Independent coalition in graphs: existence and characterization

An independent coalition in a graph $G$ consists of two disjoint sets of vertices $V_1$ and $V_2$ neither of which is an independent dominating set but whose union $V_1 \cup V_2$ is an independent dominating set. An independent coalition partition, abbreviated, $ic$-partition, in a graph $G$ is a vertex partition $\pi= \lbrace V_1,V_2,\dots ,V_k \rbrace$ such that each set $V_i$ of $\pi$ either is a singleton dominating set, or is not an independent dominating set but forms an independent coalition with another set $V_j \in \pi$. The maximum number of classes of an $ic$-partition of $G$ is the independent coalition number of $G$, denoted by $IC(G)$. In this paper we study the concept of $ic$-partition. In particular, we discuss the possibility of the existence of $ic$-partitions in graphs and introduce a family of graphs for which no $ic$-partition exists. We also determine the independent coalition number of some classes of graphs and investigate graphs $G$ of order $n$ with $IC(G)\in\{1,2,3,4,n\}$ and the trees $T$ of order $n$ with $IC(T)=n-1$.


Introduction
Let G = (V, E) denote a simple graph of order n with vertex set V = V (G) and edge set E = E(G).The open neighborhood of a vertex v ∈ V is the set N (v) = {u|{u, v} ∈ E}, and its closed neighborhood is the set N [v] = N (v) ∪ {v}.Each vertex of N (v) is called a neighbor of v, and the cardinality of N (v) is called the degree of v, denoted by deg(v) or deg G (v).A vertex v of degree 1 is called a pendant vertex or leaf, and its neighbor is called a support vertex.A vertex of degree n − 1 is called a full vertex while a vertex of degree 0 is called an isolated vertex.The minimum and maximum degree of G are denoted by δ(G) and ∆(G), respectively.For a set S of vertices of G, the subgraph induced by S is denoted by G[S].For two sets X and Y of vertices, let [X, Y ] denote the set of edges between X and Y .If every vertex of X is adjacent to every vertex of Y , we say that [X, Y ] is full, while if there are no edges between them, we say that [X, Y ] is empty.A subset V i ⊆ V is called a singleton set if |V i | = 1, and is called a non-singleton set if |V i | ≥ 2. The join G + H of two disjoint graphs G and H is the graph obtained from the union of G and H by adding every possible edge between the vertices of G and the vertices of H.We denote the family of paths, cycles, complete graphs and stars of order n by P n , C n , K n and K 1,n−1 , respectively, and the complete k-partite graph with partite sets of order n 1 , n 2 , . . ., n k , by K n 1 ,...,n k .A double star with respectively p and q leaves connected to each support vertex is denoted by S p,q .The complete graph K 3 is called a triangle, and a graph is triangle-free if it has no K 3 as an induced subgraph.The girth of a graph with a cycle is the length of its shortest cycle.For a graph G, the girth of G is denoted by g(G).For a graph G of order n, let G denote the complement of G with V (G) = V (G) and E(G) = E(K n ) − E(G) [13].
A set S ⊆ V in a graph G = (V, E) is called a dominating set if every vertex v ∈ V is either an element of S or is adjacent to an element of S. A set S ⊆ V is called an independent set if its vertices are pairwise nonadjacent.The vertex independence number, denoted by α(G), is the maximum cardinality of an independent set of G.An independent dominating set in a graph G is a set that is both independent an dominating.
A partition of the vertices of G into dominating sets (independent dominating sets) is called a domatic partition (idomatic partition).The maximum number of classes of a domatic partition (idomatic partition) of G is called the domatic number (idomatic number) of G, denoted by d(G) (id(G)).The concepts of domination and domatic partition and their variations have been studied widely in the literature.See, for example, [1,2,3,9,10,11,12].
Definition 1.1.[6] A coalition in a graph G consists of two disjoint sets of vertices V 1 , V 2 ⊂ V , neither of which is a dominating set but whose union V 1 ∪ V 2 is a dominating set.We say that the sets V 1 and V 2 form a coalition, and are coalition partners.
