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:

  1 #include <iostream>
  2 #include <cmath>
  3 #include <cstdlib>
  4 #include <time.h>
  5 using namespace std;
  6 
  7 #define DIM 1024
  8 #define NAME_LEN 256
  9 #define DM1 (DIM-1)
 10 #define _sq(x) ((x)*(x)) // square
 11 #define _cb(x) abs((x)*(x)*(x)) // absolute value of cube
 12 #define _cr(x) (unsigned char)(pow((x),1.0/3.0)) // cube root
 13 
 14 unsigned char RD(int,int,int);
 15 unsigned char GR(int,int,int);
 16 unsigned char BL(int,int,int);
 17 
 18 unsigned char RD(int t,int i,int j){
 19     switch(t)
 20     {
 21     case 0:
 22         {
 23             return (char)(_sq(cos(atan2(j-512,i-512)/2))*255);
 24         }
 25     case 1:
 26         {
 27             #define r(n)(rand()%n)
 28             static char c[1024][1024];
 29             return !c[i][j]?c[i][j]=!r(999)?r(256):RD(t,(i+r(2))%1024,(j+r(2))%1024):c[i][j];
 30         }
 31     case 2:
 32         {
 33             float x=0,y=0;
 34             int k;
 35             for(k=0;k++<256;)
 36             {
 37                 float a=x*x-y*y+(i-768.0)/512;y=2*x*y+(j-512.0)/512;
 38                 x=a;if(x*x+y*y>4)
 39                     break;
 40             }
 41             return log(k)*47;
 42         }
 43     case 3:
 44         {
 45             double a=0,b=0,c,d,n=0;
 46             while((c=a*a)+(d=b*b)<4&&n++<880)
 47             {
 48                 b=2*a*b+j*8e-9-.645411;a=c-d+i*8e-9+.356888;
 49             }
 50             return 255*pow((n-80)/800,3.);
 51         }
 52     case 4:
 53         {
 54             static double k;
 55             k+=rand()/1./RAND_MAX;
 56             int l=k;
 57             l%=512;
 58             return l>255?511-l:l;
 59         }
 60     case 5:
 61         {
 62             float s=3./(j+99);
 63             float y=(j+sin((i*i+_sq(j-700)*5)/100./DIM)*35)*s;
 64             return (int((i+DIM)*s+y)%2+int((DIM*2-i)*s+y)%2)*127;
 65         }
 66     case 6:
 67         {
 68             #define D DIM
 69             #define M m[(x+D+(d==0)-(d==2))%D][(y+D+(d==1)-(d==3))%D]
 70             #define R rand()%D
 71             #define B m[x][y]
 72             return(i+j)?256-(BL(t,i,j))/2:0;
 73         }
 74     default:
 75         {
 76             return 0;
 77         }
 78     }
 79 }
 80 unsigned char GR(int t,int i,int j){
 81     switch(t)
 82     {
 83     case 0:
 84         {
 85             return (char)(_sq(cos(atan2(j-512,i-512)/2-2*acos(-1)/3))*255);
 86         }
 87     case 1:
 88         {
 89             static char c[1024][1024];
 90             return !c[i][j]?c[i][j]=!r(999)?r(256):GR(t,(i+r(2))%1024,(j+r(2))%1024):c[i][j];
 91         }
 92     case 2:
 93         {
 94             float x=0,y=0;
 95             int k;
 96             for(k=0;k++<256;)
 97             {
 98                 float a=x*x-y*y+(i-768.0)/512;
 99                 y=2*x*y+(j-512.0)/512;
100                 x=a;
101                 if(x*x+y*y>4)
102                     break;
103             }
104             return log(k)*47;
105         }
106     case 3:
107         {
108             double a=0,b=0,c,d,n=0;
109             while((c=a*a)+(d=b*b)<4&&n++<880)
110             {
111                 b=2*a*b+j*8e-9-.645411;a=c-d+i*8e-9+.356888;
112             }
113             return 255*pow((n-80)/800,.7);
114         }
115     case 4:
116         {
117             static double k;
