The reserved dominating set is special up gradation of domination set; where in some of the vertices in the vertex set have special privilege (reserved) to appear in the Dominating set irrespective of their adjacency due to the necessity of the user. The minimum cardinality of a reserved dominating set of G is called the reserved domination number of G and is denoted by R(k) -Y(G) where k is the number of reserved vertices. In this paper reserved domination number of (LPn), (LCn), L(Sn), L(Bm,n), L(Wn) and L(F l,n ) have been found

G. Rajasekar

Rajasekar G

English

Christ University Bengaluru: Open Journal Systems

Mapana – Journal of Sciences 2023, Vol. 22, Special Issue 1, 51-62 ISSN 0975-3303|https://doi.org/10.12723/mjs.sp1.5 51   Reserved Domination Number of Line Graph Rajasekar G.*, G. Rajasekar†  Abstract The reserved dominating set is special up gradation of dominating set, such that some of the vertices in the vertex set have the special privilege (reserved) to appear in the Dominating set irrespective of their adjacency due to the necessity of the user. The minimum --cardinality of a reserved dominating set of 𝐺 is called the reserved domination number of 𝐺 and is denoted by 𝑅(𝑘) −𝛾(𝐺)where 𝑘  is the number of reserved vertices. In this paper reserved domination number of𝐿(𝑃𝑛), 𝐿(𝐶𝑛), 𝐿(𝑆𝑛), 𝐿(𝐵𝑚,𝑛), 𝐿(𝑊𝑛) and 𝐿(𝐹1,𝑛) are found  Keywords: Dominating set, reserved dominating set, reserved domination number, line graph. 1. Introduction  The Oystein Ore [1] defined that the dominating set of a graph. Rajasekar et al., [3,6] defined the reserved dominating set (𝑅𝐷𝑆) of the graph 𝐺 to be the subset 𝑆 of 𝑉, whose vertices are reserved in such a way that they must appear in the dominating set. The dominating 𝑅𝐷𝑆 with the minimum cardinality is called reserved domination number of 𝐺  and is denoted by 𝑅(𝑘) − 𝛾(𝐺)where 𝑘 is the number of reserved vertices. In [4] authors found the location domination number of line graph. Rajasekar et al. [3,5,6,7] have found the reserved domination