Generally as mentioned in the question, we can use numerical methods to estimate the edge length of a hyperbolic tiling, but these method are limited when we deal with hyperbolic apeirogonal tiling, because the edge number trends to infinity. We can still try larger and larger edge numbers to approach the exact solution, and then guess the exact solution, and then try to proof it.
We followed this way to solve the specific case of the order-4 apeirogonal tiling, and get some result. By using upper half plane model in which the calculation is the easiest.
Construction process
Given $$ f = \frac{z-1}{z+1} $$
we have $$ f^{-1} = \frac{1+z}{1-z} $$
After numerical calculation, we decided a constant $k=3 - 2 \sqrt2$, and follow below procedure, we can draw the order-4 apeirogonal tiling.
- given the positive half y-axis as the first arm $A_0$
- given four mappings
- $F_1: z \mapsto k f(z)$
- $F_2: z \mapsto f(z / k)$
- $F_3: z \mapsto k f^{-1}(z)$
- $F_4: z \mapsto f^{-1}(z / k)$
- take the 1st construction $A_1$
- $P_0 = A_0 \cup F_1(A_0) \cup F_1^2(A_0) \cup F_1^3(A_0) \cdots$
- $Q_0 = A_0 \cup F_2(A_0) \cup F_2^2(A_0) \cup F_2^3(A_0) \cdots$
- $R_0 = A_0 \cup F_3(A_0) \cup F_3^2(A_0) \cup F_3^3(A_0) \cdots$
- $S_0 = A_0 \cup F_4(A_0) \cup F_4^2(A_0) \cup F_4^3(A_0) \cdots$
- $A_1 = P_0 \cup Q_0 \cup R_0 \cup S_0$
- apply the recursive construction
- $P_n = A_n \cup F_1(A_n) \cup F_1^2(A_n) \cdots$
- $Q_n = A_n \cup F_2(A_n) \cup F_2^2(A_n) \cdots$
- $R_n = A_n \cup F_3(A_n) \cup F_3^2(A_n) \cdots$
- $S_n = A_n \cup F_4(A_n) \cup F_4^2(A_n) \cdots$
- $A_{n+1} = P_n \cup Q_n \cup R_n \cup S_n$
Visualization and explanation
$A_1$ can be visualized as below
$A_2$ can be visualized as below
Remark on generation of $Q[\sqrt 2]$
Very intuitively, we can observe a natural corollary of the construction process. If the construction process leads to a tree $T$ (if fact, it is a 4-Cayley Tree), and the vertex of the tree is $p_i$, then $x_i = Re(p_i) \in Q[\sqrt 2]$, the generation process of the tree leads to a generation of ${x_i} \in Q[\sqrt 2]$.
If it is the order-4 apeirogonal tiling, the leaf vertex of the tree should be dense on the boundary, i.e., the x-axis. So this will give a generation process that is very similar to Calkin–Wilf tree to generate rational numbers $Q$.
Question
The problem is we need to proof it is the apeirogon we need.
We tried to translate them into elementary geometry language, but the calculation is so cumbersome. We wonder any smart advanced way to calculate the case of order-4 apeirogonal tiling.