Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3s (2025) 154 https://internationalpubls.com Relatively Prime Domination Number in Quadrilateral Snake Graphs A. Anat Jaslin Jini1*, A. Jancy Vini1, B. Shoba2, S. Manikanda Prabhu2, P. Chellamani2 1Department of Mathematics, Holy Cross College (Autonomous), Nagercoil - 4, Tamilnadu, India. 2Department of Mathematics, St. Joseph’s College of Engineering, OMR, Chennai – 600 119, Tamil Nadu, India. βˆ—Corresponding author: anatjaslin@holycrossngl.edu.in Article History: Received: 23-09-2024 Revised: 04-11-2024 Accepted: 18-11-2024 Abstract: A set 𝑆 βŠ† 𝑉 is said to be relatively prime dominating set if it is a dominating set with at least two elements and for every pair of vertices 𝑒 and 𝑣 in 𝑆, (deg(𝑒) , deg(𝑣)) = 1. The minimum cardinality of a relatively prime dominating set is called relatively prime domination number and it is denoted by π›Ύπ‘Ÿπ‘π‘‘(𝐺). If there is no such pair exist, then π›Ύπ‘Ÿπ‘π‘‘(𝐺) = 0. For a finite undirected graph 𝐺(𝑉, 𝐸) and a subset  V, the switching of G by  is defined as the graph G (V, Eο‚’ ) which is obtained from G by removing all edges between  and its complement V- and adding as edges all non-edges between  and V- . This article delves into the discussion of the relatively prime domination number on quadrilateral snake graphs and their complements. The findings reveal that for quadrilateral snake graphs, the relatively prime domination number π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) equals either 2, 3 or 4. Similarly, for alternate quadrilateral snake graphs, the π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) is determined to be 2, 3 or 4. In the case of double quadrilateral snake graphs, the relatively prime domination number π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) is established as 2, 3, 4, 6 or 7, while for double alternate quadrilateral snake graphs, it is 2, 3, 4 or 5. Notably, the complements of quadrilateral, alternate quadrilateral, double quadrilateral, and double alternate quadrilateral snake graphs exhibit a relatively prime domination number of 2. Keywords: Dominating Set, Domination Number, Relatively Prime Dominating Set, Relatively Prime Dominating Number 1. Introduction By a graph G = (V, E) we mean a finite undirected graph without loops and multiple edges. The order and size of G are denoted by p and q respectively. For graph theoretical terms, we refer to Harary [2] and for terms related to domination we refer to Haynes [7]. A subset S of V is said to be a dominating set in G if every vertex in V– S is adjacent to at least one vertex in S. The domination number 𝛾(G) is the minimum cardinality of a dominating set in G. Berge [1] and Ore [6] formulated the concept of domination in graphs. It was further extended to define many other dominations related parameters in graphs. In 2017, C. Jayasekaran and A. Jancy Vini [3] have introduced the concept of relatively prime domination number in graph theory. Let G be a non – trivial graph. A set S V is said to be a relatively prime dominating set if it is a dominating set and for every pair of vertices u and v in S such that (d(u), d(v)) = 1. The minimum cardinality of a relatively prime dominating set is called the relatively prime domination number and it is denoted by rpdΞ³ (G) . Further they have introduced the concept of relatively prime dominating polynomial in [4]. Switching in graphs was introduced by Lint and Seidel [5]. For a finite undirected graph G(V, E) and a subset  V, the switching of G by  is defined as the graph Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3s (2025) 155 https://internationalpubls.com G (V, Eο‚’ ) which is obtained from G by removing all edges between  and its complement V– and adding as edges all non-edges between  and V– . For  = {v}, we write Gv instead of G{v} and the corresponding switching is called as vertex switching. In this paper we determine the relatively prime domination number v rpdΞ³ (G ) and rpdΞ³ (G) , where G is a quadrilateral snake graph. 