Isolation of k-cliques

12/28/2018 ∙ by Peter Borg, et al. ∙ University of Malta King Mongkut's University of Technology Thonburi 0

For any positive integer k and any n-vertex graph G, let ι(G,k) denote the size of a smallest set D of vertices of G such that the graph obtained from G by deleting the closed neighbourhood of D contains no k-clique. Thus, ι(G,1) is the domination number of G. We prove that if G is connected, then ι(G,k) ≤n/k+1 unless G is a k-clique or k = 2 and G is a 5-cycle. The bound is sharp. The case k=1 is a classical result of Ore, and the case k=2 is a recent result of Caro and Hansberg. Our result solves a problem of Caro and Hansberg.

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1 Introduction

Unless stated otherwise, we use small letters such as to denote non-negative integers or elements of a set, and capital letters such as to denote sets or graphs. The set of positive integers is denoted by . For , the set is denoted by . Note that is the empty set . Arbitrary sets are assumed to be finite. For a set , the set of -element subsets of is denoted by (that is, ).

If is a subset of and is the pair , then is called a graph, is called the vertex set of and is denoted by , and is called the edge set of and is denoted by . A vertex of is an element of , and an edge of is an element of . We call an -vertex graph if . We may represent an edge by . If , then we say that is a neighbour of in (and vice-versa). For , denotes the set of neighbours of in , denotes , and denotes and is called the degree of in . For , denotes (the closed neighbourhood of ), denotes the graph (the subgraph of induced by ), and denotes (the graph obtained from by deleting ). We may abbreviate to . Where no confusion arises, the subscript is omitted from any of the notation above that uses it; for example, is abbreviated to .

If and are graphs, is a bijection, and , then we say that is a copy of , and we write . Thus, a copy of is a graph obtained by relabeling the vertices of .

For , the graphs and are denoted by and , respectively. A copy of is called a complete graph or an -clique. A copy of is called an -path or simply a path.

If and are graphs such that and , then is called a subgraph of , and we say that contains .

If is a set of graphs and is a copy of a graph in , then we call an -graph. If is a graph and such that contains no -graph, then is called an -isolating set of . Let denote the size of a smallest -isolating set of . The study of isolating sets was introduced recently by Caro and Hansberg [1]. It is an appealing and natural generalization of the classical domination problem [2, 3, 4, 5, 6, 7]. Indeed, is a -isolating set of if and only if is a dominating set of (that is, ), so is the domination number of (the size of a smallest dominating set of ). In this paper, we obtain a sharp upper bound for , and consequently we solve a problem of Caro and Hansberg [1].

We call a subset of a -clique isolating set of if contains no -clique. We denote the size of a smallest -clique isolating set of by . Thus, .

If are graphs such that for every with , then are vertex-disjoint. A graph is connected if, for every , contains a path with . A connected subgraph of is a component of if, for each connected subgraph of with , is not a subgraph of . Clearly, two distinct components of are vertex-disjoint.

For , let and . We have . If , then let . If , then let be copies of such that are vertex-disjoint, and let be the connected -vertex graph given by

Thus, is the graph obtained by taking and joining (a vertex of ) to each vertex of for each .

For with , let

For , let

In Section 2, we prove the following result.

Theorem 1.1

If is a connected -vertex graph, then, unless is a -clique or and is a -cycle,

Consequently, for any and ,

A classical result of Ore [8] is that the domination number of a graph with is at most (see [4]). Since the domination number is , it follows by Lemma 2.2 in Section 2 that Ore’s result is equivalent to the bound in Theorem 1.1 for . The case is also particularly interesting; while deleting the closed neighbourhood of a -isolating set yields the graph with no vertices, deleting the closed neighbourhood of a -isolating set yields a graph with no edges. In [1], Caro and Hansberg proved Theorem 1.1 for , using a different argument. Consequently, they established that . In the same paper, they asked for the value of . The answer is given by Theorem 1.1.

Corollary 1.2

For any ,

Proof. By Theorem 1.1, for any , , and, if is a multiple of , then . Thus,

2 Proof of Theorem 1.1

In this section, we prove Theorem 1.1. We start with two lemmas that will be used repeatedly.

If a graph contains a -clique , then we call a -clique of . We denote the set by .

Lemma 2.1

If is a vertex of a graph , then .

Proof. Let be a -clique isolating set of of size . Clearly, for each . Thus, is a -clique isolating set of . The result follows.  

Lemma 2.2

If are the distinct components of a graph , then .

