1. Introduction
Let
,
E) be a simple, undirected, and connected graph.
V(
G) denotes the nonempty set of vertices of
, and
E(
) denotes the set of edges of
. For further graph theoretical terminology, we followed Bondy and Murthy [
1].
Ore [
2] was the first author to use the terms “dominating set” and “domination number” in his research. A subset
is a dominating set if for every vertex
in
V −
X′, there exists a vertex
in
X′ such that
is adjacent to
. The minimum cardinality of a dominating set in
is called the domination number of
and is denoted by
. We refer to a dominating set of minimum cardinality as
set of
. For further theoretical domination terminology, we followed Haynes et al. [
3]. Many types of domination parameters have been introduced and studied. Haynes et al. presented various types of domination and their relations with bounds. Total domination was studied by Cockayne et al. [
4] and further investigated by Bujtas et al. and Favaron et al. [
5,
6]. Harary et al. studied double domination in their research work [
7]. Paired domination was examined by Chellali et al., Gorzkowska et al., and Micheal et al. [
8,
9,
10]. Chellali et al. presented the relationship between the total and paired domination of trees [
8]. Edwards, Brigham et al., and Xinmin et al. studied the critical concepts of paired domination [
11,
12,
13]. Meenakshi and Baskar babujee studied a complementary-tree paired-domination vertex critical graph [
14]. Dunbar and Haynes [
15] started the study of domination in inflated graphs, which was further continued by other researchers [
16,
17,
18,
19]. Further, Swaminathan and Dharmalingam studied equitable domination [
20]. Paired equitable domination was introduced by Meenakshi and Baskar babujee in 2016 [
21]. The equitable domination of the complement of inflated graphs was studied by Meenakshi [
22]. Meenakshi elaborated on studies of the complement of inflated graphs of some standard graphs [
23].
The motivation of this research is to discuss the relationship between domination numbers, equitable domination, and paired equitable domination numbers in inflated graphs.
Section 2 focuses on the preliminary studies on equitable and paired equitable domination numbers of graphs.
Section 3 deals with applications of equitable and paired equitable domination of graphs in various fields.
In
Section 4, the effective results of the sum of the paired equitable domination numbers of certain types of graphs and their complements are discussed.
Section 5 provides the relationship between domination number, equitable domination, and the paired equitable domination number of the complement of inflated graphs.
Section 6 talks about the paired equitable domination of the complement of the inflated graph of the complement of
. In this paper, we considered only simple, finite, and connected graphs.
Table 1 represents the symbols used in this research and their meaning.
3. Applications
Equitable domination is a concept that arises in the field of graph theory. In graph theory, a graph consists of a set of vertices (also known as nodes) and a set of edges (also known as links) that connect pairs of vertices. Equitable domination focuses on the problem of assigning weights to the vertices of a graph in such a way that every vertex is dominated by the sum of its own weight and the weights of its neighbors. The concept of equitable domination has applications in various fields, including those outlined in the following subsections.
3.1. Network Design
Equitable domination can be applied in the design and analysis of communication networks, such as wireless networks or computer networks. By assigning appropriate weights to the vertices of the network graph, one can ensure a fair distribution of resources or coverage among the network nodes.
3.2. Social Networks
Equitable domination can be used to model and analyze social networks, where vertices represent individuals or entities and edges represent connections or relationships between them. Assigning equitable weights to the vertices can help to identify influential individuals or groups within the network, ensuring a balanced representation of influence or control.
3.3. Resource Allocation
Equitable domination can be utilized in scenarios where resources need to be allocated among a set of entities or regions. By assigning weights to the vertices representing entities or regions, one can ensure that the allocation is performed fairly and that no entity or region is significantly dominating or dominating excessively.
These are just a few examples of how equitable domination can be applied in different domains. The specific application depends on the context and problem at hand, but the underlying idea of achieving a fair distribution or coverage is common to all these applications.
Paired equitable domination is an extension of equitable domination that takes into account pairs of vertices in a graph. In this concept, the goal is to assign weights to both individual vertices and pairs of vertices in such a way that every vertex and every pair of vertices are dominated by the sum of their own weights and the weights of their neighbors.
While paired equitable domination is a relatively new concept in graph theory, it has the potential for various applications. A few possible applications are outlined in the following subsections.
3.4. Recommender Systems
Paired equitable domination can be used in recommender systems to improve the accuracy and fairness of recommendations. By assigning weights to both items and pairs of items, the system can take into account the similarity or compatibility between items when making recommendations. This can lead to more personalized and different recommendations that consider the preferences and relationships between users and items.
