Upper bounds for inverse domination in graphs

07/12/2019 ∙ by Elliot Krop, et al. ∙ Clayton State University Auburn University 0

In any graph G, the domination number γ(G) is at most the independence number α(G). The Inverse Domination Conjecture says that, in any isolate-free G, there exists pair of vertex-disjoint dominating sets D, D' with |D|=γ(G) and |D'| ≤α(G). Here we prove that this statement is true if the upper bound α(G) is replaced by 3/2α(G) - 1 (and G is not a clique). We also prove that the conjecture holds whenever γ(G)≤ 5 or |V(G)|≤ 16.

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1. Introduction

In this paper all graphs are simple. A dominating set for a graph is a set of vertices such that every vertex of either lies in or has a neighbor in . The domination number of , written , is the size of a smallest dominating set in . Note that a maximum independent set is a dominating set, so , where is the independence number of .

If a graph has no isolates and is a minimum dominating set in , then is also a dominating set in (owing to the minimality of ); this was first observed by Ore [10]. In general we say that a dominating set is an inverse dominating set for a graph if there is some minimum dominating set such that . A graph with isolates cannot have an inverse dominating set, but otherwise, given Ore’s observation, we can define the inverse domination number of a graph , written , as the smallest size of an inverse dominating set in . The Inverse Domination Conjecture asserts that for every isolate-free .

The Inverse Domination Conjecture originated with Kulli and Sigarkanti [9], who in fact provided an erroneous proof. Discussion of this error and further consideration of the conjecture first appeared in a paper of Domke, Dunbar, and Markus [3]. It has since been shown by Driscoll and Krop [4] that the weaker bound of holds in general, and Johnson, Prier and Walsh [7] showed that the conjecture itself holds whenever . Johnson and Walsh [8] have also proved two fractional analogs of the conjecture, and Frendrup, Henning, Randerath and Vestergaard [5] have shown that the conjecture holds for a number of special families, including bipartite graphs and claw-free graphs.

In this paper we prove two main results in support of the Inverse Domination Conjecture. The first is an improvement on the approximation to the conjecture.

Theorem 1.1.

If is a graph with no isolated vertices and is not a clique, then .

Note that if is a clique and , then trivially , which is why we must exclude cliques in Theorem 1.1.

Our second main result improves the range of for which the conjecture is known.

Theorem 1.2.

If is a graph with no isolated vertices and , then .

As a corollary of Theorem 1.2 we are also able to obtain the following.

Corollary 1.3.

If is a graph with no isolated vertices and , then .

It is worth noting that Asplund, Chaffee, and Hammer [2] have formulated a stronger form of the Inverse Domination Conjecture. In the strengthened version one requires, for every minimum dominating set , the existence of a dominating set with and . It is not hard to see that our proof for Theorem 1.1 also works for this stronger conjecture. However, the same is not true for Theorem 1.2, where we pick our minimum dominating set very carefully.

The rest of the paper is organized as follows. In Section 2 we introduce the notion of an independent set of representatives, or ISR, and explore the connections between ISRs and inverse domination. (In this section, we also obtain, as a corollary, the inequality for graphs without isolated vertices, where is the largest number of vertices in an induced bipartite subgraph of .) In Section 3 we prove Theorem 1.1. In Section 4 we leverage the machinery of Section 2 to prove Theorem 1.2 and Corollary 1.3.

2. ISRs and Inverse Domination

If is a collection of sets, a set of representatives for is a set such that for each . If is a graph and are subsets of , an independent set of representatives, or ISR, for is a set of representatives for the sets that is also an independent set in . A partial ISR for is an ISR for any subfamily of .

Several authors have proved various sufficient conditions guaranteeing the existence of ISRs; many of the proofs are topological in nature. See [1] for a collection of such results. A fundamental result on ISRs is the following sufficient condition due to Haxell [6]. In what follows, given a graph and a set , denotes the subgraph of induced by . Given a collection of sets and , we write for the union .

Theorem 2.1 (Haxell [6]).

Let be a graph and let be a partition of . If, for all ,

then has an independent set such that for each (that is, has an ISR).

