Perimeter of the Koch Snowflake « Reasonable Deviations

Perimeter of the Koch Snowflake

By Rod Carvalho

A couple of days ago, I wrote a post about the Koch snowflake and how to compute its area. However, the perimeter of the Koch snowflake is (in my humble opinion) more interesting than its area.

As I wrote in my previous post, we draw the Koch snowflake by starting with an equilateral triangle, and then following some simple rules ad infinitum. At each iteration $k \geq 0$ we will have a snowflake, a regular polygon of $n_k$ sides, each of length $l_k$ . As we have seen before, at each iteration, the number of sides of the snowflake is

$n_k = 4^k 3$

and the length of each side is given by

$l_k = 3^{-k} x$ .

The perimeter of the snowflake is simply

$P_k = n_k l_k = \left(\displaystyle\frac{4}{3}\right)^k P_0$ ,

where $P_0 = 3 x$ is the perimeter of the initial snowflake (i.e., the equilateral triangle whose side is $x$ ). Note that the perimeter of the snowflake is given by a geometric progression of common ratio $r = 4/3$ (which is $> 1$ ), and therefore the progression $\{P_k\}_{k=0}^{\infty}$ will diverge: the perimeter of the snowflake will tend to infinity as $k \rightarrow +\infty$ . The Koch snowflake has a finite area, but a boundary of infinite length! Wow! Ain’t that cool!?

Note also that the boundary of the Koch snowflake is continuous but not differentiable anywhere.

Tags: Fractals, Koch Snowflake

This entry was posted on December 28, 2007 at 4:17 am and is filed under Recreational Math. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

62 Responses to “Perimeter of the Koch Snowflake”

Vladimir Zakharov Says:
December 28, 2007 at 8:45 am | Reply
Fancy a fractal, the area of which will tend to zero, and the perimeter – to infinity. Do you think that’s possible?
Ali Says:
December 28, 2007 at 10:47 pm | Reply
for Koch snowflake the new triangles radiate “outward” from previous line. Consider the case that those triangles oscillate between radiating outward and radiating inward.
I think you may get a zero area with infinite perimeter.
eldila Says:
December 29, 2007 at 12:31 am | Reply
I agree that the cool part of the problem is that the Koch Snowflake has infinite perimeter while containing a finite area. However, it seemed less obvious to me that a Koch Snowflake contained a finite area as opposed to an infinite perimeter. I always assumed that all fractals had a infinite perimeter. However, this is probably not the case now that I think about it.

It would be fun if someone to could come up with a fractal that had a finite perimeter and an infinite area (if it is even possible).
rod. Says:
December 29, 2007 at 1:19 am | Reply
@ Vladimir

I have been thinking about. To be specific, I was thinking of possible modifications to the Koch snowflake algorithm so that the area would tend to zero while the perimeter would tend to infinity. So far I have found nothing interesting, but there may be fractals with zero area and infinite perimeter… I don’t know, I am an enthusiast, not a professional mathematician, LOL ;-)

@ Ali

Your suggestion is interesting, However, I am more comfortable with the idea of “adding” area than “removing” area. In any case, I will think about it. Mathematically, I may get to some results, but I’d like to visualize them first. Maybe I can write a MATLAB script that draws triangles radiating outward/inward. Could be cool.

@ eldila

I think that both the area and the perimeter are interesting. A domain with a fractal boundary is quite a cool thing: I start wondering whether one can solve simple PDEs over such domains. If I remember right, I once read something about solving wave equations over fractal domains… you could name that “how does a fractal sound?” ;-)
Vladimir Zakharov Says:
December 29, 2007 at 8:24 am | Reply
@rod

I think it’s possible if you make a modification to Koch’s snowflake, but with triangles added pointing their pikes inside the figure, not outside. Than at n->infinity you’ll get something which will have no area but will be long as hell in perimeter :) I haven’t done any calculations or even drawings yet, so it’s just a clue as of now.

@elidla

I believe that’s impossible, and most probably there already is already some theorem in mathematical analysis stating that a shape with inifinte area and finite perimeter is not possible.
Vladimir Zakharov Says:
December 29, 2007 at 8:29 am | Reply
@ Ali

Whoops, just noticed you’ve proposed the same earlier. Gotta spend more time reading comments.
Vladimir Zakharov Says:
December 29, 2007 at 10:05 am | Reply
Nope… unfortunately it won’t have zero area.. it converges to some finite one
Ali Says:
January 1, 2008 at 4:05 am | Reply
Vladimir
you are right, It might converge to a circle but not zero.

I was thinking in one D, if you start with a line and remove middle part of line and repeat the process at limit you will have infinitely many end points but no length. I wonder if there is an equivalent process in 2D.

Happy new year everyone.
rod. Says:
January 2, 2008 at 7:53 am | Reply
@ Vladimir & Ali

If the triangles “radiate” inwards, then we will be “removing” area from the initial equilateral triangle. So far, so good. The recursion formula for the original problem

$A_{k+1} = A_k + n_k a(l_{k+1})$

will now become:

$A_{k+1} = A_k - n_k a(l_{k+1})$ .

because we are “removing” area (instead of “adding”). Since $A_0 = a(x)$ , we obtain

$A_k = a(x) - \displaystyle\frac{3}{4} \displaystyle\sum_{m=1}^k 4^m a(x/3^m)$

or , in a more compact form

$A_k = a(x) \left[1 - \displaystyle\frac{3}{4} \displaystyle\sum_{m=1}^k (4/9)^m \right]$ .

