Drawing Fractals from IFS Equations

[kcurse21]

Homework Statement

If Gn is the nth stage of the construction then Gn+1 = f1(Gn) ∪ f2(Gn) ∪ f3(Gn).
Beginning with the unit interval sketch the next three stages of the construction and calculate the fractal/box counting dimension of the object as n → ∞.

Relevant Equations

Let X = [0, 1]. Consider the Iterated Function System
f1(X) = 1/3X
f2(X) = 1/3X+1/3(1,1)
f3(X) = 1/3X+2/3(1,0)

Attempt
The interval sketch is obviously a line from 0, 1.
F1 would cut the interval line by a third similar to the Cantor set.
F2 would cut the interval line by a third and then there is a transformation that moves to the point (1/3, 1/3).
F3 would cut the interval line by a third and then there is a transformation that moves to the point (2/3, 0).
Based on the cantor set, you would assume that the third would be placed underneath the interval sketch but my problem arises with f2. F3 I'm assuming you cut the interval sketch by a third at the point (2/3, 0) which would give you the other side of the cantor set per se. But I have no idea what to do with F2 and how to draw this as a whole. I will provide a sketch of the Koch curve so it is understood what I mean.

 

[Mark44]

Homework Statement:: If Gn is the nth stage of the construction then Gn+1 = f1(Gn) ∪ f2(Gn) ∪ f3(Gn).
Beginning with the unit interval sketch the next three stages of the construction and calculate the fractal/box counting dimension of the object as n → ∞.
Relevant Equations:: Let X = [0, 1]. Consider the Iterated Function System
f1(X) = 1/3X
f2(X) = 1/3X+1/3(1,1)
f3(X) = 1/3X+2/3(1,0)

What do 1/3(1,1) and 2/3(1,0) mean?

The interval sketch is obviously a line from 0, 1.
F1 would cut the interval line by a third similar to the koch curve.

So you're removing the middle third of the line segment? That seems to be what you're saying.

F2 would cut the interval line by a third and then there is a transformation that moves to the point (1/3, 1/3).

Not following this at all. Starting from the interval [0, 1], after the first step you have two subintervals [0, 1/3] and [2/3, 1]. In the Koch Snowflake construction, you replace the middle third with two segments each of length 1/3, just like in the drawing you show, the second from the top. With coordinates, the two added segments run from the point (1/3, 0) to (1/2, √3/6), and from (1/2, √3/6) to (2/3, 0).

F3 would cut the interval line by a third and then there is a transformation that moves to the point (2/3, 0).
Based on the cantor set, you would assume that the third would be placed underneath the interval sketch but my problem arises with f2. F3 I'm assuming you cut the interval sketch by a third at the point (2/3, 0) which would give you the other side of the cantor set per se.

I don't know what you mean about placing one of the thirds underneath the sketch, nor what you mean by the other side of the Cantor set.

But I have no idea what to do with F2 and how to draw this as a whole. I will provide a sketch of the Koch curve so it is understood what I mean.

 

[kcurse21]

What do 1/3(1,1) and 2/3(1,0) mean?
So you're removing the middle third of the line segment? That seems to be what you're saying.
Not following this at all. Starting from the interval [0, 1], after the first step you have two subintervals [0, 1/3] and [2/3, 1]. In the Koch Snowflake construction, you replace the middle third with two segments each of length 1/3, just like in the drawing you show, the second from the top. With coordinates, the two added segments run from the point (1/3, 0) to (1/2, √3/6), and from (1/2, √3/6) to (2/3, 0).
I don't know what you mean about placing one of the thirds underneath the sketch, nor what you mean by the other side of the Cantor set.

I'm assuming 1/3(1,1) and 2/3(1,0) means 1/3 of the point (1,1) and 2/3 of the point (1,0) but those are the exact equations I've been given.
I meant to say Cantor set not Koch curve, sorry I got confused there.
For F2 and F3 I just wrote what I thought might happen because you require to put an attempt at the answer, I'm not sure if its right, I just had a go. But from what I know the 1/3X relates to scaling the original segment and the + involves moving the point.
When I say underneath the sketch, when you draw the Cantor set, you draw a line, then underneath two lines, one in the first third and one in the last third and continue drawing underneath each split. By the other side I meant the right hand side line, as compared to the left hand side which is the F2 of the Cantor set.

 

[Mark44]

I'm assuming 1/3(1,1) and 2/3(1,0) means 1/3 of the point (1,1) and 2/3 of the point (1,0) but those are the exact equations I've been given.

A point has no length, so you can't take 1/3 of a point or 2/3 of a point.

 

[kcurse21]

A point has no length, so you can't take 1/3 of a point or 2/3 of a point.

