Let $M=\left[a_{ij}\right]\left(i,j\in 1,2,3,\cdots ,n\right)$ be an $n\times n$ matrix. The permanent of $M$ is defined as

$\text{per}M=\underset{\sigma \in {\text{Λ}}_n}{{\displaystyle \sum }}\underset{i=1}{\overset{n}{{\displaystyle \prod }}}{a}_{i\varphi \left(i\right)},$

where ${\text{Λ}}_n$ is the symmetric group of degree $n$. Valiant [1] proved that even for (0,1)-matrices, computing the permanent is $\#P$-complete.

Let $G=\left(V\left(G\right),E\left(G\right)\right)$ be a simple connected graph with $n$ vertices, and let $d_{v_i}$ denote the degree of vertex $v_i\left(i=1,2,\cdots ,n\right)$ in $V\left(G\right)$. The adjacency matrix and degree matrix of graph $G$ are denoted by $A\left(G\right)$ and $D\left(G\right)$, respectively. The Laplacian matrix of graph $G$ is defined as $L\left(G\right)=D\left(G\right)-A\left(G\right)$. Based on the Laplacian matrix of graphs, researchers began to study the Laplacian permanents of graphs. For more research on the permanent, refer to [2] - [7] . Brualdi and Goldwasser [8] obtained some results on the Laplacian permanents of bipartite graphs: the minimum and maximum values of the Laplacian permanents of bipartite graphs with $n$ vertices. They also gave bounds for $\text{per}L\left(G\right)$.

Theorem 1. ( [8] ) Let $T$ be an $n$-vertex tree. Then

$2\left(n-1\right)\le \text{per}T\le \frac{2-\sqrt{2}}{2}{\left(1+\sqrt{2}\right)}^n+\frac{2+\sqrt{2}}{2}{\left(1-\sqrt{2}\right)}^n.$

The lower bound is achieved if and only if $T$ is a star, and the upper bound is achieved if and only if $T$ is a path.

When $G$ is a tree, the minimum value of the Laplacian permanent can be obtained directly from the degrees of the vertices and the size of the matching. For the maximum value of the Laplacian permanent, it is related to the degrees of the vertices and the size of the matching. The following problems are proposed:

Problem 1. ( [8] ) For an $n$-vertex tree with maximum degree $k$, what is the maximum value of the Laplacian permanent?

Problem 2. ( [8] ) For an $n$-vertex tree with matching number $k$, what is the maximum value of the Laplacian permanent?

Problem 3. ( [8] ) For an $n$-vertex tree that is a $\left(p,q\right)$-bipartition, what is the maximum value of the Laplacian permanent?

Brualdi and Goldwasser [8] defined the Laplacian permanent $\pi \left(G\right)$ of a graph.

$\pi \left(G\right)=\frac{\text{per}L\left(G\right)}{d_{v_1}{d}_{v_2}\cdots d_{v_n}}.$

Brualdi and Goldwasser [8] first studied the Laplacian ratios of graphs and systematically studied the properties of the Laplacian ratios of graphs. They characterized the lower bounds of the Laplacian ratios of some graphs. Goldwasser [9] studied the maximum and minimum values of the Laplacian ratios of graphs with matching number $k$. The Laplacian ratio of graph plays an important role in statistics, chemistry, and communication. The Laplacian ratio can also be used to count the number of spanning trees of a graph. Brualdi and Goldwasser [8] first studied the Laplacian ratios of graphs and obtained some results on the Laplacian ratios of trees. They further proposed the following problems in 1983:

Problem 4. ( [8] ) What is the minimum value of the Laplacian ratios of $n$-vertex trees with maximum degree $k$?

Problem 5. ( [8] ) What is the minimum value of the Laplacian ratios of $n$-vertex trees with $k$-matching?

Problem 6. ( [8] ) What is the maximum value of the Laplacian ratios of $n$-vertex trees?

In the past few decades, Goldwasser [9] solved Problem 5. Currently, the study of the Laplacian ratios of graphs has attracted widespread attention. The Laplacian ratio theory is well elaborated in [10] - [16] .

We further study the Laplacian ratios of trees. Through graph transformations, we determine the graph that minimizes the Laplacian ratios among all caterpillar trees.

In this paper, in Section 2, we provide some preliminary knowledge. In Section 3, we study some Laplacian permanents of trees. We also characterize a graph transformation related to the Laplacian ratio. We obtain that for any $n$-vertex caterpillar tree $T$, there exists an $n$-vertex caterpillar tree $C_n$ such that $\pi \left(C_n\right)\le \pi \left(T\right)$. In Section 4, we provide a conclusion related to the Laplacian permanent and the Laplacian ratio.

