Abstract
M.Kriesell conjectured that there existed , such that every 5-connected graph with at least vertices can be contracted to a 5-connected graph such that . We show that this conjecture holds for vertex transitive 5-connected graphs.
1. Introduction
All graphs considered here are supposed to be simple, finite, and undirected graphs. For a connected graph , a subset is called a smallest separator if and has at least two components. Let be a -connected graph, and let be a subgraph of . Let stand for the graph obtained from by contracting every component of to a single vertex and replacing each resulting double edges by a single edge. A subgraph of is said to be -contractible if is still -connected. An edge is a -contractible edge if is -connected; otherwise, we call it a noncontractible edge. Clearly, two end-vertices of a noncontractible edge are contained in some smallest separator. A -connected graph without a -contractible edge is said to be a contraction-critical -connected graph.
Tutte’s [1] wheel theorem showed that every 3-connected graph on more than four vertices contains a 3-contractible edge. For , Thomassen and Toft [2] showed that there were infinitely many contraction-critical -regular -connected graphs. On the other hand, one can find that every 4-connected graph can be reduced to a smaller 4-connected graph by contracting at most two edges. Therefore, Kriesell [3] posted the following conjecture.
Conjecture 1. (see [3]). There exists , such that every -connected graph with at least vertices can be contracted to a k-connected graph such that .
Clearly, Conjecture 1 is true for . By Kriesell’s examples [3], Conjecture 1 fails for . Hence, it is still open for .
A smallest separator of a -connected graph is said to be trivial if has exactly two components and one of them has exactly one vertex. A 5-connected graph is essentially a 6-connected graph if every smallest separator of is trivial. In ([3]), Kriesell proved the following results.
(a)
(b)
(c)
(d)
(e)
Theorem 1. (see [3]). Every essentially 6-connected graph with at least 13 vertices can be contracted to a 5-connected graph such that .
In this paper, we will show that Conjecture 1 is true for vertex transitive 5-connected graphs. Clearly, Conjecture 1 holds for 5-connected graphs which contain a contractible edge. Hence, in order to show that Conjecture 1 holds for vertex transitive 5-connected graphs, we have to show that all vertex transitive contraction-critical 5-connected graphs have a small contractible subgraph. So, the key point of this paper is to characterize the local structure of a vertex transitive contraction-critical 5-connected graph and, then, to find the contractible subgraph of it. In the following, for convenience, a vertex transitive contraction-critical 5-connected graph will be called a -5-connected graph. For a contraction-critical 5-connected graph, there are some results on the local structure of it [4–10].
To state our results, we need to introduce some further definitions. Let be a 5-connected graph which is 5-regular. For any , we say that has one of the following four types according the graph induced by the neighborhood of (see Figures 1(a)–1(d)).(i)Type 1: (ii)Type 2: (iii)Type 3: (iv)Type 4:
Moreover, for , has if every vertex of has type .
Furthermore, we need to introduce the graph (see Figure 1(e)). One can check that is vertex transitive, and can be reduced to by contracting and .
First, we have the following results on the local structure of -5-connected graphs.
Theorem 2. Let be a -5-connected graph. If , then either or .
Theorem 3. Let be a -5-connected graph. If , then has type 1, type 2, type 3, or type 4.
Theorem 4. Let be a -5-connected graph. If has type 2, then is isomorphic to icosahedron.
Then, we will prove the following main result of the paper.
Theorem 5. Let be a 5-connected vertex transitive graph which is neither nor icosahedron, and then, can be contracted to a 5-connected such that .
The organization of the paper is as follows. Section 2 contains some preliminary results. In Section 3, we will characterize the local structure of 5-connected -graphs. In Section 4, we will prove Theorem 5.
2. Terminology and Lemma
For terms not defined here, we refer the reader to [11]. Let be a graph, where denotes the vertex set of and denotes the edge set of . Let denote the automorphism group of , and let denote the vertex connectivity of . Let denote a path on vertices. An edge joining vertices and will be written as . Let stand for the new vertex obtained by contracting the edge . For , we define . For , we define . Furthermore, let denote the subgraph induced by , and let denote the graph obtained from by deleting all the vertices of together with the edges incident with them. Let stands for the set of edge with one end in and the other end in .
