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Article

On the Strong Secure Domination Number of a Graph

by
Turki Alsuraiheed
1,
J. Annaal Mercy
2,
L. Benedict Michael Raj
2 and
Thangaraj Asir
3,*
1
Department of Mathematics, College of Science, King Saud University, Riyadh 11451, Saudi Arabia
2
Department of Mathematics, St. Joseph’s College, Affiliated to Bharathidasan University, Trichy 620002, India
3
Department of Mathematics, Pondicherry University, Puducherry 605014, India
*
Author to whom correspondence should be addressed.
Mathematics 2024, 12(11), 1666; https://doi.org/10.3390/math12111666
Submission received: 18 March 2024 / Revised: 7 May 2024 / Accepted: 21 May 2024 / Published: 27 May 2024
(This article belongs to the Special Issue Graph Theory and Applications, 2nd Edition)

Abstract

:
In this paper, we characterize trees with a strong secure domination number less than or equal to 4 and compute this parameter for certain classes of graphs. Also, we investigate bounds for the strong secure domination number of vertex gluing of two graphs.

1. Introduction

The graphs examined in this study are finite, simple, undirected, without loops, and with multiple edges. Let G = ( V , E ) be a graph and let v V ( G ) . The number of vertices adjacent to v is called the degree of v and is denoted by d e g ( v ) . The open neighborhood of v is defined as N ( v ) = { u : u v E ( G ) } . The closed neighborhood of v is defined as N [ v ] = N ( v ) { v } . Let X V ( G ) and u X . The private neighborhood p n ( u , X ) is defined by p n ( u , X ) = { v : N ( v ) X = { u } } . The external private neighborhood of a vertex is defined as e p n ( u , X ) = { v : v V \X and N ( v ) X = { u } } , also represented by e p n G ( u , X ) . A connected acyclic graph is called a tree. A tree with a vertex of full degree is referred to as a star and is denoted by K 1 , k , where k is the degree of the full-degree vertex. A vertex of degree one is known as a pendant vertex. A neighboring vertex to a pendant vertex is referred to as its support vertex. A caterpillar is a tree in which the removal of all its pendant vertices results in a path. The graph obtained by removing an edge e from a graph G is denoted by G e and this is called edge deletion. A path on n vertices and a cycle on n vertices are denoted P n and C n , respectively. The reader may refer to [1] for notation and terminology that is not defined here.
Let D V ( G ) . If every vertex that is not in D is adjacent to a vertex that is in D, then D is a dominating set. The minimum cardinality of a dominating set in G is its domination number γ ( G ) of G. A well-researched graph parameter in the literature is domination, we refer the reader to the book [2] and its references for more details.
Sampathkumar et al. [3] introduced the strong domination number in 1996. If for every v V ( G ) \D there is a vertex u D with u v E ( G ) and d e g ( u ) d e g ( v ) , then a set D of vertices in a graph G is a strong dominating set. The minimum cardinality of a strong dominating set in G is its strong domination number γ s t ( G ) of G. The results of a study on strong domination in graphs were presented in [4,5]. The concept of strong domination has, thus, been extensively studied in the literature. In [6], Cockayne et al. introduced the concept of secure domination. A dominating set D is called a secure dominating set (sds) if for every v V \D there exists some u D such that u and v are adjacent and N [ ( D \ { u } ) { v } ] = V ( G ) . Here, u is called a defender of v. The secure domination number γ s d ( G ) of G is the minimum cardinality of a secure dominating set in G. Additional findings regarding this parameter can be found in [7,8,9,10].
Inspired by the ideas of secure and strong domination, it is logical to explore both of these concepts further, beginning with the idea of strong secure domination as put forth by Annaal Mercy et al. [11]. A secure dominating set D is called a strong secure dominating set (ssds) if for every v V \D there exists u D such that u and v are adjacent and d e g ( u ) d e g ( v ) . The corresponding u is called a strong neighbor of v. The minimum cardinality of an ssds is called a strong secure domination number and is denoted by γ s s ( G ) . Note that γ s s ( G ) = n if and only if G is a totally disconnected graph.
Why should such a domination in graphs be taken into consideration? This is the scenario: A computer network consists of a number of linked devices, each of which has connections to other devices. Consider a scenario where a network administrator must set up access control lists (ACLs) to manage network traffic flow. The following guidelines are what the administrator wishes to implement:
  • Each device not explicitly allowed access to certain network resources should have at least one connected device that is granted access, ensuring connectivity and efficient data transmission.
  • If a device is persistent about using a particular resource and applies pressure to be allowed access, it can be swapped out for a connected device that has already been given permission, as long as the new configuration keeps the connectivity requirement from the initial condition.
  • It is recommended that any device that is not explicitly granted access to specific network resources be connected directly to a device that has higher access privileges. In this way, a device with greater permissions will always be accessible for potential administrative or troubleshooting needs. This implies that devices with limited access are purposefully connected to devices with more expansive access capabilities. This makes network management easier and guarantees that troubleshooting tasks can be completed successfully, which improves the overall security and resilience of the network.
The goal of the network administrator is to reduce the size of the access control lists while optimizing resource allocation and network security. This strategy is consistent with the strong secure dominating theory, which maintains network connectivity while attending to the needs of individual device access. This idea is useful in a variety of contexts, including the setting up of ATMs in multiple locations, the distribution of secret keys in cryptography, military operations, and commercial endeavors.

