绘制随机的黑白森林


66

您的任务是编写将绘制800x600黑白图像(类似于森林)的程序。

像这样(抖动的照片):

规则

  • 您不可以使用任何现有图像-您应该纯粹通过算法生成图像
  • 仅使用2种颜色-黑白(无灰度)
  • 每次程序运行时映像应该是新的-每次随机
  • 一棵树不是森林(至少要说五棵树)
  • 禁止绘制树木/森林的特殊图书馆
  • 以多数选票获胜

9
这个问题似乎不合时宜,因为它更多的是艺术比赛,而不是编程比赛。
ProgramFOX 2014年

19
@ProgramFOX是不是编程艺术?:)
Somnium

15
我希望看到一些有关此挑战的条目,但对它被搁置感到失望。
Braden Best

3
我喜欢这个挑战。并非本着这种精神的答案不会被广泛批评,那么这是什么问题呢?
cjfaure 2014年

3
@cjfaure用于不同目的,例如用于生成游戏的模型和图像。
2014年

Answers:


98

电话:3863 1144 1023 999 942 927

原始解决方案每次运行可保存2个pnm文件(在抖动之前,每个文件都附加了g)。由于前几行的抖动效果并不理想,因此存在一种破解方法,可以渲染多于所需的行,并在输出时进行裁剪。

高尔夫球解决方案具有更简单的抖动,并且仅保存抖动图像。(没有gcc -std = c11 -pedantic -Wall -Wextra的警告)

来自3个原始程序运行和1个高尔夫球版本的示例图像(最后一个图像):

例子1 例子2 例子3 例子4

高尔夫版

  #include <math.h>
  #include <time.h>
  #include <stdlib.h>
  #include <stdio.h>
  #define D float
  #define R rand()/RAND_MAX
  #define Z(a,b,c) if(10.*R>a)T(h,s,j+b+c*R,g);
  #define U a[y][x]
  #define V e[y+1][x
  #define F(x) for(i=0;i<x;++i)
  int i,x,y,W=800,H=600;unsigned char a[600][800],c;D e[601][802],r,b,k,l,m,n,f,w,
  d,q=.01;void T(D h,D s,D j,D g){r=b=0;do{j+=.04*R-.02;h+=sin(j)*q;s+=cos(j)*q;b
  +=q;g+=q;f=525/d;r=.25-g/44;m=w*f+s*f+W/2;n=2*f-h*f+H/2;f*=r;for(y=n-f-2;y<n+f+2
  ;++y)if(y>=0&&y<H)for(x=m-f-2;x<m+f+2;++x)if(x>=0&&x<W)if(k=m-x,l=n-y,f>sqrt(k*k
  +l*l))if(U>d*3)U=d*3;}while(b<10*r+12*r*R&&r>q);if(r>q){Z(2,.26,.35)Z(2,-.26,-
  .35)Z(7,-.05,.1)}}int main(){FILE* o=fopen("i","wb");srand(time(0));F(W*H){y=i/W
  ;a[y][i%W]=(y<313)?255:6e3/(2*y-H);}F(200)w=1e2*R-60,d=80.*R+5,T(0,0,1.58,0);F(W
  *H){x=i%W;y=i/W;k=U+e[y][x+1];U=-(k>0);l=(k-U)*.1;e[y][x+2]+=l*4;V]+=l*2;V+1]+=l
  *3;V+2]+=l;}fprintf(o,"P5 800 600 255 ");fwrite(a,1,W*H,o);}

原始版本

  #include <math.h>
  #include <stdio.h>
  #include <stdlib.h>

  #define W 800
  #define H 600
  #define SPEED 0.01
  #define HEIGHT 11.0

  #define R(m) ((double)(m) * rand() / RAND_MAX)
  #define RAD(deg) ((deg) / 180.0 * M_PI)
  #define LIMIT(x, min, max) ((x) < (min) ? (min) : (x) > (max) ? (max) : (x))

  void shade(void);
  void growTree(double dist, double side, double h, double s, double alpha, double grown);
  void plot(double dist, double side, double h, double s, double alpha, double diam);
  void dither(void);
  void writeImg(int dither);

  unsigned char img[H+10][W];
  double err[H+10+2][W+4];
  long tim;