Let be a smallest separator of a noncomplete connected , and the union of at least one but not of all components of is called a -fragment. A fragment of is a -fragment for some smallest separator . Let be a -fragment, and let . Clearly, , and is also a -fragment such that . A fragment with least cardinality is called an atom. For , and , we often omit the index if it is clear from the context.
Furthermore, we need some special terminologies for 5-connected graphs. Let be a fragment of , and let . Let , and . A vertex is said to be an of if both of the following two conditions hold.
A vertex is said to be an of , if is an of for some . Let (resp. ) stand for the set of admissible vertices of (resp. ). Let be an edge of , and a fragment is said to be a fragment with respect to if .
The following properties of fragment are well known (for the proof, see [12]), and we will use them without any further reference.
Lemma 1. (see [12]). Let and be two distinct fragments of ; , . Then, the following statements hold.(1)If , then (2)If and is not a fragment of G, then and (3)If , then both and are fragments of G, and
Lemma 2. (see [4]). Let be a -connected graph, and is a fragment of . Let . If , then .
Lemma 3. (see [5]). Let be a contraction-critical 5-connected graph, and then, contains a vertex such that every edge incident with is contained in some triangle.
Lemma 4. (see [6]). Let be a contraction-critical 5-connected graph. Let , and be a fragment such that , , and . If , then .
Lemma 5. (see [7]). Let be a fragment of a contraction-critical 5-connected graph such that , and let be two vertices of such that . Then, either or .
Lemma 6. Let be a vertex transitive connected graph, and then, for any two vertices and , .
Proof. Since is a vertex transitive graph, there exist such that . It follows that . Hence, is an isomorphic of and , where is the restriction of on .
Lemma 7. Let be a prime integer, and let be a vertex transitive graph with ; then, is a -regular graph.
Proof. To the contrary, we may assume that since . It follows that every atom of has at least two vertices. Since is a vertex transitive graph, then every vertex of is contained in some atom.
First, we show that any two atoms of are disjoint. Otherwise, let and be two distinguished atoms of such that . By the definition of atom, is not a fragment. Lemma 1 assures us that and . This implies that , a contradiction. Thus, any two atoms of are disjoint.
Let and be two atoms of such that . It follows that . We show that . Otherwise, suppose . If is a fragment of , then we see that , since . This contradicts the definition of atom. So, is not a fragment of . Lemma 1 assures us that and . It follows that , a contradiction. Hence, . Therefore, is the disjoint union of some atom, since any two atoms of are disjoint and every vertex of is contained in some atom. This means that is a subdivision of , and hence, . It follows that . By symmetry, we see that , which implies that , a contradiction.
Lemma 8. Let be a -5-connected graph. If does not contain as subgraph, then for any , .
Proof. Clearly, Lemma 7 assures us that is 5-regular, which implies that has an even order. Suppose that with such that adjacent to at least three vertices of . Let , and it follows . If , then has six vertices. It follows that , which implies contains , a contradiction. Hence, we may assume that . It follows that is a fragment of . By symmetry, we assume that . Now, we observe that and . Let . Let , and then . Now, the fact that is even assures us that . It follows that is a fragment of . Furthermore, we see that and .Now, Lemma 5 assures us that either or . If , then without loss of generality, assume . Therefore, is a connected graph. If , then, similarly, we have that is a connected graph. Now, since is vertex transitive, the following claims hold.
Claim 1. For any , is a connected graph.
Claim 2. For any , contains a cycle of length 4.
Proof. Since is a vertex transitive graph, we only show that has a cycle of length 4. By Claim 1, we see that for . On the other hand, we observe that , since does not contain . This implies that every member of is either adjacent to or . It follows that or . By symmetry, we may assume that . It follows that contains a cycle of length 4. Hence, for any , has a cycle of length 4.
Now, we are ready to complete the proof of Lemma 8. By Claim 2, we see that for . This implies that . Now, we see that every vertex of is adjacent to exactly one vertex of . If , say , then and , a contradiction. If , we can observe that has a vertex with a degree of at most 4, a contradiction. Hence, we may assume that . Now, Lemma 4 shows that , which implies for some , a contradiction.
Lemma 9. Let be a -5-connected graph. If has type 4, then is essentially 6-connected.
Proof. Since has type 4, we see that for any , .
Claim 3. If is a fragment of , then .
