# Iterated function system with a fractional number of contractions

Iterated function systems can be used to generate fractals. One starts off with a simple geometric figure and applies the IFS infinitely many times to obtain a fractal. For example, in the case of the Sierpinski Triangle, the IFS is given by $$\mathcal{F}(X) = \bigcup_{i=1}^3 f_i(X),$$ where each $$f$$ scales a given figure such that its length becomes half of what it was. $$3$$ is here because the Triangle is composed of $$3$$ self-similar copies of itself.

The Hausdorff dimension of self-similar fractals is in general $$\log_b a$$, where $$b$$ is the factor by which the $$f$$s contract figures. My question is, I can obtain a fractal of dimension, say, $$\log_2 \frac{5}{2}$$. Here, $$a$$ is fractional. How many contractions will the IFS have then? Is it rounded up? Is there a formula for it? I also would like to get a proof.

• A fractal dimension is just a number, and the form $D = \log_b a$ does not uniquely determine $b$ and $a$ (for example, $\log_2 4 = \log_3 9 = \log_\frac{5}{2} \frac{25}{4} = \ldots$). Commented Apr 18 at 10:28

### Some Theory

1. An IFS is typically understood to act on some ambient metric space, not a "simple geometric figure". So, for example, the Cantor set is the attractor of the IFS given by $$\{\varphi_0, \varphi_1\},$$ where $$\varphi_j : \mathbb{R} \to \mathbb{R} : x \mapsto \frac{1}{3} x + j \frac{2}{3}.$$ That is, $$\varphi_0$$ scales the entire real line by a factor of $$\frac{1}{3}$$, while $$\varphi_1$$ scales the real line by a factor of $$\frac{1}{3}$$ and then translates it $$\frac{2}{3}$$ of a unit to the right. The key observation here is that both of these maps act on the entirely real line, not some "simple geometric figure" that is taken as a starting point.

Given an IFS $$\{\varphi_j\}_{j=1}^{n}$$, where the functions act on some metric space $$X$$, it is possible to define a function $$\Phi : \mathscr{P}(X) \to \mathscr{P}(X)$$ (i.e. $$\Phi$$ acts on subsets of $$X$$). If all of the maps defining $$\Phi$$ are contractive (they scale sets by a positive factor smaller than one) there is a unique, non-empty, compact set $$F$$ such that $$\Phi(F) = F.$$ This set is called the attractor of $$\Phi$$.

Now, it is true that if $$A$$ is a non-empty subset of the ambient metric space $$X$$, then $$\lim_{n\to \infty} \Phi^n(A) = F,$$ where $$\Phi^n$$ denotes the $$n$$-fold composition of $$\Phi$$ with itself. So it is possible to start with some simple geometric figure and approximate the attractor $$F$$, but this isn't quite the right spirit.

2. Not every attractor of an IFS is a fractal. For example, the IFS $$\{ \varphi_0, \varphi_1\}$$ with maps $$\varphi_j : \mathbb{R} \to \mathbb{R} : \frac{1}{2}x + j\frac{1}{2}.$$ The attractor of this IFS is the interval $$[0,1]$$. For most definitions of "fractal", this is not a fractal.

3. If an IFS acts on $$\mathbb{R}^n$$, then each of the maps in the IFS can be written as $$\varphi_j(x) = r_j O_j x + b_j,$$ where $$r_j \in \mathbb{R}$$ is the contraction ratio, $$O_j$$ is a orthogonal transformation (basically, a rotation), and $$b_j$$ is a translation. If an IFS satisfies the open set condition, i.e. there exists an open set $$U$$ such that $$\Phi(U) \subseteq U \qquad\text{and}\qquad i\ne j \implies \phi_j(U) \cap \phi_k(U) = \varnothing,$$ then the Hausdorff dimension of the attractor is the unique real solution to the Moran equation, $$1 = \sum_{j=1}^{n} r_j^s.$$ If there is some fixed number $$r$$ such that $$r = r_j$$ for all $$j$$, this simplifies to $$1 = n r^s \implies -\log(n) = s \log(r) \implies s = -\frac{\log(n)}{\log(r)}.$$ For the Cantor set, which consists of two maps which each scale the real line by a factor of $$\frac{1}{3}$$, this gives $$s = -\frac{\log(2)}{\log(1/3)} = \frac{\log(2)}{\log(3)}.$$

All of the above is discussed in Hutchinson's seminal paper, Fractals and Self-Similarity.[1] In my opinion, this is a very readable paper—it does assume some basic knowledge of measure theory and maybe a little bit of functional analysis (the existence of an attractor is a result of the Banach fixed point theorem), but you can probably pick up enough to get through the paper if you are motivated.

### Constructing Sets

Given any $$s \in (0,1)$$, it is possible to construct a Cantor set in $$\mathbb{R}$$ with dimension $$s$$. The key is use two maps, each of which has the same contraction ratio, where this ratio is chosen so that $$s = -\frac{\log(2)}{\log(r)} \implies \log(r) = - \frac{1}{s} \log(n) \implies r = 2^{-1/s}.$$ An IFS which has an attractor with dimension $$s$$ is given by $$\varphi_j : \mathbb{R} \to \mathbb{R} : rx + j(1-r).$$

In higher dimensions, the same kind of construction works. For example, if $$s \in (n,n+1)$$ (with $$n$$ a natural number), then choose $$r = n^{-1/s}.$$ Then define $$n$$ maps, each of which has contraction ratio $$r$$, where map $$j$$ includes a translation by $$1-r$$ along the $$j$$-th axis.

[1] Hutchinson, John E., Fractals and self similarity, Indiana Univ. Math. J. 30, 713-747 (1981). ZBL0598.28011.

• Thank you for answering. You gave an example of the Cantor set, but obviously this is not limited only to Cantor sets. For example, take the Quadric cross (listed on en.wikipedia.org/wiki/List_of_fractals_by_Hausdorff_dimension). For that fractal, $a$ is $\frac{10}{3}$. It seems to me that there are $4$ contractions used and the copies are made to overlap with the starting figure. But is it possible generally to just use the ceil of $a$ as the number of contractions? I do not want to be limited to Cantor, and I want to know how to deal with the IFSs. Commented Apr 17 at 15:29
• @ArturWiadrowski The quadratic rose doesn't seem to be the attractor of an IFS. Remember that IFSes are only one way to generate sets with fractal properties. Commented Apr 17 at 17:58