用箭头上的一些方程来绘制依赖关系

Question

我想用 TIKZ 重现下图。节点t_i依赖于其相邻节点，因此在时间轴上距离越远，依赖性就越弱。

正如您在下图中看到的，我已经重现了图形的结构，但我不知道如何在箭头上建立方程。

   \documentclass[margin=5mm]{standalone}
  \usepackage{tikz}
  \usetikzlibrary{fit,positioning,arrows,automata,calc}
   \tikzset{
   main/.style={circle, minimum size = 10mm, thick, draw =black!80, node 
   distance = 10mm},
    box/.style={rectangle, draw=black!100}
   }
   \begin{document}
   \begin{tikzpicture}
   % put nodes
   \node[main,draw =red!80] (t1)   {$t_i$};
   \node[main] (t2) [right= of t1] {$t_{i+1}$};
   \node[main] (t3) [right= of t2] {$t_{i+2}$};
   \node[main] (tn) [right= of t3] {$t_{n}$};
   \node[main] (t)  [left=  of t1] {$t_{i-1}$};
   \node[main] (tt) [left=  of t]  {$t_{i-2}$};
   \node[main] (t0) [left=  of tt] {$t_{0}$};

   % make path ...
   \path (t3) -- node[auto=false]{\ldots} (tn);
   \path (tt) -- node[auto=false]{\ldots} (t0);

   % draw arrows
   \draw [->] (t1) to [out=45,in=135] (tn) node [midway,below]{$P(k=0|\lambda=i)$}; 
   \draw [->] (t1) to [out=35,in=135] (t3);
   \draw [->] (t1) to (t2);
   \draw [->] (t1) to (t);
   \draw [->] (t1) to [out=135,in=35] (tt);
   \draw [->] (t1) to [out=135,in=35] (t0);
   \end{tikzpicture} 
   \end{document}

因此，如果您有任何建议可以让这张图形更加漂亮（结构），请不要犹豫。

谢谢

编辑1 我按照@Ignasi 的建议修改了代码，得到了以下结果：

有没有一种简单的方法可以防止方程式交叉箭头，如图所示，箭头 t_i 到 t_ {i + 1}？

     \draw [->] (t1) to [out=45,in=135] node [rotate=-5,above]{\tiny 
      $P(k=n\mid\lambda=1)$} (tn);
     \draw [->] (t1) to [out=35,in=135] node [rotate=-5,above]{\tiny 
      $P(k=i+1\mid\lambda=1)$}(t3) ;

编辑2t_i根据@Ignasi的建议，我得到了以下图片。但是，我如何在和 t_ {i + 1}（t_i和） 之间添加等式t_{i-1}？

修改后的代码如下：

           \documentclass[margin=5mm]{standalone}
           \usepackage{tikz}
           \usetikzlibrary{fit,positioning,arrows,automata,calc}
           \tikzset{
            main/.style={circle, minimum size = 10mm, thick, draw                =black!80, node distance = 10mm},
            box/.style={rectangle, draw=black!100}
           }
           \begin{document}
           \begin{tikzpicture}
              % put nodes
              \node[main,draw =red!80] (t1)   {$t_i$};
              \node[main] (t2) [right= of t1] {$t_{i+1}$};
              \node[main] (t3) [right= of t2] {$t_{i+2}$};
              \node[main] (tn) [right= of t3] {$t_{n}$};
              \node[main] (t)  [left=  of t1] {$t_{i-1}$};
              \node[main] (tt) [left=  of t]  {$t_{i-2}$};
              \node[main] (t0) [left=  of tt] {$t_{0}$};

              % make path ...
              \path (t3) -- node[auto=false]{\ldots} (tn);
              \path (tt) -- node[auto=false]{\ldots} (t0);

              % draw arrows
              \draw [->] (t1) to [out=45,in=135] node [midway, above]               {$P(k=n|\lambda=i)$} (tn); 
              \draw [->] (t1) to [out=-45,in=-135] node [midway, below]               {$P(k=i+2|\lambda=i)$}(t3);
              \draw [->] (t1) to node[midway, below] {$P(k=i+1|\lambda=i)$}               (t2);
              \draw [->] (t1) to node[midway, above] {$P(k=i-1|\lambda=i)$}(t);
              \draw [->] (t1) to [out=-135,in=-45] node [midway, below]{$P(k=i-2|\lambda=i)$}(tt);
              \draw [->] (t1) to [out=135,in=35] node [midway, above]{$P(k=0|\lambda=i)$}(t0);
           \end{tikzpicture} 
           \end{document}

