Constructing an iterated function system for a unique fractal shape

[Deadstar]

Let A be the following set (NOTE: The part within the red square is NOT part of the original question, I have added that in and it will be explained)

[PLAIN]http://img695.imageshack.us/img695/1037/unleddo.png

After choosing coordinates, construct the iterated function system whose fixed point it A.


Now I find the whole iterated function system a bit odd as it is but I can usually construct ones for most fractals (Koch snowflakes and the like...) But this one is a bit different since it looks like the bottom of the fractal has a different system to the rest of it. I have included an example of the iterated function of the Koch snowflake at the end of this post to give you an idea of what I'm trying to do.

The bottom of the fractal seems like it's missing some parts and the part within the red square is what I think should be added on to the fractal to make any iterated function system work. I'm finding it difficult to explain what I think the pattern is in terms of an actual function system as I keep involving too many parts at once.

I would like to select coordinates such that the 0 occurs at the bottom left of the entire fractal and the length of the largest line is one (which bring problems as the I think the largest line should be the one I added on!).

Letting Phi be the iterated function.
$$\Phi_1(z) = z e^{i\pi/2}$$
$$\Phi_2(z) = z e^{i \pi/2} + \frac{1}{2}$$
$$\Phi_3(z) = \frac{z}{2} e^{i \pi/2} + \frac{1}{4}$$
$$\Phi_4(z) = \frac{z}{4} e^{i \pi/2} + \frac{1}{8}$$

This (in my mind at least) creates the following...

[PLAIN]http://img202.imageshack.us/img202/6851/unledxgi.png

So is this correct? Repeated iterations of this will give the fractal I'm after..?


Below is the iterated function system for the Koch snowflake.

[PLAIN]http://img860.imageshack.us/img860/6989/unledpjt.png



Thanks in advance for any answers!

 

[micromass]

Hi Deadstar!

The fractal as it is drawn can be obtained by an IFS. Here is a piece of software that is extremely handy: http://ecademy.agnesscott.edu/~lriddle/ifskit/download.htm
I've used this software and have succesfully created your fractal with it.

I think you are viewing the fractal a bit wrong. Basically, the fractal consists out of three parts: one upper left, one down left and one down right part. Furthermore the upperleft part is rotated clockwise. You seem to see the fractal as four parts...

Sadly, I cannot draw anything for you here, but I view the fractal as:

$$\begin{array}{cc}\rightarrow & \\ \uparrow & \uparrow \\ \end{array}$$

I suggest playing a bit with that program I found. If you want, I can tell you the affine transformations that generate this IFS...

 

[Deadstar]

Hi Deadstar!

The fractal as it is drawn can be obtained by an IFS. Here is a piece of software that is extremely handy: http://ecademy.agnesscott.edu/~lriddle/ifskit/download.htm
I've used this software and have succesfully created your fractal with it.

I think you are viewing the fractal a bit wrong. Basically, the fractal consists out of three parts: one upper left, one down left and one down right part. Furthermore the upperleft part is rotated clockwise. You seem to see the fractal as four parts...

Sadly, I cannot draw anything for you here, but I view the fractal as:

$$\begin{array}{cc}\rightarrow & \\ \uparrow & \uparrow \\ \end{array}$$

I suggest playing a bit with that program I found. If you want, I can tell you the affine transformations that generate this IFS...

Yeah you can tell me. Do you know how to do it using the notation I have to use? I have to say I still don't understand the basic concept of this. Why is z/3 used and why is 1/3 used for drawing the Koch snowflake? What does z/3 actually mean? It appears to mean - draw a horizontal line of length 1/3 of what the length of the previous line is.

I'll play around with that program for a bit but I'd like to know what the system I said would do. Where I put + 1/2, + 1/4 and + 1/8 perhaps it should have been + z/2, z/4 and z/8..?


Actually I do have one basic question.
Is each Phi part of the iterated function a 'seperate' thing or does one follow the other? Would it matter in which order they were numbered? (e.g in the Koch snowflake, does it matter which one is Phi_1, Phi_2, etc...)


You said I view the fractal wrong but by repeatedly drawing the 'F' shape I put on each new line created, you do create the fractal (well, not the bottom bit but I have no idea how to get that...)

 

[micromass]

Uh, well, I don't exactly follow your notation, but I think this is what you're looking for

$$\Phi_1(z)=\frac{z}{2},~\Phi_2(z)=\frac{z}{2}+\frac{1}{2},~\Phi_3(z)=\frac{e^{-i\frac{\pi}{2}}z}{2}+i$$

Yeah you can tell me. Do you know how to do it using the notation I have to use? I have to say I still don't understand the basic concept of this. Why is z/3 used and why is 1/3 used for drawing the Koch snowflake? What does z/3 actually mean? It appears to mean - draw a horizontal line of length 1/3 of what the length of the previous line is.

Yes, z/3 means that the length is 1/3 of the previous length. We add 1/3 to translate the line over a length of 1/3.