Our basic idea for using Theorem 2.1 to obtain results on inverse domination is to apply it to a specific partition of vertices outside (where is a minimum dominating set), namely to what we’ll call a standard partition.

Let be a graph and suppose that are disjoint sets of vertices where dominates . The standard partition of , subject to a given ordering of , is the partition with

where indicates those neighbors of that are in . Consider a minimum dominating set , and the standard partition of with respect to any ordering of . If this partition has an ISR, then the ISR is an independent set disjoint from that dominates . Expanding this independent set to a maximal independent set in would give an independent dominating set disjoint from , implying that . However, we cannot always find an ISR for a standard partition of . Instead, we obtain more technical results.

In the following, given disjoint sets and , we write for the set . When , we’ll denote the unique element of by .

Theorem 2.2.

Let be a graph, let be a minimum dominating set in , and let be a maximal independent set in . Let be any ordering of , and let be the standard partition of subject to this ordering. Then there exist two partial ISRs of such that and .

Proof.

Let be a graph consisting of two disjoint copies of , and let be a partition of obtained by letting each consist of both copies of each vertex in .

We will use Theorem 2.1 to obtain an ISR of . Let be any subset of , and let . We will show that .

Observe that consists of two disjoint copies of the subgraph , so that any dominating set in must dominate each of those copies. If , then let be a minimum dominating set of . We can partition into , where dominates one copy of and dominates the other copy. Without loss of generality , and since , this implies . Let be the set of vertices in corresponding to the vertices of , and let . We know that dominates , and moreover since and dominates , we see that is a dominating set of . Since , this contradicts the minimality of .

Thus has some ISR . We can partition where consists of the -vertices in one copy of and consists of the -vertices in the other copy of . Now and are each independent subsets of , and since is an ISR we see that and . ∎

As an immediate and useful corollary to Theorem 2.2, we get the following.

Corollary 2.3.

Let be a graph, let be a minimum dominating set in , and let be a maximal independent set in . Let be any ordering of , and let be the standard partition of subject to this ordering. Then has a partial ISR of size at least .

Observe that if is a minimum dominating set in a graph without isolates, then each vertex in has a neighbor in . These neighbors can be used to help build inverse dominating sets, and our first use of this will be in the following corollary.

Corollary 2.4.

Let be a graph without isolated vertices and let be a minimum dominating set in . If is the largest number of vertices in an induced bipartite subgraph of , then .

Proof.

Let be a maximal independent set in , and let be partial ISRs as in Theorem 2.2. As and are each independent and , induces a bipartite subgraph of . Since , the set dominates . Expand to a maximal set inducing a bipartite subgraph.

The maximality of implies that dominates . Let , so that dominates . Observe that still induces a bipartite graph, so that . On the other hand, each vertex has some neighbor . Augmenting by adding in such a vertex for each yields a inverse dominating set of size at most , which is at most . ∎

3. Proof of Theorem 1.1

Theorem 3.1.

Let be a graph, and let be a minimum dominating set in . There is a set such that is a dominating set in and .

Proof.

Let be a maximal independent set in , and write as . Let be the standard partition of .

Let be a largest possible partial ISR for . By Corollary 2.3, we have . Expand to a maximal independent set in . The set dominates every vertex of and at least vertices of . We now expand to dominate the rest of .

Let . Observe that is an independent set, so . Expand to a set by adding an arbitrary -neighbor of for each ; we have . Next, expand to a set by adding an arbitrary -neighbor of for each ; note that , so . As and is an integer, this implies that

Since is a dominating set in , the theorem is proved. ∎

The following lemma is more general than is necessary for proving Theorem 1.1, but stating it in this generality will be useful for later results.

Lemma 3.2.

If a graph has a minimum dominating set and an independent set such that dominates , then .

Proof.

Let and let . Expand to a maximal independent set of . Now dominates . Let be the set of vertices in not dominated by . Observe that , since by hypothesis dominates . Hence is an independent set, so that .

