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. ;-)