  int main(void)
  {
     int i;
     tim = time(0);
     srand(tim);
     shade();
     for(i = 0; i < 200; ++i)
     {
        growTree(5 + R(75), -60 + R(120), 0.0, 0.0, RAD(90), 0.0);
     }
     writeImg(0);
     dither();
     writeImg(1);
  }

  void shade(void)
  {
     int y;
     for(y = -10; y < H; ++y)
     {
        double dist = H * 3.5 / (2 * y - H);
        unsigned char color = dist / 80 * 255;
        if(y <= H / 2 || dist > 80) color = 255;
        memset(img[y+10], color, W);
     }
  }

  void growTree(double dist, double side, double h, double s, double alpha, double grown)
  {
     double diam, branchLength = 0.0;

     do
     {
        alpha += R(RAD(3)) - RAD(1.5);
        h += sin(alpha) * SPEED;
        s += cos(alpha) * SPEED;
        branchLength += SPEED;
        grown += SPEED;
        diam = (1.0 - grown / HEIGHT) * 0.5;
        plot(dist, side, h, s, alpha, diam);
     } while(branchLength < 5 * diam + R(6 * diam) && diam > 0.02);

     if(diam > 0.02)
     {
        int br = 0;

        if(R(10) > 2) br++,growTree(dist, side, h, s, alpha + RAD(15) + R(RAD(20)), grown);
        if(R(10) > 2) br++,growTree(dist, side, h, s, alpha - RAD(15) - R(RAD(20)), grown);
        if(R(10) < 2 || br == 0) growTree(dist, side, h, s, alpha - RAD(2.5) + R(RAD(5)), grown);
     }
  }

  void plot(double dist, double side, double h, double s, double alpha, double diam)
  {
     int x, y;
     double scale = H / 4.0 * 3.5 / dist;
     double x0 = side * scale + s * scale + W / 2.0;
     double y0 = H / 2.0 + 2.0 * scale - h * scale;
     diam *= scale;
     h *= scale;
     s *= scale;
     for(y = y0 - diam / 2 - 2; y < y0 + diam / 2 + 2; ++y)
     {
        if(y < -10 || y >= H) continue;
        for(x = x0 - diam / 2 - 2; x < x0 + diam / 2 + 2; ++x)
        {
           double dx, dy, d;
           if(x < 0 || x >= W) continue;
           dx = x0 - x;
           dy = y0 - y;
           d = diam / 2 - sqrt(dx * dx + dy * dy) + 0.5;
           if(d > 0)
           {
              unsigned char color = dist / 80 * 255;
              if(img[y+10][x] > color) img[y+10][x] = color;
           }
        }
     }
  }

  void dither(void)
  {
     int x0, x, y;
     for(y = -10; y < H; ++y)
     {
        for(x0 = 0; x0 < W; ++x0)
        {
           double error, oldpixel;
           unsigned char newpixel;
           if(y%2) x = W - 1 - x0;
           else x = x0;
           oldpixel = img[y+10][x] + err[y+10][x+2];
           newpixel = oldpixel > 127 ? 255 : 0;
           img[y+10][x] = newpixel;
           error = oldpixel - newpixel;
           err[y+10  ][x+1+2*(1-y%2)] += error * 7 / 48;
           err[y+10  ][x+4*(1-y%2)] += error * 5 / 48;
           err[y+10+1][x  ] += error * 3 / 48;
           err[y+10+1][x+1] += error * 5 / 48;
           err[y+10+1][x+2] += error * 7 / 48;
           err[y+10+1][x+3] += error * 5 / 48;
           err[y+10+1][x+4] += error * 3 / 48;
           err[y+10+2][x  ] += error * 1 / 48;
           err[y+10+2][x+1] += error * 3 / 48;
           err[y+10+2][x+2] += error * 5 / 48;
           err[y+10+2][x+3] += error * 3 / 48;
           err[y+10+2][x+4] += error * 1 / 48;
        }
     }
  }

  void writeImg(int dither)
  {
     FILE* fp;
     char buffer[32];
     sprintf(buffer, "%ld%s.pnm", tim, dither ? "" : "g");
     fp = fopen(buffer, "wb");
     fprintf(fp, "P5\n%d %d\n255\n", W, H);
     fwrite(&img[10][0], 1, W * H, fp);
     fclose(fp);
  }

