Abstract
In this paper, we study the properties and structure of the maximal-adjacency-spectrum unicyclic graphs with given maximum degree. We obtain some necessary conditions on the maximal-adjacency-spectrum unicyclic graphs in the set of unicyclic graphs with vertices and maximum degree and describe the structure of the maximal-adjacency-spectrum unicyclic graphs in the set. Besides, we also give a new upper bound on the adjacency spectral radius of unicyclic graphs, and this new upper bound is the best upper bound expressed by vertices and maximum degree from now on.
1. Introduction
The spectral theory of graphs was established in the 1940s and 1950s. It is a branch of mathematics that is widely applied, and itis a powerful tool for solving problems in discrete mathematics. Many of the early results were related to the relationship between the spectrum and the structural properties of a graphs [1–4]. The spectral theory of graphs has been widely used in quantum theory, chemistry, physics, computer science, the theory of communication networks, and information science. Along with the continuous research of the spectral theory of graphs, applications of the spectral theory of graphs have also been found in the fields of electrical networks and vibration theory [5, 6].
Not only has the spectral theory of graphs pushed forward and enriched research into combinatorics and graph theory but also it has been widely used in quantum theory, chemistry, physics, computer science, the theory of communication networks, and information science. The wide range of application of the spectral theory of graphs has led to the spectral theory of graphs becoming a very active field of research over the last thirty to forty years, and large numbers of results are continuously emerging.
There are many results on the adjacency spectral radius for different classes of graphs. Guo et al. [7] have studied the largest and the second largest spectral radius of trees with vertices and diameter . Guo and Shao [8] have studied the first spectral radii of graphs with vertices and diameter . Petrovic et al. [9, 10] have studied the spectral radius of unicyclic and bicyclic graphs with vertices and pendant vertices. Guo et al. [11, 12] have studied the spectral radius of unicyclic and bicyclic graphs with vertices and diameter d.
Let be a connected graph with edge set and vertex set . In this paper, we denote by a edge of , where . Denote the maximum degree of vertex of by . For convenience, we shall sometimes denote simply by . Denote the degree of vertex by . Denote by the set which consists of the vertices adjacent to vertex in . Denote by the shortest distance between vertex and vertex . Denote by the adjacency spectral radius of . If is a connected graph with vertices, where is the vertex set of , is the edge set of , and , then is called a unicyclic graph. We denote the set which consists of the unicyclic graphs with vertices and maximum degree by .
In this paper, we study further the properties and structure of the maximal-adjacency-spectrum unicyclic graphs in the set of unicyclic graphs with vertices and the maximum degree , where . Besides, we study the new upper bound on the adjacency spectral radius of unicyclic with vertices and the maximum degree .
Suppose that , then we obtain some necessary conditions about that is maximal-adjacency-spectrum unicyclic graph in using the following theorems.
Theorem 1. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , if is the Perron vector of , then the degree of the vertices that corresponds to all largest components in the component of is .
Theorem 2. Suppose that is a maximal-adjacency-spectrum unicyclic graph in and the only circle in is , is the Perron vector of ; if the component which corresponds to the vertex in satisfies that , then .
Theorem 3. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , if the only circle in is , and there exists a nonfull internal vertex in the set of , then the number of all the nonfull internal vertices of is one.
Theorem 4. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , let be the only circle in and , if there is no nonfull internal vertex in the set of , then both the following propositions are established:(1)The number of the nonfull internal vertex in is at most 2.(2)When the number of nonfull internal vertex in is 2, then there is at least one vertex with degree 2 in the two nonfull internal vertices in .
Theorem 5. Suppose that is a maximal-adjacency-spectrum unicyclic graph in . If the only circle in is , there is no nonfull internal vertex in the set of , and the number of the nonfull internal vertex in the set of is equal to or greater than 2, then the length of the circle is 3, that is, .
Theorem 6. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , if is the only circle in , then .
Theorem 7. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , and let be the vertex that corresponds to a maximum component in the component of the Perron vector of . is the rooted unicyclic graph with root node , let be the only circle in , and . Let be the shortest distance between vertex and vertex in , denote that = { is the leaf node of , and the shortest path from to neither pass nor pass }, = { is the leaf node of , and the shortest path from to either pass or pass }, then the following propositions are established:(1)If , then .(2)If , then for the arbitrary leaf node in , we all have . If , then for the arbitrary leaf node in , we all have .(3)If , then for the arbitrary leaves in , we all have .(4)If , then .(5)If and there exists a vertex with degree 2 in , then is the rooted unicyclic graph with 3 levels, and for , we all have .
Theorem 8. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , is the Perron vector of , is the vertex that corresponds to a maximum component in the component of the Perron vector of . Let be the rooted unicyclic graph with root node , then is an almost full-degree unicyclic graph with root node .
Theorem 9. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , is the Perron vector of , is the vertex that corresponds to a maximum component in the component of the Perron vector of , and is the rooted unicyclic graph with root node , has only one nonfull internal vertex . Suppose that is the only one circle in , and , denote that = { is the leaf node of , and the shortest path from to neither pass nor pass }, = { is the leaf node of , and the shortest path from to either pass or pass }. Assume that , that is, , then the following propositions are established:(1)If the distances from all leaves of to are all equal, then there exists a leaf node which makes is a pendant edge.(2)If there are two leaves in which make and , then either there exists a leaf node , which makes is a pendant edge or and .(3)If , then there exists a leaf node , which makes is a pendant edge.
Finally, we obtain the main result of this paper in the following theorem.
Theorem 10. Suppose that , and is a rooted unicyclic graph with root vertex which is the vertex that corresponds to a maximum component in the component of the Perron vector of , then is a maximal-adjacency-spectrum unicyclic graph in if and only if .
That is, we describe the structure of the maximal-adjacency-spectrum unicyclic graphs in . In addition, we give a new upper bound on the adjacency spectral radius of unicyclic graphs on the basis of Theorem 10.
In the following discussion of this paper, we assume that the maximum degree of unicyclic graphs satisfies .
2. Preliminaries
2.1. Some Basic Concepts
A rooted unicyclic graph is a simple nonlinear structure. Figure 1 shows a rooted unicyclic graph with root node .
We can divide a rooted unicyclic graph into levels according to the following principles.
The root nodes are in the first level. For instance, in the rooted unicyclic graph with root node shown in Figure 1, is in the first level. In a rooted unicyclic graph, the level which arbitrary vertex is defined as the shortest distance from the vertex to the root node adds 1; for instance, in the rooted unicyclic graph with root node shown in Figure 1, the vertices are in the second level, the vertices are in the third level, and the vertices are in the fourth level. The levels of a rooted unicyclic graph are defined as the maximum of the levels of all the vertices in the rooted unicyclic graph. For instance, the level of the rooted unicyclic graph shown in Figure 1 is 4.
Assume that is a rooted unicyclic graph, is the root node of and is an arbitrary vertex which is not equal to in . Suppose that the shortest path from to is , where , then we call is a father vertex of . For instance, in the rooted unicyclic graph with root node shown in Figure 1, the shortest path from vertex to vertex is , is a father vertex of . The shortest paths from vertex to vertex are , respectively; therefore, and are both father vertices of .
