/ / Article

Research Article | Open Access

Volume 2019 |Article ID 4271783 | https://doi.org/10.1155/2019/4271783

Jia-Bao Liu, S. N. Daoud, "Number of Spanning Trees in the Sequence of Some Graphs", Complexity, vol. 2019, Article ID 4271783, 22 pages, 2019. https://doi.org/10.1155/2019/4271783

# Number of Spanning Trees in the Sequence of Some Graphs

Revised26 Dec 2018
Accepted09 Jan 2019
Published12 Mar 2019

#### Abstract

In mathematics, one always tries to get new structures from given ones. This also applies to the realm of graphs, where one can generate many new graphs from a given set of graphs. In this work, using knowledge of difference equations, we drive the explicit formulas for the number of spanning trees in the sequence of some graphs generated by a triangle by electrically equivalent transformations and rules of weighted generating function. Finally, we compare the entropy of our graphs with other studied graphs with average degree being 4, 5, and 6.

#### 1. Introduction

Deriving closed formulas of the number of spanning trees for various graphs has attracted the attention of a lot of researchers. The importance of this research line is in fact due to the following:(1)Enumerating specific chemical isomers.(2)Counting the number of Eulerian circuits in a graph.(3)Solving some computationally hard problems such as the Steiner tree problem and traveling salesman problem.(4)Deriving formulas for different type of graphs, which can be helpful in identifying those graphs that contain the maximum number of spanning trees; such investigation has practical consequences related to network reliability [1â€“12].

The number of spanning trees of a finite connected undirected graph is an acyclic - edge spanning subgraph.

There exist various methods for finding this number. Kirchhoff [13] gave the famous matrix tree theorem: if is the diagonal matrix of the degrees of and denoting the adjacency matrix of , Kirchhoff matrix has all of its cofactors equal to .

Another method to count the complexity of a graph is using Laplacian eigenvalues. Let be a connected graph with vertices. Kelmans and Chelnoknov [14] derived the following formula:where are the eigenvalues of the Kirchhoff matrix .

Degenerating the graph through successive elimination of contraction of its edges represents the core of another way to compute the complexity of a graph [15, 16]. If is a multigraph with , then is the graph obtained from by contracting the degree until its endpoints are a single vertex. The formula for computing the number of spanning trees of a multigraph is given by This formula is beautiful but not practically useful (it grows exponentially with the size of the graph and may be as many as terms).

Many works have conceived techniques to derive the number of spanning tree of a graph which can be found in [17â€“20].

#### 2. Electrically Equivalent Transformations

Kirchhoff's motivation was a study of electrical networks: an edge-weighted graph can be regarded as an electrical network, where weights are the conductance of the respective edges. The effect conductance between two specific vertices can be written as the quotient of (weighted) number of spanning trees and the (weighted) number of so called thickets, i.e., spanning forests with exactly two components and property that each of the components contains precisely one of the vertices [21]. In the following, we list the effect of some simple transformations on the number of spanning trees. Let be an edge weighted graph, be the corresponding electrically equivalent graph, and denote the weighted number of spanning trees .(i)Parallel edges: if two parallel edges with conductances and in are merged into a single edge with conductances in , then (ii)Serial edges: if two serial edges with conductances and in are merged into a single edge with conductance in , then (iii) transformation: if a triangle with conductances , and in is changed into an electrically equivalent star graph with conductances , , and in , then (iv) transformation: if a star graph with conductances , and in is changed into an electrically equivalent triangle with conductances , , and in , then .

In mathematics one always tries to get new structures from given ones. This also applies to the realm of graphs, where one can generate many new graphs from a given set of graphs. In this work, we compute the number of spanning trees of five sequences of graphs generated by a triangle; we named it , and , respectively.

#### 3. Number of Spanning Trees of the Sequence Graphs

Consider the sequence of graphs constructed as shown in Figure 1.