The fractals I've seen in the past were matrices so I assumed it would be that, maybe it's a direction? and the 1/3 is the length of the direction? That was the part I was struggling with the most because none of the examples I have seen use that notation and I know that the first part is usually length and the second is usually a location.
Another question I completed was the same but of the Koch curve and the numbers in brackets were the point at which to take 1/3 of. For example, f2(x) = 1/3R(π/3)x + (1/3,0), was to take 1/3 of the rotation, R, at the point (1/3,0).

 

[Mark44]

and the 1/3 is the length of the direction? That was the part I was struggling with the most because none of the examples I have seen use that notation and I know that the first part is usually length and the second is usually a location.

A direction doesn't have a length. You could take 1/3 of an angle though.

Are the examples talking about vectors and points? You could have a vector of a certain length, whose tail was at a given point. If you go back to what I said in post #2, third paragraph, another way to describe what I said is this:
From point (1/3, 0), create a line segment of length 1/3 in the direction $\pi/3$. That creates the segment going up to the point (1/2, √3/6). So the starting point is (1/3, 0) and the direction is $\pi/3$. The downward sloping segment is also of length 1/3, starts at (1/2, √3/6) and goes downward with the direction angle $-\pi/3$. Both angles are measured relative to the positive horizontal axis.

Another question I completed was the same but of the Koch curve and the numbers in brackets were the point at which to take 1/3 of. For example, f2(x) = 1/3R(π/3)x + (1/3,0), was to take 1/3 of the rotation, R, at the point (1/3,0).

I don't know how to make sense of this function formula. What is x? What does 1/3R(π/3)x mean? Whatever this expression means, how do you add a point to it?

 

[kcurse21]

A direction doesn't have a length. You could take 1/3 of an angle though.
√
Are the examples talking about vectors and points? You could have a vector of a certain length, whose tail was at a given point. If you go back to what I said in post #2, third paragraph, another way to describe what I said is this:
From point (1/3, 0), create a line segment of length 1/3 in the direction $\pi/3$. That creates the segment going up to the point (1/2, √3/6). So the starting point is (1/3, 0) and the direction is $\pi/3$. The downward sloping segment is also of length 1/3, starts at (1/2, √3/6) and goes downward with the direction angle $-\pi/3$. Both angles are measured relative to the positive horizontal axis.
I don't know how to make sense of this function formula. What is x? What does 1/3R(π/3)x mean? Whatever this expression means, how do you add a point to it?

You are not creating a line segment, it says to take 1/3X to start and X is the unit interval. Each equation from then on does something to the unit interval and you have to draw the total drawing using all functions up to that one. For instance, F3 includes F2 and F1.
The example equation matches the picture of the Koch curve I posted in the original post. R is a transformation matrix, a rotation. However, I am looking for the drawings for the original equations I posted, that was just an example which uses the same notation.

The function is a function of a map. So you take 1/3 of X, which I stated in the question (applies to my question and the example) and times it by a rotation matrix (contains cos and sin) of π/3. So cos(π/3), etc. That's the function. F1= 1/3X. So in the picture, the first line is the unit sketch of X. The second line is F1, the third line is F1 and F2, the fourth line F1, F2 and F3 and so on.

I don't believe the addition is 'adding a point'. I think it moves what you have done to that point. So for instance, if I had F1= 1/3X + (2/3,0) you would take 1/3 of X but starting at 2/3. So you would have a line from 2/3->1. I'm not 100% sure but that's what it seems like to me.

 

[jim mcnamara]

From a site describing Koch curve:

An iteration building the Koch curve starts at x, advances by v, advances by v rotated by 60 degrees, advances by v rotated -60 degrees, and finally advances by another v, reaching y. x seg y produces this expansion. koch takes a list of points and expands segments between consecutive ones, producing another list.

I take seg() as a function for generating a sequence of y points. By integer iteration.

The instruction set shown is confusing to say the least. So I think the (1/3) is 60 degree rotation mentioned above.

Ian Stewart has written more than a few really good books on fractals. You might want to try the school library.

 

[kcurse21]

From a site describing Koch curve:I take seg() as a function for generating a sequence of y points. By integer iteration.

The instruction set shown is confusing to say the least. So I think the (1/3) is 60 degree rotation mentioned above.

Ian Stewart has written more than a few really good books on fractals. You might want to try the school library.

Unfortunately that's all I was given for the question and anything I have added on is all inferences. Just the homework statement and the relevant equations. I made an attempt at describing it because, firstly, I tried to do the question myself and secondly it is a requirement for posting. The Cantor set does not have any rotations and is in a similar format that's why I'm not sure if it is a rotation or not because if it was it should be in radians not written as 1/3.

