如何消除幻灯片之间方程式的跳跃？我希望，即使变量发生变化，两张幻灯片上的 也位于同一位置\leq。\geq
\documentclass{beamer}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[ngerman]{babel}
\usepackage{tikz}
\usetikzlibrary{decorations.pathmorphing}
\usepackage{tkz-euclide}
\usepackage{tikz}
\usetikzlibrary{shapes,arrows}
\usepackage{geometry}
\geometry{left = 30mm, right = 30mm}
\usepackage[it]{caption}
\usepackage{subfig}
\begin{document}
\begin{columns}[onlytextwidth]
\begin{column}{0.45\textwidth}
Given convex $MINLP$
\begin{align*}
\min z \\
y_1 + y_2 + x_1^2 + x_2^2 &\leq z \\
(x_1-2)^2-x_2 &\leq 0 \\
x_1 - 2y_1 &\geq 0 \\
x_1-x_2-3(1-y_1) &\leq 0 \\
x_1 - (1-y_1) &\geq 0 \\
x_2-y_2 &\leq 0 \\
x_1+x_2 &\leq 3y_1\\
y_1+y_2 &\leq 1\\
0 \leq x_1,x_2, &\leq 4\\
y_1,y_2 &\in \{0,1\}
\end{align*}
Choose initial integer solutions $y_1 = y_2 = 1$. Upper bound $U = \infty$.
\end{column}
\begin{column}{0.45\textwidth}
% Define block styles
\tikzstyle{decision} = [diamond, draw, fill=blue!20,
text width=7.5em, text badly centered, node distance=3cm, inner sep=0pt]
\tikzstyle{block} = [rectangle, draw, fill=blue!20,
text width=5em, text centered, rounded corners, minimum height=4em]
\tikzstyle{blockH} = [rectangle, draw, fill=red!20,
text width=5em, text centered, rounded corners, minimum height=4em]
\tikzstyle{line} = [draw, -latex']
\tikzstyle{cloud} = [draw, ellipse,fill=red!20, node distance=3cm,
minimum height=2em]
\begin{tikzpicture}[node distance = 3.5cm, auto]
% Place nodes
\node [blockH] (init) {Init};
\node [block, below of=init, node distance=3cm] (NLP) { Solve NLP($x^k_I$) or F($x^k_I$)};
\node [block, below of=NLP] (Bound) { Add linearaziation at $x^k$ to MIP};
\node [block, below of=Bound] (MIP) { Solve MIP($x^k$, $U$) };
\node [decision, right of=MIP, node distance=6cm] (Infeas) { MIP infeasible? -> Exit };
\path [line] (init) -- (NLP);
\path [line] (NLP) -- node {Update U?} (Bound);
% \path [line] (Bound) -- (Update);
\path [line] (Bound) -- (MIP);
\path [line] (MIP) -- (Infeas);
\path [line] (Infeas) |- node[near start] {No} (NLP);
\end{tikzpicture}
\end{column}
\end{columns}
\end{frame}
%Step 2
\begin{frame}
\begin{columns}[onlytextwidth]
\begin{column}{0.45\textwidth}
Fix $y_1 = y_2 = 1$ and solve NLP.
\begin{align*}
\min z \\
1 + 1 + x_1^2 + x_2^2 &\leq z \\
(x_1-2)^2-x_2 &\leq 0 \\
x_1 - 2 &\geq 0 \\
x_1-x_2-3(1-1) &\leq 0 \\
x_1 - (1-1) &\geq 0 \\
x_2-1 &\leq 0 \\
x_1+x_2 &\leq 3\\
1+1 &\leq 1\\
0 \leq x_1,x_2, &\leq 4\\
1,1 &\in \{0,1\}
\end{align*}
Optimal solution $x_1=x_2=2$, (x,y) = (2,2,1,1) is a solution to our $MINLP$.
