跨案例对齐

Question 1

我只会使用rotating其他sidewaystable功能，例如rotate环境，实际上只是为了与旧版 latex2.09 软件包兼容，该软件包graphicx是基于旧版 latex2.09 的。它只是一个非常薄的兼容性包装器，graphicx在这种情况下，失去了将旋转设置为围绕中心的选项。

对于其余部分，我只是强制了一些明确的间距，唯一棘手的一点是，大号的{比标准号的更宽。

\documentclass{article}
\usepackage{amsmath}
\usepackage{graphicx}

\begin{document}

\begin{align}
\rotatebox[origin=c]{90}{Init}
&\begin{cases}
\makebox[2.5em][l]{$P_{1,1}    $}&= P_0 \\
\makebox[2.5em][l]{$S_{1,1}    $}&= S_0                                                                         \\
\makebox[2.5em][l]{$R_{1,1}    $}&= R_0                                                                         \\
\end{cases} \\
&\begin{cases}
\sbox0{$\Bigg\{$}
\sbox2{$\big\{$}
\kern\dimexpr\wd0-\wd2\relax
\makebox[2.5em][l]{$g(P_{y,t}) $}&= u_{y,t,P}+u_{y,t,S}+u_{y,t,R}                \forall y \forall t>1          \\
\end{cases} \\
&\begin{cases}
\makebox[2.5em][l]{$P_{y,t+1}  $}&=  P_{y,t}+c_P \cdot u_{y,t,P}\cdot\Delta t   \forall y \forall t<T          \\
\makebox[2.5em][l]{$S_{y,t+1}  $}&=  S_{y,t}+c_S \cdot u_{y,t,S}\cdot\Delta t                      \forall y \forall t<T          \\
\makebox[2.5em][l]{$R_{y,t+1}  $}&=  R_{y,t}+c_R \cdot u_{y,t,R}\cdot\Delta t                      \forall y \forall t<T          \\
\end{cases} \\
&\begin{cases}
\makebox[2.5em][l]{$P_{y,1}    $}&= \gamma_P P_{y-1,T}+\gamma_W S_{y-1,T}        \forall y>1                    \\
\makebox[2.5em][l]{$S_{y,1}    $}&= 0                                                              \forall y>1  \\
\makebox[2.5em][l]{$R_{y,1}    $}&= 0                                                              \forall y>1 
\end{cases}
\end{align}

\end{document}

Answer

我只会使用rotating其他sidewaystable功能，例如rotate环境，实际上只是为了与旧版 latex2.09 软件包兼容，该软件包graphicx是基于旧版 latex2.09 的。它只是一个非常薄的兼容性包装器，graphicx在这种情况下，失去了将旋转设置为围绕中心的选项。

对于其余部分，我只是强制了一些明确的间距，唯一棘手的一点是，大号的{比标准号的更宽。

\documentclass{article}
\usepackage{amsmath}
\usepackage{graphicx}

\begin{document}

\begin{align}
\rotatebox[origin=c]{90}{Init}
&\begin{cases}
\makebox[2.5em][l]{$P_{1,1}    $}&= P_0 \\
\makebox[2.5em][l]{$S_{1,1}    $}&= S_0                                                                         \\
\makebox[2.5em][l]{$R_{1,1}    $}&= R_0                                                                         \\
\end{cases} \\
&\begin{cases}
\sbox0{$\Bigg\{$}
\sbox2{$\big\{$}
\kern\dimexpr\wd0-\wd2\relax
\makebox[2.5em][l]{$g(P_{y,t}) $}&= u_{y,t,P}+u_{y,t,S}+u_{y,t,R}                \forall y \forall t>1          \\
\end{cases} \\
&\begin{cases}
\makebox[2.5em][l]{$P_{y,t+1}  $}&=  P_{y,t}+c_P \cdot u_{y,t,P}\cdot\Delta t   \forall y \forall t<T          \\
\makebox[2.5em][l]{$S_{y,t+1}  $}&=  S_{y,t}+c_S \cdot u_{y,t,S}\cdot\Delta t                      \forall y \forall t<T          \\
\makebox[2.5em][l]{$R_{y,t+1}  $}&=  R_{y,t}+c_R \cdot u_{y,t,R}\cdot\Delta t                      \forall y \forall t<T          \\
\end{cases} \\
&\begin{cases}
\makebox[2.5em][l]{$P_{y,1}    $}&= \gamma_P P_{y-1,T}+\gamma_W S_{y-1,T}        \forall y>1                    \\
\makebox[2.5em][l]{$S_{y,1}    $}&= 0                                                              \forall y>1  \\
\makebox[2.5em][l]{$R_{y,1}    $}&= 0                                                              \forall y>1 
\end{cases}
\end{align}

