, h is a TRDF on D , and so γ t R ( D ) ≤ ω ( h ) = n . In the following, we show the second assertion.

( ⇒ ) Assume γ t R ( D ) = n . Suppose, however, that there is an ( X , W , n − 1 ) -structure with ∣ X ∣ ≤ n − 1 − ∣ W ∣ in D . If ∣ W ∣ = 0 , then ∣ V ∣ = ∣ X ∣ ≤ n − 1 , a contradiction. If 1 ≤ ∣ W ∣ ≤ n − 1 2 , then the function f = ( V \ X , X \ W , W ) is a TRDF on D of weight ω ( f ) ≤ ∣ X \ W ∣ + 2 ∣ W ∣ ≤ n − 1 − ∣ W ∣ − ∣ W ∣ + 2 ∣ W ∣ = n − 1 < n = γ t R ( D ) , a contradiction.

( ⇐ ) It is sufficient to prove that γ t R ( D ) ≥ n . Suppose, however, that γ t R ( D ) ≤ n − 1 . Let h 1 = ( V 0 , V 1 , V 2 ) be a γ t R ( D ) -function. Since γ t R ( D ) = ∣ V 1 ∣ + 2 ∣ V 2 ∣ ≤ n − 1 , we have 1 ≤ ∣ V 2 ∣ ≤ ⌊ n − 1 2 ⌋ . Furthermore, since γ t R ( D ) = ∣ V 1 ∣ + 2 ∣ V 2 ∣ ≤ n − 1 = ∣ V 0 ∣ + ∣ V 1 ∣ + ∣ V 2 ∣ − 1 , we obtain ∣ V 0 ∣ ≥ ∣ V 2 ∣ + 1 . Let W = V 2 and X = V 1 ∪ V 2 . Then, V \ X ⊆ N + ( W ) and D [ X ] has no isolated vertex by the definition of TRDF, where ∣ X ∣ = ∣ V 1 ∣ + ∣ V 2 ∣ = n − ∣ V 0 ∣ ≤ n − ∣ W ∣ − 1 , a contradiction. Consequently, γ t R ( D ) ≥ n . Hence, γ t R ( D ) = n , as desired.□

Lemma 3.3

Let k ≥ 5 be an integer and D = ( V , A ) a digraph of order n ≥ k + 1 with γ t R ( D ) ≥ k . If there exists a subset W ⊂ V with 1 ≤ ∣ W ∣ ≤ ⌊ k 2 ⌋ such that D [ V \ N + ( W ) ] has no isolated vertex, then ∣ N + [ W ] ∣ ≤ n + 2 ∣ W ∣ − k .

Proof

Let W ⊂ V with 1 ≤ ∣ W ∣ ≤ k 2 be a set such that D [ V \ N + ( W ) ] has no isolated vertex. Suppose, however, that ∣ N + [ W ] ∣ ≥ n + 2 ∣ W ∣ − k + 1 . Since D [ V \ N + ( W ) ] has no isolated vertex, the function g = ( N + ( W ) , V \ N + [ W ] , W ) is a TRDF on D . Then, γ t R ( D ) ≤ ω ( g ) = ∣ V \ N + [ W ] ∣ + 2 ∣ W ∣ ≤ n − ( n + 2 ∣ W ∣ − k + 1 ) + 2 ∣ W ∣ = k − 1 , a contradiction to γ t R ( D ) ≥ k .□

Lemma 3.4

Let k ≥ 5 be an integer and D = ( V , A ) a digraph of order n ≥ k + 1 with γ t R ( D ) ≥ k . If there exists a subset W ⊂ V with 1 ≤ ∣ W ∣ ≤ k 2 such that D [ V \ N + ( W ) ] has at least one isolated vertex, then D has an ( X , W , k ) -structure and D [ X ] has no ( X ′ , W ′ , k − ∣ W ∣ − 1 ) -structure with ∣ X ′ ∣ + ∣ W ′ ∣ ≤ k − 1 − ∣ W ∣ and W ⊆ X ′ .