2. Preliminaries Definition 2.1. A quadrilateral snake is obtained from a path π‘Ž1,π‘Ž2, …,π‘Žπ‘› by joining π‘Žπ‘– and π‘Žπ‘–+1 to new vertices 𝑏𝑖 and 𝑐𝑖 respectively and joining the vertices 𝑏𝑖 and 𝑐𝑖 for i = 1,2, …, n – 1. That is every edge is of a path is replaced by a cycle 𝐢4. Definition 2.2. An alternate quadrilateral snake is obtained from a path π‘Ž1,π‘Ž2, …,π‘Žπ‘›by joining π‘Žπ‘– and π‘Žπ‘–+1 to new vertices 𝑏𝑖 and 𝑐𝑖 respectively and joining the vertices 𝑏𝑖 and 𝑐𝑖 for 𝑖 ≑ 1(mod 2) and i ≀ n – 1 and then joining 𝑏𝑖 and 𝑐𝑖. That is every alternate edge of a path is replaced by a cycle 𝐢4. It is denoted by 𝐴(𝑄𝑛). Definition 2.3. A double quadrilateral snake is obtained from two quadrilateral snakes that have a common path. It is denoted by 𝐷(𝑄𝑛). Definition 2.4. An alternate double quadrilateral snake is obtained from two alternative quadrilateral snakes that have a common path. It is denoted by 𝐴(𝐷(𝑄𝑛)). 3. Relatively Prime Domination Number of Quadrilateral Snake Graph In this section we have discussed the relatively prime domination number for snake graphs. Theorem 3.1. Let G be a quadrilateral snake graph with p vertices, where p = 3n+1, n β‰₯ 2. Then π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2, 3 or 4. Proof: Let G be a quadrilateral snake graph with p vertices. Let the vertices in the path be 𝑣1, 𝑣2, …, 𝑣𝑝 and the vertices in the quadrilateral be 𝑒1, 𝑒2, 𝑀2, 𝑒3, 𝑀3, …, π‘’π‘šβˆ’1, π‘€π‘šβˆ’1, π‘’π‘š, π‘€π‘š. Then the degree of vertices in the path except the initial and the end vertex is 4; the degree of initial and the end vertex is 2; the degree of vertices in the quadrilateral is 2. Let v be a vertex in G. We have the following cases. Case 1: v is any vertex from {𝑒1, 𝑒2, 𝑀2, 𝑒3, 𝑀3, …, π‘’π‘šβˆ’1, π‘€π‘šβˆ’1, π‘’π‘š}. Without loss of generality, let v = 𝑒𝑖, i = 1, 2, …, m-1. Then d(𝑒𝑖) = p–3. Clearly, this vertex covers all the vertices except two vertices, say π‘€π‘–βˆ’1 and 𝑣𝑖. Then d(π‘€π‘–βˆ’1) = 1 and d(𝑣𝑖) = 1 if 𝑣𝑖 is an initial(end) vertex, otherwise d(𝑣𝑖) = 3. To cover the vertex π‘€π‘–βˆ’1 and 𝑣𝑖, either we have to take these two vertices or take a vertex which is adjacent to both π‘€π‘–βˆ’1 and 𝑣𝑖. Such a vertex always will exist, since it is a quadrilateral graph and |V| β‰₯ 6. Let the vertex be 𝑣𝑑 . Then d(𝑣𝑑) = 5 if it is an internal path vertex; otherwise d(𝑣𝑑) = 3. We have two more subcases. Case 1.1: 𝑣𝑖 is an initial(end) vertex. If d(v) is not a multiple of 5, then {𝑣, 𝑣𝑑} is a relatively prime dominating set. Hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2 in this case. If d(v) is multiple of 5, then the set {𝑣, π‘€π‘–βˆ’1, 𝑣𝑖} is our required relatively prime dominating set. Therefore, π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 3 in this case. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3s (2025) 156 https://internationalpubls.com Case 1.2: 𝑣𝑖 is not an initial(end) vertex. Then d(𝑣𝑑) = 3. Since |V| = 3n+1 and d(v) = p–3, the degree of v cannot be a multiple of 3 and so (p– 3, 3) = 1. Thus, the set {𝑣, 𝑣𝑑} is our required relatively prime dominating set. Therefore, π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2 in this case. Case 2: v is an initial or an end vertex of a path. Without loss of generality, let it be 𝑣1. Then d(v) = p-3. This vertex does not cover the two vertices 𝑒1 and 𝑣2. Then d(𝑣2) = 3 and d(𝑒1) = 1. To cover the vertices 𝑒1 and 𝑣2, two possibilities are there. Either we have to take these two vertices or a vertex which is adjacent to both 𝑒1 and 𝑣2. Such a vertex always exists, since G is a quadrilateral snake graph and |V| β‰₯ 6. Then the vertex must be 𝑒2 and d(𝑒2) = 3. Since p–3 is not a multiple of 3, we have the set {𝑣, 𝑒2} is our required relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2. Case 3: v is any internal path vertex. Without loss of generality, let it be 𝑣𝑖. Then d(𝑣𝑖) = p-5. This vertex does not cover four vertices; namely, π‘£π‘–βˆ’1, 𝑣𝑖+1, 𝑒𝑖 and 𝑀𝑖. Then d(𝑒𝑖) = d(𝑀𝑖) = 1 and d(π‘£π‘–βˆ’1) = 1 if it is an initial vertex and d(𝑣𝑖+1) = 3; similarly d(π‘£π‘–βˆ’1) = 3 and d(𝑣𝑖+1) = 1 if it is an end vertex. To cover these four vertices, either we have to take these four vertices or the vertices which are adjacent to these four vertices. Since G is a quadrilateral graph, such a vertex always exists as in Case 2. Let them be π‘€π‘–βˆ’1 and 𝑒𝑖+2 and degree of these two vertices is three and hence we cannot take these two vertices together. Suppose π‘£π‘–βˆ’1 is an initial vertex, then the set {𝑣𝑖 , π‘£π‘–βˆ’1, 𝑒𝑖, 𝑒𝑖+2} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 4. Similarly, if 𝑣𝑖+1 is an end vertex. Hence assume that neither π‘£π‘–βˆ’1 is an initial vertex nor 𝑣𝑖+1 is an end vertex. Since we cannot take the vertices π‘€π‘–βˆ’1 and 𝑒𝑖+2 together, we have to choose a vertex π‘£π‘–βˆ’1 or 𝑣𝑖+1. But both of them has degree 3. Therefore, relatively prime dominating set does not exist in this case. Theorem 3.2. Let G be an alternate quadrilateral snake graph with p vertices, where p = 4n, n β‰₯ 2. Then π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2, 3 or 4. Proof: Let G be an alternate quadrilateral snake graph with p vertices. Let the vertices in the path be 𝑣1,𝑣2,𝑣6,…,π‘£π‘š and the vertices in the quadrilateral be 𝑒1 , 𝑒2 , 𝑒3,…,π‘’π‘š. Then degree of each vertex in the path except the initial and end vertex is 3; degree of initial and end vertex is 2; degree of vertices in the quadrilateral is 2. Let v be any vertex in G. We have the following cases: Case 1: v is any vertex from {𝑒1 , 𝑒2 , …, π‘’π‘š}. Without loss of generality, we take v = 𝑒𝑖 , 𝑖 = 1,2, … , π‘š. Then d(v) = p-3. This vertex covers all the vertices of 𝐺𝑣, except the two vertices, namely 𝑣𝑖 and π‘’π‘–βˆ’1. Then d(𝑣𝑖) =1 if 𝑣𝑖 is initial or end vertex, otherwise 2. To cover the vertices 𝑣𝑖and π‘’π‘–βˆ’1,either we have to take these two vertices or choose a vertex which is adjacent to both 𝑣𝑖 and π‘’π‘–βˆ’1. Such a vertex always exists in alternate quadrilateral snake graph. Let the vertex be 𝑣𝑖+1 and d(𝑣𝑖+1) = 4. Since d(𝑣𝑖) = p-3 and |𝑉| = 4n , it cannot be multiple of 4 and hence (d(𝑣𝑖), d(𝑣𝑖+1)) = (p-3, 4) =1 and these two vertices covers all the vertices of 𝐺𝑣. Hence relatively prime dominating set is {𝑒𝑖,𝑣𝑖+1} and π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3s (2025) 157 https://internationalpubls.com Case 2: v is an initial vertex or an end vertex. Without loss of generality, let v = 𝑣1. Then d(v) = p-3. This vertex covers all the vertices except two vertices, namely 𝑒1 and 𝑣2 and d(𝑒1) =1 and d(𝑣2) = 2. To cover the vertices 𝑒1 and 𝑣2, either we have to choose these two vertices or a vertex which is adjacent to both 𝑒1 of 𝑣2. Such a vertex is always existing, since G is