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编辑

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编辑

我想要绘制所谓的 Cantor 目标：

我开始“手动”绘制许多圆圈，但花费的时间太长了。该集合是通过取中间第三个康托集合构建的：并以端点为枢轴，绕轴旋转 360 度。

有没有办法做到这一点，而不需要看似无穷无尽的圆圈？

先感谢您。

我对中间三分之一康托尔的代码

\documentclass{article}
\usetikzlibrary{decorations.fractals,math}

\begin{document}

\begin{tikzpicture}[decoration=Cantor set]
    \draw (0,0) -- (7,0) node[right] {$E_{0}$};
    \draw decorate{ (0,-1) -- (7,-1)} node[right] {$E_{1}$};
    \draw decorate{ decorate{ (0,-2) -- (7,-2)}} node[right] {$E_{2}$};
    \draw decorate{ decorate{ decorate{ (0,-3) -- (7,-3)}}} node[right] {$E_{3}$};
    \draw decorate{ decorate{ decorate{ decorate{ (0,-4) -- (7,-4)}}}} node[right] {$E_{4}$};
\end{tikzpicture}

\end{document}

答案1

康托目标（可能的近似值）

这将迭代原始图形，每次迭代旋转给定角度。循环次数越多，近似值就越接近。我每度使用 1 次迭代，因此 360 次循环。增加此数字将提高质量，但会降低编译速度。

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{decorations.fractals}

\begin{document}

\begin{tikzpicture}
  [
    decoration=Cantor set,
    deco/.pic={
      \draw [pic actions] (0,0) -- (7,0) node[right] {$E_{0}$};
      \draw [pic actions] decorate{ (0,-1) -- (7,-1)} node[right] {$E_{1}$};
      \draw [pic actions] decorate{ decorate{ (0,-2) -- (7,-2)}} node[right] {$E_{2}$};
      \draw [pic actions] decorate{ decorate{ decorate{ (0,-3) -- (7,-3)}}} node[right] {$E_{3}$};
      \draw [pic actions] decorate{ decorate{ decorate{ decorate{ (0,-4) -- (7,-4)}}}} coordinate [midway] (#1) node[right] {$E_{4}$};
    }
  ]
  \pic [blue] {deco=a};
  \foreach \i [evaluate=\i as \j using {int(\i/3.6)} ] in {1,...,359} \pic [blue!\j!magenta, rotate around={\i:(a)}] {deco};
\end{tikzpicture}

\end{document}

编辑

但是，我怀疑您实际上只想迭代手册中的最后一行代码。如果您还删除了节点，那么这将近似于同心圆：

康托目标（更可能的近似值）

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\usetikzlibrary{decorations.fractals}

\begin{document}

\begin{tikzpicture}
  [
    decoration=Cantor set,
    deco/.pic={
      \draw [pic actions] decorate{ decorate{ decorate{ decorate{ (0,0) -- (7,0)}}}} coordinate [midway] (#1);
    }
  ]
  \pic [blue] {deco=a};
  \foreach \i [evaluate=\i as \j using {int(\i/3.6)} ] in {.5,1,1.5,...,359.5} \pic [blue!\j!magenta, rotate around={\i:(a)}] {deco};
\end{tikzpicture}

\end{document}

毫无疑问还有更有效的方法！

答案2

我猜是这样的

\documentclass[border=9,tikz]{standalone}
\begin{document}
    \tikz{
        \draw[line width=729]circle(729/2pt);
        \draw[white,line width=243]circle(729/2pt);
        \draw[white,line width=81]circle(243*.5pt)circle(243*2.5pt);
        \draw[white,line width=27]circle(81*.5pt)circle(81*2.5pt)circle(81*6.5pt)circle(81*8.5pt);
        \draw[white,line width=9]circle(27*  .5pt)circle(27* 2.5pt)circle(27* 6.5pt)circle(27* 8.5pt)
                                 circle(27*18.5pt)circle(27*20.5pt)circle(27*24.5pt)circle(27*26.5pt);
    }
\end{document}

方法 2

\documentclass[border=9,tikz]{standalone}
\begin{document}
    \def\drawwhite#10;{
        \ifnum1#1=1\else
            \draw[white,line width=\a]circle(\b/4pt);
            {\pgfmathsetmacro\a{\a-\b}\pgfmathsetmacro\b{\b/3}\drawwhite#1;}
            {\pgfmathsetmacro\a{\a+\b}\pgfmathsetmacro\b{\b/3}\drawwhite#1;}
        \fi
    }
    \tikz{
        \pgfmathsetmacro\a{243}
        \pgfmathsetmacro\b{2*\a/3}
        \draw[line width=\a]circle(\a/2pt);
        \drawwhite00000000;
    }
\end{document}

如果愿意的话，可以继续添加 0。

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