将两组方程对齐并拟合到

Question

我在使用投影幻灯片时遇到了问题。我尝试解决过几个问题。

1. 如何正确缩放方程式
2. 如何将四个限制对齐到同一高度
3. 如何写两条带有变量的垂直线，如附图所示。

感谢您的任何想法！

\documentclass{beamer}

\usetheme{CambridgeUS}

\begin{document}

\begin{frame}{Primal-dual}
    


\begin{columns}[T]
        \begin{column}{.5\textwidth}
            \small
            \[
                    \text{Min} \sum_{i=1}^{m+1}\sum_{j=1}^{n+1}c_{ij}x_{ij}
                \]
                \[\scalebox{0.8}{% 
                $
                   \hskip4.5em \text{s.a.} \quad
                    \begin{aligned}[t]
                            &\sum_{j=1}^{n}x_{ij}-x_{i,n+1}= a_i  \qquad \forall i=1,\hdots,m\\
                            & \textcolor{blue}{\sum_{j=1}^{n}}x_{m+1,j}+\textcolor{magenta}{x_{m+1,n+1}}=a_{m+1}\\
                            &\sum_{i=1}^{m}x_{ij}+x_{m+1,j}= b_j  \qquad \forall j=1,\hdots,n\\
                            &-\textcolor{red}{\sum_{i=1}^{m}}x_{i,n+1}+\textcolor{magenta}{x_{m+1,n+1}}=b_{n+1}\\
                        &x_{ij}\ge 0 \qquad 
                        \left\{
                              \begin{aligned}
                                  \forall i=&1,\hdots,m\\     \forall j=&1,\hdots,n\\
                              \end{aligned}
                         \right.
                         \end{aligned}
                $}
                \]
            \normalsize
        \end{column}
        \hfill%
        \begin{column}{.5\textwidth}
            \small
             \[
                    \text{Max} \sum_{i=1}^{m+1}u_i a_i+\sum_{j=1}^{n+1}v_j b_j
                \]
                \[\scalebox{0.8}{% 
                $
                   \hskip5em \text{s.a.} \quad
                    \begin{aligned}[t]
                            & u_i+v_j \le c_{ij}\\
                            & \\
                            & -u_i-v_{n+1} \le c_{i,n+1}\\
                             & \\
                           & u_{m+1}+v_j \le c_{m+1,j}\\
                            & \\
                            & u_{m+1}+v_{n+1} \le c_{{m+1},{n+1}}\\
                  \end{aligned}
                $}
                \]
            \normalsize
    \end{column}
\end{columns}
  
\end{frame}
  
\end{document}

Answer 1

这个想法与 Werner 的使用想法相同array，但有一些小细节（以及您在答案中展示的颜色）。

\documentclass{beamer}
\usetheme{CambridgeUS}

\usepackage{mathtools,array}

\DeclareMathOperator{\Min}{Min}
\DeclareMathOperator{\Max}{Max}
\newcommand{\nominus}{\mathllap{-}}

\begin{document}

\begin{frame}{Primal-dual}

\scriptsize
\begin{equation*}
\renewcommand{\arraystretch}{2}
\begin{array}{
  @{}
  >{$}c<{$} % s.a.
  >{\displaystyle}l % main column 1
  |@{\,} % vertical line
  >{\displaystyle}c % middle column
  @{\,}| % vertical line
  >{$}c<{$} % s.a.
  >{\displaystyle}l % main column 2
  @{}
}
\multicolumn{2}{@{}>{\displaystyle}l}{
  \! \Min \sum_{i=1}^{m+1} \sum_{j=1}^{n+1}c_{ij}x_{ij}
}
&
\multicolumn{1}{c}{} % no vertical rule here
&
\multicolumn{2}{>{\displaystyle}l}{
  \! \Max \sum_{i=1}^{m+1}u_i a_i+\sum_{j=1}^{n+1}v_j b_j
}
\\
% first row
s.a. &
\sum_{j=1}^{n}x_{ij}-x_{i,n+1}= a_i  \qquad\hfill \forall i=1,\hdots,m &
u_i &
s.a. &
u_i+v_j \le c_{ij}
\\
% second row
&
\mathop{\textcolor{blue}{\sum_{j=1}^{n}}}x_{m+1,j}+\textcolor{magenta}{x_{m+1,n+1}}=a_{m+1} &
u_{m+1} &
&
\nominus u_i-v_{n+1} \le c_{i,n+1}
\\
% third row
&
\sum_{i=1}^{m}x_{ij}+x_{m+1,j}= b_j  \qquad\hfill \forall j=1,\hdots,n &
v_j &
&
u_{m+1}+v_j \le c_{m+1,j}
\\
% fourth row
&
\nominus\!\mathop{\textcolor{red}{\sum_{i=1}^{m}}}x_{i,n+1}
  +\textcolor{magenta}{x_{m+1,n+1}}=b_{n+1} &
v_{n+1} &
&
u_{m+1}+v_{n+1} \le c_{{m+1},{n+1}}
\\
% fifth row
&
\multicolumn{1}{l}{
x_{ij}\ge 0 \qquad 
  \left\{
    \begin{aligned}
    \forall i&=1,\dots,m\\
    \forall j&=1,\dots,n
    \end{aligned}
  \right.
}
\end{array}
\end{equation*}