π is either a singleton dominating set, or is not a dominating set but forms a coalition with another set V j in π.The coalition number C(G) equals the maximum order k of a c-partition of G, and a c-partition of G having order C(G) is called a C(G)-partition.
Herein we will focus on coalitions involving independent dominating sets in graphs.In other words, we will study the concepts of independent coalition and independent coalition partition which have been introduced in [6] as an area for future research.We begin with the following definitions.
Definition 1.3.An independent coalition in a graph G consists of two disjoint sets of independent vertices V 1 and V 2 , neither of which is an independent dominating set but whose union V 1 ∪ V 2 is an independent dominating set.We say the sets V 1 and V 2 form an independent coalition, and are independent coalition partners (or ic-partners).
Definition 1.4.An independent coalition partition, abbreviated ic-partition, in a graph G is a vertex partition π = {V 1 , V 2 , . . ., V k } such that every set V i of π is either a singleton dominating set, or is not an independent dominating set but forms an independent coalition with another set V j ∈ π.The independent coalition number IC(G) equals the maximum number of classes of an ic-partition of G, and an ic-partition of G having order IC(G) is called an IC(G)-partition.Definition 1.5.[8] Let G be a graph of order n with vertex set This paper is organized as follows.Section 1 is devoted to terminology and definitions.We discuss the possibility of the existence of ic-partitions in graphs and derive some bounds on independent coalition number in Section 2. In Section 3, we determine the independent coalition number of some classes of graphs.The graphs G with IC(G) ∈ {1, 2, 3, 4} are investigated in Section 4. In Section 5, we characterize triangle-free graphs G with IC(G) = n and trees T with IC(T ) = n − 1.Finally, we end the paper with some research problems.

Independent coalition partition: existence and bound
This section is divided into two subsections.In the first subsection, we show that not all graphs admit an ic-partition, and in the second subsection, we present some bounds on IC(G) whenever the graph G admits an ic-partition.

Existence
In the following definition, we construct graphs with arbitrarily large order for which no icpartition exists.Definition 2.1.Let B be the set of all graphs obtained from the complete graph K n , (n ≥ 4) with the vertices v i , (1 ≤ i ≤ n), and two additional vertices v n+1 , v n+2 such that v n+1 and v n+2 are adjacent to v n , and v n+1 is adjacent to v n−1 .Figure 1 illustrates such a graph for n = 4. Proof.Suppose, to the contrary, that G has an ic-partition π.The vertices v 1 , v 2 , . . ., v n−1 are pairwise adjacent, so they must be in different classes.Further, v n is a full vertex, so it must be in a singleton class.Since v n−1 is adjacent to all vertices except v n+2 , and {v n−1 , v n+2 } dominates G, it follows that {v n−1 } ∈ π.Further, since {v n−1 } can only form an independent coalition with {v n+2 }, it follows that {v n+2 } ∈ π.If {v n+1 } ∈ π, then π is a singleton partition.In this case, {v n+1 } has no ic-partner, a contradiction.Hence, {v n+1 } / ∈ π.It follows that π consists of a non-singleton set {v n+1 , v i } such that v i ∈ {v 1 , v 2 , . . ., v n−2 }, and n singleton sets.Assume, without loss of generality, that {v n+1 , v 1 } ∈ π.Now for each 2 ≤ i ≤ n − 2, the set {v i } has no ic-partner, a contradiction.

Bounds
Definition 1.4 implies that an ic-partition of a graph G is also a c-partition.Further, we note that an ic-partition of G is a proper coloring as well.Hence, we have the following two sharp bounds on IC(G).To see the sharpness of them, consider the complete graph K n .
Observation 2.3.Let G be a graph.If G has an ic-partition, then IC(G) ≤ C(G).Furthermore, this bound is sharp.
Observation 2.4.Let G be a graph.If G has an ic-partition, then IC(G) ≥ χ(G).Furthermore, this bound is sharp.
Given a connected graph G and an ic-partition π of it, the following theorem shows that each set in π admits at most ∆(G) ic-partners.
