The Number of Matching Equivalent for the Union Graph of Vertices and Cycles

Xiaoling Wang

School of Mathematics and Statistics, Qinghai Nationalities University, Xining, China.

DOI: 10.4236/am.2021.126032                

Abstract

For two graphs G and H, if G and H have the same matching polynomial, then G and H are said to be matching equivalent. We denote by δ (G), the number of the matching equivalent graphs of G. In this paper, we give δ (sK₁ ∪ t₁C₉ ∪ t₂C₁₅), which is a generation of the results of in [1].

Keywords

Graph, Matching Polynomial, Matching Equivalence

1. Introduction

This paper only considers finite undirected simple graphs. Let G be a graph with n vertices. The matching of G means a spanning subgraph of G, and each of its connected branches is either an isolated vertex or an isolated edge. t-matching means matching that there are t edges. Matching polynomial of graph G is defined as follows in [2]:

$\mu \left(G,x\right)={\displaystyle \underset{t\ge 0}{\sum }{\left(-1\right)}^t{\alpha }_t\left(G\right)x^{n-2t}}$, (1)

where ${\alpha }_t\left(G\right)$ is number of t-matching for G.

If graphG and graphH satisfy $\mu \left(G,x\right)=\mu \left(H,x\right)$, then we say G and H are matching equivalence, denoted as $G~H$.

Set $\delta \left(G\right)$ which denotes the number of matching equivalent graphs of all different isomorphisms for graphG, if $\delta \left(G\right)=1$, we say graphG is a unique matching. In [2], the authors gave some elegant properties of matching polynomials, and proved that to find the matching polynomial of a graph is an NP-problem. Thus, in [1], the authors studied the number of matching equivalent graphs of some vertices and some cycle-union graphs. It turns out that the problem is not simple. In this paper, we study the number of matching equivalent for the union graph of vertices and cycles, that is $\delta \left(s{K}_1\cup t_1{C}_9\cup t_2{C}_{15}\right)$.

Throughout the paper, K₁ denotes an isolated vertex, $P_n\left(n\ge 2\right)$ denotes the path including n vertices; $C_m\left(m\ge 3\right)$ denotes a cycle that including m vertices; $T_{i,j,k}$ denotes a tree that has only a 3-degree vertex, three 1-degree vertex, and the distance between this 3-degree vertex and three 1-degree vertex is $i,j,k$ separately; $D_n\left(n\ge 4\right)$ denotes a graph that produced by bonding a vertex on a triangle to an end of a path $P_{n-2}$ ; nG denotes disjoint union of n graph G.

2. Preliminaries

Lemma 2.1 [3] Suppose graph G has k connected component: $G_1,G_2,\cdots ,G_k$, then $\mu \left(G,x\right)={\displaystyle {\prod }_{i=1}^k\mu \left(G_i,x\right)}$, the roots of matching polynomials are all real numbers, denote $M\left(G\right)$ as the maximum root of $\mu \left(G,x\right)$.

Lemma 2.2 [3] Suppose G is connected graph, then $M\left(G\right)<2$ if and only if $G\in \Gamma =\left\{K_1,P_n,T_{1,1,n},T_{1,2,2},T_{1,2,3},T_{1,2,4},C_n,D_4\right\}$.

Lemma 2.3 [3] (i) $M\left(C_m\right)=M\left(T_{1,1,m-2}\right)=M\left(P_{2m-1}\right)$.

(ii) $M\left(C_6\right)=M\left(T_{1,1,4}\right)=M\left(T_{1,2,2}\right)=M\left(D_4\right)=M\left(P_{11}\right)$.

(iii) $M\left(C_9\right)=M\left(T_{1,1,7}\right)=M\left(T_{1,2,3}\right)=M\left(P_{17}\right)$.

(iv) $M\left(C_{15}\right)=M\left(T_{1,1,13}\right)=M\left(T_{1,2,4}\right)=M\left(P_{29}\right)$.