number, 2-reserved domination  *  Department of Mathematics, C. Kandaswami Naidu College for Men,    Chennai, Tamil Nadu, India; r.g.raja3007@gmail.com † Department of Mathematics, Jawahar Science College, Neyveli, Tamil      Nadu, India;  grsmaths@gmail.com Mapana - Journal of Sciences, Vol. 22, Special Issue 1  ISSN 0975-3303 52  number of graphs and reserved domination number of complement of a graphs. Throughout this paper we use the indexing set [initial value; final value: step value] where initial value is the first value of indexing set, step value is the incremented value of initial value and the final value is the maximum value that can be achieved by initial value by incrementing. Therefore  1; :1k n  implies 1, 2,3,...,k n=  and  1; :1nv v  implies  1 2 3, , ,..., .nv v v v  Further throughout the paper Reserved vertex is referred to as𝑅𝑉, Dominating Set as DS, Reserved Dominating Set as 𝑅𝐷𝑆 and Reserved Domination Number as𝑅𝐷𝑁. 2. Preliminaries Definition 2.1: [3] Reserved Domination. Let 𝐺 = (𝑉, 𝐸) be a graph. A subset 𝑆 of 𝑉 is called a Reserved Dominating Set (𝑅𝐷𝑆)of 𝐺 if (i) 𝜇 be any nonempty proper subset of 𝑆. (ii) Every vertex in 𝑉 − 𝑆is adjacent to a vertex in𝑆. The dominating set 𝑆is called a minimal reserved dominating set if no proper subset of 𝑆containing 𝜇is a dominating set. The set 𝜇 is called Reserved set. The minimum cardinality of a reserved dominating set 𝑆  of 𝐺 is called the reserved domination number of𝐺and is denoted by 𝑅 − 𝛾(𝐺). Definition 2.2: [3,6] 2-Reserved Domination. Let 𝐺 = (𝑉, 𝐸)be a graph. A subset 𝑆of 𝑉 is called a 𝑘  -reserved dominating set (𝑅𝐷𝑆)of 𝐺 if (i) 𝜇 is any nonempty proper subset of 𝑆with 𝑘 vertices. (ii) Every vertex in 𝑉 − 𝑆is adjacent to a vertex in𝑆. The dominating set 𝑆is called a minimal 𝑘 -reserved dominating set if no proper subset of 𝑆 containing 𝜇 is a dominating set. The set 𝜇 is called 𝑘 -reserved set. The minimum cardinality of a 𝑘 -reserved dominating set 𝑆 of 𝐺is called the 𝑘  -reserved domination number of𝐺and is