an alternate quadrilateral snake graph. Let the vertex be 𝑒2 and d(𝑒2) = 3. If d(v) is not a multiple of 3, the set {v,𝑒2} satisfies all the condition for being a relatively prime dominating set. Hence π›Ύπ‘Ÿπ‘π‘‘= 2 in this case. Suppose that d(v) is a multiple of 3. Since |𝑉| = 4n , p-3 is always odd. Hence {v, 𝑒1 ,𝑣2} is a relatively prime dominating set. Thus π›Ύπ‘Ÿπ‘π‘‘= 3 in this case. Case 3: v is any internal path vertex. Let it be 𝑣𝑖. Then d(𝑣𝑖) = p-4. This vertex covers all the vertices of 𝐺𝑣, except the 3 vertices, namely π‘£π‘–βˆ’1, 𝑣𝑖+1, 𝑒𝑖 and d(𝑒𝑖) = 1; d(π‘£π‘–βˆ’1) = 1 if π‘£π‘–βˆ’1 is an initial vertex, otherwise 2. Similarly d(𝑣𝑖+1) = 1 if 𝑣𝑖+1 is an initial vertex, otherwise 2. Since G is an alternate quadrilateral snake graph, let the vertex adjacent to 𝑒𝑖 and π‘£π‘–βˆ’1 be π‘’π‘–βˆ’1 and the vertex adjacent to 𝑣𝑖+1 be 𝑒𝑖+1 and 𝑣𝑖+2 and d(π‘’π‘–βˆ’1) = d(𝑒𝑖+1) = 3; d(𝑣𝑖+2) = 4. Since d(v) is a multiple of 4, we cannot take the vertex 𝑣𝑖+2. We have the following subcases. Case 3.1: π‘£π‘–βˆ’1 is an initial vertex and π‘£π‘–βˆ’1 is not an end vertex. If d(v) is not a multiple of 3, then the set {𝑣𝑖 , 𝑒𝑖 , π‘£π‘–βˆ’1¸𝑒𝑖+1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 4. If d(v) is a multiple of 3, then relatively prime dominating set does not exist. Case 3.2: 𝑣𝑖+1 is an end vertex and π‘£π‘–βˆ’1 is not an initial vertex. Same as Case 3.1. Case 3.3: Neither π‘£π‘–βˆ’1 is an initial vertex nor 𝑣𝑖+1 is an end vertex. Then d(π‘£π‘–βˆ’1) = d(𝑣𝑖+1) = 2. Therefore, we cannot choose these two vertices. Also note that the vertices which are adjacent to π‘£π‘–βˆ’1 and 𝑣𝑖+1 of degree 3 and 4. Therefore, relatively prime dominating set does not exist. Theorem 3.3. Let G be a double quadrilateral snake graph with p vertices, where p = 5n+1, n β‰₯ 2. Then π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2, 3, 4, 6 or 7. Proof: Let G be a double quadrilateral snake graph with p vertices. Let the vertices in the path be 𝑣1, 𝑣2, … , π‘£π‘š and the vertices in the upper quadrilateral be 𝑒1, 𝑒2, 𝑀2, 𝑒3, 𝑀3, … , π‘’π‘šβˆ’1, π‘€π‘šβˆ’1, π‘’π‘š and the vertices in the lower quadrilateral be π‘₯1, π‘₯2, 𝑦2, π‘₯3, 𝑦3, … , π‘₯π‘šβˆ’1, π‘¦π‘šβˆ’1, π‘₯π‘š. Then degree of each internal vertex is 6; degree of initial and end vertex is 3; degree of vertices in the upper and lower quadrilateral is 2. Let v be any vertex in G. We consider the following cases: Case 1: v is any vertex from {𝑒1, 𝑒2, 𝑀2, 𝑒3, 𝑀3, … , π‘’π‘šβˆ’1, π‘€π‘šβˆ’1, π‘’π‘š, π‘₯1, π‘₯2, 𝑦2, π‘₯3, 𝑦3, … , π‘₯π‘šβˆ’1, π‘¦π‘šβˆ’1, π‘₯π‘š}. Without loss of generality, let v = 𝑒𝑖 , 𝑖 = 1,2, … , π‘š. Then d(v) = p-3 in 𝐺𝑣. This vertex 𝑒𝑖 covers all the vertices in 𝐺𝑣 other than the two vertices which are adjacent to 𝑒𝑖 in G, namely 𝑣𝑖 , π‘€π‘–βˆ’1. Since G is a quadrilateral snake graph, the vertices 𝑣𝑖 and π‘€π‘–βˆ’1 are adjacent with a vertex π‘£π‘–βˆ’1. To cover the Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3s (2025) 158 https://internationalpubls.com vertices 𝑣𝑖 and π‘€π‘–βˆ’1, either we have to take these two vertices or which is adjacent to both 𝑣𝑖 and π‘€π‘–βˆ’1, that is, the vertex π‘£π‘–βˆ’1. Note that if n is even and odd, then d(v) is odd and even respectively. Here, d(π‘€π‘–βˆ’1) = 1; d(𝑣𝑖) = 2 if 𝑣𝑖 is a initial vertex or end vertex, otherwise 5. And d(π‘£π‘–βˆ’1) = 7 if π‘£π‘–βˆ’1 is an internal path vertex, otherwise 4. We have the following subcases. Case 1.1: 𝑣𝑖 is an initial or an end vertex. Then d(𝑣𝑖) = 2. If d(v) is odd and not a multiple of 7, then {𝑣, π‘£π‘–βˆ’1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2 in this case. If