| First, recall the Random Algorithm starts with the
fixed point of one of
the Ti. |
| For definiteness, say we start with the fixed point
(x0, y0) of T1. |
| The address of this fixed point is
111... = 1∞. (Here is the reason.) |
| Suppose the first transformation applied is Ti1, the
next Ti2, and so on. |
| What is the effect of these transformations on the
address of the point, and on the address length N region in which the point lies? |
| point | address of the point | address length N region containing the point |
| (x0, y0) | 1∞ |
1N |
| (x1, y1) = Ti1(x0, y0) |
i1111... = i1(1∞) |
i1 1N-1 |
| (x2, y2) =
Ti2(x1, y1) |
i2i1(1∞) |
i2i1(1N-2) |
| (x3, y3) =
Ti3(x2, y2) |
i3i2i1(1∞) |
i3i2i1(1N-3) |
| ... | ... | ... |
| (xN, yN) =
TiN(xN-1, yN-1) |
iN...i3i2i1(1∞) |
iN...i3i2i1 |
| (xN+1, yN+1) =
TiN+1(xN, yN) |
iN+1iN...i3i2i1(1∞) |
iN+1iN...i3i2 |
| ... | ... | ... |
|
| So we see each new transformation applied has this effect on the N-digit
address: discard the right-most digit, shift the remaining
N-1 digits one place to the right, and
insert the new address on the left. |