---
abstract: |
    The energy of a graph $G$, denoted by $E(G)$, is defined as the sum of the absolute values of all eigenvalues of $G$. Let $n$ be an even number and $\mathbb{U}_{n}$ be the set of all conjugated unicyclic graphs of order $n$ with maximum degree at most $3$. Let $S_n^{\frac{n}{2}}$ be the radialene graph obtained by attaching a pendant edge to each vertex of the cycle $C_{\frac{n}{2}}$. In \[Y. Cao et al., On the minimal energy of unicyclic Hückel molecular graphs possessing Kekulé structures, Discrete Appl. Math. 157 (5) (2009), 913–919\], Cao et al. showed that if $n\geq 8$, $S_n^{\frac{n}{2}}\ncong G\in \mathbb{U}_{n}$ and the girth of $G$ is not divisible by $4$, then $E(G)>E(S_n^{\frac{n}{2}})$. Let $A_n$ be the unicyclic graph obtained by attaching a $4$-cycle to one of the two leaf vertices of the path $P_{\frac{n}{2}-1}$ and a pendent edge to each other vertices of $P_{\frac{n}{2}-1}$. In this paper, we prove that $A_n$ is the unique unicyclic graph in $\mathbb{U}_{n}$ with minimal energy.\
    [**Keywords:**]{} Minimal energy; Unicyclic graph; Perfect matching; Characteristic polynomial; Degree\
    [**AMS Subject Classification 2000:**]{} 15A18; 05C50; 05C90; 92E10
---

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\[section\] \[lem\][Theorem]{} \[lem\][Corollary]{} \[lem\][Conjecture]{} \[lem\][Remark]{} \[lem\][Definition]{}

[**On the minimal energy of conjugated unicyclic graphs with maximum degree at most 3**]{}

=0.20in

=0.20in

[ Hongping Ma$^{1}$, Yongqiang Bai$^{1}$[^1], Shengjin Ji$^{2}$\
$^{1}$ School of Mathematics and Statistics, Jiangsu Normal University,\
Xuzhou 221116, China\
$^{2}$ School of Science, Shangdong University of Technology,\
Zibo 255049, China\
Email: hpma@163.com, bmbai@163.com, jishengjin2013@163.com ]{}

=0.27in

Introduction
============

Let $G$ be a simple graph with $n$ vertices and $A(G)$ the adjacency matrix of $G$. The eigenvalues $\lambda_{1}, \lambda_{2},\ldots, \lambda_{n}$ of $A(G)$ are said to be the eigenvalues of the graph $G$. The energy of $G$ is defined as $$E=E(G)=\sum_{i=1}^{n}|\lambda_{i}|.$$ This concept was intensively studied in chemistry, since it can be used to approximate the total $\pi$-electron energy of a molecular. Further details on the mathematical properties and chemical applications of $E(G)$, see the recent book [@LSG], reviews [@G2; @GLZ], and papers [@BB; @DM; @DMG; @GFAHG; @LSWL; @MMZ; @Z]. One of the fundamental question that is encountered in the study of graph energy is which graphs (from a given class) have minimal and maximal energies. A large of number of papers were published on such extremal problems, especially for various subclasses of trees and unicyclic graphs, see Chapter 7 in [@LSG]. A conjugated unicyclic graph is a connected graph with one unique cycle that has a perfect matching. The problem of determining the conjugated unicyclic graph with minimal energy has been considered in [@LZZ; @WCL], and Li et al. [@LZZ] proved that the conjugated unicyclic graph of order (even) $n$ with minimal energy is $U_1$ or $U_2$, as shown in Figure \[fig-minimal\]. It has been shown that $E(U_1)<E(U_2)$ by Li and Li [@LL]. Recently, results on ordering of conjugated unicyclic graphs by minimal energies have been extended in [@W; @Z]. In particular, $U_2$ is unique conjugated unicyclic graph of order $n$ with second-minimal energy.

(1784.8, 562.2)(0,0) (0,0)[![The conjugated unicyclic graphs with minimal and second-minimal energy.[]{data-label="fig-minimal"}](fig-minimal "fig:")]{} (73.92,38.20)

(150.0, 60.0)\[l\]

$U_1$

(1294.83,382.88)

(450.0, 60.0)\[l\]

$\frac{n}{2}-2$

(561.90,383.72)

(450.0, 60.0)\[l\]

$\frac{n}{2}-3$

(944.44,58.39)

(150.0, 60.0)\[l\]

$U_2$

The degree of a vertex $v$ in a graph $G$ is denoted by $d_G(v)$. Denote by $\Delta$ the maximum degree of a graph. From now on, let $n$ be an even number. Let $\mathbb{U}_{n}$ be the set of all conjugated unicyclic graphs of order $n$ with $\Delta\leq 3$. Let $G\in \mathbb{U}_{n}$, the length of the unique cycle of $G$ is denoted by $g(G)$, or simply $g$, and the unique cycle of $G$ is denoted by $C_g(G)$, or simply $C_g$. Let $S_n^{\frac{n}{2}}$ be the radialene graph obtained by attaching a pendant edge to each vertex of the cycle $C_{\frac{n}{2}}$. Wang et al. [@WCZL] showed the following results: Assume that $n\geq 6$ and $S_n^{\frac{n}{2}}\ncong G\in
 \mathbb{U}_{n}$. Then if one of the following conditions holds: (i) $\frac{n}{2}\equiv g\equiv 1$ (mod $2$) and $g\leq\frac{n}{2}$, (ii) $g\not\equiv \frac{n}{2}\equiv 0$ (mod $4$), (iii) $\frac{n}{2}\equiv g\equiv  2$ (mod $4$), and $g\leq\frac{n}{2}$, then $E(G)>E(S_n^{\frac{n}{2}})$. Y. Cao et al. [@CLLZ] improved the above results by proving the following Lemma.

[[@CLLZ]]{.nodecor}\[lem g-non-4-multiple\] If $n\geq 8$, $S_n^{\frac{n}{2}}\ncong G\in \mathbb{U}_{n}$ with $g\not\equiv  0$ [(mod $4$)]{.nodecor}, then $E(G)>E(S_n^{\frac{n}{2}})$.

Let $A_n$, $B_n$, $D_n$ and $E_n$ be the graphs shown in Figure \[fig-minimal-degree-3\].

(2792.5, 525.4)(0,0) (0,0)[![Four graphs in $\mathbb{U}_{n}$.[]{data-label="fig-minimal-degree-3"}](fig-minimal-degree-3 "fig:")]{} (185.67,39.45)

(150.0, 60.0)\[l\]

$A_n$

(374.88,419.32)

(450.0, 60.0)\[l\]

$\frac{n}{2}-2$

(847.63,38.20)

(150.0, 60.0)\[l\]

$B_n$

(1012.53,423.62)

(450.0, 60.0)\[l\]

$\frac{n}{2}-3$

(1472.33,42.99)

(150.0, 60.0)\[l\]

$D_n$

(1578.87,409.06)

(450.0, 60.0)\[l\]

$\frac{n}{2}-3$

(1578.87,409.06)

(450.0, 60.0)\[l\]

$\frac{n}{2}-3$

(2196.00,41.96)

(150.0, 60.0)\[l\]

$E_n$

(2302.54,408.04)

(450.0, 60.0)\[l\]

$\frac{n}{2}-4$

In this paper, we completely characterize the graph with minimal energy in $\mathbb{U}_{n}$ by showing the following result.

\[main thm\] $A_n$ is the unique unicyclic graph in $\mathbb{U}_{n}$ with minimal energy for $n\geq 8$.

Preliminaries
=============

In this section, we first introduce some notations and properties which are need in the sequel. Then we give some results on the energies of graphs $A_n$, $B_n$, $D_n$, $E_n$ and $S_n^{\frac{n}{2}}$.