Let be a rooted unicyclic graph, if and are two different vertices of , and have the same father vertex, then is called a brother of . If and are two different vertices in the same level in , and the father vertex of is not equal to the father vertex of , then is called a cousin of . For instance, in the rooted unicyclic graph with root node shown in Figure 1, vertex , , and have the same father vertex ; therefore, is a brother of . Besides, vertices are two vertices in the same level in , and the father vertex of is not equal to the father vertex of ; therefore, is a cousin of .
The vertices (except ) on the shortest path that connects the root node to vertex are called the direct ancestor of the vertex . For instance, in the rooted unicyclic graph with root node shown in Figure 1, the shortest path from the root node to the vertex is ; therefore, and are both the direct ancestors of .
Assume that are two different vertices in the rooted unicyclic graph , the level of the vertex in is and the level of the vertex in is , where and is not the direct ancestor of , then we call is a collateral ancestor of . For instance, in the rooted unicyclic graph with root node shown in Figure 1, the level of vertex in is 3, the level of vertex in is 2, and is not the direct ancestor of ; therefore, vertex is the collateral ancestor of vertex .
Suppose that is a rooted unicyclic graph, if vertex and vertex are both direct ancestors of vertex , and the level of vertex in is larger than the level of vertex in , then the generation of vertex to is closer than the generation of vertex to vertex . For instance, in the rooted unicyclic graph with root node shown in Figure 1, the shortest path from vertex to vertex is , vertices are both direct ancestors of , the root node is in the first level, and vertex is in the second level; therefore, the generation of vertex to is closer than the generation of vertex to vertex .
If , and for any number which satisfies , we all have that is a direct ancestor of , then is called a common direct ancestor of . For instance, in the rooted unicyclic graph with root node shown in Figure 1, is both the direct ancestor of and the direct ancestor of ; therefore, is a common direct ancestor of .
If , is a common direct ancestor of , and which is the arbitrary direct ancestor of satisfies that is either equal to or the direct ancestor of , then is called the common direct ancestor of of the nearest generation. For instance, in the rooted unicyclic graph with root node shown in Figure 1, is the common direct ancestor of of the nearest generation.
Let be two cousins of , if the common direct ancestor of and of the nearest generation is the common direct ancestor of and of the nearest generation, then is the cousin of with generation closer than . For instance, in the rooted unicyclic graph with root node shown in Figure 1, are two cousins of , and the common direct ancestor of and of the nearest generation is , the common direct ancestor of and of the nearest generation is , and is the ancestor of ; therefore, is the cousin of generation closer than .
In order to give the main results of this paper, we introduce some basic definitions and lemmas.
2.2. Some Definitions
Definition 1. Suppose that , if for any all have that establish, then is called a maximal-adjacency-spectrum unicyclic graph in .
Definition 2. Suppose that and are vertices different from each other in graph , if , and for any natural number which satisfies all have , then a path of graph is called an internal path of graph .
Definition 3. Suppose that is a unicyclic graph, and the degree of is , is called a nonfull vertex of , which means satisfies .
Definition 4. Suppose that is a simple connected graph with vertices, the vertices in with degree 1 are called the pendant point of , or call that vertex in the leaf node of ; for convenience, the leaf node of is sometimes simply called leaf node of . The edge associated with the pendant point is called pendant edge.
Definition 5. Suppose that is a unicyclic graph, and the maximum degree of is , if satisfies or , where is shown in Figure 2, and is the root node of . Or is a rooted unicyclic graph with levels more than two, and satisfies the following properties:(1)The vertex in the first level is , and is the root node of ; the vertices in the second level from left to right are .(2)Suppose that the only circle in is , and .(3)The internal vertices of are all full-degree vertices.(4)The distance from all the leaf node nodes of to is equal.Then, is called a completely full degree unicyclic graph.
Definition 6. Suppose is a unicyclic graph, and the maximum degree of is , if satisfies or that , where is shown in Figure 2, and is the root node of . Or is a completely full degree unicyclic graph with levels more than two. Or is a rooted unicyclic graph obtained from another completely full degree unicyclic graph with levels more than two by deleting some right leaf nodes. Then, is called an almost completely full degree unicyclic graph. We denote the almost completely full degree unicyclic graph with vertices and maximum degree by .
Definition 7. Suppose that is a unicyclic graph, and the maximum of is , if satisfies or , where is shown in Figure 2, and is the root node of . Or , where (where is the number which satisfies that ), , and is the root node of . or is a rooted unicyclic graph with levels, and satisfies the following properties:(1).(2)The vertex in the first level is , and is the root node of ; the vertices in the second level from left to right are .(3)Suppose that the only circle in is , and .(4)There is at most one nonfull internal vertex in .(5)When there is only one nonfull internal vertex in , this nonfull internal vertex is in the level of .Then, is an almost full degree unicyclic graph with root node .
From the definitions above, we know that completely full degree unicyclic graph is the special situation of almost completely full degree unicyclic graph. completely full degree unicyclic graph and almost completely full degree unicyclic graph are both special situations of almost full degree unicyclic graph. For convenience, in this paper, we denote completely full degree unicyclic graph, almost completely full degree unicyclic graph, almost full degree unicyclic graph by completely full-degree unicyclic graph, almost completely full-degree unicyclic graph, and almost full-degree unicyclic graph, respectively.
For instance, suppose that the unicyclic graph shown in Figure 3(a) is a rooted unicyclic graph with root node , then the rooted unicyclic graph shown in Figure 3(a) is a completely full 3 degree unicyclic graph with levels 3, maximum degree 3, and root node . Suppose that the root node of the unicyclic graph shown in Figure 3(b) is , then the rooted unicyclic graph shown in Figure 3(b) is an almost completely full 4 degree unicyclic graph with levels 4, maximum degree 4, and root node . Suppose that the root node of the unicyclic graph shown in Figure 3(c) is , then the rooted unicyclic graph shown in Figure 3(c) is an almost full 4 degree unicyclic graph with levels 4, maximum degree 4, and root node .
In order to give the main results of this paper, we give some lemmas.
(a)
(b)
(c)
2.3. Some Lemmas
Now, we give some lemmas which we use to proof the main results.
Lemma 1 (see [13]). Suppose that is a simple connected graph with vertices and maximum , is the Perron vector of , correspond to vertices , respectively, and . If , let , if is still a simple connected graph, then .
Lemma 2 (see [13]). Suppose that is a simple connected graph with vertices and maximum , is the Perron vector of , correspond to vertices , respectively, and . Suppose that . Let , and are shown in Figure 4. If is still a simple connected graph, then .
Lemma 3 (see [13]). Suppose that is a simple connected graph with vertices and maximum , is the Perron vector of , and suppose that correspond to the four different vertices , respectively, where and . Let , and are shown in Figure 5; if is still a simple connected graph, then .
Lemma 4 (see [14]). Suppose that is a simple connected graph, and is an internal path of . If , where , then .
Lemma 5 (see [13]). Suppose that is a connected graph, if , let , then .
Lemma 6. Assume that , the edge set of is , and is the only circle in . Suppose that , where , and for any natural number which satisfies , we all have , denote . Suppose that is the Perron vector of , (where ) is the set which consists of all the vertices whose components of Perron vector are equal to in , if for any arbitrary natural number which satisfies , we all have , then there exists which makes that holds.
Proof. Since for any arbitrary natural number which satisfies , we all have , we can get . Without loss of generality, we assume that have clockwise arrangement in , as shown in Figure 6, and denote .
