EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 1326-1341 ISSN 1307-5543 – ejpam.com Published by New York Business Global Perfect Isolate Domination in Graphs Cris L. Armada1,3,∗, Jamil J. Hamja2 1 Mathematics Department, College of Arts and Sciences, Cebu Normal University, Cebu City, Philippines 2 Office of the Vice Chancellor for Academic Affairs, MSU-Tawi-Tawi College of Technology and Oceanography, 7500 Tawi-Tawi, Philippines 3 Center for Research and Development, Cebu Normal University, Cebu City, Philippines Abstract. Let G = (V (G), E(G)) be a simple connected graph. A set S ⊆ V (G) is said to be a perfect isolate dominating set of G if S is a perfect dominating set and an isolate dominating set of G. The minimum cardinality of a perfect isolate dominating set of G is called perfect isolate domination number, and is denoted by γp0(G). A perfect isolate dominating set S with |S| = γp0(G) is said to be γp0-set. In this paper, the author gives a characterization of perfect isolate dominating set of some graphs and graphs obtained from the join, corona and lexicographic product of two graphs. Moreover, the perfect isolate domination number of the forenamed graphs is determined and also, graphs having no perfect isolate dominating set are examined. 2020 Mathematics Subject Classifications: 05C69,05C38,05C76 Key Words and Phrases: Perfect domination, isolate domination, perfect isolate domination 1. Introduction In 1960, the study of domination in graphs began and it became the most interesting topic in graph theory because of its application in networking. In 1990, Livingston and Stout [10], introduced the concept of perfect dominating sets of G, denoted by γp(G). They studied the existence and construction of perfect dominating sets in families of graphs arising from the interconnection networks of parallel computers. In 2014, Kwon and Lee [9] investigated some results related to perfect domination sets of Cayley graphs. In 2013, Hamid and Balamurugan [4] studied the concept of isolate domination in graphs. In 2015, Ariola [2] looked at another aspect of the isolate dominating set and characterized the lower and upper bounds of the isolate domination number and those graphs resulting from some binary operations such as join and corona. In 2016, Hamid and Balamurugan [4] extended these parameters isolate domination number γ0 and the upper isolate domination number Γ0. In 2017, Rad [11] studied the complexity of the ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4760 Email addresses: armadac@cnu.edu.ph (C. Armada), jamilhamja@msutawi-tawi.edu.ph (, J. Hamja) https://www.ejpam.com 1326 © 2023 EJPAM All rights reserved. C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1327 isolate domination in graphs, and obtain several bounds and characterizations on the isolate domination number. Furthermore, some variations and parameters of domination in graphs are studied in many classes (see [1], [8], [6] and [5]). In this paper, we introduce the concept of perfect isolate domination number. We characterize the perfect isolate dominating set and determine the exact values of perfect isolate domination number of some special graphs and graphs under some binary operations such as join, corona and lexicographic product of two graphs. Not all graphs have perfect isolate dominating set and we called them non-γp0-graphs. 