Let $G$ be a graph of order $k$. The characteristic polynomial of $A(G)$ is also called the characteristic polynomial of $G$, denoted by $\phi(G,x)=\mbox{det}(xI-A(G))=\sum_{i=0}^{k}a_i(G)x^{k-i}$. Using these coefficients of $\phi(G,x)$, the energy of $G$ can be expressed as the Coulson integral formula [@GP]: $$\begin{aligned}
 E(G)=\frac{1}{2\pi}{\large\int}_{-\infty}^{+\infty}\frac{1}{x^{2}}\ln\left[\left(\sum\limits_{i=0}^{\lfloor
\frac{k}{2}\rfloor}
(-1)^ia_{2i}(G)x^{2i}\right)^2+\left(\sum\limits_{i=0}^{\lfloor
\frac{k}{2}\rfloor} (-1)^ia_{2i+1}(G)x^{2i+1}\right)^2\right]dx.
\label{energy-1}\end{aligned}$$ Write $b_i(G)=|a_i(G)|$. Clearly, $b_0(G)=1$, $b_1(G)=0$, and $b_2(G)$ equals the number of edges of $G$. For unicyclic graphs or bipartite graphs, it can be shown [@GP; @H] that $$\begin{aligned}
 E(G)=\frac{1}{2\pi}{\large\int}_{-\infty}^{+\infty}\frac{1}{x^{2}}\ln\left[\left(\sum\limits_{i=0}^{\lfloor
\frac{k}{2}\rfloor}
b_{2i}(G)x^{2i}\right)^2+\left(\sum\limits_{i=0}^{\lfloor
\frac{k}{2}\rfloor} b_{2i+1}(G)x^{2i+1}\right)^2\right]dx.
\label{energy-2}\end{aligned}$$ By formula , it is convenient to introduce the following quasi-order relation [@SSGL]: if $G_1$ and $G_2$ are two unicyclic or bipartite graphs with $k$ vertices, then $$G_1\succeq G_2 \Leftrightarrow b_i(G_1)\geq b_i(G_2) \mbox{ for
all } i=2, \ldots, k.$$ If $G_1\succeq G_2$ and there exists some $j$ such that $b_j(G_1) > b_j(G_2)$, then we write $G_1\succ G_2$. Clearly, $G_1\succeq G_2\Rightarrow  E(G_1)\geq E(G_2)$, and $G_1\succ G_2\Rightarrow  E(G_1)>E(G_2)$. It is known [@CDS] that $b_{2i+1}(G)=0$ for a bipartite graph $G$, and $b_{2i}(G)$ equals the number of $i$-matchings of $G$ if $G$ is a tree.

[[@LZ]]{.nodecor}\[lem delete-edge\] Let $G$ be a graph whose components are all trees except at most one being a unicyclic graph.

\(1) If $G$ contains exactly one cycle $C_g$, and $uv$ is an edge on this cycle, then $$\begin{array}{lll}
b_i(G)=b_i(G-uv) + b_{i-2}(G-u-v)-2b_{i-g}(G-C_g)\ \mbox{if}\
g\equiv 0
 \  (\textnormal{mod} \ 4),\\
b_i(G)=b_i(G-uv) + b_{i-2}(G-u-v)+2b_{i-g}(G-C_g)\ \mbox{if}\
g\not\equiv 0  \ (\textnormal{mod} \ 4).
\end{array}$$ (2) If $uv$ is a cut edge of $G$, then $$b_i(G)=b_i(G-uv) +b_{i-2}(G-u-v).$$ In particular, if $uv$ is a pendent edge with pendent vertex $u$, then $$b_i(G)=b_i(G-u) + b_{i-2}(G-u-v).$$

[[@G1]]{.nodecor}\[lem tree maximal-minimal\] Let $T$ be a tree on $n$ vertices. If $T$ is different from the path $P_n$ and the star $S_n$, then $P_n\succ T\succ S_n$.

Let $T$ be a tree of order $n\geq 3$, $e=uv$ be a non-pendent edge of $T$. Denote by $T_1$ and $T_2$ the two components of $T-e$ with $u\in T_1$ and $v\in T_2$. If $T'$ is the tree obtained from $T$ by contracting the edge $e=uv$ and attaching a pendent vertex to the vertex $u$ ($=v$), we say that $T'$ is obtained from $T$ by edge-growing transformation (on edge $e=uv$), or e.g.t (on edge $e=uv$) for short [@Xu; @LGL].

[[@LGL]]{.nodecor}\[lem edge-growing transformation\] If $T'$ is obtained from $T$ by one step of e.g.t, then $T\succ T'$.

[[@SSGL]]{.nodecor}\[lem delete-edge-2\] Let $G$ be a unicyclic or bipartite graph, $uv$ be a cut edge of $G$. Then $G\succ G-uv$.

[[@SSGL]]{.nodecor}\[lem delete-vertex\] Let $G$ be a unicyclic or bipartite graph, $v$ be a non-isolated vertex in $G$, and $K_1$ be the trivial graph of order $1$. Then $G\succ (G-v)\cup K_1$.

Let $F_n$ and $H_{n+1}$ be the graphs shown in Figure \[fig-minimal-tree-degree-3\].

(1197.3, 437.5)(0,0) (0,0)[![The graphs $F_n$ and $H_{n+1}$.[]{data-label="fig-minimal-tree-degree-3"}](fig-minimal-tree-degree-3 "fig:")]{} (78.18,43.07)

(150.0, 60.0)\[l\]

$F_n$

(114.75,329.90)

(390.0, 60.0)\[l\]

$\frac{n}{2}$

(765.29,38.20)

(270.0, 60.0)\[l\]

$H_{n+1}$

(767.34,335.67)

(390.0, 60.0)\[l\]

$\frac{n}{2}$

[[@ZL]]{.nodecor}\[lem minimal-tree-degree-3\] Among conjugated trees of order $n$ with $\Delta\leq 3$, $F_n$ has minimal energy.

\[lem formular for Bn and An\] For $A_n$, $B_n$, $D_n$ and $E_n$, we have that

\(1) $
 b_{2i}(A_n) = b_{2i}(F_n) + b_{2i-2}(F_{n-2})-2 b_{2i-4}(F_{n-4})
            =  b_{2i}(H_{n-1})+2b_{2i-2}(H_{n-3}).
$

\(2) $b_{2i}(B_n) = b_{2i}(A_n) + 2b_{2i-6}(F_{n-8}).$

\(3) -1cm -2cm $$\begin{array}{lll}
 b_{2i}(A_n)  =  b_{2i}(A_{n-2})+b_{2i-2}(A_{n-2})+b_{2i-2}(A_{n-4}) \ (n\geq
 8),\\
 b_{2i}(B_n)  =  b_{2i}(B_{n-2}) + b_{2i-2}(B_{n-2})+ b_{2i-2}(B_{n-4}) \ (n\geq
 10),\\
 b_{2i}(D_n)  =  b_{2i}(D_{n-2}) + b_{2i-2}(D_{n-2})+ b_{2i-2}(D_{n-4}) \ (n\geq
 10),\\
 b_{2i}(E_n)  =  b_{2i}(E_{n-2}) + b_{2i-2}(E_{n-2})+ b_{2i-2}(E_{n-4}) \ (n\geq
 12).
\end{array}$$

\(1) By Lemma \[lem delete-edge\] (1) and (2), we can obtain that $$\begin{array}{lll}
 b_{2i}(A_n) & = & b_{2i}(F_n) + b_{2i-2}(F_{n-2})-2 b_{2i-4}(F_{n-4})\\
             & = & b_{2i}(H_{n-1})+2b_{2i-2}(H_{n-3}).
\end{array}$$

\(2) By Lemma \[lem delete-edge\] (2), we have $$\begin{array}{lll}
 b_{2i}(B_n) & = & b_{2i}(K_2 \cup A_{n-2}) + b_{2i-2}(H_{n-3})\\
             & = & b_{2i}(K_2 \cup A_{n-2}) + b_{2i-2}(F_{n-4})+
             b_{2i-4}(F_{n-6}),\\
 b_{2i}(A_n) & = & b_{2i}(K_2 \cup A_{n-2}) + b_{2i-2}(A_{n-4})\\
             & = & b_{2i}(K_2 \cup A_{n-2}) + b_{2i-2}(F_{n-4})+
             b_{2i-4}(F_{n-6})-2 b_{2i-6}(F_{n-8}).
\end{array}$$ Hence $ b_{2i}(B_n)  =  b_{2i}(A_n)+2 b_{2i-6}(F_{n-8})$.

\(3) The results directly follow from Lemma \[lem delete-edge\] (2).

\[lem Bn and An\] $B_n\succ A_n$ for $n\geq 8$.

It is easy to obtain that $$\begin{array}{lll}
 \phi(B_8,x)  =  x^8-8x^6+16x^4-8x^2,\\
 \phi(A_8,x)  =  x^8-8x^6+16x^4-6x^2.
\end{array}$$ So $B_8\succ A_8$. Suppose $n>
 8$. By Lemma \[lem formular for Bn and An\], $b_{2i}(B_n)-b_{2i}(A_n)=2 b_{2i-6}(F_{n-8})\geq 0$ and $b_{8}(B_n)>
b_{8}(A_n)$. So $B_n\succ A_n$ for $n>8$. The proof is thus complete.

\[lem En and Dn\] $D_8\succ E_8$ and $E_n\succ D_n$ for $n\geq 10$.

It is easy to obtain that $$\begin{array}{lll}
\phi(E_{8},x) =   x^8-8x^6+14x^4-8x^2+1,\\
\phi(D_{8},x) =   x^8-8x^6+15x^4-8x^2+1,\\
\phi(E_{10},x) =   x^{10}-10x^8+29x^6-31x^4+12x^2-1,\\
 \phi(D_{10},x)  =   x^{10}-10x^8+29x^6-28x^4+10x^2-1,\\
 \phi(E_{12},x)  =   x^{12}-12x^{10}+47x^8-74x^6+51x^4-14x^2+1,\\
 \phi(D_{12},x)  =   x^{12}-12x^{10}+47x^8-72x^6+46x^4-12x^2+1.
\end{array}$$ So $D_{8}\succ E_{8}$, $E_{10}\succ D_{10}$ and $E_{12}\succ D_{12}$. Suppose $n > 12$. By Lemma \[lem delete-edge\], $$\begin{array}{lll}
 b_{2i}(E_n) =  b_{2i}(E_{12}\cup F_{n-12}) + b_{2i-2}(E_{10}\cup F_{n-14}),\\
 b_{2i}(D_n)  =  b_{2i}(D_{12}\cup F_{n-12}) + b_{2i-2}(D_{10}\cup E_{F-14}).
\end{array}$$ Hence we have $E_n\succ D_n$ for $n>12$. The proof is thus complete.

\[lem Sn and Bn\] $E(S_8^{4})>E(B_8)$ and $S_n^{\frac{n}{2}}\succ B_n$ for $n\geq 10$.

