1. Introduction
In general, we use the standard terminology and notation of graph theory (see [
1]). In this paper, the graph is simple, connected, and undirected. Let
G be a connected graph with vertex set
and edge set
If there exists an edge
in the graph
then
x is a neighbor of
The notation
denotes the set of all neighbors of vertex
x in
Let
is written as the degree of vertex
x in
abbreviated as
We have
The minimum degree of
G is denoted by
A star graph
with
n vertices is a tree where one vertex is called the central vertex, which has degree
and the other
vertices have degree 1. The symbols
and
represent a path, a cycle, and a complete graph of
n vertices, respectively. The notation
represents a complete bipartite graph with
vertices.
If a graph satisfies and then the graph is a subgraph of Suppose that The induced subgraph of a graph G is formed by a subset of the vertex set and all the edges in G whose two end vertices are in
A subset
is said to be a dominating set of a graph
G if every vertex outside
D has at least one neighbor in
For more details about the domination of graphs, see [
2,
3]. A subset
is called a
p-dominating set if every vertex outside
S has at least
p neighbors in
where
p is any integer satisfying
If
then the
p-dominating set agrees with the classical dominating set. For
the 2-dominating set was studied in [
4,
5,
6,
7,
8]. The
p-domination number
is the minimum cardinality among all
p-dominating sets of
When
is also written as
For the graph
a
p-dominating set of minimum cardinality is denoted as a
-set or a
-set of
If vertex
has at least
p neighbors in the set
then we say that vertex
x is dominated by at least
p vertices in
If vertex
has fewer than
p neighbors in the set
then we say that vertex
x is not dominated by
p vertices in
By definition, for any positive integer
we know that a
p-dominating set is also an
h-dominating set. Then, a 4-dominating set is also a 3-dominating set, and we have
For some results on the
p-domination number, see [
9,
10,
11]. As a generalization of domination, Harary and Haynes [
12] introduced the concept of
k-tuple domination. For
k-tuple domination, see [
13,
14]. When
k-tuple domination is called double domination, which was studied in [
15,
16,
17].
Dominating sets and their variants have extensive applications in the real world. The literature indicates that dominating sets and dominating parameters have applications in many fields, such as road safety, vehicular ad hoc communications, disaster rescue operations, and air/land/navy defense (see [
18]). Moreover, dominating sets and their variants have been used in wireless networks [
19] and chemical graphs [
20]. From an algorithmic perspective, domination and its variants in graphs have been studied in [
19,
21,
22].
We give an application problem for the p-domination number. A company needs to promote a new product. Suppose there are n potential customers for this new product. The company needs to select some salespeople from these n potential customers. None of these salespeople will buy the new product. We refer to the rest, excluding the salespeople, as target customers. If a salesperson knows a certain target customer, they will promote the product to the customer. We represent n potential customers as n vertices. If two potential customers know each other, then the vertices corresponding to the two potential customers are adjacent. This forms a graph G with n vertices. If each target customer is approached about the product by at least p salespeople, the company needs to know the minimum number of salespeople. That is, the company needs to know the p-domination number of the graph
Bednarz and Pirga [
23] introduced the concept of proper 2-dominating sets. Similar to [
23], a proper 3-dominating set is defined as follows:
Definition 1. A subset is said to be a proper 3-dominating set if it is a 3-dominating set but not a 4-dominating set.
We use
-dset to denote a proper 3-dominating set. The proper 3-domination number
is the minimum cardinality among all proper 3-dominating sets of
For a graph
a proper 3-dominating set of cardinality
is called a
-set or a
-set of
D is a proper 3-dominating set if every vertex outside the set
D has at least three neighbors in the set
D and there exists a vertex outside the set
D that has exactly three neighbors in the set
In the 3-dominating set
if there exists a vertex outside
that is dominated by three vertices in the set
then this set
cannot be a 4-dominating set. Thus, the existence of such a vertex fundamentally distinguishes a 3-dominating set from a 4-dominating set. For the graph
G, let
For proper dominating sets related to secondary domination, see [
24,
25].
Referring to [
23], we present an example of the application of
-dset. Suppose there are
n places prone to natural disasters. In order to provide assistance to the residents in places affected by these disasters, rescue stations are to be established at
m points (distinct from the
n places). Suppose the residents in these
n places can seek help from at least three rescue stations. In order to reduce costs by minimizing the number of rescue stations, where should the rescue stations be established? This problem is equivalent to finding a minimum 3-dominating set. Furthermore, if the minimum 3-dominating set is a minimum
-dset, then it determines the weakest places from which only three rescue stations can be reached.