2. Preliminaries

First, let us display the strong secure domination number of standard graphs.
Remark 1. 
For n N , we have
(i). 
γ s s ( K 1 , n ) = n .
(ii). 
γ s s ( P n ) = 3 n 7 .
(iii). 
γ s s ( C n ) = 3 n 7 .
The next two results are helpful in the sections that follow.
Proposition 1 
([12]). Let G be a connected graph and X be a non-empty subset of V. Then, the following statements are equivalent.
(i). 
X is a secure dominating set.
(ii). 
For each u V \X, there exists v X N ( u ) such that e p n ( v , X ) N [ u ] .
(iii). 
For each u V \X, there exists v X N ( u ) such that the subgraph induced by { u , v } e p n ( v , X ) is complete.
The following proposition is obtained by using Proposition 1.
Proposition 2. 
Let G K 2 be a graph. If a support vertex u in G has k pendant neighbors, then:
(i). 
Every sds must contain either u and k 1 of its pendant neighbors or k pendant neighbors.
(ii). 
Every ssds must contain u and k 1 of its neighbors.
The existence of a graph for a given strong secure domination number is the main point of the following result.
Theorem 1. 
For any two positive integers k and n for k n , there exists a graph G of order n with γ s s ( G ) = k .
Proof. 
Let us construct a graph G with V ( G ) = { v 1 , v 2 , , v n } with γ s s ( G ) = k . When k = 1 or n 1 or n, choose G K n or K 1 , n 1 or K n ¯ , respectively. Let 1 < k n 2 . Consider the path P k = ( v 1 , v 2 , , v k ) . Now, partition the remaining n k vertices into k non-empty sets V 1 , V 2 , , V k . Now, construct a graph G from the path P k in such a way that the subgraph induced by V i { v i } is complete for all 1 i k . Then, { v 1 , v 2 , , v k } is a γ s s -set of G, and so, γ s s ( G ) = k . Let n 2 < k < n 1 . Consider a path P 2 with its vertices as v 1 and v 2 . Partition the remaining n 2 vertices into two sets V 1 = { v 3 , v 4 , , v k + 1 } and V 2 = { v k + 2 , v k + 3 , , v n } . Construct G from the path P 2 in such a way that each vertex of V 1 is adjacent to v 1 and the subgraph induced by V 2 { v 2 } is complete. Then, { v 1 , v 2 , , v k } is a γ s s -set of G. □