2. Terminology and Notation This section contains definitions that are needed for the study. Let G = (V (G), E(G)) be a simple connected graph where V (G) is a vertex-set of G and E(G) is an edge-set of G. The number of edges incident with v is called the degree of a vertex v and is denoted by deg(v). The maximum of {deg(v) : v ∈ V (G)} is denoted by ∆(G). The set of neighbors of a vertex u in G is called the open neighborhood of u in G and is denoted by NG(u) = {v ∈ V (G) : uv ∈ E(G)}. The closed neighborhood of u in G is the set NG[u] = NG(u) ∪ {u} and the closed neighborhood of a subset S of V (G) is the set NG[S] = N [S] = ∪v∈SNG[v]. The subgraph induced by a set S of vertices of a graph G is denoted by ⟨S⟩ with V (⟨S⟩) = S and E(⟨S⟩) = {uv ∈ E(G) : u, v ∈ S}, Harary in [7]. A set S ⊆ V (G) is said to be a dominating set if N [S] = V (G). A dominating set S is a minimal dominating set if no proper subset S′ ⊂ S is a dominating set. The domination number γ(G) of a graph G is the minimum cardinality of a dominating set of G. A domi- nating set S with |S| = γ(G) is said to be a γ-set. A set S ⊆ V (G) is said to be a perfect dominating set if each vertex v ∈ V (G) \ S is dominated by exactly one element in S. The minimum cardinality of a perfect dominating set of G is called perfect domination number, and is denoted by γp(G). A perfect dominat- ing set S with |S| = γp(G) is said to be a γp-set. A dominating set S ⊆ G is said to be an isolate dominating set of G if ⟨S⟩ has at least one isolated vertex. An isolate dominating set S is said to be a minimal isolate dominating set if no proper subset of S is an isolate dominating set. The minimum cardinality of a minimal isolate dominating set of G is called the isolate domination number and is denoted by γ0(G). An isolate dominating set S with |S| = γ0(G) is said to be a γ0-set. A perfect dominating set S ⊆ V (G) is said to be a independent perfect dominating set (ipds) if no two vertices in S are adjacent. The minimum cardinality of an independent perfect dominating set of G is called independent perfect domination number and is de- noted by γip(G). A perfect dominating set S with |S| = γip(G) is said to be a γip-set. C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1328 A set S ⊆ V (G) is said to be a perfect isolate dominating set of G if S is a perfect dominating set and the induced subgraph ⟨S⟩ has at least one isolated vertex. The minimum cardinality of a perfect isolate dominating set of G is called perfect isolate domination number and is denoted by γp0(G). A perfect isolate dominating set S with |S| = γp0(G) is said to be γp0-set. If a graph G has no perfect isolate dominating set, then we say that the graph G is a non-γp0-graph. Example 1. Let G be the graph in Figure 1 and S = {v1, v2, v7}. Then v1 dominates v3, v2 dominates v4, while v7 dominates v5 and v6. This shows that for all vertices v3, v4, v5, v6 are elements in V (G) \S which are dominated by exactly one vertex in S. Thus, S is a perfect dominating set of G. Observe further that v7 is an isolated vertex of the induced subgraph ⟨S⟩. Therefore, S is a perfect isolate dominating set of G and γp0(G) = γp(G) =|S|= 3. v1 v2 v3 v4 v5 v6 v7 G : Figure 1: Graph G with γp0(G) = γp(G) = 3 3. Results This section contains some known results involving the domination number, the perfect domination number and the isolate domination number. Also, it contains the the perfect isolate domination number of paths, cycles, complete graphs, fans, wheels, friendship graphs, windmill graphs and graphs resulting from some binary operations. Furthermore, some non-γp0-graphs are shown. Proposition 1. [8] For paths and cycles of order n, γ0(Pn) = γ0(Cn) = ⌈n3 ⌉. Proposition 2. [8] For Kn, Sn−1, and Wn−1 be complete, star, and wheel of n ≥ 2 vertices, respectively. Then γ0(Kn) = γ0(Sn−1) = γ0(Wn−1) = 1. Theorem 1. [8] For Kn, Sn−1, and Wn−1 be complete, star, and wheel of n ≥ 2 vertices, respectively. Then γ0(Kn) = γ0(Sn−1) = γ0(Wn−1) = 1. C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1329 Corollary 1. [3] Let G and H be connected graphs. Then γ(G+H) = { 1, γ(G) = 1 or γ(H) = 1 2, γ(G) ̸= 1 and γ(H) ̸= 1. The next four results follow directly from the definition of perfect isolate