Since for any natural number which satisfies , we all have . We know that there exist two vertices in and exist two numbers and that satisfy both , and , which make is in clockwise arrangement between and , is adjacent to , and is in clockwise arrangement between and , is adjacent to , and . From the definition of , we know . Let , then it is easy to know . From and , by Lemma 3, we have ; therefore, Lemma 6 holds.
Lemma 7. Assume that are four different vertices of , and . If is the Perron vector of , correspond to vertices , respectively, and . Let , if is still a simple connected graph, then .
Proof. Since is the Perron vector of , we can imply that , where is the adjacency matrix of . For , and is a simple connected graph; hence, we get .
Therefore, Lemma 7 holds.
3. The Properties and Structure of the Maximal-Adjacency-Spectrum Unicyclic Graphs in
3.1. The Properties of the Maximal-Adjacency-Spectrum Unicyclic Graphs in
First, we give the properties of the Perron vector of the maximal-adjacency-spectrum unicyclic graphs in , as in the following theorems:
Theorem 11. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , if is the Perron vector of , then the degree of the vertices that corresponds to all largest components in the component of is .
Proof. Assume that the proposition is not established, we suppose that there exists a vertex in such that and . Denote , it is obvious that ; hence, . We choose a vertex with degree in , suppose that . For is a unicyclic graph, it is easy to know that there are vertices in the set of such that is still a simple connected graph. Denote , then it is easy to know . Again by , by Lemma 2, we get , and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, the hypothesis is not established; therefore, Theorem 11 holds.
Theorem 12. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , and the only circle in is , is the Perron vector of , if the component which corresponds to the vertex in satisfies that , then .
Proof. Assume that the proposition is not established, we suppose that there exists a vertex in such that and . From and Theorem 11, we get . For and is a unicyclic graph, we know that there exists a path whose length is in , which makes , are difference with each other, and is the leaf node.
First, for , we have . Otherwise, suppose there exists a vertex in such that . Now according to the value of , we discuss the following two cases: Case 1. . Since , we have ; thus in the set of , there exists a vertex which satisfies such that is still a simple connected graph. Denote , then it is easy to know . By Lemma 1, we get , and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Case 2. . In this case, it is easy to know that in the set of , there exist vertices such that is still a simple connected graph. Denote , then it is easy to know . From , by Lemma 2, we can get , and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in .From the above discussion of the two cases, we know that the hypothesis is not established. Hence, for , we all have that holds.
Second, denote by , suppose that is the vertex which is nearest to in . Choose two vertices different from in , then , and . Thus, we have that there exists a natural number which satisfies such that at least one of the following two inequality groups: ① and ② holds. Without loss of generality, we assume that there exists a natural number which satisfies such that and . Then, let , then it is easy to know . By Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Therefore, Theorem 12 holds.
Lemma 8. Assume that , is the Perron vector of , and is the only circle in , , suppose that corresponds to the vertices , respectively. If , and , then there must exist such that .
Proof. Assume that the proposition is not established, then is a maximal-adjacency-spectrum unicyclic graph in . From Theorem 12, we can get ; again by Theorem 11, we get ; hence, for any natural number which satisfies , and we all have . Besides, from Theorem 12, we get that for , we all have .
Without loss of generality, we assume that have clockwise arrangement in . Since , choose , then we have . For , we can know are different from each other. From and , we know that ; again from , we can get . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Therefore, Lemma 8 holds.
Lemma 9. Assume that , where , is the circle of , , and for any natural number which satisfies , we all have . If there exists a natural number which satisfies such that , then there exists such that .
Proof. Assume that the proposition is not established, then is a maximal-adjacency-spectrum unicyclic graph in .
Form Theorems 11 and 12, we know that there exists a natural number which satisfies such that and . Now, we discuss the following two cases: Case 1. There are at least two vertices with the degree not less than 3 in . Denote , for any natural number which satisfies , we all have ; hence, for any natural number which satisfies , we all have . And for there exists a natural number which satisfies such that and there are at least two vertices with the degree not less than 3 in , then we know that there exist two natural numbers and which satisfy and such that is an internal path. Let , where . It is easy to know , and by Lemma 4, we have that holds. Suppose that is an arbitrary leaf node of . Let , then it is easy to know , and by Lemma 5, we have that holds, and thus, . Therefore, this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Case 2. There is only one vertex with degree not less than 3 in . In this case, it is easy to know that the only one vertex with degree not less than 3 must be . Suppose that , and have clockwise arrangement in , as shown in Figure 7. From Theorems 11 and 12, we get . From , we know that are different. And for , it is easy to prove that holds. Otherwise, , let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, . Let , then it is easy to know , and by Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in .By consideration of the above two cases, we know that the hypothesis is not established. Therefore, Lemma 9 holds.
Lemma 10. Assume that is a maximal-adjacency-spectrum unicyclic graph in , and is the Perron vector of . Then, for any leaf node and any vertex which is not the leaf node in , we all have that holds, where correspond to the vertices , respectively.
Proof. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , and is the Perron vector of . Let be any one of the vertices whose component is equal to in , then from Theorem 11, we know . Then, for any nonleaf node which is not in , and for the arbitrary leaf node in , we all have that holds. Otherwise, there exists a vertex which is not the vertex and is not a leaf node in , which makes . For is not the leaf node, hence , thus there exists a vertex in the set of such that still belong to . Let , then by Lemma 1, we have that holds, this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Again since , then ; thus, for any vertex which is not the leaf node in , and for any leaf node in , we all have that holds.
For the properties of the nonfull internal vertices of the maximal-adjacency-spectrum unicyclic graph in , we have the following theorems.
Theorem 13. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , if the only circle in is , and there exists a nonfull internal vertex in the set of , then the number of all the nonfull internal vertices of is one.
Proof. Assume that the proposition is not established, then suppose that is a maximal-adjacency-spectrum unicyclic graph in , the only circle in is , there exists a nonfull internal vertex in the set of , and the number of the nonfull internal vertex in is more than 1. Without loss of generality, assume that are two nonfull internal vertices in , where .
Let be the Perron vector of , correspond to the vertices , respectively. Now according to whether and adjacent or not , we discuss the following two cases: Case 1. and adjacent. For case 1, we discuss the following two subcases again according to the value of and . Subcase 1. . Since is the nonfull internal vertex in , then , choose . Let , then it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Subcase 2. . Since is the nonfull internal vertex in , then , choose . Let , then it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Case 2. and are not adjacent. Since are the nonfull internal vertices in , we have that and hold at the same time. For case 2, we discuss the following two subcases again according to the value of and . Subcase 1. . Since and is a unicyclic graph, we know that there exists such that is still a simple connected graph. Denote , then it is easy to know ; by Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Subcase 2. . For and is a unicyclic graph, we know that there exists such that is still a simple connected graph. Denote , then it is easy to know ; by Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in .Hence, from the discussions of case 1 and case 2, we can know that the hypothesis is not established; therefore, Theorem 13 holds.
Theorem 14. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , let be the only circle in , and , if there is no nonfull internal vertex in the set of , then both the following propositions are established:(1)The number of the nonfull internal vertex in is at most 2.(2)When the number of nonfull internal vertex in is 2, then there is at least one vertex with degree 2 in the two nonfull internal vertices in .
