☰

如何对齐并居中独立的 amsmath 方程式？

tag-icon • tag-icon horizontal-alignment formatting align amsmath standalone
如何对齐并居中独立的 amsmath 方程式？

我让这段代码生成以下文档。我想对齐几件事。
特别是在“度数序列”标题下：

\documentclass[varwidth,margin=2mm]{standalone}
\usepackage{amsmath,tikz}
\begin{document}
    \begin{center}
        \begin{tikzpicture}
            %%%%% SNIP %%%%%
        \end{tikzpicture}\\
        \rule[0cm]{200pt}{0.5pt}\\
        \begin{math}
            $degree sequence$\\
            $e=\{2,2,2,2,2\}\ \ \ \ \ \ \ \ \ \ \ \ c=\{2,2,2,2,2\}$\\
            \delta(e)=2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \delta(c)=2\\
            \Delta(e)=2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Delta(c)=2\\
            \rule[0cm]{200pt}{0.5pt}\\
            \alpha(e_{1})=c_{1},\ \ \alpha(e_{2})=c_{3},\\
            \alpha(e_{3})=c_{5},\ \ \alpha(e_{4})=c_{2},\\
            \alpha(e_{5})=c_{4},\ \ \alpha(e_{1})=c_{1},\\
            \rule[0cm]{200pt}{0.5pt}\\
            $e_{1}\to{}e_{2}\to{} e_{3}\to{}e_{4}\to{}
            e_{5}\to{}e_{1}\\
            $c_{1}\to{}c_{3}\to{} c_{5}\to{}c_{2}\to{}
            c_{4}\to{}c_{1}\\
            \rule[0cm]{200pt}{0.5pt}\\
        \end{math}    
    \end{center}
\end{document}

在此处输入图片描述

因此我尝试使用 align 包来对齐它，但是它却让所有东西都变得乱糟糟的：

\documentclass[varwidth,margin=2mm]{standalone}
\usepackage{amsmath,tikz}
\begin{document}
    \begin{center}
        \begin{tikzpicture}
            %%%%% SNIP %%%%%
        \end{tikzpicture}\\
        \rule[0cm]{200pt}{0.5pt}\\
            degree sequence\\
            e = \{2,2,2,2,2,2\} c = \{2,2,2,2,2,2\}\\
        \begin{align*}
            \delta(e) & = 2 & \delta(c) & = 2\\
            \Delta(e) & = 2 & \Delta(c) & = 2\\
        \end{align*}
        \begin{math}
            \rule[0cm]{200pt}{0.5pt}\\
            \alpha(e_{1})=c_{1},\ \ \alpha(e_{2})=c_{3},\\
            \alpha(e_{3})=c_{5},\ \ \alpha(e_{4})=c_{2},\\
            \alpha(e_{5})=c_{4},\ \ \alpha(e_{1})=c_{1},\\
            \rule[0cm]{200pt}{0.5pt}\\
            $e_{1}\to{}e_{2}\to{} e_{3}\to{}e_{4}\to{}
            e_{5}\to{}e_{1}\\
            $c_{1}\to{}c_{3}\to{} c_{5}\to{}c_{2}\to{}
            c_{4}\to{}c_{1}\\
            \rule[0cm]{200pt}{0.5pt}\\
        \end{math}    
    \end{center}
\end{document}

在此处输入图片描述

有没有办法解决这个问题，还是我只能这样了。我知道这只是一件小事，但如果能有更多的控制权就好了。

答案1

您想要使用tabular：

\documentclass[margin=6]{standalone}
\usepackage{amsmath,tikz,booktabs}

\begin{document}

\begin{tabular}{cc}
\begin{tikzpicture}
\draw (0,0)--(1,1);
\end{tikzpicture}
&
\begin{tikzpicture}
\draw (0,0)--(1,1);
\end{tikzpicture}
\\
\midrule
\multicolumn{2}{c}{degree sequence}\\
$\begin{aligned}
& e=\{2,2,2,2,2\} \\
& \delta(e)=2 \\
& \Delta(e)=2 \\
\end{aligned}$ &
$\begin{aligned}
& c=\{2,2,2,2,2\} \\
& \delta(c)=2 \\
& \Delta(c)=2
\end{aligned}$ \\
\midrule
\multicolumn{2}{c}{%
  $\begin{aligned}
  \alpha(e_{1})&=c_{1}, & \alpha(e_{2})&=c_{3},\\
  \alpha(e_{3})&=c_{5}, & \alpha(e_{4})&=c_{2},\\
  \alpha(e_{5})&=c_{4}, & \alpha(e_{1})&=c_{1},
  \end{aligned}$%
} \\
\midrule
\multicolumn{2}{c}{%
  $\begin{array}{@{} c *{5}{@{} >{{}}c<{{}} @{} c @{}} }
  e_{1} & \to & e_{2} & \to & e_{3} & \to & e_{4} & \to & e_{5} & \to & e_{1} \\
  c_{1} & \to & c_{3} & \to & c_{5} & \to & c_{2} & \to & c_{4} & \to & c_{1}
  \end{array}$%
} \\
\bottomrule[\lightrulewidth]
\end{tabular}

