转:用三段 140 字符以内的代码生成一张 1024×1024 的图片

原文转自 :http://blogread.cn/it/article/6899?f=wb&from=timeline&isappinstalled=1

Kyle McCormick 在 StackExchange 上发起了一个叫做 Tweetable Mathematical Art 的比赛，参赛者需要用三条推这么长的代码来生成一张图片。具体地说，参赛者需要用 C++ 语言编写 RD 、 GR 、 BL 三个函数，每个函数都不能超过 140 个字符。每个函数都会接到 i 和 j 两个整型参数（0 ≤ i, j ≤ 1023），然后需要返回一个 0 到 255 之间的整数，表示位于 (i, j) 的像素点的颜色值。举个例子，如果 RD(0, 0) 和 GR(0, 0) 返回的都是 0 ，但 BL(0, 0) 返回的是 255 ，那么图像的最左上角那个像素就是蓝色。参赛者编写的代码会被插进下面这段程序当中（我做了一些细微的改动），最终会生成一个大小为 1024×1024 的图片。

很多牛人贡献出了自己写的代码,生成了精美的图片,我觉得很有意思,自己试验了下他们的代码,摘录几段如下:

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <time.h>
using namespace std;

#define DIM 1024
#define NAME_LEN 256
#define DM1 (DIM-1)
#define _sq(x) ((x)*(x)) // square
#define _cb(x) abs((x)*(x)*(x)) // absolute value of cube
#define _cr(x) (unsigned char)(pow((x),1.0/3.0)) // cube root

unsigned char RD(int,int,int);
unsigned char GR(int,int,int);
unsigned char BL(int,int,int);