\end{document}

Question 2

这是使用该包的建议eqparbox。它需要一些编译。

第二个案例所需的调整归功于戴维斯·卡莱尔。

\documentclass{article}
\usepackage{amsmath}
\usepackage{rotating}

\usepackage{eqparbox}
\newcommand\MB[3][]{%
  \eqmakebox[#2][#1]{$#3$}
}

\begin{document}

\begin{align}
\begin{rotate}{90}
Init
\end{rotate}
&
\begin{cases}
\MB[r]{A}{P_{1,1}}  = \MB[l]{B}{P_0}                                                                 \\
\MB[r]{A}{S_{1,1}}  = \MB[l]{B}{S_0}                                                                 \\
\MB[r]{A}{R_{1,1}}  = \MB[l]{B}{R_0}                                                                 \\
\end{cases}                                                                                          \\
&
\begin{cases}
 % manually adjusted bacause of the { width difference
\sbox0{$\Bigg\{$}
\sbox2{$\big\{$}
\kern\dimexpr\wd0-\wd2\relax
\MB[r]{A}{g(P_{y,t})} = \MB[l]{B}{u_{y,t,P}+u_{y,t,S}+u_{y,t,R}}             & \forall y \forall t>1 \\
\end{cases}                                                                                          \\
&
\begin{cases}
\MB[r]{A}{P_{y,t+1}}  =  \MB[l]{B}{P_{y,t}+c_P \cdot u_{y,t,P}\cdot\Delta t} & \forall y \forall t<T \\
\MB[r]{A}{S_{y,t+1}}  =  \MB[l]{B}{S_{y,t}+c_S \cdot u_{y,t,S}\cdot\Delta t} & \forall y \forall t<T \\
\MB[r]{A}{R_{y,t+1}}  =  \MB[l]{B}{R_{y,t}+c_R \cdot u_{y,t,R}\cdot\Delta t} & \forall y \forall t<T \\
\end{cases}                                                                                          \\
&
\begin{cases}
\MB[r]{A}{P_{y,1}}    = \MB[l]{B}{\gamma_P P_{y-1,T}+\gamma_W S_{y-1,T}}     & \forall y>1           \\
\MB[r]{A}{S_{y,1}}    = \MB[l]{B}{0}                                         & \forall y>1           \\
\MB[r]{A}{R_{y,1}}    = \MB[l]{B}{0}                                         & \forall y>1 
\end{cases}
\end{align}

