#### Abstract

Let be a graph and be an integer. A subset of vertices in a graph is called a -component independent set of if each component of has order at most . The -component independence number, denoted by , is the maximum order of a vertex subset that induces a subgraph with maximum component order at most . We prove that if a tree is of order , then . The bound is sharp. In addition, we give a linear-time algorithm for finding a maximum -component independent set of a tree.

#### 1. Introduction

Let be a graph and be an integer and . We use to denote the subgraph of induced by . We call a -component independent set of if each component of has order at most . A -component independent set is maximum if contains no larger -component independent set and maximal if the set cannot be extended to a larger -component independent set. The -component independence number, denoted by , is the cardinality of a maximum -component independent set of .

On the contrary, is called a -component vertex covering of if is a -component independent set of . A -component vertex covering is minimum if contains no smaller -component vertex covering and minimal if the set cannot be contained in a smaller -component vertex covering. The -component vertex covering number, denoted by , is the cardinality of a minimum -component vertex covering of .

By the definition above, for any graph of order ,where and are the ordinary independence number and vertex covering number of .

The -component chromatic number of a graph , denoted by , is the smallest number of colours needed in -component coloring, a coloring of the vertices such that color classes are -component independent sets. The notations and come from . The notion of -component coloring is first studied by Kleinberg et al. . It was extensively studied in the past two decades . We refer to an excellent survey on this topic .

A notion, close to -component vertex covering of a graph, is called the fragmentability of a graph, which was first introduced by Edwards and McDiarmid  when they were investigating the harmonious colorings of graphs. It was further studied in [12, 13].

Proposition 1. In general, deciding is NP-hard for a graph .

Proof. Note that for any graph . If is determined by polynomial-time algorithm, then is determined by at most,additional check that whether is an independent set or not, for every with , where , contradicting the folklore that determining is NP-hard for a graph , in general.

In this note, we give a linear-time algorithm for finding a maximum -component independence number of a tree.

#### 2. An Lower Bound on for a Tree

Let be a graph and . The order of is denoted by . We use denote the set of neighbors of a vertex of . The degree of , denoted by , is the number of edges incident with in . Furthermore, the two symbols are simply denoted by and , respectively. For a subset of the vertex set of , denotes the subgraph of induced by .

Let be a tree with root . The level of a vertex in is the length of the path . Each vertex on the path is called an ancestor of , and each vertex of, which is an ancestor, is a descendant of . An ancestor or descendant of a vertex is proper if it is not the vertex itself. The immediate proper ancestor of a vertex is its predecessor or parent, denoted . Let denote the subtree of with the vertex set which consists of the set of descendants of .

Lemma 1. Let and be two integers with .

For any tree of order , there exists a vertex such that has components, each of which has order at most , but the sum of their order is at least .

In particular, every nontrivial tree has a vertex such that all its neighbors but one are leaves.

Proof. Take a vertex as the root of , thereby and of are uniquely defined for each . Let . Let with , for each .
Define a weight function , for each . If there is a vertex such that , then is the vertex, as we desired.
Otherwise, , for every vertex . It follows that has components, each of which has order at most for each . Define for each .
If there is a vertex with , then is the vertex , as we want. Otherwise, , for every vertex . It follows that has components, each of which has order at most , for each . Define a weight function , for each .
Repeat the procedure above; since is finite, there exists an integer such that there exists a vertex with . It can be seen that is the vertex we required.

Theorem 1. Let be an integer. For any tree of order , . Equivalently, . The bound is sharp.

Proof. We use induction on . If , then , and the result trivially holds. Now, assume that . By Lemma 1, there exists a vertex of as the assertion in Lemma 1. Let be all components of such that and , for each , where . By the induction hypothesis, . So,The bound is achieved by the path of order when is divisible by .

By taking in the above theorem, we have the following.

Corollary 1. (see ). For a tree of order , ; equivalently, there exists a vertex such that each component of has order at most .

A path in a vertex-colored graph is called conflict-free if there is a color used on exactly one of its vertices. A vertex-colored graph is said to be conflict-free vertex-connected if any two vertices of the graph are connected by a conflict-free path. The conflict-free vertex-connection number, denoted by , is defined as the smallest number of colours required to make conflict-free vertex connected. Li et al.  conjectured that, for a connected graph of order , . Using Corollary 1, the authors of  are able to confirm the above conjecture. We refer to , 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.

 Input: a tree with a vertex as its root. Output: a minimum -component vertex covering of . (1) (2) while , do (3) set , , for each (4) if (5) choose a vertex (6) if , set , go to Step 2 (7) if , put , and put , go to Step 4 (8) else (9) replace (10) for each , go to Step 4 (11) end if (12) end while (13) return

#### 3. Linear-Time Algorithm

Theorem 2. Every returned by the algorithm is a minimum -component vertex covering of .

Proof. We prove it by the induction on .
If , returned by the algorithm is the empty set and thus is a minimum -component vertex covering of since .
Next, assume that . Let , where is the first vertex added to by the algorithm. By the choice of the algorithm, is a vertex with the property described in the assertion in Lemma 1. Let be all components of such that and , for each , where .
By the induction hypothesis, is a minimum -component vertex covering of . Thus, is a -component vertex covering of .
Suppose is not a minimum -component vertex covering of , and let be a minimum -component vertex covering of . It is clear that . Note that is a -component vertex covering of . Thus, .
Since , . It follows that , a contradiction.

In the execution of the algorithm, each vertex is explored at most once to check whether or not. So, the running time of the algorithm is .

#### 4. Further Research

Zito  determined that the greatest number of maximum independent sets of a tree of order is,

More relevant work can be found in . Naturally, one asks the following questions:(1)What is the largest number of maximum (or maximal) -component independent sets on a tree of order ?(2)What is the largest number of maximum (or maximal) -component independent sets on a (or connected) graph of order ?(3)What is the largest number of maximum (or maximal) -component independent sets on a (or connected) graph of order vertices and edges?

#### Data Availability

No data were used to support the findings of the study.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Authors’ Contributions

All authors contributed equally and significantly in conducting this research work and writing this paper.

#### Acknowledgments

This research was supported by the Key Laboratory Project of Xinjiang (2018D04017), NSFC (no. 12061073), and Xinjiang Education Department XJEDU2019I001.