Research Article | Open Access

Lihua You, Jieshan Yang, Yingxue Zhu, Zhifu You, "The Maximal Total Irregularity of Bicyclic Graphs", *Journal of Applied Mathematics*, vol. 2014, Article ID 785084, 9 pages, 2014. https://doi.org/10.1155/2014/785084

# The Maximal Total Irregularity of Bicyclic Graphs

**Academic Editor:**Frank Werner

#### Abstract

In 2012, Abdo and Dimitrov defined the total irregularity of a graph as , where denotes the vertex degree of a vertex . In this paper, we investigate the total irregularity of bicyclic graphs and characterize the graph with the maximal total irregularity among all bicyclic graphs on vertices.

#### 1. Introduction

Let be a simple undirected graph with vertex set and edge set . For any vertices , the degree of is denoted by , and the distance is defined as the length of the shortest path between and in . Let , , and be the path, cycle, and star on vertices, respectively.

A graph is regular if all its vertices have the same degree; otherwise it is irregular. Several approaches that characterize how irregular a graph is have been proposed. In [1], Albertson defined the imbalance of an edge as and the irregularity of as More results on the imbalance and the irregularity of a graph can be found in [1–4].

Inspired by the structure and meaning of (1), Abdo and Dimitrov [5] introduced a new irregularity measure, called the total irregularity. For a graph , it is defined as

Although the two irregularity measures capture the irregularity only by a single parameter, namely, the degree of a vertex, the new measure is more superior than the old one in some aspects. For example, (2) has an expected property of an irregularity measure that graphs with the same degree sequence have the same total irregularity, while (1) does not have. Both measures also have common properties, including that they are zero if and only if is regular.

Obviously, is an upper bound of . In [6], the authors derived relation between and for a connected graph with vertices; that is, . Furthermore, they showed that for any tree .

In [5], the authors obtained the upper bound of the total irregularity among all graphs with vertices, and they showed that the star graph is the tree with the maximal total irregularity among all trees with vertices.

Theorem 1 (see [5]). *Let be a simple, undirected graph on vertices. Then,*(1)*. *(2)*If is a tree, then , with equality holds if and only if .*

In [7], the authors investigated the total irregularity of unicyclic graphs and determined the graph with the maximal total irregularity among unicyclic graphs on vertices. In [8], the authors investigated the minimal total irregularity of graphs, and they characterized the graph with the minimal, the second minimal, the third minimal total irregularity among trees, unicyclic graphs or bicyclic graphs on vertices.

Recently, Abdo and Dimitrov [9] also obtained the upper bounds on the total irregularity of graphs under several graph operations including join, lexicographic product, Cartesian product, strong product, direct product, corona product, disjunction, and symmetric difference.

In this paper, we introduce two transformations to study the total irregularity of bicyclic graphs in Section 2 and characterize the graph with the maximal total irregularity among all bicyclic graphs on vertices in Section 3.

#### 2. Some Preliminaries and Two Transformations

In this section, we introduce some definitions, notations, two transformations, and basic properties which we need to use in the proofs of the main results. Other undefined notations may refer to [10].

A bicyclic graph is a simple connected graph in which the number of edges equals the number of vertices plus one. There are two basic bicyclic graphs: -graph and -graph. An -graph, denoted by (see Figure 1), is obtained from two vertex-disjoint cycles and by connecting one vertex of and one of with a path of length (in the case of , identifying the above two vertices, see Figure 2) where and ; and a -graph, denoted by (see Figure 3), is a graph on vertices with the two cycles and having common vertices, where and .

In Figure 1, let be the common vertex of and ; and if , .

In Figure 2, let , , and .

In Figure 3, let , , if ,, and .

A rooted graph has one of its vertices, called the root, distinguished from the others. Let be a rooted tree with and root ().

Let is a rooted path with length at least one and the root is its starting vertex}, is a rooted star and the root is its center}, and and is a star}, where the rooted graph is obtained by identifying the end vertex of with the center of a star and the root of is the root of .

To identify nonadjacent vertices with of a graph is to replace these two vertices by a single vertex incident to all the edges which were incident in to either or . Let and be two graphs: and . The graph denotes the graph resulting from identifying with . Let , , and be the root of the rooted tree . Take and . For example, see Figure 4.