Proof. For each , let be a smallest -clique isolating set of . Then, is a -clique isolating set of . Thus, . Let be a smallest -clique isolating set of . For each , is a -clique isolating set of , so . We have . The result follows. 

Proof of Theorem 1.1. We use induction on . If is a -clique, then . If and is a -cycle, then . Suppose that is not a -clique and that, if , then is not a -cycle. Suppose . If , then . If , then , so . If , then , so . Now suppose . If , then . Suppose . Let . Since is connected and is not a -clique, there exists some such that . Thus, as . If , then is a -clique isolating set of , so . Suppose . Let and . Then,

and . Let be the set of components of . If , then let . If , then let . By the induction hypothesis, for each . If , then, by Lemmas 2.1 and 2.2,

Suppose . For any and any , we say that is linked to if for some . Since is connected, each member of is linked to at least one member of . One of Case 1 and Case 2 below holds.

Case 1: For each , is linked to at least two members of . Let and such that is linked to . Let be the set of members of that are linked to only. Then,

and hence, by the induction hypothesis, each member of has a -clique isolating set with .

Let and . Then, has a component with , and the other components of are the members of . Let be a -clique isolating set of of size . Since is linked to , for some . If is a -clique, then let . If and is a -cycle, then let be one of the two vertices in , and let . We have and . Let . Since the components of are and the members of , we have , and, since , is a -clique isolating set of . Thus,

(1)

Subcase 1.1: is neither a -clique nor a -cycle. Then, by the induction hypothesis. By (1), .
Subcase 1.2: is a -clique. Since and , we have . If is a -clique, then let and . If and is a -cycle, then let be the set whose members are , , and the two neighbours of in , and let . Let . Let . Then, the components of are the components of and the members of .

If has no -clique, then, since , is a -clique isolating set of , and hence

This is the case if as we then have .

Suppose that and has a -clique . We have

(2)

Thus, and . Let and . Since is a vertex of each of the -cliques and ,

(3)

We have

(4)

Let . Then, . We have that the components of are (which is a clique or a path, depending on whether a -clique or a -cycle) and the members of , (by (2)), , and, by the definition of , for each . Thus, is connected, and, if , then is neither a clique nor a -cycle.

Suppose . By the induction hypothesis, . Let be a -clique isolating set of of size . By (3), is a -clique isolating set of . Thus, , and hence, by (4), .

Now suppose . Then, , so . Recall that either is a -clique or and is a -cycle.

Suppose that is a -clique. Then, . By (3), . Suppose . Then, , and hence is a -clique isolating set of . Thus, . Now suppose . Then, by (4), and . Let be the element of , and let . Since is a vertex of each of the -cliques and , . We have and . If , then is a -clique isolating set of , and hence . Suppose . Then, , , and are the -cliques with vertex sets , , and , respectively. Thus, , and contains the -cycle with edge set . Since is not a -cycle, for some . Since , is a -clique isolating set of , and hence .

Now suppose that and is a -cycle. Then, and for some . Recall that . Let . Since and are vertices of , . We have , , (as by (2)), and . If is or , then is or . If is or , then . Thus, is a -clique reducing set of , and hence .
Subcase 1.3: is a -cycle. If , then the result follows as in Subcase 1.1. Suppose . We have for some . Let . Recall that the components of are and the members of . Thus, is connected and .

Suppose that is not a -cycle. By the induction hypothesis, has a -clique isolating set with . Since , is a -clique isolating set of , so .

Now suppose that is a -cycle. Then, is a -clique and , where . Since and is a -cycle, . We have . If , then is a -clique isolating set of . If , then , so is a -clique isolating set of . Therefore, .

Case 2: For some and some , is linked to only. Let is linked to only and is linked to only. Let and . Since , . For each , for some . Let .

For each -clique , let . If , then, for each -cycle , let be one of the two vertices in , and let . Let . Then, is a -clique isolating set of . If , then , so . If and we let , then .

Let . Then, has a component with , and the other components of are the members of . By the induction hypothesis, for each . For each , let be a -clique isolating set of of size .

If is a -clique, then let . If and is a -cycle, then let be one of the two vertices in , and let . If neither is a -clique nor and is a -cycle, then, by the induction hypothesis, has a -clique isolating set with .

Let . By the definition of and , the components of are and the members of . Thus, is a -clique isolating set of since , , and is a -clique isolating set of . Let and . We have

If is a -clique, then . If and is a -cycle, then

If neither is a -clique nor and is a -cycle, then