\end{document}

Answer

这是使用该包的建议eqparbox。它需要一些编译。

第二个案例所需的调整归功于戴维斯·卡莱尔。

\documentclass{article}
\usepackage{amsmath}
\usepackage{rotating}

\usepackage{eqparbox}
\newcommand\MB[3][]{%
  \eqmakebox[#2][#1]{$#3$}
}

\begin{document}

\begin{align}
\begin{rotate}{90}
Init
\end{rotate}
&
\begin{cases}
\MB[r]{A}{P_{1,1}}  = \MB[l]{B}{P_0}                                                                 \\
\MB[r]{A}{S_{1,1}}  = \MB[l]{B}{S_0}                                                                 \\
\MB[r]{A}{R_{1,1}}  = \MB[l]{B}{R_0}                                                                 \\
\end{cases}                                                                                          \\
&
\begin{cases}
 % manually adjusted bacause of the { width difference
\sbox0{$\Bigg\{$}
\sbox2{$\big\{$}
\kern\dimexpr\wd0-\wd2\relax
\MB[r]{A}{g(P_{y,t})} = \MB[l]{B}{u_{y,t,P}+u_{y,t,S}+u_{y,t,R}}             & \forall y \forall t>1 \\
\end{cases}                                                                                          \\
&
\begin{cases}
\MB[r]{A}{P_{y,t+1}}  =  \MB[l]{B}{P_{y,t}+c_P \cdot u_{y,t,P}\cdot\Delta t} & \forall y \forall t<T \\
\MB[r]{A}{S_{y,t+1}}  =  \MB[l]{B}{S_{y,t}+c_S \cdot u_{y,t,S}\cdot\Delta t} & \forall y \forall t<T \\
\MB[r]{A}{R_{y,t+1}}  =  \MB[l]{B}{R_{y,t}+c_R \cdot u_{y,t,R}\cdot\Delta t} & \forall y \forall t<T \\
\end{cases}                                                                                          \\
&
\begin{cases}
\MB[r]{A}{P_{y,1}}    = \MB[l]{B}{\gamma_P P_{y-1,T}+\gamma_W S_{y-1,T}}     & \forall y>1           \\
\MB[r]{A}{S_{y,1}}    = \MB[l]{B}{0}                                         & \forall y>1           \\
\MB[r]{A}{R_{y,1}}    = \MB[l]{B}{0}                                         & \forall y>1 
\end{cases}
\end{align}

\end{document}

Question 3

这是另一个选项，使用matrix of nodes来自蒂克兹包裹。

这里有一点技巧tikpage节点“手动”放置方程编号。等号的对齐并不完美，但可以matrix of nodes让您完全控制要添加到这些方程中的任何“装饰”。

代码如下：

\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usepackage{tikzpagenodes}
\usetikzlibrary{matrix,fit,decorations.pathreplacing}

\begin{document}

\begin{center}
  \begin{tikzpicture}[remember picture]
    \matrix (M)[
        matrix of math nodes,
        column 1/.style = {anchor=base east, minimum width=3.6em},
        column 2/.style = {anchor=base west},
        column 3/.style = {anchor=base west},
    ]{
      P_{1,1}    &= P_0 \\
      S_{1,1}    &= S_0 \\
      R_{1,1}    &= R_0 \\
      g(P_{y,t}) &= u_{y,t,P}+u_{y,t,S}+u_{y,t,R}            &\forall y \forall t>1 \\
      P_{y,t+1}  &= P_{y,t}+c_P \cdot u_{y,t,P}\cdot\Delta t &\forall y \forall t<T \\
      S_{y,t+1}  &= S_{y,t}+c_S \cdot u_{y,t,S}\cdot\Delta t &\forall y \forall t<T \\
      R_{y,t+1}  &= R_{y,t}+c_R \cdot u_{y,t,R}\cdot\Delta t &\forall y \forall t<T \\
      P_{y,1}    &= \gamma_P P_{y-1,T}+\gamma_W S_{y-1,T}    &\forall y>1 \\
      S_{y,1}    &= 0                                        &\forall y>1 \\
      R_{y,1}    &= 0                                        &\forall y>1 \\
    };
    % add braces around equation groups
    \foreach \top/\bot/\lab in {1/3, 4/4, 5/7, 8/10} {
        \draw[thick,decorate, decoration={mirror,brace}]
        ([yshift=-1pt]M-\top-1.north west)--([yshift=1pt]M-\bot-1.south west);
    }
    % add the init label
    \node[rotate=90] at ([xshift=-3mm]M-2-1.west){Init};
    % and now add the equation lines, using tikzpagenodes to put
    % them on the righthand margin
    \begin{scope}[overlay]
      \foreach \eq in {2,4,6,9} {
          \refstepcounter{equation}
          \node[anchor=east] at ([xshift=-3em]M-\eq-1-|current page text area.east) {(\theequation)};
      }
    \end{scope}
  \end{tikzpicture}
\end{center}

