# Drawing a hypergraph

What is the proper way to draw a hypergraph? Is there a package to do this, or is there an easy way to draw such diagrams with a common drawing method like tikz?

I'm trying to draw something like

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It is -- of course -- posible to draw diagrams like this with TikZ. However, as far as I know there is no automatic placement for things like this.

Once one gets the hang of guessing distances, it is fairly easy to reproduce graph like the example in TikZ:

\usepackage{tikz}
\usetikzlibrary{topaths,calc}

\begin{tikzpicture}
\node (v1) at (0,2) {};
\node (v2) at (1.5,3) {};
\node (v3) at (4,2.5) {};
\node (v4) at (0,0) {};
\node (v5) at (2,0.5) {};
\node (v6) at (3.5,0) {};
\node (v7) at (2.5,-1) {};

\begin{scope}[fill opacity=0.8]
\filldraw[fill=yellow!70] ($(v1)+(-0.5,0)$)
to[out=90,in=180] ($(v2) + (0,0.5)$)
to[out=0,in=90] ($(v3) + (1,0)$)
to[out=270,in=0] ($(v2) + (1,-0.8)$)
to[out=180,in=270] ($(v1)+(-0.5,0)$);
\filldraw[fill=blue!70] ($(v4)+(-0.5,0.2)$)
to[out=90,in=180] ($(v4)+(0,1)$)
to[out=0,in=90] ($(v4)+(0.6,0.3)$)
to[out=270,in=0] ($(v4)+(0,-0.6)$)
to[out=180,in=270] ($(v4)+(-0.5,0.2)$);
\filldraw[fill=green!80] ($(v5)+(-0.5,0)$)
to[out=90,in=225] ($(v3)+(-0.5,-1)$)
to[out=45,in=270] ($(v3)+(-0.7,0)$)
to[out=90,in=180] ($(v3)+(0,0.5)$)
to[out=0,in=90] ($(v3)+(0.7,0)$)
to[out=270,in=90] ($(v3)+(-0.3,-1.8)$)
to[out=270,in=90] ($(v6)+(0.5,-0.3)$)
to[out=270,in=270] ($(v5)+(-0.5,0)$);
\filldraw[fill=red!70] ($(v2)+(-0.5,-0.2)$)
to[out=90,in=180] ($(v2) + (0.2,0.4)$)
to[out=0,in=180] ($(v3) + (0,0.3)$)
to[out=0,in=90] ($(v3) + (0.3,-0.1)$)
to[out=270,in=0] ($(v3) + (0,-0.3)$)
to[out=180,in=0] ($(v3) + (-1.3,0)$)
to[out=180,in=270] ($(v2)+(-0.5,-0.2)$);
\end{scope}

\foreach \v in {1,2,...,7} {
\fill (v\v) circle (0.1);
}

\fill (v1) circle (0.1) node [right] {$v_1$};
\fill (v2) circle (0.1) node [below left] {$v_2$};
\fill (v3) circle (0.1) node [left] {$v_3$};
\fill (v4) circle (0.1) node [below] {$v_4$};
\fill (v5) circle (0.1) node [below right] {$v_5$};
\fill (v6) circle (0.1) node [below left] {$v_6$};
\fill (v7) circle (0.1) node [below right] {$v_7$};

\node at (0.2,2.8) {$e_1$};
\node at (2.3,3) {$e_2$};
\node at (3,0.8) {$e_3$};
\node at (0.1,0.7) {$e_4$};
\end{tikzpicture}


I think the code is fairly self-explanatory: First the vertice coordinates are specified. Then all the areas are drawn with the help of curve to statements. Afterwards the circles and labels are placed. The result is

This is not a perfect copy of the original image. With some more fiddling around with the coordinates and some additional points in the paths one could get closer.

