On Total Dominating Graphs
Abstract
For a graph , the total dominating graph is the graph whose vertices correspond to the total dominating sets of that have cardinality at most ; two vertices of are adjacent if and only if the corresponding total dominating sets of differ by either adding or deleting a single vertex. The graph is used to study the reconfiguration problem for total dominating sets: a total dominating set can be reconfigured to another by a sequence of single vertex additions and deletions, such that the intermediate sets of vertices at each step are total dominating sets, if and only if they are in the same component of . Let be the smallest integer such that is connected for all .
We investigate the realizability of graphs as total dominating graphs. For the upper total domination number , we show that any graph without isolated vertices is an induced subgraph of a graph such that is connected. We obtain the bounds for any connected graph of order , characterize the graphs for which either bound is realized, and determine and .
Keywords: Total domination; Total domination reconfiguration problem; Total dominating graph
AMS Subject Classification Number 2010: 05C69
1 Introduction
A total dominating set (TDS) of a graph without isolated vertices is a set such that every vertex of is adjacent to a vertex in . If no proper subset of is a TDS of , then is a minimal TDS (an MTDS) of . Every graph without isolated vertices has a TDS, since is such a set. The total domination number of , denoted by , is the minimum cardinality of a TDS. The upper total domination number of , denoted by , is the maximum cardinality of an MTDS. A TDS of size is called set of , and an MTDS of size is called a set.
For a given threshold , let and be total dominating sets of order at most of . The total dominating set reconfiguration (TDSR) problem asks whether there exists a sequence of total dominating sets of starting with and ending with , such that each total dominating set in the sequence is of order at most and can be obtained from the previous one by either adding or deleting exactly one vertex. This problem is similar to the dominating set reconfiguration (DSR) problem, which is PSPACEcomplete even for planar graphs, bounded bandwidth graphs, split graphs, and bipartite graphs, while it can be solved in linear time for cographs, trees, and interval graphs [10].
The DSR problem naturally leads to the concept of the dominating graph introduced by Haas and Seyffarth [8] as follows. If is a graph and a positive integer, then the dominating graph of is the graph whose vertices correspond to the dominating sets of that have cardinality at most , two vertices of being adjacent if and only if the corresponding dominating sets of differ by either adding or deleting a single vertex. The DSR problem therefore simply asks whether two given vertices of belong to the same component of . The HaasSeyffarth paper [8] stimulated the work of Alikhani, Fatehi and Klavžar [1], Mynhardt, Roux and Teshima [15], Suzuki, Mouawad and Nishimura [16], as well as their own followup paper [9].
The study of dominating graphs was further motivated by similar studies of graph colourings and other graph problems, such as independent sets, cliques and vertex covers – see e.g. [2, 3, 4, 13, 14] – and by a general goal to further understand the relationship between the dominating sets of a graph. Motivated by definition of dominating graph, we define the total dominating graph of as follows.
Definition 1
The total dominating graph of is the graph whose vertices correspond to the total dominating sets of that have cardinality at most . Two vertices of are adjacent if and only if the corresponding total dominating sets of differ by either adding or deleting a single vertex. For , we abbreviate to , and to .
In studying the TDSR problem, it is therefore natural to determine conditions for to be connected. We begin the study of this problem in Section 2. To this purpose we define to be the smallest integer such that is connected for all , and note that exists for all graphs without isolated vertices because is connected.
We introduce our notation in Section 1.1 and provide background material on total domination in Section 1.2. For instance, we characterize graphs such that . In Section 3 we determine and ; interestingly, it turns out that , making the only known graph for which is disconnected. The main result for cycles requires four lemmas, which we state in Section 3 but only prove in Section 5 to improve the flow of the exposition. In Section 4 we study the realizability of graphs as total dominating graphs. We show that the hypercubes and stars are realizable for all , that and are the only realizable cycles, and that and are the only realizable paths. Section 6 contains a list of open problems and questions for future consideration.