Lemma 2.4 [4] [5] (i) The matching equivalent graph of $K_1\cup C_m\left(m\ne 6,9,15\right)$ is $K_1\cup C_m,T_{1,1,m-2}$.

(ii) The matching equivalent graph of $K_1\cup C_6$ is: $K_1\cup C_6,T_{1,1,4},P_{\text{3}}\cup D_{\text{4}}$.

(iii) The matching equivalent graph of $K_1\cup C_9$ is $K_1\cup C_9,T_{1,1,7},C_{\text{3}}\cup T_{\text{1,2,3}}$.

(iv) The matching equivalent graph of $K_1\cup C_{15}$ is: $K_1\cup C_{15}~C_{\text{3}}\cup C_5\cup T_{\text{1,2,4}}$.

Lemma 2.5 [6] (i) $P_{\text{2}m+1}~P_m\cup C_{m+1},\left(m\ge 2\right)$.

(ii) $T_{1,1n}~K_1\cup C_{n+2}$.

(iii) $T_{1,2,2}~P_{\text{2}}\cup D_{\text{4}}$.

(iv) $K_1\cup C_6~P_{\text{3}}\cup D_{\text{4}}$.

(v) $K_1\cup C_9~C_{\text{3}}\cup T_{\text{1,2,3}}$.

(vi) $K_1\cup C_{15}~C_{\text{3}}\cup C_5\cup T_{\text{1,2,4}}$.

Lemma 2.6 [1] Suppose $G=s{K}_1$ or $t_1{C}_{m_1}\cup \cdots \cup t_k{C}_{m_k}$, then G is a unique matching, that is $\delta \left(G\right)=1$.

Lemma 2.7 [1] Suppose

$G=s{K}_1\cup t_1{C}_{m_1}\cup t_2{C}_{m_2}\cup \cdots \cup t_k{C}_{m_k},m_i\ne 6\left(i=1,2,\cdots ,k\right),$

then all matching equivalent graphs of G do not contain road branches.

Lemma 2.8 [1]

(i) $G=s{K}_1\cup a{P}_3\cup t{C}_6$, then all matching equivalent graphs ofG do not contain $P_{11}$ branch and also $T_{1,2,2}$ [7],

(ii) $\delta \left(s{K}_1\cup a{P}_3\cup t{C}_6\right)=\delta \left(s{K}_1\cup t{C}_6\right)$.

Lemma 2.9 [1] [7] If $m\ne 6,9,15$, then

$\delta \left(s{K}_1\cup t{C}_6\right)=\min \left\{s,t\right\}+1$. (2)

Lemma 2.10 [1] [8] If $m_i\ne 6,9,15\left(i=1,2\right)$, then

$\delta \left(s{K}_1\cup t_1{C}_{m_1}\cup t_2{C}_{m_2}\right)={\displaystyle \underset{i=0}{\overset{r}{\sum }}\min }\left\{s-i,t_1\right\}+r+1$, where

$r=\min \left\{s,t_2\right\}$. (3)

3. Main Results

Lemma 3.1 $\delta \left(s{K}_1\cup t_1{C}_3\cup t_2{C}_5\cup t_3{C}_9\right)={\displaystyle \underset{j=0}{\overset{r}{\sum }}{\displaystyle \underset{i=j}{\overset{r}{\sum }}\delta \left(s-i,t_1+\left(i-j\right),t_2\right)}}$,

where

$r=\min \left\{s,t_3\right\}$. (4)

Proof. For simplicity, denote $\delta \left(s{K}_1\cup t_1{C}_3\cup t_2{C}_5\cup t_3{C}_9\right)=\delta \left(s,t_1,t_2,t_3\right)$.

Set $H~s{K}_1\cup t_1{C}_3\cup t_2{C}_5\cup t_3{C}_6$, by lemma 2.3 (iii) and lemma 2.7, we know H contains connected component $C_9,T_{1,1,7}$ or $T_{1,2,3}$.