Basically, if we have function f₁,...,f_n in our IFS, then we can start with any shape A₀, and then we start iterating. First, we apply all the functions to get

$$A_1=f_1(A_0)\cup...\cup f_n(A_0)$$

to get the first step of our iteration. We continue doing this, so in general, we do

$$A_{k+1}=f_1(A_k)\cup...\cup f_n(A_k)$$

The figure will "converge" the the end figure A (convergence can be made precise here), and this figure has the properties

$$A=f_1(A)\cup...\cup f_n(A)$$

this is the attractors of the IFS. What if you want to go conversely? Given an attractor A, you should find transformations f₁,...,f_n such that

$$A=f_1(A)\cup...\cup f_n(A)$$

These transformations form the IFS...

I'll play around with that program for a bit but I'd like to know what the system I said would do. Where I put + 1/2, + 1/4 and + 1/8 perhaps it should have been + z/2, z/4 and z/8..?

I don't know off-hand what your equations give. But enter them in the program I gave, and you'll see!

Actually I do have one basic question.
Is each Phi part of the iterated function a 'seperate' thing or does one follow the other? Would it matter in which order they were numbered? (e.g in the Koch snowflake, does it matter which one is Phi_1, Phi_2, etc...)

The order doesn't matter.

You said I view the fractal wrong but by repeatedly drawing the 'F' shape I put on each new line created, you do create the fractal (well, not the bottom bit but I have no idea how to get that...)

Well, your drawing does give the correct view for the upper part of the fractal. But not for the lower part.

 

[Deadstar]

I now see exactly what you mean by three parts. It became much clearer as I tried to compute the area. I started with a square and gradually removed more squares a quarter the size of the previous square at each iteration. (I.e. each new square becomes an 'L' shape)

Currently trying to figure out that program works... Can't work out how to use my notation though.

 

[micromass]

You'll need to translate your complex expression to matrix expressions.

For example, $ze^{i\theta}+a+bi$ would become

$$\left(\begin{array}{cc} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\\ \end{array}\right)\left(\begin{array}{c} x\\ y\\ \end{array}\right)+\left(\begin{array}{c} a\\ b\\ \end{array}\right)$$

This is if you would like to enter the IFS algebraically. You can also enter the IFS graphically by using the Design-window. There you can put in graphically what the IFS does...

 

[Deadstar]

Uh, well, I don't exactly follow your notation, but I think this is what you're looking for

$$\Phi_1(z)=\frac{z}{2},~\Phi_2(z)=\frac{z}{2}+\frac{1}{2},~\Phi_3(z)=\frac{e^{-i\frac{\pi}{2}}z}{2}+i$$

Ok I think I follow this...

I interptret the first three iterations as...

[PLAIN]http://img849.imageshack.us/img849/6053/unledfep.png

although this is more what I think you were intending rather than my understanding of the function. Perhaps though, it should be iz/2..? If my diagram above is right, how does the function know to 'rotate clockwise' to fill in all the details and not just the parts similar to your arrow diagram.

 

[Deadstar]

You'll need to translate your complex expression to matrix expressions.

For example, $ze^{i\theta}+a+bi$ would become

$$\left(\begin{array}{cc} \cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\\ \end{array}\right)\left(\begin{array}{c} x\\ y\\ \end{array}\right)+\left(\begin{array}{c} a\\ b\\ \end{array}\right)$$

This is if you would like to enter the IFS algebraically. You can also enter the IFS graphically by using the Design-window. There you can put in graphically what the IFS does...

Great thanks I'll give that a go.

 

[micromass]

Ok I think I follow this...

I interptret the first three iterations as...

[PLAIN]http://img849.imageshack.us/img849/6053/unledfep.png

although this is more what I think you were intending rather than my understanding of the function.

Hmm, how so? Your figure looks a lot like the fractal you were attempting to describe??

Perhaps though, it should be iz/2..?

It should be -iz/2, which is exactly the expression I posted above, since $$e^{-i\pi/2}=-i$$.

If my diagram above is right, how does the function know to 'rotate clockwise' to fill in all the details and not just the parts similar to your arrow diagram.

I don't really get that...

 

[Deadstar]

Hmm, how so? Your figure looks a lot like the fractal you were attempting to describe??

I mean I know what your function is trying to do but I either don't understand the notation or I don't think it's correct (as in, correct in the notation that I'm supposed to use).

I think it should therefore be...

$$\Phi_1(z)=\frac{iz}{2}$$
$$\Phi_2(z)=\frac{iz}{2}+\frac{1}{2}$$
$$\Phi_3(z)=\frac{e^{i\frac{\pi}{2}}z}{2}+i$$

Minor changes, adding in the i's and changing -ve to a +ve in exponential but that makes more sense to me. Compare it to the Koch snowflake example I posted where z/3 represented a horizontal line and z/3 + 2/3 represented a horizontal line with a jump of 2/3. Perhaps it doesn't matter since it's all about what coordinates you choose...

Thanks for your help so far though, starting to make a bit on sense.