 |
| From the graph we see there are edges from each vertex to vertex 1,
and from each vertex to vertex 4. Consequently, 1 and 4 are the full states.
Notice A1 = T1(A) and
A4 = T4(A) are copies of A scaled by
1/2. |
| The only edges leading into vertex 2 come from vertex 1 and from vertex 4.
So the only parts of the attractor in the lower right square are
A21 =
T2(A1) and
A24 =
T2(A4). These are copies
of A scaled by 1/4. |
| The only edges leading into vertex 3 come from vertex 1, from vertex 2, and from vertex 4.
So the only parts of the attractor in the upper left square are
A31 =
T3(A1),
A34 =
T3(A4),
and A321 = T3(A21) and
A324 = T3(A24).
The first two are copies of A scaled by 1/4, the last two are copies of A
scaled by 1/8. |
| This is all. One way to see this is to note in the graph every path through a non-rome
must enter a rome after passing through only finitely many states. In this
case, these paths must be short. |
| rome → 3 → rome |
| rome → 2 → 3 → rome |
|
| |
| To find the rules for the memoryless IFS to generate this fractal, start with
the rules for the romes, T1 and T4 in this case, and form
all the allowed compositions that do not include another application of a rome rule.
These are |
| T2T1,
T2T4,
T3T1,
T3T4,
T3T2T1, and
T3T2T4 |
| The corresponding IFS table is |
| composition |
r |
s |
θ |
φ |
e |
f |
| T1 |
0.5 |
0.5 |
0 |
0 |
0 |
0 |
| T4 |
0.5 |
0.5 |
0 |
0 |
0.5 |
0.5 |
| T2T1 |
0.25 |
0.25 |
0 |
0 |
0.5 |
0 |
| T2T4 |
0.25 |
0.25 |
0 |
0 |
0.75 |
0.25 |
| T3T1 |
0.25 |
0.25 |
0 |
0 |
0 |
0.5 |
| T3T4 |
0.25 |
0.25 |
0 |
0 |
0.25 |
0.75 |
| T3T2T1 |
0.125 |
0.125 |
0 |
0 |
0.25 |
0.5 |
| T3T2T4 |
0.125 |
0.125 |
0 |
0 |
0.375 |
0.625 |
|