Abstract
A graph G is called -free if G contains no as an induced subgraph. A tree with at most m leaves is called an m-ended tree. Let be the minimum degree sum of k independent vertices in G. In this paper, it is shown that every connected -free graph G contains a spanning 6-ended tree if .
1. Introduction
We will begin with some basic definitions, notations, and terminologies used by Bondy and Murty [1]. The graph generally means a simple undirected graph G with a vertex set and an edge set . We write for the order of G. For any , let be the set of vertices adjacent to in G and . For a nonempty subset , we write
For an integer i, we define and . Let T be a tree. A vertex of degree one in T is called a leaf of T, and a vertex of degree at least three is called a branch vertex. If are vertices of T, the path in T connecting u and is unique and is denoted by . Define , and . The distance between u and in T is denoted by .
A subset is called an independent set of G if no two vertices of A are adjacent in G. The independence number of G, , is the order of the maximum independent set of G. For an integer k, we define
For an integer , a graph G is called -free if G contains no as an induced subgraph. A tree with at most m leaves is called an m-ended tree. If T is an m-ended tree in G such that is as large as possible, we call T the maximum m-ended tree of G. There are several well-known conditions ensuring that a graph G contains a spanning m-ended tree (see the survey paper [2]). In 1979, Win [3] obtained a sufficient condition related to an independent number for k-connected graphs, which confirms the conjecture of Las Vergnas [4].
Theorem 1 (Win [3]). Let G be a k-connected graph, and let . If , then G has a spanning m-ended tree.
There are many results on the degree sum conditions for a graph to contain a spanning tree with a bounded number of leaves or branch vertices. Among them are the following theorems.
Theorem 2 (Broersma and Tuinstra [5]). Let G be a connected graph with n vertices, and let . If , then G has an m-ended tree.
Theorem 3 (Gargano et al. [6]). Let k be a nonnegative integer, and let G be a connected claw-free graph. If , then G has a spanning tree with at most k branch vertices.
Theorem 4 (Kano et al. [7]). Let k be a nonnegative integer, and let G be a connected claw-free graph. If , then G has a spanning tree with at most leaves.
Theorem 5 (Kyaw [8]). Every connected -free graph with contains a spanning tree with at most 3 leaves.
Theorem 6 (Kyaw [9]). Let G be a connected -free graph.(1)If , then G has a Hamiltonian path.(2)If for an integer , then G has a spanning tree with at most k leaves.
Theorem 7 (Chen et al. [10]). Let G be a connected -free graph with n vertices. If , then G contains a spanning 4-ended tree.
In this paper, we further consider -free graphs and generalize Theorem 7 as follows.
Theorem 8. Let G be a connected -free graph. If , then G contains a spanning 6-ended tree.
Before proving Theorem 8, we first give an example to show that the condition in Theorem 8 is sharp.
Let m be a positive integer, and let G be the graph obtained from a triangle and 7 copies of by joining x to each vertex of , y to each vertex of , and z to each vertex of (see Figure 1). It is easy to see that , but each spanning tree of G has at least 7 leaves. Therefore, the condition in Theorem 8 is sharp.
2. Proof of Theorem 8
In this section, we will prove Theorem 8. First, we prove the following two lemmas.
Lemma 1. Let G be a connected graph with no spanning 6-ended tree, and let T be a maximum 7-ended tree of G. Then, G does not contain a 6-ended tree such that .
Proof of Lemma 1. By way of contradiction, suppose that there is a 6-ended tree in G such that . Since G has no spanning 6-ended tree, is not a spanning tree. Hence, . As G is connected, there are two vertices such that and . Noting that has at most 7 leaves, we conclude that is a 7-ended tree, contrary to the maximality of T. Hence, Lemma 1 is true.
Lemma 2. Let T be a tree with n vertices. For an integer , if T has at most k leaves, then it has at most branch vertices.
Proof of Lemma 2. Assume that T has x vertices of degree two, y branch vertices, and z leaves. Then, and . Hence, . This completes the proof of Lemma 2.
Based on the above two lemmas, we can now prove the main result of this paper.
Proof of Theorem 8. Let G be a connected -free graph. Suppose, by contradiction, that G does not contain a spanning 6-ended tree. As G is connected, there is a spanning tree T in G. By our hypothesis, i.e., G does not contain a spanning 6-ended tree, the number of leaves in T is at least 7. If the number of leaves in T is 7, then we have found a tree with exactly 7 leaves. If the number of leaves in T is more than 7, then we can delete some of the leaves, and hence, there always exists a 7-ended tree in G. We choose a maximum 7-ended tree T. By Lemma 1, T has exactly 7 leaves. Denote by the set of leaves of T, and let be the set of branch vertices of T. From Lemma 2, we can conclude that . By the maximality of T, we have . For , let be the unique component of containing , and let be the only neighbor of S in . Denote by the only neighbor of in S. For convenience, we let . Note that . For , let be the predecessor of x and be the successor of x. If , let be the longest path between the branch vertices. Without loss of generality, we may assume that and . For convenience, denote by the predecessor of and the successor of for all . Set . If for some and , then we call the branch vertex for .