Proof. Let be the Perron vector of .(1)Suppose that a vertex in satisfies , then from Theorem 12, we get , and from Theorem 11, we know . Hence, the number of the nonfull internal vertices in is at most 2, and then, (1) of Theorem 14 holds.(2)When the number of nonfull internal vertices in is 2, without loss of generality, we assume that are the nonfull internal vertices in . Suppose that correspond to the vertices , respectively. Assume that and , then we must have that and hold. Hence, and . Choose . Without loss of generality, we assume that , and let , then it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, the hypothesis is not established. Therefore, either or , then (2) of Theorem 14 holds.By consideration, Theorem 14 holds.
3.2. The Structure of the Maximal-Adjacency-Spectrum Unicyclic Graphs in
For the length of the circle of the maximal-adjacency-spectrum unicyclic graphs in , we have the following theorems.
Theorem 15. Suppose that is a maximal-adjacency-spectrum unicyclic graph in . If the only circle in is , there is no nonfull internal vertex in the set of , and the number of the nonfull internal vertex in the set of is equal to or greater than 2, then the length of the circle is 3, that is, .
Proof. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , the only circle in is , there is no nonfull internal vertex in the set of , and the number of the nonfull internal vertices in the set of is not less than 2. Denote , and assume that Theorem 15 is not established, then . Suppose that , and are two different nonfull internal vertices in . Let be the Perron vector of , and suppose that correspond to the vertices . Now according to the value of , we discuss the following two cases: Case 1. . In case 1, there exists a path which satisfies in . Without loss of generality, we assume that . Let , then it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Case 2. . Now according to whether the vertices and adjacent or not, we discuss the following two subcases again: Subcase 1. If and are adjacent. At this time, it is easy to know that there is a path in . Without loss of generality, we assume that . Let , then it is easy to know ; By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Subcase 2. If and are not adjacent. Suppose that . First, we must have that holds; otherwise, there must exist two adjacent nonfull internal vertices in . From the discussion of the subcase 1 in case 2, we know that it will imply a contradiction. Without loss of generality, we assume that the position relationship of the vertices in is shown in Figure 8. Second, we must have and . Otherwise, there are at least one of the two inequalities, and , holds. Without loss of generality, assume that . For , . And since , we have that is an internal path, let , where , then it is easy to know . By Lemma 4, we have that holds. Let be an arbitrary leaf node in , and let , then it is easy to know . By Lemma 5, we have that holds; hence, , and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . From and , we know that and hold. Choose . Without loss of generality, assume that . Let , then it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in .Hence, from the discussion of case 1 and case 2, we know that does not hold. Therefore, Theorem 15 holds.
Theorem 16. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , if is the only circle in , then .
Proof. Suppose that is a maximal-adjacency-spectrum unicyclic graph in and is the only circle in . Denote ; assume that Theorem 16 does not hold, then we have . From Theorems 13 and 15, we can get that there is at most one nonfull internal vertex in . Suppose that , and is the Perron vector of . By Theorem 12, we can get . Then, from , is a maximal-adjacency-spectrum unicyclic graph in , and from Lemma 8, we can know . Now, according to the value of , we discuss the following two cases: Case 1. . Suppose that , from Theorem 12, we have that holds. Without loss of generality, assume that , then we have ; again from Theorem 11, we can get . Without loss of generality, we assume that have anticlockwise arrangement in , and the position of the relationship of these vertices is shown in Figure 9. In the situation of case 1, we could prove that the following six propositions hold.
Proposition 1. For any vertex which satisfies in , we all have that holds.
For there is at most one nonfull internal vertex in , there is at most one of unequal to . Besides, ; otherwise, . Since there is at most one nonfull internal vertex in , hence there must exist an internal path in . Let , where . Then, it is easy to know , and by Lemma 4, we have that holds. Let be an arbitrary leaf node in , and let , then it is easy to know . By Lemma 5, we have that holds. Hence, , and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Then, we have , and . Therefore, for any vertex which satisfies in the set of , we all have that holds.
Proposition 2. .
Assume that is not established, and we have . Without loss of generality, assume that , then we have . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, the hypothesis is not established. Therefore, .
Proposition 3. For , we all have that holds.
And for , we all have that holds. From Proposition 1, we know . Assume that, for , we all have that the proposition is not established, then there must exists a vertex such that .
Proposition 4. and .
If is not established, then . From Proposition 1, we know . Choose , and let , then it is easy to know . From Proposition 2, we know , and by Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, the hypothesis is not established, that is, . The same reason, we can prove that holds.
Proposition 5. .
From Proposition 1, we know and . For , where is the adjacency matrix of , denote , then . From Propositions 3 and 4, we can know and ; thus, we have and . Hence, . Therefore, .
Proposition 6. and .
Assume that , then from , we know . From Proposition 1, we can get . Choose , let , and then it is easy to know . From Proposition 3, we know ; hence, . And for , then by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, we have that holds. And from Proposition 5, we get ; therefore, .
From Proposition 2, we know ; from Proposition 6, we have , and by Lemma 8, we know ; hence, . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in .
Therefore, in conclusion, case 1 implies a contradiction.
Case 2. .
In this case, from Lemma 8, we have that . Let be a vertex in such that . From Theorem 12, we can get that there exists a natural number which satisfies such that . Assume that is the number of the vertices whose components are equal to in . From , we have . Suppose that are all the vertices whose components are equal to in , and are clockwise arrangement in . Then, by Lemma 6, we can suppose that the vertices which follow the clockwise direction in are in sequence.
First, we can prove that holds.
Assume that is not established, then . Then now, we can suppose that the vertices which follow the clockwise direction in are in sequence, where are five different vertices in , and let the circle be shown as Figure 10. From the definitions of and from , we can know that, for any natural number which satisfies , we all have that and hold.
From Lemma 9, we can get . Then for , we all have that holds. Otherwise, there exists such that . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in .
Besides, we also can prove . Otherwise, ; let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in .
Choose ; let , then it is easy to know . From the above conclusion, we have . For , by Lemma 3, we have that holds. This implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, the hypothesis is not established; therefore, holds.
Second, we can prove that holds.
Assume that is not established, then ; hence, we can suppose that the vertices which follow the clockwise direction in are in sequence. From , we know that holds; thus are four different vertices. Let , it is easy to know . From the definitions of and , we know . Since , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Therefore, is not established, then we have .
Denote , from , and we have that holds. Then, we can suppose that the vertices which follow the clockwise direction in are in sequence. From and , we know that are four different vertices in , and the relationship of the vertices of is shown in Figure 11.
For the relationship between and , we must have that holds. Otherwise, there must exist one of the two equations, and , holds. Let , then it is easy to know . If , for , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . If , for , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, . Then, we have that holds.
By Lemma 9, we know ; hence, . Choose ; then from Theorem 11, we get . Let , it is easy to know . From and , by Lemma 3, we know that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in .
Hence, in conclusion, case 2 implies a contradiction.
By consideration of the discussion of case 1 and case 2, we know that when or , a contradiction is implied. Hence, is not established; therefore, and . Then, we have that Theorem 16 holds.
Suppose that is a maximal-adjacency-spectrum unicyclic graph in , then the leaf nodes of have the properties stated in the following theorem.