Proof

Let W ⊂ V with 1 ≤ ∣ W ∣ ≤ k 2 be a set such that D [ V \ N + ( W ) ] has at least one isolated vertex. Suppose that h = ( V 0 , V 1 , W ) is a TRDF on D , let X = V 1 ∪ W . Then, we have V \ X ⊆ N + ( W ) and D [ X ] has no isolated vertex by the definition of TRDF. This implies that there exists an ( X , W , k ) -structure in D .

Next, we prove that D [ X ] does not contain an ( X ′ , W ′ , k − ∣ W ∣ − 1 ) -structure with ∣ X ′ ∣ + ∣ W ′ ∣ ≤ k − 1 − ∣ W ∣ and W ⊆ X ′ by contradiction. Let D 1 = D [ X ] .

Suppose, however, that D 1 has an ( X ′ , W ′ , k − ∣ W ∣ − 1 ) -structure with ∣ X ′ ∣ + ∣ W ′ ∣ ≤ k − 1 − ∣ W ∣ and W ⊆ X ′ . Then, the function f = ( X \ X ′ , X ′ \ W ′ , W ′ ) is a TRDF on D 1 .

Let h 1 = ( V 0 h 1 , V 1 h 1 , V 2 h 1 ) be defined as follows: h 1 ( v ) = 2 for each vertex v ∈ W , h 1 ( u ) = 0 for each vertex u ∈ V \ X , and h 1 ( x ) = f ( x ) for each vertex x ∈ X \ W . This implies V 0 h 1 = ( V \ X ) ∪ ( X \ X ′ ) , V 1 h 1 = X ′ \ ( W ∪ W ′ ) , V 2 h 1 = W ∪ W ′ . If ∣ W ′ ∣ = 0 , then V 0 h 1 = V \ X , V 1 h 1 = X ′ \ W , and V 2 h 1 = W . According to the definitions of ( X , W , k ) -structure and ( X ′ , W ′ , k − ∣ W ∣ − 1 ) -structure, we have V 0 h 1 ⊆ N + ( V 2 h 1 ) and D [ V 1 h 1 ∪ V 2 h 1 ] = D [ X ′ ] = D [ X ] has no isolated vertex. Thus, h 1 is a TRDF on D . Similarly, if 1 ≤ ∣ W ′ ∣ ≤ k − ∣ W ∣ − 1 2 , then we have X \ X ′ ⊆ N + ( W ′ ) and D [ X ′ ] has no isolated vertex by the definition of ( X ′ , W ′ , k − ∣ W ∣ − 1 ) -structure. Furthermore, V \ X ⊆ N + ( W ) according to the definition of ( X , W , k ) -structure. Thus, V 0 h 1 = ( V \ X ) ∪ ( X \ X ′ ) ⊆ N + ( W ∪ W ′ ) = N + ( V 2 h 1 ) and D [ V 1 h 1 ∪ V 2 h 1 ] = D [ X ′ ] has no isolated vertex. That is, h 1 is a TRDF on D . Hence, γ t R ( D ) ≤ ω ( h 1 ) = ∣ X ′ \ ( W ∪ W ′ ) ∣ + 2 ∣ W ∪ W ′ ∣ ≤ ∣ X ′ ∣ + ∣ W ∣ + ∣ W ′ ∣ ≤ k − 1 , a contradiction to γ t R ( D ) ≥ k . This completes the proof of Lemma 3.4.□

Theorem 3.5

Let k ≥ 5 be an integer and D = ( V , A ) a digraph of order n ≥ k + 1 . Then, γ t R ( D ) ≥ k if and only if the following hold:

for any subset W ⊂ V with 1 ≤ ∣ W ∣ ≤ k 2 such that D [ V \ N + ( W ) ] has no isolated vertex, there must be ∣ N + [ W ] ∣ ≤ n + 2 ∣ W ∣ − k ;
for any subset W ⊂ V with 1 ≤ ∣ W ∣ ≤ k 2 such that D [ V \ N + ( W ) ] has at least one isolated vertex, D has an ( X , W , k ) -structure and D [ X ] has no ( X ′ , W ′ , k − ∣ W ∣ − 1 ) -structure with ∣ X ′ ∣ + ∣ W ′ ∣ ≤ k − 1 − ∣ W ∣ and W ⊆ X ′ .