It is easy to obtain that $9.65685\doteq E(S_8^{4})>E(B_8)\doteq
 9.15298$. Suppose $n\geq 10$. Since $b_{2i+1}(S_n^{\frac{n}{2}})\geq 0=b_{2i+1}(B_n)$, we only need to consider $b_{2i}(S_n^{\frac{n}{2}})$ and $b_{2i}(B_n)$. By Lemmas \[lem delete-edge\] and \[lem formular for Bn and An\], $$b_{2i}(S_n^{\frac{n}{2}}) =  \left\{\begin{array}{ll} b_{2i}(F_n) + b_{2i-2}(F_{n-4})-2
 b_{2i-\frac{n}{2}}(\frac{n}{2}K_1), &  \mbox{if}\ g\equiv 0
\ (\textnormal{mod} \ 4) \\
  b_{2i}(F_n) + b_{2i-2}(F_{n-4})+2 b_{2i-\frac{n}{2}}(\frac{n}{2}K_1), &  \mbox{if}\ g\not\equiv 0
 \ (\textnormal{mod} \ 4)
\end{array}\right.,$$ $$\begin{array}{lll}
  b_{2i}(B_n) & = &  b_{2i}(F_n) + b_{2i-2}(F_{n-2})-2b_{2i-4}(F_{n-4})+2b_{2i-6}(F_{n-8})\\
              & = &  b_{2i}(F_n) + b_{2i-2}(F_{n-4})+b_{2i-4}(F_{n-6})-b_{2i-4}(F_{n-4})+2b_{2i-6}(F_{n-8})\\
              & = &  b_{2i}(F_n) + b_{2i-2}(F_{n-4})-b_{2i-6}(F_{n-6})+b_{2i-6}(F_{n-8})\\
              & = &  b_{2i}(F_n) + b_{2i-2}(F_{n-4})-b_{2i-8}(F_{n-8})-b_{2i-8}(F_{n-10}).
\end{array}$$ Since $$b_{2i-\frac{n}{2}}(\frac{n}{2}K_1) =
\left\{\begin{array}{ll} 1, &  \mbox{if} \  2i=\frac{n}{2}\\
 0, &  \mbox{otherwise}\end{array}\right.,$$ we have $b_{2i}(S_n^{\frac{n}{2}})\geq b_{2i}(B_n)$, $b_8(S_{10}^5)>
 b_8(B_{10})$, and $b_{10}(S_n^{\frac{n}{2}})>
 b_{10}(B_n)$ when $n\geq 12$. The proof is thus complete.

In the following, we will show that $E(A_n)<E(D_n)$ by using the Coulson integral formula method, which had successfully been applied to compare the energy of two given graphs by Huo et al., see [@Huo1]-[@Huo4]. Before proving it, we prepare some results as follows.

[[@CDS]]{.nodecor}\[lem characteristic polynomial\] Let $uv$ be an edge of $G$. Then $$\phi(G,x) = \phi(G-uv,x)-\phi(G-u-v,x)-2\sum_{C\in\mathcal
 {C}(uv)}\phi(G-C,x),$$ where $\mathcal{C}(uv)$ is the set of cycles containing $uv$. In particular, if $uv$ is a pendent edge with pendent vertex $v$, then $\phi(G,x) =
 x\phi(G-v,x)-\phi(G-u-v,x)$.

[[@Z2]]{.nodecor}\[lem log inequality\] For any real number $X > -1$, we have $$\frac{X}{1 + X}\leq \log(1 +
X)\leq X.$$ In particular, $\log(1 + X) < 0$ if and only if $X < 0$.

[[@G2]]{.nodecor}\[lem energy difference\] If $G_1$ and $G_2$ are two graphs with the same number of vertices, then $$E(G_1)-E(G_2) =\frac{1}{\pi}
\int_{-\infty}^{+\infty}\log\left|\frac{\phi(G_1,ix)}{\phi(G_2,ix)}\right|dx.$$

From Lemma \[lem characteristic polynomial\], we can easily obtain the following lemma.

\[lem characteristic polynomial-An and Dn\] $ \phi(A_n,x) = (x^2-1)\phi(A_{n-2},x)-x^2\phi(A_{n-4},x)$ for $n\geq 8$, and $\phi(D_n,x) = (x^2-1)\phi(D_{n-2},x)-x^2\phi(D_{n-4},x)$ for $n\geq 10$.

By some easy calculations, we have $\phi(A_6,x) = x^6-6x^4+6x^2$, $\phi(A_8,x) = x^8-8x^6+16x^4-6x^2$, $\phi(D_6,x) = x^6-6x^4+5x^2-1$ and $\phi(D_8,x) = x^8-8x^6+15x^4-8x^2+1$. Now for convenience, we define some notations as follows: $$\begin{aligned}
 Y_1(x)=\frac{x^2-1+\sqrt{x^4-6 x^2+1}}{2}, & & Y_2(x)=\frac{x^2-1-\sqrt{x^4-6 x^2+1}}{2}, \\
 Z_1(x)=\frac{-x^2-1+\sqrt{x^4+6 x^2+1}}{2}, & & Z_2(x)=\frac{-x^2-1-\sqrt{x^4+6 x^2+1}}{2}, \\
 A_1(x)=\frac{\phi(A_8,x)-Y_2(x)\phi(A_6,x)}{(Y_1(x))^4-(Y_1(x))^2x^2}, & & A_2(x)=\frac{\phi(A_8,x)-Y_1(x)\phi(A_6,x)}{(Y_2(x))^4-(Y_2(x))^2x^2},\\
  B_1(x)=\frac{\phi(D_8,x)-Y_2(x)\phi(D_6,x)}{(Y_1(x))^4-(Y_1(x))^2x^2},
& &
B_2(x)=\frac{\phi(D_8,x)-Y_1(x)\phi(D_6,x)}{(Y_2(x))^4-(Y_2(x))^2x^2},\end{aligned}$$ $$\begin{aligned}
   f_6(x)=x^6+6x^4+6x^2, & &  f_8(x)=x^8+8x^6+16x^4+6x^2,\\
     g_6(x)=x^6+6x^4+5x^2+1, & & g_8(x)=x^8+8x^6+15x^4+8x^2+1.\end{aligned}$$ It is easy to check that $Y_1(ix)=Z_1(x)$, $Y_2(ix)=Z_2(x)$, $$\begin{aligned}
 A_1(ix)=\frac{f_8(x)+Z_2(x)f_6(x)}{(Z_1(x))^4+(Z_1(x))^2x^2}, & & A_2(ix)=\frac{f_8(x)+Z_1(x)f_6(x)}{(Z_2(x))^4+(Z_2(x))^2x^2},\\
  B_1(ix)=\frac{g_8(x)+Z_2(x)g_6(x)}{(Z_1(x))^4+(Z_1(x))^2x^2}, & & B_2(ix)=\frac{g_8(x)+Z_1(x)g_6(x)}{(Z_2(x))^4+(Z_2(x))^2x^2},\end{aligned}$$ $Z_1(x)+Z_2(x)=-x^2-1$ and $Z_1(x)Z_2(x)=-x^2$. In addition, for $x>0$, $0<\frac{Z_1(x)}{x}<1$; for $x<0$, $-1<\frac{Z_1(x)}{x}<0$.

\[lem recursive formula of characteristic polynomial-An and Dn\] For $n\geq 6$ and $x\neq 0$, the characteristic polynomials of $A_n$ and $D_n$ have the following forms: $$\phi(A_n,x) =
 A_1(x)(Y_1(x))^{\frac{n}{2}}+A_2(x)(Y_2(x))^{\frac{n}{2}}$$ and $$\phi(D_n,x) =
 B_1(x)(Y_1(x))^{\frac{n}{2}}+B_2(x)(Y_2(x))^{\frac{n}{2}}.$$

By Lemma \[lem characteristic polynomial-An and Dn\], we have that $\phi(A_n,x), \phi(D_n,x)$ satisfy the recursive formula $f(n,x)=(x^2-1)f(n-2,x)-x^2f(n-4,x)$. Therefore, the form of the general solution of the linear homogeneous recursive relation is $f(n,x)=C_1(x)(Y_1(x))^{\frac{n}{2}}+C_2(x)(Y_2(x))^{\frac{n}{2}}$. By some simple calculations, together with the initial values $\phi(A_6,x)$ and $\phi(A_8,x)$ ($\phi(D_6,x)$ and $\phi(D_8,x)$, respectively), we can get that $C_i(x)=A_i(x)$ ($C_i(x)=B_i(x)$, respectively), $i=1, 2$.

\[lem An and Dn\] $E(A_n) < E(D_n)$ for $n\geq 6$.

