EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 12, No. 4, 2019, 1371-1381 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Forcing Independent Domination Number of a Graph Cris L. Armada1, Sergio R. Canoy, Jr.2,3,∗ 1 Mathematics Department, Cebu Normal University, 6000 Cebu City, Philippines 2 Department of Mathematics and Statistics, College of Sciences and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines 3 Center for Graph Theory, Algebra, and Analysis, Premier Research Institute in Science and Mathematics, Mindanao State University-Iligan Institute of Technology, 9200 Iligan City, Philippines Abstract. In this paper, we obtain the forcing independent domination number of some special graphs. Further, we determine the forcing independent domination number of graphs under some binary operations such join, corona and lexicographic product of two graphs. 2010 Mathematics Subject Classifications: 05C69 Key Words and Phrases: Forcing, independent, domination, join, corona, lexicographic product 1. Introduction Let G = (V (G), E(G)) be a graph and v ∈ V (G). The open neighborhood of v in G is the set N(v) = {u ∈ V (G) : uv ∈ E(G)} and the closed neighborhood of v is the set N [v] = N(v) ∪ {v}. For X ⊆ V (G), the open neighborhood of X is the set N(X) = ∪v∈XNG(v) and its closed neighborhood is the set N [X] = N(X) ∪X. A set I ⊆ V (G) is an independent set of G if I ∩ N(I) = ∅. A set D ⊆ V (G) is a dominating set of G if N [D] = V (G). A set T ⊆ V (G) is an independent dominating set of G if T is both independent and dominating set. The independent domination number γi(G) of G is the minimum cardinality of an independent dominating set. If S is an independent dominating set with |S| = γi(G), then we call S a γi-set of G. A maximum independent set (α-set) is an independent set of largest possible size for a given graph G. This size, denoted by α(G), is called the independence number of G. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v12i4.3484 Email addresses: cris.armada@g.msuiit.edu.ph, armadac@cnu.edu.ph (C. Armada), serge_canoy@yahoo.com (S. Canoy) http://www.ejpam.com 1371 c© 2019 EJPAM All rights reserved. C. Armada, S. Canoy Jr. / Eur. J. Pure Appl. Math, 12 (4) (2019), 1371-1381 1372 Let I be a γi-set of a graph G. A subset D of I is said to be a forcing subset for I if I is the unique γi-set containing D. The forcing independent domination number of I is given by fγi(I) = min{|D| : D is a forcing subset for I}. The forcing independent domination number of G is given by fγi(G) = min{fγi(I) : I is a γi-set of G}. Let B be an α-set of a graph G. A subset P of B is said to be a forcing subset for B if B is the unique α-set containing P . The forcing independence number of B is given by fα(B) = min{|P | : P is a forcing