Recursive drawing, command flow

Question

I am trying to recursively draw C_k x C_k, where C_k is the k-th iteration of a Cantor set. I have a command which recursively calls itself four times. If I pass N as parameter #6, the output should be 4^N little squares. But it only works for N = 1, 2. If I do anything higher, I only get 4 squares, spaced out in the four corners. They are the right size, but there aren't enough of them.

Apparently, the problem has something do with \pgfmathtruncatemacro\C{#6 - 1}. As a test I inserted \draw (0,0) circle (#6); right before this line, and then right after this line, and the output was NOT always the same. It seems that \pgfmathtruncatemacro\C{#6 - 1} somehow changes the value of #6 (in the SAME level of recursion), but I don't see why.

\usepackage{tikz, pgf, ifthen}

\newcommand\cantorsquare[6]{ %x y x y coordinates, dissection coordinates, nesting
%the coordinates are listed in the same order as coordinates for a rectangle
    \ifthenelse{#6 = 0} %base case
    {
        \fill (#1, #2) rectangle (#3, #4);
    }
    {
        {
            %find the coordinates for the first smaller square
            \pgfmathsetmacro\A{#1 + (1 - #5) * (#3 - #1) / 2}
            \pgfmathsetmacro\B{#2 + (1 - #5) * (#4 - #2) / 2}
            \pgfmathtruncatemacro\C{#6 - 1}
            \cantorsquare{#1}{#2}{\A}{\B}{#5}{\C}
        }
        {
            \pgfmathsetmacro\A{#1 + (1 - #5) * (#3 - #1) / 2}
            \pgfmathsetmacro\B{#4 - (1 - #5) * (#4 - #2) / 2}
            \pgfmathtruncatemacro\C{#6 - 1}
            \cantorsquare{#1}{\B}{\A}{#4}{#5}{\C}
        }
        {
            \pgfmathsetmacro\A{#3 - (1 - #5) * (#3 - #1) / 2}
            \pgfmathsetmacro\B{#2 + (1 - #5) * (#4 - #2) / 2}
            \pgfmathtruncatemacro\C{#6 - 1}
            \cantorsquare{\A}{#2}{#3}{\B}{#5}{\C}
        }
        {
            \pgfmathsetmacro\A{#3 - (1 - #5) * (#3 - #1) / 2}
            \pgfmathsetmacro\B{#4 - (1 - #5) * (#4 - #2) / 2}
            \pgfmathtruncatemacro\C{#6 - 1}
            \cantorsquare{\A}{\B}{#3}{#4}{#5}{\C}
        }
    }
}

\begin{document}
    \begin{tikzpicture} \cantorsquare{0}{0}{5}{5}{0.3}{3} \end{tikzpicture}
\end{document}

Answer 1

I agree with Michael's analysis: that it is a problem of expansion. When you pass \C as the sixth element then #6 gets replaced in the macro body by \C so if \C changes then so does the ultimate value of #6. In particular, when you do \pgfmathtruncatemacro\C{#6 - 1} then you are redefining #6.

Having said that, I'm not sure where the expansion is going wrong because you've correctly grouped your recursive calls so modifications to \C in one iteration shouldn't affect others. Moreover, when I run your code then I get the answer that I expect to see. So I have a suspicion that this is an issue with an old version of PGF making a global assignment where it should make a local one. What version of PGF are you using?

I find that LaTeX3 has lots of useful stuff that helps sort out these issues of expansion. In particular, it is possible to ensure that the value of a macro is passed, not the macro itself. So here is your code rewritten in LaTeX3. By passing values rather than macros, I can avoid having to group the recursive calls.

\documentclass{article}
%\url{http://tex.stackexchange.com/q/136907/86}
\usepackage{tikz}
\usepackage{xparse}

\ExplSyntaxOn

\cs_new_nopar:Npn \cantor_sq:nnnnnn #1#2#3#4#5#6
{
  \int_compare:nTF {#6 = 0}
  {
    \fill (\fp_to_decimal:n {#1},\fp_to_decimal:n {#2}) rectangle (\fp_to_decimal:n {#3},\fp_to_decimal:n {#4});
  }
  {
    %find the coordinates for the first smaller square
    \fp_set:Nn \l_tmpa_fp {#1 + (1 - #5) * (#3 - #1) / 2}
    \fp_set:Nn \l_tmpb_fp {#2 + (1 - #5) * (#4 - #2) / 2}
    \int_set:Nn \l_tmpa_int {#6 - 1}
    \cantor_sq:nnVVnV {#1}{#2} \l_tmpa_fp \l_tmpb_fp {#5} \l_tmpa_int