Theorem 2.5.Let G be a connected graph with maximum degree ∆(G), and let π be an icpartition of G.If X ∈ π, then X is in at most ∆(G) independent coalitions.Furthermore, this bound is sharp.
Proof.Let π be an ic-partition of G, and let X be a set in π.If X is a dominating set, then it has no ic-partner.Hence, we may assume that X does not dominate G. Let x be a vertex that is not dominated by X.Now every ic-partner of X must dominate x, that is, it must contain a vertex in N [x].Hence, there are at most |N [x]| ≤ ∆(G) + 1 sets in π that can form an independent coalition with X.Now we show that X cannot form an independent coalition with ∆(G) + 1 sets.Suppose, to the contrary, that X has ∆(G)+1 ic-partners . Consider an arbitrary vertex v ∈ U (say v ∈ V i ) and an arbitrary set V j such that 1 ≤ j ≤ ∆(G) + 1 and j = i.Since X ∪ V j dominates G and [X, V i ] is empty, it follows that v has a neighbor in V j .Choosing V j arbitrarily, we conclude that deg G ′ (v) ≥ ∆(G), and so To prove the sharpness, let G be the graph that is obtained from the complete graph K n with vertices v i , (1 ≤ i ≤ n), and a path P 2 = (a, b), where b is adjacent to v 1 .Let A = {a}, B = {b} and V i = {v i }, for 1 ≤ i ≤ n.One can observe that ∆(G) = n and that the singleton partition Note that the bound presented in Theorem 2.5 does not hold for disconnected graphs.As a counterexample, consider the graph G = K 2 ∪ K 2 and the singleton partition π 1 of it.On can verify that π 1 is an ic-partition of G such that each set in π 1 has two ic-partners, while ∆(G) = 1.
The next bound relates independent coalition number of a graph to its idomatic number.As we will see in the proof of Theorem 2.6, any graph admitting an idomatic partition has an ic-partition.However, the converse is not necessarily true.For example, the singleton partition of the cycle C 5 is an ic-partition of it, while C 5 has no idomatic partition.Or the cycle C 7 has the ic-partition π = {{v 1 , v 5 }, {v 2 , v 4 }, {v 3 }, {v 6 }, {v 7 }}, while it has no idomatic partition.Theorem 2.6.Let G be a connected graph, and let r ≥ 0 be the number of full vertices of G.If G admits an idomatic partition, then IC(G) ≥ 2id(G) − r.
Proof.Let F = {v 1 , v 2 , . . ., v r } be the set of full vertices of G, and let π = {V 1 , V 2 , . . ., V id(G) } be an idomatic partition of G of order id(G).Note that each full vertex must be in a singleton set of π.Without loss of generality, assume that v i ∈ V i , for each 1 ≤ i ≤ r.It follows that for each r + 1 ≤ i ≤ k, we have |V i | ≥ 2. Now for each r + 1 ≤ i ≤ k, we partition V i into two nonempty subsets V i,1 and V i,2 .Note that no proper subset of V i is a dominating set.Thus, neither V i,1 nor V i,2 is an independent dominating set, and so V i,1 and V i,2 are ic-partners.It follows that the partition 3 Independent coalition number for some classes of graphs Let us begin this section with some routine results.
For complete multipartite graphs, the following result is obtained.
Proof.Let π = {V 1 , V 2 , . . ., V k } be the partition of G into its partite sets.Assume, without loss of generality, that the sets V i , for 1 ≤ i ≤ m, are those containing full vertices.Now for each m + 1 ≤ i ≤ k, we partition V i into two sets V i,1 and V i,2 .Observe that V i,1 and V i,2 are ic-partners, and so the partition }} is an ic-partition of G of order 2k − m.Thus, IC(G) ≥ 2k − m.Now let π ′′ be an ic-partition of G.We note that π ′′ has the following properties: • For any set S ∈ π ′′ , all vertices in S are in the same partite set of G.