3. Trees with Secure Domination Number at Most 4

In this section, we give a necessary and sufficient condition for a tree to have a secure domination number less than or equal to 4. A survey on the constructive characterization of trees using different types of domination numbers was recently provided by the authors in [13].
Let us first review the results for the trees whose strong secure dominance number is either 2 or 3.
Proposition 3 
([11]). Let T be a tree. Then:
(i). 
γ s s ( T ) = 1 if and only if T P 2 .
(ii). 
γ s s ( T ) = 2 if and only if T is either P 3 or P 4 .
Let K 1 , k and K 1 , l be two stars. A path of length t connects the centers of K 1 , k and K 1 , l . The resulting graph is represented by K k , l t .
The following theorem characterize all trees whose strong secure dominance number is 3. In what follows, the graphs’ G i s are given in Figure 1.
Theorem 2. 
Let T be a tree. Then, γ s s ( T ) = 3 if and only if T i = 1 4 G i .
Proof. 
Assume that γ s s ( T ) = 3 . Then, by Proposition 2, T can have at most three support vertices. Suppose T has exactly one support vertex. Then, the support vertex must have 3 pendant neighbors and so T K 1 , 3 . Assume that T has exactly two support vertices. Then, T is K k , l t , where k , l 1 and t 1 . By Proposition 2, we have k + l { 2 , 3 } . If k + l = 2 , then T is a path. Since γ s s ( P n ) = 3 n 7 = 3 , we obtain n { 5 , 6 , 7 } ; these graphs are listed in G 2 . If k + l = 3 , without loss of generality, assume that k = 2 and l = 1 , then T G 3 . When T has exactly three support vertices, T is isomorphic to the graph G 4 . The converse is obvious. □
The following theorem characterize all trees whose strong secure dominance number is 4. In what follows, the graphs’ F i s are given in Figure 2 and Figure 3.
Theorem 3. 
Let T be a tree. Then, γ s s ( T ) = 4 if and only if T i = 1 8 F i .
Proof. 
Assume that γ s s ( T ) = 4 . Let X be a minimum strong secure dominating set of T. Then, by Proposition 2, T has at most four support vertices. Suppose T has exactly one support vertex. Then, T is a star graph. Since γ s s ( K 1 , n ) = n , we have T K 1 , 4 = F 1 . Suppose T has exactly two support vertices. Obviously T becomes K k , l t , where k , l 1 and t 1 . By Proposition 2, we have 2 k + l 4 . If k + l = 2 , then T is a path. In this case, T is isomorphic to a graph in F 2 . Let k + l = 3 . Without loss of generality, let k = 2 and l = 1 . Let x and y be the two support vertices of T. Since | X | 4 , x and y are connected by a path of length at most 4. In this case, T is isomorphic to a graph in F 3 . If k + l = 4 , then k = 2 and l = 2 or k = 1 and l = 3 . In this case, T is isomorphic to a tree in F 4 . Next, we assume that T has exactly three support vertices, say, x , y , and z. By Proposition 2, x , y , z X and the number of pendant vertices is at most 4. Without loss of generality, let us assume that the path between x and y is the longest. By Proposition 1, no vertex in the path has either x or y as defender, with respect to X . In fact, it cannot have z as a defender. Also, the length of the path between x and y is at most 4. Therefore, d ( x , y ) 1 ; otherwise T becomes a tree with two support vertices, a contradiction. Let d ( x , y ) = 2 . In this case T is isomorphic to a graph in F 5 . When d ( x , y ) = 3 , there exist two vertices u and v between x and y. In this case, T is isomorphic to a graph in F 6 . When d ( x , y ) = 4 , T is isomorphic to a graph in F 7 . If T has exactly four support vertices, then T F 8 . The converse is obvious. □