This closed-form expression can be written simply as

$A_k = a(x) \left[1 - \displaystyle\frac{3}{5}\left(1-(4/9)^k\right)\right]$

or, in an even more compact form,

$A_k = a(x) \left[\displaystyle\frac{2}{5} + \displaystyle\frac{3}{5}\left(\frac{4}{9}\right)^k\right]$ .

Finally, we can conclude that

$\displaystyle\lim_{k \rightarrow +\infty} A_k = \displaystyle\frac{2}{5} a(x)$ ,

which means that the area of this modified Koch snowflake will not be zero.
David Says:
January 2, 2008 at 9:46 pm | Reply
Is the only goal of the radiating-inward triangles to remove area? When you are ‘done’ will you really have two perimeters? Do you think that you need to attach an outer edge to an inner edge a la moebius? Does the Koch snowflake need to be planar?
rod. Says:
January 3, 2008 at 1:01 am | Reply
David,

If the triangles radiate inwards, we will have a polygon with an ever-increasing number of sides. I don’t know what you mean by “two perimeters”.

The Koch snowflake does not need to be planar. We could generalize to 3 dimensions, which would make things really cool. For instance, instead of starting with an equilateral triangle, start with a regular tetrahedron, and then add smaller and smaller tetrahedra to it. In the end, we have a 3-D snowflake, a fractal solid, so to say.
David Says:
January 3, 2008 at 2:58 pm | Reply
Hmm, I didn’t think clearly about the radiating inward proposition. When I first read it, I imagined an alternating inward-outward scheme that generated an annulus. But now that I reconsider it, it isn’t a reasonable outcome on a lot of accounts.

Now I get it, though. Radiating inwards squeezes the area out. It’s an interesting problem. When you calculate the perimeter of a figure-eight, how do you account for the crossover point? The radiating-inward-Koch-snowflake has an infinite number of crossover points.
Shubhendu Trivedi Says:
February 27, 2008 at 4:44 am | Reply
@Vladimir

Don’t you think that a Sierpinski Triangle has zero area and an infinite perimeter?
Or did you just mean to modify the Koch curve to get such a fractal?
Shubhendu Trivedi Says:
February 27, 2008 at 5:09 am | Reply
I think a Sierpinski triangle is an example wherein you can have a fractal of zero area.
To find the area of the Sierpinski Triangle. Assume that we have a triangle of area
A0=1.
Now suppose at the first iteration we remove one of the triangles out of the four equal triangles inside the initial triangle with area A0.
So after the first iteration, the area inside the figure would be
A1=(3/4)x1;
Similarly in the second iteration we remove one quarter of each of the three triangles in the figure above.
Hence, the total area of the second iterate will be
A2=(3/4)(3/4)x1;
This you may extend till the “nth” iteration. This will get arbitrarily small as the number of iterations increases. So a Sierpinski gasket will have zero area.
Now a similar approach may be used to find out the total boundary length of the gasket. The boundary in the nth iteration will be the sum total of the perimeters of all the triangles that have remained in the gasket. Now this number will be arbitrarily large. And would tend to infinity.

Sierpinski gaskets have been shown to work very well in wireless receivers.
Shubhendu Trivedi Says:
February 27, 2008 at 6:21 am | Reply
If we take the Koch snowflake as described in the blog entry, then it’s area is about 1.6 times the area of the original triangle used to generate it. The Koch snowflake is therefore an example of a shape with a finite area enclosed within an infinite boundary as Rod pointed out.
This seems contrary to how we think, but is characteristic of many shapes in nature.
For example if we take all the arteries, veins and capillaries in the human body, then they occupy a relative small fraction of the body. Yet if we were able to lay them out end to end, we would find that the total length would come to over 60000 kilometres. This example i am quoting from an article i had copied years ago. :)
Shubhendu Trivedi Says:
May 8, 2008 at 6:35 am | Reply
Hey i came across an article in a dynamical systems text that confirms the above.
kram Says:
June 8, 2008 at 7:58 pm | Reply
I’m searching for a special kind of sierpinski gasked and koch snowflake combination…. If you use a sierpinski gasket at iteration-lvl 1 as a start of the koch flake and every triangle you add at the n-th iteration of the Koch flake is a sierpinski gasket at the same n-th iteration….

Anyone who can visualize that? :) – or even calculate the area… (The perimeter most likely is zero) – or the fractal dimension :) (will it have the Dimension of the Koch Snowflake or the dimension of the Sierpinski gasket? Or somewhere in between?)
rod. Says:
June 8, 2008 at 8:53 pm | Reply
@ kram

I like the idea of combining Koch flakes and Sierpinski triangles, but I am afraid I did not understand the scheme you proposed. Could you please explain it in greater detail?