• For any set V i ∈ π, the vertices in V i are in at most two sets of π ′′ .Hence, we have IC(G) ≤ 2k − m, and so IC(G) = 2k − m.
Case 1. π consists of a set (name A) of cardinality 3 and five singleton sets.Since γ i (P 8 ) = 3, each singleton set must be an ic-partner of A. On the other hand, Theorem 2.5 implies that A has at most two ic-partners, a contradiction.
Case 2. π consists of two sets of cardinality 2 and four singleton sets.Since γ i (P 8 ) = 3, each singleton set must be an ic-partner of a set of cardinality 2. Therefore, using Theorem 2.5, we deduce that for any two ic-partners C and D, it holds that |C ∪ D| = 3.On the other hand, v 3 and v 6 are not present in any independent dominating set of cardinality 3, a contradiction.The partition {{v 1 , v 3 , v 6 }, {v 2 , v 7 }, {v 8 }, {v 4 }, {v 5 }} is an ic-partition of P 8 .Therefore, IC(P 8 ) = 5.
Case 1. π consists of a set (name A) of cardinality 4 and five singleton sets.Since γ i (P 9 ) = 3, each singleton set must be an ic-partner of A. On the other hand, by Theorem 2.5, A has at most two ic-partners, a contradiction.
Case 2. π consists of a set (name A) of cardinality 3, a set (name B) of cardinality 2 and four singleton sets.Since γ i (P 9 ) = 3, no two singleton sets in π are ic-partners.Furthermore, by Theorem 2.5, A has at most two ic-partners, so at least two singleton sets of π must be ic-partners of B, which is impossible, as P 9 has a unique independent dominating set of cardinality 3.
Case 3. π consists of three sets of cardinality 2, and three singleton sets.We note that each singleton set in π must be an ic-partner of a set of cardinality 2, which is impossible, as P 9 has a unique independent dominating set of cardinality 3.
Finally, Let n ≥ 10.Consider the path P n with V (P n ) = {v 1 , v 2 , . . ., v n }.By Lemma 3.5 and Observation 2.3, we have IC(P n ) ≤ 6.Now we consider the sets } is an ic-partition of P n , where V 3 and V 4 are ic-partners of V 1 , and V 5 and V 6 are ic-partners of V 2 .So the proof is complete.Lemma 3.9.For any cycle C n with n ≥ 8 and n ≡ 0 (mod 2), it holds that IC(C n ) = 6.
, where V 3 and V 4 are ic-partners of V 1 , and V 5 and V 6 are ic-partners of V 2 .Hence, by Lemma 3.8 and Observation 2.3, we have Lemma 3.10.For any cycle C n with n ≥ 8 and n ≡ 0 (mod 3), it holds that IC(C n ) = 6.
and V i,2 .Observe that V i,1 and V i,2 are ic-partners.Hence, by Lemma 3.8 and Observation 2.3, we have IC(C n ) = 6.Lemma 3.11.For any cycle C n with n ≥ 8 and n ≡ 5 (mod 6), it holds that IC(C n ) = 6.
One can observe that π is an ic-partition of C n , where A 1 and A 2 are ic-partners of A, and B 1 and B 2 are ic-partners of B. Now using Lemma 3.8 and Observation 2.3, we have IC(C n ) = 6.Lemma 3.12.For any cycle C n with n ≥ 8 and n ≡ 1 (mod 6), it holds that IC(C n ) = 6.
One can observe that π is an ic-partition of C n , where A 1 and A 2 are ic-partners of A, and B 1 and B 2 are ic-partners of B. Now using Lemma 3.8 and Observation 2.3, we have IC(C n ) = 6.Theorem 3.13.For the cycle C n , Proof.If 1 ≤ n ≤ 6, then it is easy to check that IC(C n ) = n.Now assume n = 7.Consider the cycle C 7 with V (C 7 ) = {v 1 , v 2 , v 3 , v 4 , v 5 , v 6 , v 7 }.First we show that IC(C 7 ) = 6.Suppose, to the contrary, that IC(C 7 ) = 6.Let π be an IC(C 7 )-partition.We note that π consists of five singleton sets and a set of cardinality 2 (name A).By Theorem 2.5, A has at most two ic-partners.Hence, π contains two singleton sets that are ic-partners, which contradicts the fact that γ i (C 7 ) = 3.The partition {{v 1 , v 3 }, {v 5 }, {v 6 }, {v 4 , v 7 }, {v 2 }} is an ic-partition of C 7 , so IC(C 7 ) = 5.Furthermore, by Lemmas 3.9, 3.10, 3.11 and 3.12 we have IC(C n ) = 6, for n ≥ 8.