4. Bounds on the Strong Secure Domination Number of Graphs

In this section, we provide an upper and lower bound for the strong secure domination number of the rooted tree and the vertex gluing of two graphs.
A tree is said to be a rooted tree [14] if one of the vertices is distinguished as the root from the remaining. In what follows, we consider a rooted tree T k , where a root vertex is of degree and all the other vertices except the support vertices are of degree k, where k . Support vertices can be of any degree.
The following theorem gives bounds for the strong secure domination number of the rooted tree.
Theorem 4. 
For the rooted tree T k , γ s d ( T k ) γ s s ( T k ) γ s d ( T k ) + 1 .
Proof. 
Let X be a minimum secure dominating set of T k . Then, by Proposition 1 either all the pendant vertices belong to X or every support vertex and every pendant vertex, with the exception of one, belong to X. In other words, with respect to every support vertex, either the support vertex or one of its corresponding pendant vertices must be in V \ X .
If a pendant vertex is in V \ X , then its corresponding support vertex is the only defender as well a strong neighbor in X as X is the secure dominating set.
If a support vertex is in V \ X , then replacing a pendant vertex with its corresponding support vertex in X results in a strong secure dominating set. Also, note that the support vertex cannot be a defender for the root vertex as well as any internal vertex.
Suppose the root vertex belongs to V \ X , then it has a strong neighbor in X as X is a secure dominating set and its defender is an internal vertex whose degree is greater than or equal to l. In this case, γ s d ( T k ) = γ s s ( T k ) .
If the root vertex belongs to X, then it can defend at most l neighbors which are internal vertices whose degree is greater than or equal to l. Let the internal vertices be x 1 , x 2 , , x m , where m l . Since X is a secure dominating set, x 1 , x 2 , , x m have another neighbor in X. If they are strong neighbors, then γ s d ( T k ) = γ s s ( T k ) . Otherwise, remove the root vertex from X and add x 1 , x 2 , , x m in X. Then, each vertex in V \ X has a strong neighbor. Thus, γ s s ( T k ) = γ s d ( T k ) + m 1 γ s d ( T k ) + 1 .
If there exist t , 1 t m internal vertices which have strong neighbors in X and m t neighbors have no strong neighbors in X, then by adding those m t vertices in X, we obtain a strong secure dominating set. Therefore, we obtain γ s s ( T k ) γ s d ( T k ) + 1 . □
Let G 1 and G 2 be two connected graphs and let u V ( G 1 ) , v V ( G 2 ) . Then, a new graph obtained by merging u and v as a single vertex is called a 1-gluing (or vertex gluing) [15] of G 1 and G 2 and is denoted by G 1 K 1 G 2 . One can visualize any graph that has a cut vertex as a 1-gluing of two graphs. Clearly, all trees can be viewed as a 1-gluing of two trees.
The first result gives bounds on the secure domination number of the 1-gluing of two graphs.
Theorem 5. 
Let G 1 and G 2 be two connected graphs. Then, γ s d ( G 1 ) + γ s d ( G 2 ) 2 γ s d ( G 1 K 1 G 2 ) γ s d ( G 1 ) + γ s d ( G 2 ) .
Proof. 
Let S and S be γ s d sets of G 1 and G 2 , respectively. Let x be the merged vertex in G 1 K 1 G 2 . Clearly, X = S S is a dominating set of G 1 K 1 G 2 .
Case 1: Let x S S . Clearly, V ( G 1 K 1 G 2 ) X = ( V ( G 1 ) S ) ( V ( G 2 ) S ) . Let s V ( G 1 K 1 G 2 ) X . This implies that s V ( G 1 ) S or V ( G 2 ) S or both. Without loss of generality, let s V ( G 1 ) S . Since S is a secure dominating set, there exists a defender t S . By Proposition 1, the subgraph induced by e p n G 1 ( t , S ) { t , s } forms a clique in G 1 . Therefore,
e p n G 1 K 1 G 2 ( t , X ) = e p n G 1 ( t , S ) if t x e p n G 1 ( t , S ) e p n G 2 ( t , S ) if t = x .
If e p n G 1 ( t , S ) and e p n G 2 ( t , S ) , then e p n G 1 K 1 G 2 ( x , X { y } ) = e p n G 2 ( x , S ) , where y e p n G 1 ( x , S ) . Thus,
