从一个矩阵节点绘制到其他节点的箭头，并在其上方标注标签（Beamer）

Question

我有一张带有单行矩阵的幻灯片，我希望它以以下方式出现：首先出现中间的元素，周围有一个框；然后它左边和右边的元素同时出现，自然的箭头从中心元素指向它们两个，一些值写在箭头上方或下方（我会根据它们的外观选择上方或下方）。

目前，我有一些这样的代码，但有几个问题：1) 整行都应该居中，但实际上没有；2) 只显示一个箭头；3) 文本没有出现在正确的位置；4) 箭头太靠下。我希望它们离行近一点，现在看起来空隙很大。

\documentclass{beamer}
\mode<presentation>
{
  \setbeamertemplate{navigation symbols}{}
  \setbeamertemplate{caption}[numbered]
}
%%%%%%%%%
\usepackage{tikz}
\usetikzlibrary{fit,overlay-beamer-styles,tikzmark, matrix}
\usepackage{tcolorbox}

\begin{document}
\section{Proofs}         


\begin{frame}[fragile]
\frametitle{title}

\begin{tcolorbox}[left = .8 mm, right=.8 mm]
Idea: here is the idea.
\end{tcolorbox}

\begin{tikzpicture}[optimization problem/.style = {%
            rounded corners, 
            draw = green,
            thick,
            fill= white,
            inner ysep=5pt,
            inner xsep=5pt,
            align = center},
            row 1 column 2/.style={visible on = <2->}, 
            row 1 column 1/.style = {visible on = <3->}, 
            row 1 column 3/.style = {visible on =<3->}
            ]

\matrix[matrix of nodes,row sep=1em,column sep=1em,
    nodes={anchor=west}
    ](M){
$f$ & |[alt=<2>{draw=green,thick,fill=white, rounded corners}{}]|$g$ & $r$\\
};  
    \draw[->, thick, visible on=<3->] (M-1-2)[out=-90,in=-90,loop,looseness=1] (M-1-1) node[midway, above] {$t=1$}; 
   \draw[->, thick, visible on=<3->] (M-1-2) to [out=-90,in=-90,loop,looseness=1] (M-1-3) node[midway, below] {$t=0$}; 
\end{tikzpicture}
\vspace{2em}

\onslide<4->{ 
If the following inequality holds: \[a \leq b,\] then combine with result 1 + result 2 + result 3 to finish the proof.}

\vspace{2em}
\onslide<5->{
However, this is {\color{red}FALSE}; counter-example:
\[1 + 1 = 2.\]}
\end{frame}
\end{document}

Answer 1

至于您的问题：

1. 您可以使用\hfill来使 居中tikzpicture。
2. 你忘了to，不应该用于loop从 A 到 B 的箭头，如果你想沿着弯曲的路径放置一个节点，你需要把它前目标^。1

3. 要仅在特定覆盖层上获取框架，您可以使用

   |[draw=green,thick,fill=white, rounded corners,alt=<2>{opacity=1}{opacity=0,text opacity=1}]|

我还在代码中添加了一些注释。

\documentclass{beamer}
\mode<presentation>
{
  \setbeamertemplate{navigation symbols}{}
  \setbeamertemplate{caption}[numbered]
}
%%%%%%%%%
\usepackage{tikz}
\usetikzlibrary{fit,overlay-beamer-styles,tikzmark, matrix}
\usepackage{tcolorbox}

\begin{document}
\section{Proofs}         


\begin{frame}[fragile]
\frametitle{title}

\begin{tcolorbox}[left = .8 mm, right=.8 mm]
Idea: here is the idea.
\end{tcolorbox}

\hfill
\begin{tikzpicture}[optimization problem/.style = {%
            rounded corners, 
            draw = green,
            thick,
            fill= white,
            inner ysep=5pt,
            inner xsep=5pt,
            align = center},
            row 1 column 2/.style={visible on = <2->}, 
            row 1 column 1/.style = {visible on = <3->}, 
            row 1 column 3/.style = {visible on =<3->}
            ]

\matrix[matrix of nodes,row sep=1em,column sep=1em,
    nodes={anchor=west}
    ](M){
$f$ & |[draw=green,thick,fill=white, rounded
corners,alt=<2>{opacity=1}{opacity=0,text opacity=1}]|$g$ & $r$\\
};  
    \draw[->, thick, visible on=<3->] 
    (M-1-2) to[out=90,in=90,looseness=1] node[midway, above] {$t=1$}
    (M-1-1) ;  % target node needs to be last, no loop & you forgot "to"
   \draw[->, thick, visible on=<3->] (M-1-2) to [out=-90,in=-90,looseness=1]
     node[midway, below] {$t=0$} (M-1-3);  % target node needs to be last, no loop
\end{tikzpicture}\hfill\mbox{}\par
\vspace{2em}

\onslide<4->{ 
If the following inequality holds: \[a \leq b,\] then combine with result 1 + result 2 + result 3 to finish the proof.}

\vspace{2em}
\onslide<5->{
However, this is \textcolor{red}{FALSE}; counter-example:
\[1 + 1 = 2.\]}
\end{frame}
\end{document}

_{¹钛钾Z 对于直线路径非常宽容，你可以把node[pos=0.5]{...} 后目标。我有时认为，如果你总是需要把它放在前面，那会更容易……}