[Bug]: 多行公式和单行公式与上下文的间距不一样

[bottomCaiji]

发生了什么？

如题，3.8.3版本单行公式与上下文的间距非常宽，多行公式反倒很窄。
另外，多行公式的公式内行间距也很窄，希望能再加宽一些。

你期望的结果

3.7.1版本的效果就很理想。

问题复现的步骤（或者截图）

满足如下式所示之关系
\begin{equation}
	\boldsymbol{\mathcal{A}}_3^T \boldsymbol{\mathcal{P}}_3 \boldsymbol{\mathcal{A}}_3 - \boldsymbol{\mathcal{P}}_3 = -\boldsymbol{I}_2
\end{equation}
并将李雅普诺夫函数定义为
\begin{equation}
	V_3(k) = \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) 
\end{equation}
则该李雅普诺夫函数相邻两时间步的差分为
\begin{equation}
	\renewcommand\arraystretch{1.5}
	\begin{aligned}
		\Delta V_3(k+1) &= V_3(k+1) - V_3(k) \\
		&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
		&= \Delta \boldsymbol{\xi}_3^T (k) \left( \boldsymbol{\mathcal{A}}_3^T \boldsymbol{\mathcal{P}}_3 \boldsymbol{\mathcal{A}}_3 - \boldsymbol{\mathcal{P}}_3 \right) \Delta \boldsymbol{\xi}_3 (k) +
		2 \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{A}}_3^T \boldsymbol{\mathcal{P}}_3 \boldsymbol{\beta}_3 \left( u_3^{\rm cl} + \Delta d_a (k) \right) \\ 
		&\leq -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 2 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \Vert \boldsymbol{\mathcal{A}}_3 \boldsymbol{\mathcal{P}}_3 \boldsymbol{\beta}_3 \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right) \\
		&= -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 0.638 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right)
	\end{aligned}
\end{equation}
其中，

BIThesis 模板

硕博·学位论文 — graduate-thesis

操作系统或平台

Windows

LaTeX 发行版以及发行年份

TeX Live 2023

日志信息（可选）

[YDX-2147483647]

感谢反馈。我复现了你描述的现象。

完整测试用例

\documentclass[type=master]{bithesis}

\usepackage{bm}

\begin{document}

满足如下式所示之关系
\begin{equation}
	\boldsymbol{\mathcal{A}}_3^T \boldsymbol{\mathcal{P}}_3 \boldsymbol{\mathcal{A}}_3 - \boldsymbol{\mathcal{P}}_3 = -\boldsymbol{I}_2
\end{equation}
并将李雅普诺夫函数定义为
\begin{equation}
	V_3(k) = \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) 
\end{equation}
则该李雅普诺夫函数相邻两时间步的差分为
\begin{equation}
	\renewcommand\arraystretch{1.5}
	\begin{aligned}
		\Delta V_3(k+1) &= V_3(k+1) - V_3(k) \\
		&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
		&= \Delta \boldsymbol{\xi}_3^T (k) \left( \boldsymbol{\mathcal{A}}_3^T \boldsymbol{\mathcal{P}}_3 \boldsymbol{\mathcal{A}}_3 - \boldsymbol{\mathcal{P}}_3 \right) \Delta \boldsymbol{\xi}_3 (k) +
		2 \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{A}}_3^T \boldsymbol{\mathcal{P}}_3 \boldsymbol{\beta}_3 \left( u_3^{\rm cl} + \Delta d_a (k) \right) \\ 
		&\leq -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 2 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \Vert \boldsymbol{\mathcal{A}}_3 \boldsymbol{\mathcal{P}}_3 \boldsymbol{\beta}_3 \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right) \\
		&= -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 0.638 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right)
	\end{aligned}
\end{equation}
其中，

\end{document}

二分法结果：

• 单行公式间距：v3.7.1 = v3.7.4 < v3.7.5 = v3.8.4.
• 多行公式间距：v3.7.1 = v3.7.4 < v3.7.5 = v3.7.9 > v3.7.10-alpha-1 = v3.8.4.

截图

diffpdf main-v3.7.4.pdf main-v3.7.5.pdf

diffpdf main-v3.7.9.pdf main-v3.7.10-alpha-1.pdf

BIThesis> make cls; mv bithesis.cls $"tmp/reg/bithesis.cls-(git describe --tags)"

BIThesis\tmp\reg> gh release download -R BITNP/BIThesis -p bithesis.cls $TAG -O bithesis.cls-($TAG)

BIThesis\tmp\reg> let TAG = git describe --tags
rm bithesis.cls -p; ln -s bithesis.cls-($TAG) bithesis.cls; xelatex main; mv main.pdf $"main-($TAG).pdf"
diffpdf main-v3.7.9.pdf main-($TAG).pdf

• v3.7.4 / v3.7.5 是因为 39c27df chore: Make the skips for math env bigger. #472 调整公式与上下文间距的默认值（适当增加一些） #463
• v3.7.9 / v3.7.10-alpha-1 是因为 bcae36c feat: 改小数学行距至西文水平 #537 公式行距应比中文小 #500