1.1 Notation
For domination related concepts not defined here we refer the reader to [11]. The monograph [12] by Henning and Yeo is a valuable resource on total domination.
For vertices of a graph , we write if and are adjacent. A vertex such that for all is a universal vertex. We refer to a vertex of of degree as a leaf and to the unique neighbour of a leaf in as a stem, and denote the number of leaves and stems of by and , respectively. As usual, for , denotes the distance from to .
A set of cardinality is also called an set. A subset of cardinality of a set is called a subset of . The hypercube is the graph whose vertices are the subsets of an set, where two vertices are adjacent if and only if one set is obtained from the other by deleting a single element.
The disjoint union of copies of a graph is denoted by . The corona of a graph is the graph obtained by joining each vertex of to a new leaf. A generalized corona of is a graph obtained by joining each vertex of to one or more new leaves. For a graph and a subset of , we denote by the subgraph of induced by .
Remark 1.1
The set of stems of a graph is a subset of any TDS of , otherwise some leaf is not totally dominated. Hence .
The open neighbourhood of a vertex is and the closed neighbourhood of is . Let . The closed neighbourhood of is . The open private neighbourhood of a vertex relative to , denoted , consists of all vertices in the open neighbourhood of that do not belong to the open neighbourhood of any , that is, . A vertex in may belong to , in which case it is called an internal private neighbour of relative to , or it may belong to , in which case it is called an external private neighbour of relative to . The set of internal (external, respectively) private neighbours of relative to are denoted by (, respectively). Hence . These sets play an important role in determining whether a TDS is an MTDS or not.
1.2 Preliminary results
Cockayne, Dawes and Hedetniemi [5] characterize minimal total dominating sets as follows.
Proposition 1.2
[5] A TDS of a graph is an MTDS if and only if for every .
We restate Proposition 1.2 in a more convenient form for our purposes.
Corollary 1.3
Let be a TDS of a graph , the subgraph of consisting of the components of of order at least , and the set of stems of . Then is an MTDS if and only if for each .
Proof. A vertex belongs to if and only if and . Therefore, if is a component of , then and . Further, if is a stem of , then is adjacent to a vertex such that , hence . By Proposition 1.2, therefore, is an MTDS if and only if for every . But for any , each vertex in is adjacent to at least two vertices in , hence , which implies that if and only if .
Cockayne et al. [5] also established an upper bound on the total domination number, while Favaron and Henning [7] established an upper bound on the upper total domination number.
Proposition 1.4
If is a connected graph of order , then
We now characterize graphs such that .
Proposition 1.5
A connected graph of order satisfies if and only if is odd and is obtained from by joining a new vertex to at least one vertex of each .
Proof. It is clear that for any such graph . For the converse, assume is a connected graph of order such that and let be a set of . Suppose is a component of of order at least . If , then has no stems. If , then has at least as many leaves as stems, so that has at least two vertices that are not stems. In either case Corollary 1.3 implies that at least two vertices in has nonempty external private neighbourhoods, which is impossible since . Therefore each component of is a . Since only one vertex of does not belong to and is connected, the result follows.
2 Connectedness of
Haas and Seyffarth [8] showed that is disconnected whenever . In contrast, we show that any graph without isolated vertices is an induced subgraph of a graph such that is connected. We obtain the bounds for any connected graph of order , and characterize graphs that satisfy equality in either bound.
We begin with some definitions and basic results. For and total dominating sets of of cardinality at most , we write , or simply if is clear from the context, if there is a path in connecting and . The binary relation is clearly symmetric and transitive. If and , then is adjacent to a vertex in because is a TDS. Hence is a TDS. Repeating this argument shows that . More generally, if is also a TDS of cardinality at most such that , then . Repeating the same argument if is a TDS and is an MTDS shows that if for all MTDS’s of cardinality at most , then is connected. We state these facts explicitly for referencing.
Observation 2.1
Let be total dominating sets of a graph of cardinality at most .