Proof

Lemmas 3.3 and 3.4 mean that necessity holds. Here, we just show sufficiency.

Let h = ( V 0 , V 1 , V 2 ) be a γ t R ( D ) -function. Suppose, however, that γ t R ( D ) = ∣ V 1 ∣ + 2 ∣ V 2 ∣ ≤ k − 1 . Then, 1 ≤ ∣ V 2 ∣ ≤ k − 1 − ∣ V 1 ∣ 2 ≤ k 2 .

If D [ V \ N + ( V 2 ) ] has no isolated vertex, then we have ∣ N + [ V 2 ] ∣ ≤ n + 2 ∣ V 2 ∣ − k by ( 1 ) . This implies

(3.1) n ≥ ∣ N + [ V 2 ] ∣ + k − 2 ∣ V 2 ∣ .

Furthermore, γ t R ( D ) = ∣ V 1 ∣ + 2 ∣ V 2 ∣ ≤ k − 1 , then

(3.2) k − 2 ∣ V 2 ∣ ≥ ∣ V 1 ∣ + 1 .

Combining the inequalities ( 3.1 ) and ( 3.2 ) , we obtain n ≥ ∣ N + [ V 2 ] ∣ + ∣ V 1 ∣ + 1 = ∣ V 0 ∣ + ∣ V 2 ∣ + ∣ V 1 ∣ + 1 , a contradiction.

If D [ V \ N + ( V 2 ) ] has at least one isolated vertex, then let W = V 2 , X ′ = X = V 1 ∪ W , and W ′ = ∅ . It is not difficult to see that W ⊆ X , V \ X ⊆ N + ( W ) , and D [ X ] has no isolated vertex. This implies that there exists an ( X , W , k ) -structure. Furthermore, since γ t R ( D ) = ∣ V 1 ∣ + 2 ∣ V 2 ∣ = ∣ X \ W ∣ + 2 ∣ W ∣ ≤ k − 1 , we have ∣ X ′ ∣ = ∣ X ∣ ≤ k − 1 − ∣ W ∣ . Thus, ∣ X ′ ∣ + ∣ W ′ ∣ ≤ k − 1 − ∣ W ∣ . On the other hand, W ⊆ X = X ′ , D [ X ′ ] = D [ X ] has no isolated vertex, and ∣ W ′ ∣ = 0 , and then there exists an ( X ′ , W ′ , k − ∣ W ∣ − 1 ) -structure with ∣ X ′ ∣ + ∣ W ′ ∣ ≤ k − 1 − ∣ W ∣ and W ⊆ X ′ in D [ X ] , a contradiction to ( 2 ) .□

Theorem 3.6

Let k ≥ 5 be an integer and D = ( V , A ) a digraph of order n ≥ k + 1 . Then, γ t R ( D ) = k if and only if D satisfies Theorem 3.5(1) and (2) and one of the following is true:

there exists a subset W ⊂ V with 1 ≤ ∣ W ∣ ≤ k 2 such that ∣ N + [ W ] ∣ = n + 2 ∣ W ∣ − k , and D [ V \ N + ( W ) ] has no isolated vertex;
D has an ( X , W , k ) -structure with ∣ X ∣ = k − ∣ W ∣ and D [ V \ N + ( W ) ] has at least one isolated vertex.

Proof

( ⇒ ) From γ t R ( D ) = k , we see that D satisfies Theorem 3.5 ( 1 ) and ( 2 ) . Next, we prove that ( 1 ) or ( 2 ) of this theorem holds. Let h = ( V 0 , V 1 , V 2 ) be a γ t R ( D ) -function. Since ∣ V 1 ∣ + 2 ∣ V 2 ∣ = γ t R ( D ) = k , we may deduce that one of the following is true:

∣ V 2 ∣ = 0 ;
1 ≤ ∣ V 2 ∣ ≤ k 2 .