Graphs with small independent coalition number
In this section we investigate graphs G with IC(G) ∈ {1, 2, 3, 4}.We will make use of the following two lemmas.Lemma 4.1.Let G be a graph of order n containing r ≥ 1 full vertices, and let Let G be a graph containing a nonempty set of isolated vertices I.If IC(G) ≥ 3, then for any IC(G)-partition π, there is a set V r ∈ π such that V r = I.
Proof.First we show that all vertices in I are in the same set of π.Suppose, to the contrary, that there are sets V i ∈ π and V j ∈ π such that both V i and V j contain isolated vertices.Let V k ∈ π be an arbitrary set in π such that V k / ∈ {V i , V j }. (Since IC(G) ≥ 3, such a set exists).Then V k has no ic-partner, a contradiction.Now let V r be the set in π containing isolated vertices.Further, let v be an arbitrary vertex in V r , and let u ∈ V (G) be an arbitrary vertex such that u = v.If u ∈ V r , then u is not adjacent to v. Otherwise, the set in π containing u is an ic-partner of V r , which again implies that u is not adjacent to v. Hence, we have deg(v) = 0. Choosing v arbitrarily, we conclude that V r = I.
(2) If G ≃ K 2 , then we clearly have IC(G) = 2. Now assume G ≃ K n , for some n ≥ 2. Let π be an ic-partition of G.Note that no more than two sets in π contain isolated vertices, for otherwise, no two sets in π are ic-partners.Thus, |π| ≤ 2. Partitioning vertices of G into two nonempty sets yields an ic-partition of G. Hence, IC(G) = 2. Conversely, suppose that IC(G) = 2. Let π = {V 1 , V 2 } be an IC(G)-partition.If both V 1 and V 2 are singleton dominating sets, then G ≃ K 2 .Hence, we may assume that at least one of them (say V 1 ) is not a singleton dominating set.It follows that V 2 is not a singleton dominating set either, for otherwise, G is a star, and so by Observation 3.2, we have IC(G) = 3.Hence, V 1 and V 2 are ic-partners, and so Proof.Observations 3.1 and 3.2 imply that IC(K 3 ) = 3 and that IC(K 1,n−1 ) = 3, respectively.Now let G ∈ B 2 .Let I be the set of isolates vertices of G, and let {H 1 , H 2 } be a partition of G − I into its partite sets.We observe that the partition {I, H 1 , H 2 } is an ic-partition of G, so IC(G) ≥ 3. Now we show that IC(G) = 3. Suppose, to the contrary, that IC(G) ≥ 4. Let π = {V 1 , V 2 , . . ., V k } be an IC(G)-partition.By Lemma 4.2, we have I ∈ {V 1 , V 2 , . . ., V k }.Assume, without loss of generality, that I = V 1 .Now for each 2 ≤ i ≤ k, V i forms an independent coalition with V 1 , and so V i dominates H. Hence, the partition {V 2 , . . ., V k } is an idomatic partition of H, which contradicts the assumption.Hence, IC(G) = 3.Conversely, let G be a graph with IC(G) = 3, and let π = {V 1 , V 2 , V 3 } be an IC(G)-partition.We consider four cases depending on the number of full vertices of G.
Case 1. G has three full vertices.In this case, the sets V 1 , V 2 and V 3 are all singleton dominating sets, so G ≃ K 3 .