dominating set. Proposition 3. Let S ⊆ V (G) be an isolate dominating set of G. Then S is a perfect isolate dominating set if and only if for every v ∈ V (G) \ S, NG(v) ∩ S = {u} for some u ∈ S. Proposition 4. If S is a dominating set or a perfect dominating set or an isolate dominating set of G with |S| = 1, then S is a perfect isolate dominating set of G. In particular, γ(G) = γp(G) = γ0(G) = 1 if and only if γp0(G) = 1. Proposition 5. If γp(G) = c where c ∈ Z+ and if S is a γp-set of G such that ⟨S⟩ has an isolated vertex, then γp0(G) = c. Proposition 6. If γ0(G) = c where c ∈ Z+ and if S is a γ0-set of G such that every vertex v ∈ V (G) \ S is dominated by exactly one vertex in S, then γp0(G) = c. Proposition 7. Let G be any graph such that G has a perfect isolate dominating set. Then γ0(G) ≤ γp0(G). Theorem 2. For any positive integers a and b with 1 ≤ a ≤ b, there exists a connected graph G such that γ0(G) = a and γp0(G) = b. Proof. Consider the following cases: Case 1: a = b Let G be the graph shown in Figure 2. Clearly, the set S1 = {v1, v2, v3, . . . , va} is both γ0-set and γp0-set of G. Therefore, γ0(G) = γp0(G) = a = b. . . . . . . v1 v2 v3 va−1 va u1 u2 u3 ua−1 ua G : Figure 2: Graph G with γ0(G) = γp0(G) = a C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1330 Case 2:a < b. Let G be the graph shown in Figures 3 or Figure 4. Let m = b − a + 4 and j, k, l ∈ Z+. Observe that S1 = {v1, v2, v3 . . . , va−4} ∪ {z1, z2, z3, zm} is a γ0-set of G and S2 = {v1, v2, v3 . . . , va−4} ∪ {zi : i = 1, 2, . . . ,m} is a γp0-set of G. It follows that γ0(G) = |S1| = a − 4 + 4 = a and γp0(G) = |S2| = a − 4 +m = a − 4 + (b − a + 4) = b. Therefore, γ0(G) = a < b = γp0(G). . . . u1 u2 u3 ua−4 ua v1 v2 v3 va−4 G : . . . . . . ... w1 w2 wj x1 x2 xk y1y2yl z1 z2 zm−1 . . . z3 z4 z5 zm .. . Figure 3: Graph G with γ0(G) = a . . . u1 u2 u3 ua−4 ua v1 v2 v3 va−4 G : . . . . . . ... w1 w2 wj x1 x2 xk y1y2yl z1 z2 zm−1 . . . z3 z4 z5 zm .. . Figure 4: Graph G with γp0(G) = b This proves the assertion. C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1331 Corollary 2. The difference γp0- γ0 can be made arbitrarily large. The next result follows from Proposition 1 and Proposition 4. Corollary 3. Let G be a simple connected graph of order 3, that is, either G = P3 or G = C3. Then γp0(G) = 1. Theorem 3. Let G be a connected graph of order n ≥ 2. Then γp0(G) = 1 if and only if ∆(G) = n− 1. Proof. Suppose that γp0(G) = 1. Let S = {u} be the perfect isolate dominating set of G. If G is trivial, then deg(u) = 0. So, we are done. Assume that G is nontrivial. Then every vertex v ∈ V (G) \ S is adjacent to u ∈ S. Hence, deg(u) = n − 1 since |V (G)|= n. Therefore, ∆(G) = n− 1. For the converse, let ∆(G) = n− 1. Then there exists a vertex u ∈ V (G) such that deg(u) = n − 1. Since every vertex v ∈ V (G) \ {u} is dominated by exactly one vertex u, S = {u} is a perfect dominating set of G. By Proposition 4, γp0(G) = |S| = 1. The next result follows from Theorem 3 since ∆(Kn) = n− 1. Corollary 4. For the complete graph Kn, where n ≥ 1, γp0(Kn) = 1. Theorem 4. Let G be connected graph of order n ≥ 4. Then γp0(G) = 2 if and only if there exists two vertices x, y ∈ S ⊆ V (G) such that NG(x) ∩ NG(y) = ∅ and NG[x] ∪NG[y] = V (G). Proof. Suppose that γp0(G) = 2. Let S ⊆ V (G) be a perfect isolate dominating set of G. Then |S| = 2 and so, there exist two vertices x, y ∈ S which are isolated vertices in ⟨S⟩ such that each vertex v ∈ V (G)\S is dominated only by either x or y but not both. Thus, NG(x) ∩NG(y) = ∅ and NG[S] = NG[x] ∪NG[y] = V (G). For the converse, suppose that NG(x) ∩ NG(y) = ∅ and NG[x] ∪ NG[y] = V (G). It follows that the graph G cannot be dominated by 1 vertex only, that is, γp0(G) > 1 and each vertex v ∈ V (G)\S is dominated only by either x or y but not both and ⟨S⟩ has two isolated vertices. Clearly, S = {x, y} is a perfect isolate dominating set of G. Therefore, γp0(G) = |S| = 2. Theorem 5. For any path Pn of order n ≥ 1, γp0(Pn) = ⌈n 3 ⌉ . Proof. Suppose that V (Pn) = {u1, u2, . . . , un−1, un} such that deg(u1) = deg(un) = 1 and deg(ui) = 2 for all i = 2, 3, . . . , n − 1. Note that by Proposition 1, γ0(Pn) = ⌈n3 ⌉. Consider the following cases: C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1332 Case 1: Suppose that n ≡ 0 (mod 3). If n = 3, by Corollary 3, γp0(P3) = 1 = ⌈33⌉. Suppose that n > 3. Let r = n 3 and j = 1, 2, . . . , r − 1, r. Group the vertices of Pn into r disjoint subsets Sj S1 = {u1, u2, u3} S2 = {u4, u5, u6} S3 = {u7, u8, u9} S4 = {u10, u11, u12} ... Sr−1 = {un−5, un−4, un−3} Sr = {un−2, un−1, un} Clearly, the set S = {u2, u5, u8, u11, . . . , un−4, un−1} is a γ0-set of Pn since N [S] = V (Pn), ⟨S⟩ has isolated vertices and |S| = ⌈n3 ⌉ by Proposition 1. Clearly, all other vertices ui ∈ V (Pn) \ S for all i = 1, 3, 4, 6, . . . , n − 5, n − 3, n − 2, n are dominated by exactly one vertex in S. By Proposition 6, γp0(Pn) = ⌈n3 ⌉. Case 2: Suppose that n ≡ 1 (mod 3). If n = 1, then γp0(P1) = 1 = ⌈13⌉. Suppose that n = 4. Clearly, S = {u1, u4} is a γ0-set of P4 since |S| = ⌈43⌉ = 2. Clearly, N(u1) ∩ N(u4) = ∅ and N [u1] ∪ N [u4] = V (P4). By Theorem 4, γp0(P4) = 2 = ⌈43⌉. Suppose that n > 4. Let r = n+2 3 and j = 1, 2, . . . , r − 1, r. Group the vertices of Pn into r disjoint subsets Sj S1 = {u1} S2 = {u2, u3, u4} S3 = {u5, u6, u7} S4 = {u8, u9, u10} ... Sr−1 = {un−5, un−4, un−3} Sr = {un−2, un−1, un} Clearly, the set S = {u1, u4, u7, u10, . . . , un−3, un} is a γ0-set of Pn since N [S] = V (Pn), ⟨S⟩ has isolated vertices and |S| = ⌈n3 ⌉ by Proposition 1. Clearly, all other vertices ui ∈ V (Pn) \ S for all i = 2, 3, 5, 6, . . . , n − 5, n − 4, n − 2, n − 1 are dominated by exactly one vertex in S. By Proposition 6, γp0(Pn) = ⌈n3 ⌉. Case 3: Suppose that n ≡ 2 (mod 3). Suppose that n = 2. Clearly, ∆(P2) = 1. By Theorem 3, γp0(P2) = 1 = ⌈23⌉. Suppose that n > 2. Let r = n+1 3 and j = 1, 2, . . . , r−1, r. Group the vertices of Pn into r disjoint subsets Sj C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1333 S1 = {u1, u2} S2 = {u3, u4, u5} S3 = {u6, u7, u8} S4 = {u9, u10, u11} ... Sr−1 = {un−5, un−4, un−3} Sr = {un−2, un−1, un} Clearly, the set S = {u1, u4, u7, u10, . . . , un−4, un−1} is a γ0-set of Pn since N [S] = V (Pn), ⟨S⟩ has isolated vertices and |S| = ⌈n3 ⌉ by Proposition 1. Clearly, all other vertices ui ∈ V (Pn) \ S for all i = 2, 3, 5, 6, . . . , n − 5, n − 3, n − 2, n are dominated by exactly one vertex in S. By Proposition 6, γp0(Pn) = ⌈n3 ⌉. Therefore, in any case, γp0(Pn) = ⌈n3 ⌉. Theorem 6. For a cycle Cn of order n = 3 or n ≥ 6, γp0(Cn) = {⌈ n 3 ⌉ + 1, if n ≡ 2 (mod 3)⌈ n 3 ⌉ , if n ̸≡ 2 (mod 3) Proof. Suppose that V (Cn) = {u1, u2, . . . , un−1, un} such that deg(ui) = 2 for all i = 1, 2, 3, . . . , n− 1, n. By Proposition 1, γ0(Cn) = ⌈n3 ⌉. Consider the following cases: Case 1: Suppose that n ≡ 0 (mod 3). If n = 3, by Corollary 3, γp0(C3) = 1 = ⌈33⌉. Suppose that n > 3. Let r = n 3 and j = 1, 2, . . . , r − 1, r. Group the vertices of Cn into r disjoint subsets Tj T1 = {u1, u2, u3} T2 = {u4, u5, u6} T3 = {u7, u8, u9} T4 = {u10, u11, u12} ... Tr−1 = {un−5, un−4, un−3} Tr = {un−2, un−1, un} Clearly, the set S = {u2, u5, u8, u11, . . . , un−4, un−1} is a γ0-set of Cn since N [S] = V (Cn), ⟨S⟩ has isolated vertices and |S| = ⌈n3 ⌉ by Proposition 1. Clearly, all other vertices ui ∈ V (Cn) \ S for all i = 1, 3, 4, 6, . . . , n− 5, n− 3, n− 2, n are dominated by exactly one vertex in S. By Proposition 6, γp0(Cn) = ⌈n3 ⌉. C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1334 Case 2: Suppose that n ≡ 1 (mod 3). Suppose that n = 7. By Proposition 1, γ0(C7) = 3. Clearly, S = {u1, u2, u5} is a γ0-set since N [S] = V (C7), ⟨S⟩ has an isolated vertex u5 and |S| = 3. Also,u3, u4, u6, u7 ∈ V (C7) \ S are dominated by exactly one vertex in S. By Proposition 6, γp0(C7) = 3 = ⌈73⌉. Suppose that n > 7. Let r = n+2 3 and j = 1, 2, . . . , r − 1, r. Group the vertices of Cn into r disjoint subsets Tj T1 = {u1} T2 = {u2, u3, u4} T3 = {u5, u6, u7} T4 = {u8, u9, u10} ... Tr−1 = {un−5, un−4, un−3} Tr = {un−2, un−1, un} Clearly, the set S = {u1, u2, u5, u8, . . . , un−5, un−2} is a γ0-set of Cn since N [S] = V (Cn), ⟨S⟩ has isolated vertices and |S| = ⌈n3 ⌉ by Proposition 1. Clearly, all other vertices ui ∈ V (Cn) \ S for all i = 3, 4, 6, 7, . . . , n− 4, n− 3, n− 1, n are dominated by exactly one vertex in S. By Proposition 6, γp0(Cn) = ⌈n3 ⌉. Case 3: Suppose that n ≡ 2 (mod 3). Suppose that n = 8. Clearly, S = {u1, u3, u6} is a γ0-set of C8 since N [S] = V (C8) and |S| = ⌈83⌉ = 3. Note that u2 is dominated by both u1 and u3. Hence, S is not a perfect dominating set. Let T = S ∪ {u2} = {u1, u2, u3, u6}. Clearly, |T | = 4 = ⌈83⌉ + 1 and u4, u5, u7, u8 ∈ V (C8) \ T are dominated by exactly one vertex in T . Clearly, T is a γp0-set in C8 and so, γp0(C8) = 4 = ⌈83⌉ + 1. Suppose that n > 8. Let r = n+1 3 and j = 1, 2, . . . , r − 1, r. Group the vertices of Cn into r disjoint subsets Tj T1 = {u1, u2} T2 = {u3, u4, u5} T3 = {u6, u7, u8} T4 = {u9, u10, u11} ... Tr−1 = {un−5, un−4, un−3} Tr = {un−2, un−1, un} Clearly, the set S = {u1, u3, u6, u9, u12, . . . , un−5, un−2} is a γ0-set of Cn since N [S] = V (G), ⟨S⟩ has isolated vertices and |S| = ⌈n3 ⌉ by Proposition 1. Clearly, u2 is dominated by both u1 and u3. Hence, S is not a perfect dominating set. Let T = S ∪ {u2} = {u1, u2, u3, u6, . . . , un−5, un−2}. Clearly, |T | = ⌈83⌉ + 1 and all other vertices ui ∈ V (Cn) \ T for all i = 4, 5, 7, 8, . . . , n − 4, n − 3, n − 1, n are dominated by exactly one vertex in T . Clearly, T is a γp0-set in Cn and so, γp0(Cn) = ⌈n3 ⌉+ 1. C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1335 By Proposition 1 and Theorem 6, γ0(Cn) ≤ γp0(Cn) for n = 3 or n ≥ 6. Theorem 7. The cycle graphs C4 and C5 are non-γp0-graphs. Proof. Let V (C4) = {u1, u2, u3, u4} and V (C5) = {u1, u2, u3, u4, u5}. For the graph C4, clearly, the sets S1 = {u1, u2}, S2 = {u2, u3}, S3 = {u3, u4}, and S4 = {u1, u4} are perfect dominating sets (γp-sets) of C4 but none of them are isolate dominating sets since ⟨Si⟩ has no isolated vertex for all i = 1, 2, 3, 4. Also, the sets T1 = {u1, u3} and T2 = {u2, u4} are isolate dominating sets (γ0-sets) of C4 by Proposition 1 but none of them are perfect dominating sets since all vertices in V (C4) \Tj are dominated by two vertices in Tj for all j = 1, 2. Choosing 3 vertices or 4 vertices in C4 for γp0-sets is not possible since their induced subgraphs has no isolated vertex. Thus, in any case, C4 contains a perfect dominating set or an isolate dominating set but not both. Therefore, C4 is a non-γp0-graph. For the graph C5, clearly, the sets S1 = {u1, u2, u3}, S2 = {u2, u3, u4}, S3 = {u3, u4, u5}, S4 = {u4, u5, u1}, and S5 = {u5, u1, u2} are perfect dominating sets (γp-sets) of C5 but none of them are isolate dominating sets since ⟨Si⟩ has no isolated vertex for all i = 1, 2, 3, 4, 5. Also, the sets T1 = {u1, u3}, T2 = {u1, u4}, T3 = {u2, u4}, T4 = {u2, u5} and T5 = {u3, u5} are isolate dominating sets (γ0-sets) of C5 by Proposition 1 but none of them are perfect dominating sets since some vertices in V (C5)\Tj are dominated by two vertices in Tj for all j = 1, 2, 3, 4, 5. Note that the sets R1 = {u1, u3, u5}, R2 = {u1, u3, u4}, R3 = {u2, u3, u5}, R4 = {u2, u4, u5}, and R5 = {u1, u2, u4} are isolate dominating sets but none of them are perfect dominating sets since all vertices in V (C5) \ Rk are dominated by two vertices in Rk for all k = 1, 2, 3, 4, 5. Choosing 4 or 5 vertices in C5 for γp0-sets is not possible since their induced subgraphs has no isolated vertex. Thus, in any case, C5 contains a perfect dominating set or an isolate dominating set but not both. Therefore, C5 is a non-γp0-graph. 