If , then

If , then .

If for all MTDS’s of cardinality at most , then is connected.
As in the case of dominating sets, the connectedness of however does not guarantee the connectedness of . For example, consider the tree (the spider with three legs of length each) in Figure 1. This figure shows , where vertices are represented by copies of , and the total dominating sets are indicated by the solid circles. The unique set is an isolated vertex in , so is disconnected.
In the case of dominating sets it is easy to see that is disconnected whenever has at least one edge (and hence at least two minimal dominating sets). For total domination the situation is not quite as simple. A fundamental difference between domination and total domination is that every graph with at least one edge has at least two different minimal dominating sets, whereas there are many graphs with a unique MTDS. Consider, for example, the double star , which consists of two adjacent vertices and such that is adjacent to leaves and is adjacent to leaves. By Remark 1.1, and belong to any TDS of . Since is an MTDS, it is the only MTDS of . Therefore , which is connected. We show that the stems of determine whether is connected or not.
Theorem 2.2
Let be a connected graph of order with . Denote the set of stems of by . Then is connected if and only if is a TDS of .
Proof. Let be any set of . Then no subset of is a TDS and no superset of is a vertex of , hence is an isolated vertex of . Therefore is connected if and only if is the only MTDS of .
Suppose is a TDS of . Any is adjacent to a leaf, hence does not dominate . Therefore is an MTDS. This implies that no superset of is an MTDS. But by Remark 1.1, is contained in any TDS of . Consequently, is the only MTDS of , so and .
Conversely, suppose is not a TDS of . We show that has at least two MTDS’s. First assume that dominates . Then has an isolated vertex, say , which is adjacent to a leaf . Now
is an MTDS of . For another MTDS of , let be a spanning tree of , let be the subtree of obtained by deleting all leaves of and let . If , then and is a star. Say . Then is a universal vertex of and . Since there exists a vertex , and is an MTDS of distinct from . On the other hand, if , then is a TDS that does not contain any leaves of . Hence contains an MTDS distinct from .
Now assume that does not dominate and let be any MTDS of . Then there exists a vertex . Let be the component of that contains and consider the sets and . By Proposition 1.2, , hence at least one of and is nonempty. Since , for each . For each we can therefore choose a vertex adjacent to , where possibly for distinct . Let . Noting that if , then no other neighbour of belongs to , we define the set similarly. Let . By definition and (if they are defined) have no isolated vertices. If , then , and as shown in the proof of Corollary 1.3, has no isolated vertices. Therefore has no isolated vertices. Since and dominate and , respectively, dominates . This shows that is a TDS of . Since , contains an MTDS distinct from , which is what we wanted to show.
The class of graphs whose stems form a TDS includes (but is not limited to) the generalized coronas of graphs without isolated vertices. Hence any graph without isolated vertices is an induced subgraph of a graph such that is connected. The first paragraph of the proof of Theorem 2.2 implies the following result.
Corollary 2.3
The graph is disconnected if and only if has at least two MTDS’s.
If the set of stems of is a TDS, then it is the unique MTDS of , hence we also have the following corollary. The converse does not hold – for the spider in Figure 1, , which is connected, but the stems form an independent set of cardinality , which is not a TDS.
Corollary 2.4
If is a connected graph of order whose set of stems is a TDS, then is connected.
Since any TDS of cardinality greater than contains a TDS of cardinality , the following result is immediate from Observation 2.1 (and similar to [8, Lemma 4 ]).
Lemma 2.5
If and is connected, then is connected.
We now know that
for any connected graph of order , and that the first inequality is strict if and only if the stems of do not form a TDS. Equality in the upper bound is realized by graphs with total domination number equal to , as characterized in Proposition 1.5, because all these graphs also have an MTDS of cardinality different from the set described in the proof, so is disconnected. We next show that if , then .
Theorem 2.6
If is a connected graph of order such that , then .
Proof. Let be the set of stems of . Suppose first that has a unique MTDS , so that by Corollary 2.3. By Remark 1.1, is the unique MTDS of , hence . But each vertex of is adjacent to a leaf, hence . Therefore
Assume therefore that has at least two MTDS’s and let and be any two MTDS’s of . If , then by Observation 2.1, hence assume . By the hypothesis, , hence there exist distinct vertices and . By Remark 1.1, . Consider the four pairs . Suppose first that for one of these pairs , every vertex adjacent to both and has degree at least . Since we also have that , has no isolated vertices. This implies that and are TDS’s of , and we have
Hence assume that for each pair there exists a vertex such that . Then in , , and . Since , and have no common neighbours except possibly (which is adjacent to ). Therefore has no isolated vertices, which means that is a TDS of . Similarly, and are TDS’s of . Now
By Observation 2.1, .
Now let be any fixed set and any MTDS of . Then . By Observation 2.1, . By transitivity, for all MTDS’s of , so by Observation 2.1, .
To summarise, in this section we showed that