denoted by 𝑅(𝑘) − 𝛾(𝐺) where 𝑘 is the number of reserved vertices. Rajasekar & Rajasekar                                                                         RDN of Line Graph 53  Definition 2.3: Bistar Graph. A Bistar graph is the graph obtained by joining the centre (apex) vertices of two copies of 𝐾1,𝑛 by an edge and it is denoted by 𝐵𝑚,𝑛. Theorem 2.4: [3] For 𝑃𝑛,  the𝑅𝐷𝑁, 𝑅(1) − 𝛾(𝑃𝑛, 𝜇) = 1 + ⌈𝑘−23⌉ + ⌈𝑛−(𝑘+1)3⌉ , if𝜇 = 𝑣𝑘(𝑘 ∈ [1; 𝑛: 1]). Theorem 2.5: [3] For 𝐶𝑛, the 𝑅𝐷𝑁, 𝑅(1) − 𝛾(𝐶𝑛, 𝜇) = 𝛾(𝐶𝑛) = ⌈𝑛3⌉ if𝜇 =𝑣𝑘(𝑘 ∈ [1; 𝑛: 1]). Remark 2.6: [3] For 𝐾𝑛, 𝑛 ≥ 3 the 𝑅𝐷𝑁, 𝑅(1) − 𝛾(𝐾𝑛) = 1. 3. Reserved Domination Number of Line Graph Proposition 3.1: For 𝑃𝑛,  𝑅(1) − 𝛾(𝐿(𝑃𝑛)) = 𝑅(1) − 𝛾(𝑃𝑛−1), asf𝐿(𝑃𝑛) =𝑃𝑛−1. Proposition 3.2: For 𝐶𝑛 , 𝐿(𝐶𝑛) = 𝐶𝑛 and hence 𝑅(1) − 𝛾(𝐿(𝐶𝑛)) =𝑅(1) − 𝛾(𝐶𝑛). Theorem 3.3: For 𝑆𝑛 = 𝐾1,𝑛, 𝑅(1) − 𝛾(𝐿(𝑆𝑛), 𝜇) = 𝑅(1) − 𝛾(𝐾𝑛, 𝜇) = 1, if 𝜇 = 𝑒𝑘(𝑘 ∈ [1; 𝑛: 1]).  Proof: 𝑆1 = 𝐾2 = 𝑃2  and so by Proposition 3.1, 𝑅(1) − 𝛾(𝐿(𝑆1)) =𝑅(1) − 𝛾(𝐿(𝑃2)) = 𝑅(1) − 𝛾(𝑃1) = 1. For 𝑛 > 1 ,f𝐿(𝑆𝑛) ≅ 𝐾𝑛  and so 𝑅(1) − 𝛾(𝐿(𝑆𝑛), 𝜇) = 𝑅(1) − 𝛾(𝐾𝑛, 𝜇) =1. Theorem 3.4: For 𝐵𝑚,𝑛, the 𝑅𝐷𝑁, 𝑅(1) − 𝛾(𝐿(𝐵𝑚,𝑛), 𝜇) = {1, if 𝜇 = 𝑒2, if 𝜇 = {𝑒𝑢𝑘 , 𝑘 ∈ [1;𝑚: 1]  or𝑒𝑣𝑘 , 𝑘 ∈ [1; 𝑛: 1].  Proof:  Case (i): When 𝑚 = 1 = 1, 𝐵1,1 ≅ 𝑃4  and by Proposition 3.1, 𝑅(1) −𝛾(𝐿(𝐵1,1), 𝜇) = {1, if 𝜇 = 𝑒2, if 𝜇 = 𝑒𝑢1 or 𝑒𝑣1 . Case (ii): Either 𝑚 = 1for 𝑛 = 1. Without loss of generality, assume that 𝑚 > 1 and 𝑛 = 1. 𝐿(𝐵𝑚,1)is isomorphic to𝐿𝑚+1,1. Mapana - Journal of Sciences, Vol. 22, Special Issue 1  ISSN 0975-3303 54  Suppose 𝜇 = 𝑒 is 𝑅𝑉. Then 𝑒 must be in the DS and 𝑒 dominates all other vertices. Hence the required 𝑅𝐷𝑆 is {𝑒}. Thus 𝑅(1) − 𝛾(𝐿(𝐵𝑚,1), 𝜇) = 1 where 𝜇 = 𝑒.   Fig. 1: (a) ,1mB  and (b) ( ),1mL B  Suppose 𝜇 = 𝑒𝑢𝑘(𝑘 ∈ [1;𝑚: 1]) is the 𝑅𝑉. Then 𝑒𝑢𝑘 must be in the DS and 𝑒𝑢𝑘  dominates the vertices {𝑒𝑢1 , 𝑒𝑢2 , . . . , 𝑒𝑢𝑘−1 , 𝑒𝑢𝑘+1 , . . . , 𝑒𝑢𝑚} ∪{𝑒}. The remaining vertex which is not dominated by 𝑒𝑢𝑘 is 𝑒𝑣1. So 𝑒𝑣1 must be in the DS. Hence the required 𝑅𝐷𝑆 is {𝑒𝑢𝑘 , 𝑒𝑣1}. Thus 𝑅(1) − 𝛾(𝐿(𝐵𝑚,1), 𝜇) = 2 where 𝜇 = 𝑒𝑢𝑘(𝑘 ∈ [1;𝑚: 1]). Suppose 𝜇 = 𝑒𝑣1  is the 𝑅𝑉 . Then 𝑒𝑣1  must be in the DS and 𝑒𝑣1 dominates only the vertex 𝑒. The remaining vertices which aren’t dominated by 𝑒𝑢𝑘  are {𝑒𝑢1; 𝑒𝑢𝑚: 1} . To dominate the remaining vertices, choose any one of the vertex say 𝑒𝑢1 from {𝑒𝑢1; 𝑒𝑢𝑚: 1}.  