I was curious whether this fractal(?) is named/famous, or is it just another fractal?

I was playing with the idea of randomness with constraints and the fractal was generated as follows:

- Draw a point at the center of a square.
- Randomly choose any two corners of the square and calculate their center.
- Calculate the center of the last drawn point and this center point of the corners.
- Draw a new point at this location.
Not sure if this will help because I just made these rules up while in the shower, but sorry I do not have any more information or an equation.

Thank you.

**Answer**

Your image can be generated using a weighted iterated function system or IFS. Specifically, let

\begin{align}

f_0(x,y) &= (x/2,y/2), \\

f_1(x,y) &= (x/2+1,y/2), \\

f_2(x,y) &= (x/2,y/2+1), \\

f_3(x,y) &= (x/2-1,y/2), \text{ and } \\

f_4(x,y) &= (x/2,y/2-1).

\end{align}

Let $(x_0,y_0)$ be the origin and define $(x_n,y_n)$ by a random, recursive procedure:

$$(x_n,y_n) = f_i(x_{n-1},y_{n-1}),$$

where $i$ is chosen randomly from $(0,1,2,3,4)$ with probabilities $p_0=1/3$ and $p_i=1/6$ for $i=1,2,3,4$.

If we iterate the procedure $100,000$ times, we generate the following image:

This image is a solid square but the points are not uniformly distributed throughout that square. Technically, this illustrates a self-similar measure on the square.

To be a bit more clear, an invariant set of an IFS is a compact set $E\subset\mathbb R^2$ such that

$$E = \bigcup_{i=0}^4 f_i(E).$$

It’s pretty easy to see that the square with vertices at the points $(-2,0)$, $(0,-2)$, $(2,0)$, and $(0,2)$ is an invariant set for this IFS. It can be shown that an IFS of contractions always has a *unique* invariant set; thus, this square is the only invariant set for this IFS.

Let’s call this square $E$, in honor of its status as an invariant set.

We can get a deterministic understanding of the distribution of points on $E$ by thinking in terms of a mass distribution on the square (technically, a measure). Start with a uniform mass distribution throughout the square. Generate a second mass distribution on $E$ by distributing $1/3$ of the mass to $f_0(E)$ and $1/6$ of the mass to each of $f_i(E)$ for $i=1,2,3,4$. We can then iterate this procedure. The step from the original distribution to the next to the next might look like so:

The evolution of the first 8 steps looks like

**Attribution***Source : Link , Question Author : SilverSlash , Answer Author : Mark McClure*