3.5. Collaborative Filtering
Paired equitable domination can be applied in collaborative filtering, a technique used in recommendation systems and information filtering. By assigning weights to both users and pairs of users, collaborative filtering algorithms can consider the similarity and compatibility between users’ preferences when generating recommendations. This can lead to more accurate predictions and better user satisfaction.
3.6. Genetic Analysis
Paired equitable domination can be applied in genetic analysis to understand the interactions between genes or genetic variants. By assigning weights to both individual genes and pairs of genes, researchers can analyze the relationships and dependencies between genes more effectively. This can help in identifying gene interactions, predicting disease risk, or understanding complex genetic traits.
These are just a few examples of how paired equitable domination can potentially be applied. As the field of paired equitable domination evolves, more specific and tailored applications are likely to emerge in various domains in which understanding pairwise relationships is important.
Illustration of equitable domination: Assume that
G is an organization with six employees. Each employee is equally qualified or may differ by a degree. Let the employees be denoted by EM1(J) (J represents junior), EM2(J), EM3(S) (S represents senior), EM4(J), EM5(S), and EM6(J). The aim is to find the minimum number of monitors or number of employees to share the information or knowledge to all. Select EM3(S) (which monitors EM2(J) and EM4(J)) and EM6(S) (which monitors EM1(J) and EM5(J)) to monitor others, which is subject to the constraint that the number of degrees of monitoring and nonmonitoring differs by at most one. To monitor the organization, a minimum of two employees is ample (see
Figure 2).
4. Paired Equitable Domination of the Complement of the Graphs
The paired equitable domination of the complement of some classes of graphs was studied.
4.1. Theorem 1
- (i)
If = Pm is a path, then .
- (ii)
If = Cm is a path, then .
- (iii)
If = Kr,s − e, then .
- (iv)
If = Pn × Cm, then .
Proof of (i): Let = and A be a minimal PEDS of Let x and y be the end nodes of H, each of a degree of one. In , the degree of both x and y is m − 2 and the degree of the remaining nodes is m − 3. Since it is a biregular graph, choose the node x which is of degree m − 2 and any one node of degree m − 3, say w, then A = {u, w} will be a dominating set of . Also, for all and Therefore, A is a PEDS of . Hence, .
Proof of (ii): Let = and A be a minimal PEDS of . Since every node of is of degree m 3, arbitrarily choose any two nodes, say x and y, such that both are paired to each other; then A = {x, y} will be a dominating set of , since for all and Hence .
Proof of (iii): Let where e = xy and S be a PEDS of Let the degree of every node be s and the degree of every node be r. Now every node in is of degree (r + s 1) s = r 1 except and deg(x) = r. Similarly, every node in is of degree (r + s 1) r = s 1 except and deg(y) = s. Clearly, x and y are paired in , and the pair dominates all the nodes, and also, for every and for every . Hence, S = {x, y} is a minimal PEDS of
Proof of (iv): Let
H =
Pn ×
Cm, which is shown in
Figure 3, and let
V(
H) =
be the vertex set of
H. Now
and
Let and be the corresponding node of a; then deg(a) + deg(b) = rs 1. Hence, , , and Let X be a minimal PEDS of Arbitrarily choose two nodes, say x and u; both are paired in , which dominates every node of Also, for all and . Hence, X = {x, y} is a PEDS of
4.2. Theorem 2
If , then prove that .
Proof: It follows from Theorem 4.1(i) and 4.1(ii).
4.3. Theorem 3
If is an acyclic graph and is connected, then .
Proof: Using Theorem 4.1(i), this result is obvious.
4.4. Theorem 4
Prove that if
- (i)
is a graph of order m where and for all .
- (ii)
is a graph of order m where and for all
Proof of (i): Let A be a minimal PEDS of an (m/2)-regular graph H on m nodes. Construct the (m/2)-regular graph H such that each node bi in H is joined to m/2 number of nodes to the left side of bi and m/2 number of nodes to the right side of bi. Since each vertex of H is joined to m/2 vertices, select any two paired vertices, say bi and bj, where ; these two nodes are ample to dominate all the nodes in H. Further, for all and for all . Hence, .
Since H is an (m/2)-regular graph on m nodes, deg(ai) = m/2 for every Let be a minimal PEDS of . Since is an (m/2)-regular graph, every node in dominates (m/2) nodes. Hence, we can find exactly two paired nodes, say and where and , that pairwise dominate . Further, for every and for every . Hence, . Hence, .