Hence the required𝑅𝐷𝑆 is {𝑒𝑣1 , 𝑒𝑢1}. Thus 𝑅(1) − 𝛾(𝐿(𝐵𝑚,1), 𝜇) = 2 where 𝜇 = 𝑒𝑣1. Case (iii): Line graph of 𝐵𝑚,𝑛  when 𝑚, 𝑛 > 1  is isomorphic to the graph obtained due to single vertex fusion of𝐾𝑚+1 and𝐾𝑛+1. Rajasekar & Rajasekar                                                                         RDN of Line Graph 55   (a)  (b) Fig. 2: (a) 𝐵𝑚,𝑛 and (b) 𝐿(𝐵𝑚,𝑛) Suppose 𝜇 = 𝑒 is 𝑅𝑉. Then 𝑒 must be in the DS and 𝑒 dominates all other vertices. Hence the required 𝑅𝐷𝑆 set is {𝑒}. Thus 𝑅(1) − 𝛾(𝐿(𝐵𝑚,𝑛), 𝜇) = 1 where 𝜇 = 𝑒. Suppose 𝜇 = 𝑒𝑢𝑘(𝑘 ∈ [1;𝑚: 1]) is the 𝑅𝑉. Then 𝑒𝑢𝑘 must be in the DS and 𝑒𝑢𝑘  dominates the vertices {𝑒𝑢1 , 𝑒𝑢2 , . . . , 𝑒𝑢𝑘−1 , 𝑒𝑢𝑘+1 , . . . , 𝑒𝑢𝑚} ∪{𝑒} . The remaining vertices which are not dominated by 𝑒𝑢𝑘  are {𝑒𝑣1; 𝑒𝑣𝑛: 1}.  To dominate the remaining vertices, choose any one of the vertices say 𝑒𝑣1 from{𝑒𝑣1; 𝑒𝑣𝑛: 1}.  Hence the required 𝑅𝐷𝑆 is {𝑒𝑢𝑘 , 𝑒𝑣1}. Thus 𝑅(1) − 𝛾(𝐿(𝐵𝑚,𝑛), 𝜇) = 2 where 𝜇 = 𝑒𝑢𝑘(𝑘 ∈ [1;𝑚: 1]). Similarly, one can prove for the 𝑅𝑉𝜇 = 𝑒𝑣𝑘 where 𝑘 ∈ [1; 𝑛: 1]. Mapana - Journal of Sciences, Vol. 22, Special Issue 1  ISSN 0975-3303 56  Theorem 3.5: For 𝑊𝑛, 𝑅(1) − 𝛾(𝐿(𝑊𝑛), 𝜇) = ⌈𝑛+13⌉ if 𝜇 = 𝑒𝑘 or 𝑒𝑣𝑘  (𝑘 ∈[1; 𝑛: 1]). Proof: Let 𝑉(𝑊𝑛) = {𝑣, {𝑣1; 𝑣𝑛: 1}}  where 𝑑𝑒𝑔 𝑣 = 𝑛  and 𝑑𝑒𝑔 𝑣𝑘 = 3 for all 𝑘 ∈ [1; 𝑛: 1]. Label the edge 𝑣𝑣𝑘  as 𝑒𝑣𝑘 , edge 𝑣𝑘𝑣𝑘+1 as 𝑒𝑘 for 𝑘 ∈ [1; 𝑛 − 1: 1] and 𝑣1𝑣𝑛 as 𝑒𝑛 as represented in the Fig. 3.  Figure 3: 𝑊𝑛 𝐿(𝑊𝑛) is constructed as shown in Fig. 4.   Figure 4: 𝐿(𝑊𝑛) Rajasekar & Rajasekar                                                                         RDN of Line Graph 57  The induced subgraph of the sets {𝑒𝑣1; 𝑒𝑣𝑛: 1}  and {𝑒1; 𝑒𝑛: 1}  is 𝐾𝑛 and 𝐶𝑛 respectively. Case (i): Suppose 𝜇 = 𝑒𝑣𝑘(𝑘 ∈ [1; 𝑛: 1]) is 𝑅𝑉. Then 𝑒𝑣𝑘 must be in the DS and 