Filter foibles

I claim that any ``nice'' IFS, $W$ $=$ $w_1$ $\cup$ $\cdots$ $\cup$ $w_n$ (where each $w_i$ is a contraction and the $w_i$ are non-overlapping) corresponds to a choice of probabilities $p_1$, $\ldots$ $p_n$ such that if $S$ $=$ $i_1 \cdots i_k$ is a string with $i_j$ $\in$ $\{1,\ldots, n\}$, the ratio

$$\prod_{j=1}^k \left( p_{i_j}/ \text{Area}(i_1 \cdots i_k) \right) = \text{constant}^k,$$

...where Area $(i_1 \cdots i_k)$ denotes the image of the unit square under $w_{i_1} \circ \cdots \circ w_{i_k}$. In other words for all addresses of a given length, the probability that a point lies in a particular address is proportional to the area of that address. This ensures (in the limit) equal densities at all addresses of the same length.

The converse may be true, but I don't see a ready proof. The contrapositive implies that our object constructed by driving the full square IFS with a random sequence lacking the substring ``12'' cannot be created by a ``nice'' IFS whose transformations map to sub-squares $1$, $2$, $3$, and $4$. A similar argument excludes IFSs based on smaller sub-squares, strongly suggesting that our object is not a ``really nice'' fractal.

Proof Let $c_i$ be the contraction corresponding to $w_i$, then $c_i^2$ is $\vert\det(w_i)\vert$. Let $C$ $=$ $c_1^2$ $+$ $\cdots$ $+$ $c_n^2$, and let $p_i$ $=$ $c_i^2/C$. Let $U$ be a unit square containing our fixed point $\mathcal{F}$, and so the area of $w_i(U)$ is $A(i)$ $=$ $c_i^2$, so you have

$$\frac{p_i}{A(i)} = \frac{1}{C}.$$

For compositions $w_{i_1}$ $\circ$ $\cdots$ $\circ$ $w_{i_k}$, you have

$$\frac{p_{i_1} \cdots p_{i_k}}{A(i_1) \cdots A(i_k)} = \frac{1}{C^k}.$$

This is what I claimed.