1) IfH contains $C_9$, by $H=C_9\cup H_2~s{K}_1\cup t_1{C}_3\cup t_2{C}_5\cup t_3{C}_9$, we know $H_2~s{K}_1\cup t_1{C}_3\cup t_2{C}_5\cup \left(t_3-1\right)C_9$.

Such $H_2$ has a total of $\delta \left(s,t_1,t_{\text{2}},t_{\text{3}}-1\right)$.

2) IfH contains $T_{1,1,7}$, by $H=T_{1,1,7}\cup H_2~s{K}_1\cup t_1{C}_3\cup t_2{C}_5\cup t_3{C}_9$ and lemma 2.5 (ii), we know

$H_2~\left(s-1\right)K_1\cup t_1{C}_3\cup t_2{C}_5\cup \left(t_3-1\right)C_9$,

Such $H_2$ has a total of $\delta \left(s-1,t_1,t_{\text{2}},t_3-1\right)$.

3) IfH contains $T_{1,2,3}$, by $H=T_{1,2,3}\cup H_2~s{K}_1\cup t_1{C}_3\cup t_2{C}_5\cup t_3{C}_9$ and lemma 2.5(v), we get

$H_2~\left(s-1\right)K_1\cup \left(t_1+1\right)C_3\cup t_2{C}_5\cup \left(t_3-1\right)C_9$,

Such $H_2$ has a total of $\delta \left(s-1,t_1+1,t_{\text{2}},t_3-1\right)$.

4) IfH contains $C_9$ and $T_{1,1,7}$ simultaneously, such $H_2$ has a total of $\delta \left(s-1,t_1,t_{\text{2}},t_3-2\right)$.

5) IfH contains $C_9$ and $T_{1,2,3}$ simultaneously, such $H_2$ has a total of $\delta \left(s-1,t_1+1,t_{\text{2}},t_3-2\right)$.

6) IfH contains $T_{1,1,7}$ and $T_{1,2,3}$ simultaneously, such $H_2$ has a total of $\delta \left(s-2,t_1+1,t_{\text{2}},t_3-2\right)$.

7) IfH contains $C_9$, $T_{1,1,7}$ and $T_{1,2,3}$ simultaneously, such $H_2$ has a total of $\delta \left(s-2,t_1+1,t_{\text{2}},t_3-3\right)$.

Thus,

$\begin{array}{c}\delta \left(s,t_1,t_2,t_3\right)=\delta \left(s,t_1,t_{\text{2}},t_{\text{3}}-1\right)+\delta \left(s-1,t_1,t_{\text{2}},t_3-1\right)+\delta \left(s-1,t_1+1,t_{\text{2}},t_3-1\right)\\ -\delta \left(s-1,t_1,t_{\text{2}},t_3-2\right)-\delta \left(s-1,t_1+1,t_{\text{2}},t_3-2\right)\\ -\delta \left(s-2,t_1+1,t_{\text{2}},t_3-2\right)+\delta \left(s-2,t_1+1,t_{\text{2}},t_3-3\right)\end{array}$

Then,

$\begin{array}{l}\delta \left(s,t_1,t_2,t_3\right)-\delta \left(s-1,t_1,t_{\text{2}},t_3-1\right)-\delta \left(s-1,t_1+1,t_{\text{2}},t_3-1\right)\\ +\delta \left(s-2,t_1+1,t_{\text{2}},t_3-2\right)\\ =\delta \left(s,t_1,t_{\text{2}},t_{\text{3}}-1\right)-\delta \left(s-1,t_1,t_{\text{2}},t_3-2\right)-\delta \left(s-1,t_1+1,t_{\text{2}},t_3-2\right)\\ +\delta \left(s-2,t_1+1,t_{\text{2}},t_3-3\right)\end{array}$

