Exploding dots: 1

Exploding dots machines arise from patterns of dot replacement, which should, in turn, determine a “base”.

For example the replacement pattern $1 \leftarrow 2$ gives rise to a base $b$ satisfying $1\times b = 2$ , whereas the replacement pattern $2 \leftarrow 3$ gives rise to a rational base $b$ satisfying $2\times b = 3$ .

The replacement pattern $1\vert0\vert0 \leftarrow 0\vert 1\vert 1$ gives rise to an irrational base $b$ satisfying $b^2=b+1$

So for a $2 \leftarrow 1$ machine we see that the base is 2 because $1b=2$ :

On the other hand a $2\leftarrow 4$ machine also has base 2 because $2b=4$ :

So these two machines are representing positive integers as finite sums of “digits” times powers of 2.

What’s different is what counts as a digit.

In the $2 \leftarrow 1$ machine the digits are 0 and 1, while in the $2\leftarrow 4$ machine the (allowable) digits are 0, 1, 2, and 3.

This gives rise to correct, but distinct representations, but both “base 2”, (!!) coming from the ways in which the two different machines operate:

Base 10	Base 2/1	Base 4/2
1	1	1
2	10	2
3	11	3
4	100	20
5	101	21
6	110	22
7	111	23
8	1000	200

The connections between the pattern that defines an exploding dots machine, the digits that are allowable, and the algebraic number that results from the pattern, is not entirely clear.

And what happens if a pattern gives a root of a cubic as the base? Do any of the – possibly complex – roots also work equally well? And what are the allowable digits then?

For that matter, what if take the negative root of $b^2=b+1$ as the base for an exploding dots machine with pattern $1\vert0\vert0 \leftarrow 0\vert 1\vert 1$ ?