Case 2. G has two full vertices.Note that this case never occurs.Case 3. G has one full vertex.Let v 1 be the full vertex of G. Lemma 4.1 implies that Case 4. G has no full vertex.Let I be the set of isolated vertices of G. First we note that V 1 , V 2 and V 3 are not pairwise ic-partners, for otherwise, we have G ≃ K n , and so by Proposition 4.3, we have IC(G) = 2, a contradiction.Hence, π contains a set (say V 1 ) that forms an independent coalition with V 2 and V 3 , while V 2 and V 3 are not ic-partners.Therefore, each vertex in V 1 is an isolated vertex, so it follows from Lemma 4.2 that I = V 1 .Further, the sets V 2 and V 3 are independent dominating sets of We consider two cases.
Case 1. G has a full vertex.Let v 1 be a full vertex of G. Lemma 4.1 implies that IC(G−v 1 ) = 3.Thus, by Proposition 4.7, we have Case 2. G has no full vertex.First assume that G contains a nonempty set I of isolated vertices.Then, by Lemma 4.2, we have I ∈ π.Without loss of generality, assume that I = V 4 .Now for each 1 ≤ i ≤ 3, V i must form an independent coalition with I. Thus, is a 3-partite graph with id(U ) ≥ 3. Since IC(G) = 4, the case id(U ) > 3 is impossible.Hence, id(U ) = 3, and so G ∈ B 3 .Now assume that G contains no isolated vertex.Since G has neither full vertices nor isolated vertices, each set of π has either one or two ic-partners.If there is a set of π, (say V 1 ) having one ic-partner, (say V 2 ), then it follows that V 3 and V 4 are ic-partners, and so G is a bipartitie graph with partite sets V 1 ∪ V 2 and V 3 ∪ V 4 .Otherwise, assume, without loss of generality, that V 2 and V 3 are ic-partners of V 1 .It follows that V 4 has an ic-partner in {V 2 , V 3 }.By symmetry, we may assume that V 4 and V 3 are ic-partners.Then G is again a bipartitie graph with partite sets V 1 ∪ V 2 and V 3 ∪ V 4 .Now using Theorem 2.6, we have id(G) = 2, and so G ∈ B 1 .

Graphs with large independent coalition number
Our main goal in this section is to investigate structure of graphs G of order n with IC(G) = n, under specified conditions.In addition, we will characterize all trees T of order n with IC(T ) = n − 1.Let us begin with an observation that characterizes all disconnected graphs G of order n with IC(G) = n.Observation 5.1.Let G be a disconnected graph of order n.Then IC(G) = n if and only if G ≃ K s ∪ K r , for some s ≥ 1, and r ≥ 1.Now we introduce two sufficient conditions for a graph G of order n to have independent coalition number n. Observation 5.2.If G is a graph of order n with α(G) = 2, then IC(G) = n.

Triangle-free graphs G with IC(G) = n
In this subsection, we characterize graphs G of order n with g(G) = 4 and IC(G) = n.This will lead to characterization of all triangle-free graphs G of order n with IC(G) = n.We will make use the following lemmas.
Lemma 5.7.Let G be a triangle-free graph of order n with IC(G) = n.Then g(G) ≤ 6.
Proof.Let G be a graph of order n with IC(G) = n, and suppose, to the contrary, that g(G) ≥ 7. Let C ⊆ G be a cycle of order g(G).Consider an arbitrary vertex v ∈ V (C).Note that γ i (C) ≥ 3, and so {v} is not an ic-partner of any set {u} ⊂ V (C).Therefore, it must be an ic-partner of a set {u} ⊆ V (G) \ V (C).It follows that, {u} dominates V (C) \ N c [v], which implies that G contains triangles, a contradiction.Lemma 5.8.Let G be a graph of order n with g(G) = 6.Then IC(G) = n if and only if G ≃ C 6 .
Proof.Let G be a graph of order n with g(G) = 6.If G ≃ C 6 , then by Theorem 3.13, we have IC(G) = 6.Conversely, assume that IC(G) = n.Let C ⊆ G be a cycle of order 6, and suppose, to the contrary, that If {v} is an ic-partner of a set {u} ⊂ V (C), then {v} must dominate V (C) \ N c [u], which implies that G contains triangles, a contradiction.Otherwise, {v} must be an ic-partner of a set {u} ⊂ V (G) \ V (C).Now since {u, v} dominates C, it follows that G contains triangles, or induces cycles of order 4, a contradiction.