3
+1。好图。无论如何,这是流行竞赛,而不是代码高尔夫球。因此,无需打高尔夫球。:)
矢量化2014年

2
看起来不错,我有类似的想法,但是太懒了=)我认为您可以尝试随机化分支的递归深度,我认为这样看起来更自然。
瑕疵的

3
我知道我不需要打高尔夫球,但这对我来说是最有趣的部分!
Manuel Kasten 2014年

2
好漂亮
昆汀

12
此外,我在棕褐色调中加入了不变色的版本-太棒了!看到这里:i.imgur.com/lCrJCp7.jpg
DreamWarrior 2014年

42

爪哇丛林

(954打高尔夫球)

充满了深深的扭曲的灌木丛,这是一个不容易穿越的森林。

在此处输入图片说明

从根本上讲,它是分形的随机游走,并逐渐收缩着弯曲的藤蔓。我画了其中的75张,从背面的白色逐渐变为正面的黑色。然后,我抖动了整个事情,无耻地适应阿威罗伊代码在这里为。

打高尔夫球:(只是因为其他人决定这样做)

import java.awt.*;import java.awt.image.*;import java.util.*;class P{static Random rand=new Random();public static void main(String[]a){float c=255;int i,j;Random rand=new Random();final BufferedImage m=new BufferedImage(800,600,BufferedImage.TYPE_INT_RGB);Graphics g=m.getGraphics();for(i=0;i++<75;g.setColor(new Color((int)c,(int)c,(int)c)),b(g,rand.nextInt(800),599,25+(rand.nextInt(21-10)),rand.nextInt(7)-3),c-=3.4);for(i=0;i<800;i++)for(j=0;j<600;j++)if(((m.getRGB(i,j)>>>16)&0xFF)/255d<rand.nextFloat()*.7+.05)m.setRGB(i,j,0);else m.setRGB(i,j,0xFFFFFF);new Frame(){public void paint(Graphics g){setSize(800,600);g.drawImage(m,0,0,null);}}.show();}static void b(Graphics g,float x,float y,float s,float a){if(s>1){g.fillOval((int)(x-s/2),(int)(y-s/2),(int)s,(int)s);s-=0.1;float n,t,u;for(int i=0,c=rand.nextInt(50)<1?2:1;i++<c;n=a+rand.nextFloat()-0.5f,n=n<-15?-15:n>15?15:n,t=x+s/2*(float)Math.cos(n),u=y-s/2*(float)Math.sin(n),b(g,t,u,s,n));}}}

Sane原始代码:

import java.awt.Color;
import java.awt.Graphics;
import java.awt.image.BufferedImage;
import java.util.Random;
import javax.swing.JFrame;

public class Paint {

    static int minSize = 1;
    static int startSize = 25;
    static double shrink = 0.1;
    static int branch = 50;
    static int treeCount = 75;

    static Random rand = new Random();
    static BufferedImage img;

    public static void main(String[] args) {
        img = new BufferedImage(800,600,BufferedImage.TYPE_INT_ARGB);
        forest(img);
        dither(img);
        new JFrame() {
            public void paint(Graphics g) {
                setSize(800,600);
                g.drawImage(img,0,0,null);
            }
        }.show();
    }

    static void forest(BufferedImage img){
        Graphics g = img.getGraphics();
        for(int i=0;i<treeCount;i++){
            int c = 255-(int)((double)i/treeCount*256);
            g.setColor(new Color(c,c,c));
            tree(g,rand.nextInt(800), 599, startSize+(rand.nextInt(21-10)), rand.nextInt(7)-3);
        }
    }

    static void tree(Graphics g, double x, double y, double scale, double angle){
        if(scale < minSize)
            return;
        g.fillOval((int)(x-scale/2), (int)(y-scale/2), (int)scale, (int)scale);
        scale -= shrink;
        int count = rand.nextInt(branch)==0?2:1;
        for(int i=0;i<count;i++){
            double newAngle = angle + rand.nextDouble()-0.5;
            if(newAngle < -15) newAngle = -15;
            if(newAngle > 15) newAngle = 15;
            double nx = x + (scale/2)*Math.cos(newAngle);
            double ny = y - (scale/2)*Math.sin(newAngle);
            tree(g, nx, ny, scale, newAngle);
        }
    }

