MATLAB Central Newsreader - Function for Kochs snowflake

Function for Kochs snowflake

Beginner — Tue, 02 Jan 2007 15:33:02 -0500

Hi,
I'm new on using Matlab and I could need some help.
I want to create a picture of Kochs snowflake but I don´t know how.
Unforturnatly I dont even got a clue where to begin.
Can somebody help me
If you got time I´d be grateful for stepbystep instructions

Greetings

Re: Function for Kochs snowflake

Ken Davis — Tue, 02 Jan 2007 16:33:34 -0500

"Beginner" <partypingla@gmail.com> wrote in message
news:ef4a06b.-1@webcrossing.raydaftYaTP...
> Hi,
> I'm new on using Matlab and I could need some help.
> I want to create a picture of Kochs snowflake but I don´t know how.
> Unforturnatly I dont even got a clue where to begin.
> Can somebody help me
> If you got time I´d be grateful for stepbystep instructions
>
> Greetings

The MATLAB documentation has a part called Getting Started. I recommend that
you start there.

Re: Function for Kochs snowflake

Beginner — Tue, 02 Jan 2007 17:13:19 -0500

Ken - that would be the best solution but I don´t have much time so
I´d hope somebody could guide me.

Ken Davis wrote:
>
>
> "Beginner" <partypingla@gmail.com> wrote in message
> news:ef4a06b.-1@webcrossing.raydaftYaTP...
>> Hi,
>> I'm new on using Matlab and I could need some help.
>> I want to create a picture of Kochs snowflake but I don´t know
> how.
>> Unforturnatly I dont even got a clue where to begin.
>> Can somebody help me
>> If you got time I´d be grateful for stepbystep instructions
>>
>> Greetings
>
> The MATLAB documentation has a part called Getting Started. I
> recommend that
> you start there.
>
>
>

Re: Function for Kochs snowflake

Maarten van Reeuwijk — Wed, 03 Jan 2007 02:30:51 -0500

> Ken - that would be the best solution but I don´t have much time so
> I´d hope somebody could guide me.
>
> Ken Davis wrote:
>>
>>
>> "Beginner" <partypingla@gmail.com> wrote in message
>> news:ef4a06b.-1@webcrossing.raydaftYaTP...
>>> Hi,
>>> I'm new on using Matlab and I could need some help.
>>> I want to create a picture of Kochs snowflake but I don´t know
>> how.
>>> Unforturnatly I dont even got a clue where to begin.
>>> Can somebody help me
>>> If you got time I´d be grateful for stepbystep instructions

Okay Beginner, I will give you a start.

As you know, the Koch snowflake starts from a triangle. We will parameterize
this by a relative angle variable and a unit length l:

angle = [0, -2/3*pi, -2/3*pi];
l = 1;

which can be visualized as follows:

x = zeros([length(angle)+1, 1]);
y = zeros([length(angle)+1, 1]);

x(1) = 0; y(1) = 0;
phi=0;
for i=1:length(angle);
    phi = phi+angle(i);
    x(i+1) = x(i) + l * cos(phi);
    y(i+1) = y(i) + l * sin(phi);
end

plot(x,y);
axis tight equal;

This displays the basic shape. The thing you need to realize now is that the
generator for the Koch snowflake in terms of the relative angle is

phi -> [phi, pi/3,-2*pi/3, pi/3] for each phi in angle;
l -> l /3;

That is, at each refinement of the curve, the angle array increases by a
factor 4 and the length parameter l shrinks by a factor 3.

