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Welcome to Fractal Forums > Fractal Software > Programming > the simpliest naive bruteforce code for mandelbulb
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Author Topic: the simpliest naive bruteforce code for mandelbulb  (Read 14820 times)
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ker2x
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« Reply #15 on: August 01, 2011, 08:50:19 AM »
Quote from: cKleinhuis on July 29, 2011, 12:12:12 PM
the 3d buddha is emerges wink

Done  grin
Window binary and source code : http://fractals.s3.amazonaws.com/flowabulb/voxelflowabrot-0.2.zip

A video of the program :
<a href="http://www.youtube.com/v/lGLcslZqE8c&rel=1&fs=1&hd=1" target="_blank">http://www.youtube.com/v/lGLcslZqE8c&rel=1&fs=1&hd=1</a>
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
ker2x
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« Reply #16 on: August 02, 2011, 09:49:59 PM »
voxelflowabrot-0.4 : http://fractals.s3.amazonaws.com/flowabulb/voxelflowabrot-0.4.zip

Lot of code refactoring and some openGL tuning (including antialiasing) smiley
« Last Edit: August 02, 2011, 10:14:24 PM by ker2x » Logged
often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
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ker2x
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« Reply #17 on: August 07, 2011, 10:12:21 PM »
Quote from: eiffie on July 28, 2011, 05:49:58 PM
Good you caught that typo.
I assume you are already ray-marching thru z until you hit the bulb so why not add DE like this:
dr=1.0
inside iterate:
dr=dr*p*pow(r,p-1.0)+1.0
after iterate advance z by 0.5*log(r)*r/dr

It will greatly increase the speed and make experimenting with coloring much more enjoyable.

It doesn't work sad
i coded what i understood of the raymarching thingy.
If i advance by 0.1 instead of your formula, i have something.
if i use 0.5*log(r)*r/dr; i have nothing.

Can you take a look at this very ugly code and help please ? smiley

Code:
final int MAXITER = 16;
final int MAXSTEP = 255;
final float p=8.0;
final color white = color(255);

float[][] pts = new float[MAXITER+1][3];
int count = 0;
float x,y,z;


void setup() {
  size(800,800,P2D);
  frameRate(1000);
  background(0);
  stroke(255);
}

void draw() {
  loadPixels();
  count++;
  if(count > 10000) {
    count=0;
    randomSeed(millis());
  }
  for(int i = 0; i < 1000; i++) {
  x = random(-1.5,1.5);
  y = random(-1.5,1.5);
  z = -4.0;
  float dr = 1.0;
  int step =0;

  for(step = 0; step < MAXSTEP; step++) {
  int iter = 0;
  float r = 0.0;
  float nx = x;
  float ny = y;
  float nz = z;
  r=sqrt(x*x+y*y+z*z);
  float th=atan(y/x)*p;
  float ph=asin(z/r)*p;
  float u,v;

    while(iter < MAXITER && r<2.0  ) {
      float r2p = pow(r,p);
      th=atan(ny/nx)*p;
      ph=asin(nz/r)*p;
      nx=r2p*cos(ph)*cos(th)+x;
      ny=r2p*cos(ph)*sin(th)+y;
      nz=r2p*sin(ph)+z;
      r=sqrt(nx*nx+ny*ny+nz*nz);
      dr=dr*p*pow(r,p-1.0)+1.0;
      iter++;
    }
    //println(dr);
    if(r<2.0) {
        u =  x * 6.0 / (z + 4.0);
        v = y  * 6.0 / (z + 4.0);
        u = (2.0 + u) / 4.0 * width;
        v = (2.0 + v) / 4.0 * height;
        if(u > 0 && u < width && v > 0 && v < height) {
          pixels[(int)((int)u + (int)v * width)] = color(step*1,step*1,step*1);
        }
        break;
    } else {
      step++;
      z+=0.1;//*log(r)*r/dr;
    }
  }
  } 
  updatePixels();
} 

 
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
ker2x
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« Reply #18 on: August 07, 2011, 10:23:26 PM »
i just noticed that the mandelbulb formula in fragmentarium was very easy to understand and i'll try to learn from it smiley
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
ker2x
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« Reply #19 on: August 07, 2011, 10:41:42 PM »
My problem is that dr quickly become == Double.POSITIVE_INFINITY
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
Syntopia
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« Reply #20 on: August 08, 2011, 12:16:00 AM »
The 'dr'-value must be reset inside the MAXSTEP loop, just like 'r'.
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ker2x
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« Reply #21 on: August 08, 2011, 01:03:21 AM »
Quote from: Syntopia on August 08, 2011, 12:16:00 AM
The 'dr'-value must be reset inside the MAXSTEP loop, just like 'r'.