𝑒𝑣𝑘  dominates the vertices{𝑒𝑣1 , 𝑒𝑣2 , . . . , 𝑒𝑣𝑘−1 , 𝑒𝑣𝑘+1 , . . . , 𝑒𝑣𝑛} ∪{𝑒𝑘−1, 𝑒𝑘}. The remaining vertices which are not dominated by 𝑒𝑣𝑘 are {𝑒1, 𝑒2, . . . , 𝑒𝑘−2, 𝑒𝑘+1, . . . , 𝑒𝑛}. Now it is enough to find the DS for the vertices {𝑒1, 𝑒2, . . . , 𝑒𝑘−2, 𝑒𝑘+1, . . . , 𝑒𝑛} . The 𝐿(𝑊𝑛)[𝑉1]  with 𝑉1 ={𝑒1, 𝑒2, . . . , 𝑒𝑘−2, 𝑒𝑘+1, . . . , 𝑒𝑛} is 𝑃𝑛−2. Hence 𝑅(1) − 𝛾(𝐿(𝑊𝑛), 𝜇) = |{𝑒𝑣𝑘}| + 𝛾(𝑃𝑛−2)    = 1 + ⌈𝑛−23⌉ = ⌈𝑛+13⌉ where𝜇 = 𝑒𝑣𝑘(𝑘 ∈ [1; 𝑛: 1]). Case (ii): Suppose 𝜇 = 𝑒𝑘(𝑘 ∈ [1; 𝑛: 1]) is 𝑅𝑉. Then 𝑒𝑘 must be in the DS and 𝑒𝑘  dominates the vertices {𝑒𝑘−1, 𝑒𝑘+1} ∪ {𝑒𝑣𝑘 , 𝑒𝑣𝑘+1} . The remaining vertices which are not dominated by 𝑒𝑘  are {𝑒1, 𝑒2, . . . , 𝑒𝑘−2, 𝑒𝑘+2, . . . , 𝑒𝑛} ∪ {𝑒𝑣1 , 𝑒𝑣2 , . . . , 𝑒𝑣𝑘−1 , 𝑒𝑣𝑘+2 , . . . , 𝑒𝑣𝑛}. To dominate the remaining vertices from the set {𝑒𝑣1 , 𝑒𝑣2 , . . . , 𝑒𝑣𝑘−1 , 𝑒𝑣𝑘+2 , . . . , 𝑒𝑣𝑛} , choose the vertex 𝑒𝑣𝑘+3  or 𝑒𝑣𝑘−3 . Consider the vertex 𝑒𝑣𝑘+3 which dominates 𝑒𝑘+2 and 𝑒𝑘+3. Now it is enough to find the DS for the vertices {𝑒1, 𝑒2, . . . , 𝑒𝑘−2, 𝑒𝑘+4, . . . , 𝑒𝑛} . The 𝐿(𝑊𝑛)[𝑉2]  with 𝑉2 ={𝑒1, 𝑒2, . . . , 𝑒𝑘−2, 𝑒𝑘+4, . . . , 𝑒𝑛} is 𝑃𝑛−5. Hence 𝑅(1) − 𝛾(𝐿(𝑊𝑛), 𝜇) = |{𝑒𝑘}| + |{𝑒𝑣𝑘+3}| + 𝛾(𝑃𝑛−5) = 1 + 1 +⌈𝑛−53⌉ = ⌈𝑛+13⌉ where 𝜇 = 𝑒𝑘(𝑘 ∈ [1; 𝑛: 1]). Theorem 3.6: For the fan graph 𝐹1,𝑛, the reserved domination number for the different values of 𝑛 is summarized as follows: i. For 𝑛 ≡ 0(𝑚𝑜𝑑 3)  𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝜇)={    ⌈𝑛 + 13⌉ , if 𝜇 = {𝑒𝑣𝑘(𝑘 = 1,3,4,6, . . . , 𝑛 − 5, 𝑛 − 3, 𝑛 − 2, 𝑛) or𝑒𝑘(𝑘 = 1,3,6,9, . . . , 𝑛 − 9, 𝑛 − 6, 𝑛 − 3, 𝑛 − 1)⌈𝑛3⌉ , if 𝜇 = {𝑒𝑣𝑘(𝑘 = 2,5,8,11, . . . , 𝑛 − 10, 𝑛 − 7, 𝑛 − 4, 𝑛 − 1) or𝑒𝑘(𝑘 = 2,4,5,7, . . . , 𝑛 − 7, 𝑛 − 5, 𝑛 − 4, 𝑛 − 2) ii. For 𝑛 ≡ 1(𝑚𝑜𝑑 3) Mapana - Journal of Sciences, Vol. 22, Special Issue 1  ISSN 0975-3303 58  𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝜇) = ⌈𝑛3⌉ , if 𝜇 = {𝑒𝑣𝑘(𝑘 ∈ [1; 𝑛: 1]) or𝑒𝑘(𝑘 ∈ [1; 𝑛 − 1: 1])  iii. For 𝑛 ≡ 2(𝑚𝑜𝑑 3) 𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝜇) ={⌈𝑛+23⌉ , if 𝜇 = 𝑒𝑣𝑘(𝑘 = 3,6,9,12, . . . , 𝑛 − 11, 𝑛 − 8, 𝑛 − 5, 𝑛 − 2)⌈𝑛3⌉ , if 𝜇 = {𝑒𝑣𝑘(𝑘 = 1,2,4,5, . . . , 𝑛 − 4, 𝑛 − 3, 𝑛 − 1, 𝑛) or𝑒𝑘(𝑘 ∈ [1; 𝑛 − 1: 1])  Proof: 𝐹1,1 and 𝐹1,2 are isomorphic to 𝑃2 and 𝐶3 respectively.  