118             k+=rand()/1./RAND_MAX;
119             int l=k;
120             l%=512;
121             return l>255?511-l:l;
122         }
123     case 5:
124         {
125             float s=3./(j+99);
126             float y=(j+sin((i*i+_sq(j-700)*5)/100./DIM)*35)*s;
127             return (int(5*((i+DIM)*s+y))%2+int(5*((DIM*2-i)*s+y))%2)*127;
128         }
129     case 6:
130         {
131             #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
132             return RD(t,i,j);
133         }
134     default:
135         {
136             return 0;
137         }
138     }
139 }
140 unsigned char BL(int t,int i,int j){
141     switch(t)
142     {
143     case 0:
144         {
145             return (char)(_sq(cos(atan2(j-512,i-512)/2+2*acos(-1)/3))*255);
146         }
147     case 1:
148         {
149             static char c[1024][1024];
150             return !c[i][j]?c[i][j]=!r(999)?r(256):BL(t,(i+r(2))%1024,(j+r(2))%1024):c[i][j];
151         }
152     case 2:
153         {
154             float x=0,y=0;
155             int k;
156             for(k=0;k++<256;)
157             {
158                 float a=x*x-y*y+(i-768.0)/512;
159                 y=2*x*y+(j-512.0)/512;
160                 x=a;
161                 if(x*x+y*y>4)
162                     break;
163             }
164             return 128-log(k)*23;
165         }
166     case 3:
167         {
168             double a=0,b=0,c,d,n=0;
169             while((c=a*a)+(d=b*b)<4&&n++<880)
170             {
171                 b=2*a*b+j*8e-9-.645411;a=c-d+i*8e-9+.356888;
172             }
173             return 255*pow((n-80)/800,.5);
174         }
175     case 4:
176         {
177             static double k;
178             k+=rand()/1./RAND_MAX;
179             int l=k;
180             l%=512;
181             return l>255?511-l:l;
182         }
183     case 5:
184         {
185             float s=3./(j+99);
186             float y=(j+sin((i*i+_sq(j-700)*5)/100./DIM)*35)*s;
187             return (int(29*((i+DIM)*s+y))%2+int(29*((DIM*2-i)*s+y))%2)*127;
188         }
189     case 6:
190         {
191             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;}}}}}
192             return m[i][j];
193         }
194     default:
195         {
196             return 0;
197         }
198     }
199 }
200 void draw_pic(int);
201 void pixel_write(int,int,int,FILE*);
202 
203 int main(){
204     time_t t1,t2;
205     for(int i = 0;i < 7;i++)
206     {
207         t1 = clock();
208         draw_pic(i);
209         t2 = clock();
210         cout<<"use time:"<<t2 - t1<<"(ms)"<<endl;
211     }
212     return 0;
213 }
214 
215 void draw_pic(int t)
216 {
217     FILE *fp = NULL;
218     char name[NAME_LEN] = {0};
219     sprintf(name,"MathPic_%d.ppm",t);
220     cout<<"rendering "<<name<<"..."<<endl;
221     fp = fopen(name,"wb");
222     if(fp == NULL){
223         return;
224     }
225     fprintf(fp, "P6\n%d %d\n255\n", DIM, DIM);
226     for(int j=0;j<DIM;j++){
227         for(int i=0;i<DIM;i++){
228             pixel_write(t,i,j,fp);
229         }
230     }
231     fclose(fp);
232     cout<<"render "<<name<<" finished"<<endl;
233 }
234 void pixel_write(int t,int i, int j,FILE* fp){
235     static unsigned char color[3];
236     color[0] = RD(t,i,j)&255;
237     color[1] = GR(t,i,j)&255;
238     color[2] = BL(t,i,j)&255;
239     fwrite(color, 1, 3, fp);
240 }

View Code

图片效果如下