By Lemma \[lem energy difference\], we have $$E(A_n)-E(D_n) =\frac{1}{\pi}
\int_{-\infty}^{+\infty}\log\left|\frac{\phi(A_n,ix)}{\phi(D_n,ix)}\right|dx.$$ From Lemma \[lem recursive formula of characteristic polynomial-An and Dn\], we know that both $\phi(A_n,ix)$ and $\phi(D_n,ix)$ are polynomials of $x$ with all real coefficients. For convenience, we abbreviate $A_k(ix)$, $B_k(ix)$ and $Z_k(x)$ to $A_k$, $B_k$ and $C_k$ for $k=1,2$, and abbreviate $f_k(x)$ and $g_k(x)$ to $f_k$ and $g_k$ for $k=6,8$, respectively. In the following, we assume that $x\neq 0$. We distinguish two cases in terms of the parity of $n/2$.

[**Case 1.**]{} $n=4k$ $(k\geq 2)$. Notice that $Z_1Z_2=-x^2$. When $n \rightarrow \infty$, $$\begin{aligned}
\frac{\phi(A_n,ix)}{\phi(D_n,ix)} =
\frac{A_1Z_1^{\frac{n}{2}}+A_2Z_2^{\frac{n}{2}}}{B_1Z_1^{\frac{n}{2}}+B_2Z_2^{\frac{n}{2}}}
                                 =  \frac{A_2+A_1(\frac{Z_1}{x})^n}{B_2+B_1(\frac{Z_1}{x})^n}
                                \rightarrow  \frac{A_2}{B_2}.\end{aligned}$$ We will show that $$\begin{aligned}
\log\left|\frac{\phi(A_n,ix)}{\phi(D_n,ix)}\right|<\log\left|\frac{A_2(ix)}{B_2(ix)}\right|=\log\frac{A_2}{B_2}.\end{aligned}$$ Assume that $$\begin{aligned}
 2\log\left|\frac{\phi(A_n,ix)}{\phi(D_n,ix)}\right|-2\log\left|\frac{A_2}{B_2}\right|=\log\left(1+\frac{F_1(n,x)}{G_1(n,x)}\right).\end{aligned}$$ Then we get that $G_1(n,x)=(A_2(ix) \phi(D_n,ix))^2>0$ and $$\begin{aligned}
F_1(n,x) & = &
  B_2^2(A_1Z_1^{\frac{n}{2}}+A_2Z_2^{\frac{n}{2}})^2-A_2^2(B_1Z_1^{\frac{n}{2}}+B_2Z_2^{\frac{n}{2}})^2\\
   & = &
   (A_1^2B_2^2-A_2^2B_1^2)Z_1^n
    +2A_2B_2(A_1B_2-A_2B_1)(Z_1Z_2)^{\frac{n}{2}}\\
   & < &  0.\end{aligned}$$ Since $Z_1^{n}>0$, $(Z_1Z_2)^{\frac{n}{2}}=(-x^2)^{\frac{n}{2}}>0$, $A_2B_2>0$, and by some elementary calculations, we have $$\begin{aligned}
A_1B_2-A_2B_1=\frac{(-3x^8-15x^6-8x^4)\sqrt{x^4+6x^2+1}}{x^6(x^4+6x^2+1)}<0,\end{aligned}$$ and $$\begin{aligned}
A_1B_2+A_2B_1=\frac{4x^4+17x^6+20x^8+3x^{10}}{x^6(x^4+6x^2+1)}>0.\end{aligned}$$ Hence by Lemma \[lem log inequality\], $$\begin{aligned}
 \frac{1}{\pi} \int_{-\infty}^{+\infty}\log \frac{A_2(ix)}{B_2(ix)}dx
 \leq \frac{1}{\pi} \int_{-\infty}^{+\infty} \left(\frac{A_2(ix)}{B_2(ix)}-1\right)dx\doteq \frac{1}{\pi}
 (-0.8538292323)<0.\end{aligned}$$ Therefore, $$E(A_n)-E(D_n)\leq \frac{1}{\pi} \int_{-\infty}^{+\infty}\log \frac{A_2(ix)}{B_2(ix)}dx<0.$$

[**Case 2.**]{} $n=4k+2$ $(k\geq 1)$. We will show that $\log\left|\frac{\phi(A_n,ix)}{\phi(D_n,ix)}\right|$ is monotonically decreasing in $n$. Assume that $$\begin{aligned}
 2\log\left|\frac{\phi(A_{n+4},ix)}{\phi(D_{n+4},ix)}\right|-2\log\left|\frac{\phi(A_{n},ix)}{\phi(D_{n},ix)}\right|=\log\left(1+\frac{F_2(n,x)}{G_2(n,x)}\right).\end{aligned}$$ Then we can obtain that $G_2(n,x)=(\phi(A_n,ix)\phi(D_{n+4},ix))^2>0$ and $$\begin{aligned}
 F_2(n,x) & = & (\phi(A_{n+4},ix)\phi(B_n,ix))^2-(\phi(A_n,ix)\phi(D_{n+4},ix))^2\\
          & = &
          (A_1Z_1^{\frac{n}{2}+2}+A_2Z_2^{\frac{n}{2}+2})^2(B_1Z_1^{\frac{n}{2}}+B_2Z_2^{\frac{n}{2}})^2 \\
          & & -  (B_1Z_1^{\frac{n}{2}+2}+B_2Z_2^{\frac{n}{2}+2})^2(A_1Z_1^{\frac{n}{2}}+A_2Z_2^{\frac{n}{2}})^2\\
          & = & -(A_1B_2-A_2B_1)x^n(Z_1^2-Z_2^2)\cdot F_3(n,x),\end{aligned}$$ where $
F_3(n,x)=(A_1B_2+A_2B_1)(Z_1Z_2)^{\frac{n}{2}}(Z_1^2+Z_2^2)+2A_1B_1Z_1^{n+2}+2A_2B_2Z_2^{n+2}$. Since $A_1B_2-A_2B_1<0$, $Z_1^2-Z_2^2<0$ and $x^n>0$, to prove $F_2(n,x)<0$, it suffice to show that $F_3(n,x)>0$. By some elementary calculations, we can get that $$\begin{aligned}
F_3(n,x)=\frac{2f_8g_8F_4(n,x)+2f_6g_6F_5(n,x)+(f_8g_6+f_6g_8)F_6(n,x)}{x^4+6x^2+1},\end{aligned}$$ where $$\begin{aligned}
 F_4(n,x) & = &
 Z_1^{n-4}+Z_2^{n-4}-(Z_1Z_2)^{\frac{n}{2}-3}(Z_1^2+Z_2^2)\\
          & = &
          (Z_1^{\frac{n}{2}-1}-Z_2^{\frac{n}{2}-1})(Z_1^{\frac{n}{2}-3}-Z_2^{\frac{n}{2}-3}),\\
F_5(n,x) & = &
 Z_2^2Z_1^{n-4}+Z_1^2Z_2^{n-4}-(Z_1Z_2)^{\frac{n}{2}-2}(Z_1^2+Z_2^2)\\
          & = &
          (Z_1Z_2)^2(Z_1^{\frac{n}{2}-2}-Z_2^{\frac{n}{2}-2})(Z_1^{\frac{n}{2}-4}-Z_2^{\frac{n}{2}-4}),\\
F_6(n,x) & = &
 2Z_2Z_1^{n-4}+2Z_1Z_2^{n-4}-(Z_1+Z_2)(Z_1Z_2)^{\frac{n}{2}-3}(Z_1^2+Z_2^2)\\
          & = &
          Z_1Z_2[(Z_1^{\frac{n}{2}-1}-Z_2^{\frac{n}{2}-1})(Z_1^{\frac{n}{2}-4}-Z_2^{\frac{n}{2}-4})+(Z_1^{\frac{n}{2}-3}-Z_2^{\frac{n}{2}-3})(Z_1^{\frac{n}{2}-2}-Z_2^{\frac{n}{2}-2})].\end{aligned}$$ Notice that $Z_1>0$, $Z_2<0$, we have $Z_1^{k}-Z_2^{k}>0$ when $k$ is odd. On the other hand, for $k\geq 1$, we have $$\begin{aligned}
Z_1^{2k}-Z_2^{2k}=Z_1^{2k}\left[1-\left(\frac{x}{Z_1}\right)^{4k}\right]<0.\end{aligned}$$ Therefore, we have $F_4(n,x)\geq 0$, $F_5(n,x)\geq 0$, $F_6(n,x)>0$, and so $F_3(n,x)>0$ and $F_2(n,x)<0$. Hence, $$E(A_n)-E(D_n)\leq E(A_6)-E(D_6)\doteq 6.60272-7.20775<0.$$ The proof is thus complete.