Claim 1. For all with , if , then .
Proof of Claim 1. By way of contradiction, assume Claim 1 is false. SetThen, is a 6-ended tree with , which is contrary to Lemma 1. This completes the proof of Claim 1.
It follows from Claim 1 that U is an independent set. Since G is -free, .
Claim 2. For , .
Proof of Claim 2. By way of contradiction, assume that there exists a vertex for some . Then, there exist three distinct integers such that . By Claim 1, we can derive that and . As , . Set . Then, is a 6-ended tree with , which is a contradiction. This completes the proof of Claim 2.
By an argument similar to the proof of Claim 2, we can derive that , .
Claim 3. for all .
Proof of Claim 3. Suppose, on the contrary, that there are two vertices for some . Without loss of generality, assume that . By Claim 1, . Assume that for some . Since , there exists such that .
If , then . Then, is a 6-ended tree with a vertex set , which contradicts Lemma 1. Hence, . Since G is -free, it can be derived that . Set . Then, is a 6-ended tree with , which is contrary to Lemma 1. This completes the proof of Claim 3.
Claim 4. If , then for all .
Proof of Claim 4. By way of contradiction, assume that Claim 4 is false. Then, there is a vertex x such that for some with . By Claim 1, . Set . Then, is a 6-ended tree with , which contradicts Lemma 1. This completes the proof of Claim 4.
Claim 5. For all , , , , and are pairwise disjoint, where .
Proof of Claim 5. It can be easily concluded that . It follows from Claim 1 that . This completes the proof of Claim 5.
By Claim 5, we haveThis together with Claim 4 implies that
Claim 6. If for some and , then and for some .
Proof of Claim 6. By way of contradiction, assume that or . SetThen, is a 6-ended tree with , which is contrary to Lemma 1. This completes the proof of Claim 6.
For every vertex , defineFor every vertex set , defineIf , we consider the following two cases.
Case 1. .
In this case, . We choose the maximum T with 7 leaves such that(C1) is as large as possible.(C2) is as small as possible, subject to (C1).Without loss of generality, we may assume that , , and . Hence, and . SetLet be the family of graphs shown in Figure 2. Then, in Case 1, .
The following claim plays a crucial role in the proof of Case 1.
Claim 7. Let be two consecutive vertices in such that , where and . Let be the branch vertex for , where . If G has no 6-ended tree, then and .
Proof of Claim 7. By way of contradiction, assume that . Set . Then, is a 6-ended tree with , which is contrary to Lemma 1. Similarly, . This completes the proof of Claim 7.
Claim 8. Let be the branch vertex for for some and . Then, . Specifically, for all and for all .
Proof of Claim 8. Suppose, for contradiction, that for some . Set . Hence, is a 6-ended tree with , contradicting Lemma 1. Hence, . Likewise, . This completes the proof of Claim 8.
Claim 9. If for some , then .
Proof of Claim 9. By way of contradiction, suppose and . Hence, there exist two integers with such that . By Claim 8, we can see that is not the branch vertex for or . This together with , by applying Claim 7, implies that . This and Claim 8 imply that , which is a contradiction. Hence, , and similarly, if , then .
The proof of Case 1 has been divided into 6 subcases.
Subcase 1. .
Without loss of generality, we may assume that , .
We choose T such that(C3) is as large as possible, subject to (C1) and (C2).
Subclaim 1. For every vertex .
Proof of Subclaim 1. By way of contradiction, suppose that there is a vertex such that . Then, there exist three distinct integers such that . By (C2), we can see that . If , set . Hence, is a 6-ended tree with , which is contrary to Lemma 1. Thus, . Since G is -free, there exists such that or , contradicting Claim 7. This completes the proof of Subclaim 1.
Set . If for every vertex , then . Otherwise, suppose that there are s vertices in satisfying for all . Define , , and . Therefore, .
For every , . It follows from Claim 7 and (C2) that . Hence, . Since , . Thus, . Therefore,
Subclaim 2.
Proof of Subclaim 2. If , by (C2), , say . Then, . Otherwise, assume that and for some . Then by (C2), , that is, . If , the result holds. Hence, we assume that . Since , by Claims 4 and 6, for some . SetThen, is a 6-ended tree with , which contradicts Lemma 1. It follows from (5) that
Subclaim 3. .