Repeat the application of the above formula, we obtain

$\begin{array}{l}\delta \left(s,t_1,t_2,t_3\right)-\delta \left(s-1,t_1,t_{\text{2}},t_3-1\right)-\delta \left(s-1,t_1+1,t_{\text{2}},t_3-1\right)\\ +\delta \left(s-2,t_1+1,t_{\text{2}},t_3-2\right)\\ =\delta \left(s,t_1,t_2,2\right)-\delta \left(s-1,t_1,t_2,1\right)-\delta \left(s-1,t_1+1,t_{\text{2}},1\right)+\delta \left(s-2,t_1+1,t_{\text{2}},0\right)\\ =\delta \left(s,t_1,t_2,1\right)-\delta \left(s-1,t_1,t_2,0\right)-\delta \left(s-1,t_1+1,t_{\text{2}},0\right)\\ =\delta \left(s,t_1,t_2,0\right)=\delta \left(s,t_1,t_2\right)\end{array}$

Thus,

$\begin{array}{l}\delta \left(s,t_1,t_2,t_3\right)-\delta \left(s-1,t_1,t_{\text{2}},t_3-1\right)\\ =\delta \left(s-1,t_1+1,t_{\text{2}},t_3-1\right)-\delta \left(s-2,t_1+1,t_{\text{2}},t_3-2\right)+\delta \left(s,t_1,t_2\right)\\ =\delta \left(s-2,t_1+2,t_{\text{2}},t_3-2\right)-\delta \left(s-3,t_1+2,t_{\text{2}},t_3-3\right)\\ +\delta \left(s-1,t_1+1,t_2\right)+\delta \left(s-1,t_1+1,t_2\right)\\ =\cdots ={\displaystyle \underset{i=0}{\overset{r}{\sum }}\delta }\left(s-i,t_1+i,t_2\right)\end{array}$

$\delta \left(s,t_1,t_2,t_3\right)-\delta \left(s-1,t_1,t_{\text{2}},t_3-1\right)={\displaystyle \underset{i=0}{\overset{r}{\sum }}\delta }\left(s-i,t_1+i,t_2\right)$ (1')

$\delta \left(s-1,t_1,t_2,t_3-1\right)-\delta \left(s-2,t_1,t_{\text{2}},t_3-2\right)={\displaystyle \underset{i=1}{\overset{r}{\sum }}\delta }\left(s-i,t_1+\left(i-1\right),t_2\right)$ (2')

$\cdots$

$\begin{array}{l}\delta \left(s-\left(r-1\right),t_1,t_2,t_3-\left(r-1\right)\right)-\delta \left(s-r,t_1,t_{\text{2}},t_3-r\right)\\ ={\displaystyle \underset{i=r-1}{\overset{r}{\sum }}\delta }\left(s-i,t_1+i-\left(r-1\right),t_2\right)\end{array}$ (r')

$\delta \left(s-r,t_1,t_{\text{2}},t_3-r\right)={\displaystyle \underset{i=r}{\overset{r}{\sum }}\delta }\left(s-i,t_1+\left(i-r\right),t_2\right)$ (r+1')

Add (1'), (2'), $\cdots$, (r+1') together, we get

$\delta \left(s,t_1,t_{\text{2}},t_3\right)={\displaystyle \underset{j=0}{\overset{r}{\sum }}{\displaystyle \underset{i=j}{\overset{r}{\sum }}\delta }}\left(s-i,t_1+\left(i-j\right),t_2\right)$. +

Theorem 3.1

$\delta \left(s{K}_1\cup t_1{C}_9\cup t_2{C}_{15}\right)={\displaystyle \underset{j=0}{\overset{r}{\sum }}{\displaystyle \underset{i=j}{\overset{r}{\sum }}\delta \left(s-i,i-j,i-j,t_1\right)}}$. (5)

Proof. For simplicity, denote

$\delta \left(s{K}_1\cup t{C}_3\cup t^{\prime }{C}_5\cup t_1{C}_9\cup t_2{C}_{15}\right)=\delta \left(s,t,t^{\prime }{t}_1,t_2\right)$.

Suppose $H~G$, by lemma 2.3(iv) and lemma 2.7, we know H contains connected component $C_{15},T_{1,1,13}$ or $T_{1,2,4}$.

