# Colorize sides regular polygon

To illustrate what edge-coloring is, I'd like to colorize a regular polygon. I want to generate the shape using TikZ, then colorize the sides (edges) of the polygon separately, for instance by doing something like this:

\draw[red] (polygon.side 1);


Unfortunately, the world does not seem to work like this. For now my ugly workaround is to overdraw using dotted, dashed, and thick lines.

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric,positioning}

\begin{document}
\begin{tikzpicture}
\node[draw, dotted, minimum size=2cm, regular polygon,
regular polygon sides=5] (polygon) {};
\draw[thick] (polygon.corner 1) -- (polygon.corner 2);
\draw[dashed] (polygon.corner 2) -- (polygon.corner 3);
\draw[thick] (polygon.corner 4) -- (polygon.corner 5);
\draw[dashed] (polygon.corner 5) -- (polygon.corner 1);
\end{tikzpicture}
\end{document}


Thanks!

• Do you really want to use a node? – Sigur Nov 21 '18 at 23:02
• I want the simplest thing possible. I may have driven into the wrong direction. Also, why wouldn't I want a node? – Antonin Décimo Nov 21 '18 at 23:04

With regular polygon colors option, there's no need for extra commands to draw the sides. Everything is done within the node command.

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{shapes.geometric, positioning}

\begin{document}
\begin{tikzpicture}[
regular polygon colors/.style 2 args={
append after command={%
\pgfextra
\foreach \i [count=\ni, remember=\ni as \lasti (initially #1)] in {#2}{
\draw[\i] (\tikzlastnode.corner \lasti) --(\tikzlastnode.corner \ni);}
\endpgfextra
}
},
]
\node[minimum size=2cm, regular polygon,
regular polygon sides=5,
regular polygon colors={5}{red,blue,orange,green,cyan}] (polygon) {};

\node[minimum size=2cm, regular polygon,
regular polygon sides=7,
regular polygon colors={7}{red, blue, orange, green, cyan, violet, brown},
right=of polygon] (polygon2) {};
\end{tikzpicture}

\end{document}


Like this?

\documentclass[margin=1mm]{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}[thick]
\foreach \i/\c in {90/red,162/blue,234/green,306/pink,378/orange}{
\draw[\c] (\i:1cm) -- (72+\i:1cm);
}
\end{tikzpicture}
\end{document}


Edit 1: Another option, to have a better join line:

\begin{tikzpicture}[thick]
\foreach \i/\c in {90/red,162/blue,234/green,306/pink,378/orange}{
\fill[\c] (\i:1cm) -- (72+\i:1cm) -- (0,0) -- cycle;
}
\fill[white] (90:.95cm)--(162:.95cm)--(234:.95cm)--(306:.95cm)--(378:.95cm)--cycle;
\end{tikzpicture}


Edit 2: just for fun.

\begin{tikzpicture}
\foreach \r in {1,.9,...,.1}{
\pgfmathparse{30 * \r}%
\pgfmathsetmacro\percent{\pgfmathresult}
\foreach \i/\c in {90/red,162/blue,234/green,306/pink,378/orange}{
\fill[\c!\percent] (\i:\r cm) -- (72+\i:\r cm) -- (0,0) -- cycle;
}}
\end{tikzpicture}


Here is a proposal that defines a style polygon side which can be used e.g. like \draw[red,polygon side=2]; to draw the second side red.

\documentclass[tikz,border=3.14mm]{standalone}
\usetikzlibrary{shapes.geometric}

\begin{document}
\begin{tikzpicture}[get next vertex/.code n args={2}{%
\pgfmathparse{int(1+mod(#1,5))}
\xdef#2{\pgfmathresult}},
polygon side/.style={get next vertex={#1}{\tmpnextn},
insert path={(polygon.corner #1)-- (polygon.corner \tmpnextn)}}
]
\node[draw, dotted, minimum size=2cm, regular polygon,
regular polygon sides=5] (polygon) {};
\foreach \X [count=\Y] in {red,blue,orange,green,cyan}
{\draw[\X,polygon side=\Y];}
\end{tikzpicture}
\end{document}


Note that as of now the number of corners (or sides) as well as the name of the object (or node), polygon, are hard coded. One could of course use pgfkeys to store them in some keys, and if you ask me to I'll be happy to add this to this answer.