To facilitate the proofs in the following sections, we introduce some basic concepts in this section.

Let $G$ be a graph. If $G$ has a $u,v$-path, then the distance between $u$ and $v$ (denoted by $d_G\left(u,v\right)$, or simply $d\left(u,v\right)$) is the length of the shortest $u,v$-path. The diameter is defined as ${\max }_{u,v\in V\left(G\right)}d\left(u,v\right)$. $P_k$ is a path with diameter $k-1$. A caterpillar is a tree that consists of a central path with leaves attached to it. A broom graph $B\left(n,k\right)$ is obtained by attaching $n-k$ pendant vertices to one end of a path $P_k$ with $k$ vertices. $S_n$ is the star graph with $n$ vertices. $L_S\left(G\right)$ denotes the Laplacian matrix of graph $G$ after removing the rows and columns corresponding to the vertices in $S\left(S\subseteq V\left(G\right)\right)$. In particular, if $S=\left\{v\right\}$ and $v\in V\left(G\right)$, then $L_{\left\{v\right\}}\left(G\right)$ can be abbreviated as $L_v\left(G\right)$. Let $A$ be an $n\times n$ matrix. $A\left(i,j\right)$ denotes the $\left(n-2\right)\times \left(n-2\right)$ submatrix obtained by deleting the $i$-th and $j$-th rows and columns of $A$. $A_{ij}$ denotes the $\left(n-1\right)\times \left(n-1\right)$ submatrix obtained by deleting the $i$-th row and column of $A$. Let $Q_n$ be the $n\times n$ submatrix obtained by deleting the first row and column of $L\left(P_n\right)$.

Lemma 2. Let $A$ be an $n\times n$ matrix and $i,j\in \left\{1,2,\cdots ,n\right\}$. Then

$\text{per}A=\underset{i}{{\displaystyle \sum }}{a}_{ij}\text{per}{A}_{ij}\text{and}\text{per}A=\underset{j}{{\displaystyle \sum }}{a}_{ij}\text{per}{A}_{ij}.$

Lemma 3. ( [8] ) Let $T$ be a tree. Then

$\pi \left(T\right)=\underset{k=0}{\overset{\lfloor \frac{n}{2}\rfloor }{{\displaystyle \sum }}}{\pi }_k\left(T\right)$

Lemma 4. ( [8] ) Let $k\ge 2$ and $r\ge 1$ be integers. Then

$\text{per}L\left(B\left(n,k\right)\right)=\left(2n-2k+1\right)\text{per}{Q}_{k-1}+\text{per}{Q}_{k-2},$

$\text{per}{Q}_k=\frac{1}{2}{\left(1+\sqrt{2}\right)}^k+\frac{1}{2}{\left(1-\sqrt{2}\right)}^k,$

and

$\text{per}{P}_k=\frac{2-\sqrt{2}}{2}{\left(1+\sqrt{2}\right)}^k+\frac{2+\sqrt{2}}{2}{\left(1-\sqrt{2}\right)}^k.$

Lemma 5. ( [16] ) Let $P_{k+1}=u_1{u}_2\cdots u_i\cdots u_{k+1}$ be a path. Let $T_{n,k,i}$ be the tree obtained by attaching $n-k-1$ pendant vertices to vertex $u_i$ of path $P_{k+1}$. Then

$\text{per}L\left(T_{n,k,i}\right)=\text{per}L\left(P_{k+1}\right)+2\left(n-k-1\right)\text{per}{Q}_{i-1}\text{per}{Q}_{k-i+1}.$

Lemma 6. ( [11] ) A tree is a caterpillar if and only if it contains no $Y$-structure (see Figure 1 ).

Lemma 7. Let $R_n^2$ and $R_n^{i-1}$ be the graphs shown in Figure 2 . Then

(i) $\text{per}L\left(R_n^2\right)=\frac{216-127\sqrt{2}}{2}{\left(1+\sqrt{2}\right)}^{n-5}+\frac{216+127\sqrt{2}}{2}{\left(1-\sqrt{2}\right)}^{n-5}$,

(ii) $\text{per}L\left(R_n^{i-1}\right)=\frac{4-\sqrt{2}}{2}{\left(1+\sqrt{2}\right)}^{n-2}+\frac{4-\sqrt{2}}{2}{\left(1-\sqrt{2}\right)}^{n-2}+{\left(1+\sqrt{2}\right)}^{n-4i+1}+$ ${\left(1-\sqrt{2}\right)}^{n-4i+1}$.