\end{document}

在此处输入图片描述

对“度数序列”块进行不同的对齐：

\documentclass[margin=6]{standalone}
\usepackage{amsmath,tikz,booktabs,array}

\begin{document}

\begin{tabular}{cc}
\begin{tikzpicture}
\draw (0,0)--(1,1);
\end{tikzpicture}
&
\begin{tikzpicture}
\draw (0,0)--(1,1);
\end{tikzpicture}
\\
\midrule
\multicolumn{2}{c}{degree sequence}\\
$\begin{gathered}
  e=\{2,2,2,2,2\} \\
  \begin{aligned}
  \delta(e)&=2 \\
  \Delta(e)&=2
  \end{aligned}
\end{gathered}$ &
$\begin{gathered}
  c=\{2,2,2,2,2\} \\
  \begin{aligned}
  \delta(c)&=2 \\
  \Delta(c)&=2
  \end{aligned}
\end{gathered}$ \\
\midrule
\multicolumn{2}{c}{%
  $\begin{aligned}
  \alpha(e_{1})&=c_{1}, & \alpha(e_{2})&=c_{3},\\
  \alpha(e_{3})&=c_{5}, & \alpha(e_{4})&=c_{2},\\
  \alpha(e_{5})&=c_{4}, & \alpha(e_{1})&=c_{1},
  \end{aligned}$%
} \\
\midrule
\multicolumn{2}{c}{%
  $\begin{array}{@{} c *{5}{@{} >{{}}c<{{}} @{} c @{}} }
  e_{1} & \to & e_{2} & \to & e_{3} & \to & e_{4} & \to & e_{5} & \to & e_{1} \\
  c_{1} & \to & c_{3} & \to & c_{5} & \to & c_{2} & \to & c_{4} & \to & c_{1}
  \end{array}$%
} \\
\bottomrule[\lightrulewidth]
\end{tabular}

\end{document}

在此处输入图片描述

答案2

align并alignat允许您对齐。您可以将东西放在与图形宽度相同的节点中。

\documentclass[tikz,margin=2mm]{standalone}
\usetikzlibrary{positioning,calc}
\usepackage{amsmath}
\begin{document}
\begin{tikzpicture}[Bullet/.style={circle,draw,fill=black,scale=0.75}]
\begin{scope}[local bounding box=top]
\begin{scope}[local bounding box=left]
 \node[Bullet,label=left :{$e_1$}] (E1) at (0,2) {} ;
 \node[Bullet,label=above:{$e_2$}] (E2) at (1,3) {} ;
 \node[Bullet,label=right:{$e_3$}] (E3) at (2,2) {} ;
 \node[Bullet,label=right:{$e_4$}] (E4) at (2,0) {} ;
 \node[Bullet,label=left :{$e_5$}] (E5) at (0,0) {} ;
 \draw[thick] (E1)--(E2)--(E3)--(E4)--(E5)--(E1) {} ;
\end{scope}
 \begin{scope}[local bounding box=right,xshift=4cm]
 \node[Bullet,label=left :{$c_1$}] (C1) at (0,2) {} ;
 \node[Bullet,label=above:{$c_2$}] (C2) at (1,3) {} ;
 \node[Bullet,label=right:{$c_3$}] (C3) at (2,2) {} ;
 \node[Bullet,label=right:{$c_4$}] (C4) at (2,0) {} ;
 \node[Bullet,label=left :{$c_5$}] (C5) at (0,0) {} ;
 \draw[thick] (C1)--(C3)--(C5)--(C2)--(C4)--(C1) {} ;
\end{scope}
\end{scope}
\path let \p1=($(top.east)-(top.west)$) in
node[below=of top,align=center,text width=\x1]{
\rule[0cm]{\x1}{0.5pt}
degree sequence
\begin{alignat*}{2}
 e&=\{2,2,2,2,2\}& c&=\{2,2,2,2,2\}\\
 \delta(e)&=2 &\delta(c)&=2\\
 \Delta(e)&=2&\Delta(c)&=2
\end{alignat*}
\rule[0cm]{\x1}{0.5pt}
\begin{alignat*}{2}
 \alpha(e_{1})&=c_{1},\quad&\alpha(e_{2})&=c_{3},\\
 \alpha(e_{3})&=c_{5},&\alpha(e_{4})&=c_{2},\\
 \alpha(e_{5})&=c_{4},&\alpha(e_{1})&=c_{1},
\end{alignat*}
\rule[0cm]{\x1}{0.5pt}
\begin{align*}
e_{1}&\to{}e_{2}\to{} e_{3}\to{}e_{4}\to{}
            e_{5}\to{}e_{1}\\
            c_{1}&\to{}c_{3}\to{} c_{5}\to{}c_{2}\to{}
            c_{4}\to{}c_{1}
\end{align*}
\rule[0cm]{\x1}{0.5pt}};
\end{tikzpicture}
\end{document}

在此处输入图片描述

答案3

使用嵌套array：

\documentclass[margin=2mm]{standalone}
\usepackage{amsmath,animate}
\usepackage{graphicx}
\usepackage{booktabs}
\usepackage{tikz}

\begin{document}
$
\begin{array}{@{}c @{}}
\begin{array}{cc}
\tikz\node[draw]{\includegraphics[width=31mm]{example-image-duck}};
                    & \tikz\node[draw]{\includegraphics[width=31mm]{example-image-duck}};\\
   \midrule
\multicolumn{2}{c}{$degree sequence$}               \\[1ex]
e =\{2,2,2,2,2\}    & c =\{2,2,2,2,2\}              \\
\delta(e)=2         & \delta(c)=2                   \\
\Delta(e)=2         & \Delta(c)=2                   \\
        \end{array}                                 \\    
    \midrule
\alpha(e_{1})=c_{1},\quad   \alpha(e_{2})=c_{3},    \\
\alpha(e_{3})=c_{5},\quad   \alpha(e_{4})=c_{2},    \\
\alpha(e_{5})=c_{4},\quad   \alpha(e_{1})=c_{1},    \\
    \midrule
e_{1}\to e_{2}\to e_{3}\to e_{4}\to e_{5}\to e_{1}  \\
c_{1}\to c_{3}\to c_{5}\to c_{2}\to c_{4}\to c_{1}  \\
    \midrule
\end{array}
$
\end{document}

相反，tikzpicture我使用您的代码示例图像。

在此处输入图片描述

相关内容