d(v) is odd and a multiple of 7, then the set {𝑣, 𝑣𝑖 , π‘€π‘–βˆ’1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 3 in this case. If d(v) is even and not multiple of 7, then the set {𝑣, π‘£π‘–βˆ’1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2. If d(v) is even and multiple of 7, then relatively prime dominating set does not exist in this case. Case 1.2: 𝑣𝑖 is neither an initial nor an end vertex. Then d(𝑣𝑖) = 5. Since d(v) = p-3 and |V| = 5n+1, degree of v cannot be a multiple of 5. Suppose that π‘£π‘–βˆ’1 is an initial or end vertex. Then d(π‘£π‘–βˆ’1) = 4. If d(v) is odd, then the set {𝑣, π‘£π‘–βˆ’1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2. If d(v) is even and not a multiple of 4, then the set {𝑣, π‘£π‘–βˆ’1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2. If d(v) is even and multiple of 4, then the set {𝑣, 𝑣𝑖 , π‘€π‘–βˆ’1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 3. Case 2: v is an initial vertex or an end vertex. Without loss of generality, let v = 𝑣1. Then d(v) = p-4 in 𝐺𝑣. Then the vertex 𝑣1 covers all the vertices except the three vertices, namely, 𝑒1, π‘₯1 and 𝑣2 and d(𝑒1) = d(π‘₯1) = 1; d(𝑣2) = 5. If n is even and odd, then d(v) = p-4 is even and odd respectively. To cover the vertices 𝑒1, π‘₯1 and 𝑣2, either we have to choose these vertices or a vertex which are adjacent to 𝑒1, π‘₯1 and 𝑣2. Note that there is no vertex which is adjacent to these three vertices, since G is a double quadrilateral snake graph. But 𝑒1, 𝑣2 and π‘₯1, 𝑣2 are connected by a vertex. They are 𝑒2 and π‘₯2 and d(𝑒2) = d(π‘₯2) = 3. Note that, since |V| = 5n+1 and d(v) = p-4, degree of v cannot be a multiple of 5. If d(v) is odd and not a multiple of 3, then the set {𝑣, 𝑒2, π‘₯1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 3. If d(v) is odd and a multiple of 3, then the set {𝑣, 𝑒1, π‘₯1, 𝑣2} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 4. If d(v) is even and not multiple of 3, then the set {𝑣, 𝑒2, π‘₯1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 3. If d(v) is even and multiple of 3, then the set {𝑣, 𝑒1, π‘₯1, 𝑣2} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 4. Case 3: v is any internal path vertex. Without loss of generality, let it be 𝑣𝑖 , 𝑖 = 2,3, … , π‘š βˆ’ 1. Then d(𝑣𝑖) = p-7 in 𝐺𝑣. Then the vertex 𝑣𝑖 covers all the vertices except the six vertices, namely 𝑒𝑖 , 𝑀𝑖, π‘₯𝑖 , 𝑦𝑖, π‘£π‘–βˆ’1, 𝑣𝑖+1. Then d(𝑒𝑖) = d(𝑀𝑖) = d(π‘₯𝑖) = d(𝑦𝑖) = 1 and d(π‘£π‘–βˆ’1) = 2 if it is an initial vertex, otherwise 5. Similarly for the vertex 𝑣𝑖+1. To cover these six vertices, either we have to take these six vertices or vertices which are adjacent to these six vertices. As in case 2, the vertices which are adjacent to 𝑒𝑖 and π‘£π‘–βˆ’1, 𝑀𝑖 and 𝑣𝑖+1, π‘₯𝑖 and π‘£π‘–βˆ’1, 𝑦𝑖 and 𝑣𝑖+1 are π‘€π‘–βˆ’1, 𝑒𝑖+1, π‘₯π‘–βˆ’1, 𝑦𝑖+1 respectively. Then d(π‘€π‘–βˆ’1) = d(𝑒𝑖+1) = d(π‘₯π‘–βˆ’1) = d(𝑦𝑖+1) = 3. To obtain a relatively prime dominating set, we can take only one of these four vertices. We have the following subcases. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3s (2025) 159 https://internationalpubls.com Case 3.1: π‘£π‘–βˆ’1 is an initial vertex. Then d(π‘£π‘–βˆ’1) = 2. Note that d(v) cannot be a multiple of 5. If d(v) is odd and not a multiple of 3, then the set {𝑣, 𝑒𝑖 , π‘₯𝑖, π‘£π‘–βˆ’1, 𝑒𝑖+1, 𝑦𝑖} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 6. If d(v) is odd and a multiple of 3, then the set {𝑣, 𝑒𝑖 , π‘₯𝑖 , 𝑀𝑖, 𝑦𝑖¸𝑣𝑖+1, π‘£π‘–βˆ’1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 7. Case 3.2: 𝑣𝑖+1 is an end vertex Same as Case 3.1. Case 3.3: π‘£π‘–βˆ’1 is an initial vertex and 