1) IfH contains $C_{15}$, by $H=C_{15}\cup H_2~s{K}_1\cup t_1{C}_9\cup t_2{C}_{15}$ we know $H_2~s{K}_1\cup t_1{C}_9\cup \left(t_2-1\right)C_{15}$. Such $H_2$ has a total of $\delta \left(s,0,0,t_{\text{1}},t_{\text{2}}-1\right)$.

2) IfH contains $T_{1,1,13}$, by $H=T_{1,1,13}\cup H_2~G=s{K}_1\cup t_1{C}_9\cup t_2{C}_{15}$ and lemma 2.5(ii), we get $H_2~\left(s-1\right)K_1\cup t_1{C}_9\cup \left(t_2-1\right)C_{15}$. Such $H_2$ has a total of $\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-1\right)$.

3) IfH contains $T_{1,2,4}$, by $H=T_{1,2,4}\cup H_2~G=s{K}_1\cup t_1{C}_9\cup t_2{C}_{15}$ and lemma 2.5(vi), we get $H_2~\left(s-1\right)K_1\cup C_3\cup C_5\cup t_1{C}_9\cup \left(t_2-1\right)C_{15}$, such $H_2$ has a total of $\delta \left(s-1,1,1,t_{\text{1}},t_2-1\right)$.

4) If H contains $C_{15}$ and $T_{1,1,13}$, such $H_2$ has a total of $\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-2\right)$.

5) IfH contains $C_{15}$ and $T_{1,2,4}$, such $H_2$ has a total of $\delta \left(s-1,1,1,t_{\text{1}},t_2-2\right)$.

6) If H contains $T_{1,1,13}$ and $T_{1,2,4}$, such $H_2$ has a total of $\delta \left(s-2,1,1,t_{\text{1}},t_2-2\right)$.

7) If H contains $C_{15}$, $T_{1,1,13}$ and $T_{1,2,4}$, such $H_2$ has a total of $\delta \left(s-2,1,1,t_{\text{1}},t_2-3\right)$.

Then

$\begin{array}{c}\delta \left(s,0,0,t_{\text{1}},t_{\text{2}}\right)=\delta \left(s,0,0,t_{\text{1}},t_{\text{2}}-1\right)+\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-1\right)\\ +\delta \left(s-1,1,1,t_{\text{1}},t_2-1\right)-\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-2\right)\\ -\delta \left(s-1,1,1,t_{\text{1}},t_2-2\right)-\delta \left(s-2,1,1,t_{\text{1}},t_2-2\right)\\ +\delta \left(s-2,1,1,t_{\text{1}},t_2-3\right)\end{array}$

Thus,

$\begin{array}{l}\delta \left(s,0,0,t_{\text{1}},t_{\text{2}}\right)-\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-1\right)-\delta \left(s-1,1,1,t_{\text{1}},t_2-1\right)\\ +\delta \left(s-2,1,1,t_{\text{1}},t_2-2\right)\\ =\delta \left(s,0,0,t_{\text{1}},t_{\text{2}}-1\right)-\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-2\right)-\delta \left(s-1,1,1,t_{\text{1}},t_2-2\right)\\ +\delta \left(s-2,1,1,t_{\text{1}},t_2-3\right)\end{array}$

Repeat the application of the above formula, we obtain

$\begin{array}{l}\delta \left(s,0,0,t_{\text{1}},t_{\text{2}}\right)-\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-1\right)-\delta \left(s-1,1,1,t_{\text{1}},t_2-1\right)\\ +\delta \left(s-2,1,1,t_{\text{1}},t_2-2\right)\\ =\delta \left(s,0,0,t_{\text{1}},2\right)-\delta \left(s-1,0,0,t_{\text{1}},1\right)-\delta \left(s-1,1,1,t_{\text{1}},1\right)+\delta \left(s-2,1,1,t_{\text{1}},0\right)\\ =\delta \left(s,0,0,t_{\text{1}},1\right)-\delta \left(s-1,0,0,t_{\text{1}},0\right)-\delta \left(s-1,1,1,t_{\text{1}},0\right)\\ =\delta \left(s,0,0,t_{\text{1}}\right)\end{array}$