4. The Perfect Isolate Dominating set in the Join of Graphs This section contains results when the join G+H has a γp0-set or has no γp0-set and its perfect isolate domination number. The join of two graphs G and H, denoted by G + H, is the graph with V (G+H) = V (G) ∪ V (H) and E(G+H) = E(G) ∪E(H) ∪ {uv : u ∈ V (G), v ∈ V (H)}. Theorem 8. Let G and H be any graphs with γ(G) = 1 or γ(H) = 1. Then γp0(G+H) = 1. Proof. By Corollary 1, γ(G+H) = 1. By Proposition 4, γp0(G+H) = 1. The next result follows from Theorem 8 since γ(K1) = γ(P2) = γ(P3) = γ(C3) = 1. C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1336 Corollary 5. The following are graphs having γp0(G) = 1. i. Star graph Sn = K1 +Kn, n ≥ 2 ii. Fan graph Fn = K1 + Pn, n ≥ 2 iii. Wheel graph Wn = K1 + Cn , n ≥ 3 iv. Friendship graph Fn = K1 + nP2, n ≥ 2 v. Windmill graph Wm n = K1 +mKn−1, n ≥ 3 and m ≥ 2 vi. Complete bipartite graph Km,n = Km +Kn, either m = 1 or n = 1. vii. Generalized Fan Graph Fm,n = Km + Pn, m ≥ 1 and n = 2, 3 viii. Generalized Wheel Graph Wm,3 = Km + C3, m ≥ 1. Theorem 9. Let G and H be any graphs with γ(G) ≥ 2 and γ(H) ≥ 2. Then the graph G+H is a non-γp0-graph. Proof. By Corollary 1, γ(G+H) = 2. Let S be a γp0-set of G+H. By Proposition 4, γp0(G+H) > 1. Thus, |S| > 1. Suppose that x, y ∈ S. Consider the following cases: Case 1: x, y ∈ S ⊆ V (G). Clearly, for all z ∈ V (H), z is dominated in G + H by the two vertices x and y in S . Thus, S is not a perfect dominating set of G+H. Thus, S is not a γp0-set of G+H. This is a contradiction. Similarly, if x, y ∈ S ⊆ V (H), then S is not γp0-set of G+H. This is a contradiction. Therefore, having two vertices in V (G) or in V (H) is not possible for a set S to be a γp0-set of G+H. Case 2: x, y ∈ S such that x ∈ V (G) and y ∈ V (H). Clearly, the induced subgraph ⟨S⟩ has no isolated vertex since xy ∈ E(G+H). Hence, S is not an isolate dominating set of G+H. Thus, S is not a γp0-set of G+H, a contradiction. Furthermore, adding a vertex in S which is either from V (G) or V (H) is not possible by case 1. Hence, in any case, S is not a γp0-set of G + H. Therefore, the graph G + H is a non-γp0-graph if γ(G) ≥ 2 and γ(H) ≥ 2. The next result follows from Theorem 9. Corollary 6. The following graphs are non-γp0-graphs . i. Complete bipartite graph Km,n = Km +Kn, m ≥ 2 and n ≥ 2. ii. Generalized Fan Graph Fm,n = Km + Pn, m ≥ 2 and n ≥ 4 iii. Generalized Wheel Graph Wm,n = Km + Cn, m ≥ 2 and n ≥ 4 C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1337 5. The Perfect Isolate Dominating set in the Corona of Graphs This section contains results when the corona G + H has a γp0-set or has no γp0-set and its perfect isolate domination number. The corona of graphs G and H, G ◦ H, is the graph obtained by taking one copy of G and |V (G)| copies of H, and then joining the ith vertex of G to every vertex of the ith copy of H. For every v ∈ V (G), denote by Hv the copy of H whose vertices are attached one by one to the vertex v. Subsequently, denote by v + Hv the subgraph of the corona G ◦H corresponding to the join ⟨{v}⟩+Hv, v ∈ V (G), Harary in [7]. Theorem 10. Let G be a connected graph and H be any graph. Then a subset S of V (G ◦H) is a perfect isolate dominating set of G ◦H if and only if for every v ∈ V (G), S = ⋃ v∈V (G) Sv where Sv is a minimal dominating set of Hv and γ(H) = 1. Proof. Let S be a perfect isolate dominating set of G ◦ H. Suppose that S = V (G). Clearly, S is a perfect dominating set since N [S] = V (G◦H) and every vertex in V (Hv) is dominated by exactly one vertex v ∈ V (G) but