Since is a minimum dominating set of and has no isolated vertices, each vertex of has a neighbor outside of . Let be the vertex set obtained from by adding in, for each , a neighbor of outside of . Now is a dominating set in and . Hence . ∎

The proof of Theorem 1.1 now follows easily. If has a minimum dominating set that is independent, then we can choose to vacuously meet the hypothesis of Lemma 3.2, and hence . Otherwise, , so by Theorem 3.1, we have

4. Proof of Theorem 1.2

Our proof of Theorem 1.2 relies on a careful choice of minimum dominating set. For shorthand, it will be convenient to speak of the independence number of a dominating set to refer to the independence number of the induced subgraph , and likewise to write for . We will consider a dominating set in a graph to be optimal if it is of minimum size and, among minimum-size dominating sets, has greatest independence number and, subject to that, has the fewest edges in the induced subgraph . In order to build inverse dominating sets in a graph , we previously used the fact that any vertex in a minimum dominating set has a neighbor in (provided is isolate-free). In some arguments, it is helpful if such a neighbor is private with respect to ; that is, if we are able to choose with . In fact, the choice of a private neighbor for is always possible when is a minimum dominating set, unless is isolated in . The following lemma tells us that if is optimal, we can improve on this.

Lemma 4.1.

Let be an isolate-free graph and let be an optimal dominating set in . If is not an isolated vertex in , then has at least private neighbors with respect to .

Proof.

Let be the subgraph of induced by the private neighbors of . We in fact show . Suppose to the contrary that has a dominating vertex . Let . Every vertex of is either itself, hence dominated by , or a private neighbor of , hence dominated by , or a vertex of that is not a private neighbor of , hence dominated by . Thus, is a dominating set. Furthermore, as was a private neighbor of , the vertex is an isolated vertex in . In particular, for any maximum independent set in , we see that is also a maximum independent set in , so has at least as large an independence number as did. As is isolated in but was not isolated in , we see that has fewer edges than , contradicting the optimality of . ∎

Lemma 4.2.

Let be an isolate-free graph and let be an optimal dominating set in . Suppose that the number of isolates in is . Then either has an independent set such that dominates , or all of the following are true:

  1. ,

  2. , and

  3. .

Proof.

Assuming that has no such independent set , we prove each part of the conclusion separately.

(1) If is an independent set, then taking gives the desired independent set. Hence , and we may choose a vertex that is not isolated in . If , then letting be a private neighbor of and taking gives the desired independent set.

Hence we may assume that . Let be an ordering of with independent, and let be the standard partition of with respect to this ordering.

If there is a pair of nonadjacent vertices , , then taking yields an independent set such that dominates . Otherwise, there is a complete bipartite graph between and . Taking and to be private neighbors of and respectively, we see that is a dominating set in having independence number , contradicting the optimality of .

(2) Let be the set of isolated vertices in . Notice that if , then is a dominating set of size less than , which is impossible. Hence, . We count as well as and then apply Lemma 4.1, which implies that .

(3) Suppose . By (1) we get , so in fact . Moreover, by (1), this means that , so .

Write with isolated in , and let be the standard partition of with respect to this ordering. Suppose first that there is an independent set in hitting at least three of the sets . Then define to be ; note is independent. Out of the four vertices in , at most one is not dominated by . However, if such a vertex exists, then we can add it to as well, without violating independence. Thus, we may assume that no such set exists.

If the pair is joined by a complete bipartite graph, then we may take and to be private neighbors of and respectively. Now

is a dominating set of of size containing the independent set

of size , contradicting the optimality of .

Otherwise, there is a pair of nonadjacent vertices and . Since, by assumption, this pair cannot be extended to an independent set that also hits one of the sets or , we see that dominates . Thus

is a dominating set in containing the independent set

contradicting the optimality of . ∎

In the remainder of the section we will prove the inverse domination conjecture for graphs with . In light of the following lemma, it will suffice to prove the conjecture for graphs with domination number exactly .

Lemma 4.3.

Let be a positive integer. If for every isolate-free graph with , then for every isolate-free graph with .

Proof.

Let be an isolate-free graph with , and let . Let be the disjoint union of and copies of . Now , so by hypothesis, . In particular, in we can choose a minimum dominating set and a second disjoint dominating set with . Observe that and must each contain one vertex from every added copy of . Hence, letting and , we see that and . Furthermore, and are dominating sets in . Hence, . ∎

We wish to strengthen the conclusion of Theorem 2.2 by eliminating the maximal independent set inside , and instead finding a pair of ISRs that jointly dominate the entire minimum dominating set . When and , we are able to do this.

