TikZ：用于前置语法的路径式语法倒置树

Question

这当然还不是一个完整的答案，只是想告诉你，我不同意你对这个软件包的怀疑。在我看来，这是一个手册写得非常好的软件包，这就是为什么我相信你可以用它实现你想要的东西。这个答案的关键要素是一个宏

\Connect{<start>}{<end>}{<level>}{<label>}

在哪里start，end 和 level都是整数。其用法由以下 MWE 说明。

\documentclass[border=3.14mm,standalone]{standalone}
\usepackage{mathtools}
\usepackage{tikz-dependency}
\usetikzlibrary{intersections} % not used ... YET
\newcounter{depaths}
\newcommand{\Connect}[5][]{
\stepcounter{depaths}
\draw[name path=dep connect \thedepaths](\wordref{2}{#2}) -- ++ (0,-#4) coordinate (aux-\thedepaths-1)  -| 
(\wordref{2}{#3}) coordinate[pos=0.5] (aux-\thedepaths-2);
\draw[-latex,shorten >=1pt] (aux-\thedepaths-1) -- (aux-\thedepaths-2);
\ifnum#3>#2
\node[anchor=south east] at (aux-\thedepaths-2) {#5};
\else
\node[anchor=south west] at (aux-\thedepaths-2) {#5};
\fi
\foreach \X in {1,...,#4}
{\fill (\wordref{2}{#2}) ++ (0,-\X) circle(1.5pt); 
\fill (\wordref{2}{#3}) ++ (0,-\X) circle(1.5pt);}
\pgfmathtruncatemacro{\DeltaX}{abs(#3-#2)}
\ifnum\DeltaX>1
\foreach \X in {#2,...,#3}
\fill (\wordref{2}{\X}) ++ (0,-#4) circle(1.5pt);
\fi
}
\begin{document}
\begin{dependency}
\begin{deptext}[column sep=1em]
the \& girl \& for \& wh- \& om \& the \& man \& bought \& the \& book\\
A \& O \& U \& U \& O \& A \& O \& I \& A \& O\\
\end{deptext}
\Connect{1}{2}{1}{$\overleftarrow{O}$}
\Connect{4}{2}{6}{$\overleftarrow{O}$}
\Connect{4}{3}{1}{$\overleftarrow{O}$}
\end{dependency}

\end{document}

附录：这是一实验版本在交叉点上绘制圆弧。如您所见，有相当多的\typeouts，这远非完美，但也许是一个开始。更准确地说，只要满足以下条件之一，就不会绘制圆弧：

第一条路径的起点和终点就是第二条路径的起点和终点。
第一条路径的结束位置与第二条路径的开始或结束位置相同。
路径具有相同的级别（=宏的第三个参数\Connect{4}{3}{1}）。

很容易看出，如果满足任何一个条件，就很可能不会有弧，但以下两条路径也不会有弧：

******************
 |   | |
 *---*-*
     | |
     *-*

虽然很麻烦，但也可以弥补这个漏洞。这取决于这是否会真的也习惯于某个地方，我可能会获得足够的动力来做到这一点，也使宏更加Ti钾Zy，即与 pgfkeys 一起工作。

\documentclass[border=3.14mm,standalone]{standalone}
\usepackage{mathtools}
\usepackage{tikz-dependency}
\usetikzlibrary{intersections} 
\newcounter{depaths}
\newif\ifIntersect
\tikzset{every picture/.append style={execute at begin picture={%
\xdef\LstPaths{}% may be unnecessary
\setcounter{depaths}{0}}}}
\newcommand{\Connect}[5][]{
\stepcounter{depaths}%\typeout{new\space path\space\thedepaths}
\draw[name path=dep connect \thedepaths-tmp](\wordref{2}{#2}) -- ++ (0,-#4) coordinate (aux-\thedepaths-1)  -| 
(\wordref{2}{#3}) coordinate[pos=0.5] (aux-\thedepaths-2);
\draw[-latex,shorten >=1pt] (aux-\thedepaths-1) -- (aux-\thedepaths-2);
\ifnum#3>#2
  \node[anchor=south east] at (aux-\thedepaths-2) {#5};
\else
  \node[anchor=south west] at (aux-\thedepaths-2) {#5};
\fi
\foreach \X in {1,...,#4}
{\fill (\wordref{2}{#2}) ++ (0,-\X) circle(1.5pt); 
\fill (\wordref{2}{#3}) ++ (0,-\X) circle(1.5pt);}
\pgfmathtruncatemacro{\DeltaX}{abs(#3-#2)}
\ifnum\DeltaX>1
\foreach \X in {#2,...,#3}
\fill (\wordref{2}{\X}) ++ (0,-#4) circle(1.5pt);
\fi
\ifnum\thedepaths>1  
  \foreach \Y [count=\X] in \LstPaths
  {\Intersectfalse
  %\typeout{checking\space intersection\space with\space path\space \X}
  \path[name intersections={of=dep connect \thedepaths-tmp and dep
  connect \X-tmp,total=\t}] \pgfextra{%\typeout{\t:\Y}
  \ifnum\t>0
  \global\Intersecttrue
  %\typeout{paths\space\X\space and\space\thedepaths \space\space have\space common\space points}
  \fi};  
  \foreach \V [count=\W] in \Y
  {\ifcase\W % no 0
  \or % 1 boring
  \or % 2
   \ifnum\V=#2
    %\typeout{case2-2}
    \global\Intersectfalse
   \fi
   \ifnum\V=#3
    %\typeout{case2-3}
    \global\Intersectfalse
   \fi
  \or
   \ifnum\V=#3
    %\typeout{case3-3}
    \global\Intersectfalse
   \fi
   \ifnum\V=#2
    %\typeout{case3-2}
    \global\Intersectfalse
   \fi
  \or
   \ifnum\V=#4
    %\typeout{case4}
    \global\Intersectfalse
   \fi
  \fi}  
  \ifIntersect
   \node[circle,fill=white,outer sep=0pt,minimum size=10pt] (aux-int) at (intersection-1){};
   \draw (aux-int.south) -- (aux-int.north);
   \draw[white,double=black] (aux-int.west) to[out=60,in=120] (aux-int.east);
   %\typeout{real\space intersection\space of\space path\space \thedepaths\space with\space path\space \X}
  \fi
  }
\fi
\ifnum\thedepaths=1
  \xdef\LstPaths{{\thedepaths,#2,#3,#4}}
\else
  \xdef\LstPaths{\LstPaths,{\thedepaths,#2,#3,#4}}
\fi
}
\begin{document}
\begin{dependency}
\begin{deptext}[column sep=1em]
the \& marmot \& for \& wh- \& om \& the \& duck \& bought \& the \& cake\\
A \& O \& U \& U \& O \& A \& O \& I \& A \& O\\
\end{deptext}
\Connect{1}{2}{1}{$\overleftarrow{O}$}
\Connect{4}{2}{6}{$\overleftarrow{O}$}
\Connect{3}{8}{4}{$\overleftarrow{3}$}
\Connect{5}{3}{1}{$\overleftarrow{O}$}
\end{dependency}
\end{document}

Answer 1