subset for B}. The forcing independence number of G is given by fα(G) = min{fα(B) : B is an α-set of G}. Chartrand et. al [3] initiated the investigation on the relation between forcing and domination concepts in 1997 and used the term "forcing domination number". Independent domination under some binary operations such as corona and composition is studied by Canoy [2]. In 2013, Larson et. al [5] investigated the forcing independence number. In 2018, Canoy et. al [1] investigated the forcing domination number of graphs under some binary operations. Let G and H be two graphs. The join of G and H, denoted by G + H is the graph with vertex set V (G + H) = V (G) ∪ V (H) and edge set E(G+H) = E(G)∪E(H)∪ {uv : u ∈ V (G), v ∈ V (H)}. The corona G ◦H of G and H is the graph obtained by taking one copy of G and |V (G)| copies of H, and then forming the join 〈{v}〉+Hv = v +Hv, where Hv is a copy of H, for each v ∈ V (G). The lexicographic product (or composition) G[H] of G and H is the graph with V (G[H]) = V (G)× V (H), and (u, u′)(v, v′) ∈ E(G[H]) if and only if either uv ∈ E(G) or u = v and u′v′ ∈ E(H). Note that for each ∅ 6= C ⊆ V (G) × V (H), the G-projection and H-projection of C are, respectively, the sets CG = {x ∈ V (G) : (x, a) ∈ C for some a ∈ V (H)} and CH = {a ∈ V (H) : (y, a) ∈ C for some y ∈ V (G)}. Observe that any non-empty subset C of V (G)×V (H) can be written as C = ∪x∈S({x}×Tx) ⊆ V (G[H]), where S = CG ⊆ V (G) and Tx = {a ∈ CH : (x, a) ∈ C} for each x ∈ S. 2. Known Results Theorem 2.1. [4] For any graph G, d n 1+∆e ≤ γi(G) ≤ n−∆. Theorem 2.2. [4] The independent domination number of a cycle Cn with n ≥ 3, a path Pn with n ≥ 1 and complete bipartite graph Kr,s are given by i. γi(Pn) = γi(Cn) = dn 3 e, ii. γi(Kr,s) = min{r, s}. Theorem 2.3. [2] Let G be a connected graph and H be any graph. Then C ⊆ V (G ◦ H) is an independent dominating set in G ◦ H if and only if C ∩ V (G) is an independent set in G and C ∩ V (v+Hv) is an independent dominating set in v+Hv for every v ∈ V (G). C. Armada, S. Canoy Jr. / Eur. J. Pure Appl. Math, 12 (4) (2019), 1371-1381 1373 Theorem 2.4. [2] Let G be a connected graph of order n and H any graph with γi(H) 6= 1. If C ⊆ V (G ◦H) is a minimum independent dominating set in G ◦H, then C ∩ V (G) is a maximum independent set in G. Theorem 2.5. [2] Let G and H be nontrivial connected graphs. A subset C = ∪x∈S({x} × Tx) of V (G[H]), is an independent dominating set in G[H] if and only if S is an independent dominating set in G and Tx is an independent dominating set in H for every x ∈ S. Corollary 2.6. [2] Let G and H be nontrivial connected graphs. Then γi(G[H]) = γi(G)γi(H). Corollary 2.7. [2] Let G be a connected graph and Kn the complete graph of order n ≥ 1. Then γi(G[Kn]) = γi(G). 