e p n G 1 K 1 G 2 ( t , X { y } ) = e p n G 1 ( t , S ) if t x e p n G 2 ( t , S ) if t = x .
So, the subgraph induced by e p n G 1 K 1 G 2 ( t , X { y } ) ) { s , t } forms a clique in G 1 K 1 G 2 . By Proposition 1, X { y } is a secure dominating set of G 1 K 1 G 2 . So, γ s d ( G 1 K 1 G 2 ) | X | + 1 | S | + | S | = γ s d ( G 1 ) + γ s d ( G 2 ) .
If either e p n G 1 ( x , S ) = or e p n G 2 ( x , S ) = , then e p n G 1 K 1 G 2 ( t , X ) = e p n G 1 ( t , S ) or e p n G 2 ( t , S ) . In this case, e p n G 1 K 1 G 2 ( t , X ) { s , t } is a clique in G 1 K 1 G 2 for every s V ( G 1 K 1 G 2 ) X . Thus, by Proposition 1, t is a defender of s V ( G 1 K 1 G 2 ) X and X is a secure dominating set of G 1 K 1 G 2 . Hence, γ s d ( G 1 K 1 G 2 ) | X | = | S | + | S | 1 = γ s d ( G 1 ) + γ s d ( G 2 ) 1 .
Case 2: When x S S , e p n G 1 K 1 G 2 ( t , X ) = e p n G 1 ( t , S ) or e p n G 2 ( t , S ) for every defender t X . So, for each s V ( G 1 K 1 G 2 ) X , either s V ( G 1 ) S or V ( G 2 ) S . Since S and S are secure dominating sets, there exists t S S such that t is a defender of s. By Proposition 1, either e p n G 1 ( t , S ) { s , t } or e p n G 2 ( t , S ) { s , t } forms a clique in G 1 or G 2 for every s V ( G 1 K 1 G 2 ) X . Therefore, e p n G 1 K 1 G 2 ( t , X ) { s , t } forms a clique in G 1 K 1 G 2 for every s V ( G 1 K 1 G 2 ) X . Hence, X = S S is a secure dominating set of G 1 K 1 G 2 , and so, γ s d ( G 1 K 1 G 2 ) | X | = | S | + | S | γ s d ( G 1 ) + γ s d ( G 2 ) .
Now, we prove the lower bound. Let X be a γ s d set of G 1 K 1 G 2 . Let X = X X , where X V ( G 1 ) and X V ( G 2 ) . Suppose that x X ; clearly, X and X are dominating sets of G 1 and G 2 , respectively. Also, e p n G 1 K 1 G 2 ( t , X ) = e p n G 1 ( t , X ) or e p n G 2 ( t , X ) for every t X . Note that V ( G 1 K 1 G 2 ) X = ( V ( G 1 ) X ) ( V ( G 2 ) X ) . We prove that X and X are secure dominating sets of G 1 and G 2 , respectively. For s V ( G 1 ) X , s V ( G 1 K 1 G 2 ) X . Since X is a secure dominating set of G 1 K 1 G 2 , s has a defender t in X, and so, by Proposition 1, e p n G 1 K 1 G 2 ( t , X ) { s , t } is a clique in G 1 K 1 G 2 . So, e p n G 1 ( t , X ) { s , t } forms a clique in G 1 . By proposition 1, X is a secure dominating set of G 1 . Similarly X is a secure dominating set of G 2 . Therefore, γ s d ( G 1 K 1 G 2 ) = | X | = | X | + | X | 1 , and so, γ s d ( G 1 K 1 G 2 ) γ s d ( G 1 ) + γ s d ( G 2 ) 1 .
Assume that x X . Suppose x has two defenders with respect to X in G 1 K 1 G 2 , one in X and another in X . Since X is a secure dominating set, for every s V ( G 1 ) X there exists t X such that e p n G 1 ( t , X ) { t , s } is a clique in G 1 . By Proposition 1, X is a secure dominating set of G 1 . Similarly X is a secure dominating set of G 2 . So, γ s d ( G 1 K 1 G 2 ) = | X | = | X | + | X | γ s ( G 1 ) + γ s ( G 2 ) .
Suppose x has a defender in exactly one of X or X , say in X . Then, X { x } and X { x } are secure dominating sets in G 1 and G 2 , respectively. Therefore, γ s ( G 1 K 1 G 2 ) γ s ( G 1 ) + γ s ( G 2 ) 2 . □
The bounds given in Theorem 5 are sharp. For instance, consider the graph P 7 = P 3 K 1 P 5 , where x is one of the pendant vertices of the path P 3 and P 5 . Here, 3 = γ s d ( P 7 ) = γ s d ( P 5 ) + γ s d ( P 3 ) 2 . Furthermore, take a look at the graph in Figure 4 for the archiving upper bound in Theorem 5.
We are now in the position to prove the following result. A bound for the strong secure domination number of 1-gluing of two graphs is provided by the following theorem.
Theorem 6. 
Let G 1 and G 2 be two connected graphs. Then, γ s s ( G 1 ) + γ s s ( G 2 ) 2 γ s s ( G 1 K 1 G 2 ) γ s s ( G 1 ) + γ s s ( G 2 ) + 1 .
Proof. 
Let S and S be two γ s s sets of G 1 and G 2 , respectively, and let x be the common vertex of G 1 K 1 G 2 . Let X = S S . Clearly, X { x } is a strong dominating set of G 1 K 1 G 2 .
