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. ;-)

Deep Inelastic Scattering Part 1

3 weeks ago

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