Proposition 1 (Lemma 1 [
23])
. Suppose k is an integer satisfying In any connected graph, each vertex of degree belongs to a -dominating set, where i is any integer satisfying According to Proposition 1, in any connected graph, each vertex of degree 1 and 2 belongs to a 3-dominating set. Next, this paper considers what properties of graphs have a proper 3-dominating set, as well as the proper 3-domination number of some special graphs.
For convenience, we give the list of symbols in
Table 1.
2. Main Results
Lemma 1. Let G be a connected graph with n vertices, where If there exists a vertex v in the graph G satisfying then G has a -dset.
Proof. Let then Since then So, D is a -dset. □
Lemma 2. Let G be a connected graph with n vertices, where If then G has a -dset.
Proof. Let Let x be the vertex of minimum degree in the graph G, and Assume that Let It can be seen that the vertices in D are and all vertices in Then, For any vertex in we consider the number of its neighbors in Since then where That is, the vertex has no more than neighbors outside Since we have where Furthermore, we have So, and D is a -dset. □
Corollary 1. Let G be a connected graph, and let and be connected. For the graph suppose x is a vertex of minimum degree. For any vertex x is not adjacent to If then G has a -dset.
Proof. We have Let Assume that Let Then, Similar to the proof of Lemma 2, we have where Moreover, Then, D is a -dset. □
Theorem 1. Let G be a connected graph with n vertices, where If then G has a -dset.
Proof. If then there exists a vertex v in the graph G satisfying Then, according to Lemma 1, G has a -dset. If according to Lemma 2, G has a -dset. □
Referring to the definition of the joint product of two graphs
in [
26], we give the following definition.
Definition 2. The joint product of two graphs G,H, denoted by is a graph with vertex set and edge set
For definitions of the two fundamental graph products (strong and lexicographic), see [
27,
28]. Next, we repeat these definitions.
Definition 3. The strong product of two graphs G,H, denoted by is a graph with vertex set and and are adjacent if and only if one of the following conditions holds:
(1) and ;
(2) and ;
(3) and .
Definition 4. The lexicographic product of two graphs denoted by is a graph with vertex set and and are adjacent if and only if one of the following conditions holds:
(1) and ;
(2) .
Corollary 2. Let be two connected graphs with vertex numbers , respectively, where Then, each graph of , and has a -dset.
Proof. Suppose that and Since G is a connected graph, any vertex u in G satisfies Moreover, So, the vertex u in satisfies Similarly, any vertex v in H satisfies The vertex v in satisfies Then, any vertex x in satisfies That is, Then, according to Theorem 1, has a -dset. Next, we consider and For any vertex or assume that and
For according to Definition 3, is adjacent to , , and Therefore, for any vertex in , the degree of is greater than or equal to So, has a -dset.
For according to Definition 4, is adjacent to , , and Therefore, for any vertex in , the degree of is greater than or equal to So, has a -dset. □
Theorem 2. Let G be a connected graph. Then, G has a -dset if and only if one of the following conditions holds:
(1) There is a vertex u that satisfies ;
(2) There is a vertex v that satisfies Let and Moreover, for any vertex holds.
Proof. If G has a vertex u satisfying according to Lemma 1, the conclusion clearly holds. Next, we consider the case where G has a vertex v that satisfies Since then For any vertex we have Furthermore, we have Then, D is a -dset.
For a graph we assume that any vertex u of G satisfies If any vertex w satisfies then G does not have a -dset. Since the graph G has a -dset, there exist vertices in the graph G with degree greater than or equal to Suppose that is a -dset of We have Then, for any vertex we have Moreover, there exists a vertex v that satisfies where Let and Then, we have and We have Moreover, for any vertex we have □
Theorem 3. Let G be a connected graph. If there exists a vertex v in G satisfying the following conditions, then G has a -dset:
(1) and ;
(2) For any vertex x is not adjacent to
Proof. Assume that and where Since Since suppose that where j is any integer that satisfies If there is a vertex and according to Lemma 1, G has a -dset. Next, we consider the case where where j is any integer satisfying So, we consider where j is any integer satisfying
If then If and then where Let Then, For any j satisfying since has at least three neighbors within That is, for any j satisfying Furthermore, we have Then, v has three neighbors within So, D is a -dset. □
Theorem 3 requires this condition to be satisfied. That is, for a vertex v satisfying any two neighbors x and y of v are not adjacent. If G is a tree, then this condition holds. In Theorem 4, we consider the case where G is a tree.