\end{document}

Answer

这是另一个选项，使用matrix of nodes来自蒂克兹包裹。

这里有一点技巧tikpage节点“手动”放置方程编号。等号的对齐并不完美，但可以matrix of nodes让您完全控制要添加到这些方程中的任何“装饰”。

代码如下：

\documentclass{article}
\usepackage{amsmath}
\usepackage{tikz}
\usepackage{tikzpagenodes}
\usetikzlibrary{matrix,fit,decorations.pathreplacing}

\begin{document}

\begin{center}
  \begin{tikzpicture}[remember picture]
    \matrix (M)[
        matrix of math nodes,
        column 1/.style = {anchor=base east, minimum width=3.6em},
        column 2/.style = {anchor=base west},
        column 3/.style = {anchor=base west},
    ]{
      P_{1,1}    &= P_0 \\
      S_{1,1}    &= S_0 \\
      R_{1,1}    &= R_0 \\
      g(P_{y,t}) &= u_{y,t,P}+u_{y,t,S}+u_{y,t,R}            &\forall y \forall t>1 \\
      P_{y,t+1}  &= P_{y,t}+c_P \cdot u_{y,t,P}\cdot\Delta t &\forall y \forall t<T \\
      S_{y,t+1}  &= S_{y,t}+c_S \cdot u_{y,t,S}\cdot\Delta t &\forall y \forall t<T \\
      R_{y,t+1}  &= R_{y,t}+c_R \cdot u_{y,t,R}\cdot\Delta t &\forall y \forall t<T \\
      P_{y,1}    &= \gamma_P P_{y-1,T}+\gamma_W S_{y-1,T}    &\forall y>1 \\
      S_{y,1}    &= 0                                        &\forall y>1 \\
      R_{y,1}    &= 0                                        &\forall y>1 \\
    };
    % add braces around equation groups
    \foreach \top/\bot/\lab in {1/3, 4/4, 5/7, 8/10} {
        \draw[thick,decorate, decoration={mirror,brace}]
        ([yshift=-1pt]M-\top-1.north west)--([yshift=1pt]M-\bot-1.south west);
    }
    % add the init label
    \node[rotate=90] at ([xshift=-3mm]M-2-1.west){Init};
    % and now add the equation lines, using tikzpagenodes to put
    % them on the righthand margin
    \begin{scope}[overlay]
      \foreach \eq in {2,4,6,9} {
          \refstepcounter{equation}
          \node[anchor=east] at ([xshift=-3em]M-\eq-1-|current page text area.east) {(\theequation)};
      }
    \end{scope}
  \end{tikzpicture}
\end{center}

\end{document}

Question 4

我最终融合了几个答案并添加了额外的空间，因为我不喜欢其中一个括号比其他括号窄：

\documentclass{article}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{eqparbox}
\newcommand\MB[3][]{%
  \eqmakebox[#2][#1]{$#3$}
}

\begin{document}

{\small\allowdisplaybreaks
\begin{align}
\rotatebox[origin=c]{90}{\footnotesize \parbox{1.8cm}{\centering Initial \\ Conditions}}
&\left\{
\begin{aligned} \label{equ:mod1_init}
\MB[r]{A}{P_{1,1}}  = \MB[l]{B}{P_0}                                                                 \\
\MB[r]{A}{S_{1,1}}  = \MB[l]{B}{S_0}                                                                 \\
\MB[r]{A}{R_{1,1}}  = \MB[l]{B}{R_0}                                                                 \\
\end{aligned}
\right.           \\
\rotatebox[origin=c]{90}{\footnotesize \parbox{1.8cm}{\centering Mass\\ Conservation}}
&\left\{
\begin{aligned} \label{equ:mod1_masscons}
 % manually adjusted bacause of the { width difference
\sbox0{$\Bigg\{$}
\sbox2{$\big\{$}
\kern\dimexpr\wd0-\wd2\relax
&&\\
\MB[r]{A}{g(P_{y,t})} &= \MB[l]{B}{u_{y,t,P}+u_{y,t,S}+u_{y,t,R}}             & \forall y \forall t>1 \\
&&\\
\end{aligned}
\right.           \\
\rotatebox[origin=c]{90}{\footnotesize Integration}
&\left\{
\begin{aligned} \label{equ:mod1_integration}
\MB[r]{A}{P_{y,t+1}}  &=  \MB[l]{B}{P_{y,t}+c_P \cdot u_{y,t,P}\cdot\Delta t} & \forall y \forall t<T \\
\MB[r]{A}{S_{y,t+1}}  &=  \MB[l]{B}{S_{y,t}+c_S \cdot u_{y,t,S}\cdot\Delta t} & \forall y \forall t<T \\
\MB[r]{A}{R_{y,t+1}}  &=  \MB[l]{B}{R_{y,t}+c_R \cdot u_{y,t,R}\cdot\Delta t} & \forall y \forall t<T \\
\end{aligned}      
\right. \\
\rotatebox[origin=c]{90}{\footnotesize Continuity}
&\left\{
\begin{aligned} \label{equ:mod1_continuity}
\MB[r]{A}{P_{y,1}}    &= \MB[l]{B}{\gamma_P P_{y-1,T}+\gamma_W S_{y-1,T}}     & \forall y>1           \\
\MB[r]{A}{S_{y,1}}    &= \MB[l]{B}{0}                                         & \forall y>1           \\
\MB[r]{A}{R_{y,1}}    &= \MB[l]{B}{0}                                         & \forall y>1 
\end{aligned}
\right.
\end{align}
}

