On the k-Component Independence Number of a Tree

Let G be a graph and k≥ 1 be an integer. A subset S of vertices in a graph G is called a k-component independent set of G if each component of G[S] has order at most k. +e k-component independence number, denoted by αkc (G), is the maximum order of a vertex subset that induces a subgraph with maximum component order at most k. We prove that if a tree T is of order n, then αk(T)≥ (k/(k + 1))n. +e bound is sharp. In addition, we give a linear-time algorithm for finding a maximum k-component independent set of a tree.


Introduction
Let G � (V(G), E(G)) be a graph and k ≥ 1 be an integer and S ⊆ V(G). We use G[S] to denote the subgraph of G induced by S ⊆ V(G). We call S a k-component independent set of G if each component of G [S] has order at most k. A k-component independent set is maximum if G contains no larger k-component independent set and maximal if the set cannot be extended to a larger k-component independent set. e k-component independence number, denoted by α k c (G), is the cardinality of a maximum k-component independent set of G.
On the contrary, By the definition above, for any graph G of order n, where α(G) and β(G) are the ordinary independence number and vertex covering number of G. e k-component chromatic number of a graph G, denoted by χ k c (G), is the smallest number of colours needed in k-component coloring, a coloring of the vertices such that color classes are k-component independent sets. e notations α k c (G) and χ k c (G) come from [1]. e notion of k-component coloring is first studied by Kleinberg et al. [2]. It was extensively studied in the past two decades [3][4][5][6][7][8][9]. We refer to an excellent survey on this topic [10].
A notion, close to k-component vertex covering of a graph, is called the fragmentability of a graph, which was first introduced by Edwards and McDiarmid [11] when they were investigating the harmonious colorings of graphs. It was further studied in [12,13].
additional check that whether S is an independent set or not, for every S ⊆ V(G) with ⌊a/k⌋ ≤ |S| ≤ a, where a � α k c (G), contradicting the folklore that determining α(G) is NP-hard for a graph G, in general.

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In this note, we give a linear-time algorithm for finding a maximum k-component independence number of a tree. For any tree T of order n, there exists a vertex v such that T − v has d(v) − 1 components, each of which has order at most k, but the sum of their order is at least k.

An Lower Bound on
In particular, every nontrivial tree T has a vertex v such that all its neighbors but one are leaves.
Proof. Take a vertex r ∈ V(T) as the root of T, thereby p(x) and l(x) of x are uniquely defined for each Define a weight function w(u) � d(u) − 1, for each u ∈ V q−1 . If there is a vertex u q−1 ∈ V q−1 such that w(u q−1 ) ≥ k, then u q−1 is the vertex, as we desired.
Otherwise, w(u) ≤ k − 1, for every vertex u ∈ V q−1 . It follows that T − u has d(u) − 1 components, each of which has order at most k for each u ∈ V q−2 . Define w(x): � p(y)�x (w(y) + 1) for each x ∈ V q−2 .
If there is a vertex u q−2 ∈ V q−2 with w(u q−2 ) ≥ k, then u q−2 is the vertex v, as we want. Otherwise, w(u) ≤ k − 1, for every vertex u ∈ V q−2 . It follows that T − u has d(u) − 1 components, each of which has order at most k, for each u ∈ V q−3 . Define a weight function w(x): � p(y)�x (w(y) + 1), for each x ∈ V q−3 .
Repeat the procedure above; since n ≥ k + 1 is finite, there exists an integer i ∈ 0, 1, . . . , q − 1 such that there exists a vertex u q−i ∈ V q−i with w(u q−i ) ≥ k. It can be seen that u q−i is the vertex we required. □ Theorem 1. Let k ≥ 1 be an integer. For any tree T of order n, β k c (T) ≤ (n/(k + 1)). Equivalently, α k c (T) ≥ (k/(k + 1))n. e bound is sharp.
Proof. We use induction on n. If n ≤ k, then β k (T) � 0, and the result trivially holds. Now, assume that n ≥ k + 1. By Lemma 1, there exists a vertex v of T as the assertion in Lemma 1. Let T 1 , . . . , T d−1 , T d be all components of T − v such that e bound is achieved by the path P n of order n when n is divisible by k + 1.
□ By taking k � ⌊n/2⌋ in the above theorem, we have the following.
A path in a vertex-colored graph is called conflict-free if there is a color used on exactly one of its vertices. A vertexcolored graph is said to be conflict-free vertex-connected if any two vertices of the graph are connected by a conflict-free path. e conflict-free vertex-connection number, denoted by vcfc(G), is defined as the smallest number of colours required to make G conflict-free vertex connected. Li et al. [15] conjectured that, for a connected graph G of order n, vcfc(G) ≤ vcfc(P n ). Using Corollary 1, the authors of [14] are able to confirm the above conjecture. We refer to [16][17][18], for more recent results, on conflict-free vertex-connection of graphs. Next we give a linear time algorithm (Algorithm 1) for finding minimum k-component vertex covering of a tree.

Linear-Time Algorithm
Theorem 2. Every C returned by the algorithm is a minimum k-component vertex covering of T.

Proof. We prove it by the induction on v(T).
If v(T) ≤ k, C returned by the algorithm is the empty set and thus is a minimum k-component vertex covering of where v 1 is the first vertex added to C by the algorithm. By the choice of the algorithm, v 1 is a vertex with the property described in the assertion in Lemma 1. Let T 1 , . . . , T d−1 , T d be all components of T − v 1 such that d−1 i�1 v(T i ) ≥ k and v(T i ) ≤ k, for each i ∈ 1, . . . , d − 1 { }, where d � d(v 1 ). By the induction hypothesis, v 2 , . . . , v t is a minimum k-component vertex covering of T d . us, C is a k-component vertex covering of T.
Suppose C is not a minimum k-component vertex covering of T, and let C * be a minimum k-component vertex covering of T. It is clear that v 1 ∉ C * . Note that C * * � (C * ∩ V(T d )) ∪ v 1 is a k-component vertex covering of T. us, |C * ∩ V(T d )| ≥ | v 2 , . . . , v t | � |C| − 1.