Here is the entire code (iters is the number of times you would like to
refine the shape - do not set iters very high as the length of angle goes
as 4^iters !!!!):

angle = [0, -2/3*pi, -2/3*pi];
l = 1;
iters=0;

for i=1:iters
    l = l/3;
    angle1 = zeros([4*length(angle),1]);
    for j=1:length(angle)
      % insert the Koch generator here
    end
    angle = angle1;
end

x = zeros([length(angle)+1, 1]);
y = zeros([length(angle)+1, 1]);

x(1) = 0; y(1) = 0;
phi=0;
for i=1:length(angle);
    phi = phi+angle(i);
    x(i+1) = x(i) + l * cos(phi);
    y(i+1) = y(i) + l * sin(phi);
end

plot(x,y);
axis tight equal;

You only need to add one line of code to make it work! Good luck,

Maarten

--
===================================================================
Maarten van Reeuwijk dept. of Multiscale Physics
Phd student Faculty of Applied Sciences
maarten.vanreeuwijk.net Delft University of Technology

Re: Function for Kochs snowflake

Beginner — Wed, 03 Jan 2007 03:43:45 -0500

Maarten . Dank u!
I really appriciate your help.
/Ulrika

Okay Beginner, I will give you a start.

As you know, the Koch snowflake starts from a triangle. We will
parameterize
this by a relative angle variable and a unit length l:

angle = [0, -2/3*pi, -2/3*pi];
l = 1;

which can be visualized as follows:

x = zeros([length(angle)+1, 1]);
y = zeros([length(angle)+1, 1]);

x(1) = 0; y(1) = 0;
phi=0;
for i=1:length(angle);
    phi = phi+angle(i);
    x(i+1) = x(i) + l * cos(phi);
    y(i+1) = y(i) + l * sin(phi);
end

plot(x,y);
axis tight equal;

This displays the basic shape. The thing you need to realize now is
that the
generator for the Koch snowflake in terms of the relative angle is

phi -> [phi, pi/3,-2*pi/3, pi/3] for each phi in angle;
l -> l /3;

That is, at each refinement of the curve, the angle array increases
by a
factor 4 and the length parameter l shrinks by a factor 3.

Here is the entire code (iters is the number of times you would like
to
refine the shape - do not set iters very high as the length of angle
goes
as 4^iters !!!!):

angle = [0, -2/3*pi, -2/3*pi];
l = 1;
iters=0;

for i=1:iters
    l = l/3;
    angle1 = zeros([4*length(angle),1]);
    for j=1:length(angle)
      % insert the Koch generator here
    end
    angle = angle1;
end

x = zeros([length(angle)+1, 1]);
y = zeros([length(angle)+1, 1]);

x(1) = 0; y(1) = 0;
phi=0;
for i=1:length(angle);
    phi = phi+angle(i);
    x(i+1) = x(i) + l * cos(phi);
    y(i+1) = y(i) + l * sin(phi);
end

plot(x,y);
axis tight equal;

You only need to add one line of code to make it work! Good luck,

Maarten

--
===================================================================
Maarten van Reeuwijk dept. of Multiscale Physics
Phd student Faculty of Applied Sciences
maarten.vanreeuwijk.net Delft University of Technology

Re: Function for Kochs snowflake

Ulrika — Wed, 03 Jan 2007 08:49:31 -0500

YEEES
I´ve got it!

% insert the Koch generator here
angle1(4*j-3:4*j) = [angle(j), pi/3, -2*pi/3, pi/3];

Now is the diffucult part left. Try to understand what the diffrent
parts of the code does.

Maarten did a great job ealier but I need even more help... I need to
undestand the diffrent parts of the code. So if anyone can explain it
in a basic, basic way, you make my day!
/Ulrika

