6

I want to draw the first orders of a Hilbert curve and a Lebesgue curve. In order to draw a Hilbert curve I use the following tikZ code:

\documentclass{article}
\usepackage{tikz}

\usetikzlibrary{lindenmayersystems}
\begin{document}

\pgfdeclarelindenmayersystem{Hilbert curve}{
  \rule{L -> +RF-LFL-FR+}
  \rule{R -> -LF+RFR+FL-}}

\begin{tabular}{cc}
\begin{tikzpicture}
    \shadedraw [bottom color=white, top color=white, draw=black]
    [l-system={Hilbert curve, axiom=L, order=4, step=8pt, angle=90}]
    lindenmayer system;
\end{tikzpicture}
\end{tabular}
\end{document}

enter image description here

Then I tried to arrange the first orders side by side and I got this:

\documentclass{article}
\usepackage{tikz}

\usetikzlibrary{lindenmayersystems}
\begin{document}

\pgfdeclarelindenmayersystem{Hilbert curve}{
  \rule{L -> +RF-LFL-FR+}
  \rule{R -> -LF+RFR+FL-}}

\begin{tabular}{cc}
\begin{tikzpicture}
\foreach \i in {1,...,4}{
   \begin{scope}[xshift=2*\i cm,yshift=0cm,rotate=0]
    \shadedraw [bottom color=white, top color=white, draw=black]
    [l-system={Hilbert curve, axiom=L, order=\i, step=8pt, angle=90}]
    lindenmayer system;
    \end{scope}
    }
\end{tikzpicture}
\end{tabular}
\end{document}

enter image description here

But they are not really good arranged. How can I arrange them in such a way, that they have the same size with a constant distance to each other?

I would appreciate any help!

2 Answers 2

7

This seems to do the trick : scale each by its size, computed in terms of each \i

By math, your drawings have size :

value of \i      relative size       formula
      1                  1           2^1 - 1
      2                  3           2^2 - 1
      3                  7           2^3 - 1
      4                 15           2^4 - 1

enter image description here

\documentclass[border=2pt]{standalone}
\usepackage{tikz}

\usetikzlibrary{lindenmayersystems}
\begin{document}

\pgfdeclarelindenmayersystem{Hilbert curve}{
  \rule{L -> +RF-LFL-FR+}
  \rule{R -> -LF+RFR+FL-}}

  \begin{tikzpicture}[scale=10]
    \foreach \i in {1,...,4}{
      \begin{scope}[xshift=.5*\i cm,yshift=0cm,rotate=0, scale = 1/(2^(\i)-1)]
        \shadedraw [bottom color=white, top color=white, draw=black]
        [l-system={Hilbert curve, axiom=L, order=\i, step=8pt, angle=90}]
        lindenmayer system;
      \end{scope}
    }
  \end{tikzpicture}
\end{document}

Cheers,

3
  • Very clever. I like that solution!
    – Jan
    Commented Jan 11, 2017 at 19:54
  • I couldn't live on math ;-)
    – Jan
    Commented Jan 11, 2017 at 20:08
  • Is it possible to add a colour gradient to the final curve? Each smallest line segment will have the same colour. The very first line segment will start from a given colour, say red, and the very last segment will have a different colour, say green. Commented Apr 22, 2023 at 18:56
2

No Lindenmayer system, only recursive.

Compile here: http://asymptote.ualberta.ca/

Only the static version!

You can find its animation (similarity) in your question.

path Hilbertcurve(pair A, pair B, int ite=1){
path[] g;
if (ite == 1){ 
  g.push((A+(B-A)/4)--(xpart(A+(B-A)/4),ypart(A+3*(B-A)/4))--
              (A+3*(B-A)/4)--(xpart(A+3*(B-A)/4),ypart(A+(B-A)/4)));
  }
  else {
  g.push(rotate(-90,(A+(B-A)/4))*reverse(Hilbertcurve(A,(A+B)/2, ite-1)));
  g.push(Hilbertcurve((A.x,ypart((A+B)/2)),(xpart((A+B)/2),B.y), ite-1));
  g.push(Hilbertcurve((A+B)/2,B, ite-1));
  g.push(rotate(90,(xpart(A+3*(B-A)/4),ypart(A+(B-A)/4)))*reverse(Hilbertcurve((xpart((A+B)/2),A.y),(B.x,ypart((A+B)/2)), ite-1)));
}
return operator --(... g);
}
unitsize(1cm);
draw(Hilbertcurve((0,0),(4,4),4));
shipout(bbox(2mm,invisible));

enter image description here

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