3. Main Results The next two results follow directly from the definition of forcing independent domination and Theorem 2.1. Remark 3.1. Let G be a graph. Then (i) fγi(G) = 0 if and only if G has a unique γi-set. (ii) fγi(G) = 1 if and only if G has at least two γi-sets and there exists a vertex which is contained in exactly one γi-set of G. Remark 3.2. Let G be any graph of order n. Then 0 ≤ fγi(G) ≤ n−∆. Theorem 3.3. Let G be a connected graph. Then fγi(G) = γi(G) if and only if for every γi-set S of G and for each x ∈ S, there exists yx ∈ V (G)\S such that [S\{x}] ∪ {yx} is a γi-set of G. Proof. Suppose that fγi(G) = γi(G). Let S be a γi-set of G. Then, by assumption, γi(G) = |S| = fγi(G), that is, S is the only forcing subset for itself. Let x ∈ S. Since S\{x} is not a forcing subset for S, there exists a yx ∈ V (G)\S such that [S\{x}] ∪ {yx} is a γi-set of G. Conversely, suppose that every γi-set S′ of G satisfies the given condition. Let S be a a γi-set of G such that fγi(G) = fγi(S). Suppose further that S has a forcing subset P with |P | < |S|, that is, S = P ∪K, where K = {x ∈ S : x /∈ P}. Pick x ∈ K. By assumption, there exists yx ∈ V (G)\S such that [S\{x}] ∪ {yx} = T is a γi-set of G. Hence, T = P ∪ R, where R = [K\{x}] ∪ {yx}, is a γi-set containing P , a contradiction. Hence, S is the only forcing subset for S. Therefore, fγi(G) = |S| = γi(G). Theorem 3.4. For any complete graph Kn with n ≥ 1 vertices, fγi(Kn) = { 0, n = 1, 1, n > 1. C. Armada, S. Canoy Jr. / Eur. J. Pure Appl. Math, 12 (4) (2019), 1371-1381 1374 Proof. Each vertex of Kn forms a γi-set of Kn, so by Remark 3.1, the result follows. Theorem 3.5. For any path Pn with n ≥ 1 vertices, fγi(Pn) = { 0, if n = 1 or n ≡ 0(mod 3), 1, otherwise. Proof. Suppose that Pn = [u1, u2, . . . , un]. By Theorem 2.2(i), γi(Pn) = dn 3 e. Note that P1 ∼= K1. Then fγi(P1) = 0 by Theorem 3.4. Next, let n ≥ 2 and consider the following cases: Case 1: n ≡ 0(mod 3) Let S = {u2, u5, u8, . . . , un−1} = {u3k−1 : k = 1, 2, . . . , n 3 }. Clearly, S ∩ N(S) = ∅ and |S| = dn 3 e. Since S is the only γi-set of Pn, fγi(Pn) = 0 by Remark 3.1(i). Case 2: n ≡ 1(mod 3) Let S1 = {u1} ∪ {u3k : k = 1, 2, . . . , n−1 3 }, S2 = {u1} ∪ {u3k+1 : k = 1, 2, . . . , n−2 3 } and Sk,p = {u1} ∪ {u3k+1 : k = 1, 2, . . . , n−4 3 } ∪ {u3p : k < p ≤ n−1 3 }. Then for all i ∈ {1, 2} and for all k, p with k ∈ {1, 2, . . . , n−4 3 } and k < p ≤ n−1 3 , Si ∩N(Si) = ∅ and |Si| = dn 3 e, that is, Si and Sk,p are γi-sets of Pn and u3 ∈ S1\(S2 ∪ Sk,p). Let S be a γi-set of Pn such that u2 ∈ S. Then u3 ∈ S1\S. Since S1 is the only γi-set containing u3, by Remark 3.1(ii), fγi(S1) = 1 = fγi(Pn). Case 3: n ≡ 2(mod 3) Then the set S1 = {u1, u4, u7, . . . , un−1} = {u3k+1 : k = 0, 1, . . . , n−2 3 } is the only γi-set of Pn that contains u1. Since S2 = {u2, u4, u7, . . . , un−1} is