\end{column}
\begin{column}{0.45\textwidth}
% Define block styles
\tikzstyle{decision} = [diamond, draw, fill=blue!20,
text width=7.5em, text badly centered, node distance=3cm, inner sep=0pt]
\tikzstyle{block} = [rectangle, draw, fill=blue!20,
text width=5em, text centered, rounded corners, minimum height=4em]
\tikzstyle{blockH} = [rectangle, draw, fill=red!20,
text width=5em, text centered, rounded corners, minimum height=4em]
\tikzstyle{line} = [draw, -latex']
\tikzstyle{cloud} = [draw, ellipse,fill=red!20, node distance=3cm,
minimum height=2em]
\begin{tikzpicture}[node distance = 3.5cm, auto]
% Place nodes
\node [block] (init) {Init};
\node [blockH, below of=init, node distance=3cm] (NLP) { Solve NLP($x^k_I$) or F($x^k_I$)};
\node [block, below of=NLP] (Bound) { Add linearaziation at $x^k$ to MIP};
\node [block, below of=Bound] (MIP) { Solve MIP($x^k$, $U$) };
\node [decision, right of=MIP, node distance=6cm] (Infeas) { MIP infeasible? -> Exit };
\path [line] (init) -- (NLP);
\path [line] (NLP) -- node {Update U?} (Bound);
% \path [line] (Bound) -- (Update);
\path [line] (Bound) -- (MIP);
\path [line] (MIP) -- (Infeas);
\path [line] (Infeas) |- node[near start] {No} (NLP);
\end{tikzpicture}
\end{column}
\end{columns}
\end{frame}
\end{document}
答案1
为什么要使用geometry包装？您应该记住，beamer幻灯片的实际尺寸是 128 x 96 毫米，因此您需要\geometry{left = 30mm, right = 30mm}截断一半的幻灯片宽度！因此，所有内容都移到了右侧，由于空间太小，这会带来问题。
编辑（1）：
- 删除
geometry包并定义文本宽度 - 重新绘制你的图形，以便它们适合幻灯片
- 不要使用过时的语法
tikzpicure（而是right=of (...)使用right=of (...)，而是tikzstyle使用tikzset，...） - 在序言中集中样式（代码将变得更短）
编辑（2）：
- 如果右列（第二列）的图片在两个框架上相等，那么您可以考虑将两个框架合并为一个，并使用
\only<..>{ ...}揭示机制 - 将方程式放入
array环境中，并在两个数组中定义相等的行高和第一列宽 - 添加
\vphantom{$y_1$}文本“给定凸 MINLP。” - 将两个数组添加
\setlength\arraycolsep{2pt}到两个数组，使水平距离\leq相等。 - 使第一列中的图片高于文本和公式
经过所有这些测量后，图片和位置leq将在框架中固定，但是由于第一张和第二张幻灯片上的方程式不同，它们的外观（宽度）将发生变化。新代码是：
\documentclass{beamer}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
%\usepackage[ngerman]{babel}
\usepackage{tikz}
\usetikzlibrary{arrows,positioning,shapes.geometric}
\usepackage{array}
\tikzset{
node distance = 9mm and 5mm,
auto,
decision/.style = {diamond, draw, fill=blue!20,
text width=5em, align=flush center, inner sep=-2pt},
block/.style = {rectangle, draw, fill=blue!20,
text width=5em, align=center, rounded corners,
minimum height=4em},
blockH/.style = {rectangle, draw, fill=red!20,
text width=5em, align=center, rounded corners,
minimum height=4ex},
line/.style = {draw, -latex'},
cloud/.style = {draw, ellipse, fill=red!20, minimum height=2em}
}
\begin{document}
\begin{frame}
\begin{columns}[b,onlytextwidth]
\begin{column}{0.5\textwidth}
\only<1>{
\vphantom{$y_1$}Given convex MINLP.\vspace*{-1ex}
\setlength\arraycolsep{2pt}
\[\begin{array}{@{\rule[-1ex]{0pt}{3ex}}>{\raggedleft\arraybackslash$}p{9em}<{$}l}
\min z & \\
y_1 + y_2 + x_1^2 + x_2^2 & \leq z \\
(x_1-2)^2-x_2 & \leq 0 \\
x_1 - 2y_1 & \geq 0 \\
x_1-x_2-3(1-y_1) & \leq 0 \\
x_1 - (1-y_1) & \geq 0 \\
x_2-y_2 & \leq 0 \\
x_1+x_2 & \leq 3y_1 \\
y_1+y_2 & \leq 1 \\
0 \leq x_1,x_2, & \leq 4 \\
y_1,y_2 & \in \{0,1\}
\end{array}\]
Choose initial integer solutions $y_1 = y_2 = 1$. Upper bound $U = \infty$.
}
\only<2>{
Fix $y_1 = y_2 = 1$ and solve NLP.\vspace*{-1ex}
\setlength\arraycolsep{2pt}
\[\begin{array}{@{\rule[-1ex]{0pt}{3ex}}>{\raggedleft\arraybackslash$}p{9em}<{$}l}
\min z \\
1 + 1 + x_1^2 + x_2^2 & \leq z \\
(x_1-2)^2-x_2 & \leq 0 \\
x_1 - 2 & \geq 0 \\
x_1-x_2-3(1-1) & \leq 0 \\
x_1 - (1-1) & \geq 0 \\
x_2-1 & \leq 0 \\
x_1+x_2 & \leq 3\\
1+1 & \leq 1\\
0 \leq x_1,x_2 & \leq 4\\
1,1 & \in \{0,1\}
\end{array}\]
Optimal solution $x_1=x_2=2$, $(x,y)=(2,2,1,1)$ is a solution
to our MINLP.
