# Fill an ellipse with random dots

For a paper, I must draw an ellipse, which is filled by random dots in a coordinate system.

Of course it isn't a problem to draw an ellipse and the coordinate system. Also I know the function "rand" to create random dots in combination with "only marks". How is it possible to create these (random) dots in an ellipse?

• – Jake
Commented Dec 16, 2015 at 21:05

## 2 Answers

Here's a very, very easy one. Some (or, in worst case, all) points might be invisible as they are hidden. You can alter \pgfmathsetseed to get another distribution or comment it out to get a new distribution on every run.

## Code

\documentclass[tikz, border=2mm]{standalone}

\begin{document}

\begin{tikzpicture}

\draw (0,0) ellipse (4 and 2);
\clip (0,0) ellipse (4 and 2);
\pgfmathsetseed{24122015}
\foreach \p in {1,...,50}
{ \fill (4*rand,2*rand) circle (0.05);
}

\end{tikzpicture}

\end{document}


## Output

Edit 1: I played around with it a little, as I wanted to keep all points inside of the ellipse. The second version generates a random x value and then computes a random y value in such a way that it is inside. As there is less "y space" for extremal values of x, the points cluster at the end points of the major axis. The third is an ellipse in polar form: first, an angle is randomly chosen and the radius is computed to be on the inside of the ellipse. Here, the points cluster at the minor axis and the center, a so called Gorthaur1-Distribution.

## Code

\documentclass[tikz, border=2mm]{standalone}

\begin{document}

\begin{tikzpicture}
\draw (0,0) ellipse (4 and 2);
\clip (0,0) ellipse (4 and 2);
\pgfmathsetseed{24122015}
\foreach \p in {1,...,1000}
{ \fill[black]  (4*rand,2*rand) circle (0.05);
}
\end{tikzpicture}

\begin{tikzpicture}
\draw (0,0) ellipse (4 and 2);
\clip (0,0) ellipse (4 and 2);
\pgfmathsetseed{24122015}
\foreach \p in {1,...,1000}
{ \pgfmathsetmacro{\x}{4*rand}
\pgfmathsetmacro{\y}{rand*0.5*sqrt(16-pow(\x,2))}
\fill[black]    (\x,\y) circle (0.05);
}
\end{tikzpicture}

\begin{tikzpicture}
\fill[inner color=black, outer color=yellow!20!black] (0,0) ellipse (4 and 2);
\clip (0,0) ellipse (4 and 2);
\pgfmathsetseed{24122015}
\foreach \p in {1,...,1000}
{ \pgfmathsetmacro{\t}{360*rnd}
\pgfmathsetmacro{\r}{rnd*4*2/(sqrt(pow(2*cos(\t),2)+pow(4*sin(\t),2)))}
\pgfmathsetmacro{\c}{abs(\r)/4*100}
\fill[yellow!\c!red]    (\t:\r) circle (0.05);
\typeout{\t, \r, \c}
}
\end{tikzpicture}

\end{document}


## Output

1: I totally made this up. Gorthaur is a name Sauron, the villain in the lord of the rings, used in earlier ages ;-) As I usually watch the trilogy with my family over christmas I felt this little joke was in order.

• Thank you very much! You are a master! I search for this! Commented Dec 16, 2015 at 21:01
• @danielg: The clip command is very useful. Hoever, all commands following it will be clipped. To keep the effect local, you can put the clip and relevant commands in a scope. Commented Dec 16, 2015 at 21:05

Whenever there's a question about filling something with random dots, there has to be an answer using JLDiaz's Poisson Disc Sampling code:

\documentclass{article}
\usepackage{tikz}
\usepackage{poisson}
\begin{document}
\edef\mylist{\poissonpointslist{8}{4}{0.1}{20}}
\begin{tikzpicture}
\begin{scope}
\clip (4,2) ellipse (4 and 2);
\foreach \x/\y in \mylist {
\fill (\x,\y) circle(1pt);
}
\end{scope}
\draw (4,2) ellipse (4 and 2);
\end{tikzpicture}
\end{document}


And here's the approach from Filling specified area by random dots in TikZ: 150 equally distributed dots that all lie in the ellipse:

The equation for the top half of the ellipse is

sqrt( 2^2 * (1 - x^2/(4^2) )

2 * asin(x/4)


and the scaling factor is 4/pi.

\documentclass[tikz, border=2mm]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\pgfmathsetseed{3}

\begin{document}

\begin{tikzpicture}
\begin{axis}[
hide axis,
axis equal,
declare function={a(\x) = sqrt( 2^2 * (1 - \x^2/(4^2) );},
declare function={b(\x) = -sqrt( 2^2 * (1 - \x^2/(4^2) );},
declare function={f(\x) = 2 * rad(asin(x/4)) * 4 / pi;}
]
\addplot [only marks, samples=150, domain=-4:4] ({f(x) },{rand * ( a(f(x)) - b(f(x)) ) / 2} );
\draw (0,0) ellipse [x radius=4, y radius=2];
\end{axis}

\end{tikzpicture}

\end{document}


Here's the code for an ellipse with x radius=2 and y radius=1:

\documentclass[tikz, border=2mm]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}
\pgfmathsetseed{3}

\begin{document}

\begin{tikzpicture}
\begin{axis}[
%hide axis,
xmin=-4, xmax=4,
axis equal,
declare function={a(\x) = sqrt( 1^2 * (1 - \x^2/(2^2) );},
declare function={b(\x) = -sqrt( 1^2 * (1 - \x^2/(2^2) );},
declare function={f(\x) = 2 * rad(asin(x/4)) * 2 / pi;}
]
\addplot [only marks, samples=150, domain=-4:4] ({f(x) },{rand * ( a(f(x)) - b(f(x)) ) / 2} );
\draw (0,0) ellipse [x radius=2, y radius=1];
\end{axis}

\end{tikzpicture}

\end{document}

• However, that does not appear truly random. With that many dots one would expect some clustering and overlaps, right? Commented Dec 16, 2015 at 20:43
• It's not "truly random" in that the points aren't independent, but I think it's prettier than completely random points. It's definitely not "ordered" (i.e. there's no grid structure). Whether this distribution is useful probably depends on the exact use case.
– Jake
Commented Dec 16, 2015 at 20:58
• Yeah, definitely looks nice, I'll remember this for future applications :D Commented Dec 16, 2015 at 21:00
• @Jake Thank you for your answer! Your solution is also good and simple. But I have a little problem with the scaling factor, which you write here. Because I change your code with sqrt( 1^2 * (1 - x^2/(2^2) ) at the specific places (I want to create an ellipse with xRadius = 2 and yRadius = 1). Also I change the scaling factor to 2 / pi, but I get an error. Latex said: ! Missing number, treated as zero ! He write the "addplot - line" as the mistake place. Where is my mistake? If you want, I can post this minimal code. Commented Dec 17, 2015 at 22:53
• @danielg: I've edited my answer to show how to get an ellipse with x radius=2 and y radius=1. Let me know if that helps
– Jake
Commented Dec 18, 2015 at 7:19