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使用 tikz 制作的混乱流程图

tag-icon • tag-icon tikz-pgf beamer
使用 tikz 制作的混乱流程图

我有一张投影幻灯片：

\documentclass{beamer}

\usepackage{tikz}
\usetikzlibrary{shapes,arrows}

\tikzstyle{decision} = [diamond, draw, fill=blue!20, text width=4.5em, 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{line} = [draw, -latex]

\begin{document}

\begin{frame}{Simulated Annealing (SA)}
  \begin{center}
    \resizebox{0.4 \linewidth}{!}{%
      \begin{tikzpicture}[node distance = 2cm, auto]
        \node[block]                  (init){Init $n=0$, $T_0$, and $S_0$};
        \node[block, below of=init]   (nbrh){$S_{n+1}=N(S_n)$};
        \node[decision, below of=nbrh](ovgt){$f(S_{n+1}) \le f(S_n)$};
        \node[block, below of=ovgt]   (accp){Accept $S_{n+1}$};
        \node[decision, right of=ovgt](rand){Accept with $P = e^{-\frac{\Delta f}{t_n}}$};
        \node[block, right of=nbrh, anchor=west]   (rejj){Reject $S_{n+1}$};
        \node[block, below of=accp]   (incr){$T_{n+1} = K(T_n)$ and $n=n+1$};
        \node[block, below of=incr]   (stop){Stop};
        \node[decision, left of=stop] (stcd){Stop?};

        \path[line] (init) --          (nbrh);
        \path[line] (nbrh) --          (ovgt);
        \path[line] (ovgt) -- node{yes}(accp);
        \path[line] (ovgt) -- node{no} (rand);
        \path[line] (rand) -- node{no} (rejj);
        \path[line] (rejj) --          (nbrh);
        \path[line] (rand) |- node{yes}(accp);
        \path[line] (accp) --          (incr);
        \path[line] (incr) --          (stcd);
        \path[line] (stcd) -- node{yes}(stop);
        \path[line] (stcd) |- node{no} (nbrh);
      \end{tikzpicture}%
    }%
  \end{center}
\end{frame}

\end{document}

这使：

捕捉精彩瞬间

我对输出结果有些不满意：

  1. 简单的方程式$S_{n+1}=N(S_n)%应该在一行上。文本可以小一点。同样，$T_{n+1}=K(T_n)$应该在一行上，'和' 可以单独成行，也$n=n+1$应该单独成行。判决中的不平等也应该在同一行。

  2. 连接右侧“接受”决策和“拒绝”操作的垂直“否”线不是完全垂直的。此外，此决策比左侧的决策更大。

  3. 这两个决定之间的差距很小。

  4. 不等式判定下的“接受”操作与上面的判定重叠。相反，该判定与上面的操作相距甚远。

  5. 最后，我对底部的两个元素不满意。我更希望将“停止？”决定直接放在下方，将“停止”操作放在右侧。不幸的是，我不知道如何在“停止？”的“否”中放置 2 个弯角，这意味着该线实际上会穿过上面的所有内容。我试过了，|-|但-|-不起作用。

我之前问过一个问题这里关于将整个 tikzpicture 缩放到幻灯片的高度，但是提出的解决方案对我来说不起作用。它仍然在底部被切断。问题可能与我的实际 beamer 演示文稿使用 Hannover 主题有关，因此幻灯片中可用的空间较少？我已经通过0.4 \linewidth临时使用解决了这个问题。

答案1

我的解决方案与 Altermundus 的上次编辑类似（我太慢了），但我解决了仅在一行中包含方程式的问题。此外，\resizebox我没有采用，而是更喜欢\scalebox从graphicx包中获取。

以下是代码：

\documentclass{beamer}
\usepackage{lmodern}
\usepackage{graphicx}

\usepackage{tikz}
\usetikzlibrary{shapes,arrows,positioning}

\tikzset{decision/.style={diamond, draw, fill=blue!20, text width=4.5em, text badly centered, inner sep=0pt}}
\tikzset{block/.style={rectangle, draw, fill=blue!20, text width=5em, text centered, rounded corners,
 minimum width=3.5cm}}
\tikzset{line/.style={draw, -latex}}

\begin{document}

\begin{frame}{Simulated Annealing (SA)}
\begin{center}
\scalebox{0.6}{
      \begin{tikzpicture}[node distance=2.3cm]
        \node[block]                  (init){Init $n=0$, $T_0$, and $S_0$};
        \node[block, below of=init]   (nbrh){\footnotesize{$S_{n+1}=N(S_n)$}};
        \node[decision, below of=nbrh](ovgt){\footnotesize{$f(S_{n+1}) \le f(S_n)$}};
        \node[block, below of=ovgt]   (accp){Accept $S_{n+1}$};
        \node[decision, right=2cm of ovgt](rand){Accept with $P = e^{-\frac{\Delta f}{t_n}}$};
        \node[block,  above of=rand]   (rejj){Reject $S_{n+1}$};
        \node[block, below of=accp]   (incr){\scriptsize{$T_{n+1} = K(T_n)$} \\[1ex] and\\ \scriptsize{$n=n+1$}};       
        \node[decision, below of=incr] (stcd){Stop?};
        \node[block, right= 2cm of stcd]   (stop){Stop};
        % invisible node helpful later
        \node[left=1cm of accp,scale=0.05](inv){};