    static void dither(BufferedImage img) {
        for (int i=0;i<800;i++)
            for (int j=0;j<600;j++) {
                double lum = ((img.getRGB(i, j) >>> 16) & 0xFF) / 255d;
                if (lum <= threshold[rand.nextInt(threshold.length)]-0.2)
                    img.setRGB(i, j, 0xFF000000);
                else
                    img.setRGB(i, j, 0xFFFFFFFF);
            }
    }

    static double[] threshold = { 0.25, 0.26, 0.27, 0.28, 0.29, 0.3, 0.31,
            0.32, 0.33, 0.34, 0.35, 0.36, 0.37, 0.38, 0.39, 0.4, 0.41, 0.42,
            0.43, 0.44, 0.45, 0.46, 0.47, 0.48, 0.49, 0.5, 0.51, 0.52, 0.53,
            0.54, 0.55, 0.56, 0.57, 0.58, 0.59, 0.6, 0.61, 0.62, 0.63, 0.64,
            0.65, 0.66, 0.67, 0.68, 0.69 };

}

多一个?好的!抖动程度有所降低,因此前面的黑人更加讨人喜欢。

在此处输入图片说明

不幸的是,抖动未显示出葡萄藤层的精细细节。这是灰度版本,仅供比较:

在此处输入图片说明


1
真好 很高兴我的代码对您有所帮助;)
Averroes 2014年

1
在要求抖动的规则中,我什么也没读。我认为未经修改的版本看起来更漂亮。
拉尔斯·艾伯特

4
@Lars抖动是不需要的,但是:只有黑白可以使用,不允许灰度值....
曼努埃尔卡斯滕

31

Javascript + HTML-不打高尔夫

@Manuel Kansten算法的javascript移植-令人惊讶的是这些树看起来多么好。

只是做一些不同的事情,我将图像绘制为彩色,然后在最后一步将其抖动为黑白。

我不知道为什么,但是我的森林对曼努埃尔的森林没有那么黑暗,也没有那么可怕的尊重。

使用JSfiddle进行测试或运行下面的新代码段。那不是很快。要有耐心,注意森林的生长。

森林1 森林1色

森林2 森林2色

W=800
H=600
canvas.width = W;
canvas.height = H;

var ctx = canvas.getContext("2d");

R=function(m) { return m * Math.random()};
RAD=function(deg) { return deg / 180 * Math.PI};
LIMIT=function(x, min, max) {return x < min ? min : x > max ? max : x};
var SPEED = 0.01, HEIGHT = 11.0;

// Ground
var grd = ctx.createLinearGradient(0,0,0,H);
grd.addColorStop(0,"#88ccff");
grd.addColorStop(0.45,"#ffffee");
grd.addColorStop(0.5,"#80cc80");
grd.addColorStop(1,"#001100");
ctx.fillStyle = grd;
ctx.fillRect(0,0, W,H);


Plot = function(dist, side, h, s, alpha, diam)
{
    var x, y, a1,a2,scale = H/4 * 3.5 / dist, 
        x0 = side * scale + s * scale + W/2,
        y0 = H/2 + 2.5*scale - h*scale;
    
    k = dist
    if (diam > 0.05) {
        red = k*3|0;     
        green = k|0;
        a1=alpha+1
        a2=alpha-1
    }
    else
    {
        green= 80+(1-diam)*k*2|0;
        red = k|0;
        a1=0;
        a2=2*Math.PI;
    }
    diam *= scale;
    h *= scale;
    s *= scale;
    ctx.beginPath();
    ctx.arc(x0,y0,diam/2, a1,a2);//lpha-1, alpha+1);//0,2*Math.PI);
    ctx.fillStyle = 'rgb('+red+','+green+',0)';
    ctx.fill();
}

Grow = function(dist, side, h, s, alpha, grown)
{
    var diam, branchLength = 0.0;
    diam = (1.0 - grown / HEIGHT) * 0.5;
    do
    {
        alpha += R(RAD(3)) - RAD(1.5);
        h += Math.sin(alpha) * SPEED;
        s += Math.cos(alpha) * SPEED;
        branchLength += SPEED;
        grown += SPEED;
        diam = (1.0 - grown / HEIGHT) * 0.5;
        Plot(dist, side, h, s, alpha, diam);
    } while(branchLength < 5 * diam + R(6 * diam) && diam > 0.02);

    if (diam > 0.02)
    {
        var br = 0;

        if(R(10) > 2) br++,Grow(dist, side, h, s, alpha + RAD(15) + R(RAD(20)), grown);
        if(R(10) > 2) br++,Grow(dist, side, h, s, alpha - RAD(15) - R(RAD(20)), grown);
        if(R(10) < 2 || br == 0) Grow(dist, side, h, s, alpha - RAD(2.5) + R(RAD(5)), grown);
    }
}

trees=[]
for(i = 0; i < 300; ++i) trees.push({ z: 1+R(70), s:R(120)-60 });
trees.sort( function (a,b) { return a.z - b.z} );