also a γi-set of Pn, it follows from Remark 3.1(ii) that fγi(S1) = 1 = fγi(Pn). Theorem 3.6. For any cycle Cn with n ≥ 3 vertices, fγi(Cn) = { 1, if n = 4 or n ≡ 0(mod 3), 2, otherwise. Proof. Suppose that Cn = [u1, u2, . . . , un, u1]. By Theorem 2.2(i), γi(Cn) = dn 3 e. If n = 4, then S1 = {u1, u3} and S2 = {u2, u4} are the only γi-sets of C4. Since S1 contains an element which is not in S2, fγi(S1) = fγi(C4) = 1 by Remark 3.1(ii). Next, let n ≥ 3, where n 6= 4, and consider the following cases: Case 1: n ≡ 0(mod 3) Let I1 = {u3k : k = 1, 2, . . . , n 3 }, I2 = {u3k+1 : k = 0, 1, . . . , n−3 3 }, and I3 = {u3k+2 : k = 0, 1, . . . , n−3 3 }. Then for all j ∈ {1, 2, 3}, Ij ∩N(Ij) = ∅ and |Ij | = dn 3 e. Thus, I1, I2 and I3 are the only γi-sets of Cn. Clearly, I1 contains an element which is not in I2 and I3, by Remark 3.1(ii), fγi(I1) = fγi(Cn) = 1. C. Armada, S. Canoy Jr. / Eur. J. Pure Appl. Math, 12 (4) (2019), 1371-1381 1375 Case 2: n ≡ 1(mod 3) Then clearly, S = {u1, u3}∪{u3k+2 : k = 1, 2, . . . , n−4 3 } is a γi-set of Cn. Clearly, S contains u1 and u5 where d(u1, u3) = d(u3, u5) = 2. Replacing any of the vertices u3k+2 (k 6= 1) to form another γi-set is not possible since d(u5, u8) = d(u1, un−2) = d(u3k+2, u3k+5) = 3 for all k ∈ {2, 3, . . . , n−7 3 }. Since u1 is also contained in the γi-set S′ = {u1, un−1} ∪ {u3k : k = 1, 2, . . . , n−4 3 }, no vertex of Cn is contained in a unique γi-set. Thus, fγi(S) ≥ 2. Clearly, {u1, u5} is uniquely contained in S. Therefore, fγi(S) = 2 = fγi(Cn). Case 3: n ≡ 2(mod 3) Suppose that n = 5. The γi-sets of C5 are S1 = {u1, u3}, S2 = {u1, u4}, S3 = {u2, u4}, S4 = {u2, u5} and S5 = {u3, u5}. Clearly, for each ui ∈ Sj where i, j ∈ {1, 2, 3, 4, 5}, there exists uk ∈ V (C5)\Sj such that [Sj\{ui}] ∪ {uk} is a γi-set of G. By Theorem 3.3, fγi(C5) = 2. Now, suppose that n > 5. Then S = {u1} ∪ {u3k : k = 1, 2, . . . , n−2 3 } is a γi-set of Cn. Clearly, S contains u1 and u3 where d(u1, u3) = 2. Since u1 is also contained in the γi-set S′ = {u1, u4}∪{u3k : k = 2, 3, . . . , n−2 3 }, no vertex of Cn is contained in a unique γi-set. Thus, fγi(S) ≥ 2. Since {u1, u3} is a forcing subset for S, fγi(S) = 2 = fγi(Cn). Theorem 3.7. Let G and H be any graphs. Then S0 ⊆ V (G+H) is a γi-set of G+H if and only if one of the following holds: (i) S0 is a γi-set of G and γi(G) < γi(H) (ii) S0 is a γi-set of H and γi(H) < γi(G) (iii) S0 is either a γi-set of G or H, and γi(H) = γi(G). In particular, γi(G+H) = min{γi(G), γi(H)}. Proof. Clearly, S ⊆ V (G+H) is an independent dominating set of G+H if and only if either S is an independent dominating set of G or S is an independent dominating set of H. In particular, γi(G+H) = min{γi(G), γi(H)}. Hence, S0 is a γi-set of G+H if and only if one of (i), (ii), and (iii) holds. Theorem 3.8. For any graphs G and H with γi(G) = γi(H), fγi(G+H) = { 1, if either G or H has a unique γi-set, min{fγi(G), fγi(H)}, otherwise. Proof. By Theorem 3.7, γi(G+H) = γi(G) = γi(H). Suppose that either G or H has a unique γi-set. W.l.o.g., suppose that G has a unique γi-set, say S. Then by Corollary 3.7, S and the γi-sets of H are γi-sets of G+H. Clearly, for any x ∈ S, {x} is uniquely contained in S and not in any γi-set of G+H. By Remark 3.1(ii), fγi(S) = 1 = fγi(G+H). C. Armada, S. Canoy Jr. / Eur. J. Pure Appl. Math, 12 (4) (2019), 1371-1381 1376 Suppose that both G and H have no unique γi-sets. We may assume that fγi(G) ≤ fγi(H). Since every γi-set of G and H is a γi-set of G + H, fγi(G) ≥ fγi(G + H). Now, let S0 be a γi-set of G + H such that fγi(G + H) = fγi(S0). If S0 ⊆ V (G), then S0 is a γi-set of G. Hence, fγi(G + H) = fγi(S0) ≥ fγi(G). If S0 ⊆ V (H), then S0 is a γi-set of H. Hence, fγi(G+H) = fγi(S0) ≥ fγi(H) ≥ fγi(G). Hence, in any case, fγi(G+H) = fγi(G). Theorem 3.9. For any graphs G and H with γi(G) 6= γi(H), fγi(G+H) =  0, if γi(G) < γi(H) and G has a unique γi-set or if γi(H) < γi(G) and H has a unique γi-set, fγi(G), if γi(G) < γi(H) and G has no unique γi-sets, fγi(H), if γi(H) < γi(G) and H has no unique γi-sets. Proof. Suppose that γi(G) < γi(H). By Theorem 3.7, γi(G + H) = γi(G). Suppose that G has a unique γi-set, say S. Then by Corollary 3.7, S is the only γi-set of G+H. By Remark 3.1(i), fγi(G + H) = 0. Now, suppose that G has no unique γi-sets. By Corollary 3.7, the γi-sets of G are also the γi-sets of G+H. Thus, fγi(G+H) = fγi(G). Similarly, if γi(H) < γi(G), then fγi(G + H) = 0 whenever H has a unique γi-set, and fγi(G+H) = fγi(H) whenever H has no unique γi-sets. The next results are direct consequences of Theorem 3.8 and Theorem 3.9 Corollary 3.10. For any graph H, fγ(K1 +H) = { 1, γi(H) = 1, 0, γi(H) > 1. Corollary 3.11. For the complete bipartite graph Kn,m such that n,m ≥ 1, fγi(Kn,m) = { 0, n 6= m, 1, n = m. Corollary 3.12. For the generalized fan Fn,m = Kn + Pm, where n ≥ 1 and m ≥ 2, fγi(Fn,m) = { 0, if either n < dm 3 e or n > dm 3 e with m ≡ 0(mod 3), 1, if either n = dm 3 e or n > dm 3 e with m 6≡ 0(mod 3). Corollary 3.13. For the fan Fn = K1 + Pn, where n ≥ 2, fγi(Fn) = { 0, n > 3, 1, n ≤ 3. C. Armada, S. Canoy Jr. / Eur. J. Pure Appl. Math, 12 (4) (2019), 1371-1381 1377 Corollary 3.14. For the generalized wheel Wn,m = Kn + Cm, where n ≥ 1 and m ≥ 3, fγi(Wn,m) =  0, if n < dm 3 e 1, if either n = dm 3 e or n > dm 3 e with m = 4 or m ≡ 0(mod 3), 2, if n > dm 3 e with m 6= 4 or m 6≡ 0(mod 3). Corollary 3.15. For the wheel Wn = K1 + Cn, where n ≥ 3, fγi(Wn) = { 0, n > 3, 1, n = 3. The following results are restatements of Theorems 2.3 and 2.4. Theorem 3.16. Let G be a connected graph of order n and