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 [1]. The notion of -component coloring is first studied by Kleinberg et al. [2]. It was extensively studied in the past two decades [39]. We refer to an excellent survey on this topic [10].

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 [11] 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 [14]). 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. [15] conjectured that, for a connected graph of order , . Using Corollary 1, the authors of [14] are able to confirm the above conjecture. We refer to [1618], 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 [19] determined that the greatest number of maximum independent sets of a tree of order is,

More relevant work can be found in [2022]. 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.