𝑣𝑖+1 is an end vertex. Then |V| must be 11 and hence d(v) =4. Hence the set {𝑣, 𝑒𝑖 , π‘₯𝑖 , π‘£π‘–βˆ’1, 𝑒𝑖+1, 𝑦𝑖} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 6. Case 3.4: Neither π‘£π‘–βˆ’1 is an initial vertex nor 𝑣𝑖+1 is an end vertex. Then d(π‘£π‘–βˆ’1) = d(𝑣𝑖+1) = 5. Since degree of these two vertices are 5, we cannot take these vertices together. So we consider the vertices which are adjacent to 𝑣𝑖+1, namely 𝑒𝑖+1, π‘₯𝑖+1, 𝑀𝑖+1, 𝑦𝑖+1, 𝑣𝑖+2. Then d(𝑒𝑖+1) = d(π‘₯𝑖+1) = d(𝑀𝑖+1) = d(𝑦𝑖+1) = 3 and d(𝑣𝑖+2) = 4 if 𝑣𝑖+2 is an end vertex, otherwise 7. If d(v) is odd and not a multiple of 3, then the set {𝑣, 𝑒𝑖 , π‘₯𝑖 , π‘£π‘–βˆ’1, 𝑒𝑖+1, 𝑦𝑖} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 6. Suppose that d(v) is odd and a multiple 3. Here we cannot choose a vertex of degree 3. The only possibility is choose the vertex 𝑣𝑖+2. If 𝑣𝑖+2 is an end vertex, then the set {𝑣, 𝑒𝑖 , π‘₯𝑖 , π‘£π‘–βˆ’1, 𝑀𝑖, 𝑦𝑖 , 𝑣𝑖+2} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 7. If 𝑣𝑖+2 is not an end vertex and d(v) is not a multiple of 7, then set {𝑣, 𝑒𝑖 , π‘₯𝑖 , π‘£π‘–βˆ’1, 𝑀𝑖, 𝑦𝑖, 𝑣𝑖+2} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 7. If d(v) is odd and a multiple of 3 and 7, then relatively prime dominating set does not exist. If d(v) is even and not a multiple of 3, then the set {𝑣, 𝑒𝑖 , π‘₯𝑖 , π‘£π‘–βˆ’1, 𝑀𝑖, 𝑒𝑖+1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 6. Suppose that d(v) is even and a multiple of 3. As said above, we cannot take the vertices of degree 3. We can cover the vertices 𝑒𝑖 , 𝑀𝑖, π‘£π‘–βˆ’1, π‘₯𝑖 , 𝑦𝑖. We have only one vertex to cover is 𝑣𝑖+1. If 𝑣𝑖+2 is an end vertex, then the set {𝑣, 𝑒𝑖 , π‘₯𝑖, π‘£π‘–βˆ’1, 𝑀𝑖, 𝑦𝑖 , 𝑣𝑖+2} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 7. If 𝑣𝑖+2 is not an end vertex and d(v) is not a multiple of 7, then set {𝑣, 𝑒𝑖 , π‘₯𝑖 , π‘£π‘–βˆ’1, 𝑀𝑖, 𝑦𝑖, 𝑣𝑖+2} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 7. If d(v) is odd and a multiple of 3 and 7, then relatively prime dominating set does not exist. Theorem 3.4. Let G be a double alternate quadrilateral snake graph with p vertices, where p = 6n, n β‰₯ 2. Then π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2, 3, 4 or 5. Proof: Let G be a double alternate quadrilateral snake graph with p vertices. Let the vertices in the path be 𝑣1,𝑣2, …,π‘£π‘š, where 𝑣1 and π‘£π‘š denote the initial and end vertex respectively. Let the vertices in the upper quadrilateral be 𝑒1,𝑒2,…, π‘’π‘š and the vertices in the lower quadrilateral be 𝑀1,𝑀2,…,π‘€π‘š. Then degree of each internal path vertex is 4; degree of initial and end vertex is 3; degree of vertices in the quadrilateral is 2. Let v be any vertex in G. We have the following cases. Case 1 : v is any vertex from {𝑒1,𝑒2,…, π‘’π‘šβˆ’1, 𝑀1,𝑀2,…,π‘€π‘šβˆ’1}. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3s (2025) 160 https://internationalpubls.com Without loss of generality, let v = 𝑒𝑖 , 𝑖 = 1,2, … , π‘š. Then d(v) = p–3. This vertex covers all the vertices in 𝐺𝑣 except the two vertices, namely 𝑣𝑖 and 𝑒𝑖+1 or 𝑣𝑖 and π‘’π‘–βˆ’π‘–. Without loss of generality, let us take 𝑣𝑖 and 𝑒𝑖+1. To find the relatively prime dominating set, we have to cover these two vertices. Either we have to choose these two vertices or a vertex which is adjacent to both the vertices 𝑣𝑖 and 𝑣𝑖+1. Such a vertex always exists, since G is a double alternative quadrilateral snake graph and let the vertex be 𝑣𝑖+1. Then d(𝑣𝑖+1) = 4, if 𝑣𝑖+1 is an end vertex, otherwise 4. Since |V| = 6n, d(v) is always odd and it is a multiple of 3. We consider the following subcases. Case 1.1: 𝑣𝑖 is an initial vertex. Then