This kind of brainstorming is much easier when one is in meatspace, talking to other people in front of a white board. In cyberspace, conveying one’s ideas in an efficient manner is much harder.
Shubhendu Trivedi Says:
June 8, 2008 at 10:32 pm | Reply
If i understood you right, then i have thought about it two three times.

However one thing to be noted it that in a Sierpinski gasket, the triangles formed are two dimensional “whole” laminas. The idea for a koch curve would be very difficult to apply to laminas in practice. The koch curve developed here is for a line basically and these have been combined to form a triangle. The situation changes when you take a lamina.
However i have never thought of trying to model it. If we make assumptions to simplify the above problem, then in first thought i was of the opinion that if at all tried the structure will shrink from all sides and therefore the area will approach zero.

I have never given serious thought beyond some visualization in the brain. I think after i get back home. This deserves very serious contemplation and a model in NetLogo.
Shubhendu Trivedi Says:
June 8, 2008 at 10:45 pm | Reply
I thought about it a little more. And i think i jumped the gun in the above comment. It is not necessarily a lamina.

I now think it is possible. And very much so. :D
Apologies.
Shubhendu Trivedi Says:
June 8, 2008 at 10:57 pm | Reply
And when i said “lamina”, it was in the following context. I was not trying to say that it had to be necessarily a plate kind of an object.

If you consider a koch curve as described above. Consider one side. The V figure formed looks like a triangle but is not. It has two sides only. So we can not apply the L-Systems rules for the Sierpinski gasket to just that figure per say. So i said lamina as it would still be a complete triangle though realizing it would be rather difficult LOL.

I think after playing around with it. It should be possible.
However i am also hoping i understood your idea correctly.
kram Says:
June 9, 2008 at 3:52 pm | Reply
I think, you did… though, I’m sure, the way I meant it wont go to an area of zero:

I didn’t mean to use a perfect sierpinski gasket for every triangle on the koch flake, but for the very first triangle, I’d use the sierpinski gasket at iteration 1 (where there are three triangles, around a hole which together shape the outline of a bigger triangle…)
And for the three triangles, added in the next iteration of the Koch flake, use a sierpinski gasket of the next iteration aswell…

So, the final Koch snowflake would be made of an infinite amount of sierpinski gaskets, somewhere between the iteration-levels 1 (in the middle) and infinite (towards the frame)…
________________

I also tried another approach of getting Koch Snowflake where you also get some kind of interior… (This doesn’t use any sierpinski gasket):

If you rotate the first triangle by 180°, you get the iteration-lvl 2 of the Koch flake.
There, you have six outer triangles, around a hexagon. If you split them away from the central hexagon, you can rotate them again by 180°, which gives you the iteration level 3.
It looks like six Koch snowflakes of iteration level 2, building the frame of one of level three. Each of the level 2 flakes also is built by six level 1 flakes. (And each level 1 flake is made from a six-pointed star, or if you look more closely, by a hexagon, surrounded by six triangles)

- I dunno, if it’s clear what I want to express, so I’ll up an image of one made by hand to the iteration level 4. :)
kram Says:
June 9, 2008 at 5:47 pm | Reply
Ok, I just noticed: The inner “koch snowflake” isn’t a real one: It’s ends aren’t triangles but hexagons… – though, it’s still a snowflake… Maybe even a better approximation of a “perfect” snowflake :) – I also didn’t just do 4 but 5 iterations now :)

http://img292.imageshack.us/my.php?image=multiflake5512id5.png
kram Says:
June 14, 2008 at 11:02 am | Reply
No thoughts anymore, anyone?
rod. Says:
June 15, 2008 at 3:28 am | Reply
@ kram,

Sorry for the late reply. I have been trying to find the time to comment, but you and Shubhendu gave me too much info to digest, and now my brain has crashed ;-)

OK, a bit more seriously now:

That image you included is pretty cool! How did you generate that image? Manually, or automatically?

I like your idea of building Koch snowflakes out of Sierpinsky gaskets — quite an original thought IMHO. Apart from the fact that such a snowflake would be quite aesthetically pleasing, I like the fact that the “density” of points decreases as one travels away from the center of the snowflake. I am not sure if you understand what I am saying here.

Unfortunately, I don’t understand your 2nd idea (the one depicted in that image). Could you please explain it one step at a time once again? If you have images depicting the flake at each intermediate step, it would make things much easier for me. Like I said, these things are much easier to understand in front of a whiteboard… oh well.
kram Says:
June 15, 2008 at 3:37 pm | Reply
Yeah, I also think, it’s an original idea, especially as I can’t find any images of this via googling or similar…

I manually did that because I’ve no idea how an L-system or a mathematical description would look like… Geometrically, it’s pretty simple.

I did it in Blender, a Free OpenSource 3D-program… ( http://www.blender.org/ )

Pretty simple to do, but getting more and more time consuming. While the first step takes about 0.1 secs, later you’d need hours xD

I created an unfilled, regular triangle and rotated it, so that one of it’s peeks points exactly upwards. (Iteration #1)

Then, I copied the triangle and rotated it by 180°. Now, I got that typical hexagram six pointed star.