Proof. By expanding the first row and column of $\text{per}{R}_n^2$, we obtain the permanent of the Laplacian matrix of $R_n^2$.

$L\left(R_n^2\right)=\left[\begin{array}{l}\begin{array}{cccccccccccc}3& -1& 0& -1& 0& 0& 0& -1& 0& \cdots & 0& 0\\ -1& 2& -1& & & & & & & & & \\ 0& -1& 1& & & & & & & & & \\ -1& & & 2& -1& & & & & & & \\ 0& & & -1& 2& -1& & & & & & \\ 0& & & & -1& 2& -1& & & & & \\ 0& & & & & -1& 1& & & & & \\ -1& & & & & & & 2& -1& & & \\ 0& & & & & & & -1& 2& & & \\ ⋮& & & & & & & & & \ddots & & \\ 0& & & & & & & & & & 2& -1\\ 0& & & & & & & & & & -1& 1\end{array}\hfill \end{array}\right].$

By Lemma 2, the permanent of $L\left(R_n^2\right)$ can be expanded as

$\text{per}L\left(R_n^2\right)=3\times 3\times \text{per}{Q}_4\text{per}{Q}_{n-7}+\text{per}{Q}_4\text{per}{Q}_{n-7}+3\times \text{per}{Q}_3\text{per}{Q}_{n-7}+3\times \text{per}{Q}_4\text{per}{Q}_{n-8}$.

By Lemma 4,

$\begin{array}{c}\text{per}L\left(R_n^1\right)=3\times 3\times \left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^4+\frac{1}{2}{\left(1-\sqrt{2}\right)}^4\right)\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{n-7}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{n-7}\right)\\ +\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^4+\frac{1}{2}{\left(1-\sqrt{2}\right)}^4\right)\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{n-7}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{n-7}\right)\\ +3\times \left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^4+\frac{1}{2}{\left(1-\sqrt{2}\right)}^4\right)\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{n-7}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{n-7}\right)\\ +3\times \left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^4+\frac{1}{2}{\left(1-\sqrt{2}\right)}^4\right)\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{n-8}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{n-8}\right)\\ =\frac{216-127\sqrt{2}}{2}{\left(1+\sqrt{2}\right)}^{n-5}+\frac{216+127\sqrt{2}}{2}{\left(1-\sqrt{2}\right)}^{n-5}\end{array}$.

By expanding the first row and column of $\text{per}{R}_n^{i-1}$, we obtain the permanent of the Laplacian matrix of $R_n^{i-1}$.

$L\left(R_n^2\right)=\left[\begin{array}{l}\begin{array}{ccccccccccc}3& -1& 0& -1& 0& \cdots & 0& 0& \cdots & 0& 0\\ -1& 2& -1& & & & & & & & \\ 0& -1& 1& & & & & & & & \\ -1& & & 2& -1& & & & & & \\ 0& & & -1& 2& & & & & & \\ ⋮& & & & & \ddots & & & & & \\ 0& & & & & & 2& -1& & & \\ 0& & & & & & -1& 1& & & \\ ⋮& & & & & & & & \ddots & & \\ 0& & & & & & & & & 2& -1\\ 0& & & & & & & & & -1& 1\end{array}\hfill \end{array}\right].$

By Lemma 2, the permanent of $L\left(R_n^{i-1}\right)$ can be expanded as

$\text{per}L\left(R_n^2\right)=3\times 3\times \text{per}{Q}_{2i-2}\text{per}{Q}_{n-2i-1}+\text{per}{Q}_{2i-2}\text{per}{Q}_{n-2i-1}+3\times \text{per}{Q}_{2i-3}\text{per}{Q}_{n-2i-1}+3\times \text{per}{Q}_{2i-2}\text{per}{Q}_{n-2i-2}$.

By Lemma 4,

$\begin{array}{c}\text{per}L\left(R_n^1\right)=3\times 3\times \left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{2i-2}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{2i-2}\right)\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{n-2i-1}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{n-2i-1}\right)\\ +\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{2i-2}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{2i-2}\right)\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{n-2i-1}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{n-2i-1}\right)\\ +3\times \left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{2i-3}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{2i-3}\right)\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{n-2i-1}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{n-2i-1}\right)\\ +3\times \left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{2i-2}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{2i-2}\right)\left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{n-2i-2}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{n-2i-2}\right)\\ =\frac{4-\sqrt{2}}{2}{\left(1+\sqrt{2}\right)}^{n-2}+\frac{4-\sqrt{2}}{2}{\left(1-\sqrt{2}\right)}^{n-2}+{\left(1+\sqrt{2}\right)}^{n-4i+1}+{\left(1-\sqrt{2}\right)}^{n-4i+1}\end{array}$.

