I use \shade [ball color=black!60!red] (0,0) circle (4pt);
command inside tikz image to represent a vertex of a graph. Now I am using some operation where two vertices are merged. So is it possible to have a shaded ball with two sided color?
I have gone through the following link, but it is a a circle type. I need a ball type. So is it possible? Please help.
\documentclass[10pt]{article}
\usepackage{pgf,tikz}
\usepackage{mathrsfs}
\usetikzlibrary{arrows}
\pagestyle{empty}
\begin{document}
\definecolor{qqqqff}{rgb}{0.,0.,1.}
\definecolor{ffqqqq}{rgb}{1.,0.,0.}
\begin{figure}
\centering
\begin{tikzpicture}[ultra thick,x=1.0cm,y=1.0cm]
\draw (-1.,3.)-- (-1.,1.);
\draw (1.,1.)-- (3.,1.);
\draw (3.,1.)-- (2.,3.);
\draw (2.,3.)-- (1.,1.);
\draw (6.,3.)-- (5.,1.);
\draw (5.,1.)-- (7.,1.);
\draw (7.,1.)-- (6.,3.);
\draw (5.99,5.02)-- (5.99,3.02);
\draw (-1.2,3.5) node[anchor=north west] {$u$};
\draw (1.84,3.5) node[anchor=north west] {$v$};
\draw (6,3.5) node[anchor=north west] {$u+v$};
\begin{scriptsize}
\shade [ball color=ffqqqq] (-1.,3.) circle (5pt);
\shade [ball color=ffqqqq] (-1.,1.) circle (5pt);
\shade [ball color=qqqqff] (1.,1.) circle (5pt);
\shade [ball color=qqqqff] (3.,1.) circle (5pt);
\shade [ball color=qqqqff] (2.,3.) circle (5pt);
\shade [ball color=qqqqff] (6.,3.) circle (5pt);
\shade [ball color=qqqqff] (5.,1.) circle (5pt);
\shade [ball color=qqqqff] (7.,1.) circle (5pt);
\shade [ball color=ffqqqq] (5.99,5.02) circle (5pt);
\shade [ball color=ffqqqq] (5.99,3.02) circle (5pt);
\end{scriptsize}
\end{tikzpicture}
\caption{I am trying to color the vertex $u+v$ half red - half blue}
\end{figure}
\end{document}
Best Answer
Here is an answer with semi-sphaire over your original sphaire: (The idea was from: This answer)
Output: