A few lines of code, code monkey second great artist! -Programmer, programming-IT information
Original title of the three paragraphs within 140 characters of code generation for a 1024x1024 picture, IT information Edit modify, as appropriate.
Kyle McCormick on the StackExchange has launched a competition called the Tweetable Mathematical Art, contestants need to push with three long code to produce a picture. Specifically, participants need to be written in C++ language RD, GR, BL three functions, each function is limited to 140 characters. Each function receives two integers I and j parameters (0 ≤ I, j ≤ 1023), and then returns an integer from 0 to 255, represents a (I, j) color value for the pixel . For example, if RD (0, 0) and GR (0, 0) returns 0, BL (0, 0) returns is 255, so the upper-left pixel of the image is blue. Contestants write code that will be inserted into the following programs (I did some minor changes), will eventually generate a picture of size 1024x1024.
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);
}
I have selected some of my favorite works and put them in below and share.
First of all is the one from Martin b u ttner works:
Its code is as follows:
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);
}
Also works from Martin b u ttner:
This is by far the number one works for the time being. Its code is as follows:
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];
}
This picture below is still from the hand of Martin b u ttner:
It is hard to imagine, Mandelbrot fractal images could only draw with so little code:
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 also produced a picture of the Mandelbrot set, and is just different, which is characterised by somewhere local amplification results of the Mandelbrot set:
Its code is as follows:
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);
}
This is another work of Manuel Kasten:
Generate code that this picture is very interesting: functions rely on static variables to control the process of painting, there is no use I and j parameters!
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;
}
This is from the githubphagocyte works:
Its code is as follows:
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;
}
This is another piece from the githubphagocyte:
This is a picture of the diffusion-limited aggregation model, the program takes a lot of time to run. Code is very interesting: clever use of macro definitions, breaking the boundaries between functions and function, c code combined with the word limit will be used.
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=f2){B=f-1;} else{++e%=4;d=e;if(!c[e]){B=0; M=1;}}}}} return m[i][j];
}
This last picture is from Eric Tressler:
This is the Feigenbaum obtained by logistic map bifurcation diagram. And as before, corresponding code also skillfully uses the macro definitions to save characters:
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;kD/2){e=a;o=(E*x);c[e][o]+=0.01;}}}}} R C(c[j][i])*i/D;
}
几行代码，程序猿秒变大艺术家！ -
程序员,编程 - IT资讯
原标题《用三段 140 字符以内的代码生成一张 1024×1024 的图片》，IT资讯编辑酌情修改。
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。
// 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);
}
我选了一些自己比较喜欢的作品，放在下面和大家分享。
首先是一个来自Martin Büttner的作品：
它的代码如下：
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);
}
同样是来自Martin Büttner的作品：
这是目前暂时排名第一的作品。它的代码如下：
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];
}
下面这张图片仍然出自Martin Büttner之手：
难以想象，Mandelbrot分形图形居然可以只用这么一点代码画出：
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：
它的代码如下：
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);
}
这是Manuel Kasten的另一作品：
生成这张图片的代码很有意思：函数依靠static变量来控制绘画的进程，完全没有用到i和j这两个参数！
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;
}
这是来自githubphagocyte的作品：
它的代码如下：
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;
}
这是来自githubphagocyte的另一个作品：
这是一张使用diffusion-limited aggregation模型得到的图片，程序运行起来要耗费不少时间。代码很有意思：巧妙地利用宏定义，打破了函数与函数之间的界限，三段代码的字数限制便能合在一起使用了。
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=f2){B=f-1;}else{++e%=4;d=e;if(!c[e]){B=0;M=1;}}}}}return m[i][j];
}
最后这张图来自Eric Tressler：
这是由logistic映射得到的Feigenbaum分岔图。和刚才一样，对应的代码也巧妙地利用了宏定义来节省字符：
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;kD/2){e=a;o=(E*x);c[e][o]+=0.01;}}}}}R C(c[j][i])*i/D;
}