Well, it still doesn't work  cry
But at least one insane error is patched  grin

here is the current code :

Code:
final int MAXITER = 16;
final int MAXSTEP = 255;
final float p=8.0;
final color white = color(255);

float[][] pts = new float[MAXITER+1][3];
int count = 0;
float x,y,z;


void setup() {
  size(800,800,P2D);
  frameRate(1000);
  background(0);
  stroke(255);
}

void draw() {
  loadPixels();
  count++;
  if(count > 10000) {
    count=0;
    randomSeed(millis());
  }
  for(int i = 0; i < 1000; i++) {
  x = random(-1.5,1.5);
  y = random(-1.5,1.5);
  z = -4.0;
 
  int step =0;

  for(step = 0; step < MAXSTEP; step++) {
  int iter = 0;
  float r = 0.0;
  float dr = 1.0;
  float nx = x;
  float ny = y;
  float nz = z;
  r=sqrt(x*x+y*y+z*z);
  float th=atan(y/x)*p;
  float ph=asin(z/r)*p;
  float u,v;
 
    while(iter < MAXITER /*&& r<2.0*/  ) {
      float r2p = pow(r,p);
      th=atan2(ny,nx)*p;
      ph=asin(nz/r)*p;
      nx=r2p*cos(ph)*cos(th)+x;
      ny=r2p*cos(ph)*sin(th)+y;
      nz=r2p*sin(ph)+z;
      r=sqrt(nx*nx+ny*ny+nz*nz);
      dr=dr*p*pow(r,p-1.0)+1.0; 
      iter++;
    }
   
    if(r<2.0) {
        println(dr);
        u =  x * 6.0 / (z + 4.0);
        v = y  * 6.0 / (z + 4.0);
        u = (2.0 + u) / 4.0 * width;
        v = (2.0 + v) / 4.0 * height;
        if(u > 0 && u < width && v > 0 && v < height) {
          pixels[(int)((int)u + (int)v * width)] = color(step*1,step*1,step*1);
        }
        break;
    } else {
      step++;
      z+=0.1;
      z+=0.5*log(r)*r/dr; //<-nothing
    }
  }
  } 
  updatePixels();
} 



the output that display dr whrn r < 2.0 :
11.36559
3.330166
12.572722
10.215517
5.687963
28.261217
43.640415
3.3310628
3.3878212
13.518621
3.347372
10.360928
7.386033
...
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
ker2x
Fractal Molossus
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Posts: 795


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« Reply #22 on: August 08, 2011, 01:15:08 AM »
Well, obviously i should display println(0.5*log(r)*r/dr); to see how big was supposed to be my last step :

4.1891032E-4
0.0011521813
5.504626E-4
-0.003734893
-0.051313702
1.649025E-4
-0.015488651
2.312548E-6
-0.0359119
-0.013137315
-0.0013213182
-0.036753275
-1.0157118E-4
-3.4222467E-7
-0.0076081255
-0.014232793
-0.03867302
-2.776414E-4
-8.763797E-6
-0.039671708
-0.0022190188
9.5214637E-4
-6.9133566E-5
-0.029464567
-2.9889583E-7
-0.0017719215
-0.027600588
-0.011584057
-0.00846173
-6.32731E-5
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
ker2x
Fractal Molossus
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Posts: 795


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« Reply #23 on: August 08, 2011, 01:20:50 AM »
if i replace :
if(r<2.0)
by :
if(0.5*log(r)*r/dr > -10e-20 && 0.5*log(r)*r/dr < 10e-20)
i have a correct something that look like a buddhabrot.


but i still can't do z+=0.5*log(r)*r/dr;
« Last Edit: August 08, 2011, 01:22:26 AM by ker2x » Logged
often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
eiffie
Guest
« Reply #24 on: August 09, 2011, 05:31:23 PM »
One problem I see with the code is the order you are calculating r and dr. Switch them.
dr=dr*...   //you want to calc using the old r
then
r=sqrt(...