##### 2.1. Transformation

*-Transformation*. Let be a bicyclic graph obtained from () with rooted trees attached, and let be one of the maximal degree vertices of and let be any pendent vertex of which is adjacent to vertex . Let be the graph obtained from by deleting the pendent edge and adding a pendent edge . We call the transformation from to an -transformation on . For example, see Figure 5.

*Remark 2. *Note that in Figure 5, the edge . In fact, for any .

Lemma 3. *Let be a bicyclic graph obtained from with rooted trees attached, and let be the graph obtained from by -transformation as above. Then, .*

*Proof. *Note that after the -transformation on , only the degrees of and have been changed; namely, , , and for any . Clearly,
because for any integer , . Then,

By Lemma 3 and the definition of -transformation, we have the following result.

Lemma 4. *Let be a bicyclic graph obtained from with rooted trees attached.*(1)*If one cannot get a new graph from by -transformation, then there exists some rooted tree such that , where .*(2)*Let be the graph obtained from by repeating -transformation, and one cannot get other graphs from by repeating -transformation. Then there exists some rooted tree such that , where and .*

##### 2.2. Transformation

*-Transformation.* Let be a bicyclic graph and , where , let be the root of the unique rooted tree , let be a vertex of and let , , …, be the pendent vertices adjacent to . Let be the graph obtained from by deleting the pendent edges , , …, and adding pendent edges , , …, . We call the transformation from to a transformation on . For example, see Figures 6–8.

Lemma 5. *Let be a bicyclic graph where , let be the root of the unique rooted tree , let be a vertex of and let , , …, be the pendent vertices adjacent to . Let be the graph obtained from by -transformation (e.g., see Figure 7). Then, .*

*Proof. *For convenience, let . Note that the root of the unique rooted tree, , is not necessarily different with . Clearly, we know that only the degrees of and have been changed after the -transformation; namely, , , and for any vertex . Let . Note that ; then we complete the proof by the following three cases.*Case **1* . In this case, . It implies that the -transformation on is transformation times on . Then, by using Lemma 3 times. *Case **2* *, *. The proof is similar to the proof of Case 1; thus, we omit it.*Case **3* *, *. In this case, vertex is one of the maximal degree vertices of ; namely, for any vertex . Then,

Now we discuss as follows. *Subcase **3.1* *, *. Because and for any , then
*Subcase **3.2* *, *. Because for any , then

Combining the above two subcases, we have

By and for any and the above arguments, we have

Let () be a bicyclic graph, let as defined in Figures 1–3, and let be the root of . Take

Clearly, .

Lemma 6. *Let be a bicyclic graph, where and , let be the root of the unique rooted tree , and let , , …, be the pendent vertices adjacent to . Let be the graph obtained from by -transformation e.g., see Figures 6 and 8. Then,*(1)*.*(2)*, and the equality holds if and only if .*

*Proof. *Clearly, , since .

By the definition of -transformation, is obvious. Now we show that holds.

If when , namely, , then and have the same degree sequence; thus they have the same total irregularity: .

Hence, we assume in the following. Note that only the degrees of vertices and have been changed after the -transformation; namely, , , and for any vertex . Let . Note that ; then we complete the proof by the following two cases.*Case **1* . In this case, vertex is one of the maximal degree vertices of ; namely, for any vertex and for any . Then,

By and and the above arguments, we have
*Case **2* . In this case, . It implies that the -transformation on is transformation time on . Then, by Lemma 3.

By the proof of Lemma 6, we obtain the following result.

Corollary 7. *Let . Then,*(1)*, where .*(2)*, , .*(3)*.*

#### 3. The maximal Total Irregularity of Bicyclic Graphs

In this section, we will obtain the upper bound of the total irregularity among all bicyclic graphs on vertices and characterize the extremal graph.

Observe that any bicyclic graph is obtained from an -graph or a -graph (possibly) by attaching trees to some of its vertices. Denoted by is the set of all bicyclic graphs on vertices. Obviously, consists of three types of graphs: first type, denoted by , is the set of those graphs each of which is an -graph, , with trees attached when ; second type, denoted by , is the set of those graphs each of which is an -graph, , with trees attached when ; third type, denoted by , is the set of those graphs each of which is a -graph, , with trees attached. Then, .

##### 3.1. The Graph with the Maximal Total Irregularity in

In this subsection, the bicyclic graph with the maximal total irregularity in is determined.

Let be positive integers with and ; . Clearly, the degree sequence of is . By simple calculation, we have

Theorem 8. *Let be positive integers with and and let be a bicyclic graph on vertices obtained from with rooted trees attached. Then, , and the equality holds if and only if .*

*Proof. *If , then , and the rooted trees are trivial; namely, with . Then, and .

If , we complete the proof by the following three cases.*Case **1* * where *. Let be the graph obtained from by repeating -transformation, and we cannot get a new graph from by -transformation. Then there exists a rooted tree such that where and by Lemma 4. Let be the root of the rooted tree .*Subcase **1.1* * where *. If , then and .

If , then we can get a new graph by -transformation on ; thus by Lemma 6.*Subcase **1.2* * where *. If , let be the graph obtained from by -transformation. Then by Lemma 5. Repeating -transformation on until we cannot get a new graph, the resulting graph is . By Lemma 4, and thus .

If and , let be the graph obtained from by -transformation; then and by Lemma 5.

If and , Let be the graph obtained from by -transformation and let be the graph obtained from by repeating -transformation until we cannot get a new graph from by -transformation. Then by Lemmas 4 and 5, we know that and .*Case **2* * where *.

The proof is similar to the proof of Subcase 1.1; thus, we omit it.*Case **3* * where *.

The proof is similar to the proof of Subcase 1.2; thus, we omit it.

Combining the above arguments, we complete the proof.

Theorem 9. *Let be positive integers with .*(1)*If , then .*(2)*If , then .*

*Proof. *Clearly, the proofs of and are similar; we only show that holds. Let ; by formula (13), we have .

Theorem 10. *Let be a positive integer, and let be a bicyclic graph on vertices. Then and the equality holds if and only if (see Figure 9).*

*Proof. *Let be positive integers with and and let be a bicyclic graph on vertices obtained from with rooted trees attached. Then by Theorem 8, we have , and the equality holds if and only if . Now we complete the proof by the following three cases.*Case **1* . Then, .*Case **2* * and *. Then by using Theorem 9 times, we have .*Case **3* * and *. The proof is similar to Case 2; thus, we omit it.

##### 3.2. The Graph with the Maximal Total Irregularity in

In this subsection, the bicyclic graph with the maximal total irregularity in is determined.

Let be positive integers with , , and ; or . Clearly, the degree sequence of is . By simple calculation, we have

Theorem 11. *Let be positive integers with , , and and let be a bicyclic graph on vertices obtained from with rooted trees attached. Then, , and the equality holds if and only if or .*

*Proof. *If , then , and the rooted trees are trivial; namely, with . Then, and .

If , we complete the proof by the following three cases.*Case **1* * where *. Let be the graph obtained from by repeating -transformation, and we cannot get a new graph from by -transformation. Then there exists a rooted tree such that , where and by Lemma 4. Let be the root of the rooted tree .*Subcase **1.1* * where *. If , then and .

If , then and .

If and , then we can get a new graph by -transformation on ; thus by Lemma 6.*Subcase **1.2* * where *. If , let be the graph obtained from by -transformation. Then, by Lemma 5. Repeating -transformation on until we cannot get a new graph, the resulting graph is . By Lemma 4, and thus .

If and , let be the graph obtained from by -transformation; then , and by Lemma 5.

If and , let be the graph obtained from by -transformation and let be the graph obtained from by repeating -transformation until we cannot get a new graph from by -transformation. Then by Lemmas 4 and 5, we know that and .*Case **2* * where **. *The proof is similar to the proof of Subcase 1.1; thus, we omit it.*Case **3* * where *. The proof is similar to the proof of Subcase 1.2; thus, we omit it.

Combining the above arguments, we complete the proof.

Theorem 12. *Let be positive integers with and . Then,*(1)*If , then .*(2)*If , then .*(3)*If , then .*

*Proof. *Clearly, the proofs of , , and are similar; we only show that holds. Let ; then by formula (14), we have

Theorem 13. *Let be a positive integer and let be a bicyclic graph on vertices. Then, and the equality holds if and only if (see Figure 10).*

*Proof. *Let be positive integers with , , and and let be a bicyclic graph on vertices obtained from with rooted trees attached. Then by Theorem 11, we have , and the equality holds if and only if or . Note that the results of Theorem 12 also hold for ; then we only consider the case . Now we complete the proof by the following three cases.*Case **1* * and *. Then, .*Case **2* *, **, and *. Then by (1), (2), and (3) of Theorem 12, we have