Proof of (ii): Let A be a PEDS of (m/2)-regular graph H on m nodes. Using Theorem 4.4(i) for the graph H, (m/2)-regular graph on m nodes, . Hence, .
5. Relation between Domination Number and Equitable and Paired Equitable Domination Number of Complement of Inflated Graphs
In this section, we study the equitable domination of the complement of the inflated graph of some standard graphs like the path, cycle, complete graph, and complete bipartite graph. Let G be a connected graph, and its complement of an inflated graph is denoted by .
5.1. Theorem 1
If G is a path of order n where , then (i) , and (ii) .
Proof of (i): Let G = Pn. The number of nodes in Pn is n, and is the number of nodes in . Let X′ be a minimum equitable dominating set of It consists of 2n 2 nodes among exactly two nodes, say r and , of degree 2n 4. The node r is not adjacent to exactly one node, say y of degree 2n 5; select arbitrarily any two nodes, say r and s, to dominate 2n − 2 nodes. Further, ; for every there exists a node such that , and hence, .
The domination number of Pn is .
Clearly, ; this equality holds good when n = 4.
Proof of (ii): Let X′ be a minimum equitable dominating set of Since the nodes r and s are paired in the case of path and cycle and also both nodes are of degree 2n 4, therefore for every such that and hence and this completes the proof.
This theorem is true for a cycle graph of order n where .
5.2. Theorem 2
If = Kn is a complete graph of order n, , then (i) , and (ii) .
Proof of (i): Let G = Kn and . Since G is complete, . Hence, . Each vertex in produces a clique in whose nodes are Let be a minimum equitable dominating set of It consists of n2 − n nodes; deg(xi) = n(n − 2) for every I = 1 to n2 − n. Select a node, say x, in that is adjacent to exactly n(n − 2) nodes. To dominate n2 − n nodes, we need at least two nodes. Arbitrarily select the nodes, say r and s, to dominate 2n nodes. Further, ; for every there exists node such that . Hence, . Clearly, , hence
Proof of (ii): Let G = Kn. Let X′ be a minimum equitable dominating set of .
Since every node is of the same degree n(n 2), consider any two nodes, say r and s, are paired with each other; further, ; for every there exists a node such that and hence and this completes the proof.
5.3. Theorem 3
If G = Km,n, then (i) , and (ii) if .
Proof of (i): Let G = Km,n. Now where and . Since each node is of degree n and each is of degree m, the inflated graph consists of m complete graphs of order n and n complete graphs of order m. Let X′ be a minimum equitable dominating set of The number of nodes in Km,n is mn, and is the number of nodes in . Let be the set of nodes of degree n(2m 1) 1, which contains mn nodes, and be the set of nodes of degree m(2n 1) 1, which contains mn nodes. Since it contains 2mn nodes, deg(xi) = n(2m 1) 1 for every i = 1 to n and deg(yi) = m(2n 1) 1 for every i = 1 to m. To dominate 2mn nodes, arbitrarily select one node r from of the same degree and one more, say s, from the remaining mn nodes of the same degree from . Further, ; for every there exists a node such that and hence . Clearly, , and therefore
Proof of (ii): Let
X′ be a minimum equitable dominating set of
If
, then this follows from
Section 5.2. Suppose
. Let
n =
m + 1, then
deg(
xi) = 2
m2 +
m 2 for every
i = 1 to
n and
deg(
yi) =
2m2 +
m 1 for every
i = 1 to
m.
contains
m(
n + 1) nodes, and we need two nodes to dominate it. Arbitrarily choose paired nodes
r and
s; both are ample to dominate the nodes
m(
n + 1). Further,
; for every
, there exists a node
such that
and hence
. Clearly,
, which completes the proof.
5.4. Theorem 4: If G = Pn × Cm Then
Proof: Let
G =
Pn ×
Cm, which is shown in
Figure 3, and let
=
be the node set of
. Now
and
.
The number of nodes in G is 2m + m(n 2), and is the number of nodes in Let X′ be a minimum equitable dominating set of Since it contains 4mn − mn nodes, deg(xi) = 3 for every i = 1 to 2m and deg(yj) = 2 for every j = 1 to m(n − 2). To dominate 4mn 4m nodes, we need at least two nodes. Arbitrarily select the nodes, say r and s, to dominate these nodes such that ; for every there exists a node such that , hence the proof.
5.5. Theorem 5
If G is a (k1, k2) biregular graph with m as the number of nodes of degree k1 and n as the number of nodes of degree k2 = k1 + 1, then .