Theorem 17. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , and let be the vertex that corresponds to a maximum component in the component of the Perron vector of . is the rooted unicyclic graph with root node , let be the only circle in , and . Let be the shortest distance between vertex and vertex in , denote that = { is the leaf node of , and the shortest path from to neither pass nor pass }, = { is the leaf node of , and the shortest path from to either pass or pass }, then the following propositions are established:(1)If , then .(2)If , then for the arbitrary leaf node in , we all have . If , then for the arbitrary leaf node in , we all have .(3)If , then for the arbitrary leaves in , we all have .(4)If , then .(5)If , and there exists a vertex with degree 2 in , then is the rooted unicyclic graph with 3 levels, and for , we all have .
Proof. Assume that is a maximal-adjacency-spectrum unicyclic graph in , and suppose that is the vertex that corresponds to a maximum component in the component of the Perron vector of , and is a rooted unicyclic graph with root node . Suppose that is the only circle in , and , let be the Perron vector of .(1)Assume that is not established, then there exists a vertex such that . Suppose that the shortest path from to is , where . By Lemma 1, it is easy to prove that holds. And from Lemma 10, we get . Let , it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, . Therefore, (1) of Theorem 17 holds.(2)Assume that , and there exists a leaf node in such that . Without loss of generality, we assume that the shortest path from to is , where . From , we know , then we have . Let , then it is easy to know . If , then by Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, . Let , it is easy to know . Since , and from Lemma 10, we know that holds. By Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, when , for any leaf node in , we all have that holds. Similarly, when , for any leaf node in , we all have that holds. Therefore, (2) of theorem holds.(3)Assume that there exist two leaf nodes in such that . Suppose that , then we have . Without loss of generality, we assume that , then we have that and hold. Let be the shortest path from to and be the shortest path from to . According to whether there is a public edge between and or not, we discuss the following two cases: Case 1. There is no public edge between and . We denote the path by and the path by , where ; are shown in Figure 12. Besides, we denote ; by the definition of , we know . For is the leaf node in and is the nonleaf node in , then by Lemma 10, we have . From and , we know that there exists a natural number that satisfies , which make and . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Case 2. There is a public edge between and . From that , and there exists public edge between and , we know that there must exist a leaf node neither equal to nor equal to , which makes the shortest path from to have no public edge with . Denote the shortest path from the leaf node to by , and then the length of is . Otherwise, the difference in length between and at least one of is greater than 1. From the discussion of case 1, we know that this will lead to a contradiction. Besides, there will be . Otherwise, , it will imply that the difference in lengthen between and is more than 1, and has no public edge with . From the discussion of case 1, we know that this will lead to a contradiction. Suppose that are shown in Figure 13, where is and is , where , and there exists such that, for , we all have . Let be , where is denoted by , and denote . Now, we consider the paths and . For is the leaf node in and is the nonleaf node in , then by Lemma 10, we get . If there exists a natural number that satisfies , which make that both and hold. Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, for any natural number which satisfies , , and cannot hold meanwhile. And for , then we have that, for any natural number which satisfies , we all have . Now we consider the paths and . For is the leaf node in and is the nonleaf node in , by Lemma 10, we get . Since , then there exists a natural number that satisfies , which make that both and hold. Let , then it is easy to know . From and , by Lemma 3, we know that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . From the discussion of case 1 and case 2, we know that the hypothesis is not established. Therefore, (3) of Theorem 17 holds.(4)Assume that the proposition is not established, then there exist vertices and such that . From , we know . Without loss of generality, we assume that the shortest path from to is and the shortest path from to is , where . Then, we have that and hold. From , we know that the shortest path from to must pass one of and . Without loss of generality, we assume that the shortest path from to pass , then . By (3) of Theorem 17, we know . Since is the leaf node in and is the nonleaf node in , then by Lemma 10, we get . Let , it is easy to know . From , if , then by Lemma 3, we have that holds. This implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, . By mathematical induction, we get , then we have that holds. We also can prove that holds. Otherwise, . For , let , it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Let , it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, the hypothesis is not established; therefore, (4) of Theorem 17 holds.(5)Assume that is not a rooted unicyclic graph with levels 3, and since , then is a rooted unicyclic graph with levels not less than 4. For there exists a vertex with degree 2 in , without loss of generality, we assume that . From , the level of is not less than 4 and (4) of Theorem 17, and we know that there exists a vertex such that . Suppose that the shortest path from to is , where . Then, from , we know that holds, and from , we know that holds. Let , from and , we know . And by Lemma 1, it is easy to prove . For , then by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, the hypothesis is not established. Therefore, is a rooted unicyclic graph with levels 3. Next, we prove that, for , we all have . Otherwise, assume that there exists a vertex such that . And for is a rooted unicyclic graph with levels 3, then we have that holds. Suppose that the shortest path from to is , where , is neither equal to nor equal to . From that there exists a vertex with degree 2 in , without loss of generality, assume that . Then, from Lemma 1, it is easy to prove that holds. By Lemma 10, we can get . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, the hypothesis is not established. Therefore, for , we all have that holds. In conclusion, (5) of Theorem 17 holds.For the relationship between the maximal-adjacency-spectrum unicyclic graphs in and the almost full-degree unicyclic graphs, we have the following theorem.
Theorem 18. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , is the Perron vector of , and is the vertex that corresponds to a maximum component in the component of the Perron vector of . Let be the rooted unicyclic graph with root node , then is an almost full-degree unicyclic graph with root node .
Proof. Suppose that the only circle in is . From Theorem 16, we get . By Theorem 12, we know , then let . Denote = { is the leaf node of , and the shortest path from to neither pass nor pass }, = { is the leaf node of , and the shortest path from to either pass or pass }.