S is not an isolate dominating set since G is connected, a contradiction. Hence, S ̸= V (G). Thus, there exists v ∈ V (G) ⊆ V (G ◦H) such that v /∈ S, S contain a vertex or vertices that dominates Hv. Let Sv be a dominating set of Hv and Sv ⊆ S. Also, suppose that there exists y ∈ V (G) such that y ∈ S and vy ∈ E(G) ⊆ E(G ◦H). Clearly, v ∈ V (G ◦H) \ S is dominated by y and a vertex in Sv, a contradiction since S is a perfect dominating set. Since y is arbitrary, for all y ∈ V (G), y must not be an element in S, that is, S must not contain a vertex in V (G). Hence, S = ∪v∈V (G)Sv such that Sv is a dominating set in Hv. Suppose that Sv contains two or more vertices, say x and z, where x, z ∈ V (Hv). Thus, v ∈ V (G ◦H) \ S is dominated by x and z in Sv ⊆ S, a contradiction since S is a perfect dominating set. Thus, |Sv| = 1, and so, Sv is a minimal dominating set of Hv and γ(H) = 1. Conversely, suppose that for every v ∈ V (G), S = ∪v∈V (G)Sv where Sv is a minimal dominating set of Hv and γ(H) = 1. Clearly, N [S] = V (G ◦ H) and for every vertex v ∈ V (G) = V (G ◦H) \ S, v is dominated by exactly one vertex in Sv ⊆ S. Thus, S is a perfect dominating set. Also, since any two vertices in S is not adjacent in G ◦H, ⟨S⟩ has isolated vertices. Thus, S is also an isolate dominating set. Therefore, S ⊆ V (G ◦H) is a perfect isolate dominating set of G ◦H. The next two results follow from Theorem 10. Corollary 7. Let G be a connected graph of order n and H be any graph where γ(H) = 1. Then γp0(G ◦H) = n. Corollary 8. Let G be a connected graph and H be any graph where γ(H) ≥ 2. Then G ◦H is a non-γp0-graph. C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1338 The next two results follow from Corollary 7. Corollary 9. Let G be a connected graph and Kn be a complete graph. Then γp0(G ◦Kn) = |V (G)|. Corollary 10. Let G be a connected graph and H be any graph described in Corollary 5. Then γp0(G ◦H) = |V (G)|. The next result follows from Corollary 8. Corollary 11. Let G be a connected graph and H be any graph described in Corollary 6. Then G ◦H is non-γp0-graphs. The next result follows from Corollary 7 and Corollary 8. Corollary 12. Let G be a connected graph. Then for n ≥ 4, G ◦ Pn and G ◦ Cn are non-γp0-graphs and for n < 4, γp0(G ◦ Pn) = γp0(G ◦ C3) = |V (G)|. 6. The Perfect Isolate Dominating set in the Lexicographic Product of Graphs This section contains results when the lexicographic product G[H] has a γp0-set or has no γp0-set and its perfect isolate domination number. The lexicographic product or composition of two graphs G and H is the graph G[H] with vertex set V (G[H]) = V (G) × V (H) and edge set E(G[H]) satisfying the following conditions: (x, u)(y, v) ∈ E(G[H]) if and only if xy ∈ E(G) or x = y and uv ∈ E(H). Any subset C of V (G[H]) can be expressed as C = ⋃ x∈S ({x} × Tx) where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S. S is called as G-projection of C and ∪x∈STx is called as the H-projection of C. Theorem 11. Let G and H be connected nontrivial graphs. A subset C = ⋃ x∈S ({x} × Tx) of V (G[H]) where S ⊆ V (G) and Tx ⊆ V (H) for each x ∈ S, is a perfect isolate dominating set of G[H] if and only if S is an independent perfect dominating set (ipds) and Tx is a dominating set of H with |Tx| = 1 for all x ∈ S. Proof. Suppose that C is a perfect isolate dominating set of G[H]. Let u ∈ V (G) \ S. Pick any v ∈ V (H). Since C is a perfect dominating set, there exists (y, z) ∈ C such that NG[H](u, v) ∩ C = {(y, z)}. This implies that NG(u) ∩ S = {y}. This implies that every u ∈ V (G) \ S is dominated by exactly one vertex in S and so, S is a perfect dominating set. Suppose that p, q ∈ S such that pq ∈ E(G). Since H is connected, there exists vertices C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1339 m,n ∈ V (H) such that mn ∈ E(H), (p,m)(q,m) ∈ E(G[H]) and (p,m), (q,m) ∈ C, a contradiction since the vertices (p, n) and (q, n) are both dominated by the two vertices (p,m) and (q,m) in C. Hence, for any two vertices p, q ∈ S, pq /∈ E(G). Thus, S is an independent set. Let x ∈ S and suppose that |Tx| ≥ 2. Let c, d ∈ Tx such that c ̸= d. Thus, (x, c), (x, d) ∈ C. Let e ∈ V (G) such that xe ∈ E(G). Since S is independent, e /∈ S. Clearly, (e, c) /∈ C and (e, c) is dominated by the two vertices (x, c) and (x, d) in C, a contradiction since C is a perfect dominating set. Hence, |Tx| = 1 for each x ∈ S. Furthermore, since S is an independent dominating set and C is a dominating set, Tx is a dominating set in H for each x ∈ S. Conversely, suppose that S is an independent perfect dominating set (ipds) and Tx is a dominating set of H with |Tx| = 1 for all x ∈ S. Let Tx = {t} such that t dominates all other vertices of H. Let (p, q) ∈ V (G[H]) \ C and consider the following cases: Case 1: p /∈ S. Since S is a perfect dominating set, there exists x ∈ S such that S ∩NG(p) = {x}. Then {(x, t)} = NG[H]((p, q))∩C, that is, the point (p, q) ∈ V (G[H])\C is dominated by exactly one vertex (x, t) in C. Hence, C is a perfect dominating set. Case 2: p ∈ S. Since Tx = {t} and S is a perfect dominating set, {(p, t)} = NG[H]((p, q))∩C, that is, the point (p, q) ∈ V (G[H]) \ C is dominated by exactly one vertex (p, t) in C. Hence, C is a perfect dominating set. Since S is independent set and Tx is a dominating set with |Tx| = 1, C is also in- dependent set and so, ⟨C⟩ contains isolated vertices. Accordingly, C is a perfect isolate dominating set of G[H]. The next two results follow from Theorem 11. Corollary 13. Let G and H be connected graphs. Then γp0(G[H]) =  γp0(G), if H = K1 γp0(H), if G = K1 γip(G), if G has an ipds and γ(H) = 1. Corollary 14. Let G and H be connected graphs. Then G[H] is a non-γp0-graph if and only if one of the two cases is satisfied: (i) γ(H) ≥ 2 (ii) γ(H) = 1 and G has no independent perfect dominating set. The next two results follow from Corollary 13. C. Armada, J. Hamja / Eur. J. Pure Appl. Math, 16 (2) (2023), 1326-1341 1340 Corollary 15. Let G be a connected graph such that G has an ipds and Kn be a complete graph. Then γp0(G[Kn]) = { γp0(G), if n = 1 γip(G), if n > 1. Corollary 16. Let G be a connected graph such that G has an ipds and H be any graph described in Corollary 5. Then γp0(G[H]) = γip(G). The next result follows from Corollary 14 (ii). Corollary 17. Let G be a connected graph such that G has no ipds and H be any graph. Then G[H] is non-γp0-graphs. The next result follows from Corollary 14 (i). Corollary 18. Let G be a connected graph and H be any graph described in Corollary 6. Then G[H] is non-γp0-graphs. The next result follows from Corollary 13 and Corollary 14. Corollary 19. Let G be a connected graph such that G has an ipds. Then for n ≥ 4, G[Pn] and G[Cn] are non-γp0-graphs and for n = 2, 3, γp0(G[Pn]) = γp0(G[C3]) = γip(G). 7. Conclusion The paper has introduced the concept of perfect isolate dominating sets of some graphs resulting from some binary operations such as join, corona, and lexicographic product of two graphs. The existence of the perfect isolate dominating sets of some graphs and some binary operations are examine because not all graphs have this set. For future investigation, the authors recommend to explore this parameter to determine the exact values of some graphs and some binary operations that have not discussed in the study. Moreover, look for the relationship with other parameters of domination which are related to this parameter. Acknowledgements The authors would like to thank the anonymous referees for their comments and sug- gestions which led to the betterment of the paper. 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