I connected the two triangles to make one out of them. There now are 6 smaller triangles with side lenght a/3, with a hexagon in the middle. (Iteration #2)

Those six smaller triangles where rotated again by 180°, each and the whole mesh was merged once again. (Iteration #3)

The more detailed you get, the slower it is and the more likely you make a mistake :S – As said, I managed to do this up to iteration #5.

I’ll upload an other, probably more explanatory image, soon :)
kram Says:
June 15, 2008 at 3:57 pm | Reply
http://i154.photobucket.com/albums/s244/kram1032/multiflake_iteration_1_to_4_800_-1.png
http://i154.photobucket.com/albums/s244/kram1032/multiflake_iteration_1_to_4_800_600.png

First one just shows the first 5 iterations, while the second link gives a bit of explanation, how to get the next iteration…
rod. Says:
June 16, 2008 at 4:06 pm | Reply
@ kram

After visualizing those images, I perfectly understand your idea. Wow, that’s a pretty sexy snowflake! And it seems so simple to build (the rules that one needs to follow are simple, I mean).

I don’t know if someone had already invented such a fractal, but I really like it. To the best of my knowledge, you’re the one who invented it. Would you like to write a small essay on it, with images and all? The perfect thing would be to write a short paper on it in $\LaTeX$ , and include all the images on it.

Then, I would be happy to write a post on this blog linking to your paper, so it would gain as much visibility as possible. Perhaps uploading it to arXiv.org would be a good idea too, since the papers are time-stamped and it would help you in case any disputes on authorship would arise.

I am serious. What do you think of my idea? I think that fractal is really cool, and if your idea is original, it would be a shame not to share it with the world.
kram Says:
June 16, 2008 at 5:58 pm | Reply
I would love to :D – though I can’t get too mathematical… Could you then add your thoughts how to do it?

I can also try to visualize the other fractal, though it’ll be a bit more of an effort…
Do you know any good free LATEX writer? ^^ I once found one but it turned out to be pretty hard to use… I’d love to use the freedom of Latex…. currently I’m stuck with MS Word 2003… – Or Adobe Acrobat so that I could do pdfs…

(I also never wrote anything like that. Let’s see what I can come up with… :D)
kram Says:
June 16, 2008 at 8:26 pm | Reply
Phew, pretty hard, really^^ – The first two iterations are very simple (especially the first one as it equals a sierpinski gadget at iteration 1^^)
I remade the third several times now, but I think, I’ve removed all the errors. :) I guess, until I finish all 5 steps, it’ll take some days…
rod. Says:
June 17, 2008 at 1:58 am | Reply
@ kram

Cool! :) If you could then write that short paper explaining your idea (and with images and stuff), I would be happy to review it. Linking to your paper from this blog would help it gain visibility, and hopefully that would attract more reviewers.

Visualizing the hybrid Koch-Sierpinsky fractal should definitely be a “pain”, i.e., take a hell lot of computer time. In any case, I humbly suggest you write a paper for each fractal, to keep things separated and organized. If you would be willing to provide the source code (to use with blender), that would help too since some people (me included) like to simulate and visualize stuff on computer as a means to gain deeper understanding of things.

As for $\LaTeX$ editors, I use MikTek which is (kind of) free. It allows you to generate PDF files from TEX files, which is great. I loathe MS Word for technical writing. If you need intermediate reviews, let me know. You can reach me at this email.

Cheers,

-R
kram Says:
June 17, 2008 at 5:27 pm | Reply
Oh well, if I HAD any source code^^ Blender itself is opensource (with the source code found on the page, I linked above) but I didn’t write any scripts to visualize the fractals in blender. I built them bit for bit manually. With the modified Flake, it wasn’t actually hard – just time consuming. The sierpinski-flake is far more complex and more time consuming…
(Blender isn’t ideal for fractal simulation. Anything beyond around 0.01 mm doesn’t work anymore. If I had a script, it would be done far faster, though… – I could easily go for iteration 7 before my computer starts crashing and the precision would also be far higher)

I can just provide images and some info about how I built it. Not too much^^

Thanks for sharing MikTex, I’ll look into it :)
rod. Says:
June 17, 2008 at 5:42 pm | Reply
@ kram

Building images pixel by pixel is definitely sub-optimal. I will look for alternative tools that could be used. Something like OpenGL might be useful. Writing a C++ program or a Python script to generate fractals automatically should be possible. I will give it a try and keep you posted.
kram Says:
June 17, 2008 at 8:08 pm | Reply
You didn’t understand, I’m afraid. I don’t build pixel by pixel, but triangle by triangle – that’s probably also suboptimal as there should be a formula to get the shape…

Blender basically is a 3D modeller and renderer. No way to do that pixel by pixel per hand^^
It uses OpenGL, btw.

