Tikz Fractal - Uniform Cantor Set

Question

To go along with my other fractals: Tikz Fractal - Cantor Dust and Tikz Fractal - Menger Sponge, which you lovely people have helped my create, I would like to construct a "uniform Cantor set".

The construction is as follows:

Take the unit interval [0,1] and at each stage replace each interval with (a fixed number) n intervals of length less than |I|/n, where |I| is the length of the interval, and where an end point of the each of the subintervals coincides with the end point of its 'father' interval.

Here is a picture to try to make my shoddy explanation a little clearer:

The standard middle third Cantor set is where n=2 and |I|=1/3:

Short of working out all of the length and spacings, how can I construct this "automatically"? My thinking is that I should use a linedenmayer system but I have not done this for line segments before.

Answer 1

With lindenmayer system.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{lindenmayersystems}
\pgfdeclarelindenmayersystem{cantor set}{
  \rule{F -> FfF}
  \rule{f -> fff}
}
\begin{document}
\begin{tikzpicture}
  \foreach \order in {0,...,4}
    \draw[yshift=-\order*10pt]  l-system[l-system={cantor set, axiom=F, order=\order, step=100pt/(3^\order)}];
\end{tikzpicture}
\end{document}

A Cantor set with a division into three bits is a trivial extension of the existing one:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{lindenmayersystems}
\pgfdeclarelindenmayersystem{cantor set}{
  \rule{F -> FfFfF}
  \rule{f -> fffff}
}
\begin{document}
\begin{tikzpicture}
  \foreach \order in {0,...,4}
    \draw[yshift=-\order*10pt]  l-system[l-system={cantor set, axiom=F, order=\order, step=100pt/(5^\order)}];
\end{tikzpicture}
\end{document}

Answer 2

You can use the Cantor set decoration from the decorations.fractals library. An example is given in the manual:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.fractals}
\begin{document}
\begin{tikzpicture}[decoration=Cantor set,very thick]
  \draw decorate{ (0,0) -- (3,0) };
  \draw decorate{ decorate{ (0,-.5) -- (3,-.5) }};
  \draw decorate{ decorate{ decorate{ (0,-1) -- (3,-1) }}};
  \draw decorate{ decorate{ decorate{ decorate{ (0,-1.5) -- (3,-1.5) }}}};
\end{tikzpicture}
\end{document}

Answer 3

A solution based on boxes and rules:

\documentclass{article}

\makeatletter
\newcommand*{\gen}[1]{%
  \hrule height 5mm\relax
  \ifnum#1>0 %
    \expandafter\@firstofone
  \else
    \expandafter\@gobble
  \fi
  {%
    \kern5mm\relax
    \hbox to \hsize{%
      \vbox{%
        \hsize=.3333\hsize
        \gen{\numexpr#1-1}%
      }\hfill
      \vbox{%
        \hsize=.3333\hsize
        \gen{\numexpr#1-1}%
      }%
    }%
  }%
}
\newcommand*{\genpic}[2]{%
  \begin{minipage}{#1}%
    \gen{#2}%
  \end{minipage}%
}
\makeatother

\begin{document}
  \noindent
  \genpic{\linewidth}{8}
\end{document}

Answer 4

I know nothing about Lindenmayer systems, but this is easy to do using TeX recursivity.

The output

the PNG below doesn't look great but the PDF is alright.

The code starts getting slow from n=11 or so on.

The code

\documentclass[12pt,tikz]{standalone}
\begin{document}
\begin{tikzpicture}[xscale=20,yscale=2]
  \def\gen|#1|
  {
    \if 0#1
      \path[fill] (0,0) rectangle (1,.8) ;
    \else
      \begin{scope}[xscale=1/3,yshift=-1cm]
        \pgfmathtruncatemacro{\k}{#1-1}
        \gen|\k|
        \begin{scope}[xshift=2cm]
          \gen|\k|
        \end{scope}
      \end{scope}
    \fi
  }

  \foreach \k in {0,...,11}
  {
    \gen|\k|
  }

\end{tikzpicture}
\end{document}