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Recently in the world of the Internet this very fancy coloured pattern appeared:

Simil-Delaunay pattern

I was thinking if it is possible to generate something similar with TikZ. I saw this first example here in Drawing unstructured grids with Tikz

\documentclass[tikz,border=5]{standalone}
\begin{document}
\begin{tikzpicture}
\foreach \i [evaluate={\ii=int(\i-1);}] in {0,...,7}{
  \foreach \j [evaluate={\jj=int(\j-1);}] in {0,...,7}{
    \coordinate [shift={(\j,\i)}] (n-\i-\j) at (rand*180:1/4+rnd/8);
\ifnum\i>0
  \draw [help lines] (n-\i-\j) -- (n-\ii-\j);
\fi
\ifnum\j>0
  \draw [help lines] (n-\i-\j) -- (n-\i-\jj);
  \ifnum\i>0
    \pgfmathparse{int(rnd>.5)}
    \ifnum\pgfmathresult=0
      \draw [help lines] (n-\i-\j) -- (n-\ii-\jj);
    \else%
      \draw [help lines] (n-\ii-\j) -- (n-\i-\jj);
    \fi%
  \fi
\fi
}}
\end{tikzpicture}
\end{document}

That produces this pattern:

Delaunay tesselation B/W]

The big problem here is to add some random coloring with light shading. Is there some way in TikZ to specify gradients inside area defined by nodes?

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4
  • They are triangles with solid/shaded colors so yes it is possible.
    – percusse
    Jun 15, 2016 at 16:15
  • Why are you asking? Just curiosity or do you want to do it? If so, what have you tried? Very nearly almost all questions should include a Minimal Working Example or a Minimal Non-Working Example (if Not Working is the problem the question asks about). We have a number of procrastinators. Eventually, one of them will probably answer this if anybody knows what a Delaunay tesselation is when it's at home. I think they ought not answer. I also think I ought not answer such questions but sometimes yield to temptation anyway. Luckily, I'm off the hook this time as I've no idea where Delaunay lives.
    – cfr
    Jun 16, 2016 at 0:42
  • 3
    I've just added an idea on how to produce a Delaunay tesselation of random geometric 2D points, the problem is how to produce a nice shading.
    – linello
    Jun 16, 2016 at 8:39
  • A lot of shoulds again...
    – percusse
    Jun 16, 2016 at 11:19

1 Answer 1

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10

You could shade it if you make the paths closed.

This is based on Mark Wibrow's code, as quoted in the question.

% answer to https://tex.stackexchange.com/questions/314918/generate-coloured-fancy-delaunay-patterns-in-tikz, modifying code by Mark Wibrow at https://tex.stackexchange.com/a/260652/
\documentclass[tikz,border=10pt,multi,rgb]{standalone}
% xcolor manual: 34
\definecolorseries{colours}{hsb}{grad}[hsb]{.575,1,1}{.987,-.234,0}
\resetcolorseries[12]{colours}
\tikzset{%
  set my colour/.code={%
    \colorlet{mycolour}{colours!!+},
  },
  my colour/.style={%
    set my colour,
    bottom color=mycolour,
    top color=mycolour!50,
    fill opacity=.5,
  },
}
\begin{document}
\begin{tikzpicture}
  \foreach \i [evaluate={\ii=int(\i-1)}, remember=\i as \ilast] in {0,...,7}{
    \foreach \j [evaluate={\jj=int(\j-1)}, remember=\j as \jlast] in {0,...,7}{
      \coordinate [shift={(\j,\i)}] (n-\i-\j) at (rand*180:1/4+rnd/8);
      \ifnum\i>0
        \path [my colour] (n-\i-\j) -- (n-\ii-\j) -- (n-\ilast-\j) -- cycle;
      \fi
      \ifnum\j>0
        \path [my colour] (n-\i-\j) -- (n-\i-\jj) -- (n-\i-\jlast) -- cycle;
        \path [my colour] (n-\i-\jlast) -- (n-\i-\jj) -- (n-\i-\j) -- cycle;
        \ifnum\i>0
           \path [my colour] (n-\i-\j) -- (n-\i-\jj) -- (n-\ii-\j) -- cycle;
           \path [my colour] (n-\ilast-\j) -- (n-\ilast-\jj) -- (n-\i-\j) -- cycle;
          \pgfmathparse{int(rnd>.5)}
          \ifnum\pgfmathresult=0
            \path [my colour] (n-\ilast-\jlast) -- (n-\ilast-\jj) -- (n-\ii-\jj) -- cycle;
            \path [my colour] (n-\ilast-\j) -- (n-\ilast-\jj) -- (n-\ii-\jj) -- cycle;
            \path [my colour] (n-\i-\jlast) -- (n-\i-\jj) -- (n-\ii-\jj) -- cycle;
            \path [my colour] (n-\i-\j) -- (n-\i-\jj) -- (n-\ii-\jj) -- cycle;
            \path [my colour] (n-\ii-\j) -- (n-\i-\jj) -- (n-\i-\j) -- cycle;
          \else
            \path [my colour] (n-\ii-\j) -- (n-\i-\jj) -- (n-\i-\j) -- cycle;
            \path [my colour] (n-\i-\j) -- (n-\i-\jj) -- (n-\ii-\jj) -- cycle;
            \path [my colour] (n-\i-\jlast) -- (n-\i-\jj) -- (n-\ii-\jj) -- cycle;
            \path [my colour] (n-\ilast-\j) -- (n-\ilast-\jj) -- (n-\ii-\jj) -- cycle;
            \path [my colour] (n-\ilast-\jlast) -- (n-\ilast-\jj) -- (n-\ii-\jj) -- cycle;
          \fi
        \fi
      \fi
    }
  }
\end{tikzpicture}
\end{document}

shaded

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5
  • You are an artist.
    – AlexG
    Jun 16, 2016 at 12:04
  • Wow this is really astonishing! In order to choose a specific color palette, should I work on the \definecolourseries command? Additionally, how to define gradients with two light and dark extremes of the same color inside each triangular area?
    – linello
    Jun 16, 2016 at 12:16
  • It's things like this that continually amaze me with what LaTeX can do. Great work @cfr!
    – Richard
    Jun 16, 2016 at 12:32
  • @linello Yes, take a look at xcolor's documentation for details of how to use colour series. This is one I created for something else and I really just used it here by way of demonstration. I find the documentation a bit difficult to apply, but it gets easier once you've got an example. The !! takes the current colour. The !!+ increments the counter afterwards. The 12 gives the number of steps.
    – cfr
    Jun 16, 2016 at 14:13
  • @linello For the gradient, you would probably be best defining a custom gradient. If you just want 3 colours, you can say left color= , right color= , middle color= or top... bottom... middle (note that middle must come last - order matters). But for 4 colours, I think you need a custom shading. This is a TikZ/PGF matter, as opposed to xcolor.
    – cfr
    Jun 16, 2016 at 14:15

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