let H be any graph. Then C ⊆ V (G◦H) is an independent dominating set in G◦H if and only if C = A∪( ⋃ v∈V (G)\A Sv), where A is an independent set (may be empty) of G and Sv is an independent dominating set of Hv for all v ∈ V (G)\A. Theorem 3.17. Let G be a connected graph of order n and let H be any graph with γi(H) = 1. Then C is a γi-set of G ◦ H if and only if C = A ∪ ( ⋃ v∈V (G)\A Sv) where A is an independent set of G and Sv is a γi-set of Hv for each v ∈ V (G)\A. In particular, γi(G ◦H) = n. Theorem 3.18. Let G be a connected graph of order n and let H be any graph with γi(H) ≥ 2. Then C is a γi-set of G ◦ H if and only if C = A ∪ ( ⋃ v∈V (G)\A Sv) where A is a maximum independent set of G and Sv is a γi-set of Hv for each v ∈ V (G)\A. In particular, γi(G ◦H) = α(G) + [n− α(G)]γi(H). Theorem 3.19. Let G be a connected graph of order n and let H be any graph with γi(H) = 1. Then fγi(G ◦H) = { γi(G), if H has a unique γi-set, n, otherwise. Proof. Since γi(H) = 1, by Theorem 3.17, γi(G ◦H) = n. Suppose that H has a unique γi-set, say P = {x}. For each v ∈ V (G), let Pv = {xv} ⊆ V (Hv) be such that 〈P 〉 ∼= 〈Pv〉, where 〈P 〉 is the subgraph induced by P . Let S = T ∪ U where T is a γi-set of G and U = {xv ∈ V (Hv) : xv ∈ Pv ∀v ∈ V (G)\T}. Clearly, S is a γi-set of G ◦ H. Since H has a unique γi-set and any vertex v ∈ V (G)\T is adjacent to a vertex in T , no element in U can be replaced by any vertex in v + V (Hv) for all v ∈ V (G)\T to form another γi-set of G ◦H. Hence, T is uniquely contained in S, that is, T is a forcing subset for S. C. Armada, S. Canoy Jr. / Eur. J. Pure Appl. Math, 12 (4) (2019), 1371-1381 1378 Therefore, fγi(S) ≤ |T |. Suppose that there exists B S such that fγi(S) = |B| < |T |. Suppose that B ∩ U 6= ∅, say w ∈ B ∩ U . Let v ∈ V (G)\T such that w = xv ∈ V (Hv). Pick y ∈ T ∩NG(v) and let T ′ = T\{y} and U ′ = U ∪ {xy}. Then S′ = T ′ ∪ U ′ is a γi-set of G ◦ H with S′ 6= S and B ⊆ S′, a contradiction. Hence, B T . Let z ∈ T\B and let Sz = (T\{z}) ∪ (U ∪ {xz}) where 〈{xz}〉 ∼= 〈x〉 and xz ∈ V (Hz). Then Sz is a γi-set of G ◦H, Sz 6= S and B ⊆ Sz. This is a contradiction since B is a forcing subset for S. Therefore, fγi(S) = |B| = |T | = γi(G). Now, suppose that H does not have a unique γi-set. Let C be a γi-set of G◦H and let S be a forcing subset for C. By Theorem 3.17, C = A∪ ( ⋃ v∈V \A Sv) where A is an independent set of G and Sv is a γi-set of Hv for each v ∈ V (G)\A. Suppose that S 6= C, say w ∈ C\S. Let z ∈ V (G) such that w ∈ V (z +Hz). If w = z, then w ∈ A. Let A′ = A\{w} and let Sw = {xw} be a γi-set of Hw. Then by Theorem 3.17, C ′ = A′ ∪ ( ⋃ v∈V \A′ Sv) is a γi-set of G ◦H with C ′ 6= C and S ⊆ C ′. If w 6= z, then Sz = {w} is a γi-set of Hz. Let S∗z = {w′} be a γi-set of Hz with w 6= w′. Then C∗ = A ∪ ( ⋃ v∈V \(A∪{z}) Sv) ∪ S∗z is a γi-set of G ◦H with C∗ 6= C and S ⊆ C∗. In either case, we get a contradiction. Thus, S = C and fγi(C) = |C| = n. Consequently, fγi(G ◦H) = n. Theorem 3.20. Let G be a connected graph of order n and let H be any graph with γi(H) 6= 1. Then fγi(G ◦H) = { fα(G), if H has a unique γi-set [n− α(G)][fγi(H)], if H has no unique γi-sets. In particular, fγi(G ◦H) = 0 if G has a unique α-set and H has a unique γi-set. Proof. Since γi(H) ≥ 2, by Theorem 3.18, γi(G ◦H) = α(G) + [n− α(G)]γi(H). Let T be a maximum independent set of G. Then |T | = α(G) and |V (G)\T | = n − α(G). Consider the following cases: Case 1: Suppose that H has a unique γi-set, say R. For each v ∈ V (G), let Rv ⊆ V (Hv) such that 〈Rv〉 ∼= 〈R〉. Suppose that G has a unique α-set, say D. Then by Theorem 3.18, C = D∪( ⋃ v∈V (G)\D Rv) is the unique γi-set of G◦H. Thus, by Remark 3.1(i), fγi(G◦H) = 0. Suppose that G does not have a unique α-set. Let A be an α-set of G and let DA be a forcing subset for A such that fα(G) = fα(A) = |DA|. Let C = A ∪ ( ⋃ v∈V (G)\A Rv). Then, by Theorem 3.18, C is a γi-set of G ◦H. Since each Hv has a unique γi-set Rv, it follows that DA is a forcing subset for C. Thus, fγi(G ◦H) ≤ fγi(C) ≤ |DA| = fα(G). C. Armada, S. Canoy Jr. / Eur. J. Pure Appl. Math, 12 (4) (2019), 1371-1381 1379 Next, let C0 be a γi-set of G ◦ H such that fγi(G ◦ H) = fγi(C0). Then C0 = A0 ∪ ( ⋃ v∈V (G)\A0 Rv), where A0 is an α-set of G. Let S be a forcing subset for C0 such that fγi(C0) = |S|. Since each Hv has a unique γi-set Rv, S ⊆ A0. Since S is a forcing subset for C0, S must be a forcing subset for the α-set A0. Thus, fγi(G ◦H) = fγi(C0) = |S| ≥ fα(A0) ≥ fα(G). Therefore, fγi(G ◦H) = fα(G). Case 2: Suppose that H does not have a unique γi-set. Let Q be a γi-set of H with fγi(H) = fγi(Q) and let PQ be a forcing subset for Q with fγi(Q) = |PQ|. For each v ∈ V (G), let Qv ⊆ V (Hv) and PQv ⊆ Qv such that 〈Qv〉 ∼= 〈Q〉 and 〈PQv〉 ∼= 〈PQ〉. Let AQ be an α-set of G. Then by Theorem 3.18, CQ = AQ ∪ ( ⋃ v∈V (G)\AQ Qv) is a γi-set of G ◦ H. Let S = ⋃ v∈V (G)\AQ PQv . Then S is a forcing subset for CQ. Thus, fγi(G ◦H) ≤ fγi(CQ) ≤ |S| = [n− α(G)]fγi(H). Next, let C ′ be a γi-set of G ◦ H such that fγi(G ◦ H) = fγi(C ′). Then by Theorem 3.18, let C ′ = A′ ∪ ( ⋃ v∈V (G)\A′ Rv), where A′ is an α-set of G and Rv is a γi-set of Hv for each v ∈ V (G)\A′. Let S′ be a forcing subset for C ′ such that fγi(C ′) = |S′|. Suppose that there exists w ∈ V (G)\A′ such that S′ ∩ Rw = Sw is not a forcing subset for Rw. Let R′w be a γi-set of Hw with R′w 6= Rw. Then C” = A′ ∪ ( ⋃ v∈V (G)\(A′∪{w}) Rv) ∪ R′w is a γi-set of G ◦H with C” 6= C ′ and S′ ⊆ C”, a contradiction. Thus, Sv = S′ ∩ Rv is a forcing subset for Rv for each v ∈ V (G)\A′. Note that Sv 6= ∅ for each v ∈ V (G)\A′. Let S0 = ⋃ v∈V (G)\A′ Sv. Then fγi(G ◦H) = |S′| ≥ |S0| = ∑ v∈V (G)\A′ |Sv| ≥ ∑ v∈V (G)\A′ fγi(H) = [n− α(G)]fγi(H). Therefore, fγi(G ◦H) = [n− α(G)]fγi(H). Theorem 3.21. Let G and H be connected graphs. Then fγi(G[H]) = { fγi(G), if H has a unique γi-set, [γi(G)][fγi(H)], if H has no unique γi-sets. In particular, fγi(G[H]) = 0 if G and H have unique γi-sets. Also, fγi(G[H]) = γi(G[H]) if fγi(H) = γi(H). C. Armada, S. Canoy Jr. / Eur. J. Pure Appl. Math, 12 (4) (2019), 1371-1381 1380 Proof. By Corollary 2.6, γi(G[H]) = γi(G)γi(H). Consider the following cases: Case 1: Suppose that H has a unique γi-set, say R. Let S be a γi-set of G and let U be a forcing subset for S such that fγi(G) = fγi(S) = |U |. By Theorem 2.5, C = S × R is a γi-set of G[H]. Now, since U is a forcing subset for S, Uc = U × {c} is a forcing subset for C for each c ∈ R. Hence, for each c ∈ R, fγi(G[H]) ≤ fγi(C) ≤ |Uc| = |U | = fγi(G). Let C0 = S0 × R be a γi-set of G[H] such that fγi(G[H]) = fγi(C0). By Theorem 2.5, S0 is a γi-set of G. Let Q0 be a forcing subset for C0 with fγi(C0) = |Q0|. Let Q0 = ∪x∈K [{x} × Tx], where K ⊆ S0 and Tx ⊆ R for all x ∈ K. Since Q0 is a forcing subset for C0, it follows that K is a forcing subset for S0. Choose any x ∈ K and a ∈ Tx. Then Qa = K × {a} ⊆ Q0. Thus, fγi(G[H]) = fγi(C0) = |Q0| ≥ |Qa| = |K| ≥ fγi(S0) ≥ fγi(G). Therefore, fγi(G[H]) = fγi(G). In particular, if G has a unique γi-set, then fγi(G[H]) = 0. Case 2: Suppose that H does not have a unique γi-set. Let R0 be a γi-set of H and T0 be a forcing subset for R0 such that fγi(H) = fγi(R0) = |T0|. Let S0 be a γi-set of G. For each x ∈ S0, let Tx = T0 and Rx = R0. By Theorem 2.5, C = ∪x∈S0 [{x} × Rx] is a γi-set of G[H]. Then C0 = ∪x∈S0 [{x} × Tx] = S0 × T0 is a forcing subset for C. Hence, fγi(G[H]) ≤ fγi(C0) ≤ |C0| = |S0 × T0| = γi(G)fγi(H). Next, let C = ∪x∈S [{x} × Tx] be a γi-set of G[H] and let D be a forcing subset for C such that fγi(G[H]) = fγi(C) = |D|. Then by Theorem 2.5, S is a γi-set of G and Tx is a γi-set of H for each x ∈ S. Let D = ∪x∈K [{x} × Ex] where K ⊆ S and Ex ⊆ Tx for each x ∈ S. Suppose that K 6= S, say v ∈ S\K. Let T ′v be a γi-set of H with T ′v 6= Tv. Then C ′ = ∪x∈S\{v}[{x} × Tx] ∪ {v} × T ′v is a a γi-set of G[H] and D ⊆ C ′ 6= C, a contradiction. Thus, K = S and since D be a forcing subset for C, Ex must be a forcing subset for Tx for each x ∈ S. Hence, fγi(G[H]) = |D| = ∑ x∈S |Ex| ≥ γi(G)fγi(H). Therefore, fγi(G[H]) = γi(G)fγi(H). In particular, if fγi(H) = γi(H), then fγi(G[H]) = γi(G)γi(H) = γi(G[H]). Since the complete graph Kn has no unique γi-sets and fγi(Kn) = 1 except when n = 1, the following result is immediate from Theorem 3.21. Corollary 3.22. Let G be a connected graph and Kn the complete graph of order n ≥ 1. 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