[YDX-2147483647]

\renewcommand\arraystretch{1.5}

@bottomCaiji 为什么专门这样设置呢？它应该只影响矩阵，实测对你的多行公式完全没有影响啊。

[bottomCaiji]

\renewcommand\arraystretch{1.5}

@bottomCaiji 为什么专门这样设置呢？它应该只影响矩阵，实测对你的多行公式完全没有影响啊。

试了一下没有效果，就忘记删了……

[YDX-2147483647]

好。如果你只想恢复到 v3.7.1 那样，可以先像下面这样设置。

\BITSetup{
    style = {
        mathBelowDisplaySkip = 3pt,
        mathAboveDisplaySkip = 3pt,
    }
}

\SetMathEnvironmentSinglespace{1.3}  % 用 zhlineskip 增大多行公式行距

[bottomCaiji]

好。如果你只想恢复到 v3.7.1 那样，可以先像下面这样设置。

\BITSetup{
style = {
mathBelowDisplaySkip = 3pt,
mathAboveDisplaySkip = 3pt,
}
}

\SetMathEnvironmentSinglespace{1.3} % 用 zhlineskip 增大多行公式行距

ok，非常感谢

[Aaron-Gp]

从这张图可以明显看出来，zhls 设置的只是文本和公式的行间距，并不改变公式在文本间的显示逻辑，其实就是把公式整体当做一行，然后在两倍行间距内居中摆放公式

\documentclass{ctexart}
\usepackage[bodytextleadingratio=5,restoremathleading=true]{zhlineskip}
\usepackage{amsmath}
\usepackage{lua-visual-debug}

\begin{document}

一般书强调“克氏符不是张量”是对的，本书则应强调
\begin{equation}
  [u, v](f) = u(v(f)) - v(u(f)),
\end{equation}
“克氏符是坐标系依赖的张量”。
\begin{equation}
  \begin{aligned}
    \Delta V_3(k+1) &= V_3(k+1) - V_3(k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&\leq -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 2 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \Vert \boldsymbol{\mathcal{A}}_3 \boldsymbol{\mathcal{P}}_3 \boldsymbol{\beta}_3 \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right) \\
	&= -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 0.638 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right)
  \end{aligned}
\end{equation}
读者常问：你们为什么非把克氏符说成张量不可？

\end{document}

[YDX-2147483647]

在两倍行间距内

我不完全认同。如果公式用两倍行间距排不下，应该存在某个阈值决定到底占几行。例如下面这份文档，两页的多行公式行数不同，但上方间距一样。

\documentclass{ctexart}
\usepackage[bodytextleadingratio=5,restoremathleading=true]{zhlineskip}
\usepackage{amsmath}
\usepackage{lua-visual-debug}

\begin{document}

一般书强调“克氏符不是张量”是对的，本书则应强调
\begin{equation}
  [u, v](f) = u(v(f)) - v(u(f)),
\end{equation}
“克氏符是坐标系依赖的张量”。
\begin{equation}
  \begin{aligned}
    \Delta V_3(k+1) &= V_3(k+1) - V_3(k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&\leq -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 2 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \Vert \boldsymbol{\mathcal{A}}_3 \boldsymbol{\mathcal{P}}_3 \boldsymbol{\beta}_3 \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right) \\
	&= -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 0.638 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right)
  \end{aligned}
\end{equation}
读者常问：你们为什么非把克氏符说成张量不可？

\clearpage

一般书强调“克氏符不是张量”是对的，本书则应强调
\begin{equation}
  [u, v](f) = u(v(f)) - v(u(f)),
\end{equation}
“克氏符是坐标系依赖的张量”。
\begin{equation}
  \begin{aligned}
    \Delta V_3(k+1) &= V_3(k+1) - V_3(k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&= \Delta \boldsymbol{\xi}_3^T (k+1) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k+1) - \Delta \boldsymbol{\xi}_3^T (k) \boldsymbol{\mathcal{P}}_3 \Delta \boldsymbol{\xi}_3 (k) \\
	&\leq -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 2 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \Vert \boldsymbol{\mathcal{A}}_3 \boldsymbol{\mathcal{P}}_3 \boldsymbol{\beta}_3 \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right) \\
	&= -\Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2^2 + 0.638 \Vert \Delta \boldsymbol{\xi}_3 (k) \Vert_2 \left( \vert u_3^{\rm cl} \vert + \bar{\varepsilon}_4 \right)
  \end{aligned}
\end{equation}
读者常问：你们为什么非把克氏符说成张量不可？

\end{document}

[Aaron-Gp]

CTeX-org/ctex-kit#735 (comment)

两行内居中排版，超过两行时与上下文间隔一个 skip ？

[Aaron-Gp]

所以应该只需要调整 skip 就可以了? 看起来像是 (2*行间距-单行文本大小-单行公式大小）/ 2

自上至下分别是单行公式，split，aligned。(这张图我是手调的，但是应该是这个效果吧。）