unsigned char RD(int t,int i,int j){
    switch(t)
    {
    case 0:
        {
            return (char)(_sq(cos(atan2(j-512,i-512)/2))*255);
        }
    case 1:
        {
            #define r(n)(rand()%n)
            static char c[1024][1024];
            return !c[i][j]?c[i][j]=!r(999)?r(256):RD(t,(i+r(2))%1024,(j+r(2))%1024):c[i][j];
        }
    case 2:
        {
            float x=0,y=0;
            int k;
            for(k=0;k++<256;)
            {
                float a=x*x-y*y+(i-768.0)/512;y=2*x*y+(j-512.0)/512;
                x=a;if(x*x+y*y>4)
                    break;
            }
            return log(k)*47;
        }
    case 3:
        {
            double a=0,b=0,c,d,n=0;
            while((c=a*a)+(d=b*b)<4&&n++<880)
            {
                b=2*a*b+j*8e-9-.645411;a=c-d+i*8e-9+.356888;
            }
            return 255*pow((n-80)/800,3.);
        }
    case 4:
        {
            static double k;
            k+=rand()/1./RAND_MAX;
            int l=k;
            l%=512;
            return l>255?511-l:l;
        }
    case 5:
        {
            float s=3./(j+99);
            float y=(j+sin((i*i+_sq(j-700)*5)/100./DIM)*35)*s;
            return (int((i+DIM)*s+y)%2+int((DIM*2-i)*s+y)%2)*127;
        }
    case 6:
        {
            #define D DIM
            #define M m[(x+D+(d==0)-(d==2))%D][(y+D+(d==1)-(d==3))%D]
            #define R rand()%D
            #define B m[x][y]
            return(i+j)?256-(BL(t,i,j))/2:0;
        }
    default:
        {
            return 0;
        }
    }
}
unsigned char GR(int t,int i,int j){
    switch(t)
    {
    case 0:
        {
            return (char)(_sq(cos(atan2(j-512,i-512)/2-2*acos(-1)/3))*255);
        }
    case 1:
        {
            static char c[1024][1024];
            return !c[i][j]?c[i][j]=!r(999)?r(256):GR(t,(i+r(2))%1024,(j+r(2))%1024):c[i][j];
        }
    case 2:
        {
            float x=0,y=0;
            int k;
            for(k=0;k++<256;)
            {
                float a=x*x-y*y+(i-768.0)/512;
                y=2*x*y+(j-512.0)/512;
                x=a;
                if(x*x+y*y>4)
                    break;
            }
            return log(k)*47;
        }
    case 3:
        {
            double a=0,b=0,c,d,n=0;
            while((c=a*a)+(d=b*b)<4&&n++<880)
            {
                b=2*a*b+j*8e-9-.645411;a=c-d+i*8e-9+.356888;
            }
            return 255*pow((n-80)/800,.7);
        }
    case 4:
        {
            static double k;
            k+=rand()/1./RAND_MAX;
            int l=k;
            l%=512;
            return l>255?511-l:l;
        }
    case 5:
        {
            float s=3./(j+99);
            float y=(j+sin((i*i+_sq(j-700)*5)/100./DIM)*35)*s;
            return (int(5*((i+DIM)*s+y))%2+int(5*((DIM*2-i)*s+y))%2)*127;
        }
    case 6:
        {
            #define A static int m[D][D],e,x,y,d,c[4],f,n;if(i+j<1){for(d=D*D;d;d--){m[d%D][d/D]=d%6?0:rand()%2000?1:255;}for(n=1
            return RD(t,i,j);
        }
    default:
        {
            return 0;
        }
    }
}
unsigned char BL(int t,int i,int j){
    switch(t)
    {
    case 0:
        {
            return (char)(_sq(cos(atan2(j-512,i-512)/2+2*acos(-1)/3))*255);
        }
    case 1:
        {
            static char c[1024][1024];
            return !c[i][j]?c[i][j]=!r(999)?r(256):BL(t,(i+r(2))%1024,(j+r(2))%1024):c[i][j];
        }
    case 2:
        {
            float x=0,y=0;
            int k;
            for(k=0;k++<256;)
            {
                float a=x*x-y*y+(i-768.0)/512;
                y=2*x*y+(j-512.0)/512;
                x=a;
                if(x*x+y*y>4)
                    break;
            }
            return 128-log(k)*23;
        }
    case 3:
        {
            double a=0,b=0,c,d,n=0;
            while((c=a*a)+(d=b*b)<4&&n++<880)
            {
                b=2*a*b+j*8e-9-.645411;a=c-d+i*8e-9+.356888;
            }
            return 255*pow((n-80)/800,.5);
        }
    case 4:
        {
            static double k;
            k+=rand()/1./RAND_MAX;
            int l=k;
            l%=512;
            return l>255?511-l:l;
        }
    case 5:
        {
            float s=3./(j+99);
            float y=(j+sin((i*i+_sq(j-700)*5)/100./DIM)*35)*s;
            return (int(29*((i+DIM)*s+y))%2+int(29*((DIM*2-i)*s+y))%2)*127;
        }
    case 6:
        {
            A;n;n++){x=R;y=R;if(B==1){f=1;for(d=0;d<4;d++){c[d]=M;f=f<c[d]?c[d]:f;}if(f>2){B=f-1;}else{++e%=4;d=e;if(!c[e]){B=0;M=1;}}}}}
            return m[i][j];
        }
    default:
        {
            return 0;
        }
    }
}
void draw_pic(int);
void pixel_write(int,int,int,FILE*);

int main(){
    time_t t1,t2;
    for(int i = 0;i < 7;i++)
    {
        t1 = clock();
        draw_pic(i);
        t2 = clock();
        cout<<"use time:"<<t2 - t1<<"(ms)"<<endl;
    }
    return 0;
}

void draw_pic(int t)
{
    FILE *fp = NULL;
    char name[NAME_LEN] = {0};
    sprintf(name,"MathPic_%d.ppm",t);
    cout<<"rendering "<<name<<"..."<<endl;
    fp = fopen(name,"wb");
    if(fp == NULL){
        return;
    }
    fprintf(fp, "P6
%d %d
255
", DIM, DIM);
    for(int j=0;j<DIM;j++){
        for(int i=0;i<DIM;i++){
            pixel_write(t,i,j,fp);
        }
    }
    fclose(fp);
    cout<<"render "<<name<<" finished"<<endl;
}
void pixel_write(int t,int i, int j,FILE* fp){
    static unsigned char color[3];
    color[0] = RD(t,i,j)&255;
    color[1] = GR(t,i,j)&255;
    color[2] = BL(t,i,j)&255;
    fwrite(color, 1, 3, fp);
}

图片效果如下