#### Abstract

For a given graph , graph is called -saturated if does not contain a copy of , but has a copy of , where . The saturation number, denoted by , is the minimum size of a graph with order in all -saturated graphs. For , this paper considers the saturation number of linear forests and describes the extremal graph.

#### 1. Introduction

In this article, we only deal with simple graph. Usually, the path and the complete graph with vertices are denoted by and , respectively. For terminology and notations not undefined in this paper, the reader can refer to [1].

For a given graph , graph is called -saturated if does not contain a copy of , but has a copy of , where . The famous Turán number [2], denoted by , is the maximum number size of graphs in all -saturated graphs with size . As a complement, saturation number, denoted by , is the minimum size of graphs in all -saturated graphs with size . We use which denotes the set of graphs with a minimum number size in .

In 1964, Erdős et al. introduced the notion of the saturation number and gave the saturation number of in [3]. Kászonyi et al. gave the general upper bound of in [4], when is a kind of forbidden graphs. Then, for a wider range of graphs , saturation number of has been studied by many scholars, for example, -edge-connected graph [5], cliques [6, 7], complete bipartite graphs [8, 9], nearly complete graphs [10], books [11], cycles [12–17], trees [18, 19], and forests [20–22]. The reader can see summary of known results in [23].

In [24], Bushaw et al. gave the Turán number for the linear forest. Corresponding to that, Chen et al. concentrated on the saturation numbers in [20]. They obtained an interesting set of results; some of those are shown as the following results.

Theorem 1 (see [20]). (i)* Letbe positive integers such that sufficiently large; then,and the extremal graph is*(ii)

*(iii)*

*Let**and**be positive integers such that**is sufficiently large and**; then,**and the extremal graph is*

*Let**be positive integers such that it is sufficiently large; then,**and the extremal graph is**Let be obtained from by attaching disjoint triangles to one vertex of and attaching disjoint triangles to another vertex of (see Figure 1).*

*In this paper, we prove the following theorem.*

Theorem 2. *Suppose**and**are positive integers such that**. Then,**and**for**.**In what follows, Section 2 lists several useful results and study property of -saturated graphs. In Section 3, we give the proof of Theorem 2.*

#### 2. Preliminaries

Firstly, we introduce some notations. For a graph , the edge set and vertex set are denoted by and . The number of edges and the order of are denoted by and , respectively. Let , , and be the degree of , where . Given a subgraph of and , let , , , and . We simple denote and if . When no confusion occurs, we simple identify a subgraph of with . For example, is simple and is denoted by sometimes.

For a graph , let and denote the number of edges in the maximum matching of by . Suppose ; let . We list the following useful results.

Lemma 1. *(see [25]). Letbe a graph; then,*

Lemma 2. *(see [20]). Supposeare integers. Letbe a-saturated graph. Ifand, then.*

*Next, we give some properties of -saturated graphs.*

Lemma 3. *S upposeis-saturated with; then,. Moreover, ifand, letbe any copy ofin; then,*

*Proof. *Suppose for the sake of contradiction that . Since and is -saturated, has a containing . Let , with replaced by , and we can get a new in , a contradiction. Thus, .

Since and is -saturated, then has a containing and assumes . If there exists , with replaced by in , we can obtain a new in graph , a contradiction. Therefore,Lemma 3 is true.

Lemma 4. *Suppose**is**-saturated such that**and all the nontrivial components are with at least**vertices. Let**such that**and**. Then,**. Moreover, if**, then**.**Additionally, there does not exist a degree 2 vertex**such that**.*

*Proof. *By Lemma 2, we know . Since all the nontrivial components are with at least 4 vertices, then the triangle induced by is not a component of . Hence, we can assume for some vertex . Consequently, .

If , it is enough to show . If , has a containing , assume . Assume , replacing by , and we obtain a in . Hence, . If or or , then are selected as a in , and we can replace it with , which implies there is a in . So, we have and and . This together with and implies that and is selected as a in . Now, we can claim that is selected as a in . Otherwise, both and are selected as a in . Replacing in with , we obtain a copy of in , a contradiction. If is selected as a middle edge in , then contains a , a contradiction. If is selected as an end edge in , then contains a , a contradiction.

If there exists a degree 2 vertex such that , then select an isolated vertex and add the edge . Let be a copy of the graph in . No matter is selected as an edge in or in , since , we get , which contradicts Lemma 3.

Lemma 5. *Let**be a**-saturated graph and**. Let**be an induced subgraph of**. If**or**, then**.*

*Proof. *Suppose, for the sake of contradiction, that . Select an isolated vertex and add the edge . Since is -saturated, then has a containing and assumes .

First, consider the case . Suppose and . For any , we have , where . Otherwise, and are selected as a . However, this can be replaced with in , and then, we obtain a in , a contradiction. Hence, is selected as a in . We can choose as and then replace in with ; we obtain a in , a contradiction.

Now, consider the case . Suppose is the center of the star and . Since is -saturated with , by Lemma 3, . Then, we have for . Hence, and for all , which contradicts Lemma 4. Lemma 5 is true.

#### 3. Proof of Theorem 2

Before the proof of Theorem 2, we discuss the property of -saturated graph which has no components of order 2, 3, and 4.

Lemma 6. *Suppose**is**-saturated, but not**-saturated graph. Let**,**be the nontrivial components of**and**. If**,**and**, we have**.**Moreover, if the equality**holds, then**, where**and**.*

*Proof. *Since is -saturated, has a , where . On the contrary, must contain a , since is not -saturated.