Blender’s script language is python. I’m still struggling with the sierpinski snowflake… The modified Koch Snowflake isn’t much of a problem.
Shubhendu Trivedi Says:
June 17, 2008 at 11:43 pm | Reply
I seem to have missed a lot of action all the while i have been away.
The modified snow-flake is cool!
However i am still thinking of the koch curve-sierpinski triangle combination. That seems rather tricky.
kram Says:
June 18, 2008 at 1:38 pm | Reply
It is…
I now decided to fake a bit. For a demonstration of the appearance, it should be enough :)
kram Says:
June 18, 2008 at 8:47 pm | Reply
Hum… I think, I’m pretty close to get the area of the sierpinski flake…
The only thing I’d need is to combine your area-formula of the koch flake with the sierpinski-gasket-area-formula which is:

(sqrt(3)a²)/2^(2k+2), for start triangle length a and iteration-count k…

I think, if I use one of your sums, you had there and multiply the value inside that sum, it should show the result… (as each triangle counts one iteration less…)
rod. Says:
June 19, 2008 at 2:46 am | Reply
@ kram

I have been thinking of that hybrid flake which is built like a Koch snowflake, except that we add Sierpinski triangles to it (instead of filled triangles). Let’s call it the Koch-Sierpinski snowflake.

On a previous post, I wrote about how one can compute the area of a Koch snowflake. Building up on that, it should not be too hard to compute the area of the Koch-Sierpinski snowflake.

Let Sierpinski-1 denote the Sierpinski triangle at iteration 1, that is, we have 3 filled equilateral triangles, and 1 hollow equilateral triangle (so to say). Sierpinski-2 denotes the Sierpinski triangle at iteration 2 (where we have 9 filled triangles of equal size, plus 4 hollow triangles), and so and so on. The Sierpinski triangle at iteration $k$ will be denoted Sierpinski- $k$ triangle, or Sierpinski triangle of order $k$ .

Let $x$ be the side of an equilateral triangle, then $a(x) = \frac{\sqrt{3}}{4} x^2$ is its area. The area of a Sierpinski- $k$ triangle is $(3/4)^k a(x)$ .

The way I see things, the Koch-Sierpinski snowflake can be built as follows:

a) start with a Sierpinski-1 triangle of side $x$ and area $A_1 = \frac{3}{4} a(x)$ .

b) analogously to the Koch snowflake, add $3$ Sierpinski-2 triangles of side $x/3$ to the original triangle described in a). The total area is now $A_2 = A_1 + 3 \left(\frac{3}{4}\right)^2 a(x/3)$ .

c) add now $4 \times 3 = 12$ Sierpinski-3 triangles of side $x/3^2$ . The total area is $A_3 = A_2 + 4.3 \left(\frac{3}{4}\right)^3 a(x/3^2)$

and, at each iteration, keep adding Sierpinki triangles of increasing order. After some algebraic manipulation, I got to this expression for the total area of the Koch-Sierpinski flake at each iteration:

$\displaystyle A_{k+1} = \displaystyle A_k + \left(\frac{3}{4}\right)^2 3^{-k} a(x)$ .

Taking the limit, I get

$\lim A_k = \displaystyle \left(\frac{33}{32}\right) a(x)$ .

I will re-check everything to see if I made silly mistakes. Did you get to similar results?
kram Says:
June 19, 2008 at 3:36 pm | Reply
I didn’t get that far, yet…
Looks very good though :)

Actually, I yesterday tried to do your Koch-area-calcs on my own and the result was slightly different for some reason… Though, I don’t know where I’ve done a mistake – IF there’s one….

What I did is:
A(0)=sqrt(3)a²/4 (a is side lenght, so it should be your x)
A(1) = sqrt(3)a²/4+3/4*sqrt(3)*(a/3)² (simplification gave me, A(1) = A(0) + a²/12)
A(2) then was A(1)+sqrt(3)a²/324

My sum was:
inf
A(k)=sigma sqrt(3)/4*a²/3^k
k=0

didn’t double-check, though, whether I’ve got an error in this…
kram Says:
June 19, 2008 at 3:41 pm | Reply
Oh, I think, I see my mistake… I got a wrong count of triangle-number-increase…
kram Says:
June 19, 2008 at 6:57 pm | Reply
Ok, I now got a formula, too…
My result is
62*sqrt(3)/5*a²

Hum, quite a bit different… Your result (when adding the defined a(x)) would be:

33*sqrt(3)/128 a²… (or in your case x². I’m more used to name the side of a triangle “a”…)
kram Says:
June 19, 2008 at 7:00 pm | Reply
:S Just found a mistake :(

Gotta proof it multiple times now, then, when I’m sure, I’ll post the correct result and my way of doing it. :)
kram Says:
June 19, 2008 at 8:05 pm | Reply
Ok, I’m getting closer to your result, I think…

I now am at 105/128 * sqrt(3) a²….

Though, I also looked at the possible limits… And my result is higher than the allowed maximum :S not much but that’s still too much… (I got a²*pi/3 as a maximum, by finding the height of the whole shape and using it as diameter of a circle…)

Maybe you’re able to find the error?