Main results
============

Let $G$ be a graph in $\mathbb{U}_{n}$ with vertex set $V$ and the unique cycle $C_g$. We use $M_G$ to denote one arbitrary selected prefect matching of $G$. Let $x, y\in V$. Denote by $d_G(x,y)$ ($d_G(x,C_g)$, respectively) the distance between vertex $x$ and $y$ ($C_g$, respectively). Define $d=d(G)=\max_{x\in V(G)}\{d_G(x,C_g)\}$, $V_1=V_1(G)=\{x\in V|d_G(x,C_g)=d(G)\}$, and $t=t(G)=|V_1(G)|$. Clearly, the vertices in $V_1$ are pendent vertices when $d(G)\geq 1$.

\[lem girth at least 8-d equals 0\] Let $n\equiv  0$ [(mod $4$)]{.nodecor}, $n\geq 8$. Then $C_n\succ B_n$.

By Lemma \[lem delete-edge\], $$b_{2i}(C_n) =  \left\{\begin{array}{ll} b_{2i}(P_n) + b_{2i-2}(P_{n-2}), &  \mbox{if}\ 2i\neq n \\
   b_{2i}(P_n) + b_{2i-2}(P_{n-2})-2, &  \mbox{if}\  2i=n
\end{array}\right.,$$ and $$b_{2i}(B_n)=b_{2i}(G_1) + b_{2i-2}(K_2\cup F_{n-4})-2b_{2i-4}(K_2\cup F_{n-6}),$$ where $G_1$ is the tree of order $n$ obtained by attaching a path with $4$ edges to one of the two vertices of degree $2$ of $F_{n-4}$. By Lemmas \[lem tree maximal-minimal\] and \[lem delete-edge-2\], $P_n\succ G_1$, $P_{n-2}\succ F_{n-2}\succ K_2\cup
 F_{n-4}$. On the other hand, we have $b_{n}(C_n)=b_{n}(B_n)$ and $b_{4}(C_n)>b_{4}(B_n)$. Hence $C_n\succ B_n$.

\[lem girth at least 8-d equals 1\] Let $G\in \mathbb{U}_{n}$, $g(G)\equiv  0$ [(mod $4$)]{.nodecor}, $g(G)\geq 8$, and $d(G)=1$. Then $G\succ B_n$.

It is easy to see that $n\geq 10$ and $t(G)$ is even. If $t=g$, then $G=S_n^{\frac{n}{2}}$, and so $G\succ
 B_n$ by Lemma \[lem Sn and Bn\]. So in the following, we suppose $2\leq t\leq g-2$. Let $C_g=y_1y_2\ldots y_g$, and $x_{k_1}y_{k_1}$, $x_{k_2}y_{k_2}$, …, $x_{k_t}y_{k_t}$ ($1=k_1< k_2<\cdots < k_t$) be all the edges outside $C_G$. Then there must exist an index $k_i$ such that $k_{i}+1, k_{i}+2$ (mod $g$) are not in the set $\{k_1, k_2, \ldots, k_t\}$, that is $y_{k_{i}+1}y_{k_{i}+2}\in M_G$. Without loss of generality, we assume that $k_i=1$. Since $G$ has a perfect matching, we have that $4\leq k_2 < k_3 < \cdots < k_{t}\leq g$, and $k_2-3$ is odd, $k_{i+1}-k_i$ is odd for $2\leq i\leq t-1$, and $g-k_{t}$ is even.

By Lemma \[lem delete-edge\], we have $ b_{2i}(G) = b_{2i}(G-x_1) + b_{2i-2}(G-x_1-y_1)$, and $$b_{2i}(G-x_1)  =  \left\{\begin{array}{ll}b_{2i}(G-x_1-y_1y_2)+ b_{2i-2}(G-x_1-y_1-y_2), &  \mbox{if}\ 2i\neq g \\
b_{2i}(G-x_1-y_1y_2)+ b_{2i-2}(G-x_1-y_1-y_2)-2, &  \mbox{if}\ 2i=g
\end{array}\right..$$ Note that $G-x_1-y_1$ is a conjugated tree of order $n-2$ with $\Delta\leq 3$, by Lemma \[lem minimal-tree-degree-3\], $G-x_1-y_1\succeq F_{n-2}$. Denote $T_1=G-x_1-y_1y_2$ and $T_2=G-x_1-y_1-y_2$. Notice that if $k_{i+1}-k_i=2k+1>1$, then from $T_1$, we can obtain a different tree $T_2$ with $\Delta \leq 3$ by carrying out $k$ steps of e.g.t. Therefore we can finally get $H_{n-1}$ from $T_1$ by carrying out e.g.t repeatedly, if necessary. By Lemma \[lem edge-growing transformation\], we have $T_1\succeq H_{n-1}$. Similarly, we can obtain that $T_2\succeq H_{n-3}$. By Lemma \[lem formular for Bn and An\], we have $$b_{2i}(B_n)= b_{2i}(H_{n-1}) +
 2b_{2i-2}(H_{n-3})+2 b_{2i-6}(F_{n-8}).$$ Hence for $2i\neq g$, we have $$\begin{array}{lll}
  b_{2i}(G) & \geq &  b_{2i}(H_{n-1}) + b_{2i-2}(F_{n-2})+b_{2i-2}(H_{n-3})\\
              & = & b_{2i}(H_{n-1}) + 2b_{2i-2}(H_{n-3})+b_{2i-4}(F_{n-4})\\
              & = & b_{2i}(B_n)+b_{2i-4}(F_{n-4})-2b_{2i-6}(F_{n-8})\\
              & = &
              b_{2i}(B_n)+b_{2i-4}(F_{n-6})+b_{2i-8}(F_{n-8})+b_{2i-8}(F_{n-10})\\
              & \geq & b_{2i}(B_n),
\end{array}$$ and $$b_{g}(G)  \geq  b_{g}(B_n)+b_{g-4}(F_{n-6})+b_{g-8}(F_{n-8})+b_{g-8}(F_{n-10})-2\geq  b_{g}(B_n),$$ since $g\leq n-2$, $b_{g-4}(F_{n-6})\geq 1$ and $b_{g-8}(F_{n-8})\geq 1$. On the other hand, it is obvious that $b_{4}(G) >  b_{4}(B_n)$. Thus $G\succ B_n$.

\[lem girth at least 8-d equals 2\] Let $G\in \mathbb{U}_{n}$, $g(G)\equiv  0$ [(mod $4$)]{.nodecor}, $g(G)\geq 8$, and $d(G)=2$. Then $G\succ B_n$.

Let $C_g=z_1z_2\ldots z_g$. We apply induction on $t$. Suppose $t=1$. Assume that $d_G(x_1,z_1)=2$ and $x_1y_1z_1$ be a path with length $2$ in $G$. Then $d_G(y_1)=2$ and $x_1y_1\in M_G$. Since $G$ has a perfect matching, either $z_1z_2\in M_G$ or $z_1z_g\in M_G$. We may assume that $z_1z_2\in M_G$. Let $E_1=\{z_{i_1}y_{i_1}, z_{i_2}y_{i_2}, \ldots, z_{i_k}y_{i_k}\}$ ($i_1<i_2<\cdots <i_k$) be the set of all edges in $M_G\setminus x_1y_1$ outside $C_G$. If $E_1$ is not empty, then we have that $3\leq i_1 < i_2 < \cdots < i_{k}\leq g$, and $i_1-3$ is even, $i_{j+1}-i_j$ is odd for $1\leq j\leq k-1$, and $g-i_{k}$ is even. By Lemma \[lem delete-edge\], we have $$\begin{array}{lll}
 b_{2i}(G) & = &  b_{2i}(G-x_1) + b_{2i-2}(G-x_1-y_1)\\
           & = &  b_{2i}(G-x_1-y_1) + b_{2i-2}(G-x_1-y_1-z_1) +b_{2i-2}(G-x_1-y_1).
\end{array}$$ Denote $G_1=G-x_1-y_1$ and $G_2=G-x_1-y_1-z_1$. Then $G_1\in\mathbb{U}_{n-2}$ with $d(G_1)\leq
 1$. By Lemmas \[lem girth at least 8-d equals 0\] and \[lem girth at least 8-d equals 1\], we have $G_1\succ B_{n-2}$. By an argument similar to the proof in Lemma \[lem girth at least 8-d equals 1\], we can obtain $G_2\succeq H_{n-3}$. Therefore by Lemmas \[lem delete-edge\] and \[lem formular for Bn and An\], $$\begin{array}{lll}
 b_{2i}(G) & \geq &  b_{2i}(B_{n-2}) + b_{2i-2}(B_{n-2})+ b_{2i-2}(H_{n-3}) \\
           & = & b_{2i}(B_{n-2}) + b_{2i-2}(B_{n-2})+ b_{2i-2}(H_{n-5})+ 2b_{2i-4}(H_{n-7}) + 2b_{2i-6}(F_{n-8})\\
           & = & b_{2i}(B_{n-2}) + b_{2i-2}(B_{n-2})+ b_{2i-2}(A_{n-4}) + 2b_{2i-6}(F_{n-8}) \\
           & = & b_{2i}(B_{n-2}) + b_{2i-2}(B_{n-2})+ b_{2i-2}(B_{n-4}) + 2b_{2i-6}(F_{n-8})-2b_{2i-8}(F_{n-12}) \\
           & = & b_{2i}(B_{n-2}) + b_{2i-2}(B_{n-2})+ b_{2i-2}(B_{n-4}) + 2b_{2i-6}(F_{n-10})+2b_{2i-8}(F_{n-10}) \\
           & = & b_{2i}(B_{n}) + 2b_{2i-6}(F_{n-10})+2b_{2i-8}(F_{n-10}) \\
           & \geq & b_{2i}(B_{n}).
\end{array}$$ Since $G_1\succ B_{n-2}$, there exist $i$ such that $b_{2i}(G)>b_{2i}(B_{n})$. Hence $G\succ B_n$ for $t=1$.