}
\end{column}
\begin{column}{0.48\textwidth}
\begin{tikzpicture}
% Place nodes
\node [blockH] (init) {Init};
\node [block,below=of init] (NLP) {Solve NLP($x^k_I$) or F($x^k_I$)};
\node [block,below=of NLP] (Bound) {Add linearaziation at $x^k$ to MIP};
\node [block,below=of Bound] (MIP) {Solve MIP($x^k, U$)};
\node [decision,right=of MIP] (Infeas){MIP infeasible?\\ -> Exit};
%
\path[line] (init) edge (NLP)
(NLP) edge node {Update U?} (Bound)
(Bound) edge (MIP)
(MIP) edge (Infeas)
(Infeas) |- node[near start] {No} (NLP);
\end{tikzpicture}
\end{column}
\end{columns}
\end{frame}
\end{document}
原始代码（第一次编辑后）是：
\documentclass{beamer}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
%\usepackage[ngerman]{babel}
\usepackage{tikz}
\usetikzlibrary{arrows,decorations.pathmorphing,shapes}
\usepackage{tkz-euclide}
%\usepackage{tikz} <-- second time ...
%\usepackage{geometry}
%\geometry{left = 30mm, right = 30mm}
\usepackage[it]{caption}
\usepackage{subfig}
\tikzset{
node distance = 7mm and 5mm,
auto,
decision/.style = {diamond, draw, fill=blue!20,
text width=7.5em, align=flush center, inner sep=-2pt},
block/.style = {rectangle, draw, fill=blue!20,
text width=5em, align=center, rounded corners,
minimum height=4em},
blockH/.style = {rectangle, draw, fill=red!20,
text width=5em, align=center, rounded corners,
minimum height=4ex},
line/.style = {draw, -latex'},
cloud/.style = {draw, ellipse, fill=red!20, minimum height=2em}
}
\begin{document}
\begin{frame}
\begin{columns}[onlytextwidth]
\begin{column}{0.48\textwidth}
Given convex\\
$MINLP$
\begin{align*}
\min z & \\
y_1 + y_2 + x_1^2 + x_2^2 & \leq z \\
(x_1-2)^2-x_2 & \leq 0 \\
x_1 - 2y_1 & \geq 0 \\
x_1-x_2-3(1-y_1) & \leq 0 \\
x_1 - (1-y_1) & \geq 0 \\
x_2-y_2 & \leq 0 \\
x_1+x_2 & \leq 3y_1 \\
y_1+y_2 & \leq 1 \\
0 \leq x_1,x_2, & \leq 4 \\
y_1,y_2 & \in \{0,1\}
\end{align*}
Choose initial integer solutions $y_1 = y_2 = 1$. Upper bound $U = \infty$.
\end{column}
\begin{column}{0.5\textwidth}
\begin{tikzpicture}
% Place nodes
\node [blockH] (init) {Init};
\node [block,below=of init] (NLP) {Solve NLP($x^k_I$) or F($x^k_I$)};
\node [block,below=of NLP] (Bound) {Add linearaziation at $x^k$ to MIP};
\node [block,below=of Bound] (MIP) {Solve MIP($x^k$, $U$)};
\node [decision,right=of MIP] (Infeas){MIP infeasible? -> Exit};
%
\path[line] (init) edge (NLP)
(NLP) edge node {Update U?} (Bound)
% (Bound) edge (Update)
(Bound) edge (MIP)
(MIP) edge (Infeas)
(Infeas) |- node[near start] {No} (NLP);
\end{tikzpicture}
\end{column}
\end{columns}
\end{frame}
%Step 2
\begin{frame}
\begin{columns}[onlytextwidth]
\begin{column}{0.48\textwidth}
Fix $y_1 = y_2 = 1$\\
and solve NLP.
\begin{align*}
\leq z \\
1 + 1 + x_1^2 + x_2^2 & \leq z \\
(x_1-2)^2-x_2 & \leq 0 \\
x_1 - 2 & \geq 0 \\
\hphantom{_1}
x_1-x_2-3(1-1) & \leq 0 \\
x_1 - (1-1) & \geq 0 \\
x_2-1 & \leq 0 \\
x_1+x_2 & \leq 3\\
1+1 & \leq 1\\
0 \leq x_1,x_2, & \leq 4\\
1,1 & \in \{0,1\}
\end{align*}
Optimal solution $x_1=x_2=2$, (x,y) = (2,2,1,1) is a solution
to our $MINLP$.
\end{column}
\begin{column}{0.5\textwidth}
\begin{tikzpicture}
% Place nodes
\node [blockH] (init) {Init};
\node [block, below=of init, node distance=3cm] (NLP) { Solve NLP($x^k_I$) or F($x^k_I$)};
\node [block, below=of NLP] (Bound) { Add linearaziation at $x^k$ to MIP};
\node [block, below=of Bound] (MIP) { Solve MIP($x^k$, $U$) };
\node [decision, right=of MIP] (Infeas) { MIP infeasible? -> Exit };
%
\path [line] (init) -- (NLP);
\path[line] (init) edge (NLP)
(NLP) edge node {Update U?} (Bound)
% (Bound) edge (Update)
(Bound) edge (MIP)
(MIP) edge (Infeas)
(Infeas) |- node[near start] {No} (NLP);
\end{tikzpicture}
\end{column}
\end{columns}
\end{frame}
\end{document}
上面的代码给出：