Proof of Subclaim 3. If there are three distinct integers in such that , then since G is -free, then , which contradicts (C2).
If for , by (C2), . Without loss of generality, set . Since , by (C2) and Claims 1 and 6, we have . Also by Claim 6, . Moreover, for all .
If for some , set . Then by Claim 7, . If there is a vertex such that , then . Otherwise, for every , . By Claim 7, . Specifically, . By applying Claim 7, we get a contradiction. Hence, . If for some , then . Suppose, for contradiction, and . Then by (C3), . Set , which contradicts (C2). This together with (5) implies that the result holds.
It follows from Subclaims 2-3 thatwhich implies that , which is a contradiction. This completes the proof of Subcase 1.
Claim 10. If and there is a vertex , then .
Proof of Claim 10. Assume that there is a vertex such that . Then, there exist three distinct integers such that . Set . Then, is a maximum tree with 7 leaves. Since are branch vertices, by (C2), and . This implies that we can find a tree such that . This completes the proof of Claim 10.
Thus, in the following subcases, we suppose that, for every vertex , .
If for every vertex , then . Otherwise, suppose that there are s vertices in P satisfying for all . Define , , and . Therefore, .
It follows from Claim 7, Claim 8, and (C2) that . Since , by applying Claim 7 with and (C2), we can derive that either or there exists some such that for all . In the latter case, combining with , we can conclude that contradict Claim 7. Thus, , and hence, , where .
If , similar arguments apply to and then . Thus,If , by Claim 7, we can deduce that for all . Hence,In the following, if , we choose T such that(C4) is as large as possible, subject to (C1) and (C2).
Subcase 2. .
Without loss of generality, we may assume that and . If , by (C2), .
Since , there are three integers in , say , such that or for some . Hence, and for all . (i) . In this case, and for all . If , by Claim 6 and (C2), we can derive that . Hence, . (ii) . Then, , since otherwise is a 6-ended tree with , which contradicts Lemma 1. If , then assume that and for some . By (C2), . Set Then, is a 6-ended tree with , which contradicts Lemma 1. Thus, . Hence, .(iii). Then by similar arguments, . If , then we claim that . If , then the above claim is true. Hence, . Assume and . By similar arguments, we have . Thus, . Since , by Claims 1, 4, and 6, , where . SetThen, is a 6-ended tree with , which contradicts Lemma 1. Hence, . Above all, we haveIf there are three distinct integers such that , then . Since , we have for some , which contradicts Claim 7.
From (C4), for all . If , then .
If , there is an integer such that . By symmetry, set . Then, and . Otherwise, if , set . If , then . Set . Then, is a 6-ended tree with , which contradicts Lemma 1. Moreover, . This together with the proof of (14) implies that .
In either case, we can getwhich implies that , which is a contradiction. This completes the proof of Subcase 2.
Subcase 3. .
In this case, T satisfies and . Without loss of generality, we may assume that and .
If , then . Otherwise, we can find a tree satisfying Subcase 2. Analysis similar to that of Subcase 2 shows thatIf for some , then set . Then, . If , the result is true. Hence, . Suppose, for contradiction, for some . By (C4), we have . Since G is -free, for some . SetThen, is a 6-ended tree with , which contradicts Lemma 1. Hence, if for some , then .
If for some , then set . Then, and , that is, .
If for some , then set . Since , for some , say . Hence, . By similar arguments, we can derive that if , then . Hence,Above all, we haveThis together with (20) shows thatwhich implies that , which is a contradiction. This completes the proof of Subcase 3.
Subcase 4. .
In this case, T satisfies and . Without loss of generality, we may assume that , .
Subclaim 4. for all . Similarly, for all and for all .
Proof of Subclaim 4. By way of contradiction, suppose that for some . Set . Then, , which is a contradiction to Subcase 2. Similarly, suppose that or for some . Set or . Then, , which is a contradiction to Subcase 3. This completes the proof of Subclaim 4.
Subclaim 5. .
Proof of Subclaim 5. It follows from (C4) thatIt , then . By way of contradiction, assume that and . By (25), . Set . Then, , which contradicts (C2).
If for , then . Hence, . By Claim 7, . Moreover, . Otherwise, assume that . Then, . Set . is a 6-ended tree with , which is contrary to Lemma 1. Hence, . By a similar argument as in the proof of (10), we haveThus, . This completes the proof of Subclaim 5.
Subclaim 6. .
Proof of Subclaim 6. From (C4), we haveIf such that , then we assert that . By way of contradiction, assume that and for some . By (34), . Set . Then, , which is a contradiction to Subcase 3. Hence, our assertion holds.