Draw = function()
{
    t = trees.pop();
    if (t)
    {
        Grow(t.z, t.s, 0, 0, RAD(90), 0);
        setTimeout(Draw, 100);
    }
    else 
    {
        var e,c,d,p,i,l, img = ctx.getImageData(0,0,W,H);
        l = img.data.length;
        for (i = 0; i < l-W*4-4; i+=4)
        {
            c = (img.data[i]+img.data[i+1])/2|0
            c = img.data[i]
            d = c > 120 + R(16) ? 255 : 0
            e = c - d;
            img.data[i]=img.data[i+1]=img.data[i+2]=d
            c = (img.data[i+4]+img.data[i+5])/2|0
            
            c = LIMIT(c + ((e*7)>>4),0,255)
            img.data[i+4]=img.data[i+5]=img.data[i+6]=c
            p = i+W*4
            c = (img.data[p-4]+img.data[p-3])/2|0
            c = LIMIT(c + ((e*3)>>4),0,255)
            img.data[p-4]=img.data[p-3]=img.data[p-2]=c
            c = (img.data[p]+img.data[p+1])/2|0
            c = LIMIT(c+ ((e*5)>>4),0,255)
            img.data[p]=img.data[p+1]=img.data[p+2]=c
            c = (img.data[p+4]+img.data[p+5]*2)/3|0
            c = LIMIT(c + (e>>4),0,255)
            img.data[p+4]=img.data[p+5]=img.data[p+6]=c
    
        }
        bwcanvas.width = W;
        bwcanvas.height = H;
        var bwx = bwcanvas.getContext("2d");
        bwx.putImageData(img,0,0);
    }
}

setTimeout(Draw, 10);
<canvas id='bwcanvas'  width="2" height="2"></canvas>
<canvas id='canvas'  width="2" height="2"></canvas>


2
我认为这是地面。曼努埃尔(Manuel)的树木融入地面,营造出朦胧,模糊的外观。您的地平面更轻,外观更加通风,对比度更高。此外,Manuel图片中的纯色天空和距离会逐渐消失,有助于营造出朦胧的雾气或阴霾的外观。(请介意,我希望您的图片看起来与众不同。)
Ilmari Karonen 2014年

是的,这是曼努埃尔(Manuel)深林的果园。
shadowtalker 2014年

23

上下文自由艺术3(1133)

CF是矢量图形渲染语言,因此我无法避免抗锯齿。我通过在同一位置绘制正方形数次(可变N)来解决该问题。通过在随机位置绘制小方块来完成雾化。

startshape main

W = 80*0.6
H = 60*0.6

N = 3

CF::Background = [ b -1 ]
CF::Size = [ x 0 y -20 s W H ]
CF::Color = 0
CF::ColorDepth = 16
CF::MinimumSize = 0.6

shape main {
  transform [ z 0 y (H/2) b 1 ]
  loop 30 [ z -1 y -2 ] {
    loop 200000 []
      SQUARE1 [ s (0.1/3) x -W..W y -H..H z -0.5..0.5 ]
  }

  transform [ b -1 z 3 ]
  loop 14 [[ s 1.1 y -0.8..-1 s 0.6 z -3 ]] {
    loop 14 [] tree [ x (-30..-20) z (-3..3) ]
    loop 14 [] tree [ x (20..30) z (-3..3) ]
  }
}

shape tree {
  branch [ ]
}

shape branch
rule 7 {
  transform [ s (1..2) 1]
  SQUARE1 [ ]

  branch [ y (0.2..0.3) x (-0.05..0.05) s 0.994 r (-6..6) z (-0.3..0.3)  ]
  branch1 [ b 0.001 z -2 r -20..20 ]
}
rule 0.001 { }
rule 0.3 { branch [ r 4..20 ] }
rule 0.3 { branch [ r -4..-20 ] }

shape branch1
rule 90 { }
rule { branch [ r -22..22 s 0.8..1 ] }

path SQUARE1 {
  MOVETO( 0.5,  0.5)
  LINETO(-0.5,  0.5)
  LINETO(-0.5, -0.5)
  LINETO( 0.5, -0.5)
  CLOSEPOLY()
  loop N [] FILL()[]
}

shape S {
  SQUARE [ a -1 ]
  loop 1000 [ ] SQUARE [ x (-0.5..0.5) y (-0.5..0.5) s 0.01..0.001 ]
}