Our next result can be established almost the same way as Lemma 5.8, so we state it without proof.Lemma 5.9.Let G be a graph of order n with g(G) = 5.Then IC(G) = n if and only if G ≃ C 5 .
In order to characterize graphs G of order n with IC(G) = n and g(G) = 4, we need the following definitions.Definition 5.10.Let K 0 represent a bipartite graph with partite sets H 1 = {v 1 , v 2 , v 3 , v 4 } and H 2 = {u 1 , u 2 , u 3 , u 4 } such that for each 1 ≤ i ≤ 4, v i is adjacent to all vertices in H 2 , except u i .(see Figure 2).Definition 5.11.Let K represent a family of 4-partite graphs with partite sets Figure 3 illustrates such a graph for k = 3.  Case 2. {x} and {z} are not ic-partners.Let {e} be an ic-partner of {x}.Since {x, e} dominates G and z is not adjacent to x , it must be adjacent to e. Let A = N (x) \ {y, t} and B = N (e) \ {z}.It is not difficult to verify that A ∩ B = ∅.Now if A = ∅, then {z} cannot form an independent coalition with any other set, so A = ∅.Let {f } ⊆ A be an ic-partner of {z}.We note that if a set {g} forms an independent coalition with {y}, then g ∈ B. Further, if a set {h} forms an independent coalition with {t} then h ∈ B. Let {g} and {h} be ic-partners of {y} and {t}, respectively.Observe that {g} = {h}.Now let A ′ = A \ {f } and B ′ = B \ {g, h}.There exist the following subcases.
Subcase 2.2.A ′ = ∅ and B ′ = ∅.Let v ∈ B ′ .One can verify that {v} cannot form an independent coalition with any other set.Thus, this case is impossible.
Subcase 2.3.A ′ = ∅ and B ′ = ∅.Let v ∈ A ′ .One can verify that {v} cannot form an independent coalition with any other set.Thus, this case is impossible.• G[{x, y, z, t, e, f, g, h}] is a bipartite graph with partite sets V 1 = {x, z, g, h} and V 2 = {y, t, e, f }, which is isomorphic to K 0 , ] is a bipartite graph with partite sets A ′ and B ′ such that deg Hence, G ∈ K and the proof is complete.

Trees T with IC(T ) = n − 1
The following theorem characterizes all trees T of order n with IC(T ) = n − 1.
Theorem 5.14.Let T be a tree of order n.Then IC(T ) = n−1 if and only if T ∈ {P 5 , P 6 , S 1,2 , K 1,3 }.Proof.By Theorem 3.6, we have IC(P 5 ) = 4 and IC(P 6 ) = 5.Further, by Observation 3.2, we have IC(K 1,3 ) = 3 and by Observation 3.3, we have IC(S 1,2 ) = 4. Conversely, let T be a tree of order n with IC(T ) = n − 1, where x is a leaf , and y is the support vertex of x.Define A = N (y) \ {x} and B = V (G) \ ({x, y} ∪ A).Further, let π be an IC(T )-partition.Note that π contains a set of cardinality 2 (say V 1 = {u, v}) and n − 2 singleton sets.Since x and y are adjacent, we have V 1 = {x, y}.Note as well that any set in π must be an ic-partner of the set containing x, or the set containing y, to dominate x.We consider two cases.
Case 2. B = ∅.Since T is connected, we have A = ∅.We divide this case into some subcases.Subcase 2.1.u ∈ A and v ∈ B. We first show that |A| = 1.Suppose, to the contrary, that |A| ≥ 2. Then there is a vertex z ∈ A such that z = u.Since z and y are adjacent, {z} cannot be

Proposition 4 . 3 .
Let G be a graph of order n.Then 1. IC(G) = 1 if and only