\end{frame}
  
\end{document}

\nominus左侧附加一个减号，但我们需要修复彩色总和旁边的间距。这些命令\multicolumn会覆盖默认对齐。

我们\qquad\hfill确保“所有”符号前的间距，以及条件的右对齐。

请注意，\scriptsize必须要走出equation*环境。

Answer 2

由于有许多对齐和垂直规则，我使用以下方式设置array：

\documentclass{beamer}

\usepackage{amsmath,array,graphicx}

\begin{document}

\begin{frame}

\resizebox{\linewidth}{!}{$
  \renewcommand{\arraystretch}{2}
  \begin{array}{ r >{\displaystyle}l | c | r @{} >{{}}l }
    & \multicolumn{1}{c}{\displaystyle
      \min \sum_{i = 1}^{m + 1} \sum_{j = 1}^{n + 1} c_{ij} x_{ij}
    } & \multicolumn{1}{c}{
      \renewcommand{\arraystretch}{1}\begin{tabular}{@{} c @{}}
      variable \\ dual
    \end{tabular}} & \multicolumn{2}{c}{\displaystyle
      \max \sum_{i = 1}^{m + 1} u_i a_i + \sum_{j = 1}^{n + 1} v_j b_j
    } \\
    \text{s.t.} &
      \sum_{j = 1}^n x_{ij} - x_{i, n + 1} = a \quad \forall i = 1, \dotsc, m &
        u_i &
        u_i + v_j &\leq c_{ij} \\
    & \sum_{j = 1}^n x_{m + 1, j} + x_{m + 1, n + 1} = a_{m + 1} &
        u_{m + 1} &
        -u_i - v_{n + 1} &\leq c_{i, n + 1} \\
    & \sum_{i = 1}^m x_{ij} + x_{m + 1, j} = b_j \quad \forall j = 1, \dotsc, n &
        v_j &
        u_{m + 1} + v_j &\leq c_{m + 1, j} \\
    & -\sum_{i = 1}^m x_{i, n + 1} + x_{m + 1, n + 1} = b_{n + 1} &
        v_{n + 1} &
        u_{m + 1} + v_{n + 1} &\leq c_{m + 1, n + 1} \\
    & \multicolumn{2}{l}{
      \renewcommand{\arraystretch}{1.5}
        x_{ij} \geq 0 \quad \left\{ \begin{array}{@{} l @{}}
          \forall i = 1, \dotsc, m \\
          \forall j = 1, \dotsc, n
        \end{array}\right.
      }
  \end{array}
$}

