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Is anyone able to help me draw the following fractal in Tikz? enter image description here

Here we start of with the unit square, break it into 16 equal squares, and remove 12 of them - leaving 4 arranged as seen. The process is repeated on the remaining squares, and so on.

I am pretty sure I need to use something similar to Nesting fractal tikz decorations but I am not good enough at TeX to compile the code myself.

Thank you in advance.

My code so far:

 \newcommand{\dust}[1]{
     \foreach \i in {1,...,#1}{decorate\{}
     (0,0)--(1,1)
     \foreach \i in {1,...,#1}{\}};

      SOMETHING IN HERE (DON'T KNOW)
 }

 \begin{tikzpicture}
     \path (0,0) pic {dust=5};
 \end{tikzpicture}
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3
  • What have you tried so far? Please show us your code of what you have attempted.
    – Huang_d
    Jul 10, 2017 at 16:16
  • Welcome to TeX.SX! Please expand your code to a minimal working example (MWE). Reproducing will be much easier when we see compilable code, starting with \documentclass and ending with \end{document}.
    – Bobyandbob
    Jul 10, 2017 at 21:00
  • Instead of decorations, you could use a normal tikzpicture as in tex.stackexchange.com/questions/228106/fractals-in-tikz?rq=1
    – John Kormylo
    Jul 10, 2017 at 21:00

2 Answers 2

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11

It is a nice Lindenmayer system exercise.

\documentclass[border=9,tikz]{standalone}
\usetikzlibrary{lindenmayersystems}
\begin{document}
\pgfdeclarelindenmayersystem{Cantor dust}{
    \symbol{S}{\pgfpathrectangle{\pgfpointorigin}{\pgfpoint{\pgflsystemcurrentstep}{\pgflsystemcurrentstep}}}
    \symbol{U}{\pgftransformyshift{\pgflsystemcurrentstep}}
    \symbol{R}{\pgftransformxshift{\pgflsystemcurrentstep}}
    \rule{S -> [UUSURRS][RSRRUS]}
    \rule{U -> UUUU}
    \rule{R -> RRRR}
}
\tikz;
\tikz\fill[lindenmayer system={Cantor dust,axiom=S,step=320pt,order=1}]lindenmayer system;
\tikz\fill[lindenmayer system={Cantor dust,axiom=S,step= 80pt,order=2}]lindenmayer system;
\tikz\fill[lindenmayer system={Cantor dust,axiom=S,step= 20pt,order=3}]lindenmayer system;
\tikz\fill[lindenmayer system={Cantor dust,axiom=S,step=  5pt,order=4}]lindenmayer system;
\end{document}

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4
  • This is exactly what I wanted :) thank you very much!
    – JSharpee
    Jul 11, 2017 at 13:40
  • @Symbol I am trying to adapt this code to make the rules change the image to 5 squares of equal size instead of 4. I want 4 of the squares to be in the corners and 1 in the center. Could you help me please?
    – PercyF2519
    Jul 17, 2017 at 13:15
  • @PercyF2519 The rules should be \rule{S -> S[U[US][RS]][RRSUUS]}\rule{U -> UUU}\rule{R -> RRR}.
    – Symbol 1
    Jul 17, 2017 at 15:17
  • Thank you, I did manage to work out the code but mine isn't quite and neat as yours. Thanks again.
    – PercyF2519
    Jul 19, 2017 at 12:28
5

I've just discovered this topic, which was related to another, more recent one. It can be considered as an application of recursive programming, for which MetaPost is usually well suited. So here is a MetaPost solution, for whom it may interest. 

def cantor_dust(expr x, y, d, n) = 
    if n > 0:
        cantor_dust(x+.25d, y, .25d, n-1);
        cantor_dust(x+.75d, y+.25d, .25d, n-1);
        cantor_dust(x+.5d, y+.75d, .25d, n-1);
        cantor_dust(x, y+.5d, .25d,  n-1);
    else: fill (x, y) -- (x+d, y) -- (x+d, y+d) -- (x, y+d) -- cycle; fi
enddef;
beginfig(1);
    for i = 0 upto 4:
        draw image(cantor_dust(0, 0, 4cm, i)) shifted (4.25cm*i, 0);
    endfor;
endfig;
end.

enter image description here

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