在此处输入图片说明

使用不同数字的更多渲染 在此处输入图片说明 在此处输入图片说明 在此处输入图片说明


1
我知道没有抖动选项,但是输出仍然应该仅为黑白。
Geobits,2014年

1
您找不到抖动的方法吗?实际上,这不是有效的答案。
edc65

1
您不必抖动,但只需要使用两种颜色-黑色和白色。使用灰度不符合规则。该问题已被编辑以消除歧义。我建议您使用某种方法来实现此目的,无论是否抖动,以便您可以在更多否决票到达之前编辑答案。
trichoplax 2014年

就像是一个建议,以防万一:您可以使用当前的透明矩形方法来获得黑白图像,而不是部分透明,而应在某种网格图案中使用完全透明,以便仅每个其他网格单元是透明的。
trichoplax 2014年

1
现在,这更接近问题的实质。
trichoplax

19

C:301

该程序以PGM格式创建简单的抽象图像。您可以使用GIMP打开它。

int x,y,i,d,w;srand(time(NULL));unsigned char p[480000];FILE *f=fopen("a.pgm","w");fprintf(f,"P5\n800 600\n1\n");i=480000;while(i--)p[i]=i>240000|(i%800+i/800&3)!=0;i=100;while(i--){d=(11000-i*i)/99;y=300+1100/d;x=rand()%800;while(y--){w=300/d;while(w--)p[y*800+w+x]=0;}}fwrite(p, 1, 480000, f);fclose(f);

这是一个示例运行:生成的图像


34
这看起来更像是条码,而不是森林:)
Sylwester 2014年

6
倒挂的林木
Fabricio

5
向下滚动时,我认为浏览器无法正确渲染图像。如果您的条形码通过条形码扫描仪扫描后得到该挑战的URL,那么我将给您很多积分。除了不会是一个随机森林。
西北太平洋

6
我扫描了它,然后我的条形码扫描仪对我诅咒。谢谢。
PlasmaHH 2014年

1
@nwp不过,您可以从UPC数据库中获取随机代码。可扫描但随机的:)
Geobits

18

具有JAVA的IFS

该解决方案使用迭代功能系统(IFS)来描述一棵(原型)树。IFS被应用100次(=林)。在对每棵树进行油漆(种植到森林中)之前,将IFS进行适当的更改(随机行走样式)。因此,每棵树看起来都略有不同。

图片来自随机种子:

  • -824737443
  • -1220897877
  • -644492215
  • 1133984583

无需抖动。

import java.awt.Graphics;
import java.awt.image.BufferedImage;
import java.util.Random;
import javax.swing.*;

public class IFS {
    static Random random=new Random();
    static int BLACK = 0xff000000;
    static int treeCount = 100;
    static Random rand = new Random();
    static int Height = 600;
    static int Width = 800;
    static BufferedImage img = new BufferedImage(Width, Height, BufferedImage.TYPE_INT_ARGB);

    static double[][] ifs=new double[][] {//Tree 3 {; Paul Bourke  http://ecademy.agnesscott.edu/~lriddle/ifskit/gallery/bourke/bourke.ifs
       {0.050000,  0.000000,  0.000000,  0.600000,  0.000000,  0.000000,  0.028000},
       {0.050000,  0.000000,  0.000000, -0.500000,  0.000000,  1.000000,  0.023256},
       {0.459627, -0.321394,  0.385673,  0.383022,  0.000000,  0.600000,  0.279070},
       {0.469846, -0.153909,  0.171010,  0.422862,  0.000000,  1.100000,  0.209302},
       {0.433013,  0.275000, -0.250000,  0.476314,  0.000000,  1.000000,  0.555814 /*Paul Bourke has: 0.255814*/},
       {0.421325,  0.257115, -0.353533,  0.306418,  0.000000,  0.700000,  0.304651 /*Paul Bourke has: 0.204651*/},
    };

