IFS with Memory

Four copies of a 1-step memory picture in a 2-step memory picture

Note the vertical lines at x = 1/3 and x = 2/3.
Next, note the horizontal lines at y = 1/3 and y = 2/3.
These lines were discussed in lines whose endpoints belong to 2-cycles.
For example, the horzontal lines have endpoints
(31)infinity (42)infinity
(13)infinity (24)infinity
and are generated by allowing the transitions
3 → 1, 4 → 1, 3 → 2, 4 → 2, 1 → 3, 2 → 3, 1 → 4, and 2 → 4.
From our study of which 2-step memory configurations correspond to 1-step memory configurations, we see we need these 32 transitions
* → 3 → 1, * → 4 → 1, * → 3 → 2, * → 4 → 2, * → 1 → 3, * → 2 → 3, * → 1 → 4, and * → 2 → 4.
These transitions are encoded in the table on the left, yet the table on the right generates the picture above.
 
The table that generates this picture does not contain the table that generates these lines. How is this possible?
Think a bit, then look here for a hint, how to deal with a similar situation.

Return to Exercise 1 patterns.