Therefore 𝑅(1) − 𝛾 (𝐿(𝐹1,1)) = 1 and 𝑅(1) − 𝛾 (𝐿(𝐹1,2)) = 1. For 𝑛 > 2 , let 𝑉(𝐹1,𝑛) = {𝑣, {𝑣1; 𝑣𝑛: 1}}  where 𝑑𝑒𝑔 𝑣 = 𝑛 , 𝑑𝑒𝑔 𝑣1 =𝑑𝑒𝑔 𝑣𝑛 = 2  and 𝑑𝑒𝑔 𝑣𝑘 = 3  for all 2 ≤ 𝑘 ≤ 𝑛 − 1 . Label the edge 𝑣𝑘𝑣𝑘+1 as 𝑒𝑘 and 𝑣𝑣𝑘 as 𝑒𝑣𝑘 as shown in Fig. 5.  Figure 5: 𝐹1,𝑛 In the graph 𝐹1,𝑛, edge adjacency is given as follows: i. 𝑒1 is adjacent to 𝑒𝑣1 , 𝑒𝑣2 and 𝑒2. ii. 𝑒𝑛−1 is adjacent to 𝑒𝑣𝑛−1 , 𝑒𝑣𝑛 and 𝑒𝑛−2. iii. For 1 < 𝑘 < 𝑛 − 1, 𝑒𝑘 is adjacent to 𝑒𝑣𝑘 , 𝑒𝑣𝑘+1 , 𝑒𝑘−1 and 𝑒𝑘+1. iv. 𝑒𝑣1 is adjacent to 𝑒𝑣2 , 𝑒𝑣3 , . . . , 𝑒𝑣𝑛−1 , 𝑒𝑣𝑛 and 𝑒1. v. 𝑒𝑣𝑛 is adjacent to 𝑒𝑣1 , 𝑒𝑣2 , . . . , 𝑒𝑣𝑛−2 , 𝑒𝑣𝑛−1 and 𝑒𝑛−1. vi. For 1 < 𝑘 < 𝑛 , 𝑒𝑣𝑘  is adjacent to 𝑒𝑣1 , 𝑒𝑣2 , . . . , 𝑒𝑣𝑘−1 , 𝑒𝑣𝑘+1 , . . . , 𝑒𝑣𝑛−1 , 𝑒𝑣𝑛 , 𝑒𝑘−1 and 𝑒𝑘. Rajasekar & Rajasekar                                                                         RDN of Line Graph 59  Since adjacency matrix of line graph is nothing but the incidence matrix of the given graph, the graph 𝐿(𝐹1,𝑛) is obtained from edge adjacency of 𝐹1,𝑛 as shown in Fig. 6.  Figure 6: 𝐿(𝐹1,𝑛) The induced subgraph of {𝑒𝑣1; 𝑒𝑣𝑛: 1}  is a complete graph with 𝑛 vertices and the induced subgraph of {𝑒1; 𝑒𝑛−1: 1} is a path of length 𝑛 − 1. Case (i): For 𝑛 ≡ 0(𝑚𝑜𝑑 3). Sub case (i): Suppose 𝑒𝑣1 is the 𝑅𝑉. Then 𝑒𝑣1 must be in the DS and 𝑒𝑣1 dominates the vertices {𝑒𝑣2; 𝑒𝑣𝑛: 1} ∪ {𝑒1}. The remaining vertices which are not dominated by 𝑒𝑣1 are {𝑒2; 𝑒𝑛−1: 1}. Now it is enough to find the DS for the vertices {𝑒2; 𝑒𝑛−1: 1}. The 𝐿(𝐹1,𝑛)[𝑉1] with 𝑉1 = {𝑒2; 𝑒𝑛−1: 1} is 𝑃(𝑛−1)−1 = 𝑃𝑛−2. Hence 𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝑒𝑣1) = |{𝑒𝑣1}| + 𝛾(𝑃𝑛−2) = 1 + ⌈𝑛−23⌉ = ⌈𝑛+13⌉. Similarly, the same result is obtained for the 𝑅𝑉  𝑒𝑣𝑘  where𝑘 =3,4,6, . . . , 𝑛 − 5, 𝑛 − 3, 𝑛 − 2, 𝑛.  