I wasn’t able to fully pin down a sum: The iteration number 1 didn’t fit in the increasing count of triangles…

So, my current formula is:

Area(iteration 1) + rest:

inf
3*sqrt(3)a²/16 + sigma 3*2^(2k-2)*sqrt(3)*(a/3^(k-1))²/4*(3/4)^(k+1)
k=1

(3/4)^(k+1) comes from reducing the area of each triangle in the sierpinski part…

sqrt(3)*(a/3^(k-1))²/4 is the area of a whole triangle, which gets added around the flake… (each triangle of iteration k has a side-length of a/3^(k-1))

3*2^(2k-2) is, how many triangles we have at which iteration. After a bit of testing, this looked like this:

1
3 (3*2^0)
12 (3*2^2)
48 (3*2^4)
…
rod. Says:
June 19, 2008 at 10:59 pm | Reply
@ kram

I used $x$ for the side of the initial triangle (we could call it the Sierpinski-0 triangle) and $a(x)$ for the area of the Sierpinski-0 triangle of side $x$ solely for consistency purposes (that’s the notation in my post on the area of the Kock flake).

Let’s denote the area of the Koch-Sierpinski flake by $A$ . In my previous comment, I stated that

$A = \displaystyle\frac{33}{32} a(x) = \displaystyle\frac{33 \sqrt{3}}{128} x^2$ .

I don’t know whether this is the correct value. I have checked my derivation and it seems to be correct. Instead of focusing on correctness, let’s focus on admissibility. The area of the Koch-Sierpinski flake should be less than the area of the standard Kock flake if we start from the same initial triangle of side $x$ . This is self-evident: we build the Koch-Sierpinski flake by removing “regions” of the Kock flake, thus we’re “stealing” area. The area of the Koch flake is $\frac{8}{5} a(x)$ , and an admissible value of $A$ will satisfy the following inequality

$A < \displaystyle\frac{8}{5} a(x)$ .

I said $A = (33/32) a(x)$ , which respects the inequality. Therefore this value of $A$ is an admissible one. The last couple of values of $A$ that you proposed don’t seem to be admissible.

Of course, I may have made logical and algebraic errors along the way.
kram Says:
June 20, 2008 at 12:49 pm | Reply
Yeah, taking the original koch flake’s area is even better.
I uses the circumcircle of the flake. And my result was even bigger than the area of that circle :S
So I guess, yours is at least more if not entirely correct.

Btw: I had big problems installing MikTex. I managed to do so, yesterday but it doesn’t create any shortcut to the desktop… Now, I need to find out how to start it xD
During the weekend, I’ll also have more time to finish the Koch-Sierpinski-Snowflake (Let’s introduce a short :) KSS)
rod. Says:
June 21, 2008 at 12:28 am | Reply
@ kram

Yup, using the circumcircle should work too. The area of the standard Koch circle should give a tighter upper bound, though.

In a previous comment, I computed the area of the KSS using an iterative method. I tried another method which seems to yield the same result: count the number of Sierpinsky- $k$ triangles for all $k \geq 1$ , compute the area of each Sierpinsky- $k$ triangle, and sum their areas for $k \in \{1,2,3,\ldots\}$ . Let’s see how it goes…

Sierpinski-1 triangles:

number of triangles = $1$

side of each triangle = $x$

area of each triangle = $\displaystyle\left(\frac{3}{4}\right) a(x)$

Sierpinski-2 triangles:

number of triangles = $3 \times 4^0$

side of each triangle = $\displaystyle\frac{x}{3}$

area of each triangle = $\displaystyle\left(\frac{3}{4}\right)^2 a\left(\displaystyle\frac{x}{3^1}\right)$

Sierpinski-3 triangles:

number of triangles = $3 \times 4^1$

side of each triangle = $\displaystyle\frac{x}{3^2}$

area of each triangle = $\displaystyle\left(\frac{3}{4}\right)^3 a\left(\displaystyle\frac{x}{3^2}\right)$

Sierpinski-4 triangles:

number of triangles = $3 \times 4^2$

side of each triangle = $\displaystyle\frac{x}{3^3}$

area of each triangle = $\displaystyle\left(\frac{3}{4}\right)^4 a\left(\displaystyle\frac{x}{3^3}\right)$

Sierpinski- $k$ triangles: (where $k \geq 2$ )

number of triangles = $3 \times 4^{k-2}$

side of each triangle = $\displaystyle\frac{x}{3^{k-1}}$

area of each triangle = $\displaystyle\left(\frac{3}{4}\right)^k a\left(\displaystyle\frac{x}{3^{k-1}}\right)$

-/-

Let the area of the KSS be $A$ , then

$A = \displaystyle\left(\frac{3}{4}\right) a(x) + \displaystyle\sum_{k=2}^{\infty} \left(3 \times 4^{k-2} \displaystyle\left(\frac{3}{4}\right)^k a\left(\displaystyle\frac{x}{3^{k-1}}\right)\right)$ .

Note that $a\left(\displaystyle\frac{x}{3^{k-1}}\right) = \displaystyle\frac{1}{3^{2 k - 2}} a(x) = \displaystyle\frac{3^2}{3^{2 k}} a(x)$ .

Therefore

$A = \displaystyle\left(\frac{3}{4}\right) a(x) + \displaystyle\sum_{k=2}^{\infty} \left(\frac{3^3}{4^2}\right) \left(\frac{1}{3}\right)^k a(x)$ .