I currently do not have access to the library due to current circumstances. I have tried to search similar questions online but I haven't seen anything written with this notation and I'm struggling to come up with the drawing. I also need to draw by hand, not coding as presented in the picture attached.

 

[haruspex]

You are not creating a line segment, it says to take 1/3X to start and X is the unit interval. Each equation from then on does something to the unit interval and you have to draw the total drawing using all functions up to that one. For instance, F3 includes F2 and F1.
The example equation matches the picture of the Koch curve I posted in the original post. R is a transformation matrix, a rotation. However, I am looking for the drawings for the original equations I posted, that was just an example which uses the same notation.

The function is a function of a map. So you take 1/3 of X, which I stated in the question (applies to my question and the example) and times it by a rotation matrix (contains cos and sin) of π/3. So cos(π/3), etc. That's the function. F1= 1/3X. So in the picture, the first line is the unit sketch of X. The second line is F1, the third line is F1 and F2, the fourth line F1, F2 and F3 and so on.

I don't believe the addition is 'adding a point'. I think it moves what you have done to that point. So for instance, if I had F1= 1/3X + (2/3,0) you would take 1/3 of X but starting at 2/3. So you would have a line from 2/3->1. I'm not 100% sure but that's what it seems like to me.

As far as I can make out, we are working in 2D. We start with just the interval [0,1] on the y axis; next, we elevate the middle third by 1/3 so that it runs from (1/3,1/3) to (2/3,1/3).
(Not sure what happens at the boundaries. Maybe we should say all the intervals are half open. Or maybe it doesn't matter if we get multiple y values for the same x, etc.)
The notation

f1(X) = 1/3X​

f2(X) = 1/3X+1/3(1,1)​

f3(X) = 1/3X+2/3(1,0)​

might be clearer as

f1(X) = 1/3X​

f2(X) = 1/3X+(1/3,1/3)​

f3(X) = 1/3X+(2/3,0)​

 

[kcurse21]

As far as I can make out, we are working in 2D. We start with just the interval [0,1] on the y axis; next, we elevate the middle third by 1/3 so that it runs from (1/3,1/3) to (2/3,1/3).
(Not sure what happens at the boundaries. Maybe we should say all the intervals are half open. Or maybe it doesn't matter if we get multiple y values for the same x, etc.)
The notation

f1(X) = 1/3X​

f2(X) = 1/3X+1/3(1,1)​

f3(X) = 1/3X+2/3(1,0)​

might be clearer as

f1(X) = 1/3X​

f2(X) = 1/3X+(1/3,1/3)​

f3(X) = 1/3X+(2/3,0)​

I think you are correct and that is what the notation means, but it does seem really weird that the question would be written like that. My guess now is that it is correct in saying you have a union of the intervals (0, 1/3) @ y=0, (1/3, 2/3) @ y=1/3 and (2/3, 1) @ y=0. So you are elevating the middle third. Then to continue drawing this, every line you have you elevate the middle third. Something like the image I attached (please excuse the lack of scale).

Then you would have number of edges as 1, 3, 9, 27, length 1, 1/3, 1/9, 1/27 and is the last one iterations? as in 0, 1, 2, 3, for each drawing respectively.

 

[haruspex]

I think you are correct and that is what the notation means, but it does seem really weird that the question would be written like that. My guess now is that it is correct in saying you have a union of the intervals (0, 1/3) @ y=0, (1/3, 2/3) @ y=1/3 and (2/3, 1) @ y=0. So you are elevating the middle third. Then to continue drawing this, every line you have you elevate the middle third. Something like the image I attached (please excuse the lack of scale).

Then you would have number of edges as 1, 3, 9, 27, length 1, 1/3, 1/9, 1/27 and is the last one iterations? as in 0, 1, 2, 3, for each drawing respectively.View attachment 270113

Yes, those sketches look right.
What do you get for the box count dimension?

 

[kcurse21]

Yes, those sketches look right.
What do you get for the box count dimension?

So that would be d=log3/-log(1/3)=1?

 

[haruspex]

So that would be d=log3/-log(1/3)=1?

Yes, I think so.

 

[willem2]

It's a bit more interesting to start with the square with x and y between 0 and 1 and see what happens to that after 3 generations.

Or even better, a triangle between the fixed points of the transformations

This is just a very open version of a Sierpienski triangle. If you'd use f_i(X) = (1/2)X + Xi instead of (1/3)X you'd get a connected Sierpienski triangle.

Images produced with IFS construction kit by Larry Riddle
https://larryriddle.agnesscott.org/ifskit/