    public static void main(String[] args) {
        int seed=random.nextInt();
        //seed=-1220897877;
        random=new Random(seed);
        for (int t = 0; t < treeCount; t++) {
            for (int i = 0; i < ifs.length; i++) {
                for (int j = 0; j < ifs[0].length; j++) {
                    ifs[i][j]=R(ifs[i][j]);
                }
            }
            tree(random.nextDouble(), 0.1*random.nextDouble());
        }
        JFrame frame = new JFrame(""+seed) {
            public void paint(Graphics g) {
                setSize(800,600);
                g.drawImage(img,0,0,null);
            }
        };
        frame.setDefaultCloseOperation(WindowConstants.EXIT_ON_CLOSE);
        frame.setVisible(true);
    }

    private static void tree(double x0, double dist) {
        double y0=Math.atan(dist+0.01);
        double scale=Math.atan(0.01)/y0;
        double x=0;
        double y=0;
        for (int n = 0; n < 200000/Math.pow(20*dist+1, 8); n++) {
            int k = select(ifs);
            double newx=ifs[k][0]*x + ifs[k][1]*y + ifs[k][2];
            double newy=ifs[k][3]*x + ifs[k][4]*y + ifs[k][5];
            x=newx;
            y=newy;
            newx= Width*(0.5*scale*newx+x0);
            newy= Height*((1-0.5*scale*newy)-y0-0.1);
            if (0<=newx && newx<Width && 0<=newy && newy<Height) {
                img.setRGB((int)newx, (int)newy, BLACK);
            }
        }
    }

    private static double R(double x) {
        return (1+ 0.01*random.nextGaussian())*x;
    }

    private static int select(double[][] ifs) {
        int k;
        double sum=0;
        for(k=0; k<ifs.length; k++) {
            sum+=ifs[k][6];
        }
        double r=sum*random.nextDouble();
        sum=ifs[0][6];
        for(k=0; k<ifs.length-1 && r>sum; k++) {
            sum+=ifs[k+1][6];
        }
        return k;
    }
}

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最后一张图片看起来最好。但是我认为您应该尝试减少树木的数量,因为现在总有黑色无法辨认的区域,但是在边缘看起来更好。
2014年

“美在旁观者的眼中。” 我测试了很多变化。最后,我想出了treeCount = 100。欢迎其他所有人复制和更改我的解决方案。
Bob Genom 2014年

15

我注意到这里明显缺少针叶树,所以我用Python一起砍了一些东西。

from PIL import Image
import random

#Generates the seed for a tree
def makeSeed(y):
    random.seed()
    seed_x = random.randint(10, 590)
    seed_y = y
    width = random.randint(5, 10)
    height = random.randint(width*5, width*30)

    return (seed_x, seed_y, width, height)

#Grows the vertical components
def growStems(seed_data, pixel_field):
    seed_x = seed_data[0]
    seed_y = seed_data[1]
    width = seed_data[2]
    height = seed_data[3]
    for x in range(seed_x, seed_x+width):
        for y in range(seed_y-height, seed_y):
            pixel_field[x][y] = (0, 0, 0)
            #Dithering
            if seed_y > 300 and seed_y < 320:
                if (x+y)%2==0:
                    pixel_field[x][y] = (255, 255, 255)
            elif seed_y >= 320 and seed_y < 340:
                if (x+y)%4==0:
                    pixel_field[x][y] = (255, 255, 255)
            elif seed_y >= 340 and seed_y < 360:
                if (x+y)%8==0:
                    pixel_field[x][y] = (255, 255, 255)

    return pixel_field

#Grows the horizontal components
def growBranches(seed_data, pixel_field):
    seed_x = seed_data[0]
    seed_y = seed_data[1]
    width = seed_data[2]
    height = seed_data[3]
    branch_height = seed_y-height
    branch_width = width
    branch_length = 2
    max_prev = branch_length
    branches = []
    while(branch_height >= seed_y-height and branch_height < seed_y-(3*width) and branch_length < height/3):
        branches.append((branch_height, branch_width, branch_length))
        branch_height+= 4
        branch_length+=2
        #Gives the conifer unevenness to make it look more organic
        if random.randint(0,110) > 100 and branch_length > max_prev:
            max_prev = branch_length
            branch_length -= branch_length/4
    max_length = height/3