Proof: Since G is a biregular graph, the result is obvious.
6. Paired Equitable Domination of the Complement of the Inflated Graph of Complements
In this section, we discuss the paired equitable domination of the complement of . The complement of is denoted by .
Example: Let
G be the graph shown in
Figure 4a and its complement is
; the inflated graph of
and complement of
are shown in
Figure 4b,
Figure 5, and
Figure 6, respectively.
The paired equitable dominating set of = {ec, db}. Hence, .
6.1. Theorem 1
Let G be either a regular graph or a biregular graph of order with ; then .
Proof: Let G be a connected graph of order m with and be an inflated graph of complement of G. Let D be a minimal equitable dominating set of . Construct the graph .
Case(i): Let G be an s-regular graph where . The degree of every node in is m (1 + r). Now there are m[m (1 + s)] number of nodes in . Every node in is adjacent to exactly m[m (1 + s)] [m (1 + s)] = (n − 1 s)(m 1) number of nodes. Since (m − 1 s)(m 1) < m[m (1 + s)], choose at least two nodes, enough to dominate m[m (1 + s)] number of nodes in . Also, E and Hence, .
Case(ii): Let G be a biregular graph (k1, k2) with n1 number of nodes of degree k1 and n2 number of nodes of degree k2 such that n1 + n2 = m. Construct the graphs and . There are n1 number nodes of degree (m − 1 k1) and n2 number nodes of degree (m − 1 k2) in and n1(m − 1 k1) + n2(m − 1 k2) number nodes in . The nodes of degree (m − 1 k2) are adjacent to exactly {n1(m − 1 k1) + n2(m − 1 k2) (m − 1 k2)} 1 = {n1(m − 1 k1) + (m − 1 k2) (n2 1)} 1 nodes, and the nodes of degree (m − 1 k1) are adjacent to exactly {n1(m − 1 k1) + n2(n − 1 k2) (m − 1 k1)} 1 = {m − 1 k1)(n1 1) + n2(m − 1 k2)} 1 nodes.
Since the number of nodes n1(m − 1 k1) + n2(m − 1 k2) in is adjacent to either {n1(m − 1 k1) + (m − 1 k2) (n2 1)} 1 or {(m − 1 k1)(n1 1) + n2(n − 1 k2)} 1, two nodes are ample to dominate all the nodes. Select two nodes, say x and y, such that both are paired to each other. Furthermore, E and Hence,
6.2. Theorem 2
For any connected graph G of order , then where M is the number of different degrees present in .
Proof: Let ni be the number of nodes of degree mi, i = 1, 2, …, m, where n1 + n2 + … nm = m. Construct the graphs and . Let X be the minimal dominating set of . There are ni nodes of degree (m − 1 − mi), i = 1, 2, …, m in and n1(m − 1 k1) + n2(m − 1 k2)+ …+ nm(m − 1 km) vertices in . The nodes of degree (m − 1 k1) are adjacent to exactly {[n1(m − 1 k1) + n2(m − 1 k2) + … + nm(m − 1 km)] (m − 1 k1)} − 1 = [(m − 1 k1)(n1 1) + n2(m − 1 k2)+ … + nm(m − 1 km)] 1 number of nodes. The nodes of degree (m − 1 k2) are adjacent to exactly {[n1(m − 1 k1) + n2(m − 1 k2)+ … + nm(m − 1 km)] (m − 1 k2)} 1 = [n1(m − 1 k1) + (m − 1 k2)(n2 1) … + nm(m − 1 kn)] 1 number of nodes. Likewise, the nodes of degree (m − 1 km) are adjacent to exactly {[n1(m − 1 k1) + n2(m − 1 k2) + … + nm(m − 1 kn)] (m − 1 km)} 1 = [n1(m − 1 k1) + n2(m − 1 k2) + … + m − 1 km)(nm 1)] 1 number of nodes.
Since the nodes of degree (m − 1 k1) are adjacent to {[n1(m − 1 k1) + n2(m − 1 k2)+ … + nm(m − 1 km)] (m − 1 k1)} 1 number of nodes, (m − 1 ki) are paired with {[n1(m − 1 k1) + n2(m − 1 k2)+ … + nm(m − 1 km)] (m − 1 ki)} 1 for i = 1, 2, …, m number of nodes. To pairwise dominate the nodes in , select one of the nodes from each distinct degree, say (m − 1 ki), i = 1, 2, …, m; we obtain m number of nodes to pairwise dominate the nodes in . Since one node, say ri from different degrees, then , E() and Hence,