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That looks great! Thanks for your help! – Moor Xu Aug 6 '10 at 22:00
That's amazing. Nice. – Neil G Aug 11 '10 at 22:21

I have to confess myself not particularly enamored of the example given: it looks messy to me. I find it hard to separate the edge labels from the vertex labels, and the extra wiggliness of the edges is there for no good reason. I would imagine that for it to be worth drawing such a diagram, it cannot be too complicated (imagine drawing the hypergraph corresponding to the powerset of even 3 elements). So here's an attempt at making a cleaner diagram. Again, it uses TikZ, but by making the diagram more stylised, I can get away with some tricks. In particular, note that the edges are done by drawing really thick lines between the vertices.

\documentclass{article}
\pagestyle{empty}
\usepackage{tikz}

\tikzstyle{vertex} = [fill,shape=circle,node distance=80pt]
\tikzstyle{edge} = [fill,opacity=.5,fill opacity=.5,line cap=round, line join=round, line width=50pt]
\tikzstyle{elabel} =  [fill,shape=circle,node distance=30pt]

\pgfdeclarelayer{background}
\pgfsetlayers{background,main}

\begin{document}
\begin{tikzpicture}
\node[vertex,label=above:$$v_1$$] (v1) {};
\node[vertex,right of=v1,label=above:$$v_2$$] (v2) {};
\node[vertex,right of=v2,label=above:$$v_3$$] (v3) {};
\node[vertex,below of=v1,label=above:$$v_4$$] (v4) {};
\node[vertex,right of=v4,label=above:$$v_5$$] (v5) {};
\node[vertex,right of=v5,label=above:$$v_6$$] (v6) {};
\node[vertex,below of=v5,label=above:$$v_7$$] (v7) {};

\begin{pgfonlayer}{background}
\draw[edge,color=yellow] (v1) -- (v2) -- (v3);
\begin{scope}[transparency group,opacity=.5]
\draw[edge,opacity=1,color=green] (v3) -- (v5) -- (v6) -- (v3);
\fill[edge,opacity=1,color=green] (v3.center) -- (v5.center) -- (v6.center) -- (v3.center);
\end{scope}
\draw[edge,color=red,line width=40pt] (v2) -- (v3);
\draw[edge,color=purple] (v4) -- (v4);
\end{pgfonlayer}

\node[elabel,color=yellow,label=right:$$e_1$$]  (e1) at (-3,0) {};
\node[elabel,below of=e1,color=red,label=right:$$e_2$$]  (e2) {};
\node[elabel,below of=e2,color=green,label=right:$$e_3$$]  (e3) {};
\node[elabel,below of=e3,color=purple,label=right:$$e_4$$]  (e4) {};
\end{tikzpicture}
\end{document}


Here's the picture:

For more complicated graphs, I would see if I could get graphviz or something like that to generate it, and then get TikZ to render it.

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Your webserver seems to have a problem. – Caramdir Aug 6 '10 at 11:03
For non-mathematicians, could someone give a hint as to the standard 'style' of these things? I like the order in Andrew's approach, but wonder if this is what is expected. – Joseph Wright Aug 6 '10 at 11:15
@Caramdir: Yeah, power cut. Unfortunately, my open-id is hosted there as well so I couldn't log on here. Seems to be back now. @Joseph Wright: I've never heard of a 'hypergraph' before, but it's not my area. The Wikipedia page suggests a couple of ways of representing one; I would prefer drawing it like an ordinary graph with some vertices "verticies" and others "edges" (see near bottom of Wikipedia page). Then one could use graphviz to design it and TikZ to render it. – Loop Space Aug 6 '10 at 12:23
Not my area either but the problem of drawing graphs is apparently difficult enough to merit at least 18 "International Symposia on Graph Drawing" (graphdrawing.org) – Caramdir Aug 6 '10 at 13:01
I'm not sure what is the standard style for hypergraphs, but this looks like a great representation for the hypergraph -- much more readable. Thanks a lot! – Moor Xu Aug 6 '10 at 22:03