for any connected graph of order ,
(1) 
The lower bound in (1) is realized if and only if has exactly one MTDS, i.e., if and only if the stems of form a TDS.

The upper bound in (1) is realized if and only if , i.e., if and only if is odd and is obtained from by joining a new vertex to at least one vertex of each .
3 Determining for paths and cycles
Our aim in this section is to show (in Theorem 3.9) that and if . Similar techniques can be used to show that and if or . We need four lemmas (Lemmas 3.5 – 3.8) to obtain the result for cycles. To enhance the logical flow of the paper, we only state the lemmas in this section and defer their proofs to Section 5.
It is easy to determine the total domination numbers of paths and cycles.
Observation 3.1
[12, Observation 2.9] For ,
The upper total domination number for paths was determined by Dorbec, Henning and McCoy [6].
Proposition 3.2
[6] For any ,
The proof of the following proposition on the upper total domination number of cycles can be found in the appendix.
Proposition 3.3
For any ,
Let . When discussing subsets of the arithmetic in the subscripts is performed modulo . We mention some obvious properties of minimal total dominating sets of .
Remark 3.4
Let be an MTDS of . Then

each component of is either or ;

each or component is preceded and followed by exactly two consecutive vertices of ;

does not contain three consecutive vertices of .
Using the next four lemmas, we show in Theorem 3.9 that, with the single exception of , . We only state the lemmas here; their proofs are given in Section 5. The first lemma concerns MTDS’s that induce or components.
Lemma 3.5
Let .

If is an MTDS such that contains a component, then is connected, in , to an MTDS without components.

If is an MTDS such that contains two consecutive components, then is connected, in , to an MTDS with fewer components.

If is an MTDS such that contains at least one and at least one component but no components, then is connected, in , to an MTDS that has no components.
The next lemma concerns MTDS’s that induce only components. For brevity we refer to such an MTDS as a MTDS. For a MTDS , each component is followed by one or two vertices not belonging to . We refer to these components as and components, respectively. An MTDS is called a maximum MTDS if is a MTDS of maximum cardinality.
Lemma 3.6
Let be a MTDS of .

is a maximum MTDS if and only if has at most two components.

If has at least one component and is any MTDS such that , then .
We next consider .
Lemma 3.7
Suppose and ; say . (By Observation 3.1, .) Then

is disconnected;

if , then has a MTDS such that and has four components;

