| One can be generated by a regular IFS with a finite number of transformations, |
| one by a regular IFS with infinitely many transformations, and |
| one cannot be generated by a regular IFS at all. |
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| Central to solving this problem is seeing if the shape contains a complete, scaled copy of the whole shape. This happens if at least one of the Ti is a rome; that is, if Ti can immediately follow all four of the Tj. (Think "All Tj lead to Ti," reminiscent of, "All roads lead to rome.") |
| The middle and right shapes certainly contain complete scaled copies of themselves; the left shape does not. |
| The middle shape can be generated by a regular IFS with finitely many transformations. |
| The right shape can be generated by a regular IFS with infintely many transformations. |
| The left shape cannot be generated by any regular IFS. |
| Motivated by these examples, here are the results. Proofs can be found here. This paper is an elaboration of a student project in from the Autumn, 1998, Math 190 class at Yale. THere are three cases. |
| A If there are no romes, the shape cannot be generated by a regular IFS. |
| B If (i) there is a rome, |
| (ii) there is a sequence of allowed transitions from a rome to every non-rome, and |
| (iii) there are no cycles of allowed combinations of non-romes, |
| then the shape can be generated by a regular IFS with finitely many transformations. (By a cycle we mean a sequence that starts and ends with the same transformation.) |
| C If (i) there is a rome, |
| (ii) there is a sequence of allowed transitions from a rome to every non-rome, and |
| (iii) there is at least one cycle of allowed combinations of non-romes, |
| then the shape can be generated by a regular IFS with infinitely many transformations. |
Return to Restricted IFS.