Beginner wrote:
>
>
> Maarten . Dank u!
> I really appriciate your help.
> /Ulrika
>
> Okay Beginner, I will give you a start.
>
> As you know, the Koch snowflake starts from a triangle. We will
> parameterize
> this by a relative angle variable and a unit length l:
>
> angle = [0, -2/3*pi, -2/3*pi];
> l = 1;
>
> which can be visualized as follows:
>
> x = zeros([length(angle)+1, 1]);
> y = zeros([length(angle)+1, 1]);
>
> x(1) = 0; y(1) = 0;
> phi=0;
> for i=1:length(angle);
> phi = phi+angle(i);
> x(i+1) = x(i) + l * cos(phi);
> y(i+1) = y(i) + l * sin(phi);
> end
>
> plot(x,y);
> axis tight equal;
>
> This displays the basic shape. The thing you need to realize now is
> that the
> generator for the Koch snowflake in terms of the relative angle is
>
> phi -> [phi, pi/3,-2*pi/3, pi/3] for each phi in angle;
> l -> l /3;
>
> That is, at each refinement of the curve, the angle array increases
> by a
> factor 4 and the length parameter l shrinks by a factor 3.
>
> Here is the entire code (iters is the number of times you would
> like
> to
> refine the shape - do not set iters very high as the length of
> angle
> goes
> as 4^iters !!!!):
>
> angle = [0, -2/3*pi, -2/3*pi];
> l = 1;
> iters=0;
>
> for i=1:iters
> l = l/3;
> angle1 = zeros([4*length(angle),1]);
> for j=1:length(angle)
> % insert the Koch generator here
> end
> angle = angle1;
> end
>
> x = zeros([length(angle)+1, 1]);
> y = zeros([length(angle)+1, 1]);
>
> x(1) = 0; y(1) = 0;
> phi=0;
> for i=1:length(angle);
> phi = phi+angle(i);
> x(i+1) = x(i) + l * cos(phi);
> y(i+1) = y(i) + l * sin(phi);
> end
>
> plot(x,y);
> axis tight equal;
>
> You only need to add one line of code to make it work! Good luck,
>
> Maarten
>
> --
> ===================================================================
> Maarten van Reeuwijk dept. of Multiscale Physics
> Phd student Faculty of Applied Sciences
> maarten.vanreeuwijk.net Delft University of Technology

Re: Function for Kochs snowflake

Maarten van Reeuwijk — Thu, 04 Jan 2007 21:04:04 -0500

Ulrika wrote:

> YEEES
> I´ve got it!
>
> % insert the Koch generator here
> angle1(4*j-3:4*j) = [angle(j), pi/3, -2*pi/3, pi/3];

Yep that is it!

> Now is the diffucult part left. Try to understand what the diffrent
> parts of the code does.

It is not so difficult. All you have to do is make some sketches of the data
representations at different levels and an elementary trigonometrical
calculation to calculate the next point.

Hint: by removing the semicolons at the end you can see how the algorithm
works.

Good luck, Maarten

--
===================================================================
Maarten van Reeuwijk dept. of Multiscale Physics
Phd student Faculty of Applied Sciences
maarten.vanreeuwijk.net Delft University of Technology

Re: Function for Kochs snowflake

Ulrika — Thu, 04 Jan 2007 18:13:20 -0500

Maarten - the hint about the ; was vey useful.
Dank U
/Ulrika

Maarten van Reeuwijk wrote:
>
>
> Ulrika wrote:
>
>> YEEES
>> I´ve got it!
>>
>> % insert the Koch generator here
>> angle1(4*j-3:4*j) = [angle(j), pi/3, -2*pi/3, pi/3];
>
> Yep that is it!
>
>> Now is the diffucult part left. Try to understand what the
> diffrent
>> parts of the code does.
>
> It is not so difficult. All you have to do is make some sketches of
> the data
> representations at different levels and an elementary
> trigonometrical
> calculation to calculate the next point.
>
> Hint: by removing the semicolons at the end you can see how the
> algorithm
> works.
>
> Good luck, Maarten
>
> --
> ===================================================================
> Maarten van Reeuwijk dept. of Multiscale Physics
> Phd student Faculty of Applied Sciences
> maarten.vanreeuwijk.net Delft University of Technology
>

On another note

Mikkel Guldberg — Wed, 16 Apr 2008 07:01:25 -0400

Hi Guys

While we are on the subject i was wondering if anyone could point me to or give me a proof that the perimeter goes towards infinity and the "inside" of the star is finite...

Thanks in advance
Mikkel