Proof. Suppose . If , then has three neighbors in , a contradiction. So, we may assume . It follows that . Let be the path of . It follows that . If , then Lemma 4 implies that . Hence, either or . This is a contradiction, since . Hence, we may assume that . If , then we see that , a contradiction. So, we may assume . It follows that , which contradicts the fact that has an even order. Hence, Claim 3 holds.
Claim 4. If is a fragment of , then .
Proof. We first show that is a connected graph. Otherwise, let be a component of such that has exactly one vertex. It follows that is a fragment of cardinality 2, a contradiction. Next, we show that is a path. Suppose is a cycle, then a simple calculation shows that . This implies that one vertex of , say , has exactly one neighbor in . Now, we find that is a fragment of cardinality 2, a contradiction.
Let be the path of , and let . Without the loss of generality, let . Subclaim 1. and . Proof. Notice that has type 4; we find that . If , then we find that , a contradiction. Hence, . By symmetry, . Without the loss of generality, we may assume that . Now, if , then is a cycle of , a contradiction. Therefore, , which implies that . Subclaim 2. . Proof. If , then has a triangle, a contradiction. It follows that . Similarly, we have . Furthermore, if , then we find that there is a cycle of length four in , a contradiction. Thus, . It follows that .Now, we are ready to complete the proof of Claim 4. Focusing on , we find that since is connected. By Subclaim 2, we may assume that . Now, we find that there is a cycle of length four in , a contradiction.
Claim 5. For every smallest separator , has exactly two components.
Proof. Otherwise, assume that has at least three components. Let , , and be three connected components of . Subclaim 3. For any , , and . Proof. Let , let . Without the loss of generality, we may assume that . Now, we find that , since has type 4. Suppose . If , then the fact that is connected shows that has three neighbors in , which contradicts the fact that has type 4. So, we have . It follows that and .By Subclaim 3, , which implies that is a cycle of length 5. Hence, we see that . Furthermore, by Subclaims 3 and 4, for each . Focusing on , we find that , which implies that . Recall that , and Lemma 4 shows that . Without the loss of generality, assume that . This implies that , which contradicts Subclaim 3. Hence, Claim 5 holds.
Next, we assume that is not essentially 6-connected. It follows that there is a fragment such that and . Let is a fragment such that and , and let . By Claims 3 and 4, we see that . Let and . Let , and let . Now, since is vertex transitive, every vertex of is contained in some member of . Therefore, there exist such that . Next, we will analyse the local structure of and .
Claim 6. If , then .
Proof. Suppose and . Now, Lemma 1 assures us that . It follows thatThis contradicts the choice of .
Claim 7. if and only if .
Proof. Suppose . Now, Lemma 1 assures us that . If , then we see thatThis contradicts the choice of . Hence, we see that . By symmetry, we see that implies .
Claim 8. .
Proof. Suppose . By Claim 6, we know that . Hence, is a fragment of . By the choice of , we know that . Furthermore, since , Lemma 1 assures us that .
If , then Claim 7 assures us that . Furthermore, implies that . Hence, we find that , a contradiction.
So, we may assume . Then, Claim 7 assures us that . Hence, both and are fragments of . By the choice of and , we know that . It follows that .
Since , Lemma 1 assure us that . If , then , a contradiction.
Therefore, let . It follows that . Since and , Lemma 1 assures us that . This implies that . Let . Now, , since is a fragment. Similarly, we find that , , and . Now, we find that has at least two components, a contradiction.
Claim 9. and .
Proof. Suppose . By Claim 7, we see that . Hence, both and are fragments of . By the choice of , we see that . It follows that .
If , then . Hence, we see that , a contradiction.
Hence, we may assume that . Then, by the choice of , we know that . It follows that . Hence, we find that , a contradiction.
Now, we are ready to complete the proof of the Lemma. By Claims 8 and 9, we find that and . Now, we find that , since . It follows that . Now, Lemma 1 implies that . It follows that , a contradiction.
3. The Local Structure of TCC-5-Connected Graphs
In this section, since Lemma 7 holds, all TCC-5-connected graphs were supposed to be 5-regular and have an even order.
Theorem 6. Let be a TCC-5-connected . If , then either or .
Proof. Recall that has an even order. It follows that either or . If , then . So, we may assume . It follows that has a fragment of cardinality 2. Let be a fragment of . Let , and let .
Claim 10. .
Proof. Otherwise, we find that . It follows that . On the other hand, . It follows that , a contradiction.