Computing the geometric series we get

$\displaystyle\sum_{k=2}^{\infty} \left(\frac{3^3}{4^2}\right) \left(\frac{1}{3}\right)^k = \left(\frac{3^3}{4^2}\right) \left(\frac{1}{1- \frac{1}{3}}\right) \left(\frac{1}{3}\right)^2 = \frac{9}{32}$

and finally

$A = \displaystyle\left(\frac{3}{4} + \displaystyle\frac{9}{32}\right) a(x) = \displaystyle\left(\frac{24+9}{32}\right) a(x) = \displaystyle\left(\frac{33}{32}\right) a(x)$ .

This value of $A$ is the same as the one I computed before.
kram Says:
June 21, 2008 at 6:41 pm | Reply
Ah I see :)

Ok, can you help me starting MikTex? Which of the various executables inside the MikTex-Folders starts the app?

I hate Open Source Software for being such unclear, sometimes :( – though, else it’s just great :D
Shubhendu Trivedi Says:
June 22, 2008 at 3:49 am | Reply
If you have installed MikTeX and TeXnic center. Open TeXnic center, if you are opening it for the first time. You would need to first create an output profile. Details for which will appear at the start-up.
After you have done so you may start with creating a source file.
It is very simple and you can get started within an hour only.

for the same you can refer to:
http://www.ctan.org/tex-archive/info/lshort/english/lshort.pdf

And if you are an absolute beginner and intend to use LaTeX for limited use then this will also be useful.
http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php
kram Says:
June 22, 2008 at 11:08 am | Reply
OH, I see… I forgot the TeXnic center xD :oops:

Thanks for your help :)
I hope, I’ll have two articles about the fractals soon.
Hendrik Jan Says:
June 29, 2008 at 7:05 am | Reply
If the only goal is to create a fractal with zero area and infinite boundary, then that’s not really difficult, is it? You could proceed as follows.
Start with a square. Replace every side with a side on which you put a square 1/3-rd the size, so that
______
becomes something like this
__
__| |__ (phew, bad ascii art :)

I.e. the construction is like the Koch snowflake, except that you start with a square instead of a triangle and you attach a smaller square to each side instead of a smaller triangle.
Now make sure the smaller squares face *into* the larger square instead of pointing outward. The 4-side square will, after 1 iteration, be a 20-sided figure.
Now repeat the procedure with those 20 sides, replacing each with a 5-segment inward pointing square, again 1/3rd the size of the previous step.

(Another way (probably easier :) to view the procedure is: take 3×3 tiles, number them left to right, top to bottom from 1 to 9. Remove tiles 2, 4, 6, 8. Repeat by dividing the 5 remaining tiles into 3×3 smaller tiles, again removing the even-numbered tiles.)

Iterating these steps will create a shape with non-overlapping boundaries, infinite perimeter (since each step multiplies the perimeter by 5/3) and area zero (since each step takes away 4/9th of the entire area and the limit is thus (5/9)^inf = 0).

You could follow the same procedure with the standard Koch snowflake — add triangles pointing inward instead of outward — but if you draw he first few iterations you’ll notice the procedure with a square as starting point is more straightforward.

But maybe this isn’t what you all were after and I’m missing something…

BTW, it’s clear that my proposed shape consists of a lot of unconnected points in the limit-case (a kind of 2 dimensional Cantor dust). An nice question would be: if the original square has corner-coordinates (0,0), (1,0), (0,1) and (1,1), how can you characterize the coordinates of the points that form the fractal — i.e. the left-over points. Clearly the corners of the original square are never removed, but they’re obviously not the only ones.
rod. Says:
June 29, 2008 at 7:16 am | Reply
@ Hendrik Jan

I really like your idea :-)

I spent some time thinking of Koch snowflakes radiating inwards, but it was not easy to assess whether ther boundaries were crossing each other. Your idea is indeed much more straightforward.

Maybe that fractal has been invented already, maybe not. In any case, until proven otherwise, you’re the creator. We can call it the “Hendrik flake” for now.

I’d like to see that fractal. It probably looks quite sexy. Do you think you could generate some images of the Hendrik flake?
Sune Kristian Jakobsen Says:
June 29, 2008 at 11:14 am | Reply
“An nice question would be: if the original square has corner-coordinates (0,0), (1,0), (0,1) and (1,1), how can you characterize the coordinates of the points that form the fractal — i.e. the left-over points.”

Look at the coordinates in trinary: In the n’th iteration a point is removed if one and only one of the coordinates has 1 as its n’th digit. Some point can be represented in more than one way ((0.1,0.1)=(0.0222…,0.1)=(0.1,0.0222…) =(0.0222…,0.0222…)) and some these points is the border of the fractal. If you want the border to belong to the fractal, the characterization would be: A point (x,y) is in the fractal if and only if x and y can be writing in trinary, such that the n’th digit in x is 1 if and only if the n’th digit in y is 1.
kram Says:
June 29, 2008 at 4:35 pm | Reply
what you say sounds a bit like
http://en.wikipedia.org/wiki/Space-filling_curve
kram Says:
June 29, 2008 at 4:39 pm | Reply
Oh, sorry it doesn’t sound so^^ missread :)
Though, those space-filling curves might be what we want^^ (if you don’t use the infinite form which fills any room of any dimension above 1, but stop the iterations at a late point, you have near infinite boundaries without any area. – though, if you goo to infinity, you *suddenly* have infinite boundaries AND infinite area xD)
Hendrik Jan Says:
June 30, 2008 at 6:07 am | Reply
@Rod: no, this is not a new fractal and hence doesn’t deserve my name. Besides, it’s too obvious anyway (imo), so it probably doesn’t deserve anyone’s name :).

