Some progress...

If you read the previous posts on my tiling printing algorithms you'll understand the reason for this post: I made some progress that unraveled some serious knots in my stomach.

{ 4, 2, 4 } ->

{ 6, 2, 4, 3, 3, 2, 4, 2, 3, 2, 3, 3, 4, 4, 3, 3, 4 } ->

The second white polygon is made from the following points :
\begin{array}{cc}
-\frac{\sqrt{3}}{2} & \frac{1}{2} \\
-\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
0 & -1 \\
\frac{\sqrt{3}}{2} & -\frac{1}{2} \\
\frac{\sqrt{3}}{2} & \frac{1}{2} \\
\frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) \\
\frac{1}{2} \left(1+\sqrt{3}\right) & \frac{1}{2} \left(3+\sqrt{3}\right) \\
\frac{\sqrt{3}}{2} & \frac{3}{2}+\sqrt{3} \\
0 & 1+\sqrt{3} \\
-\frac{\sqrt{3}}{2} & \frac{3}{2}+\sqrt{3} \\
\frac{1}{2} \left(-1-\sqrt{3}\right) & \frac{1}{2} \left(3+\sqrt{3}\right) \\
\frac{1}{2} \left(-1-\sqrt{3}\right) & \frac{1}{2} \left(1+\sqrt{3}\right) \\
\end{array}

Ready to enter the next level of the problem. ;-)