Now according to , we discuss the following three cases: Case 1. There exist nonfull internal vertices in and exist nonfull internal vertices in . For case 1, according to whether there exist the vertices with degree 2 in or not, we discuss the following two subcases: Subcase 1. There exist nonfull internal vertices in and exist the vertices with degree 2 in . For subcase 1, according to the number of the vertices with degree 2 in , we continue to discuss the following two subcases: Subcase 1.1. There are at least two vertices with degree 2 in . In this case, it is easy to know ; by (1) of Theorem 17, we know ; thus, , where is shown in Figure 2. Hence, is an almost full-degree unicyclic graph with root node . Subcase 1.2. There is only one vertex with degree 2 in . In this case, it is easy to know . For there is only one vertex with degree 2 in , then by (5) of Theorem 17, we know that is a rooted unicyclic graph with levels 3. And for , we all have that holds; hence, it is easy to know that is an almost full-degree unicyclic graph with root node . Subcase 2. There exist nonfull internal vertices in , and the degree of all the vertices in is not equal to 2. In this case, it is easy to know that the degree of all the vertices in are more than 2. From that, there exist nonfull internal vertices in , and from Theorem 14, we know that there is only one nonfull internal vertex in . Without loss of generality, assume that is the nonfull internal vertex in , then we have that and holds. And by Theorem 13, we know that there is only one nonfull internal vertex in . Then, it is easy to know that and hold. Suppose , then . Assume that is not established, then we have . Now choose a vertex which satisfies and in . Choose , then . Otherwise, , let , from and is the nonfull internal vertex in , we know . And by Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Let , it is easy to know . From , and by Lemma 10, we get ; then by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, the hypothesis is not established, and then we have . From , we get . And for , by (3) of Theorem 17, we have that holds. Thus, , from , and from that there is only one nonfull internal vertex in , we know that is an almost full-degree unicyclic graph with root node . By consideration of the discussion of subcase 1 and subcase 2 in case 1, we know that if there exist nonfull internal vertices in , and there exist nonfull internal vertices in , then is an almost full-degree unicyclic graph with root node . Case 2. There exist nonfull internal vertices in , and there is no nonfull internal vertex in . Since there exist nonfull internal vertices in , and there is no nonfull internal vertex in , then there exist nonfull internal vertices out of the circle of . By Theorem 13, we know that there only one nonfull internal vertex in . We denote by , let be the only nonfull internal vertex in . Denote that ; obviously, . Choose a path in which is the starting point and a leaf node of is the terminal point, and passes . Suppose that is , where . From the definition of , we know , then we have that is the only nonfull internal vertex in . Assume that is not an almost full-degree unicyclic graph with root node . If , then from that there is only one nonfull internal vertex in , and from Theorem 17, we have that, in , there exists a path in which is the starting point and a leaf node of is the terminal point. Besides, has no public edge with , and the length of is . Suppose that is , where is the leaf node, then the relationship between and is shown in Figure 14. If , then from that, there is only one nonfull internal vertex in and from Theorem 17, we know that, in , there exists a path in which is the starting point and a leaf node of is the terminal point. Besides, has no public edge with , and from Theorem 17, we know that the length of is not less than . Hence, suppose that is , and the relationship between and is shown in Figure 15. If , from that there is only one nonfull internal vertex in , and from Theorem 17, we know that, in , there exists a path in which is the starting point and a leaf node of is the terminal point. Besides, has no public edge with , and from Theorem 17, we know that the length of is not less than . Hence, suppose that is , and the relationship between and is shown in Figure 16. Now according to the relationship between and , we discuss the following two subcases: let or be the subcase 1, and let be the subcase 2. Subcase 1. If , let , then it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . If , and for , we know that there exists a natural number that satisfies , which make that and hold. Let , it is easy to know . From and , and by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, subcase 1 of case 2 implies a contradiction. Subcase 2. In subcase 2, according to whether is the leaf node or not, we discuss the following two subcases: Subcase 2.1. is the leaf node. In this case, by Lemma 10, we have . If , and for , let , it is easy to know . By Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, . Let , it is easy to know , and by Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Subcase 2.2. is not the leaf node. In this case, if there exists a natural number which satisfies such that , then let , it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, for arbitrary vertex which satisfies , we all have . Specially, we have . From and , we know that there exists a natural number that satisfies , which makes both and hold. Let , it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, in subcase 2, no matter whether is leaf node or not, it will imply a contradiction. By consideration of the subcase 1 and subcase 2 in case 2, we know that the hypothesis is not established. Then, we know that when there exists nonfull internal vertices in , and there exists no nonfull internal vertices in , is an almost full-degree unicyclic graph with root node . Case 3. There is no nonfull internal vertex in , then there are only full-degree vertices and leaf nodes in . From Theorem 17, we know that is an almost full-degree unicyclic graph with root node . In conclusion, for each of the cases, is still an almost full-degree unicyclic graph with root node , then we have that Theorem 18 holds.Assume that is a maximal-adjacency-spectrum unicyclic graph in , and there is only one nonfull internal vertex in , then from the position of the nonfull internal vertex in , we have the following conclusion, as in the following theorem.
Theorem 19. Suppose that is a maximal-adjacency-spectrum unicyclic graph in , is the Perron vector of , is the vertex that corresponds to a maximum component in the component of the Perron vector of , and is the rooted unicyclic graph with root node , has only one nonfull internal vertex . Suppose that is the only one circle in , and , denote that = { is the leaf node of , and the shortest path from to neither pass nor pass }, = { is the leaf node of , and the shortest path from to either pass or pass }. Assume that , that is , then the following propositions are established:(1)If the distances from all leaves of to are all equal, then there exists a leaf node which makes is a pendant edge.(2)If there are two leaves in which make and , then either there exists a leaf node which makes is a pendant edge or and .(3)If , then there exists a leaf node which makes is a pendant edge.
Proof. From Theorem 18, we know that is an almost full-degree unicyclic graph with root node ; suppose that is a rooted unicyclic graph with levels , then is in the level.(1)From , we know ; thus, is a rooted unicyclic graph with the levels not less than 3. First, we have that and holds. Otherwise, there is at least one of the two equations, and , holds. From , we know that there is at most one of the two equations which are and holds. Hence, there is only one of the two equations which are and holds. By (5) of Theorem 17, we get that, for , we all have that holds; hence, . Then, we can get , and this implies a contradiction with the distances from all leaves of to are all equal; hence, and . Assume that (1) of Theorem 19 is not established, then there exists such that is a pendant edge of . Let be the shortest path from to , where , , or . Without loss of generality, we assume that , from , we know . For the range of the value of , we must have that holds. Otherwise, , since the distances from all the leaf nodes in to are equal, then for , we all have . Choose ; let be the shortest path from to , where and . By Lemma 10, we have that holds. From and , we get , and by , we have . In addition, for the relationship of and , we must have . Otherwise, , let , and it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Thus, the hypothesis does not hold; that is, holds. Since for any two leaf nodes in , we all have . And since , where , then for , we all have that holds. Choose , then we have that holds. Without loss of generality, we assume that the shortest path from to is , where and . Since is the nonfull internal vertex, then we have . Otherwise, , let , and it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, . For , hence and have the meaning. For the relationship between and , we have that holds. Otherwise, , let , and it is easy to know . Since and , by Lemma 3, we have that holds, this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . By mathematics induction, we can prove that, for any natural number which satisfies , we all have that holds. From Lemma 10, we can get . For the relationship between and , we have . Otherwise, , let , and it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . By mathematics induction, we can prove that, for any natural number which satisfies , we all have that holds. Hence, we have . For the relationship between and , we have . Otherwise, . Let , and it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Thus, we have that holds, let , and it is easy to know . By Lemma 7, we have that holds, this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . In conclusion, the hypothesis does not hold; hence, (1) of Theorem 19 holds.(2)Assume that (2) of Theorem 19 does not hold, then the nonfull internal vertex in should satisfy that there exists such that is a pedant edge. For there exist two leaf nodes in such that and . Hence by Theorem 17, we know ; thus, is in the third last level of , and this implies a contradiction with is an almost full-degree unicyclic graph with root node . Hence, the hypothesis does not hold, and then, (2) of Theorem 19 holds.(3)Since and , by Theorem 17, we get . And it is easy to know ; hence, . From (5) of Theorem 17, we get that and hold. Assume that (3) of Theorem 19 does not hold, since and , the nonfull internal vertex in should satisfy that there exists such that is a pedant edge of . From , and , we have that holds. Now we will prove . Assume that does not hold, then we have , by , we know that there exists such that . Let the shortest path from to be , where , and , and let the shortest path from to be , where , and is either or . Without loss of generality, we assume that , and it is easy to know ; otherwise, . Let , then it is easy to know , and by Lemma 1, we have that holds. This implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, . For , thus, we get . And for is the leaf node of , is the nonleaf node of , and then by Lemma 10, we have . Thus, we get that and hold. Let , then it is easy to know . By Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Thus, the hypothesis does not hold; hence, . Suppose that the shortest path from to is , where , , and = or = . Without loss of generality, we assume that . From , we can get . Since , there exists such that . Suppose that the shortest path from to is , where , and . For the relationship between and , we have . Otherwise, . Let , then it is easy to know . And by Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . In addition, we also can prove ; otherwise, . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, . By mathematics induction, we can prove that, for any natural number which satisfies , we all have that holds. By Lemma 10, we can get . For the relationship between and , we have . Otherwise, . Let , then it is easy to know . From and , by Lemma 3, we have that holds. This implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . By mathematics induction, we can prove that, for any natural number which satisfies , we all have that holds. Hence, we have . Besides, for the relationship between and , we have . Otherwise, . Let , then it is easy to know . By and , we can get . And for , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Let , then it is easy to know . From and , we can get . By Lemma 7, we have , and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . In conclusion, the hypothesis does not hold, and hence, (3) of Theorem 19 holds.Finally, we give the structure of the maximal-adjacency-spectrum unicyclic graphs in the set of unicyclic graphs given the number of vertices and the maximum degree; in the following Theorem, we describe the structure of the maximal-adjacency-spectrum unicyclic graphs in .
Theorem 20. Suppose that , and is a rooted unicyclic graph with root vertex which is the vertex that corresponds to a maximum component in the component of the Perron vector of , then is a maximal-adjacency-spectrum unicyclic graph in if and only if .
Proof. Necessity, suppose that , is a maximal-adjacency-spectrum unicyclic graph in , let be the Perron vector of , and be the vertex that corresponds to a maximum component of . Suppose that is a rooted unicyclic graph with root node ; from Theorem 18, we have that is an almost full-degree unicyclic graph with root node .
Denote the only circle in by ; from Theorem 16, we have . Again by Theorem 12, we could suppose that . Denote = { is the root node of , and the shortest path from to neither pass nor pass }, = { is the root node of , and the shortest path from to either pass or pass }.
When the level of is 2, obviously, .
When the level of is 3, if there is no nonfull internal vertex in , then it is easy to know that holds. If there exists nonfull internal vertex in , then according to the degree of the nonfull internal vertex in , we discuss the following two cases: Case 1. There is one nonfull internal vertex in with degree 2. In this case, we can prove . Otherwise, ; that is, we have that holds, and thus, . From (1) of Theorem 17, we know , then is a rooted unicyclic graph with levels 2, and this implies a contradiction with the level of is 3. Without loss of generality, we assume that ; hence, . And for there is one nonfull internal vertex with degree 2 in , hence , by (5) of Theorem 17, we get that, for , we all have that holds. Therefore, . Case 2. The degree of all the nonfull internal vertices in is not less than 3. In this case, from Theorem 14, we can get that there is only one nonfull internal vertex in . Without loss of generality, we assume that is the nonfull internal vertex. For the degree of all the nonfull internal vertices in are not less than 3, the level of is 3, and ; we have . Then, by (4) of Theorem 17, we get . We can prove . Otherwise, there exists such that . Suppose that the shortest path from to is , where , and is neither nor . According to the relationship between and , we have . Otherwise, , let , and it is easy to know . By Lemma 1, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Let , then it is easy to know . Again by Lemma 10, we get . And we have proved ; hence, by Lemma 3, we have that holds. This implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, . From that and there is only one nonfull internal vertex in , we know that holds. In conclusion, when the level of is 3, we have that holds.Now we discuss the case in which the level of is not less than 4.
Suppose that , from the level of is not less than 4, we know .
Denote = {, and the shortest path from to pass neither nor }, = {, and the shortest path from to pass either or }.
From Theorem 18, we get that is an almost full-degree unicyclic graph, then if there exists nonfull internal vertex in , we have that the nonfull internal vertex in the second last level of . And since the level of is not less than 4, if there exists nonfull internal vertex in , then the nonfull internal vertex must belong to the set of . Again from Theorem 13, we get that there is only one nonfull internal vertex in . Thus, when the level of is not less than 4, there is at most one nonfull internal vertex in .
When the level of is not less than 4, we discuss the following three cases according to the structure of . Case 1. There is no nonfull internal vertex in . At this time, we can discuss the following three subcases: Subcase 1. When for arbitrary leaf nodes in , we all have . At this time, it is easy to know that holds. Subcase 2. When there exist two leaf nodes in such that and . At this time, from Theorem 17, we get . And for , we all have that holds. Subcase 3. When . At this time, from Theorems 17 and 19, we know that the following two conclusions hold. ① . ② all have that holds. Case 2. There is only one nonfull internal vertex in , and one of the brothers of is leaf node. Now, it is easy to know that there exist two leaf node nodes in such that , then according to case (2), we discuss the following two subcases: Subcase 1. When there exist two leaf nodes in , which make and . At this time, from Theorems 17 and 19, we know that the following three conclusions hold: ① ; ② all have that holds; ③ the nonfull internal vertex . Subcase 2. When . From Theorems 17 and 19, we know that the following three conclusions hold: ① . ② all have that holds. ③ The nonfull internal vertex . Case 3. There is only one nonfull internal vertex in , and all the brothers of are full-degree vertices. According to case (3), we could discuss the following three subcases: Subcase 1. When arbitrary leaf nodes in all have . At this time, from Theorem 19, we get that the nonfull internal vertex , and all have that holds. And for there is only one nonfull internal vertex in , then we have that, for , we all have ; hence, holds. Subcase 2. When there two exist leaf nodes in , which make and . At this time, from Theorem 19, we get that the nonfull internal vertex . Since there exist two leaf node nodes such that and in , by Theorem 17, we get . Subcase 3. When . At this time, from Theorem 19, we get . Since , by (3) and (4) of Theorem 17, we can get . And for there is only one nonfull internal vertex in , and ; hence, for , we all have . In the following, we call the subcases (3.2) and (3.3)” case 3″ and the subcases (1.2), (1.3), (2.1), and (2.2) “case 4”.Since , we have that is a rooted unicyclic graph with levels .
Now according to the notation of the vertices in , we make the following instructions: ① except , if , then let express one of the vertices in the bottom layer of the first of . ② If , we denote the ancestor of , which is in the bottom layer of the first of , (note: can equal to any one of the vertex , and does not express that is in the first last level of and does not express that is in the second last level of . The above notations in ① and ② are still established when is changed for any other vertex, which is not ).
Assume that , then the structure of must be as the following case 3 or case 4 shown. Case 3. From the discussion of subcases 3.2 and 3.3, we know that there must exist a leaf node and an nonfull internal vertex in (it is easy to know that is in the second last level of , denote ), which make satisfy the following properties: ① at least has one full-degree brother. ② The common direct ancestor of and of the nearest generation is in the bottom layer of the first of , and . We choose a full-degree brother of , denote the common direct ancestor of and of the nearest generation is , then the relationship of the vertices in case 1 is shown in Figure 17, where in Figure 17, is a brother of and . If , then let , it is easy to know . And by Lemma 1, we can get , and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, . And from Theorem 20, we know . Let , and it is easy to know that . If , from , and by Lemma 3, we have that holds. This implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in ; hence, . Let , then it is easy to know that . From and , by Lemma 3, we have that holds. This implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, the hypothesis is not established; therefore, . Case 4. For is an almost full-degree unicyclic graph with root node , , and from the discussion of subcase 1.2, subcase 1,3, subcase 2.1, and subcase 2.2, we know that there must exist two leaf nodes in the second level of , which satisfy the following properties. There is a cousin or brother of the nonleaf vertex which is of the nearest generation of , there also exists a cousin or brother of the nonleaf vertex which is of nearest generation of . This makes that (where ) is the common direct ancestor of and of the nearest generation, (where ) is the common direct ancestor of and of the nearest generation, is the common direct ancestor of of the nearest generation, and and satisfy that , and . Without loss of generality, we assume that . Then, the relationship of the vertices in case 2 is shown as Figure 18. From the method of marking the index of the vertices in , we know , and . First, by Lemma 10, it is easy to know . If , then . Since , there is a natural number that satisfies , which make and . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . If , since , there exists a natural number that satisfies , which makes and . Let , then it is easy to know . From and , by Lemma 3, we have that holds, and this implies a contradiction with is a maximal-adjacency-spectrum unicyclic graph in . Hence, the hypothesis is not established; therefore, . In conclusion, if is a maximal-adjacency-spectrum unicyclic graph in , and the level of is not less than 4, we have that holds.By considering the discussion of all the cases according to the level of , we know that if , is a unicyclic graph in which the vertex corresponding to a maximum component in the component of the Perron vector of is the root node, and is a maximal-adjacency-spectrum unicyclic graph in , then .