□

Lemma 8. Let $T_{n,k,6}$ and $T_{n,k,2i}$ be two trees. Then

(i)

$\text{per}L\left(T_{n,k,6}\right)=\left(41n-41k-12+\frac{41\sqrt{2}}{2}\right){\left(1+\sqrt{2}\right)}^{k-5}+\left(41n-41k-12-\frac{41\sqrt{2}}{2}\right){\left(1-\sqrt{2}\right)}^{k-5}$,

(ii) $\begin{array}{c}\text{per}L\left(T_{n,k,2i}\right)=\left(\frac{\left(n-k-1\right){\left(2\sqrt{2}-3\right)}^{2i-1}+n-k+\sqrt{2}-1}{2}\right){\left(1+\sqrt{2}\right)}^k\\ +\left(-\frac{\left(n-k-1\right){\left(2\sqrt{2}+3\right)}^{2i-1}+n-k-\sqrt{2}-1}{2}\right){\left(1-\sqrt{2}\right)}^k\end{array}$.

Proof. By Lemma 5,

$\text{per}L\left(T_{n,k,6}\right)=\text{per}L\left(P_{k+1}\right)+2\left(n-k-1\right)\text{per}{Q}_5\text{per}{Q}_{k-5}$,

$\text{per}L\left(T_{n,k,2i}\right)=\text{per}L\left(P_{k+1}\right)+2\left(n-k-1\right)\text{per}{Q}_{2i-1}\text{per}{Q}_{k-2i+1}$.

By Lemma 4,

$\begin{array}{c}\text{per}L\left(T_{n,k,6}\right)=\text{per}L\left(P_{k+1}\right)+2\left(n-k-1\right)\text{per}{Q}_5\text{per}{Q}_{k-5}\\ =\frac{2-\sqrt{2}}{2}{\left(1+\sqrt{2}\right)}^{k+1}+\frac{2+\sqrt{2}}{2}{\left(1-\sqrt{2}\right)}^{k+1}\\ +2\left(n-k-1\right)\text{per}{Q}_5\times \left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{k-5}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{k-5}\right)\\ =\left(41n-41k-12+\frac{41\sqrt{2}}{2}\right){\left(1+\sqrt{2}\right)}^{k-5}\\ +\left(41n-41k-12-\frac{41\sqrt{2}}{2}\right){\left(1-\sqrt{2}\right)}^{k-5},\end{array}$

$\begin{array}{c}\text{per}L\left(T_{n,k,2i}\right)=\text{per}L\left(P_{k+1}\right)+2\left(n-k-1\right)\text{per}{Q}_{2i}\text{per}{Q}_{k-2i+1}\\ =\frac{2-\sqrt{2}}{2}{\left(1+\sqrt{2}\right)}^{k+1}+\frac{2+\sqrt{2}}{2}{\left(1-\sqrt{2}\right)}^{k+1}\\ +2\left(n-k-1\right)\times \left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{2i-1}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{2i-1}\right)\\ \times \left(\frac{1}{2}{\left(1+\sqrt{2}\right)}^{n-2i+1}+\frac{1}{2}{\left(1-\sqrt{2}\right)}^{n-2i+1}\right)\\ =\left(\frac{\left(n-k-1\right){\left(2\sqrt{2}-3\right)}^{2i-1}+n-k+\sqrt{2}-1}{2}\right){\left(1+\sqrt{2}\right)}^k\\ +\left(-\frac{\left(n-k-1\right){\left(2\sqrt{2}+3\right)}^{2i-1}+n-k-\sqrt{2}-1}{2}\right){\left(1-\sqrt{2}\right)}^k.\end{array}$

□

Definition 1. Let $T$ be a tree containing a $Y$-structure, where $u$ is the central vertex of the $Y$-structure, and $v_1,v_2,\cdots ,v_t,w_1,w_2,\cdots ,w_s\in N\left(v\right)$, where $w_1,w_2,\cdots ,w_s$ are pendant vertices in $T$, and $v_1,v_2,\cdots ,v_t$ are non-pendant vertices in $T$. $T\left[v_i\to u;4\right]$ is obtained by contracting the edges $u{v}_i\left(1\le i\le t\right)$ and then attaching all vertices $v_1,v_2,\cdots ,v_t$ to $u$. We call $T\left[v_i\to u;4\right]$ the graph obtained from $T$ by Graph Transformation 1.