Sub case (ii): Suppose 𝑒𝑣2 is the𝑅𝑉. Then 𝑒𝑣2 must be in the DS and 𝑒𝑣2  dominates the vertices {𝑒𝑣1 , 𝑒𝑣3 , . . . , 𝑒𝑣𝑛−1 , 𝑒𝑣𝑛} ∪ {𝑒1, 𝑒2} . The remaining vertices which are not dominated by 𝑒𝑣2 are {𝑒3; 𝑒𝑛−1: 1}. Now it is enough to find the DS for the vertices {𝑒3; 𝑒𝑛−1: 1}. The 𝐿(𝐹1,𝑛)[𝑉2] with 𝑉2 = {𝑒3; 𝑒𝑛−1: 1} is 𝑃(𝑛−1)−2 = 𝑃𝑛−3. Hence 𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝑒𝑣2) = |{𝑒𝑣2}| + 𝛾(𝑃𝑛−3) = 1 + ⌈𝑛−33⌉ = ⌈𝑛3⌉. Mapana - Journal of Sciences, Vol. 22, Special Issue 1  ISSN 0975-3303 60  The same result holds for VR 𝑒𝑣𝑘  where 𝑘 = 5,8,11, . . . , 𝑛 − 10, 𝑛 −7, 𝑛 − 4, 𝑛 − 1. Sub case (iii): Suppose 𝑒1 is 𝑅𝑉 . Then 𝑒1 must be in the DS and 𝑒1 dominates the vertices 𝑒2, 𝑒𝑣1and𝑒𝑣2. The remaining vertices which are not dominated by 𝑒1 are {𝑒𝑣3; 𝑒𝑣𝑛: 1} ∪ {𝑒3; 𝑒𝑛−1: 1}. To dominate the remaining vertices from the set{𝑒𝑣3; 𝑒𝑣𝑛: 1}, choose the vertex 𝑒𝑣4which also dominated 𝑒3 and 𝑒4.  Now it is enough to find the DS for the vertices {𝑒5; 𝑒𝑛−1: 1}.  The 𝐿(𝐹1,𝑛)[𝑉3] with 𝑉3 = {𝑒5; 𝑒𝑛−1: 1} is 𝑃(𝑛−1)−4 = 𝑃𝑛−5. Hence 𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝑒1) = |{𝑒1}| + |{𝑒𝑣4}| + 𝛾(𝑃𝑛−5) = 2 + ⌈𝑛−53⌉ = ⌈𝑛+13⌉. Similarly, we get same result for the VR 𝑒𝑘 where 𝑘 = 3,6,9, . . . , 𝑛 −9, 𝑛 − 6, 𝑛 − 3, 𝑛 − 1. Sub case (iv): Suppose 𝑒2 is 𝑅𝑉. Then 𝑒2 has to be in the DS and 𝑒2 dominates the vertices 𝑒1, 𝑒3, 𝑒𝑣2 and 𝑒𝑣3 . The remaining vertices which are not dominated by 𝑒2  are {𝑒𝑣1 , 𝑒𝑣4 , . . . , 𝑒𝑣𝑛−1 , 𝑒𝑣𝑛} ∪{𝑒4; 𝑒𝑛−1: 1} . To dominate the remaining vertices from the set {𝑒𝑣1 , 𝑒𝑣4 , . . . , 𝑒𝑣𝑛−1 , 𝑒𝑣𝑛}, choose the vertex 𝑒𝑣5which also dominates 𝑒4 and 𝑒5. Now it is enough to find the DS for the vertices {𝑒6; 𝑒𝑛−1: 1}.  The graph 𝐿(𝐹1,𝑛)[𝑉4] with 𝑉4 = {𝑒6; 𝑒𝑛−1: 1} is 𝑃(𝑛−1)−5 = 𝑃𝑛−6. Hence𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝑒2) = |{𝑒2}| + |{𝑒𝑣5}| + 𝛾(𝑃𝑛−6) = 2 + ⌈𝑛−63⌉ =⌈𝑛3⌉. The same result holds for VR 𝑒𝑘 where 𝑘 = 4,5,7, . . . , 𝑛 − 7, 𝑛 − 5, 𝑛 −4, 𝑛 − 2. Hence  Rajasekar & Rajasekar                                                                         RDN of Line Graph 61  𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝜇)={    ⌈𝑛 + 13⌉  where 𝜇 = {𝑒𝑣𝑘(𝑘 = 