Theorem 4. Let G be a tree with n vertices, where Then, G has a -dset if and only if there exists a vertex v in G satisfying one of the following conditions:
(1) ;
(2) and
Proof. If the tree G has a vertex v that satisfies according to Lemma 1, the conclusion clearly holds. Next, we consider the case where the tree G has a vertex v that satisfies and For any vertex since G is a tree, x is not adjacent to Then, according to Theorem 3, G has a -dset.
For a tree suppose that every vertex v of G satisfies If every vertex v satisfies then G does not have a -dset. Since the tree G has a -dset, there exist vertices in the tree G with degree greater than or equal to Suppose D is a -dset of Then, we have If any vertex v satisfies and then v is dominated by at least four vertices in D, where So, G does not have a -dset. Thus, if the tree G has a -dset, the tree G has a vertex v that satisfies or a vertex v that satisfies and □
Theorem 5. Let G be a connected graph with n vertices, where If G has a -dset, then
Proof. Suppose that D is a -set of Then, By definition, we have . If D is a -set of then Next, consider the case where D is not a -set of If D is not a -set of every vertex in is dominated by at least four vertices in Then, D is a 4-dominating set. Furthermore, since and we have Then, D is a -set of Since , suppose that is a -set of Then, Next, we prove that
If , then each vertex in is dominated by at least four vertices in If the vertex in is dominated by at least four vertices in So, the graph G does not have a -dset. Suppose that We have That is, If we add s vertices to then where s is an integer satisfying Since each vertex in is dominated by at least four vertices in each vertex in is dominated by at least four vertices in So, G does not have a -dset. Then, Therefore, there exists at least one vertex u in D satisfying We consider three cases.
Case 1:
Since u is adjacent to at least three vertices in Suppose that u is adjacent to three vertices in Let Since every vertex in is dominated by at least four vertices in every vertex in is dominated by at least three vertices in Moreover, u is dominated by three vertices in So, is a -dset. We have That is,
Case 2:
Since u is adjacent to at least two vertices in Suppose that u is adjacent to two vertices in Let Then, u is dominated by three vertices in and every vertex in is dominated by at least three vertices in So, is a -dset. We have That is,
Case 3:
Suppose that Then, is a 3-dominating set, which contradicts D as a -set. So, we have Since we assume that u is adjacent to where Let Then, u is dominated by three vertices in and every vertex in is dominated by at least three vertices in So, is a -dset, and □
We give examples of graphs that satisfy and
Example 1. The graph is shown in Figure 1. The set is a -set and also a -set. Furthermore, the set is a -set. We have So, Example 2. The graph is shown in Figure 2. The set is a -set and also a -set. Furthermore, the set is a -set. We have So, Example 3. The graph is shown in Figure 3. The set is a -set and also a -set. Furthermore, the set is a -set. We have So, Referring to [
27,
29], we give the following definition.
Definition 5. The corona product of two graphs G,H, denoted by is the graph obtained by taking a copy of G and copies of H and joining the i-th vertex of G to every vertex in the i-th copy of
We consider the corona product
of the two graphs
and
, as shown in
Figure 4.
Theorem 6. Let be two connected graphs with vertex numbers , respectively, where and If H has a -dset, then also has a -dset, and
Proof. Let and Let D be a -set of H. We can assume that Then, where Every vertex in is dominated by at least three vertices in and at least one vertex is dominated by three vertices in
For let denote the i-th copy of where each vertex of is connected to the i-th vertex of Suppose that The vertex in G is labeled as the vertex in where Then, Let Then, We have Then, every vertex in is dominated by at least three vertices in and at least one vertex is dominated by three vertices in is adjacent to . Then, is dominated by j vertices in where Then, is a -dset of So, □
Suppose that
G is a connected graph with
m vertices, where
Let
H be a star graph with four vertices, that is,
H is
. We have
Then, the upper bound of
can be reached. We have
. For example, in
Figure 4,
and the 15 larger nodes constitute a
-set.
Theorem 7. Let be positive integers. Then, the following statements apply:
- (1)
- (2)
- (3)
- (4)
- (5)
Proof. (1) For is obvious, so we omit the proof.
(2) Let and Let and If then is a -set. We consider Let D be a -set. Since at least one vertex is dominated by three vertices in we have or Suppose that . Then, each vertex in is dominated by at least three vertices in Since D is a -set, we have Similarly, suppose that . Then, we have So, Let Then, D is a -set.
