使用方程式打破投影仪中的框架

Question

我有以下 MWE：

\documentclass[10pt]{beamer}
\usetheme{Warsaw}
\useoutertheme{infolines}
\usepackage[english]{babel}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{array}
\usepackage{color,colortbl}
\allowdisplaybreaks[1] %to allow automatic page breaks for formulas written with \alignat


\begin{document}

\begin{frame}[allowframebreaks]{Mathematical model}

 \textbf{Modelo matemático:}


\begin{alignat}{2}
 \mbox{maximize  } z_1  =  & \sum_{i \in V} \sum_{k \in V} \left( p_{ki} + q_i \right)~y_{ki}    \label{Obj1} \\
  \hspace{-0.5cm}\mbox{minimize  } z_2  = &  \sum_{(i,j) \in A} c_{ij}~x_{ij}    \label{Obj2} \\
 \mbox{subject to} &
                   && \nonumber \\
& \sum_{j:(1,j) \in A} x_{1j} = 1, &&  j \in V    \label{Restricao1} \\
& \sum_{i:(i,1) \in A} x_{i1} = 1, &&  i \in V    \label{Restricao2} \\
& \sum_{j \in V} x_{ij} - \sum_{k \in V} y_{ki} = 0, \qquad &&  i \in V    \label{Restricao3} \\
& \sum_{i \in V} x_{ij} - \sum_{k \in V} y_{kj} = 0, &&  j \in V    \label{Restricao4} \\
& x_{ij} - y_{1j} = 0, &&  j \in V    \label{Restricao5} \\
& x_{ij} +1 \geq y_{ki} + y_{(k+1)j},  &&  (i,j) \in A, \quad k \in V    \label{Restricao6} \\
& w_j \geq w_0 + e_{1j} \cdot x_{1j} - M \cdot (1-x_{1j}) &&  (1,j) \in A    \label{Restricao7} \\
& w_j \geq w_i + \left( e_{ij} + d_i \right)\cdot x_{ij} - M \cdot (1-x_{ij}) \quad \quad && (i,j) \in A, \quad i \neq 1    \label{Restricao8} \\
& \sum_{t \in T} \phi_{it} - \sum_{k \in V} y_{ki} = 0 &&  i \in V    \label{Restricao9} \\
& z_{it} \leq M \cdot \phi_{it}  &&  i \in V, \quad t \in T    \label{Restricao10} \\
& w_i - \sum_{t \in T} z_{it} = 0 &&  i \in V    \label{Restricao11} \\
& a_{it} \cdot \phi_{it} \leq z_{it} \leq b_{it} \cdot \phi_{it}  &&  i \in V, \quad t \in T    \label{Restricao12}\\
& \alpha_t \leq \sum_{i \in B} \phi_{it} \leq \beta_t &&  t \in T    \label{Restricao13}\\
& \sum_{i \in R_t} \phi_{it} =1 &&  t \in T    \label{Restricao14}\\
& \sum_{i \in H_t} \phi_{it} =1  &&  t \in T\setminus{|T|}    \label{Restricao15}\\
&  \sum_{i \in V} y_{ki} \geq \sum_{i \in V} y_{(k+1)i}  && k \in V\setminus \{n\}      \label{Restricao16}\\
&  \sum_{i \in V} y_{ki} \leq 1  && k \in V     \label{Restricao17}\\
&  \sum_{k \in V} y_{ki}  \leq 1  &&  i \in V    \label{Restricao18}\\
& x_{ij} \in \{0,1\} && (i,j) \in A     \label{Restricao19}\\
& y_{ki} \in \{0,1\} && i \in V, \quad k \in V     \label{Restricao20}\\
& \phi_{it} \in \{0,1\} &&  i \in V, \quad t \in T   \label{Restricao21}\\
& w_{i} \geq 0 && i \in V     \label{Restricao22}\\
& z_{it} \geq 0 && i \in V, \quad t \in T.     \label{Restricao23}
\end{alignat}
\end{frame}
\end{document}