Assume now that $t\geq 2$ and the assertion holds for smaller values of $t$. Let $V_1(G)=\{x_1, x_2, \ldots, x_t\}$, and for each $i\in \{1, 2, \ldots,t \}$, $x_iy_iz_{k_i}$ ($1=k_1<k_2<\cdots<k_t\leq g$) be a path with length $2$ in $G$. Then $d_G(y_i)=2$ and $x_iy_i\in M_G$. For convenience, let $k_{t+1}=k_1$. We consider the following two cases:

[**Case 1.**]{} There exist two indices $k_i$ and $k_{i+1}$ such that $z_{k_i}z_{k_{i+1}}$ is an edge on $C_g$. Without loss of generality, we assume that $k_i=1, k_{i+1}=2$. Now let $G'$ be the graph obtained from $G$ by deleting the edge $z_2y_2$ and adding one new edge $y_2y_1$. Then $G'\in \mathbb{U}_{n}$, and $d(G')=3$.

[**Claim 1.**]{} $G\succ G'$.

By Lemma \[lem delete-edge\], we have $$\begin{array}{lll}
  b_{2i}(G) & = &  b_{2i}(G-x_2) + b_{2i-2}(G-\{x_2,y_2\}),\\
  b_{2i}(G') & = &  b_{2i}(G'-x_2) + b_{2i-2}(G'-\{x_2,y_2\}),
\end{array}$$ and $$\begin{array}{lll}
b_{2i}(G-x_2)  & = & b_{2i}(G-\{x_2,x_1\})+ b_{2i-2}(G-\{x_2,x_1,y_1\})\\
               & = &
                b_{2i}(G-\{x_2,x_1,y_2\})+b_{2i-2}(G-\{x_2,x_1,y_2,z_2\})\\
               &  &
               +b_{2i-2}(G-\{x_2,x_1,y_1,y_2\})+b_{2i-4}(G-\{x_2,x_1,y_1,y_2,z_2\}),\\
b_{2i}(G'-x_2)  & = & b_{2i}(G'-\{x_2,x_1\})+ b_{2i-2}(G'-\{x_2,x_1,y_1\})\\
               & = &
                b_{2i}(G'-\{x_2,x_1,y_2\})+
                2b_{2i-2}(G'-\{x_2,x_1,y_2,y_1\}).
\end{array}$$ Denote $G_3=G-\{x_2,x_1,y_2,z_2\}$, $G_4=G-\{x_2,x_1,y_1,y_2,z_2\}$ and $G_5=G-\{x_2,x_1,y_2,y_1\}$. Since $G-\{x_2,y_2\}=G'-\{x_2,y_2\}$, $G-\{x_2,x_1,y_2\}=G'-\{x_2,x_1,y_2\}$, and $G-\{x_2,x_1,y_1,y_2\}=G'-\{x_2,x_1,y_2,y_1\}$, we have $$b_{2i}(G)- b_{2i}(G') =
        b_{2i-2}(G_3)+b_{2i-4}(G_4) -b_{2i-2}(G_5).$$ Furthermore, $$\begin{array}{lll}
b_{2i-2}(G_5) & = & b_{2i-2}(G_5-\{z_1z_2\})+b_{2i-4}(G_5-\{z_1,z_2\})-2b_{2i-g}(G_5-C_g)\\
              & = &
              b_{2i-2}(G_5-\{z_1z_2,z_2z_3\})+b_{2i-4}(G_5-\{z_1z_2\}-\{z_2,z_3\})\\
            &  &     +b_{2i-4}(G_5-\{z_1,z_2\})-2b_{2i-g}(G_5-C_g)\\
           & = &  b_{2i-2}(G_5-\{z_2\})+b_{2i-4}(G_5-\{z_2,z_3\})\\
             &  &
             +b_{2i-4}(G_5-\{z_1,z_2\})-2b_{2i-g}(G_5-C_g),\\
b_{2i-2}(G_3) & = & b_{2i-2}(G_3-\{y_1\})+b_{2i-4}(G_3-\{y_1,z_1\})\\
              & = & b_{2i-2}(G_5-\{z_2\})+b_{2i-4}(G_5-\{z_1,z_2\}).
\end{array}$$ Note that $G_5-\{z_2,z_3\}=G_4-z_3$. Hence $G_4\succ [G_5-\{z_2,z_3\}]\cup
 K_1$ by Lemma \[lem delete-vertex\]. Therefore $$b_{2i}(G)- b_{2i}(G') = b_{2i-4}(G_4) - b_{2i-4}(G_5-\{z_2,z_3\})+2b_{2i-g}(G_5-C_g)\geq 0,$$ and $$b_{6}(G)- b_{6}(G') = b_{2}(G_4)-b_{2}(G_5-\{z_2,z_3\})>0.$$ Thus the result $G\succ G'$ holds.

Now it suffices to prove that $G'\succeq B_n$. By Lemma \[lem delete-edge\], we have $$\begin{array}{lll}
  b_{2i}(G') & = &  b_{2i}(G'-x_2) + b_{2i-2}(G'-\{x_2,y_2\})\\
             & = &  b_{2i}(G'-\{x_2,y_2\}) + b_{2i-2}(G'-\{x_2,y_2,y_1\})+ b_{2i-2}(G'-\{x_2,y_2\})\\
             & = &  b_{2i}(G'-\{x_2,y_2\}) + b_{2i-2}(G'-\{x_2,y_2,y_1,x_1\})+
             b_{2i-2}(G'-\{x_2,y_2\}).
\end{array}$$ Denote $G_6=G'-\{x_2,y_2\}$ and $G_7=G'-\{x_2,y_2,y_1,x_1\}$. Then it is easy to see that $G_6\in\mathbb{U}_{n-2}$, $G_7\in\mathbb{U}_{n-4}$, and for $i=6,7$, we have $d(G_i)\leq 1$ or $d(G_i)=2$ and $t(G_i)<t(G)$. Therefore by Lemmas \[lem girth at least 8-d equals 0\], \[lem girth at least 8-d equals 1\] and the induction hypothesis, $G_6\succ B_{n-2}$, $G_7\succ B_{n-4}$ and so we have $G'\succ B_n$.

[**Case 2.**]{} For the indices $1=k_1<k_2<\ldots<k_t\leq g$, we have $k_{i+1}-k_i\geq 2$ for $1\leq i\leq t-1$ and $k_t\leq g-2$. Since $G$ has a perfect matching, we may assume that $z_1z_2\in M_G$, and $z_{k_i}w_{k_i}\in M_G$ for $2\leq i\leq t$, where $w_{k_i}\in\{z_{k_i-1},z_{k_i+1}\}$. Similarly, we have $$b_{2i}(G) =  b_{2i}(G-x_1-y_1) + b_{2i-2}(G-x_1-y_1-z_1) + b_{2i-2}(G-x_1-y_1),$$ and $$G-x_1-y_1\succ B_{n-2}.$$ Denote $T=G-x_1-y_1-z_1$. Let $T_1$ be the tree obtained from $T$ by deleting the edge $x_{i}y_{i}$ and adding one new edge $x_iw_{{k_i}}$ ($2\leq i\leq t$), we say that $T_1$ is obtained from $T$ by Operation I.

[**Claim 2.**]{} $T\succ T_1$.

By Lemma \[lem delete-edge\], we have $$\begin{array}{lll}
 b_{2i}(T)  & = & b_{2i}(T-x_i)+ b_{2i-2}(T-\{x_i,y_i\}), \\
 b_{2i}(T_1)  & = & b_{2i}(T_1-x_i)+b_{2i-2}(T_1-\{x_i,w_{{k_i}}\}).
\end{array}$$ Note that $T-x_i=T_1-x_i$, and $T_1-\{x_i,w_{{k_i}}\}$ is isomorphic to a proper subgraph of $T-\{x_i,y_i\}$, then $T-\{x_i,y_i\}\succ T_1-\{x_i,w_{{k_i}}\}$ by Lemma \[lem delete-edge-2\]. Therefore $T\succ T_1$.

