#### Abstract

A set of vertices in a graph is a total dominating set of if every vertex of is adjacent to some vertex in . The minimum cardinality of a total dominating set of is the total domination number of . The graph is total domination edge-critical, or , if for every edge in the complement of , . If is and , we say that is . In this paper, we show that every graph with and has a Hamilton cycle.

#### 1. Introduction

All graphs we considered here are finite, undirected, and simple. We refer to [1] for unexplained terminology and notations. Let be a graph with vertex set and edge set . For a vertex of , the open neighborhood of is the set , and closed neighborhood of is . For a set , its open neighborhood is the set , and its closed neighborhood is the set . The complement of has vertex set and edge set . The subgraph of induced by the nonempty vertex subset will be denoted by , and set . We use to denote the number of components of . An independent set of the graph is a set of vertices, no two of which are adjacent. The independence number of , denoted by , is the order of a maximum independent set in . The connectivity of , denoted by , is defined as the order of the minimum vertex subset whose deletion resulting in is disconnected. We often simply write for and for , respectively. A Hamilton path of is a path passing exactly once through every vertex of . A Hamilton cycle is a closed Hamilton path. We refer to [24] for some recent results on the Hamiltonian properties of graphs.

The domination theory of graphs is an important part of graph theory because of its wide range of applications and theoretical significance [5, 6]. A subset is called a dominating set of a graph if for every , either or is adjacent to a vertex in . The domination number of , denoted by , is the minimum cardinality of a dominating set. The graph is said to be domination edge-critical, or , if for every edge , . If is and , we say that is . A set of vertices in a graph is a total dominating set of if every vertex of is adjacent to some vertex in . The minimum cardinality of a total dominating set of is the total domination number of . We refer to [712] for some recent results on the domination number and total domination number of . As introduced in [13], a graph is total domination edge-critical, or , if for every edge , . If is and , we say that is . We refer to [1418] for some results on graphs.

In [11], Pan et al. proved that the total edge domination problem is NP-complete for bipartite graphs with maximum degree 3. In [19], Furuya and Matsumoto showed that the order of a connected 3-edge-critical planar graph is at most 23.

For the sake of convenience, we write (or if . In general, for subsets and of , if every vertex in has a neighbor in , we say that dominates and write .

The study of total domination edge-critical graphs was initiated in [13]. It was shown in [13] that the addition of an edge to a graph can change the total domination number by at most two. If is graph, for any edge . Since for any graph , if is , then for any edge . Also, observe that any graph with is connected. In [14], Asplund et al. constructed some families of graphs with diameter 2 and determined which graphs have complements that are graphs.

We also use the following results on graphs.

Theorem 1 (see [13]). For any graph and nonadjacent vertices and , at least one of the following holds:(1) dominates (2)There exists such that dominates , but not , and we write (3)There exists such that dominates , but not , and we write

Theorem 2 (see [13]). A graph with a cut vertex is a graph if and only if is adjacent to a pendent vertex , and for and ,(1) is complete and (2) is complete and (3)Every vertex in is adjacent to vertices in , and every vertex in is adjacent to at least one vertex in

The following result is immediate from the above theorem.

Corollary 1. If is a graph, then if and only if ; if and only if .

In [20], we prove the following results.

Theorem 3 (see [20]). If is a graph, then , with equality only if .

Theorem 4 (see [20]). Let be a graph with . If , then is Hamiltonian.

In this paper, our main result is as follows.

Theorem 5. Let be a graph with . If , then is Hamiltonian.

Combining Theorems 35, we have the following.

Theorem 6. Every graph with is Hamiltonian.

The remainder of this paper is organized as follows. In Section 2, we list some useful notations and lemmas. The proof of our main result is presented in Section 3.

#### 2. Useful Lemmas

In this section, we list some useful notations and lemmas.

For a graph , if , then contains a cut vertex. By Theorem 2, there exists a pendant vertex adjacent to such cut vertex, which means that . If , then it is clear that . Thus, we have that for a graph , if . It follows that for a graph , if if . Since 3tEC graph G that we consider is connected, , and it does not contain isolated vertices, that is, . It implies that for a graph , if . Corollary 1. Hence, we may assume . Let be a cycle. We denote by the cycle with a given orientation and by the cycle with the reverse orientation. If , then denotes the consecutive vertices of from to in the direction specified by . The same vertices, in the reverse order, are given by . We use to denote the successor of and to denote its predecessor.

Let be a graph and a longest cycle of . Suppose has no Hamilton cycle and is any component of . We set

Let with . denotes a longest -path with the internal vertices in . The following Lemmas 14 can be found in [21], which play an important role in our proof.

Lemma 1 (see [21]). Let with . Then, there is no path in , connecting and ( and , resp.).

Lemma 2 (see [21]). For any , both and are independent sets.

Lemma 3 (see [21]). Let with . For any vertex , if , then .

Lemma 4 (see [21]). Let and with . For any vertex , if , then . Similarly, for any vertex , if , then .

In [22], Simmons obtained the following result.

Lemma 5 (see [22]). Let be a graph with and a cut vertex of . Then, .

We give the following useful lemma.

Lemma 6. If is an independent set of vertices of a graph , then there exist an ordering of the vertices of and a path in such that for .

Proof. First, we show the first part of the statement. Since is an independent set of , it is a clique of . For any two vertices , since , is not a total dominating set of . By Theorem 1, if there exists a vertex with , then we orient from to , and if there exists a vertex with , we orient it from to in . Now, becomes a tournament under this orientation. By Rédei theorem [1], let be a directed Hamilton path of . For an integer , we consider and . By our convention, there exists a neighbor of such that . Thus, .
Now, we show that , for any with for each . Since , . On the contrary, since and , . This implies that .
Since and , . This proves is a path in .

