How is this interesting fractal generated?

Question

Can someone help me with the steps through which this fractal was generated. My first observation was that we are making a vertical line at half the distance but notice that reconstructions on the rightmost vertical line is different from reconstructions on horizontal line.

Answer 1

The bottom left quadrant is a 50% downscaled copy of the whole. So is the bottom right quadrant. The top right quadrant is a downscaled copy rotated 90 degrees anticlockwise. This can be formalized as the fixed point of an iterated function system of similarities, for example the bottom right part corresponds the transformation $$\begin{pmatrix}x & y\end{pmatrix} \mapsto \frac{1}{2}\begin{pmatrix} x & y \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 1 & 0 \end{pmatrix}$$

Answer 2

Here's Mathematica code to implement the IFS:

F[{x_, y_}, r_] := Which[r == 1, {x, y}/2, r == 2, {x + 1, y}/2,
                         r == 3, {-y + 2, x + 1}/2]
ListPlot[FoldList[F, {.5, .5}, RandomInteger[{1, 3}, 10^5]],
         AspectRatio -> Automatic]

Here's what it does. F is the piecewise function $$f(x,y;r) = \begin{cases} (\frac{x}{2},\frac{y}{2}) & r = 1 \\ (\frac{x+1}{2}, \frac{y}{2}) & r = 2 \\ (\frac{-y+2}{2}, \frac{x+1}{2}) & r = 3 \end{cases}$$ that takes a point $(x,y)$ and a number $r \in \{1,2,3\}$ and outputs a new point depending on the value of $r$. Then FoldList takes a list of $10^5$ random integers in $\{1,2,3\}$, and uses this to generate iterates of the initial point $(0.5, 0.5)$. Then ListPlot plots the list of iterates.

Why does this work? It is the result of the contraction mapping theorem. The fractal comprises the three mappings described in $f$ above.