Let . By symmetry, we may assume that . We find that has at least three neighbors in . Hence, Lemma 8 implies that contains . It follows that contains a triangle.
Claim 11. .
Proof. Suppose . Notice that for , , we see that . Therefore, , which implies that does not contain a triangle, a contradiction.
Now, we observe that is 2-regular. Hence, is a cycle of length 5. Now, by symmetry, we may assume that and . It follows that since is a cycle of length 5. Therefore, we have .
Lemma 10. Let be a -5-connected with . If contains as a subgraph, then has type 1.
Proof. Since is a vertex transitive graph, we know that every vertex of is contained in a . Let be a vertex of , and let . Without the loss of generality, let .
Claim 12. .
Proof. We only show that , and the other one can be handled similarly. Otherwise, by symmetry, we may assume . Let . Let , and it follows that . Furthermore, recall that , . If , then is a separator of order 4, a contradiction. Thus, . Therefore, is a fragment of . Furthermore, since , we see that . Let , and let , where . Similarly, is a fragment of such that . Furthermore, we have .
Notice that and , we see that . Now, we see that , , , , , and .
Now, since , we find that . Let . Clearly, is a fragment with . Notice that , , and . Now, Lemma 4 implies that . It follows that either or has two neighbors in . By symmetry, let have two neighbors in . It follows that , a contradiction. Hence, Claim 12 holds.
Claim 13. .
Proof. We only show that , and the other one can be handled similarly. Otherwise, by symmetry, we may assume that . Now, by Claim 12, . Let and . Clearly, and . Now, since is 5-connected, we observe that . Therefore, is a fragment of . Subclaim 4. . Proof. Suppose , and then, is fragments of . Furthermore, we see that is a connected graph, and this implies that for any , is a connected graph. Let , and it follows , where . Now, since is 5-connected, we see that . We observe that , , , , , and . Furthermore, we see that and . Notice that is connected, and we see that for every vertex of , is connected.Now, since and , the fact shows that . Furthermore, is a fragment.
If , then has at least two components, a contradiction. Therefore, and .
On the other hand, by Claim 12, , and . This fact implies that either or .
By symmetry, we may assume . Now, since , we see that has only one vertex of degree 3. On the other hand, we find that has two vertex of degree 3, and this implies that , a contradiction. This contradiction shows that . By symmetry, . Hence, Subclaim 4 holds. Subclaim 5. . Proof. Suppose . Let be a graph which is got from the path by adding the edge . Clearly, . Now, since , we find that . This implies that . Let . Furthermore, has a triangle, since has a triangle. Therefore, is a triangle. It follows that has a Hamilton cycle. This implies that , a contradiction. Thus, Subclaim 5 holds.By Subclaims 4 and 5, has two components, and one of them has exactly one vertex. If , then , a contradiction. So, assume that , which implies that . By Claim 12 and Subclaim 5, . Now, we see that which contains is contained in . Hence, is a connected graph, a contradiction. Thus, Claim 13 holds.
By Claim 13 and Lemma 3, , and hence, has type 1. Therefore, has type 1.
Theorem 7. Let be a TCC-5-connected . If , then has type 1, type 2, type 3, or type 4.
Proof. If contains , then Lemma 10 assures us that has type 1. So, we may assume that does not contain . Hence, Lemma 8 assures us that for any , . Now, Lemma 3 assures us that has either type 2 or type 3 or type 4.
Theorem 8. Let be a TCC-5-connected . If has type 2, then is isomorphic to icosahedron.
Proof. Let , and let be the cycle of . Furthermore, let . Since has type 2, we may assume that is a cycle of . Let , and then and . Now, is a cycle of . Let . If , then, since has type 2, . This implies that , a contradiction. Hence, we have . Since does not contain , we see that . Now, we observe that is the cycle of . Let , then, similarly, we can show that and . Let , and then and . It follows that is icosahedron.
4. Proof of Theorem 5
Since is 5-regular, we see that is even. If has a contractible edge, then we are done. Therefore, in the rest of the paper, we may assume that is a contraction-critical 5-connected graph. Hence, by Theorem 1, we can assume that . By Theorem 2, we see that has type 1, type 2, type 3, or type 4. Next, we complete the proof of Theorem 5 by showing that the following lemmas are true.
Lemma 11. Let be a -5-connected-graph. Let , be a path of , and . If , then .