#### 3. Proof of Theorem 5

Let be a graph and a longest cycle of . Suppose has no Hamilton cycle and is any component of . We consider the following two cases separately to complete the proof.

Case 1. .
Suppose has no Hamilton cycle, and let be a longest cycle of . Since , . By Lemma 2, . If , then . This contradicts . Thus,

Claim 1. If , then for each .

Proof. First, we show that if , then ; otherwise, we suppose that . Since , by Lemma 2, we have and . Since is an independent set, where , we have . This implies that there is some vertex such that , which contradicts Lemma 4. Hence, we have if . Similarly, if , then .

Claim 2. For any with and with , ; for any with and with , .

Proof. Otherwise, is a cycle longer than , a contradiction. By symmetry, for any with and with , .

Claim 3. For each vertex , and for each vertex , where .

Proof. Without loss of generality, we show that, for each vertex . By contradiction, assume that is the last vertex in which is not adjacent to ; thus, . If , then is a cycle longer than , a contradiction. We have . Thus, is an independent set of four vertices which contradicts , where .

Claim 4. .

Proof. Suppose to the contrary that is another component of . Take . Since , we have ; otherwise, is an independent set of four vertices. Similarly, . By Lemma 1, we may assume . If , then is a cycle not less than , and is a component of , but , , which contradicts (2). Hence, we have . By the maximality of , we have . By Claims 1, 2, and 3, we have ; thus, is an independent set of four vertices which contradicts . Thus, we have .

Claim 5. , where .

Proof. By Lemma 5 and Claim 4, we have . If , then . Consider and , , where . is not a domination set of since cannot dominate . By Theorem 1, there exists a vertex such that or . If , then and , but this is impossible since is not dominated. Thus, , we have , and hence,Consider and , . cannot dominate since is not dominated; by Theorem 1, there exists a vertex such that or . If , then , but this is impossible since is not dominated. Thus, , we have , and hence,Consider and , , and cannot dominate since is not dominated; by Theorem 1, there exists a vertex such that or . If , in order to dominate , then , but this is impossible because of (3). If , then , but this is impossible because of (4). Thus, we get .

Claim 6. .

Proof. By Claim 5, we have . By Claims 2 and 3, we have . Assume ; by Lemma 1, for . We now show that . If , then cannot dominate since is not dominated, and there exists a vertex such that or . If , then dominates , so . We have and , but by Lemma 3, a contradiction. If , then , and is a cycle longer than , a contradiction. Thus, . Hence, .
By Claim 6, we have for . Otherwise, we suppose that ; then, is a cycle longer than , a contradiction. By symmetry, we have . Consider and , cannot dominate , and without loss of generality, there exists a vertex such that ; in order to dominate , then and . By Claims 2 and 3, , and hence, is a cycle longer than , a contradiction. Thus, has a Hamilton cycle.

Case 2. .
Let be a graph and a longest cycle of . Suppose has no Hamilton cycle and is a component of . Set , and define as before; then, . By Lemma 2, . We have since . Thus, .
Considering and , where , by Lemma 2, is an independent set. Let be the path of defined in Lemma 6. Since , by Lemma 6, we have and for ; then, and . Thus, and . We may assume , . Thus, we can obtain that for . Take . Since and , we have ; then, , that is,

Claim 7. .

Proof. If , then for . Since , . By Lemma 2, ; thus, . Since , there exists a component of such that and . By Lemma 1, there is no path in connecting and , so , which implies that , a contradiction. Thus, we get .

Claim 8. There exist a vertex and a vertex such that , where is the vertex specified in (5).

Proof. For any , , and cannot dominate ; thus, there exists a vertex such that or . For with , , if there exist vertices such that and , then we claim that . Otherwise, we set . Since and , , but , and we have ; this is impossible; thus, . For every , if there exists a vertex such that , then and , but this is impossible since and . Thus, there exist a vertex and a vertex such that .
By symmetry, we may assume that . Obviously, .

Claim 9. If , then .

Proof. Since , for , and by Lemma 1, we can assume that . Since and is an independent set, we have .
If , then we show that . Otherwise, we suppose ; then, and . Since and , . By Lemma 3, we have for . If , we have by Lemma 4; hence, . Note that and ; thus, there exists a vertex such that and , but is a cycle longer than , a contradiction. Hence, . is an independent set of with vertices; hence, , again a contradiction.
If , then . Otherwise, we have and . This implies that there exists a vertex such that and , which contradicts Lemma 4. Thus, for , and .
If and , then we have , and hence, . Since , . By Lemma 4, we get for . If , then is a cycle longer than , a contradiction. Hence, . is an independent set of vertices; hence, , again a contradiction. Thus, we have and .
By symmetry, if , then for any .
By Claim 9, we have , . Thus, by Lemma 2, we have and which imply . Since , by Lemma 3, we have for , and for . Thus, we have . Since , by Lemma 2, we can see that, for any vertex , is impossible. For any , considering and , , and cannot dominate . So, there exists a vertex such that or . If for any , for , there exist vertices such that and , then by the same analogous argument to the proof of Claim 8, we have . For every , if there exists a vertex such that , then we have , but this is impossible since and . Thus, there exist a vertex and a vertex such that . Assume, without loss of generality, that , say . Since and , , which contradicts Claim 9.
The proof of Theorem 5 is completed. We show that every graph with and has a Hamilton cycle. Combining Theorems 35, we show that every graph with is Hamiltonian.

#### Data Availability

No data were used to support this study.

#### Conflicts of Interest

The authors declare that they have no conflicts of interest.

#### Acknowledgments

This work was supported by the Key Scientific Research Projects of Xinjiang Institute of Technology (no. 2017010501004).