A set of equations is given as follows:
$$ x^{\Omega}+y^{\Omega}=1 $$
$$ (1-x)^{\Omega}+y^{\Omega}=1 $$
$$ x^\Omega+(1-y)^\Omega=1 $$
$$ (1-x)^\Omega+(1-y)^\Omega=1. $$
$(\Omega\subset\Bbb Q) =\{1.1,1.2,1.3,...,N\},$ where $N\in\Bbb N$. In the equations, $x,y \in \Bbb A(0,1),$ where A is the set of algebraic numbers.
I was able to calculate the cardinality of $\Omega:$
$|\Omega|=(20(N-1))^2.$ This is the number of intersection points in the inner grid.
Copy the structure $K$ times, which includes the original, and rotate each structure such that all corner points are equally spaced apart. For example when $K=2,$ you would rotate the copied structure by $\pi/4$ radians. When $K=3,$ you would rotate the first copy by $\pi/4$ and the second copy by $\pi/8.$ When $K=3$ you would rotate the first copy by $\pi/4$ the second copy by $\pi/8$ and the third copy by $\pi/16.$
The rotation scheme is as follows:
$$ A=(y-\frac{1}{2})\cos(z)+(x-\frac{1}{2})\sin(z)+\frac{1}{2} $$
$$ B=(x-\frac{1}{2})\cos(z)-(y-\frac{1}{2})\sin(z)+\frac{1}{2}. $$
Then we have the first copy of the original structure as: $$ A^\Omega +B^\Omega=1 $$
$$ (1-A)^\Omega +B^\Omega=1 $$
$$ A^\Omega +(1-B)^\Omega=1 $$
$$ (1-A)^\Omega +(1-B)^\Omega=1. $$
The amounts of rotation for each structure can be expressed by the series:
$$ \pi\sum_{s=2}^K \frac{1}{2^s}=\pi (\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^K}).$$
As you can see, for $K=2$ total structures, the sum is $\frac{\pi}{4}$ which tells you to rotate the first copy by that amount.
Taking infinite structures, the angle of rotation goes to zero and the sum of all rotations converges to $\frac{\pi}{2}:$
$$ \pi\sum_{s=2}^\infty \frac{1}{2^s}=\frac{\pi}{2}. $$
How many total intersections occur for a given $K$ and $N$?
How many intersections of degree $D$ are there for a given $K$ and $N$? What is $\max\{D\}$ for a given $K$ and $N?$
This is $K=4$ and the number of curves is about $392.$
This is $K=1$ and the number of curves is about $60.$