Suppose is a in with . Choose which satisfies the following:(i) is as large as possible(ii)Subject to (i), is maximalityObserve that . By the choice of , is an independent set or an empty set. As is -saturated, we haveWithout loss of generality, we may assume .

*Claim 1. **.*

*Proof. *Suppose, for the sake of contradiction, that . Since and , we have for every . Without loss of generality, we can assume , where .

As is connected, there exists a path from to avoiding all vertices of for some . Again, has a , which, combining the edges in , creates a in , a contradiction. Claim 1 is true.

It follows from Claim 1 that . Hence, in the rest of the proof of Lemma 6, is denoted by . Clearly, .

*Claim 2. **For each**, we have**.*

*Proof. *Suppose for the sake of contradiction that there exists such that . This together with (4) implies thatWe are now ready to show thatOtherwise, , since . It follows from (6) that we can assume . By Lemma 2, . Then, is a copy of in . However, by Lemma 4, we have , and hence, , which contradicts the choice of . Then, we have (7) which holds.

Combining (6) and (7), there exist such that and . Then, the subgraph will have a , which, combining the edges in , creates a in . Since , we obtain a contradiction. Claim 2 is true.

*Claim 3. *T*here does not exist**such that*

*Proof. *Suppose for the sake of contradiction that there exist such that and . By Claim 2, we may assume . However, the subgraphs and both contain , which, combining the edges in , creates a in , a contradiction. Hence, Claim 3 is true.

*Claim 4. **For each**, we have**or**.*

*Proof. *Suppose, for the sake of contradiction, that Claim 4 is false. Then, by Claim 3 and the symmetry, we may assume andThen, by Claim 3, we have

*Case 1. * and .

Combining (9), without loss of generality, in this case, we may assume and . Now, we consider , where .

If , then subgraphs and both contain , which, combining the edges in , creates a in , a contradiction. Hence, we have . Together with Claim 2 and (10), we obtainWithout loss of generality, we may assume .

If , assume . In order to avoid a in , we get that or and , which contradicts Lemma 5. Hence, . This together with and (11) implies that and . As the argument of , we have . Hence, . Then, we can conclude thatAnd, by the choice of , . Hence, by Lemma 2, . Combining (12) implies that . If , by Lemma 2, . Hence, . If , we also have . This together with , , , and implies that , a contradiction.

*Case 2. * or .

Combining (9), without loss of generality, in this case, we may assume . This combining and (4), we get and . Assume , where . By our assumption , we get and . Now, we consider , where .

If , then the subgraphs and both contain , which, combining the edges in , creates a in , a contradiction. Hence, we have . This together with Claim 2 and (9) implies that , for any . Without loss of generality, we may assume . Then, we have . So, by the choice of , (5), and , we get . Hence,In order to avoid a in , we get that . This together with and implies that and . Similarly, we have . Therefore, and for any . Hence, combining implies thatAnd, by the choice of , we have . Then, by Lemma 2, , which implies thatRecall that and . Together with (13)–(15), we can get that , a contradiction. Claim 4 is true.

*Claim 5. **If there exists**such that**, then we have**. Moreover, the equality**holds if and only if**, where**and**.*

*Proof. *Without loss of generality, we may assume . Since , we have . By Claim 4 and the symmetry, we complete the Claim 5 by following two cases. Case 1. . Since , by Lemma 2, and . By the choice of , . Also, by Lemma 2, . Now, we consider , where . It follows from Claim 3 and our assumption that , where . This together with Claim 4 implies that , where . If , assume . In order to avoid a in , we get that or and , which contradicts Lemma 5. Hence, . Since , we get and . By the similar argument, we have , and hence, . Therefore, for each , we can conclude that or . Together with , we have . Recall that , , and , we can get that . And, if , then , where . However, if , is not a -saturated graph, since the addition of an edge between an isolated vertex of and the vertex of degree does not result in a . Hence, ; Claim 5 holds. Case 2. .Since , by Lemma 2, and . By the choice of , we have . Also, by Lemma 2, . Now, we consider , where . It follows from Claim 3 and our assumption that .

If there exists such that , we can claim that . Otherwise, assume . In order to avoid a in , we get that or and , which contradicts Lemma 5. Then, as the similar argument in the Case 1, we can obtain , and if , then , where and . Then, , for all . Without loss of generality, we may assume . If , assume . In order to avoid a in , we get that or and , which contradicts Lemma 5. Hence, . Since , we get and . By the similar argument, we have , and hence, . Therefore, and for any . Hence,This together with , , and (16) implies that . This completes the proof of Claim 5.

If there exists such that , by Claim 5, the Lemma is proven by Claim 5. Hence, we can assume for each . On the contrary, by (4) and Claim 4, we can assume . Hence, there exists for any . And, by (5), for . Hence, . Since , we get . And, if the equality holds, then and , which contradicts Lemma 4. Lemma 6 is true.

Lemma 7. *Suppose**is**-saturated and**-saturated with**. Let**be the nontrivial components of**and**. If**and**with**, we have**.*

*Proof. *Suppose, for the sake of contradiction, that . Since is -saturated, there exists a in for any . Hence, . If , must contain a copy of . Since and , it is clearly that has a , which contradicts is -saturated. So, we have . By Lemma 1, we obtainThen, we can choose a maximal subset which satisfiesSuppose are the components of . We have the following claims.

*Claim 6. **is the complete graph, where**.*

*Proof. *Suppose for the sake of contradiction that