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是否有一种巧妙的方法可以自动绘制/标记框图中多条路径的交叉点？

tag-icon • tag-icon tikz-pgf diagrams tkz-euclide intersections
是否有一种巧妙的方法可以自动绘制/标记框图中多条路径的交叉点？

我正在尝试使用 tikz 绘制框图，并希望匹配有关线/路径交叉的某种约定。约定规定，标记/半圆必须位于垂直运行的线上。为此，我编辑了一个命令重叠满足我的需求。使用\overlap{master path}{slave path}从属路径将在与主路径的交叉处得到一个半圆。

在我的 MWE 中，该命令有效，但无法区分主/从线在交叉点是水平还是垂直运行。

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,matrix,positioning}
\usetikzlibrary{arrows}
\usetikzlibrary{intersections}
%
\usepackage{tkz-euclide}
%
\newcommand{\overlap}[2]{
    \path[name intersections={of= #1 and #2,sort by=#2, name=punkt, total=\t},/utils/exec={\global\let\t=\t}];
    \ifnum \t>0
    \foreach \s in {1,...,\t}{          
        % circle for intersections
        \path[name path global/.expanded =whitecirc-\s,draw=none,line width=0pt]
        (punkt-\s)circle(5.6705pt);
        
        % name intersections
        \path[name intersections={of=#2 and whitecirc-\s,sort by=#2,name=wb,total=\numwb},
        /utils/exec={\global\let\numwb=\numwb}];
        \ifnum \numwb>1

        % inner white circle
        \path[draw=white,fill=white, line width=0.2pt]
        (punkt-\s)circle(5.35pt); 
        
        % intersections if circle with other line
        \path[name intersections={of=#1 and whitecirc-\s,sort by=#2,name=wa}];

        % re-draw 2nd line (slave path)
        \draw[shorten <=-2pt,shorten >=-2pt](wa-1.center)--(wa-2.center);

        % draw arc
        \coordinate (A) at (wb-2);
        \coordinate (B) at (wb-1);
        \tkzInterCC[R](A,2.4mm)(B,2.4mm)
        \tkzGetPoints{X}{Y}
        \coordinate (Z) at ($(X)!0.5!(Y)$);

        % thicker white arc for hiding small part of slave path
        \tkzDrawArc[color=white,ultra thick, line cap=butt,shorten <=0.4pt,shorten >=0.4pt](Z,B)(A)
        % normal Arc (master path)
        \tkzDrawArc[color=black,line cap=butt](Z,B)(A)
        \fi
    }
    \fi
}
%
\begin{document}
%
\begin{tikzpicture}[scale=1.0,>=latex']
\draw[fill=white] (0,0) rectangle (6,9);
%
\foreach \i in {3,4,5} {
\coordinate (w\i) at (1,\i); \coordinate (e\i) at (5,\i);
\draw[->,name path=path\i] (w\i) --(e\i);
}
%
\foreach \i in {1,2,6,7,8} {
\node[circle,draw,align=center,fill=black,inner sep=0pt, minimum size = 3pt] (w\i) at(1,\i) {};
}
%
\coordinate (h1) at ($(w3)!0.4!(e3)$);\coordinate (h2) at ($(w3)!0.6!(e3)$);
\draw[->,name path=path2] (w2) -| (4,6) |- (2,6) |- (1,8);
\draw[->,name path=path1] (w1) -| (3,5) |- (w7);
%
\overlap{path2}{path1}
\overlap{path3}{path1}
\overlap{path4}{path1}
\overlap{path5}{path1}
%
\overlap{path3}{path2}
\overlap{path4}{path2}
\overlap{path5}{path2}
%
\node [draw,fill=red,fill opacity=0.4,rectangle,minimum size=1cm,anchor=center] at (2,7) {};
%
\end{tikzpicture}
\end{document}

不区分垂直和水平主/从段

总的来说，我不知道是否有更聪明的方法来自动重绘交叉点. 假设命名每条路径path1,path2,path3,...,pathN并搜索 2 个\foreach循环内的每个交叉点可能是一种方法。

\foreach \x in {1,2,3,4} 
\foreach \y in {2,3,4,5} 
    {
        \if\ifnum\x>\y T\else\ifnum\x=\y T\else F\fi\fi T%
        %  TRUE
        \break foreach
        \else
        %  FALSE
        \overlap{path\y}{path\x};
        \fi
    }

但是使用这个循环会导致另一个问题，即让圆圈随机翻转。 使用 foreach 循环时翻转半圆

答案1

有几种方法shorten，例如postaction知道交点后使用。

在此处输入图片描述

\documentclass[tikz,border=5mm]{standalone}
\begin{document}
\begin{tikzpicture}[>=stealth]
\def\r{.35}
\draw[fill=pink] (-1,-1) rectangle (1,1);

\draw[<-] (-2,0)--(-\r,0) arc(180:0:\r)--(2,0)
--++(-90:2-\r) arc(90:-90:\r)--++(-90:1-\r);
\fill[pink]  (0,\r) circle(2pt);        
\fill[white] (2+\r,-2) circle(2pt);
\draw[<-] (-2,2)--++(0:2)--++(-90:4)--++(0:4)--++(-90:1);
\end{tikzpicture}   
\end{document}

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