Let $T'$ be the tree obtained from $T$ by deleting $t-1$ edges $x_{2}y_{2}, x_{3}y_{3},\ldots, x_ty_t$ and adding $t$ new edges $x_{2}w_{{k_2}}, x_{3}w_{{k_3}},\ldots, x_tw_{{k_t}}$. Then from $T$ we can obtain $T'$ by applying Operation I $t-1$ times. By Claim 2, we have $T\succ T'$. Clearly, $T'$ is a tree with $\Delta\leq 3$. Now we can assume that $z_{j_1}, z_{j_2},\ldots, z_{j_{l}}$ ($2<j_1 < j_2 < \cdots < j_{l}\leq g$) are all vertices with degree $3$ in $T'$. Then we have $j_1-2$ is odd, $j_{i+1}-j_i$ is odd for $1\leq i\leq l-1$, and $g-j_{l}$ is even. By an argument similar to the proof in Lemma \[lem girth at least 8-d equals 1\], we can obtain $T'\succeq H_{n-3}$. Therefore we have $T\succ H_{n-3}$ and so similar to the above case $t=1$, we can finally obtain $G\succ B_n$.

The proof is thus complete.

Let $G$ be a graph in $\mathbb{U}_{n}$ with $g(G)\equiv  0$ [(mod $4$)]{.nodecor}, $d(G)\geq 3$. Suppose $C_g=z_1z_2\ldots z_g$, $x_1,y_1\in V_1(G)$, and $x_1x_2 x_3\ldots x_dz_1$, $y_1y_2x_3 \ldots x_dz_1$ be two paths with length $d$ in $G$. For convenience, denote $x_{d+1}=z_1$. If $G'$ is the graph obtained from $G$ by deleting two edges $x_3y_2$, $y_2y_1$ and adding two new edges $y_1x_1$ and $y_2x_2$, then we say that $G'$ is obtained from $G$ by Operation II. Clearly, $G'\in\mathbb{U}_{n}$.

\[lem Operation 2\] Let $G$ be defined as above. If $G'$ is obtained from $G$ by Operation II, then $G\succ G'$.

By Lemma \[lem delete-edge\], we have $$\begin{array}{lll}
  b_{2i}(G) & = &  b_{2i}(G-y_1) + b_{2i-2}(G-\{y_1,y_2\})\\
            & = & b_{2i}(G-\{y_1,y_2\})+ b_{2i-2}(G-\{y_1,y_2,x_3\}) +
            b_{2i-2}(G-\{y_1,y_2\}),\\
  b_{2i}(G') & = &  b_{2i}(G'-y_1) + b_{2i-2}(G'-\{y_1,x_1\})\\
             & = &   b_{2i}(G'-\{y_1,y_2\})+ b_{2i-2}(G'-\{y_1,y_2,x_2) +
             b_{2i-2}(G'-\{y_1,x_1\}).
\end{array}$$ It is easy to see that $G'-\{y_1,y_2\}=G-\{y_1,y_2\}\cong
 G'-\{y_1,x_1\}$. Let $G_1=G-\{x_1,x_2,y_1,y_2\}$. Then $$\begin{array}{lll}
 b_{2i-2}(G-\{y_1,y_2,x_3\})  & = &  b_{2i-2}((G_1-x_3)\cup K_2)\\
                             & = &  b_{2i-2}(G_1-x_3)+b_{2i-4}(G_1-x_3),\\
 b_{2i-2}(G'-\{y_1,y_2,x_2\})  & = & b_{2i-2}(G_1)\\
                             & = &  b_{2i-2}(G_1-x_3)+b_{2i-4}(G_1-\{x_3,x_4\}).\\
\end{array}$$ Since $G_1-x_3\succ (G_1-\{x_3,x_4\})\cup K_1$ by Lemma \[lem delete-vertex\], we have $b_{2i}(G)\geq b_{2i}(G')$ and $b_{6}(G)>
 b_{6}(G')$. Thus the result $G\succ G'$ holds.

Let $G$ be a graph in $\mathbb{U}_{n}$ with $g(G)\equiv  0$ [(mod $4$)]{.nodecor}, $d(G)=3$. Suppose $C_g=z_1z_2\ldots z_g$, $x_1,y_1\in V_1(G)$, and $x_1x_2 x_3z_1$, $y_1y_2y_3z_i$ ($2\leq i\leq g$) be two paths with length $3$ in $G$, where $d_G(x_3)=d_G(y_3)=2$. If $G'$ is the graph obtained from $G$ by deleting two edges $y_1y_2$, $y_2y_3$ and adding two new edges $y_1x_1$ and $y_2x_2$, then we say that $G'$ is obtained from $G$ by Operation III. Clearly, $G'\in\mathbb{U}_{n}$.

\[lem Operation 3\] Let $G$ be defined as above. If $G'$ is obtained from $G$ by Operation III, then $G\succ G'$.

The proof is similar to that of Lemma \[lem Operation 2\].

\[thm girth at least 8\] Let $G\in \mathbb{U}_{n}$, $g(G)\equiv  0$ [(mod $4$)]{.nodecor}, $g(G)\geq 8$. Then $G\succ B_n$.

We apply induction on $d$. As the case $d\leq 2$ was proved by Lemmas \[lem girth at least 8-d equals 0\], \[lem girth at least 8-d equals 1\], and \[lem girth at least 8-d equals 2\], we now suppose that $d\geq 3$ and the assertion holds for smaller values of $d$.

Let $C_g=z_1z_2\ldots z_g$. Assume that $d_G(x_1,z_1)=d$ and $x_1x_2 \ldots x_dz_1$ be a path with length $d$ in $G$. Then $d_G(x_2)=2$ and $x_1x_2\in M_G$. For convenience, denote $x_{d+1}=z_1$. Let $G_1=G-\{x_1,x_2\}$ and $G_2=G-\{x_1,x_2,x_3\}$. By Lemma \[lem delete-edge\], we have $$\begin{aligned}
b_{2i}(G) =  b_{2i}(G_1) + b_{2i-2}(G_1)+ b_{2i-2}(G_2).\label{coefficient formular}
  \end{aligned}$$ Note that $$\begin{aligned}
 b_{2i}(B_n)  = b_{2i}(B_{n-2})+b_{2i-2}(B_{n-2})+b_{2i-2}(B_{n-4}) \ (n\geq 10).\label{coefficient formular-Bn} \end{aligned}$$

Now we prove the result for the given $d$ by induction on $t$.

[**Case 1.**]{} Suppose $t=1$.

[**Subcase 1.1.**]{} $d_G(x_3)=d_G(x_4)=2$.

Note that $x_3x_4\in M_G$. It is easy to see that $G_1\in \mathbb{U}_{n-2}$, $G_2-x_4\in \mathbb{U}_{n-4}$, $d(G_1)<d$ and $d(G_2-x_4)<d$. So by the induction hypothesis, $G_1\succeq
  B_{n-2}$, $G_2-x_4\succeq B_{n-4}$. By Lemma \[lem delete-vertex\], we have $G_2\succ (G_2-x_4)\cup K_1$, and so $G_2\succ B_{n-4}\cup K_1$. It follows from Eqs. and that $G\succ B_n$.

[**Subcase 1.2.**]{} $d_G(x_3)=2, d_G(x_4)=3$ and $d>3$.

Note that $x_3x_4\in M_G$. Suppose $y_2\not\in\{x_3,x_5\}$ is a neighbor of $x_4$, and $y_1y_2\in M_G$. Since $t=1$, we have $d_G(y_1)=1$ and $d_G(y_2)=2$. Then we have $G_2-y_1\in \mathbb{U}_{n-4}$ since $(M_G\setminus(\{x_1x_2,x_3x_4,y_1y_2\})\cup\{y_2x_4\}$ is a perfect matching of $G_2-y_{1}$. Therefore similar to Subcase 1.1, we have $G_1\succeq B_{n-2}$ and $G_2\succ B_{n-4}\cup
   K_1$, and so $G\succ B_n$.

[**Subcase 1.3.**]{} $d_G(x_3)=2, d_G(x_4)=3$ and $d=3$, i.e., $x_{4}=z_1$.

Note that $x_3x_4\in M_G$. Since $G$ has a perfect matching, $g(G)\equiv  0$ [(mod $4$)]{.nodecor}, there exist $k\geq 1$ pendent edges $z_{i_1}y_{i_1},\ldots,z_{i_k}y_{i_k}$ ($2\leq i_1<\ldots<i_k\leq g$) such that $k$ is odd and $z_{i_k}y_{i_k}\in M_G$. Then we have $G_2-y_{i_1}\in \mathbb{U}_{n-4}$, since $$(M_G\setminus\{x_1x_2,x_3x_4,z_{i_1}y_{i_1},z_2z_3,z_4z_5,\ldots,z_{i_1-2}z_{i_1-1}\})\cup\{z_1z_2,z_3z_4,\ldots,z_{i_1-1}z_{i_1}\}$$ is a perfect matching of $G_2-y_{i_1}$. Therefore similarly, we have $G_1\succeq B_{n-2}$ and $G_2\succ B_{n-4}\cup K_1$, and so $G\succ B_n$.

[**Subcase 1.4.**]{} $d_G(x_3)=3$.