Case 1: Let x S S . When e p n G 1 ( x , S ) and e p n G 2 ( x , S ) , as in case 1 of Theorem 5, one can prove that X { y } , where y e p n G 1 ( x , S ) is a secure dominating set of G 1 K 1 G 2 . Therefore, X { y } is a strong secure dominating set of G 1 K 1 G 2 . When either e p n G 1 ( x , S ) = or e p n G 2 ( x , S ) = , we have that X is a secure dominating set of G 1 K 1 G 2 , refer to case 1 of Theorem 5. So, γ s s ( G 1 K 1 G 2 ) | X { y } | | X | + 1 = | S | + | S | = γ s s ( G 1 ) + γ s s ( G 2 ) .
Case 2: Let x S S . By case 2 of Theorem 5, X is a secure dominating set of G 1 K 1 G 2 and X { x } is a strong secure dominating set of G 1 K 1 G 2 . Hence, γ s s ( G 1 K 1 G 2 ) | X | + 1 = | S | + | S | + 1 = γ s s ( G 1 ) + γ s s ( G 2 ) + 1 .
Let us prove the lower bound. Let X be a γ s s -set of G 1 K 1 G 2 and X = X X , where X V ( G 1 ) and X V ( G 2 ) . Let x X . As in the proof of Theorem 5, X and X are secure dominating sets. Moreover, X and X are strong secure dominating sets of G 1 and G 2 , respectively. Therefore, γ s s ( G 1 K 1 G 2 ) γ s s ( G 1 ) + γ s s ( G 2 ) 1 . Assume that x X . Suppose x has two defenders in X, one in X and another in X , then X and X are secure dominating sets. If x has no strong neighbor in X but in X , then X { x } and X are strong secure dominating sets of G 1 and G 2 , respectively. Similarly, if x has no strong neighbor in X but in X , then X { x } and X are strong secure dominating sets. Therefore, γ s s ( G 1 K 1 G 2 ) = | X | = | X | + | X | γ s s ( G 1 ) + γ s s ( G 2 ) + 1 .
If x has no defender in X but in X , then X { x } and X { x } become strong secure dominating sets of G 1 and G 2 , respectively. Therefore, γ s s ( G 1 K 1 G 2 ) γ s s ( G 1 ) + γ s s ( G 2 ) 2 . Similarly, one can prove that x has no defender in X but in X . □
For the lower bound given in Theorem 5, consider the graph P 14 = P 10 K 1 P 5 , where x is one of the pendant vertices of the path P 10 and P 5 . Here, γ s s ( P 10 ) = 5 , γ s s ( P 5 ) = 3 , and γ s s ( P 14 ) = 6 . Moreover, the upper bound provided in Theorem 5 is equal to the graph in Figure 5.
Now, Theorem 5 gives us the following result.
Remark 2. 
For n , m N , we have
(i). 
γ s s ( K m K 1 K n ) = γ s s ( K m ) + γ s s ( K n ) .
(ii). 
γ s s ( K m K 1 K 1 , n ) = γ s s ( K m ) + γ s s ( K 1 , n ) .
(iii). 
γ s s ( C m K 1 C n ) = 2 if n = m = 3 γ s s ( C m ) + γ s s ( C n ) 1 otherwise .
Let us close this paper by comparing the strong secure domination number of a graph and the strong secure domination number of its edge removal graph.
Theorem 7. 
Let G K 2 be a graph and e = u v be any edge of G. Then, γ s s ( G ) 1 γ s s ( G e ) γ s s ( G ) + d e g ( u ) + d e g ( v ) 2 .
Proof. 
Let S be an ssds of G. Suppose both u and v S and u defends some vertices u V \S such that d e g ( u ) = d e g ( u ) . Then, S = { S \ { u , v } N ( u ) N ( v ) } is an ssds for G e . Suppose neither u nor v belongs to S. Then, removing an edge does not make any changes. In this case, S is a γ s s -set for G e . If either u S or v S , say u S . In this case, by adding an edge which is incident with v to S gives an ssds for G e . Hence, γ s s ( G e ) γ s s ( G ) + d e g ( u ) + d e g ( v ) 2 .
Let S be an ssds for G e . If u , v S , then adding an edge e does not make any difference, so that S is a γ s s set for G. Suppose either u S or v S , say u S . Then, S = S { v } is an ssds for G. Therefore, γ s s ( G ) = γ s s ( G e ) + 1 . Suppose u S and v S . Without loss of generality, let d e g ( u ) d e g ( v ) . Then, S { v } gives an ssds for G. Therefore, γ s s ( G ) = γ s s ( G e ) + 1 . Hence, γ s s ( G ) 1 γ s s ( G e ) . □
For the lower bound of Theorem 7, consider the graph G given in Figure 6. Note that γ s s ( G ) = 7 and γ s s ( G e ) = γ s s ( P 7 ) + γ s s ( P 7 ) = 3 + 3 = γ s s ( G ) 1 . For the upper bound, consider the graph P 7 . Here, γ s s ( P 7 e ) = 4 = 3 + 1 + 2 2 = γ s s ( P 7 ) + d e g ( u ) + d e g ( v ) 2 , where e = u v with u being a pendant vertex and v a vertex adjacent to u in P 7 .