    \fp_set:Nn \l_tmpa_fp {#1 + (1 - #5) * (#3 - #1) / 2}
    \fp_set:Nn \l_tmpb_fp {#4 - (1 - #5) * (#4 - #2) / 2}
    \int_set:Nn \l_tmpa_int {#6 - 1}
    \cantor_sq:nVVnnV {#1} \l_tmpb_fp \l_tmpa_fp {#4}{#5} \l_tmpa_int

    \fp_set:Nn \l_tmpa_fp {#3 - (1 - #5) * (#3 - #1) / 2}
    \fp_set:Nn \l_tmpb_fp {#2 + (1 - #5) * (#4 - #2) / 2}
    \int_set:Nn \l_tmpa_int {#6 - 1}
    \cantor_sq:VnnVnV \l_tmpa_fp {#2}{#3} \l_tmpb_fp {#5} \l_tmpa_int

    \fp_set:Nn \l_tmpa_fp {#3 - (1 - #5) * (#3 - #1) / 2}
    \fp_set:Nn \l_tmpb_fp {#4 - (1 - #5) * (#4 - #2) / 2}
    \int_set:Nn \l_tmpa_int {#6 - 1}
    \cantor_sq:VVnnnV \l_tmpa_fp \l_tmpb_fp {#3}{#4}{#5} \l_tmpa_int
  }
}

\cs_generate_variant:Nn \cantor_sq:nnnnnn {nnVVnV,nVVnnV,VnnVnV,VVnnnV}

\DeclareDocumentCommand \cantorsquare { m m m m m m }
{
  \cantor_sq:nnnnnn {#1}{#2}{#3}{#4}{#5}{#6}
}

\ExplSyntaxOff

\begin{document}
    \begin{tikzpicture}
 \cantorsquare{0}{0}{5}{5}{0.3}{3}
 \end{tikzpicture}

    \begin{tikzpicture}
 \cantorsquare{0}{0}{5}{5}{0.3}{6}
 \end{tikzpicture}
\end{document}

That the small rectangles don't look uniform is an artefact of the rendering as can be seen by looking at a larger zoom.

Answer 2

I do not know the details of the packages you are using so it is hard for me to be categorical but it seems you are tripped by the fact that macros passed as arguments are not expanded when the call is performed, but only later when the replacement text where they appear is processed. This may break your recursion. To verify this, you may execute your code with \tracingmacros=1.

You can use registers to force expansion, replacing your usage

\cantorsquare{#1}{#2}{\A}{\B}{#5}{\C}

by

\begingroup
\let\cantorsquare=0\relax % Freeze expansion in edef
\count0=\A\relax
\count2=\B\relax
\count4=\C\relax
\toks0={#1}%
\toks2={#2}%
\toks4={#5}%
\edef\next{%
  \cantorsquare{\the\toks0}{\the\toks2}{\the\count0}{\the\count2}{\the\toks4}{\the\count4}%
}%
\expandafter
\endgroup
\next

It prepares a “call” to cantorsquare where \A etc. are replaced by their replacement text — assuming these are numbers. You should edit accordingly each of your four calls.

Note that TeX is a very special computer language, because it does not work with functions and the like, rather, it offers you the possibility to write the sequel of the program—this is what macros are for. If you want to write sophisticated programs in TeX, it is much easier to learn about registers, \edef etc.

Does it help?

Edit: As a side note, I definitely recommand you to have a look at METAPOST if you are interested in programatically generating graphics! It is an awesome piece of software—which just as fun to program as TeX is.

Answer 3

EDIT: Added lualatex version below.

As already pointed out, expansion is the issue, and is easily corrected. However, here is a different solution that requires the very latest PGF/TikZ CVS version.