the set {𝑣, 𝑣𝑖 , 𝑒𝑖+1} is a relatively prime dominating set, since (p-3, 4) = 1. Therefore, π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 3. Case 1.2: 𝑣𝑖 is not an initial vertex. Consider the vertex 𝑣𝑖+1. If 𝑣𝑖+1 is an end vertex, then the set {𝑣, 𝑣𝑖+1} is a relatively prime dominating set and hence . Otherwise, we have degree of 𝑣𝑖+1 is 5. If d(v) is not a multiple of 5, then the set {𝑣, 𝑣𝑖+1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 2. If d(v) is not multiple of 5, then relatively prime dominating set does not exist. Case 2: v is an initial vertex or an end vertex. Without loss of generality, let v = 𝑣1. Then d(𝑣1) = p–4. This vertex covers all the vertices of 𝐺𝑣 except three vertices, namely 𝑒1, 𝑀1, and 𝑣2. Note that there is no vertex which covers all these three vertices. But the vertices 𝑒1 and 𝑣2, 𝑀1 and 𝑣2 are connected by a vertex, namely 𝑒2 and 𝑀2. Then d(𝑒2) = d(𝑀2) = 3. Since d(𝑣1) = p-4, it cannot be multiple of 3. Hence the set {𝑣1, 𝑒2, 𝑀1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 3. Case 3: v is anyone of internal path vertex. Without loss of generality, let v = 𝑣𝑖 , 𝑖 = 2,3, … , π‘š βˆ’ 1. Then d(v) = p–5. This vertex covers all the vertices of 𝐺𝑣 except four vertices, namely, 𝑒𝑖 , 𝑀𝑖, π‘£π‘–βˆ’1 and 𝑣𝑖+1. Then d(𝑒𝑖) = d(𝑀𝑖) = 1, d(π‘£π‘–βˆ’1) = 2 if π‘£π‘–βˆ’1 is an initial vertex, otherwise 3. Similarly, d(𝑣𝑖+1) = 2 if 𝑣𝑖+1 is an end vertex, otherwise 3. We consider the following subcases. Case 3.1: π‘£π‘–βˆ’1 is an initial vertex. Note that d(v) = p-5 is always odd and not multiple of 3. As said in case 2, the vertex which is adjacent to 𝑒𝑖 and π‘£π‘–βˆ’1 is π‘’π‘–βˆ’1, is of degree 3 and the vertex adjacent to the vertex 𝑣𝑖+1 in the paths is 𝑣𝑖+2, is of degree 5, if 𝑣𝑖+2 is not an end vertex, otherwise 4. If 𝑣𝑖+2 is an end vertex, then the set {𝑣𝑖 , 𝑣𝑖+2, π‘’π‘–βˆ’1, 𝑀𝑖} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 4. If 𝑣𝑖+2 is not an end vertex and d(v) is not a multiple of 5, then the set {𝑣𝑖 , 𝑣𝑖+2, π‘’π‘–βˆ’1, 𝑀𝑖} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 4. If 𝑣𝑖+2 is not an end vertex and d(v) is a multiple of 5, then the set {𝑣𝑖 , 𝑒𝑖 , 𝑀𝑖, π‘£π‘–βˆ’1, 𝑣𝑖+1} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 5. Case 3.2: Same as Case 3.1. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3s (2025) 161 https://internationalpubls.com Case 3.3: Neither π‘£π‘–βˆ’1 is an initial vertex nor 𝑣𝑖+1 is an end vertex. Then d(π‘£π‘–βˆ’1) = d(𝑣𝑖+1) = 3. So, we cannot take these two vertices together. Consider the vertices adjacent to π‘£π‘–βˆ’1 and 𝑣𝑖+1 in the path. Let them be π‘£π‘–βˆ’2 and 𝑣𝑖+2. Then d(π‘£π‘–βˆ’2) = 4 if it is an initial vertex, otherwise 5. Then the set {𝑣𝑖 , π‘£π‘–βˆ’2, 𝑒𝑖+1, 𝑀𝑖} is a relatively prime dominating set if π‘£π‘–βˆ’2 is an initial vertex and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 4. If π‘£π‘–βˆ’2 is not an initial vertex, 𝑣𝑖+2 is not an end vertex and d(v) is not a multiple of 5, then the set {𝑣𝑖 , π‘£π‘–βˆ’2, 𝑒𝑖+1, 𝑀𝑖} is a relatively prime dominating set and hence π›Ύπ‘Ÿπ‘π‘‘(𝐺𝑣) = 4. Otherwise, relatively prime dominating set does not exist. 