Proof. Suppose . Let be a smallest separator of , and let be a -fragment. Clearly, and . Let . Clearly, and . Furthermore, is a fragment of such that . Since is a complete graph, either or , a contradiction. Hence the lemma holds.
Lemma 12. Let be a TCC-5-connected graph such that . If has type 1, then can be contracted to a 5-connected such that .
Proof. By the definition of type 1, we know that contains as a subgraph. Since is vertex transitive graph, every vertex of is contained in some . Let be a vertex of , and let . Furthermore, without the loss of generality, suppose .
Since has type 1, we may let , , and . Clearly, , , and are all different to each other since has type 1.
Let , and let . Now, we see that and , since , and are all different to each other.
If either or , then we are done. So, by Lemma 11, we may assume that and . Let be a smallest separator of , and let be a -fragment. Since , we see that and . Furthermore, we can observe that .
Claim 14. .
Proof. Suppose . Let . It follows that and . Furthermore, . Recall that ; then, either or . Without loss of generality, we may assume that . Then, is a fragment of such . It follows that . Since , we see that . It follows that . Similarly, . Hence, . Now, Lemma 2 assures us that , a contradiction.
By Claim 14, without the loss of generality, let and .
Let , and then, and . Furthermore, is a -fragment and . Clearly, and .
Similarly, we may assume has a fragment such that and . Furthermore, we may assume that and .
Focusing on and , we see that , , , and . If , then, since , we see that , a contradiction. Hence, we may assume . By symmetry, let .
Claim 15. and .
Proof. Suppose . Since , Lemma 1 assures us that and . If , then is a fragment. On the other hand, we find that . Now, Lemma 2 assures us that and . Thus, . It follows that . Now, Lemma 1 assures us that , a contradiction. Hence, . Now, we find that . This implies that . Since and , Lemma 1 implies that and . Hence, we see that , which implies that is a fragment of . It follows that , a contradiction. Hence, we have , and, similarly, .
If , then Lemma 1 assures us that is a fragment of . Since every vertex of has type 1, we see that . Now, Lemma 2 assure us that . This implies that and . Now, Lemma 1 assures us that . Thus, we may assume that . Since , we see that , a contradiction. Hence, we may assume that . It follows that and . Furthermore, and . Now, Lemma 1 assures us that . Hence, we see that is a fragment of . Now, similarly, we see that , a contradiction. Hence, we see that either or .
Lemma 13. Let be a -5-connected graph such that . If has type 3, then can be contracted to a 5-connected such that .
Proof. Clearly, does not contain . Suppose has a fragment of cardinality two, say . Since is 5-regular, we see that . Hence, we see that . This contradicts Lemma 8. Hence, every fragment of contains either one vertex or at least three vertices. Let be a vertex of such that . Let be a path of . Furthermore, let , , and . Since has type 3, we see that and .
Let and . By Lemma 11, we have and . If either or is 5-connected, then we are done. So we may assume and .
Clearly, and . For , let be a smallest separator of and be a -fragment. Since and , we see that every component of has at least two vertices, where . Furthermore, and . Let , where . It follows that , . Clearly, either or , .
Claim 16. For a smallest separator of , the following holds.(1) and (2)If either or , then one component of has exactly one vertex
Proof. (1)By symmetry, we only show that , and the other one can be handled similarly. Suppose , which implies that . Let be a -fragment. Since , we see that . Hence, without the loss of generality, let . Notice the fact that , and we see that has at least two vertices. Now, the fact that assure us that . Similarly, we have . Hence, . Now Lemma 2 assures us that , a contradiction.(2)Suppose . Furthermore, suppose every component of has at least two vertices. Let be a fragment of . Since , we see that . Without the loss of generality, let . Since both and have at least two vertices, it follows that and have at least three vertices.Notice that . If , then . Now, Lemma 2 assures us that , a contradiction. So, we may assume that . Similarly, we see that . Since , we see that . Hence, by symmetry, we may assume and . It follows that and . Thus, Lemma 4 assures us that . Now, we see that , since . Notice that , and Lemma 2 assures us that , a contradiction. Hence, we see that one component of has exactly one vertex. Similarly, we see that the fact assures us that one component of has exactly one vertex.
Claim 17. , .
