方程环境、方程编号括号均为右手系

Question

经过一些更改后，我能够成功编译thesis.tex发行版中包含的文件（稍后我会列出）。

关于您的问题，我发现方程式数字中的所有括号都没有问题。

以防万一，我已将样本纸的样本包含在amsmath包裹中。文件名：\testmath.tex检查其他分隔符。未发现任何问题。

变化

（1）字体。我没有字体David CLM，但我发现DavidCLM字体已经在我的系统目录中。（可能是 David Libre 以前项目中的。请参阅在 LuaLaTeX 或 XeLaTeX 中我可以用哪些希伯来字体替代 David？）

因此我注释了第 133 行iitthesis.cls并 \newfontfamily在序言中定义了指向正确位置的。

（2）在文件中， my-thesis-specific.tex我将改为\renewcommand\lips@dolipsum{ ，以\newcommand\lips@dolipsum{消除 ! LaTeX Error: Command \lips@dolipsum undefined。

笔记从官方网站获取的完整套件https://github.com/eyalroz/technion-iit-thesis/而 Overleaf 没有这个问题。（2020 年 3 月 7 日修复）

（3）我删除了一条错误信息（来自倒数第二段之前的段落\abstractHebrew{）注释\include{front/abstract}。其他希伯来语文本排版正常。见最后一张图

等式 3.1

这是新的thesis.tex

\documentclass[PhD,noabbrevs]{misc/iitthesis}

\include{misc/thesis-fields}
\include{front/personal-acks}
 %\include{front/abstract}
\include{front/abbrevs}

\usepackage{misc/iitthesis-extra}

\include{misc/my-general}
\include{misc/my-thesis-specific}

\usepackage{multibib}
\newcites{pubinfo}{Acknowledgement page references}
\def\iitthesisextramultibibdefs{}
        
\usepackage{amsmath} % added to test eq. samples <<<<<<<<<<<<<<<<<<<<<<<<<<<<

    %\newfontfamily\hebrewfont[Script=Hebrew]{David CLM} % changed: line 133 of iitthesis.cls <<<<<<<<<<<
    \newfontfamily\hebrewfont[% added <<<<<<<<<<<<<<<<<<<<<
    Path =C:/Windows/Fonts/,%
    Script=Hebrew,%
    Language = Hebrew ,
    Extension =.ttf,%
    UprightFont =*-Medium,%
    BoldFont =*-Bold,%
    ItalicFont =*-MediumItalic,%
    BoldItalicFont =*-BoldItalic%
    ]{DavidCLM} 

\begin{document}
    

\makefrontmatter
\include{main/intro} 
\include{main/prelims}
\include{main/mainchap1}

\include{main/short_testmath} % added <<<<<<<<<<<<<<<

\include{main/conclusion}
\appendix
\include{main/appendix1}
\makebackmatter 
\end{document}

这是文件short_testmath.tex

%% file short_testmath.tex

\newcommand{\wt}{\widetilde}
\newcommand{\wh}{\widehat}

\section{Introduction}

This paper contains examples of various features from \AmS-\LaTeX{}.

Sample Paper for the \textbf{amsmath} Package.

 File name: \textbf{testmath.tex}

\section{Enumeration of Hamiltonian paths in a graph}

Let $\mathbf{A}=(a_{ij})$ be the adjacency matrix of graph $G$. The
corresponding Kirchhoff matrix $\mathbf{K}=(k_{ij})$ is obtained from
$\mathbf{A}$ by replacing in $-\mathbf{A}$ each diagonal entry by the
degree of its corresponding vertex; i.e., the $i$th diagonal entry is
identified with the degree of the $i$th vertex. It is well known that
\begin{equation}
\det\mathbf{K}(i|i)=\text{ the number of spanning trees of $G$},
\quad i=1,\dots,n
\end{equation}


Let $C_{i(j)}$ be the set of graphs obtained from $G$ by attaching edge
$(v_iv_j)$ to each spanning tree of $G$. Denote by $C_i=\bigcup_j
C_{i(j)}$. It is obvious that the collection of Hamiltonian cycles is a
subset of $C_i$. Note that the cardinality of $C_i$ is $k_{ii}\det
\mathbf{K}(i|i)$. Let $\wh X=\{\hat x_1,\dots,\hat x_n\}$.


\begin{equation}\label{H-cycles}
\biggl(\prod^n_{\,j=1}\hat x_j\biggr)H_c=\frac{1}{2}\hat k_{ij}\det
\wh{\mathbf{K}}(i|i),\qquad i=1,\dots,n.
\end{equation}

Then, it follows that
%\begin{lem}\label{lem-det}
\begin{equation}\label{detprod}
\prod_{i\in\mathbf{n}}
\biggl(\sum_{\,j\in\mathbf{n}}b_{ij}\hat y_j\biggr)
=\biggl(\prod_{\,i\in\mathbf{n}}\hat y_i\biggr)\det\mathbf{B}.
\end{equation}


\begin{equation}\label{detB}
\det\mathbf{B}=
\sum^n_{l =0}\sum_{I_l \subseteq n}
\prod_{i\in I_l}(b_{ii}-\lambda_i)
\det\mathbf{B}^{(\lambda)}(I_l |I_l ),
\end{equation}


\begin{equation}
\mathbf{K}(t,t_1,\dots,t_n)
=\begin{pmatrix} D_1t&-a_{12}t_2&\dots&-a_{1n}t_n\\
-a_{21}t_1&D_2t&\dots&-a_{2n}t_n\\
\hdotsfor[2]{4}\\
-a_{n1}t_1&-a_{n2}t_2&\dots&D_nt\end{pmatrix},
\end{equation}

\begin{equation}\label{j:mark}
\begin{split}
H_c=&
\frac{n_1!\,n_2!\,n_3!}
{n_1+n_2+n_3}\sum_i\left[\binom{n_1}{i}
\binom{n_2}{n_3-n_1+i}\binom{n_3}{n_3-n_2+i}\right.\\
&+\left.\binom{n_1-1}{i}
\binom{n_2-1}{n_3-n_1+i}
\binom{n_3-1}{n_3-n_2+i}\right].\end{split}
\end{equation}

Answer 1