        \path[line] (init) --          (nbrh);
        \path[line] (nbrh) --          (ovgt);
        \path[line] (ovgt) -- node[left]{yes}(accp);
        \path[line] (ovgt) -- node[above]{no} (rand);
        \path[line] (rand) -- node[right]{no} (rejj);
        \path[line] (rejj) --          (nbrh);
        \path[line] (rand) |- node[below]{yes}(accp);
        \path[line] (accp) --          (incr);
        \path[line] (incr) --          (stcd);
        \path[line] (stcd) -- node[below]{yes}(stop);
       \path[-,draw] (stcd) -| node{} (inv.north);
       \path[line]{} (inv.north) |- node[above]{no} (nbrh);
      \end{tikzpicture}%
}      
\end{center}
\end{frame}

\end{document}

这使：

在此处输入图片描述

答案2

如果使用该 positioning库，则可以避免一些问题。您需要使用选项on grid，您可以局部修改node distance。菱形节点的宽度与矩形的宽度不同，因此如果您想要垂直边缘，则需要将矩形放置在菱形上方。

\documentclass{scrartcl}

\usepackage{tikz}
\usetikzlibrary{shapes,arrows,positioning}

\tikzstyle{decision} = [diamond, draw, fill=blue!20, text width=4.5em, text badly centered, node distance=3cm, inner sep=0pt,on grid]
\tikzstyle{block} = [rectangle, draw, fill=blue!20, text width=5em, text centered, rounded corners, minimum height=4em,on grid]
\tikzstyle{line} = [draw, -latex]

\begin{document}

  \begin{center}
    \resizebox{0.4 \linewidth}{!}{%
      \begin{tikzpicture}[node distance = 2cm, auto]
        \node[block]                                (init) {Init $n=0$, $T_0$, and $S_0$};
        \node[block, below= of init]                (nbrh) {$S_{n+1}=N(S_n)$};
        \node[decision, below= of nbrh]             (ovgt) {$f(S_{n+1}) \le f(S_n)$};
        \node[block, below=2.5cm of ovgt]           (accp) {Accept $S_{n+1}$};
        \node[decision, right= 3.5cm of ovgt]       (rand) {Accept with $P = e^{-\frac{\Delta f}{t_n}}$};
        \node[block, above=3cm of rand]             (rejj) {Reject $S_{n+1}$};
        \node[block, below= of accp]                (incr) {$T_{n+1} = K(T_n)$ and $n=n+1$};
        \node[decision, below=2.5cm of incr]              (stcd) {Stop?};  
        \node[block, right=3cm of stcd]                (stop) {Stop};


        \path[line] (init) --          (nbrh);
        \path[line] (nbrh) --          (ovgt);
        \path[line] (ovgt) -- node{yes}(accp);
        \path[line] (ovgt) -- node{no} (rand);
        \path[line] (rand) -- node{no} (rejj);
        \path[line] (rejj) --          (nbrh);
        \path[line] (rand) |- node{yes}(accp);
        \path[line] (accp) --          (incr);
        \path[line] (incr) --          (stcd);
        \path[line] (stcd) -- node{yes}(stop);
        \path[line] (stcd) -- ++(-2,0) |- node[pos=.25]{no}  (nbrh);
      \end{tikzpicture}%
    }%
  \end{center}


\end{document}

在此处输入图片描述

答案3

您可能会发现使用\matrix[matrix of nodes]很容易，因为这样您就可以移动节点。这样就剩下绘制路径了，无论如何这都是很无聊的。

为了禁止数学模式中的换行，您可以将\binoppenalty和设置\relpenalty为高值。

上面的例子（我在绘制路径时删减了一些角落）：

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{matrix,shapes,arrows}

\tikzset{
  % re-usability is the key to happiness
  >=triangle 45,
  flowchart/.style={
    matrix of nodes,
    nodes=block,
    row sep=2\normalbaselineskip,
    column sep=3em,
  },
  block/.style={
    %font=\scriptsize, % <- that'd change font size
    draw,
    fill=blue!20,
    minimum height=3\normalbaselineskip,
    text width=7em,
    text badly centered,
    rounded corners,
    execute at begin node={
      \hskip0pt
      \binoppenalty=10000 % make it bad to break math line
      \relpenalty=10000},
  },
  decision/.style={% inherits from block via matrix
    diamond,
    sharp corners,
    inner sep=1pt,
  }
}

\begin{document}
\begin{tikzpicture}
  \matrix[flowchart] (fc) {
    Init $n=0$, $T_0$, and $S_0$ \\
    $S_{n+1}=N(S_n)$ & Reject $S_{n+1}$ \\
    |[decision]| $f(S_{n+1})\leq f(S_n)$ &
    |[decision]| Accept with $P=e^{-\frac{\Delta f}{t_n}}$ \\
    Accept $S_{n+1}$ \\
    $T_{n+1}=K(T_n)$ and $n=n+1$ \\
    |[decision]| Stop? & Stop \\
  };
  \begin{scope}[->]
    \draw (fc-1-1) -- (fc-2-1) edge (fc-3-1);
    \draw (fc-2-1) -- (fc-3-1) edge node[above] {no} (fc-3-2);
    \draw (fc-3-2) edge node[right] {no} (fc-2-2) |- node[below] {yes} (fc-4-1);
    \draw (fc-2-2) -- (fc-2-1);
    \draw (fc-3-1) -- node[right] {yes} (fc-4-1);
    \draw (fc-4-1) -- (fc-5-1) edge (fc-6-1);
    \draw (fc-6-1) edge node[above] {yes} (fc-6-2) -- +(-3,0) node[below] {no} |- (fc-2-1);
  \end{scope}
\end{tikzpicture}
\end{document}

在此处输入图片描述

（第一个无分支有一个小缺陷，我将其留作练习;-)。）

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