Proof. We only show that , and the other one can be handled similarly. Suppose . It follows that . Let be a component of , . As , it follows that either or . Similarly, we see that either or . Without the loss of generality, let . It follows that , , and .
It follows that is a smallest separator of . Let , and is a fragment such that . If , then either or . It follows that , a contradiction. Hence, we have . Notice that , and we see that . It follows that .
Hence, we see that is a smallest separator of such that , but both and have cardinality at least two, which contradicts Claim 16.
By Claim 17, without the loss of generality, we may assume that . It follows that . Let be a -fragment. Without the loss of generality, let . On the other hand, by Claim 17, either or . Furthermore, since every component of has at least two vertices, we see that every component of has at least two vertices, where . This implies that every component of has at least three vertices for each . We will complete the proof of the lemma according the following two cases.
Case 1. .
It follows that . Let be a -fragment. Without the loss of generality, suppose .
Now, we see that , , , and .
Claim 18. .
Proof. Otherwise, assume that . Notice that and , and we see that and since . Without the loss of generality, let . Furthermore, we see that and . If , then Lemma 1 assures us that . It follows that . Now, Lemma 1 assures us that and . This implies that , a contradiction.
Thus, we may assume that and, similarly, . Similar to the argument of the last paragraph, we have and . It follows that and . Hence, , a contradiction. Hence, Claim 18 holds.
By Claim 18 and Lemma 1, we see that is a fragment and . Now, Claim 16 implies that . Therefore, . If , it follows that , which implies that , a contradiction. So, we may assume that . Similarly, we may assume that .
Claim 19. and .
Proof. Clearly, is a fragment. It follows that . Hence, and , since .
Now, if , then the fact assures us that . Notice that and , and we have .
Now, Lemma 1 assures us that and . Hence, , a contradiction. This contradiction implies that . Similarly, .
Now, we are ready to complete the proof of Case 1. Notice that , , and , and we see that . Similarly, . Now, Lemma 1 assures us that . It follows that , a contradiction.
Case 2. .
It follows that . Let be a -fragment. Without the loss of generality, suppose . Now, we see that , , , and .
Claim 20. .
Proof. Suppose . If , then . Now, Lemma 1 assures us that . This implies that , which implies that . Now, Lemma 1 assures us that , which implies that , a contradiction.
Thus, we may assume that . Similarly, . Thus, both and are fragments of . Without loss of generality, we may assume . Therefore, and . Notice that , and then either or . Without the loss of generality, let . Now, we find that . Now, Lemma 2 assures us that . Thus, . This contradicts the fact that has type 3.
Now, we are ready to complete the proof of Case 2. By Claim 20, we have that is a fragment, and . This contradicts Claim 16.
Lemma 14. Let be a TCC-5-connected graph such that . If has type 4, then can be reduced to a 5-connected such that .
Proof. Let be a vertex of such that . Let be a path of . Now, Lemma 10 assures us that does not contain . By Lemma 9, every fragment of has cardinality one or at least four.
Claim 21. Either or .
Proof. Suppose and . Let and , where . If , then has at least two components, a contradiction. Thus, . Let . We may assume is the path of . Now, we see that . Let such that . It follows that is a map from to . It follows that either or .
If , then . Now, since , we see that , a contradiction. So, we may assume that . It follows that . Now, since , we see that , a contradiction.
Now, by symmetry, assume that . Let and . Furthermore, let and be the paths of and , respectively. Hence, we may let , and is the path of . Furthermore, we have and .
Let . Lemma 11 assures us that . Suppose . Let be a smallest separator of . Clearly, we observe that .
Claim 22. .
Proof. Suppose . Let . Then, is a smallest separator of . Let be a fragment of which contains . It follows that . Now, Lemma 9 shows that . Clearly, . Now, since , we see that , which is a contradiction. Hence, we may assume that . If , let . Then, is smallest separator . Let be a -fragment which contains . It follows that . Now, Lemma 9 shows that . Clearly, . Hence, is adjacent to either or , a contradiction. Hence, .
Let . Let be a component of , . As , or . Similarly, or . Without the loss of generality, let . Then, and , .
Hence, is a smallest separator of . Let be a fragment. Clearly, . This implies that . Now, Lemma 9 shows that . So, . Recall that , and we find that . This contradicts the fact that .
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 NSFC (no. 11961051) and Natural Sciences Foundation of Guangxi Province (no. 2018GXNSFAA050117).