Since $t=1$, $x_3x_4\not\in M_G$, we assume that $x_3y_1\in M_G$. Then $d_G(y_1)=1$, $d_{G_2}(y_1)=0$ and $G_2-y_1\in \mathbb{U}_{n-4}$. Therefore similarly, we have $G_1\succ B_{n-2}$ and $G_2\succeq B_{n-4}\cup K_1$, and so $G\succ B_n$.

[**Case 2.**]{} Assume now that $t\geq 2$ and the assertion holds for smaller values of $t$. Note that $G_1\in \mathbb{U}_{n-2}$ with $d(G_1)=d$ and $t(G_1)<t$. By the induction hypothesis, $G_1\succeq B_{n-2}$.

[**Subcase 2.1.**]{} $d_G(x_3)=d_G(x_4)=2$.

The proof is similar to that of Subcases 1.1.

[**Subcase 2.2.**]{} $d_G(x_3)=3$ and $x_3x_4\not\in M_G$.

The proof is similar to that of Subcases 1.4.

[**Subcase 2.3.**]{} $d_G(x_3)=2$, $d_G(x_4)=3$, and $d>3$.

Suppose $y_2\not\in\{x_3,x_5\}$ is a neighbor of $x_4$, and $y_1y_2\in M_G$. If $d_G(y_2)=3$, let $y_3\not\in\{x_4,y_1\}$ be a neighbor of $y_2$, and $y_4y_3\in M_G$. Then we have $d_G(y_1)=d_G(y_4)=1$, $d_G(y_3)=2$, and $G_2-y_1\in \mathbb{U}_{n-4}$, since $(M_G\setminus(\{x_1x_2,x_3x_4,y_1y_2\})\cup\{y_2x_4\}$ is a perfect matching of $G_2-y_{1}$. Therefore similarly, we have $G_2\succ B_{n-4}\cup K_1$, and so $G\succ B_n$.

[**Subcase 2.4.**]{} $d_G(x_3)=3$ and $x_3x_4\in M_G$.

Suppose $y_2\not\in\{x_2,x_4\}$ is a neighbor of $x_3$, and $y_1y_2\in
 M_G$. Then we have $d_G(y_1)=1$ and $d_G(y_2)=2$. Let $G'$ be the graph obtained from $G$ by Operation II. It follows from Lemma \[lem Operation 2\] that $G\succ G'$. Similarly, we have $$b_{2i}(G') =  b_{2i}(G'-\{y_1,x_1\}) + b_{2i-2}(G'-\{y_1,x_1\})+ b_{2i-2}(G'-\{y_1,x_1,x_2,y_2\}),$$ and $G'-\{y_1,x_1\}\succ B_{n-2}$, and $G'-\{y_1,x_1,x_2,y_2\}\succeq B_{n-4}$. Therefore $G'\succ B_n$.

[**Subcase 2.5.**]{} $d_G(x_3)=2$ and $d=3$.

Now $x_4=z_1$. Since $t\geq 2$, suppose that $y_1\in V_1(G)$ and $y_1y_2y_3z_i$($i\neq 2$) be a path with length $3$ in $G$. Then by the above subcases, we may assume that $d_G(y_3)=2$. Let $G'$ be the graph obtained from $G$ by Operation III, then we have $G\succ G'$ by Lemma \[lem Operation 3\]. And similar to the Subcase 2.4, we have $G'\succ B_n$.

The proof is thus complete.

(1831.4, 291.0)(0,0) (0,0)[![The graphs considered in Theorem \[thm girth 4-A\_n\].[]{data-label="fig-lemma-An"}](fig-lemma-An "fig:")]{} (135.76,38.20)

(150.0, 60.0)\[l\]

$I_1$

(808.17,41.93)

(150.0, 60.0)\[l\]

$I_2$

(1507.92,50.63)

(150.0, 60.0)\[l\]

$I_3$

\[thm girth 4-A\_n\] Let $G\in \mathbb{U}_{n}$, $g(G)=4$, $G\not\cong A_n$. If there are just two edges of $M_G$ in $C_4$, then $G\succ A_n$.

We apply induction on $d$. Suppose $G\not\cong A_n$ is a graph in $\mathbb{U}_{n}$ with $g=4$, and there are just two edges of $M_G$ in $C_4$. For $d\leq 1$, there is nothing to prove. Suppose $d=2$. Then $G$ is isomorphic to one of the following graphs $B_8$, $I_1$, $I_2$ and $I_3$, as shown in Figure \[fig-lemma-An\]. By Lemma \[lem Bn and An\], we have $B_8\succ A_8$. It is easy to obtain that $$\begin{array}{lll}
\phi(I_{1},x) & = &  x^{8}-8x^6+16x^4-9x^2,\\
 \phi(I_{2},x) & = &  x^{10}-10x^8+30x^6-34x^4+12x^2,\\
 \phi(I_{3},x) & = &  x^{12}-12x^{10}+48x^8-84x^6+64x^4-16x^2,\\
  \phi(A_8,x) & = & x^8-8x^6+16x^4-6x^2,\\
 \phi(A_{10},x) & = &   x^{10}-10x^8+30x^6-28x^4+6x^2,\\
 \phi(A_{12},x) & = &  x^{12}-12x^{10}+48x^8-74x^6+40x^4-6x^2.
 \end{array}$$ Hence $I_{1}\succ A_{8}$, $I_{2}\succ A_{10}$, and $I_{3}\succ A_{12}$. Now suppose that $d\geq 3$ and the assertion holds for smaller values of $d$. By an argument similar to the proof of Theorem \[thm girth at least 8\], we can obtain $G\succ A_n$.

(2482.2, 296.0)(0,0) (0,0)[![The graphs considered in Theorem \[thm girth 4-D\_n\].[]{data-label="fig-lemma-Dn"}](fig-lemma-Dn "fig:")]{} (171.40,54.93)

(150.0, 60.0)\[l\]

$I_4$

(780.67,48.28)

(150.0, 60.0)\[l\]

$I_5$

(1409.27,38.20)

(150.0, 60.0)\[l\]

$I_6$

(2040.66,39.44)

(150.0, 60.0)\[l\]

$I_7$

\[thm girth 4-D\_n\] Let $G\in \mathbb{U}_{n}$, $g(G)=4$, $G\not\cong D_n$. If there is just one edge of $M_G$ in $C_4$, then $G\succ D_n$.

We apply induction on $d$. For $d\leq 1$, there is nothing to prove. Suppose $d=2$. Then $G$ is isomorphic to $I_4$, as shown in Figure \[fig-lemma-Dn\]. It is easy to obtain that $$\begin{array}{lll}
\phi(I_{4},x) & = &  x^{10}-10x^8+29x^6-32x^4+12x^2-1,\\
 \phi(D_{10},x) & = &  x^{10}-10x^8+29x^6-28x^4+10x^2-1.
\end{array}$$ Hence $I_4\succ D_{10}$.

Now suppose that $d\geq 3$ and the assertion holds for smaller values of $d$. We use the same notations as in Theorem \[thm girth at least 8\]. Then the proof is similar to that of Theorem \[thm girth at least 8\]. We can divide two cases $x_dz_1\in M_g$ and $x_dz_1\not\in M_g$ to proceed. The difference is that we need to prove the result $G\succ D_n$ for the case: $d=3$, $t=1$, $d_G(x_3)=2$ and $z_2z_3\in
 M_4$. Since $G\not\cong D_n$, $G$ is isomorphic to one of the following graphs $I_5$, $I_6$ and $I_7$, as shown in Figure \[fig-lemma-Dn\]. It is easy to obtain that $$\begin{array}{lll}
\phi(I_{5},x) & = &  x^{10}-10x^8+30x^6-33x^4+11x^2-1,\\
\phi(I_{6},x) & = &  x^{10}-10x^8+30x^6-33x^4+12x^2-1,\\
\phi(I_{7},x) & = &  x^{12}-12x^{10}+48x^8-83x^6+62x^4-16x^2+1,\\
 \phi(D_{12},x) & = &  x^{12}-12x^{10}+47x^8-72x^6+46x^4-12x^2+1.
\end{array}$$ Hence $I_5\succ D_{10}$, $I_6\succ D_{10}$ and $I_7\succ D_{12}$.

\[thm girth 4-E\_n\] Let $G\in \mathbb{U}_{n}$, $g(G)=4$, $G\not\cong E_n$. If there are no edges of $M_G$ in $C_4$, then $G\succ E_n$.

The proof is similar to that of Theorem \[thm girth at least 8\].

0.5cm [**Proof of Theorem \[main thm\].**]{} Notice that $E_8\cong S_8^{4}$. The proof follows directly from Lemma \[lem g-non-4-multiple\], Theorems 3.6-3.9, 2.8-2.10, and 2.16. 0.5cm

[**Acknowledgments**]{} 0.4cm

The first author is supported by NNSFC (Nos. 11101351 and 11171288) and NSF of the Jiangsu Higher Education Institutions (No. 11KJB110014); the second author is supported by NNSFC (No. 11101351); and the third author is supported by NNSFC (Nos. 11326216 and 11301306).

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[^1]: Corresponding author.