其结果不理想：

我想将我的数学模型（在环境中alignat）分解为两个或更多框架。我尝试使用allowframebreaks选项，但没有用。有人能帮我吗？

Answer 1

我不会依赖自动框架中断。我进一步建议您将 23 个限制放在双列column环境中。这样您就可以在两个框架中排版所有 25 个方程式。

旁白：你确定把所有 25 个方程式都讲给听众是个好主意吗？你认为在演讲结束后，听众们会记住哪些方程式（如果有的话！）？

\documentclass[10pt]{beamer}
\usetheme{Warsaw}
\useoutertheme{infolines}
\usepackage[english]{babel}
%\usepackage[utf8]{inputenc} % that's the default nowadays
\usepackage[T1]{fontenc}
\usepackage{mathtools,amssymb,array,xcolor,colortbl}

\begin{document}

\begin{frame}{Modelo matemático I}
\begin{align}
\max z_1  &= \sum_{i \in V} \sum_{k \in V} ( p_{ki} + q_i )\,y_{ki}  \label{Obj1} \\
\min z_2  &= \sum_{\mathclap{(i,j) \in A}} c_{ij}\,x_{ij}            \label{Obj2} 
\end{align}

subject to
\small
\begin{columns}[t]
\column{0.45\textwidth}
\begin{align}
&\sum_{\mathclap{j:(1,j) \in A}} x_{1j} = 1, \quad j\in V          \label{Restricao1} \\
&\sum_{\mathclap{i:(i,1) \in A}} x_{i1} = 1, \quad i\in V          \label{Restricao2} \\
& \sum_{j \in V} x_{ij} - \sum_{k \in V} y_{ki} = 0, \quad i \in V \label{Restricao3}
\end{align}

\column{0.55\textwidth}
\begin{align}
& \sum_{i \in V} x_{ij} - \sum_{k \in V} y_{kj} = 0, \quad j \in V \label{Restricao4} \\[\jot]
& x_{ij} - y_{1j} = 0, \quad   j \in V                             \label{Restricao5} \\[3\jot]
& x_{ij} +1 \geq y_{ki} + y_{(k+1)j},                              \label{Restricao6} \\
&\qquad  (i,j) \in A, \ k \in V   \notag  
\end{align}
\end{columns}
\end{frame}

\begin{frame}{Modelo matemático II}
\small
\begin{columns}
\column{0.52\textwidth}
\begin{align}
& w_j \geq w_0 + e_{1j}  x_{1j} - M  (1-x_{1j}),        \label{Restricao7} \\
& \qquad (1,j) \in A  \notag\\
& w_j \geq w_i + ( e_{ij} + d_i ) x_{ij}                \label{Restricao8} \\
& \qquad  - M  (1-x_{ij}) (i,j), 
  \quad i\in A, \ i \neq 1 \notag\\
& \sum_{t \in T} \phi_{it} - \sum_{k \in V} y_{ki} = 0, 
  \quad   i \in V                                       \label{Restricao9}  \\
& z_{it} \leq M  \phi_{it},  \quad   i \in V, \ t \in T \label{Restricao10} \\
& w_i - \sum_{t \in T} z_{it} = 0, \quad   i \in V      \label{Restricao11} \\
& a_{it}  \phi_{it} \leq z_{it} \leq b_{it}  \phi_{it}, 
  \quad   i \in V, \ t \in T                            \label{Restricao12} \\
& \alpha_t \leq \sum_{i \in B} \phi_{it} \leq \beta_t,
  \quad   t \in T                                       \label{Restricao13} \\
& \sum_{i \in R_t} \phi_{it} =1, \quad   t \in T        \label{Restricao14}
\end{align}

\column{0.48\textwidth}
\begin{align}
& \sum_{i \in H_t} \phi_{it} =1, \quad t\in T\setminus{|T|}   \label{Restricao15}\\
& \sum_{i \in V} y_{ki} \geq \sum_{i \in V} y_{(k+1)i},
  \ k \in V \setminus \{n\}                                   \label{Restricao16}\\
& \sum_{i \in V} y_{ki} \leq 1,  \quad k \in V                \label{Restricao17}\\
& \sum_{k \in V} y_{ki} \leq 1,  \quad i \in V                \label{Restricao18}\\
& x_{ij} \in \{0,1\}, \quad  (i,j) \in A                      \label{Restricao19}\\
& y_{ki} \in \{0,1\}, \quad  i \in V, \ k \in V               \label{Restricao20}\\
& \phi_{it} \in \{0,1\}, \quad   i \in V, \ t \in T           \label{Restricao21}\\
& w_{i} \geq 0, \quad  i \in V                                \label{Restricao22}\\
& z_{it} \geq 0, \quad  i \in V, \ t \in T.                   \label{Restricao23}
\end{align}
\end{columns}

