# Is this a valid example of a non-euclidean Sierpinski attractor?

I am learning the basic concepts about the Chaos Game (I did a previous question about the same topic here), the method to create fractals elaborated by professor Michael Barnsley.

The basic example (euclidean-cartesian) requires three attractor vertices, $ABC$, and as Wikipedia explains: "The fractal is created by iteratively creating a sequence of points, starting with the initial random point, in which each point in the sequence is a given fraction of the distance between the previous point and one of the vertices of the polygon; the vertex is chosen at random in each iteration."

The sequence of points $(x,y)$ will produce the pattern of the Sierpinski attractor:

When I saw that construction, I wondered if there was another way of obtaining an attractor with two variables and three attractor points, and thought about polar coordinates. I did not find any references, so I did my own version of the Chaos Game for polar coordinates.

In this version, the points are $(\theta, r)$, (the angle in radians and the radius). And the three attractor points are $A=(0,0)$,$B=(\frac{5\pi}{4},1)$ and $C=(\frac{7\pi}{4},10^4)$.

This is the result for $10^7$ iterations:

The Sierpinski attractor appears, but there is a distortion due to the use of polar coordinates. In the case of the polar coordinates the point $(\theta,r)$ has two options to arrive to the attractor point: using the clockwise angle or the counterclockwise angle. Only selecting the shortest path (in this case the smallest angle of both options) the Sierpinski attractor is shown. It looks like a (sorry if I am not using the appropriated words) "non-euclidean" Sierpinski attractor.

This is another example locating the attractor points in the same axis: $A=(0,0)$,$B=(\pi-\frac{\pi}{8},1)$ and $C=(\frac{7\pi}{4},10^4)$.

I would like to share the following questions:

1. According to the results, can be said that the Sierpinski attractor in the polar coordinates pattern is a "non-euclidean" version of the attractor? or it is only a "euclidean" distortion of the original attractor?

2. Are there papers or web references about the Chaos Game in non-cartesian coordinates? and non-euclidean versions of it?

All the references I have found are euclidean-cartesian, and I think it would be interesting finding other "flavors" of the Chaos Game. Thank you!

Update

As requested, here is how the algorithm works:

1. Start from a random point in polar coordinates $(\theta_1,r_1)$ (angle and radius, the angle is in radians).

2. Generate a random value in $\{1,2,3\}$.

3. If the value is $1$, the attractor point is $(\theta_a,r_a)=(0,0)$. The next point of the sequence is a point $(\theta_2,\frac{r_1}{2})$ where $\theta_2$ will be the angle at half distance from the current angle $\theta_1$ to the attractor angle $\theta_a$ using the shortest path. For instance if the current point has $\pi \lt \theta_1 \le 2 \pi$ the shortest path to $\theta_a = 0$ is counterclockwise.

4. If the value is $2$, the attractor point is $(\theta_a,r_a)=(\frac{5\pi}{4},1)$. The next point of the sequence is a point $(\theta_2,r_2)$ where $\theta_2$ will be the angle at half distance from the current angle to the attractor angle using the shortest path (as explained in point 3) and $r_2$ will be a radius with length $r_1+\frac{r_a-r_1}{2}$ if $r_1 \lt r_a(=1)$ or length $r_1-\frac{r_1-r_a}{2}$ if $r_1 \ge r_a(=1)$ (means that the new radius $r_2$ is half the way from $r_1$ to the radius $r_a$ of the attractor).

5. Finally if the value is $3$, the attractor point is $(\theta_a,r_a)=(\frac{7\pi}{4},10^4)$ and the calculation of the new point $(\theta_2,r_2)$ is the same way as in step (4).

6. Iterate from step (2) as much as possible -now the current point is $(\theta_2,r_2)$-.

(*) The graph is made converting $(x,y)=(cos(\theta)\cdot r,sin(\theta)\cdot r)$

• It is not really clear what your algorithm is doing. Could you give a concrete example? Such as how would your algorithm determine which point comes after (0,5) assuming that it is going towards C? – BSteinhurst Jul 22 '15 at 8:48
• @BSteinhurst thanks for having a look! I have added the algorithm at the end of the question. I hope it will be easy to follow. Just let me know any doubts. :) – iadvd Jul 22 '15 at 9:24

## 1 Answer

What you have is the Chaos Game on the metric space $S^{1} \times [0,\infty)$ with the metric $d((\theta_1,x_1),(\theta_2,x_2)) = d_{S^1}(\theta_1,\theta_2) + |x_1-x_2|$. The distance in $S^1$ is given by the smallest angle measure between $\theta_1$ and $\theta_2$ (this is actually a scaled Euclidean metric on the unit circle itself.

The Chaos Game is about constructing a random sequence of points in a metric space where subsequent points are determined by randomly choosing one of the attractors and, in this case at least, moving half the distance to the attractor. This gives the next point in the sequence.

Since the metric space you have defined is not the Euclidean metric on $\mathbb{R}^2$ it is fair to say this is a non-Euclidean implementation of the Chaos Game. I would be careful to say that you went beyond just using polar coordinates here when you implicitly changed how you define "half the distance." In $\mathbb{R}^2$ given any two points the actual distance between them is the same whether in Cartesian or polar coordinates.

Sadly, I do not have any references for you off the top of my head but the key words to look for would be "Chaos game in metric spaces." Actually I just Googled that and the first paper that came up was from Barnsley The Chaos Game on a General Iterated Function System which looks promising.

Let me know if you need me to expand on why metric spaces are the right place to think about the Chaos game and I will get back to this later today.

• thank you for the reference and the comment. I am happy to see that the implementation I did has sense as non-Euclidean. I wanted to obtain the equivalent to the Chaos Game in polar coordinates and it worked this way. I will try to look for "Chaos game in metric spaces". Clearly all this field is so "new" (in terms of years since it was started) that all the references basically finish at professor Barnsley! Thanks again for the time you took for this. Regards. – iadvd Jul 22 '15 at 23:59