So,

$\begin{array}{l}\delta \left(s,0,0,t_{\text{1}},t_{\text{2}}\right)-\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-1\right)\\ =\delta \left(s-1,1,1,t_{\text{1}},t_2-1\right)-\delta \left(s-2,1,1,t_{\text{1}},t_2-2\right)+\delta \left(s,0,0,t_{\text{1}}\right)\\ =\delta \left(s-2,2,2,t_{\text{1}},t_2-2\right)-\delta \left(s-3,2,2,t_{\text{1}},t_2-3\right)\\ +\delta \left(s,0,0,t_{\text{1}}\right)+\delta \left(s-1,1,1,t_{\text{1}}\right)\end{array}$

$\begin{array}{l}=\delta \left(s-3,3,3,t_{\text{1}},t_2-3\right)-\delta \left(s-4,3,3,t_{\text{1}},t_2-4\right)+\delta \left(s,0,0,t_{\text{1}}\right)\\ +\delta \left(s-1,1,1,t_{\text{1}}\right)+\delta \left(s-2,2,2,t_{\text{1}}\right)\\ =\cdots ={\displaystyle \underset{i=0}{\overset{r}{\sum }}\delta }\left(s-i,i,i,t_1\right)\end{array}$

$\delta \left(s,0,0,t_{\text{1}},t_{\text{2}}\right)-\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-1\right)={\displaystyle \underset{i=0}{\overset{r}{\sum }}\delta }\left(s-i,i,i,t_1\right)$ (1'')

$\delta \left(s-1,0,0,t_{\text{1}},t_{\text{2}}-1\right)-\delta \left(s-2,0,0,t_{\text{1}},t_{\text{2}}-2\right)={\displaystyle \underset{i=0}{\overset{r}{\sum }}\delta }\left(s-i,i-1,i-1,t_1\right)$ (2'')

$\cdots$

$\begin{array}{l}\delta \left(s-\left(r-1\right),0,0,t_1,t_2-\left(r-1\right)\right)-\delta \left(s-r,0,0,t_1,t_{\text{2}}-r\right)\\ ={\displaystyle \underset{i=r-1}{\overset{r}{\sum }}\delta }\left(s-i,i-\left(r-1\right),i-\left(r-1\right),t_1\right)\end{array}$ (r'')

$\delta \left(s-r,0,0,t_1,t_{\text{2}}-r\right)={\displaystyle \underset{i=r}{\overset{r}{\sum }}\delta }\left(s-i,i-r,i-r,t_1\right)$ (r+1'')

Add (1''), (2''), $\cdots$, (r+1'') together, we get:

$\delta \left(s,0,0,t_{\text{1}},t_{\text{2}}\right)={\displaystyle \underset{j=0}{\overset{r}{\sum }}{\displaystyle \underset{i=j}{\overset{r}{\sum }}\delta }}\left(s-i,i-j,i-j,t_1\right)$. +

Characterizing all graphs determined by a graph polynomial is an important subject in algebraic graph theory, among them, matching polynomial is considered to be a better algebraic tool. It is NP difficult to completely characterize the matched equivalent graphs of a class of graphs. In this paper, we study the number of matching equivalent graphs of some points and some cycli-union graphs, that is to say we calculate

$\delta \left(s{K}_1\cup t_1{C}_9\cup t_2{C}_{15}\right)={\displaystyle \underset{j=0}{\overset{r}{\sum }}{\displaystyle \underset{i=j}{\overset{r}{\sum }}\delta \left(s-i,i-j,i-j,t_1\right)}}$.

Acknowledgements

This work is supported by the National Natural Science Foundation of China (No. 11561056), the Natural Science Foundation of Qinghai Province (2016-ZJ-914), Teaching Reform Research Project of Qinghai Nationalities University (2021-JYQN-005).

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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