Theorem 9. Let $T$ and $T\left[v_i\to u;1\right]$ be the graphs shown in Figure 3 . Then $\pi \left(T\right)>\pi \left(T\left[v_i\to u;1\right]\right)$.

Proof. Let ${\mathcal{T}}_k\left(T\right)$ be the set of all $k$-matchings of $T$, and let ${\mathcal{T}}_k\left(T\left[v_i\to u;4\right]\right)$ be the set of all $k$-matchings of $T\left[v_i\to u;4\right]$. We can partition the $k$-matchings of $T\left[v_i\to u;4\right]$ into two types: those that do not contain vertex $u$, denoted by ${\mathcal{T}}_{k,u}^{-}\left(T\left[v_i\to u;4\right]\right)$, and those that contain vertex $u$, denoted by ${\mathcal{T}}_{k,u}^{+}\left(T\left[v_i\to u;4\right]\right)$. Clearly, ${\mathcal{T}}_k\left(T\left[v_i\to u;4\right]\right)={\mathcal{T}}_{k,u}^{+}\left(T\left[v_i\to u;4\right]\right)\cup {\mathcal{T}}_{k,u}^{-}\left(T\left[v_i\to u;4\right]\right)$. Let the edges incident to $u$ in $T\left[v_i\to u;4\right]$ be labeled as $h_1,h_2,\cdots ,h_p,e_1,e_2,\cdots ,e_{s+t}$, where $h_1,h_2,\cdots ,h_p$ are non-pendant edges, and $e_1,e_2,\cdots ,e_{s+t}$ are pendant edges. Let ${h^{\prime }}_1,{h^{\prime }}_2,\cdots ,{h^{\prime }}_p,{e^{\prime }}_1,{e^{\prime }}_2,\cdots ,{e^{\prime }}_{s+t}$ be the corresponding edges in $T$ (without applying Graph Transformation 4). Let ${\mathcal{T}}_k\left(T\right)$ be the set of all $k$-matchings of $T$ that contain ${h^{\prime }}_1,{h^{\prime }}_2,\cdots ,{h^{\prime }}_p,{e^{\prime }}_1,{e^{\prime }}_2,\cdots ,{e^{\prime }}_{s+t}$, denoted by ${\mathcal{T}}_k^{+}\left(T\right)$, and those that do not contain ${h^{\prime }}_1,{h^{\prime }}_2,\cdots ,{h^{\prime }}_p,{e^{\prime }}_1,{e^{\prime }}_2,\cdots ,{e^{\prime }}_{s+t}$, denoted by ${\mathcal{T}}_k^{-}\left(T\right)$. Clearly, ${\mathcal{T}}_k\left(T\right)={\mathcal{T}}_k^{+}\left(T\right)\cup {\mathcal{T}}_k^{-}\left(T\right)$. Due to the structure of $T$ and $T\left[v_i\to u;4\right]$, for any $\mu \in {\mathcal{T}}_{k,u}^{-}\left(T\left[v_i\to u;4\right]\right)$, there exists a ${\mu }^{\prime }\in {\mathcal{T}}_k^{-}\left(T\right)$ such that $d\left({\mu }^{\prime }\right)=d\left(\mu \right)$. We further partition ${\mathcal{T}}_{k,u}^{+}\left(T\left[v_i\to u;4\right]\right)$ into two subsets: ${\mathcal{T}}_{k,u}^{++}\left(T\left[v_i\to u;4\right]\right)$ (those $k$-matchings that contain one of $h_1,h_2,\cdots ,h_p$) and ${\mathcal{T}}_{k,u}^{+-}\left(T\left[v_i\to u;4\right]\right)$ (those $k$-matchings that contain one of $e_1,e_2,\cdots ,e_{s+t}$). Clearly, ${\mathcal{T}}_{k,u}^{+}\left(T\left[v_i\to u;4\right]\right)={\mathcal{T}}_{k,u}^{++}\left(T\left[v_i\to u;4\right]\right)\cup {\mathcal{T}}_{k,u}^{+-}\left(T\left[v_i\to u;4\right]\right)$. Similarly, we partition ${\mathcal{T}}_k^{+}\left(T\right)$ into two subsets: ${\mathcal{T}}_k^{++}\left(T\right)$ (those $k$-matchings that contain one of ${h^{\prime }}_1,{h^{\prime }}_2,\cdots ,{h^{\prime }}_p$) and ${\mathcal{T}}_k^{+-}\left(T\right)$ (those $k$-matchings that contain one of ${e^{\prime }}_1,{e^{\prime }}_2,\cdots ,{e^{\prime }}_{s+t}$). Clearly, ${\mathcal{T}}_k^{+}\left(T\right)={\mathcal{T}}_k^{++}\left(T\right)\cup {\mathcal{T}}_k^{+-}\left(T\right)$. Due to the structure of $T$ and $T\left[v_i\to u;4\right]$, we have $d\left({h^{\prime }}_i\right)>d\left(h_i\right)\left(1\le i\le p\right)$. Therefore, for any $\mu \in {\mathcal{T}}_{k,u}^{++}\left(T\left[v_i\to u;4\right]\right)$, there exists a ${\mu }^{\prime }\in {\mathcal{T}}_k^{++}\left(T\right)$, such that $d\left({\mu }^{\prime }\right)>d\left(\mu \right)$. Now consider ${\mathcal{T}}_{k,u}^{+-}\left(T\left[v_i\to u;4\right]\right)$ and ${\mathcal{T}}_k^{+-}\left(T\right)$. For $T\left[v_i\to u;4\right]$,

