# 有没有一段代码，让你觉得人类的智慧也可以璀璨无比？

#### 《用三段 140 字符以内的代码生成一张 1024×1024 的图片》

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。

`// NOTE: compile with g++ filename.cpp -std=c++11  #include <iostream> #include <cmath> #include <cstdlib> #define DIM 1024 #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 GR(int,int); unsigned char BL(int,int);  unsigned char RD(int i,int j){    // YOUR CODE HERE } unsigned char GR(int i,int j){    // YOUR CODE HERE } unsigned char BL(int i,int j){    // YOUR CODE HERE }  void pixel_write(int,int); FILE *fp; int main(){     fp = fopen("MathPic.ppm","wb");     fprintf(fp, "P6/n%d %d/n255/n", DIM, DIM);     for(int j=0;j<DIM;j++)         for(int i=0;i<DIM;i++)             pixel_write(i,j);     fclose(fp);     return 0; } void pixel_write(int i, int j){     static unsigned char color[3];     color[0] = RD(i,j)&255;     color[1] = GR(i,j)&255;     color[2] = BL(i,j)&255;     fwrite(color, 1, 3, fp); }`

`unsigned char RD(int i,int j){ return (char)(_sq(cos(atan2(j-512,i-512)/2))*255); }  unsigned char GR(int i,int j){ return (char)(_sq(cos(atan2(j-512,i-512)/2-2*acos(-1)/3))*255); }  unsigned char BL(int i,int j){ return (char)(_sq(cos(atan2(j-512,i-512)/2+2*acos(-1)/3))*255); }`

`unsigned char RD(int i,int j){ #define r(n)(rand()%n) static char c[1024][1024];return!c[i][j]?c[i][j]=!r(999)?r(256):RD((i+r(2))%1024,(j+r(2))%1024):c[i][j]; }  unsigned char GR(int i,int j){ static char c[1024][1024];return!c[i][j]?c[i][j]=!r(999)?r(256):GR((i+r(2))%1024,(j+r(2))%1024):c[i][j]; }  unsigned char BL(int i,int j){ static char c[1024][1024];return!c[i][j]?c[i][j]=!r(999)?r(256):BL((i+r(2))%1024,(j+r(2))%1024):c[i][j]; }`

`unsigned char RD(int i,int j){ 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; }  unsigned char GR(int i,int j){ 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; }  unsigned char BL(int i,int j){ 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; }`

Manuel Kasten 也制作了一个 Mandelbrot 集的图片，与刚才不同的是，该图描绘的是 Mandelbrot 集在某处局部放大后的结果：

`unsigned char RD(int i,int j){ 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.); }  unsigned char GR(int i,int j){ 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); }  unsigned char BL(int i,int j){ 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); }`

`unsigned char RD(int i,int j){ static double k;k+=rand()/1./RAND_MAX;int l=k;l%=512;return l>255?511-l:l; }  unsigned char GR(int i,int j){ static double k;k+=rand()/1./RAND_MAX;int l=k;l%=512;return l>255?511-l:l; }  unsigned char BL(int i,int j){ static double k;k+=rand()/1./RAND_MAX;int l=k;l%=512;return l>255?511-l:l; }`

`unsigned char RD(int i,int j){ 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; }  unsigned char GR(int i,int j){ 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; }  unsigned char BL(int i,int j){ 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; }`

`unsigned char RD(int i,int j){ #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(i,j))/2:0; }  unsigned char GR(int i,int j){ #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(i,j); }  unsigned char BL(int i,int j){ 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]; }`

`unsigned char RD(int i,int j){ #define A float a=0,b,k,r,x #define B int e,o #define C(x) x>255?255:x #define R return #define D DIM R BL(i,j)*(D-i)/D; }  unsigned char GR(int i,int j){ #define E DM1 #define F static float #define G for( #define H r=a*1.6/D+2.4;x=1.0001*b/D R BL(i,j)*(D-j/2)/D; }  unsigned char BL(int i,int j){ F c[D][D];if(i+j<1){A;B;G;a<D;a+=0.1){G b=0;b<D;b++){H;G k=0;k<D;k++){x=r*x*(1-x);if(k>D/2){e=a;o=(E*x);c[e][o]+=0.01;}}}}}R C(c[j][i])*i/D; }`