The setting:
We consider the plane $\mathbb R^2$ with its canonical euclidean structure. The canonic base is written $(e_1,e_2) = ((1,0)^\top,(0,1)^\top)$. Let's consider $I = [a,b]e_1$ (with $a<b$ two real numbers) a segment parallele to $e_1$. We end with fixing $m\in \mathbb N^*$ and $\alpha \in (0,\pi/4)$.
We consider the following transformation:
$T$ denotes the transformation that replaces any segment $J$ of length $L$ with $m$ segment of length $L/(m\cos(\alpha))$ making the angle $\alpha$ with the segment $J$. See the following figure:
(thus, the dotted lines above make $\pi/2$ angles with $J$. And $T(J)$ is red, and here $m=3$).
The construction is then given by the following algorithm:
Let consider $E_0 = I$. Suppose that $E_n$ is already constructed. And at each semgent $J$ that makes $E_n$, we apply $T$. Then, $E_{n+1} = \bigcup_{\substack{J\subset E_n \\ J \text{ segment}}} T(J)$. For $m = 2$ and $n = 5$ we have $E_5$ represented in black in the following figure:
(where $I$ is in blue and $E_1,...,E_4$ are in gray).
The problem:
Find an IFS that allows expressing the limit set as $n\to \infty$ as its attractor.
I feel like this IFS will have to be composed of $m$ different contraction. I tried to express it with a rotation of angle $\alpha$ and a translation, but the (not-a-)solution I came with only worked with $E_1$.
Is that possible to do? If it is not, what is the best way to express mathematically such a construction? (in order to get easily the Hausdorff dimension for example).
(I must insist on the fact that $T$ replaces the segment!)
In an IFS, there are multiple transformation functions $T_i$, each operating on the whole (there is no need to decompose into segments).
Without loss of generality, assume $I = [0,1]$, otherwise each transformation $T_i$ below can be affine-conjugated with a map $\phi : I \to [0,1]$.
For $i \in \{ 0, 1, \ldots, m - 1\}$ let $$T_i(\mathbf{v}) = \frac{1}{m \cos \alpha}\begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix}\mathbf{v} + \frac{i}{m}\begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
Then $$E_{n+1} = \bigcup_{i \in \{0, 1, \ldots, m - 1\}} T_i(E_{n})$$ which is contracting (a condition for having a finite limit) when $m \cos \alpha > 1$.
The similarity dimension is then $s$ solving $$m \left(\frac{1}{m \cos \alpha}\right)^s = 1$$ and it shouldn't be too hard to verify that this is equal to the Hausdorff dimension (via open set condition).
Here primitive illustration of the process with $m = 2$ and $\alpha = 10\deg$: the region of the white square is repeatedly sampled and textured onto the two smaller squares in red and cyan, you can hopefully see the emergence of the fixed point fractal attractor:
"Most" starting distributions $E_0$ will give the same limit set $E_\infty$.