Sufficiency, suppose that , is a rooted unicyclic graph in which the vertex corresponding to a maximum component in the component of the Perron vector of is the root node, and . It is easy to know , then there must exist maximal-adjacency-spectrum unicyclic graph in . Suppose that is a maximal-adjacency-spectrum unicyclic graph in , and let be a rooted unicyclic graph in which the vertex corresponding to a maximum component in the component of the Perron vector of is the root node. Then by the necessity of Theorem 20, we know . And for , then ; thus, is a maximal-adjacency-spectrum unicyclic graph in .
From Theorem 20, we know that if we regard two isomorphic graphs as one graph, then there is only one maximal-adjacency-spectrum unicyclic graph in , and the maximal-adjacency-spectrum unicyclic graphs in is .
4. A New Upper Bound on the Adjacency Spectral Radius of the Unicyclic Graphs
In the following, on the basis of Theorem 20, we give a new upper bound on the adjacency spectral radius of the unicyclic graphs.
4.1. A New Upper Bound on the Adjacency Spectral Radius of the Unicyclic Graphs
About the upper bound on the adjacency spectral radius of the unicyclic graphs, Hu [3] has given that if , then one upper bound of is .
Suppose that , where . Let be the only circle in , and suppose that . Then, is a forest composed of , where are the rooted trees with the roots , respectively.
Denote .
Rojo [15] has given another upper bound on the adjacency spectral radius of the unicyclic graphs through the following theorem.
Theorem 21 (see [15]). Assume that , if and , or , then we have .
In order to give a new upper bound on the adjacency spectral radius of the unicyclic graphs, we first introduce some definitions and lemmas.
Definition 8. For the given natural number and , which satisfy . We suppose that is a unicyclic graph with the maximum degree , and let be the only circle in , be the vertex which corresponds to a maximum component in the components of the Perron vector of , be the rooted unicyclic graph with root node , and satisfies the following properties:① is an almost completely full-degree unicyclic graph with the maximum degree .②.③There is no nonfull internal vertex in .④Let the set of the vertices of which is the only circle in be , denote = { is the leaf node of , and the shortest path from to neither pass nor pass }, = { is the leaf node of , and the shortest path from to either pass or pass }, then we have that hold.Then, we call is a completely Bethe unicyclic graph with the maximum degree , the length of the circle is 3, and the adjacency spectral radius is not less than .
Definition 9. According to the given natural number and , which satisfy . Let be a completely Bethe unicyclic graph with the maximum degree , the length of the circle is 3, and the adjacency spectral radius is not less than . If for any completely Bethe unicyclic graph with the maximum degree , the length of the circle is 3 and the adjacency spectral radius is not less than , we all have that holds, then we call is the minimum completely Bethe unicyclic graph with the maximum degree , the length of the circle is 3, and the adjacency spectral radius is not less than . We denote the minimum completely Bethe unicyclic graph with the maximum degree , the length of the circle is 3, and the adjacency spectral radius is not less than by .
Through the direct calculation, we have , where denotes the smallest positive integral which is not less than .
For convenience, in the following proof process, now we give another notation of .
Denote , it is easy to know , and the orderly array is only determined by . Besides, it is easy to know that if there exist the natural number and that satisfy , which make the natural number and satisfy and . Then, is only determined by the orderly array ; that is, there exist the natural number and that satisfy , which make that the natural number and satisfy . Thus, the orderly array can only determine the structural of which satisfies and . Hence, we can denote by . It is easy to know that if denote , then we have that holds.
Lemma 11. (see [16]). Suppose that , where , if denote , , then we have that holds.
Lemma 12. Suppose that , where , if denote , then when and , or when , we have that holds.
Proof. By the definition of and Theorem 21, it is easy to prove that Lemma 12 holds.
Now we give a new upper bound on the adjacency spectral radius of the unicyclic graphs, that is, the following theorem.
Theorem 22. Suppose that , then the following holds:(1)When and , we have that holds, and the necessary and sufficient condition for the equal sign establishes is , that is .(2)When and , we have that holds, where is the maximum real root of the equation , and the necessary and sufficient condition for the equal sign is , that is, . Through the calculation by Matlab, we get .(3)When and , or when , we all have that holds, where .
Proof. (1) When and , it is easy to know ; by the definition of and Theorem 20, it is easy to have that holds. From Theorem 20, we know that the equal sign establishes if and only if . If we denote , then from Lemma 11, we know that holds. Hence, we have , through direct calculation, we get ; thus, we have that holds.
From the above proof process, we know that the necessary and sufficient condition for the equal sign in the inequality establish is , that is, .
(2) When and , it is easy to know ; by the definition of and by Theorem 20, it is easy to have that holds, and from Theorem 20, we know that the equal sign establishes if and only if . If denoted , then from Lemma 11, we know that holds. Hence, we have , through the direct calculation, we get that is the maximum real root of the equation . Denote the maximum real root of the equation by ; thus, we have that holds.
From the above proof process, we know that the necessary and sufficient condition for the equal sign in the inequality established is , that is, .
(3) When and , or when , denote , then we have . By the definition of and Theorem 20, we have that holds. Again by Lemma 12, we get ; hence, we have that holds, where .
4.2. The Comparison of the Results in Theorems 21 and 22
Choose , where is shown in Figure 19. By Lemma 11, we have that holds, and by Theorem 22, we get . Obviously, we have that holds; that is, the upper bound of that Theorem 22 gives is better than the one that Theorem 21 gives.
Actually, when and the length of the only circle in is 3, the upper bound of that Theorem 22 gives is either equal to the one that Theorem 21 gives or better than the one that Theorem 21 gives.
Choose , where is shown in Figure 20. By Theorem 21, we have that holds, and by Theorem 22, we can get . Obviously, we have that holds; that is, the upper bound of that Theorem 22 gives is better than the one that Theorem 21 gives.
Notice that when and the length of the only circle in is not less than 4, the upper bound of that Theorem 22 gives may not be better than the one that Theorem 21 gives.
In conclusion, sometimes, the upper bound on the adjacency spectral radius of the unicyclic graphs that Theorem 22 gives is better than the one that Theorem 21 gives.
Data Availability
The data used to support the findings of this study are included within the article.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This research was supported by the Scientific Research Foundation of Huaqiao University (10HZR26) and the Natural Science Foundation of Fujian Province (Z0511028).