4. Relatively Prime Domination Number on Complement of Quadrilateral Snake Graph In this section we have shown that the relatively prime domination number for complement of quadrilateral type graphs is 2. Theorem 4.1. Let G be a quadrilateral snake graph with p vertices. Then for p is even, π›Ύπ‘Ÿπ‘π‘‘(οΏ½Μ…οΏ½) = 2. Proof: Let G be a quadrilateral snake graph with p vertices. Let the vertices be 𝑣1,𝑣2, … 𝑣𝑝. Since degree of each vertex in the quadrilateral snake graph G is either 2 or 4, degree of each vertex in the complement of quadrilateral graph οΏ½Μ…οΏ½ is either p–3 or p–5. Note that if either n is even or odd, then p– 3 and p–5 are always odd and hence (p–3, p–5) = 1. Consider a vertex which has degree p–3. Then this vertex, say 𝑣𝑖 covers all the vertices of οΏ½Μ…οΏ½ except the vertices, say π‘£π‘˜ and 𝑣𝑙. In order to find a relatively prime dominating set, choose a vertex of degree p–4, say 𝑣𝑗 which has adjacency with the vertices π‘£π‘˜ and 𝑣𝑙. Note that such a vertex is always exist, since |V| β‰₯ 6. Since these two vertices 𝑣𝑖 and 𝑣𝑗 satisfies the conditions for being a relatively prime dominating set, we have π›Ύπ‘Ÿπ‘π‘‘(οΏ½Μ…οΏ½) = 2. Theorem 4.2. For any alternate quadrilateral snake graph G, π›Ύπ‘Ÿπ‘π‘‘(οΏ½Μ…οΏ½) = 2 Proof: Let G be alternate quadrilateral snake graph with p vertices .Let the vertices in the path be 𝑣1,𝑣2,… ,𝑣𝑝. Since degree of each vertex in an alternate quadrilateral snake graph is either 2 or 3, degree of each vertex in the complement of alternate quadrilateral snake graph is either p-3 or p-4. Choose a vertex of degree p-3, say 𝑣𝑖. This vertex cover all the vertices of οΏ½Μ…οΏ½ except two vertices, namely π‘£π‘˜ and 𝑣𝑙. Now, choose a vertex of degree p-4, say 𝑣𝑗 such that it has adjacent with the vertices π‘£π‘˜ and 𝑣𝑙. Such a vertex always exists, since|𝑉| β‰₯ 6. Since these two vertices 𝑣𝑖 and 𝑣𝑗 satisfies the conditions for being a relatively prime dominating set, we have π›Ύπ‘Ÿπ‘π‘‘(οΏ½Μ…οΏ½) = 2. Theorem 4.3. For any double quadrilateral snake graph G, π›Ύπ‘Ÿπ‘π‘‘(οΏ½Μ…οΏ½) = 2. Proof: Let G be a double quadrilateral snake graph with p vertices. Let the vertices be 𝑣1,𝑣2, … 𝑣𝑝. We know that, in the double quadrilateral snake graph, degree of each vertex is either 2,3 or 6. Hence in the complement graph οΏ½Μ…οΏ½, degree of each vertex is either p–3, p–4 or p–7. Choose a vertex of degree p–3, say 𝑣𝑖. Since this vertex cover all the vertices of οΏ½Μ…οΏ½ except two vertices say, π‘£π‘˜ and 𝑣𝑙, we have to choose a vertex of degree n – 4 such that it has adjacency with the two vertices π‘£π‘˜ and 𝑣𝑙. Such a vertex always exists, since |V| β‰₯ 6. Let the vertex which has degree p–4 be 𝑣𝑗 . Now, clearly the two vertices 𝑣𝑖 and 𝑣𝑗 cover all the vertices of οΏ½Μ…οΏ½ and (d(𝑣𝑖),d(𝑣𝑗)) = (p–3, p–4) = 1. Therefore, relatively prime dominating set is {𝑣𝑖, 𝑣𝑗} and hence π›Ύπ‘Ÿπ‘π‘‘(οΏ½Μ…οΏ½) = 2. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 32 No. 3s (2025) 162 https://internationalpubls.com Theorem 4.4. For any double alternate quadrilateral snake graph G, π›Ύπ‘Ÿπ‘π‘‘(οΏ½Μ…οΏ½) = 2. Proof: Let G be a double alternate quadrilateral snake graph with p vertices. Let the vertices in G be 𝑣1, 𝑣2, … , 𝑣𝑝. Then degree of each vertex in the complement of double alternative quadrilateral snake graph οΏ½Μ…οΏ½ is either p-3 , p-4 or p-5, since degree of each vertex in a double alternate quadrilateral snake graph is 2, 3 or 4. Consider a vertex which has degree p-3, say 𝑣𝑖. This vertex covers all the vertices of 𝐺 Μ…except two vertices, say π‘£π‘˜ and 𝑣𝑙. Now, choose a vertex degree p-4 such that it has adjacency with the vertices π‘£π‘˜ and 𝑣𝑙 . Such a vertex is always possible, since |𝑉| > 6. Let the vertex which has degree p-4 be 𝑣𝑗 . Hence the relatively prime dominating set is {𝑣𝑖 , 𝑣𝑗} and the relatively prime domination number is 2. 5. Conclusion Dominations in graph theory is a wide area with more applications to real life which helps the researchers to get more ideas to manage the problems in real life situation. 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