also typically we bail at r<2.0 so I am not sure this works well for your purpose.
I was confusing myself there! Glad it works for you.
« Last Edit: August 09, 2011, 10:34:25 PM by eiffie » Logged
ker2x
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Posts: 795


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« Reply #25 on: August 09, 2011, 07:29:56 PM »
It works \o/
Thank you  grin



Code:
final int MAXITER = 16;
final int MAXSTEP = 255;
final float p=8.0;
final color white = color(255);

float[][] pts = new float[MAXITER+1][3];
int count = 0;
float x,y,z;


void setup() {
  size(800,800,P2D);
  frameRate(1000);
  background(0);
  stroke(255);
}

void draw() {
  loadPixels();
  count++;
  if(count > 10000) {
    count=0;
    randomSeed(millis());
  }
  for(int i = 0; i < 10000; i++) {
  x = random(-1.5,1.5);
  y = random(-1.5,1.5);
  z = -4.0;
  
  int step =0;

  for(step = 0; step < MAXSTEP && z < 4.0; step++) {
  int iter = 0;
  float r = 0.0;
  float dr = 1.0;
  float nx = x;
  float ny = y;
  float nz = z;
  r=sqrt(x*x+y*y+z*z);
  float th=atan(y/x)*p;
  float ph=asin(z/r)*p;
  float u,v;
  
    while(iter < MAXITER && r<2.0  ) {
      float r2p = pow(r,p);
      th=atan2(ny,nx)*p;
      ph=asin(nz/r)*p;
      nx=r2p*cos(ph)*cos(th)+x;
      ny=r2p*cos(ph)*sin(th)+y;
      nz=r2p*sin(ph)+z;
      dr=dr*p*pow(r,p-1.0)+1.0;  
      r=sqrt(nx*nx+ny*ny+nz*nz);
      iter++;
    }
    
    if(0.5*log(r)*r/dr < 0.00001) {
        
        u =  x * 6.0 / (z + 4.0);
        v = y  * 6.0 / (z + 4.0);
        u = (2.0 + u) / 4.0 * width;
        v = (2.0 + v) / 4.0 * height;
        if(u > 0 && u < width && v > 0 && v < height) {
          pixels[(int)((int)u + (int)v * width)] = color(step*1,step*1,step*1);
        }
        break;
    } else {
      step++;
      z+=0.5*log(r)*r/dr;
    }
  }
  }  
  updatePixels();
}  

  
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
A Noniem
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Posts: 38
« Reply #26 on: August 09, 2011, 08:23:13 PM »
If I were you i'd make a ray tracer (if you know how it works, beginning from scratch is hard). It's more difficult, but the results are a lot nicer and the rendering time is a lot less.
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ker2x
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« Reply #27 on: August 10, 2011, 12:56:46 PM »
Quote from: A Noniem on August 09, 2011, 08:23:13 PM
If I were you i'd make a ray tracer (if you know how it works, beginning from scratch is hard). It's more difficult, but the results are a lot nicer and the rendering time is a lot less.

Step by step, i want to understand what i'm doing.
I took a few hours to translate some code to OpenCL, not yet optimized but it's much much faster already  grin

However, i will keep the Monte Carlo method, i like it  embarrass
I'm not good enough to write the best fractal software, and i don't have enough time for that anyway, i want to learn and have fun.

Yes, i'll write a raytracer, with openCL.
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
ker2x
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Posts: 795


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« Reply #28 on: August 10, 2011, 11:01:12 PM »
some improvments  grin

It's much nicer when the fractal is white and more step = more shadow   embarrass
(instead of a black fractal with color = number of step)
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often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
ker2x
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Posts: 795


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« Reply #29 on: August 10, 2011, 11:12:46 PM »
and some fake lightning colors based on X,Y,Z position


« Last Edit: August 10, 2011, 11:22:52 PM by ker2x » Logged
often times... there are other approaches which are kinda crappy until you put them in the context of parallel machines
(en) http://www.blog-gpgpu.com/ , (fr) http://www.keru.org/ ,
Sysadmin & DBA @ http://www.over-blog.com/
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