In (3)–(5), let and where
(3) For is a -set.
(4) Suppose that any two distinct vertices and If then is adjacent to in If then is adjacent to in If then So is adjacent to in Then, in graph is adjacent to Let For any vertex and is adjacent to and Therefore, is a -set.
(5) Suppose that any two distinct vertices and If then is adjacent to in If then is adjacent to in Let For any vertex and is adjacent to and Therefore, for is a -set. □
Theorem 8. Suppose that are star graphs, where m and n are positive integers that satisfy and Let and Let and Then, Proof. Suppose that D is a -set of If and then is a -set. If and then is a -set. If and , suppose that . Then, there exists at least one vertex in that is not dominated by three vertices in Similarly, if there exists at least one vertex in that is not dominated by three vertices in Suppose that and . Then, there exists at least one vertex in that is not dominated by two vertices in Thus, there exists at least one vertex in that is not dominated by three vertices in Similarly, if and there exists at least one vertex in that is not dominated by three vertices in So, Then, is a -set. □
In Lemma 3 and Theorem 9 below, the notation represents a cycle of length i, where . Suppose that and Let and Let and
Lemma 3. Let be positive integers. Then, Proof. Suppose that D is a -set of Let and Suppose that and If there exists one vertex x in that is not dominated by vertices in , then x is not dominated by three vertices in Let and Then, there exists one vertex y in that is not dominated by vertices in So, y is not dominated by three vertices in
Let and Suppose that If , then, since every vertex has there exists at least one vertex in that is not dominated by three vertices in Let and Suppose that Then, is not dominated by three vertices in If and , suppose that Then, is not dominated by three vertices in So,
If and , we have Suppose that If , then for any vertex we have Then, each vertex in is not dominated by three vertices in If and , then there exists at least one vertex in that is not dominated by two vertices in So, there exists at least one vertex in that is not dominated by three vertices in If and , then there exists at least one vertex in that is not dominated by one vertex in , and there exists at least one vertex in that is not dominated by three vertices in So,
If and , we have Suppose that If , then each vertex in is not dominated by three vertices in If and , then there exists at least one vertex in that is not dominated by two vertices in So, there exists at least one vertex in that is not dominated by three vertices in If and , then there exists at least one vertex in that is not dominated by one vertex in , and there exists at least one vertex in that is not dominated by three vertices in So, □
Theorem 9. Let be positive integers. Then, Proof. Suppose that D is a -set of
Case 1:
Then, is a -set.
Case 2:
According to Lemma 3, we have If and then is a -set. If and then is a -set.
Case 3:
According to Lemma 3, we have Then, is a -set.
Case 4:
According to Lemma 3, we have Then, is a -set. □
In Lemma 4 and Theorem 10 below, the notation represents a path of i vertices, where . Suppose that and Let and Let and
Lemma 4. Let be positive integers. Then, Proof. Suppose that D is a -set of Let and Suppose that and If there exists one vertex x in that is not dominated by vertices in , then x is not dominated by three vertices in Let and Then, there exists one vertex y in that is not dominated by vertices in So, y is not dominated by three vertices in
Let Suppose that If For any vertex we have so there exists at least one vertex in that is not dominated by three vertices in If and , there exists at least one vertex in that is not adjacent to the vertex in Then, there exists at least one vertex in that is not dominated by three vertices in Similarly, if and then there exists at least one vertex in that is not dominated by three vertices in So,
If or , we have Suppose that If then there exists at least one vertex in that is not dominated by three vertices in If and then there exists at least one vertex in that is not dominated by two vertices in Then, there exists at least one vertex in that is not dominated by three vertices in If and then there exists at least one vertex in that is not dominated by one vertex in Then, there exists at least one vertex in that is not dominated by three vertices in So,
If , we have Suppose that If Then, there exists at least one vertex in that is not dominated by three vertices in If and then there exists at least one vertex in that is not dominated by two vertices in Then, there exists at least one vertex in that is not dominated by three vertices in If and then there exists at least one vertex in that is not dominated by one vertex in Then, there exists at least one vertex in that is not dominated by three vertices in So, □
Theorem 10. Let be positive integers. Then, Proof. Suppose that D is a -set of
Case 1:
Then, is a -set.
Case 2:
According to Lemma 4, we have If and then is a -set. If and then is a -set.
Case 3:
Then, is a -set.
Case 4:
According to Lemma 4, we have Then, is a -set.
Case 5:
If and then is a -set. If and then is a -set.
Case 6:
According to Lemma 4, we have Then, is a -set. □