Extraction of strings

The distribution of digits here differs from the previous case. Before filtering, each $1$ has probability $1/4$ of being followed by the string $2$, probability $(1/4)\times (1/16)$ of being followed by the string $122$, and in general $(1/4)\times (1/16)^k$ of being followed by $k$ $1$s and $(k+1)$ $2$s. So, the number of $1$s is reduced by:

$$\frac{1}{4}\sum_{n=0}^{\infty} \frac{1}{16^n} = \frac{4}{15}.$$

A symmetrical argument applies to the $2$s, so after filtering there are $11/15$ as many $1$s and $2$s as in the unfiltered sequence. The area of sub-square $1$ equals that of sub-squares $3$ and $4$, and the area of sub-square $2$ is $3/4$ that of sub-square $3$, so we'd like $(11/15)p_1$ $=$ $p_3$, and $(11/15)p_2$ $=$ $(3/4)p_3$. Since the $p_i$ sum to $1$, you have

$$2p_3 + \frac{11}{15} \times \frac{4}{3} p_3 + \frac{15}{11} p_3 = 1$$

$$\Longrightarrow \frac{44}{193} = p_3, \frac{45}{193} = p_2, \frac{60}{193} = p_1.$$

Plugging these values into the Fractal Sequence applet gives a fairly uniform density of length $1$ addresses (although sub-square $2$ seems slightly sparse -- a bug perhaps?).

For the same reason as the truncation method, it is not possible to ``tweak'' the $p_i$ so as to have a uniform density of points in each of the addresses of length $2$. Indeed, this fact can be used to prove that the resulting object is not a ``really nice'' fractal.