1
$\begingroup$

I want to justify why the Chaos game works to produce Sierpinski triangle. I use a theorem taken from Massopust Interpolation and Approximation with Splines and Fractals.

Suppose that $(X,d)$ is a compact metric space and $(X,F,P)$ is an IFS with probabilities. Futher assume that $m \in P(X)$ is the invariant fractal measure. Let $x_0 \in X$ be arbitrary and let $x_k = f_i(x_{k-1})$ for $k \in \mathbb{N}$ where $f_i \in F$ is chosen with probability $p_i \in P$.

Then, for almost all random sequences $\{x_k\}$, the following equality holds: $$m(A) = \lim_{k \to \infty} \frac{N(A \cap \{x_l:l = 0,1,\ldots,k)\})}{k+1}$$ for all $A \in B(H(X))$ with $m(fr(A)) = 0$ and where $N(B)$ denotes the number of points in set $B$.

The right hand side of the equation represents the fraction of points that lie on set $A$. So if I choose the IFS generating the Sierpinski triangle such as in this video I would need that $\mu(\mathcal{a}) \sim 1$ where $\mathcal{a}$ is the Sierpinski triangle.

I tried to compute by hand the invariant fractal measure for the IFS that produces the Sierpinski triangle but I was obtaining a wrong results. Is that the right way to go? How can I justify that the chaos game produces the Sierpinski triangle in the limit using this theorem?


Definitions

Given a IFS with probabilities, $(X,\{f_i\})$ formed by a compact metric space $(X,d)$ and a finite number of contractive mappings $f_i:X \to X$ and a set of probabilities $p_i > 0$ with $\sum p_i = 1$, the measure $\mu$ such that $\mu = \sum p_i \cdot \mu \circ f_i ^{-1}$ is called $p$-balanced measure or invariant fractal measure.

Notes

This question gives a formal answer to Sierpinski Triangle and the Chaos Game.

$\endgroup$

1 Answer 1

0
$\begingroup$

The following is a small correction to Massopust Interpolation and Approximation with Splines and Fractals.

Relation between the fractal generated by the IFS $A$ and the invariant measure $m$

If the involved probabilities are strictly positive $A = supp \; m$.

Support of a Borel probability measure on a compact metric space has measure 1

  1. Since $X$ is a compact metric space, it is separable.

  2. Borel probability measures on separable metric spaces have full measure.

Recall $P(X)$ is the set of measures $\mu$ on $(X,B(X))$ such that $\mu(X) = 1$. This makes the elements of $P(X)$ Borel probability measures. So we can write:

$$1 = \lim_{k \to \infty} \frac{N(A \cap \{x_l:l = 0,1,\ldots,k)\})}{k+1}$$ for all $A \in B(H(X))$ with $m(fr(A)) = 0$. As we wanted. It would be interesting to comment on the conditions $A \in B(H(X))$ and $m(fr(A)) = 0$. To see if they are satisfied in the Sierpinsky triangle.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.