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)Letandbe positive integers such thatis sufficiently large and; then,and the extremal graph is(iii)Letbe positive integers such that it is sufficiently large; then,and the extremal graph isLet 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. Supposeandare positive integers such that. Then,andfor.
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. Supposeis-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. Supposeis-saturated such thatand all the nontrivial components are with at leastvertices. Letsuch thatand. Then,. Moreover, if, then.
Additionally, there does not exist a degree 2 vertexsuch 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. Letbe a-saturated graph and. Letbe an induced subgraph of. Ifor, 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. Supposeis-saturated, but not-saturated graph. Let,be the nontrivial components ofand. If,and, we have.
Moreover, if the equalityholds, then, whereand.
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. There does not existsuch 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 haveor.
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 existssuch that, then we have. Moreover, the equalityholds if and only if, whereand.
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. Supposeis-saturated and-saturated with. Letbe the nontrivial components ofand. Ifandwith, 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