5. Conclusions

In networking systems and protection strategies, domination in graphs plays a vital role. Now, the strong secure domination in the networking systems will be more beneficial in building stronger protection strategies.

Author Contributions

Conceptualization, T.A. (Thangaraj Asir); validation, J.A.M. and L.B.M.R.; methodology, T.A. (Thangaraj Asir); investigation, L.B.M.R.; writing—original draft preparation, J.A.M.; writing—review and editing, T.A. (Thangaraj Asir); supervision, L.B.M.R. and T.A. (Thangaraj Asir); project administration, T.A. (Turki Alsuraiheed); funding acquisition, T.A. (Turki Alsuraiheed). All authors have read and agreed to the published version of the manuscript.

Funding

This work was funded by the Researchers Supporting Project at King Saud University, Riyadh, Saudi Arabia under grant no. RSPD2024R934. The first author, therefore, acknowledge with thanks the technical and financial support provided by King Saud University.

Data Availability Statement

Data are contained within the article.

Acknowledgments

The referee provided insightful feedback on the initial manuscript draft, which the authors appreciate.

Conflicts of Interest

The authors declare that there are no conflicts of interests.

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Figure 1. Trees with γ s s ( T ) = 3 .
Figure 1. Trees with γ s s ( T ) = 3 .
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Figure 2. Trees with γ s s ( T ) = 4 .
Figure 2. Trees with γ s s ( T ) = 4 .
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Figure 3. Trees with γ s s ( T ) = 4 .
Figure 3. Trees with γ s s ( T ) = 4 .
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Figure 4. γ s d ( G 1 K 1 G 2 ) = γ s d ( G 1 ) + γ s d ( G 2 ) .
Figure 4. γ s d ( G 1 K 1 G 2 ) = γ s d ( G 1 ) + γ s d ( G 2 ) .
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Figure 5. γ s s ( G 1 K 1 G 2 ) = γ s s ( G 1 ) + γ s s ( G 2 ) + 1 .
Figure 5. γ s s ( G 1 K 1 G 2 ) = γ s s ( G 1 ) + γ s s ( G 2 ) + 1 .
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Figure 6. Graphs with γ s s ( G e ) = γ s s ( G ) 1 .
Figure 6. Graphs with γ s s ( G e ) = γ s s ( G ) 1 .
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Alsuraiheed, T.; Mercy, J.A.; Raj, L.B.M.; Asir, T. On the Strong Secure Domination Number of a Graph. Mathematics 2024, 12, 1666. https://doi.org/10.3390/math12111666

AMA Style

Alsuraiheed T, Mercy JA, Raj LBM, Asir T. On the Strong Secure Domination Number of a Graph. Mathematics. 2024; 12(11):1666. https://doi.org/10.3390/math12111666

Chicago/Turabian Style

Alsuraiheed, Turki, J. Annaal Mercy, L. Benedict Michael Raj, and Thangaraj Asir. 2024. "On the Strong Secure Domination Number of a Graph" Mathematics 12, no. 11: 1666. https://doi.org/10.3390/math12111666

APA Style

Alsuraiheed, T., Mercy, J. A., Raj, L. B. M., & Asir, T. (2024). On the Strong Secure Domination Number of a Graph. Mathematics, 12(11), 1666. https://doi.org/10.3390/math12111666

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