| Re: Fractal in IDL! [message #68371] |
Thu, 29 October 2009 11:32 |
rtk
Messages: 22
Registered: September 2008
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Junior Member |
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On Oct 28, 9:15 pm, sathya <sathya.s...@gmail.com> wrote:
> Hi,
>
> I am trying to create a fractal set in IDL. I know that a Mandelbrot
> set follows the below mentioned function,i.e.
>
> f(x) = x^2 - c
>
> where, x - is a complex number, c - constant
> and the range for x-axis is [-1.5, 1.5].
>
> In a similar way, can anyone gime me the function for anyother fractal
> or Koch snowflake. If I am not wrong, does it follow the given
> function?
>
> a =( 1/2 )+ ( i/SQRT(12) )
>
> Thanks,
> Sathya!
Perhaps the easiest way to draw fractals of various kinds is to use
the IFS ('Iterated Function System') approach. Email me if you want
IDL code that can do that.
Also, the Sierpinski triangle is perhaps the simplest of all fractals
to generate. Here's BASIC code from the 80s to do it. I leave
translation to IDL as an exercise for the reader :)
10 HOME:HGR2:HCOLOR=3: REM This just turns on the graphics and sets
the color
20 X(1)=0:Y(1)=191:X(2)=140:Y(2)=0:X(3)=278:Y(3)=191: REM Triangle
corners
30 X=X(1):Y=Y(1): REM A starting point
40 N = INT(3*RND(1))+1: X=INT(0.5*(X+X(N))): Y=INT(0.5*(Y+Y(N))): REM
The magic is here
50 HPLOT X,Y:GOTO 40: REM Plot the new point and continue
Have fun!
Ron
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| Re: Fractal in IDL! [message #68377 is a reply to message #68371] |
Thu, 29 October 2009 07:01  |
pgrigis
Messages: 436
Registered: September 2007
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Senior Member |
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On Oct 29, 10:00 am, Paolo <pgri...@gmail.com> wrote:
> On Oct 28, 11:15 pm, sathya <sathya.s...@gmail.com> wrote:
>
>> Hi,
>
>> I am trying to create a fractal set in IDL. I know that a Mandelbrot
>> set follows the below mentioned function,i.e.
>
>> f(x) = x^2 - c
>
>> where, x - is a complex number, c - constant
>> and the range for x-axis is [-1.5, 1.5].
>
>> In a similar way, can anyone gime me the function for anyother fractal
>> or Koch snowflake. If I am not wrong, does it follow the given
>> function?
>
>> a =( 1/2 )+ ( i/SQRT(12) )
>
> That doesn't make any sense.
>
> To draw the snowflake, draw a triangle first.
>
> Then on all sides "_____" of the triangle, replace the straight
> line by a line with a triangle that sticks out in the
Line break should be here: middle "__/\__" .
> Iterate ad infinitum.
>
> Ciao,
> Paolo
>
>
>
>> Thanks,
>> Sathya!
>
>
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| Re: Fractal in IDL! [message #68378 is a reply to message #68377] |
Thu, 29 October 2009 07:00 |
pgrigis
Messages: 436
Registered: September 2007
|
Senior Member |
|
|
On Oct 28, 11:15 pm, sathya <sathya.s...@gmail.com> wrote:
> Hi,
>
> I am trying to create a fractal set in IDL. I know that a Mandelbrot
> set follows the below mentioned function,i.e.
>
> f(x) = x^2 - c
>
> where, x - is a complex number, c - constant
> and the range for x-axis is [-1.5, 1.5].
>
> In a similar way, can anyone gime me the function for anyother fractal
> or Koch snowflake. If I am not wrong, does it follow the given
> function?
>
> a =( 1/2 )+ ( i/SQRT(12) )
That doesn't make any sense.
To draw the snowflake, draw a triangle first.
Then on all sides "_____" of the triangle, replace the straight
line by a line with a triangle that sticks out in the middle "__/
\__" .
Iterate ad infinitum.
Ciao,
Paolo
>
> Thanks,
> Sathya!
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| Re: Fractal in IDL! [message #68380 is a reply to message #68378] |
Thu, 29 October 2009 02:58 |
sathya
Messages: 2
Registered: October 2009
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Junior Member |
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On Oct 29, 5:39 am, Dav_Poreh <d.po...@gmail.com> wrote:
> On 28 Okt., 19:15, sathya <sathya.s...@gmail.com> wrote:
>
>
>
>> Hi,
>
>> I am trying to create afractalset inIDL. I know that a Mandelbrot
>> set follows the below mentioned function,i.e.
>
>> f(x) = x^2 - c
>
>> where, x - is a complex number, c - constant
>> and the range for x-axis is [-1.5, 1.5].
>
>> In a similar way, can anyone gime me the function for anyotherfractal
>> or Koch snowflake. If I am not wrong, does it follow the given
>> function?
>
>> a =( 1/2 )+ ( i/SQRT(12) )
>
>> Thanks,
>> Sathya!
>
> Hi Sathya!
> Could you please give us the code inIDLfor first function (x^2-c).
> and show how we could do that?
> Cheers
Hi
here is the code!
pro frac,xRange,yRange,RESULT=res
;device,PSEUDO_COLOR=8,DECOMPOSED=0
if n_elements( xRange) eq 0 then xRange = [ -1.0, 2.3]
if n_elements( yRange) eq 0 then yRange = [ -1.3, 1.3]
iter = 255
xS = 640
yS = 512
xD = float(xRange[1]-xRange[0])
yD = float(yRange[1]-yRange[0])
xStep = xD / xS
yStep = yD / yS
xStartVec = lindgen( xS) * xStep + xRange[0]
yStartVec = lindgen( yS) * yStep + yRange[0]
constArr = complex( rebin( xStartVec, xS, yS),$
rebin( transpose(yStartVec), xS, yS))
valArr = complexarr( xS, yS)
res = intarr( xS, yS)
oriIndex = lindgen( long(xS) * yS)
for i = 0, iter-1 do begin
valArr = valArr^2 - constArr
whereIn = where( abs( valArr) le 4.0d, COMPLEMENT=whereOut)
if whereIn[0] eq -1 then break
valArr = valArr[ whereIn]
constArr = constArr[ whereIn]
if whereOut[0] ne -1 then begin
res[ oriIndex[ whereOut]] = i+1
oriIndex = oriIndex[ whereIn]
endif
endfor
loadct,15
tv,res
end
here, is the code which uses the function I mentioned above!
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| Re: Fractal in IDL! [message #68382 is a reply to message #68380] |
Thu, 29 October 2009 02:39 |
d.poreh
Messages: 406
Registered: October 2007
|
Senior Member |
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On 28 Okt., 19:15, sathya <sathya.s...@gmail.com> wrote:
> Hi,
>
> I am trying to create a fractal set in IDL. I know that a Mandelbrot
> set follows the below mentioned function,i.e.
>
> f(x) = x^2 - c
>
> where, x - is a complex number, c - constant
> and the range for x-axis is [-1.5, 1.5].
>
> In a similar way, can anyone gime me the function for anyother fractal
> or Koch snowflake. If I am not wrong, does it follow the given
> function?
>
> a =( 1/2 )+ ( i/SQRT(12) )
>
> Thanks,
> Sathya!
Hi Sathya!
Could you please give us the code in IDL for first function (x^2-c).
and show how we could do that?
Cheers
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