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. |
|
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. |