\end{frame}
\end{document}

Answer 2

您应该使用@Mico 的答案，您将通过手动构建的框架获得更好的结果。

---

只是为了完整性：理论上你可以使用该allowdisplaybreaks选项来允许在数学内容中进行分页：

\documentclass[10pt]{beamer}
\usetheme{Warsaw}
\useoutertheme{infolines}
\usepackage[english]{babel}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{array}
\usepackage{color,colortbl}
\allowdisplaybreaks[1] %to allow automatic page breaks for formulas written with \alignat


\begin{document}

\begin{frame}[allowframebreaks,allowdisplaybreaks]{Mathematical model}

 \textbf{Modelo matemático:}


\begin{alignat}{2}
 \mbox{maximize  } z_1  =  & \sum_{i \in V} \sum_{k \in V} \left( p_{ki} + q_i \right)~y_{ki}    \label{Obj1} \\
  \hspace{-0.5cm}\mbox{minimize  } z_2  = &  \sum_{(i,j) \in A} c_{ij}~x_{ij}    \label{Obj2} \\
 \mbox{subject to} &
                   && \nonumber \\
& \sum_{j:(1,j) \in A} x_{1j} = 1, &&  j \in V    \label{Restricao1} \\
& \sum_{i:(i,1) \in A} x_{i1} = 1, &&  i \in V    \label{Restricao2} \\
& \sum_{j \in V} x_{ij} - \sum_{k \in V} y_{ki} = 0, \qquad &&  i \in V    \label{Restricao3} \\
& \sum_{i \in V} x_{ij} - \sum_{k \in V} y_{kj} = 0, &&  j \in V    \label{Restricao4} \\
& x_{ij} - y_{1j} = 0, &&  j \in V    \label{Restricao5} \\
& x_{ij} +1 \geq y_{ki} + y_{(k+1)j},  &&  (i,j) \in A, \quad k \in V    \label{Restricao6} \\
& w_j \geq w_0 + e_{1j} \cdot x_{1j} - M \cdot (1-x_{1j}) &&  (1,j) \in A    \label{Restricao7} \\
& w_j \geq w_i + \left( e_{ij} + d_i \right)\cdot x_{ij} - M \cdot (1-x_{ij}) \quad \quad && (i,j) \in A, \quad i \neq 1    \label{Restricao8} \\
& \sum_{t \in T} \phi_{it} - \sum_{k \in V} y_{ki} = 0 &&  i \in V    \label{Restricao9} \\
& z_{it} \leq M \cdot \phi_{it}  &&  i \in V, \quad t \in T    \label{Restricao10} \\
& w_i - \sum_{t \in T} z_{it} = 0 &&  i \in V    \label{Restricao11} \\
& a_{it} \cdot \phi_{it} \leq z_{it} \leq b_{it} \cdot \phi_{it}  &&  i \in V, \quad t \in T    \label{Restricao12}\\
& \alpha_t \leq \sum_{i \in B} \phi_{it} \leq \beta_t &&  t \in T    \label{Restricao13}\\
& \sum_{i \in R_t} \phi_{it} =1 &&  t \in T    \label{Restricao14}\\
& \sum_{i \in H_t} \phi_{it} =1  &&  t \in T\setminus{|T|}    \label{Restricao15}\\
&  \sum_{i \in V} y_{ki} \geq \sum_{i \in V} y_{(k+1)i}  && k \in V\setminus \{n\}      \label{Restricao16}\\
&  \sum_{i \in V} y_{ki} \leq 1  && k \in V     \label{Restricao17}\\
&  \sum_{k \in V} y_{ki}  \leq 1  &&  i \in V    \label{Restricao18}\\
& x_{ij} \in \{0,1\} && (i,j) \in A     \label{Restricao19}\\
& y_{ki} \in \{0,1\} && i \in V, \quad k \in V     \label{Restricao20}\\
& \phi_{it} \in \{0,1\} &&  i \in V, \quad t \in T   \label{Restricao21}\\
& w_{i} \geq 0 && i \in V     \label{Restricao22}\\
& z_{it} \geq 0 && i \in V, \quad t \in T.     \label{Restricao23}
\end{alignat}
\end{frame}
\end{document}