\documentclass[border=0.125cm]{standalone}
\usepackage{tikz}
\usetikzlibrary{math}

\begin{document}

\begin{tikzpicture}

\tikzmath{
    function cantorsquare(\x, \y, \s, \f, \d) {
        if (\d == 0) then {
            { \fill (\x, \y) rectangle ++(\s, \s); };
        } else {
            \u1 = \f*\s;
            \u2 = (1-\f)*\s;
            cantorsquare(\x, \y, \u1, \f, \d-1);
            cantorsquare(\x+\u2, \y, \u1, \f, \d-1);
            cantorsquare(\x, \y+\u2, \u1, \f, \d-1);
            cantorsquare(\x+\u2, \y+\u2, \u1, \f, \d-1);
        };
    };
    \S = 5;
    for \d in {0,...,5}{
        \x = int(\d/3) * (\S+1);
        \y = mod(\d,3) * (\S+1);
        { \draw (\x-0.1, -\y-0.1) rectangle ++(\S+0.2,\S+0.2); };
        cantorsquare(\x, -\y, \S, 1/3, \d);
    };
}

\end{tikzpicture}

\end{document}

And here is a version using lualatex which does not require the latest CVS, and is much faster.

\documentclass[border=0.125cm]{standalone}
\usepackage{tikz}

\directlua{
function cantorsquare(x, y, s, f, d)
    local u1, u2
    if d == 0 then
        tex.print("\noexpand\\fill (" .. x .. "," .. y .. ") rectangle ++(" .. s .. "," .. s .. ");")
    else
        u1 = f*s
        u2 = (1-f)*s;
        cantorsquare(x, y, u1, f, d-1)
        cantorsquare(x+u2, y, u1, f, d-1)
        cantorsquare(x, y+u2, u1, f, d-1)
        cantorsquare(x+u2, y+u2, u1, f, d-1)
    end
end
}

\begin{document}

\begin{tikzpicture}

\foreach \d [evaluate={\S=5; \x=int(\d/3)*(\S+1); \y=mod(\d,3)*(\S+1);}] in {0,...,5}{
    \draw (\x-0.1, -\y-0.1) rectangle ++(\S+0.2,\S+0.2); 
    \directlua{cantorsquare(\x, -\y, \S, 1/3, \d)}
}
\end{tikzpicture}

\end{document}

The result is the same as before.

Answer 4

I don’t know the exact specification of this thing but here’s a recursive attempt with append after command (a previous version used path picture but that only works good with rectangular shapes).

Code

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\tikzset{
  if/.code n args={3}{% forest.sty
    \pgfmathparse{#1}%
    \ifnum\pgfmathresult=0 \pgfkeysalso{#3}\else\pgfkeysalso{#2}\fi},
  cantorsquare/.style n args={3}{% #1 = current leve, #2 = max level, #3 = factor
    if={#1==#2}{cantorsquare end}{cantorsquare go on={#1+1}{#2}{#3}}},
  every cantorsquare/.style={shape=rectangle, outer sep=+0pt, ultra thin},
  cantorsquare end/.style={every cantorsquare, fill},
  cantorsquare go on/.style n args={3}{
    every cantorsquare,
    append after command={
      \pgfextra{\let\myTikzlastnode\tikzlastnode}
      let \p{@dim}=($(\myTikzlastnode.north east)-(\myTikzlastnode.south west)$) in
        [next nodes/.style={inner sep=+0pt, outer sep=+0pt, minimum width=(#3)*\x{@dim},
                                   minimum height=(#3)*\y{@dim}, cantorsquare={#1}{#2}{#3}}]
        node [next nodes, above right] at (\myTikzlastnode.south west) {}
        node [next nodes, above left]  at (\myTikzlastnode.south east) {}
        node [next nodes, below right] at (\myTikzlastnode.north west) {}
        node [next nodes, below left]  at (\myTikzlastnode.north east) {}}}}
\begin{document}
\foreach \lev in {1, ..., 5, 4, 3, 2}{
\begin{tikzpicture}
\node[cantorsquare/.expanded={1}{\lev}{.3333}, minimum size=2cm] {};
\end{tikzpicture}}
\foreach \lev in {1, ..., 5, 4, 3, 2}{
\begin{tikzpicture}[every cantorsquare/.append style={shape=circle}, 
                    every cantorsquare/.append style=draw]
\node[cantorsquare/.expanded={1}{\lev}{.3333}, minimum size=2cm] {};
\end{tikzpicture}}
\end{document}

Output