$\frac{1}{d\left(e_1\right)}+\frac{1}{d\left(e_2\right)}+\cdots +\frac{1}{d\left(e_{s+t}\right)}=\frac{s+t}{d_{v_1}+d_{v_2}+\cdots +d_{v_t}+s}.$

For $T$,

$\frac{1}{d\left({e^{\prime }}_1\right)}+\frac{1}{d\left({e^{\prime }}_2\right)}+\cdots +\frac{1}{d\left({e^{\prime }}_{s+t}\right)}=\frac{1}{s+t}\left(\frac{1}{d_{v_1}}+\frac{1}{d_{v_2}}+\cdots +\frac{1}{d_{v_t}}+s\right).$

Since $\frac{d_{v_1}+d_{v_2}+\cdots +d_{v_t}+1\times s}{s+t}\ge \sqrt[s+t]{d_{v_1}{d}_{v_2}\cdots d_{v_t}\times 1^s}\ge \frac{s+t}{\frac{1}{d_{v_1}}+\frac{1}{d_{v_2}}+\cdots +\frac{1}{d_{v_t}}+\frac{1}{1}\times s}$, we have $\frac{1}{s+t}\left(\frac{1}{d_{v_1}}+\frac{1}{d_{v_2}}+\cdots +\frac{1}{d_{v_t}}+s\right)\ge \frac{s+t}{d_{v_1}+d_{v_2}+\cdots +d_{v_t}+s}$. Therefore, $\underset{\mu \in {\mathcal{T}}_k^{+-}\left(T\right)}{{\displaystyle \sum }}\frac{1}{d\left(\mu \right)}\ge \underset{\mu \in {\mathcal{T}}_{k,u}^{+-}\left(T\left[v_i\to u;4\right]\right)}{{\displaystyle \sum }}\frac{1}{d\left(\mu \right)}$. Thus, we have ${\pi }_k\left(T\right)>{\pi }_k\left(T\left[v_i\to u;4\right]\right)$, and hence $\pi \left(T\right)>\pi \left(T\left[v_i\to u;4\right]\right)$. □

Theorem 10. For any $n$-vertex caterpillar tree $T$, there exists an $n$-vertex caterpillar tree $C_n$ such that $\pi \left(C_n\right)\le \pi \left(T\right)$.

Proof. Since $T$ is a caterpillar tree, by Lemma 6, $T$ must contain a $Y$-structure. By applying Graph Transformation 1, we can reduce the number of $Y$-structures in $T$. If $T$ still contains a $Y$-structure, we can apply Graph Transformation 1 again to transform $T$ into a graph $C_n$ without any $Y$-structure. By Lemma 6, $C_n$ is a caterpillar tree, and $\pi \left(C_n\right)\le \pi \left(T\right)$. □

In this paper, we studied the Laplacian permanents and the Laplacian ratios of trees. We characterized some Laplacian permanents of trees. We also obtained a graph transformation related to the Laplacian ratio. In future research, we will further study the Laplacian permanents and the Laplacian ratios of graphs.