According to the construction, the number of total vertices and edges are and , . It is clear that the average degree of is in the large limit which is .

Theorem 1. For , the number of spanning trees in the sequence of the graph is given by

Proof. We use the electrically equivalent transformation to transform to . Figure 2 illustrates the transformation process from to .
Using the properties given in Section 2, we have the following the transformations:Combining these eleven transformations, we have Further where ,
Its characteristic equation is which have two roots and . Subtracting these two roots into both sides of , we getLet . Then by (7) and (8), we get and .
Therefore ThusUsing the expression and denoting the coefficients of and as and , we have Thus, we obtainwhere , and , . By the expression and using (12) and (13), we haveThe characteristic equation of (16) is which have two roots and . The general solutions of (16) are Using the initial conditions , and , yieldsIf , it means that is without any electrically equivalent transformation. Plugging (18) into (15), we haveWhen , which satisfies (19). Therefore the number of spanning trees in the sequence of the graph is given bywhere Inserting (21) into (20), we obtain the result.

#### 4. Number of Spanning Trees of the Sequence Graphs

Consider the sequence of graphs constructed as shown in Figure 3.

According to the construction, the number of total vertices and edges are and , .

It is clear that the average degree of is in the large limit which is .

Theorem 2. For , the number of spanning trees in the sequence of the graph is given by

Proof. We use the electrically equivalent transformation to transform to . Figure 4 illustrates the transformation process from to .
Using the properties given in Section 2, we have the following transformations:Combining these nine transformations, we have Further where , . Its characteristic equation is , which have two roots and . Subtracting these two roots into both sides of , we getLet . Then by (26) and (27), we get and
Therefore Thus Using the expression and denoting the coefficients of and as and we haveThus, we obtainwhere and . By the expression and (31) and (32), we have The characteristic equation of (35) is which have two roots and . The general solutions of (35) are ; .
Using the initial conditions , and , yieldsIf , it means that is without any electrically equivalent transformation. Plugging (36) into (34), we haveWhen , which satisfies (37). Therefore the number of spanning trees in the sequence of the graph is given bywhere Inserting (39) into (38) we obtain the result.

#### 5. Number of Spanning Trees of the Sequence Graphs

Consider the sequence of graphs constructed as shown in Figure 5.

According to the construction, the number of total vertices and edges are , , . It is clear that the average degree of in the large limit is

Theorem 3. For , the number of spanning trees in the sequence of the graph is given by

Proof. We use the electrically equivalent transformation to transform to . Figure 6 illustrates the transformation process from to .
Using the properties given in Section 2, we have the following the transformations:Combining these eleven transformations, we have Further where , .
Its characteristic equation is which has two roots and . Subtracting these two roots into both sides of , we getLet . Then by (44) and (45), we get and
Therefore . ThusUsing the expression and denoting the coefficients of and as and , we haveThus, we obtainwhere and . By the expression and (48) and (49), we haveThe characteristic equation of (52) is which have two roots and . The general solutions of (52) are ; .
Using the initial conditions and yieldsIf , it means that is without any electrically equivalent transformation. Plugging (53) into (51), we haveWhen , which satisfies (54). Therefore the number of spanning trees in the sequence of the graph is given bywhere Inserting (56) into (55), we obtain the result.

#### 6. Number of Spanning Trees of the Sequence Graphs

Consider the sequence of graphs constructed as shown in Figure 7.

According to the construction, the number of total vertices and edges are , , .

It is clear that the average degree of in the large limit is .

Theorem 4. For , the number of spanning trees in the sequence of the graph is given by .

Proof. We use the electrically equivalent transformation to transform to . Figure 8 illustrates the transformation process from to .
Using the properties given in Section 2, we have the following the transformations:Combining these nine transformations, we have Further where , .
Its characteristic equation is which have two roots and . Subtracting these two roots into both sides of , we getLet .. Then by (60) and (61), we get and
Therefore . ThusUsing the expression