Suppose that (i) holds. Obviously, we have ∣ V 0 ∣ = 0 , and then ∣ V 1 ∣ = ∣ V ∣ = k , a contradiction to n ≥ k + 1 .

We now suppose that (ii) holds and distinguish two cases as follows.

Case 1: D [ V \ N + ( V 2 ) ] has no isolated vertex.

Let W = V 2 . It is easy to see that 1 ≤ ∣ W ∣ ≤ ⌊ k 2 ⌋ and D [ V \ N + ( W ) ] has no isolated vertex. By Theorem 3.5 ( 1 ) , we have ∣ N + [ W ] ∣ ≤ n + 2 ∣ W ∣ − k . It follows that ∣ V 1 ∣ = n − ( ∣ V 0 ∣ + ∣ W ∣ ) = n − ∣ N + [ W ] ∣ ≥ k − 2 ∣ W ∣ = ∣ V 1 ∣ according to γ t R ( D ) = k , and hence ∣ N + [ W ] ∣ = n + 2 ∣ W ∣ − k , ( 1 ) holds.

Case 2: D [ V \ N + ( V 2 ) ] has at least one isolated vertex.

Let W = V 2 . It is easy to see that 1 ≤ ∣ W ∣ ≤ k 2 and D [ V \ N + ( W ) ] has at least one isolated vertex. Since h is a γ t R ( D ) -function, we obtain V \ ( V 1 ∪ W ) ⊆ N + ( W ) and D [ V 1 ∪ W ] has no isolated vertex. Furthermore, since γ t R ( D ) = ∣ V 1 ∣ + 2 ∣ W ∣ = k , ∣ V 1 ∣ + ∣ W ∣ = k − ∣ W ∣ . It implies that there is a ( V 1 ∪ W , W , k ) -structure with ∣ V 1 ∪ W ∣ = k − ∣ W ∣ in D . Let X = V 1 ∪ W , then ( 2 ) holds.

( ⇐ ) By Theorem 3.5, we have γ t R ( D ) ≥ k . Thus, it suffices for us to show that γ t R ( D ) ≤ k . If ( 1 ) holds, then the function g 0 = ( N + ( W ) , V \ N + [ W ] , W ) is a TRDF on D . Thus, γ t R ( D ) ≤ ω ( g 0 ) = ∣ V \ N + [ W ] ∣ + 2 ∣ W ∣ = n − ( n + 2 ∣ W ∣ − k ) + 2 ∣ W ∣ = k . If ( 2 ) holds, it is not difficult to verify that g 1 = ( V \ X , X \ W , W ) is a TRDF on D . Note ∣ X ∣ = k − ∣ W ∣ , then we have γ t R ( D ) ≤ ω ( g 1 ) = ∣ X \ W ∣ + 2 ∣ W ∣ = k − 2 ∣ W ∣ + 2 ∣ W ∣ = k . Consequently, γ t R ( D ) = k .□

Acknowledgment

We would like to thank the anonymous referee for a thorough and helpful reading of the article.

Funding information: X. Zhang: The research is partially supported by the Fundamental Research Program of Shanxi Province (20210302123202). R. Li: The research is partially supported by the Youth Foundation of Shanxi Province (201901D211197).
Author contributions: All authors contributed equally to the writing of this article. All authors read and approved the final manuscript.
Conflict of interest: The authors state that there is no conflict of interest.
Data availability statement: The raw/processed data required to reproduce these findings cannot be shared at this time as the data also forms part of an ongoing study.

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Received: 2022-08-23

Revised: 2023-03-08

Accepted: 2023-03-09

Published Online: 2023-04-01

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2022-0575

Keywords for this article

total Roman dominating function; total Roman domination number; digraph

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