\end{frame}

\end{document}

如果您希望对左侧列进行一些额外的对齐，可以选择以下选项：

\documentclass{beamer}

\usepackage{amsmath,array,graphicx}

\begin{document}

\begin{frame}

\resizebox{\linewidth}{!}{$
  \renewcommand{\arraystretch}{2}
  \begin{array}{ r >{\displaystyle}r @{} >{{}}l | c | r @{} >{{}}l }
    & \multicolumn{2}{c}{\displaystyle
      \min \sum_{i = 1}^{m + 1} \sum_{j = 1}^{n + 1} c_{ij} x_{ij}
    } & \multicolumn{1}{c}{
      \renewcommand{\arraystretch}{1}\begin{tabular}{@{} c @{}}
      variable \\ dual
    \end{tabular}} & \multicolumn{2}{c}{\displaystyle
      \max \sum_{i = 1}^{m + 1} u_i a_i + \sum_{j = 1}^{n + 1} v_j b_j
    } \\
    \text{s.t.} &
      \sum_{j = 1}^n x_{ij} - x_{i, n + 1} &= a \quad \forall i = 1, \dotsc, m &
        u_i &
        u_i + v_j &\leq c_{ij} \\
    & \sum_{j = 1}^n x_{m + 1, j} + x_{m + 1, n + 1} &= a_{m + 1} &
        u_{m + 1} &
        -u_i - v_{n + 1} &\leq c_{i, n + 1} \\
    & \sum_{i = 1}^m x_{ij} + x_{m + 1, j} &= b_j \quad \forall j = 1, \dotsc, n &
        v_j &
        u_{m + 1} + v_j &\leq c_{m + 1, j} \\
    & -\sum_{i = 1}^m x_{i, n + 1} + x_{m + 1, n + 1} &= b_{n + 1} &
        v_{n + 1} &
        u_{m + 1} + v_{n + 1} &\leq c_{m + 1, n + 1} \\
    & x_{ij} &\multicolumn{1}{@{} l @{}}{
      \renewcommand{\arraystretch}{1.5}
      {}\geq 0 \quad \left\{ \begin{array}{@{} l @{}}
        \forall i = 1, \dotsc, m \\
        \forall j = 1, \dotsc, n
      \end{array}\right.
    }
  \end{array}
$}

\end{frame}

\end{document}

Answer 3

我对自己的问题找到了一个新的解决方案，其结果几乎令人满意。

\documentclass{beamer}

\usetheme{CambridgeUS}

\begin{document}

\begin{frame}{Primal-dual}
    
\begin{equation*}
\scriptsize
\begin{aligned}[c]
        \text{Min} & \sum_{i=1}^{m+1}\sum_{j=1}^{n+1}c_{ij}x_{ij}\\
        \text{s.a.} \quad
        &\sum_{j=1}^{n}x_{ij}-x_{i,n+1}= a_i  \qquad \forall i=1,\hdots,m\\
        & \textcolor{blue}{\sum_{j=1}^{n}}x_{m+1,j}+\textcolor{magenta}{x_{m+1,n+1}}=a_{m+1}\\
        &\sum_{i=1}^{m}x_{ij}+x_{m+1,j}= b_j  \qquad \forall j=1,\hdots,n\\
         -&\textcolor{red}{\sum_{i=1}^{m}}x_{i,n+1}+\textcolor{magenta}{x_{m+1,n+1}}=b_{n+1}\\
        &x_{ij}\ge 0 \qquad 
            \left\{
                \begin{aligned}
                    \forall i=&1,\hdots,m\\     \forall j=&1,\hdots,n\\
                \end{aligned}
            \right. 
\end{aligned}
\quad
\begin{vmatrix}
    \phantom{}\\
    u_i \\
    \phantom{}\\
    \phantom{}\\
    u_{m+1} \\
    \phantom{}\\
    \phantom{}\\
    v_j \\
    \phantom{}\\
    \phantom{}\\
    v_{n+1}\\
    \phantom{}\\
\end{vmatrix}
\quad
\begin{aligned}[c]
    \text{Max} & \sum_{i=1}^{m+1}u_i a_i+\sum_{j=1}^{n+1}v_j b_j\\      
    & \phantom{}\\
  \text{s.a.} \quad
        & u_i+v_j \le c_{ij}\\
        & \\
        -& u_i-v_{n+1} \le c_{i,n+1}\\
         & \\
        & u_{m+1}+v_j \le c_{m+1,j}\\
        & \\
        & u_{m+1}+v_{n+1} \le c_{{m+1},{n+1}}\\
        & \phantom{}\\
        & \phantom{}\\
        & \phantom{}\\
        & \phantom{}\\
\end{aligned}
\normalsize
\end{equation*}

\end{frame}
  
\end{document}