\end{document}

Answer

我最终融合了几个答案并添加了额外的空间，因为我不喜欢其中一个括号比其他括号窄：

\documentclass{article}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage{eqparbox}
\newcommand\MB[3][]{%
  \eqmakebox[#2][#1]{$#3$}
}

\begin{document}

{\small\allowdisplaybreaks
\begin{align}
\rotatebox[origin=c]{90}{\footnotesize \parbox{1.8cm}{\centering Initial \\ Conditions}}
&\left\{
\begin{aligned} \label{equ:mod1_init}
\MB[r]{A}{P_{1,1}}  = \MB[l]{B}{P_0}                                                                 \\
\MB[r]{A}{S_{1,1}}  = \MB[l]{B}{S_0}                                                                 \\
\MB[r]{A}{R_{1,1}}  = \MB[l]{B}{R_0}                                                                 \\
\end{aligned}
\right.           \\
\rotatebox[origin=c]{90}{\footnotesize \parbox{1.8cm}{\centering Mass\\ Conservation}}
&\left\{
\begin{aligned} \label{equ:mod1_masscons}
 % manually adjusted bacause of the { width difference
\sbox0{$\Bigg\{$}
\sbox2{$\big\{$}
\kern\dimexpr\wd0-\wd2\relax
&&\\
\MB[r]{A}{g(P_{y,t})} &= \MB[l]{B}{u_{y,t,P}+u_{y,t,S}+u_{y,t,R}}             & \forall y \forall t>1 \\
&&\\
\end{aligned}
\right.           \\
\rotatebox[origin=c]{90}{\footnotesize Integration}
&\left\{
\begin{aligned} \label{equ:mod1_integration}
\MB[r]{A}{P_{y,t+1}}  &=  \MB[l]{B}{P_{y,t}+c_P \cdot u_{y,t,P}\cdot\Delta t} & \forall y \forall t<T \\
\MB[r]{A}{S_{y,t+1}}  &=  \MB[l]{B}{S_{y,t}+c_S \cdot u_{y,t,S}\cdot\Delta t} & \forall y \forall t<T \\
\MB[r]{A}{R_{y,t+1}}  &=  \MB[l]{B}{R_{y,t}+c_R \cdot u_{y,t,R}\cdot\Delta t} & \forall y \forall t<T \\
\end{aligned}      
\right. \\
\rotatebox[origin=c]{90}{\footnotesize Continuity}
&\left\{
\begin{aligned} \label{equ:mod1_continuity}
\MB[r]{A}{P_{y,1}}    &= \MB[l]{B}{\gamma_P P_{y-1,T}+\gamma_W S_{y-1,T}}     & \forall y>1           \\
\MB[r]{A}{S_{y,1}}    &= \MB[l]{B}{0}                                         & \forall y>1           \\
\MB[r]{A}{R_{y,1}}    &= \MB[l]{B}{0}                                         & \forall y>1 
\end{aligned}
\right.
\end{align}
}

\end{document}

跨案例对齐

答案1

答案2

答案3

答案4

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