Driven IFS with Forbidden Combinations

Infinite IFSability

From the graph we see there are edges from each vertex to vertex 1, and from each vertex to vertex 2. Consequently, 1 and 2 are the full states. Notice A1 = T1(A) and A2 = T2(A) are copies of A scaled by 1/2.
The only edges leading into vertex 4 come from vertex 2 and from vertex 4. So the parts of the attractor in the upper right square are A42 = T4(A2) (A scaled by 1/4), A442 = T4(A42) (A scaled by 1/8), A4442 = T4(A442) (A scaled by 1/16), ..., and so on, for an infinite cascade of smaller and smaller copies of A.
The only edges leading into vertex 3 come from vertex 2 and from vertex 4. So the only parts of the attractor in the upper left square are A32 = T3(A2) (A scaled by 1/4), A342 = T3(A42) (A scaled by 1/8), A3442 = T3(A442) (A scaled by 1/16), ..., and so on, for an infinite cascade of smaller and smaller copies of A.
Unlike the previous example, here we have arbitrarily long paths through non-romes. These are responsible for the infinite cascades of smaller and smaller copies of A.
2 → 3 → rome
2 → 4 → 3 → rome
2 → 4 → 4 → 3 → rome
2 → 4 → 4 → 4 → 3 → rome
2 → 4 → 4 → 4 → 4 → 3 → rome
..., and
2 → 4 → rome
2 → 4 → 4 → rome
2 → 4 → 4 → 4 → full
2 → 4 → 4 → 4 → 4 → rome
2 → 4 → 4 → 4 → 4 → 4 → rome
...

Return to representing IFS with memory by IFS without memory.