Sierpinski pentagon: Difference between revisions
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* [http://ecademy.agnesscott.edu/~lriddle/ifs/pentagon/pentagon.htm Sierpinski pentagon]
<br><br>
=={{header|C}}==
The Sierpinski fractals can be generated via the [http://mathworld.wolfram.com/ChaosGame.html Chaos Game]. This implementation thus generalizes the [[Chaos game]] C implementation on Rosettacode. As the number of sides increases, the number of iterations must increase dramatically for a well pronounced fractal ( 30000 for a pentagon). Requires the [http://www.cs.colorado.edu/~main/bgi/cs1300/ WinBGIm] library.
<lang C>
/*Abhishek Ghosh, 24th September 2017*/
#include<graphics.h>
#include<stdlib.h>
#include<stdio.h>
#include<math.h>
#include<time.h>
#define pi M_PI
int main(){
time_t t;
double side, **vertices,seedX,seedY,windowSide = 500,sumX=0,sumY=0;
int i,iter,choice,numSides;
printf("Enter number of sides : ");
scanf("%d",&numSides);
printf("Enter polygon side length : ");
scanf("%lf",&side);
printf("Enter number of iterations : ");
scanf("%d",&iter);
initwindow(windowSide,windowSide,"Polygon Chaos");
vertices = (double**)malloc(numSides*sizeof(double*));
for(i=0;i<numSides;i++){
vertices[i] = (double*)malloc(2 * sizeof(double));
vertices[i][0] = windowSide/2 + side*cos(i*2*pi/numSides);
vertices[i][1] = windowSide/2 + side*sin(i*2*pi/numSides);
sumX+= vertices[i][0];
sumY+= vertices[i][1];
putpixel(vertices[i][0],vertices[i][1],15);
}
srand((unsigned)time(&t));
seedX = sumX/numSides;
seedY = sumY/numSides;
putpixel(seedX,seedY,15);
for(i=0;i<iter;i++){
choice = rand()%numSides;
seedX = (seedX + (numSides-2)*vertices[choice][0])/(numSides-1);
seedY = (seedY + (numSides-2)*vertices[choice][1])/(numSides-1);
putpixel(seedX,seedY,15);
}
free(vertices);
getch();
closegraph();
return 0;
}
</lang>
=={{header|Haskell}}==
For universal solution see [[Fractal tree#Haskell]]
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