all sets belong to the same component of and all MTDS’s of cardinality or belong to the same component of .
Our final lemma concerns small cycles.
Lemma 3.8
If and , then .
Theorem 3.9
For , . In all other cases, .
Proof. Since , Lemma 3.7 implies that is disconnected and then the first part of Lemma 3.7 implies that . By Lemma 3.8, the theorem is true for and . Hence assume .
Let be any MTDS of . By Lemma 3.5 (possibly applied several times), if has a or component, then there exists a MTDS such that . Thus we may assume that is a MTDS. If , then has at least one component. If , then and, by Lemma 3.7, all MTDS’s of cardinality or belong to the same component of . Moreover, any MTDS of cardinality has a component. In either case repeated application of Lemma 3.6 shows that all MTDS’s belong to the same component of . The result follows from Observation 2.1, Corollary 2.3 and Lemma 2.5.
The proof of the following result for paths is similar and omitted. Note that for , , which explains the difference between and . The result is trivial for , while the result for follows from Corollary 2.4.
Theorem 3.10
and if or .
4 Realizability of graphs as total dominating graphs
One of the main problems in the study of total dominating graphs is determining which graphs are total dominating graphs. Since if and only if , in studying graphs such that for a given graph we restrict our investigation to graphs without components (and also without isolated vertices, so that is defined).
As noted in [1, 15] for the dominating graph of a graph of order , the total dominating graph is similarly a subgraph of the hypercube (provided and has no isolated vertices) and is therefore bipartite. Since any subset of of cardinality at least is a TDS of and since is vertex transitive, for some . We show in Corollary 4.2 that itself is realizable as the total dominating graph of several graphs, and in Corollary 4.2 that stars , , are realizable. Again the set of stems plays an important role.
In the last two results of the section we determine the realizability of paths and cycles.
Theorem 4.1
Let be any graph of order , , without isolated vertices and let be a generalized corona of having exactly leaves. For each such that , is the subgraph of corresponding to the collection of all subsets, , of an set.
Proof. Every vertex of is a stem of . By Remark 1.1, is contained in any TDS of . Since has no isolated vertices, is an MTDS of . As shown in the proof of Theorem 2.2, is the only MTDS of . For any set of leaves of , is a TDS of . Moreover, for any sets and of leaves, if and only if each and is obtained from by adding or deleting exactly one vertex. The result now follows from the definitions of and .
Corollary 4.2
Let be any graph of order , , without isolated vertices and let be a generalized corona of having exactly leaves. For every integer ,


.
Proof. By Theorem 4.1, is the subgraph of corresponding to the collection of all subsets of an set. Hence .
By Theorem 4.1, is the subgraph of corresponding to the empty set and all singleton subsets of an set. Hence .
We mentioned above that for a graph without isolated vertices and , is a subgraph of . The strategy used in the proof of Theorem 4.1 enables us to be a little more specific in many cases.
Proposition 4.3
Let be a connected graph of order having stems. For any , is a subgraph of .
In particular, is a subgraph of in which has degree .
Proof. Let be the set of stems of . By Remark 1.1, is contained in any TDS of . Hence all TDS’s of are subsets of that contain , and there are such sets. This shows that is a subgraph of for any .
Now consider . For , has an isolated vertex if and only if . Therefore is a TDS of if and only if , which implies that in . Let be any TDS of ; necessarily, . There are at most supersets of of cardinality that are TDS’s, and at most subsets of of cardinality that are TDS’s. Hence in , .
Concerning the realizability of cycles, it is easily seen that , , and, if is the graph obtained by joining two leaves of , then . We show that , are the only cycles realizable as total domination graphs.
Proposition 4.4
There is no graph of order such that for some integer .
For , there is no graph such that for some integer .
Proof. Suppose to the contrary that . Let be a set of . Then in . Since each superset of is a TDS of , has exactly two supersets of cardinality . This implies that , i.e., . But we know that (Proposition 1.4) and so , which is a contradiction.
Now suppose that , where . By , has order and . Say is the cycle . Since , each has cardinality or .
First assume for some . Then and we also have . Since has degree in , has exactly two TDS’s of cardinality . By Remark 1.1, has exactly two vertices that are not stems. Since (and has no components), consists of two stems and two leaves, i.e., . But , contradicting .
We may therefore assume that and for each . But then, by definition of adjacency in , for values of and for the other values of . Since , has at least six subsets of cardinality , which implies that . Therefore , and , and each of the six subsets of is a TDS. By Remark 1.1, has no stems and hence no leaves. Let