If such that , then and hence . Otherwise, assume that or . Then, and . SetThen, is a 6-ended tree with , which contradicts Lemma 1. By the proof of (14), we have .
Above all, we can get that . This completes the proof of Subclaim 6.
Similar consideration as Subcase 2 shows thatCombining with Subclaims 5 and 6, (5), and (29), we havewhich implies that , which is a contradiction. This completes the proof of Subcase 4.
Subcase 5. .
In this case, T satisfies and . Without loss of generality, we may assume that , . We choose T such that(C5) is as small as possible, subject to (C1) and (C2).It is easy to check that for all . If for some , then set . We assert that . By way of contradiction, assume that and hence , since otherwise we can find a tree satisfying Subcase 4. If , the assertion holds. Hence, we assume that . Since , by Claims 1, 4, and 6, we have for some . SetThen, is a 6-ended tree with , which contradicts Lemma 1. Hence,If , by (C2), . Moreover, we have . By way of contradiction, assume that and for some . Set . Then, or , which is a contradiction. Hence, by (5),From (C5), for all . Thus, if , thenIf , then assume that for some . Then, , , and . Hence, by (14),In either case, (34) holds. Therefore,which implies that , which is a contradiction. This completes the proof of Subcase 5.
Subcase 6. .
If , then similar arguments as in the proof of Subcase 4 show thatIf and for some , then . Otherwise, we can find a tree satisfying .
If and for some , then we can find a tree satisfying . Hence, we can assume that for all .
If , then for and for . If , by similar arguments, we have . Hence, . If , then . Otherwise, we assume that . If , then set , where . Then, is a 6-ended tree with , which contradicts Lemma 1. Thus, . Set , which is contrary to (C2). Therefore, by (14), we can derive thatApplying similar arguments to in the reverse order, we haveHence,Therefore,If, then for and for . If , by (C2) and (C4), . If for some , by (C2), . Hence, , , and . Thus,Similarly, if for some , then , , and . Hence,If for some , then it follows from (C4) that . Then,If , then by similar arguments, . Thus, . Therefore,which implies that , which is a contradiction.
If , then for some and . By similar arguments, we haveSimilarly,If , then . Hence, by (14), . Similarly, . This together with (46)–(49) implies thatwhich implies that , which is a contradiction.
If , we haveFor , if , then . Thus,By similar arguments, (40) holds. Therefore,which implies that , which is a contradiction. This completes the proof of Subcase 6. Since Subcases 1–6 are proved, Case 1 is proved.
Case 2. .
Let be the family of graphs shown in Figure 3. Hence, . Without loss of generality, we may assume that and .
We choose the maximal tree T with 7 leaves such that(C6) is as small as possible, subject to (C1) and (C2).(C7) is as large as possible, subject to (C1), (C2), and (C6).
Claim 11. For every vertex , .
Proof of Claim 11. By way of contradiction, assume that for some and . Set . Then, or . In either case, we get a contradiction. This completes the proof of Claim 11.
From Claim 11, we haveThe proof of Case 2 has been divided into two subcases.
Subcase 7. .
In this case, T satisfies(i).(ii), , subject to (i).Without loss of generality, we may assume that (see Figure 3).
Subclaim 7. .
Proof of Subclaim 7. Suppose, contrary to our claim, that for some . Setwhere is the successor of on . Then, , which is a contradiction to Case 1. This completes the proof of Subclaim 7.
Subclaim 8. for all .
Proof of Subclaim 8. By way of contradiction, assume that for some . Set . Then, , which is a contradiction.
If for some , then . If for , then , since otherwise we can find a tree satisfying Case 1, which is a contradiction. Then, and . Otherwise, set . Then, contradicts Lemma 1 or (C6). Hence, by (5) and (14), we haveThis together with Subcase 4 and (55)–(58) implies thatwhich implies that , which is a contradiction. This completes the proof of Subcase 7.
Subcase 8. Either or .
Subclaim 9. and .
Proof of Subclaim 9. On the contrary, suppose that for some . Set . Hence, or . In either case, we get a contradiction. Similarly, . This completes the proof of Subclaim 9.
By Subclaim 9, we can see thatSimilarly,Combining with (55)–(62), we havewhich implies that , which is a contradiction. This completes the proof of Subcase 8. Since Subcases 7-8 are proved, Case 2 is proved.
If , since , there exist two integers such that . Hence, and . Set . Then, has exactly two branch vertices, which is a contradiction to Case 1. This completes the proof of Theorem 8.
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 is supported by the Foundation of Henan Department of Science and Technology (182102310830), the Foundation of Henan University of Engineering (D2016018), and the Foundation of Henan Educational Committee (20A110016).