@kram: yes, there is some similarity with the Peano curves. A difference (and a significant one in this context) is that the Peano curves are not closed and so have no area.
I don’t agree with your “infinite area” remark though. First, like I just said, since the curves are not closed, they don’t enclose any area. If you made a closed variant (not too hard — attach 4 Hilbert curves (see the wikipedia page you linked to before) end to end so they form a cross, and then iterate), then the iterations wouldn’t necessarily grow in size. Instead, the overall size can be kept the same, replacing line-segments with ever smaller line-segments, so the initial ‘hollow’ cross shape would become completely filled but still occupy a finite portion of the plane.

@ Sune: it’s indeed clear that powers of 3 play a role here. Using ternary coordinate notation is a clever way to capture that. I remember reading some time (= years) ago a similar approach in describing the points in Cantor dust (from a line segment, remove the middle third; repeat with the remaining 2 segments, ad infinitum).
More precisely: the sides of the original square (in the fractal I proposed) are replaced with this Cantor dust, exactly. The same holds for all sides of all “sub squares” formed along the way. So there’s a clear connection there.
I’m not sure I completely follow your explanation though. Have to give it some more thought.
The fact that I’ve just been ill with fever for 2 days, slept for 5 hours this night and have 4 year old twins running around here right now doesn’t help either :-) I’ll give it some more thought, but you may well be right…

One last remark: Sune wondered about the border being part of the fractal or not. In the limiting case (which *is* the fractal), there’s no border, but just a collection of unconnected points. Those unconnected points form the fractal.
Hendrik Jan Says:
June 30, 2008 at 6:53 am | Reply
My idea indeed isn’t original, as I already wrote earlier . Here it is: http://mathworld.wolfram.com/BoxFractal.html (including pictures, so that i don’t have to generate any myself :).

Using a triangle with inward pointing sub-triangles can be found here: http://mathworld.wolfram.com/KochAntisnowflake.html
As can be seen there, this one does *not* have an area that tends to zero, but to 2/5 (if the original triangle had area 1).
Hendrik Jan Says:
July 2, 2008 at 8:36 am | Reply
After having pondered the points belonging to the box fractal a bit: Sune is indeed right, even though I would formulate it a bit differently: If the x and y coordinates of a point are 0.x1 x2 x3… and 0.y1 y2 y3, where xi and yi are the i-th digits in the trinary expansion, then a point belongs to the fractal if for all i, xi + yi is even. In practice this means that corresponding trinary digits should be either 0-0, 0-2, 2-0, 2-2 or 1-1.

The only exception is when a number end in a 1 (followed by just zeros). In that case 0.xxx…1 can be written as 0.xxx…0222… and the above rule needs to apply to at least one of these notations.
That means that (0.100.. , 0.100… ) is in the fractal, as per the “sum is even” rule, but (0.100…, 0.200… ) is also in the set, since the coordinates can be written as (0.0222…, 0.200… ) to which the “sum is even” rule again applies.
kram Says:
July 2, 2008 at 8:47 pm | Reply
Having troubles again :S
I tried several times ’till now…
Though, I can’t finish nor render the already existing properly, because of some system problems… I guess, I’ll have to reinstall my system… :S
Tanya Says:
July 8, 2008 at 5:19 pm | Reply
The perimeter of the first equilateral triangle:the perimeter after one construction is 3:4. All the ratios are the same after each construction. The Pattern of the ratios are such that the perimeter of-
1st traingle to 2nd = 3:4
2nd triangle to 3rd = 9:16 (square of previous ratio)
3rd triangle to 4th = 27: 64 (9×3:4×16=sum of both the previous ratios)

The ratios of the perimeters go on like this.
kram Says:
August 7, 2008 at 5:36 pm | Reply
Ok… New System but the error stayed…
Sorry, this project died for now… No idea if it’ll be revived. (Though, I hope so. :) Maybe with an L-system which can do just what I want to get…)
Shubhendu Trivedi Says:
August 12, 2008 at 3:56 pm | Reply
nope the project did not die ;)

The discussion in comments has got me thinking of a number of similar fractals, i will write about them soon, i have been trying to get them on NetLogo. The project actually got revived.
erikpan Says:
November 30, 2008 at 5:09 pm | Reply
# eldila Says:

It would be fun if someone to could come up with a fractal that had a finite perimeter and an infinite area (if it is even possible).

I thought fractals had to have infinite perimeter – if they didn’t, surely it implies there is some smallest scale where they behave well, hence they wouldn’t be a fractal? Or maybe I’m wrong… It seems like the formula of the fractal dimension would be bound to give a whole number if this were the case?

Reasonable Deviations

Perimeter of the Koch Snowflake

62 Responses to “Perimeter of the Koch Snowflake”

Leave a Reply