Lemma 4.4.

Let be an optimal dominating set in an isolate-free graph . Suppose that , that , and that has no isolated vertices. Then there is an ordering of and a pair of independent sets and such that is an ISR for and is an ISR for , where is the standard partition of with respect to this ordering.

Proof.

Choose so that is an independent set, if possible. (Thus, only if is a clique.) Note that since , the set contains a maximal independent set in , hence dominates . This implies that there are at least edges from to the rest of .

First we argue that there is an independent set with . If not, then and are joined by a complete bipartite graph. Let and be private neighbors of and respectively. Observe that is a dominating set of . Furthermore, there are no edges between and . This implies that , contradicting the optimality of . (Note that since .)

Now, since is a minimal dominating set of , there is some vertex not dominated by . As dominates , we have . Choose to be a neighbor of in . Let , and let and be the remaining vertices of , ordered arbitrarily. Observe that is an ISR for in the standard partition of with respect to this ordering. It remains to find the desired .

We claim that there are nonadjacent vertices each with . If not, then and are joined by a complete bipartite graph. Let and be private neighbors of and respectively. Now is a dominating set in . Furthermore, since is a dominating set in , there are at least two edges in the cut , while by contrast there are no edges joining with . Hence , contradicting the optimality of . (Again since .) ∎

Theorem 4.5.

If is an isolate-free graph with , then has a minimum dominating set such that .

Proof.

Let be an optimal dominating set in . By Lemma 3.2 and by parts (1) and (3) of Lemma 4.2, we may assume that and that has no isolated vertices. In particular, since is not an independent set, we have , a fact we will use later.

By Lemma 4.4, we see that there is an ordering of and a pair of independent sets such that is an ISR for and is an ISR for , where is the standard partition of for the given ordering. Among all such pairs , choose and to minimize the number of edges from to .

If has a partial ISR of size , then we immediately get the desired conclusion: taking to be such an ISR, we see that dominates all of except possibly for a single vertex , so we win by letting (or ) and applying Lemma 3.2.

Thus, has no partial ISR of size , which implies that is a maximal partial ISR of this family, and so dominates .

Let be the set of vertices in that are not dominated by . If then we immediately have the desired conclusion, as is an inverse dominating set of size . Thus we may assume that is a nonempty subset of , and in particular, .

Write with . We claim that if intersects for some , then the corresponding vertex is not adjacent to any vertex of . Otherwise, let , and let . Now is an ISR of and, since is not dominated by , there are fewer edges between and than there were between and . This contradicts the choice of , establishing the claim.

In particular, the above claim implies that , since if , then taking distinct , we see that is a partial ISR of having size , contradicting our earlier claim that the largest such partial ISR has size .

Let be the unique index in . Let . We next claim that any vertex of not dominated by is adjacent to all of . Otherwise, let be such a vertex that is not adjacent to all of , with .

Let be a vertex of not adjacent to . If , then let . Now is an independent set, since and neither nor is dominated by (by choice of and because ). As this implies that is a partial ISR of having size , contradicting the earlier claim that the largest such ISR has size . If instead , then taking gives the same contradiction.

Hence, any vertex of not dominated by is adjacent to all of . If there is any vertex of not dominated by , then let be such a vertex; now is an inverse dominating set of size , where , and we are done. Hence, we may assume that dominates .

In this case, let . Since dominates , we see that is a dominating set of . Since , the set is an independent set: if were adjacent to , this would imply , contradicting , and likewise for . This contradicts the optimality of . ∎

Corollary 4.6.

If then .

Proof.

Let be some graph with , and let be an optimal dominating set in . Let be the number of isolated vertices in . By Lemma 3.2, there cannot be any independent set such that dominates , so by Lemma 4.2, we have:

  1. , and

  2. .

By Theorem 4.5 and Lemma 4.3, we have , so that . If then (2) yields . Otherwise, , and then (2) combined with (3) yields . ∎

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