Cyclic Driven IFS
The simplest repeated sequence is 11111...
= 1infinity. Starting with (0.5, 0.5),
applying T1
repeatedly produces a sequence of points converging to the
lower left corner of the unit square, as pictured below. This is no surprise:
T1(x, y) = (x/2, y/2), so the sequence of points is
(1/2, 1/2),
(1/4, 1/4), (1/8, 1/8),
(1/16, 1/16), and so on. The limit of this sequence is is a single
point (obvious from the picture - but we'll use this idea to understand the next
picture, too). It has address 11111... = 1infinity.
Moreover, this is the fixed point of T1. Say
(x*, y*) is the point with address 1infinity. Then
T1(x*, y*) has address
1(1infinity) = 1infinity. That is,
T1(x*, y*) = (x*, y*),
i.e., (x*/2, y*/2) = (x*, y*), so (x*, y*) =
(0, 0).
By similar arguments we see cycles 2infinity,
3infinity, and 4infinity converge to the points
(1, 0), (0, 1), and (1,1), respectively.
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| 3infinity
| 4infinity
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| 1infinity
| 2infinity
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The next simplest repeated sequence is
121212... = (12)infinity.
Applying T1 and T2 alternately produces a sequence of points
converging to two points along the x-axis. Click on the picture to see the animation
generating it.
These points have addresses (12)infinity
and (21)infinity. To see
this, recall the relation between the address and the order in which the transformations
are applied. The sequence 121212... gives points in regions with addresses
and so on.
Click here for an animation illustrating this.
Alternate points in the sequence have addresses
1, 121, 12121, ... , and 21, 2121, ... .
The first converges to (12)infinity, the second to (21)infinity.
Say the first is (x1, y1), the second
(x2, y2).
T2(x1, y1) has address
2(12)infinity = (21)infinity,
so T2(x1, y1) =
(x2, y2). Similarly,
T1(x2, y2) =
(x1, y1). Combining these, we see
T1T2(x1, y1) =
(x1, y1), and
T2T1(x2, y2) =
(x2, y2). From the first we obtain
(x1, y1) =
T1T2(x1, y1)
= T1(x1/2 + 1/2, y1/2)
= (x1/4 + 1/4, y1/4).
So (x1, y1) =
(1/3, 0). A similar argument
gives (x2, y2) = (2/3, 0).
| How do the limiting points depend on the choice of (0.5, 0.5)
as starting point? Not at all. Click the picture on the right for an illustration.
Here we follow the sequences generated from three different starting points, all with
vertices selected in the order 121212... = (12)infinity. Blue lines for one starting
point, green for the second, red for the third.
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Here are some sequences of points generated by other cycles.
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| (13)infinity
| (14)infinity
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| limit points of (14)infinity
| What cycle is this?
|
The general pattern should be clear. For example, the repeating
pattern (123)infinity produces a sequence converging to three points,
one with address (321)infinity, one with address
(132)infinity, and one with address
(213)infinity. Click the left picture for an
animation. The right shows the three points to which these sequences converge, along with
their length 3 addresses.
The first is the fixed point of
T1T2T3, the second of
T2T3T1, the third of
T3T1T2. Solving the fixed point
equations gives the coordinates of each point.
Suppose the pattern (312)infinity. The first
few points are shown on the left. Click the picture for an animation. THe right
shows the limiting pattern. Does it look familiar? It should; it is the pattern
generated by (123)infinity.
To understand why this is so, write the first several terms of
both sequences
| 123123123123123123123123123123...
| | 312312312312312312312312312312...
|
The second sequence is just the first shifted two terms to the
left. The first sequence is the same as the second, but starting from
T2T1(0.5, 0.5) instead of from (0.5, 0.5).
As with the fractals generated by regular IFS, here the final pattern does not depend on the
starting point. We prefer to start with (0.5, 0.5) because this is the
"most neutral" choice.
Can you find an order of cycling through 1,2, and 3 that converges to
a different triple of points? What about (213)infinity?
The number of points to which the sequence converges is the
length of the (shortest) repeated pattern. (The reason for "shortest" is that
the sequences (123)infinity and
(123123)infinity converge to the same three points.)
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