1,3,4,6, . . . , 𝑛 − 5, 𝑛 − 3, 𝑛 − 2, 𝑛) or𝑒𝑘(𝑘 = 1,3,6,9, . . . , 𝑛 − 9, 𝑛 − 6, 𝑛 − 3, 𝑛 − 1)⌈𝑛3⌉  where 𝜇 = {𝑒𝑣𝑘(𝑘 = 2,5,8,11, . . . , 𝑛 − 10, 𝑛 − 7, 𝑛 − 4, 𝑛 − 1) or𝑒𝑘(𝑘 = 2,4,5,7, . . . , 𝑛 − 7, 𝑛 − 5, 𝑛 − 4, 𝑛 − 2) Case (ii): For 𝑛 ≡ 1(𝑚𝑜𝑑 3). 𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝜇) = ⌈𝑛3⌉  where 𝜇 = {𝑒𝑣𝑘(𝑘 ∈ [1; 𝑛: 1]) or𝑒𝑘(𝑘 ∈ [1; 𝑛 − 1: 1]) Case (iii): For 𝑛 ≡ 2(𝑚𝑜𝑑 3). 𝑅(1) − 𝛾(𝐿(𝐹1,𝑛), 𝜇)={  ⌈𝑛 + 23⌉  where 𝜇 = 𝑒𝑣𝑘(𝑘 = 3,6,9,12, . . . , 𝑛 − 11, 𝑛 − 8, 𝑛 − 5, 𝑛 − 2)⌈𝑛3⌉  where 𝜇 = {𝑒𝑣𝑘(𝑘 = 1,2,4,5, . . . , 𝑛 − 4, 𝑛 − 3, 𝑛 − 1, 𝑛) or𝑒𝑘(𝑘 ∈ [1; 𝑛 − 1: 1]) References [1]. [1] Ore, O., Theory of Graphs, Amer. Math. Soc. Colloq. Publ., 38 (Amer. Math. Soc., Providence, RI), 1962. [2]. [2] Gross, J. T. and Yellen, J. Graph Theory and Its Applications, 2nd ed. Boca Raton, FL: CRC Press, pp. 20 and 265, 2006. [3]. [3] G. Rajasekar and G. Rajasekar, Reserved Domination Number of Graphs, Turkish Online Journal of Qualitative Inquiry (TOJQI), Vol.12 (2021), Issue.6, pp. 9199-9209. [4]. [4] G. Rajasekar and K. Nagarajan (2019) Location domination number of line graph, Journal of Discrete Mathematical Sciences and Cryptography, 22:5, 777-786, DOI:10.1080/09720529.2019.1681694. [5]. [5] Dr.G. Rajasekar and G. Rajasekar, Reserved Domination Number of some Graphs, Turkish Journal of Computer and Mathematics Education, Vol.12 No.11 (2021), pp. 2166-2181. [6]. [6] G. Rajasekar and Govindan Rajasekar, 2-reserved domination number of graphs, Advances and Applications Mapana - Journal of Sciences, Vol. 22, Special Issue 1  ISSN 0975-3303 62  in Discrete Mathematics 27(2) (2021), 249-264. DOI:10.17654/DM027010249. [7]. [7] G. Rajasekar and G. Rajasekar, Reserved Domination Number of Complement of a Graphs, South East Asian J. of Mathematics and Mathematical Sciences, vol. 18 (2022), pp. 173-180.   

Reserved Domination Number of Line Graph

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Reserved Domination Number of Line Graph

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