Answer 1

可以分发链接以避免方程式重叠。

\documentclass[margin=5mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{fit,positioning,arrows,automata,calc}
\tikzset{
   main/.style={circle, minimum size = 10mm, thick, 
        draw =black!80, node distance = 10mm},
   box/.style={rectangle, draw=black!100}
}
\begin{document}
\begin{tikzpicture}
   % put nodes
   \node[main,draw =red!80] (t1)   {$t_i$};
   \node[main] (t2) [right= of t1] {$t_{i+1}$};
   \node[main] (t3) [right= of t2] {$t_{i+2}$};
   \node[main] (tn) [right= of t3] {$t_{n}$};
   \node[main] (t)  [left=  of t1] {$t_{i-1}$};
   \node[main] (tt) [left=  of t]  {$t_{i-2}$};
   \node[main] (t0) [left=  of tt] {$t_{0}$};

   % make path ...
   \path (t3) -- node[auto=false]{\ldots} (tn);
   \path (tt) -- node[auto=false]{\ldots} (t0);

   % draw arrows
   \draw [->] (t1) to [out=45,in=135] node [midway, above]{$P(k=n|\lambda=i)$} (tn); 
   \draw [->] (t1) to [out=-45,in=-135] node [midway, below]{$P(k=i+2|\lambda=i)$}(t3);
   \draw [->] (t1) to (t2);
   \draw [->] (t1) to (t);
   \draw [->] (t1) to [out=-135,in=-45] node [midway, below]{$P(k=i-2|\lambda=i)$}(tt);
   \draw [->] (t1) to [out=135,in=35] node [midway, above]{$P(k=0|\lambda=i)$}(t0);
\end{tikzpicture} 
\end{document}

编辑2的答案：

t增加和侧状态之间的距离：

\documentclass[margin=5mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{fit,positioning,arrows,automata,calc}
\tikzset{
   main/.style={circle, minimum size = 10mm, thick, 
        draw =black!80, node distance = 10mm},
   box/.style={rectangle, draw=black!100}
}
\begin{document}
\begin{tikzpicture}
   % put nodes
   \node[main,draw =red!80] (t1)   {$t_i$};
   \node[main] (t2) [right= 3cm of t1] {$t_{i+1}$};
   \node[main] (t3) [right= of t2] {$t_{i+2}$};
   \node[main] (tn) [right= of t3] {$t_{n}$};
   \node[main] (t)  [left= 3cm of t1] {$t_{i-1}$};
   \node[main] (tt) [left=  of t]  {$t_{i-2}$};
   \node[main] (t0) [left=  of tt] {$t_{0}$};

   % make path ...
   \path (t3) -- node[auto=false]{\ldots} (tn);
   \path (tt) -- node[auto=false]{\ldots} (t0);

   % draw arrows
   \draw [->] (t1) to [out=45,in=135] node [midway, above]{$P(k=n|\lambda=i)$} (tn); 
   \draw [->] (t1) to [out=-45,in=-135] node [midway, below]{$P(k=i+2|\lambda=i)$}(t3);
   \draw [->] (t1) to  node [midway, above] {$P(k=i+1|\lambda=i)$}(t2);
   \draw [->] (t1) to  node [midway, above] {$P(k=i-1|\lambda=i)$} (t);
   \draw [->] (t1) to [out=-135,in=-45] node [midway, below]{$P(k=i-2|\lambda=i)$}(tt);
   \draw [->] (t1) to [out=135,in=35] node [midway, above]{$P(k=0|\lambda=i)$}(t0);
\end{tikzpicture} 
\end{document}

Answer 2

\documentclass[margin=5mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{fit,positioning,arrows,automata,calc}
\tikzset{
  main/.style={circle, minimum size = 10mm, thick, draw =black!80, node 
    distance = 10mm},
  box/.style={rectangle, draw=black!100}
}
\begin{document}
\begin{tikzpicture}
  % put nodes
  \node[main,draw =red!80] (t1)   {$t_i$};
  \node[main] (t2) [right= of t1] {$t_{i+1}$};
  \node[main] (t3) [right= of t2] {$t_{i+2}$};
  \node[main] (tn) [right= of t3] {$t_{n}$};
  \node[main] (t)  [left=  of t1] {$t_{i-1}$};
  \node[main] (tt) [left=  of t]  {$t_{i-2}$};
  \node[main] (t0) [left=  of tt] {$t_{0}$};

  % make path ...
  \path (t3) -- node[auto=false]{\ldots} (tn);
  \path (tt) -- node[auto=false]{\ldots} (t0);

  % draw arrows
  \draw [->] (t1) to [out=75,in=135]  node [midway,above]{$P(k=n|\lambda=i)$} (tn); 
  \draw [->] (t1) to [out=35,in=135] node [midway,above]{$P(k=i + 2|\lambda=i)$}  (t3);
  \draw [->] (t1) to (t2);
  \draw [->] (t1) to (t);
  \draw [->] (t1) to [out=145,in=45] node [midway,above]{$P(k= i- 2|\lambda=i)$} (tt);
  \draw [->] (t1) to [out=105,in=45] node [midway,above]{$P(k=0|\lambda=i)$}  (t0);
\end{tikzpicture} 
\end{document}