    for x in range(seed_x-max_length, seed_x+max_length):
        for y in range(seed_y-height, seed_y):
            for branch in branches:
                bh = branch[0]
                bw = branch[1]
                bl = branch[2]
                #Establishing whether a point is "in" a branch
                if x >= seed_x-bl+(width/2) and x <= seed_x+bl+(width/2):
                    if x > 1 and x < 599:
                        if y >= bh-(bw/2) and y <= bh+(bw/2):
                            if y < 400 and y > 0:
                                pixel_field[x][y] = (0, 0, 0)
                                #Dithering
                                if seed_y > 300 and seed_y < 320:
                                    if (x+y)%2==0:
                                        pixel_field[x][y] = (255, 255, 255)
                                elif seed_y >= 320 and seed_y < 340:
                                    if (x+y)%4==0:
                                        pixel_field[x][y] = (255, 255, 255)
                                elif seed_y >= 340 and seed_y < 360:
                                    if (x+y)%8==0:
                                        pixel_field[x][y] = (255, 255, 255)

    return pixel_field


def growTrees(n):
    pixel_field = [[(255, 255, 255) for y in range(400)] for x in range(600)]
    #Create the ground
    for i in range(600):    
        for j in range(400):
            if pixel_field[i][j]==(255,255,255) and j > 300:
                if (i+j)%2 == 0:
                    pixel_field[i][j]=(0,0,0)
    seed_ys=[]
    #Generates seeds for the trees and orders them back to front to make the dithering work
    for t in range(n):
        seed_ys.append(random.randint(300,390))
    seed_ys.sort()

    for s in range(len(seed_ys)):
        seed= makeSeed(seed_ys[s])
        pixel_field = growStems(seed, pixel_field)
        pixel_field = growBranches(seed, pixel_field)
    return pixel_field

def makeForest():
    forest = growTrees(25)
    img = Image.new( 'RGB', (600,400), "white") # create a new black image
    pixels = img.load() # create the pixel map
    for i in range(img.size[0]):    # for every pixel:
        for j in range(img.size[1]):
            if pixels[i,j]==(255,255,255) and j > 300:
                if (i+j)%2 == 0:
                    pixels[i,j]=(0,0,0)
            pixels[i,j] = forest[i][j] # set the colour accordingly

    img.save("Forest25.jpg")

if __name__ == '__main__':
    makeForest()

5棵树的森林 十棵树的森林 25棵树的森林

这是我第一次参加Code Golf,这很有趣!


1
看起来不错!我喜欢。
TonySniper 2014年

5

这个答案并不像我希望的那么漂亮,但这是我正在研究的更多3D想法的垫脚石,我非常喜欢实际模拟哪些树可以获取资源的想法。在此处输入图片说明

package forest;

import java.awt.BorderLayout;
import java.awt.Canvas;
import java.awt.Dimension;
import java.awt.Graphics;
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import java.util.Random;

import javax.imageio.ImageIO;
import javax.swing.JFrame;
import javax.swing.JPanel;

public class Forest extends Canvas{
    private int[] heights = new int[800];
    private BufferedImage buffered_image;
    File outputFile = new File("saved.png");
    Random r = new Random();
    public Forest() {
        buffered_image = new BufferedImage(800, 600,
                BufferedImage.TYPE_INT_RGB);
        for( int j = 0; j < 800; j++){
            heights[j] = -10000;
            for(int k = 0; k < 600; k++){
                buffered_image.setRGB(j, k, 0xFFFFFF);
            }
        }
        for(int i = 0; i < 7; i ++){
            heights[r.nextInt(800)] = 0;
        }

        this.setPreferredSize(new Dimension(800, 600));
        this.setSize(new Dimension(800, 600));
        for( int i = 0; i < 200000; i++){
            int x = r.nextInt(798) + 1;
            heights[x] =  Math.min(599, heights[x - 1] == heights[x + 1] ? heights[x] : Math.max(Math.max(heights[x - 1], heights[x]),heights[x + 1]) + 1);
            buffered_image.setRGB(x, Math.min(599, 600 - heights[x]), 0);
        } 

        try {

            ImageIO.write(buffered_image, "png", outputFile);
        } catch (IOException e) {

        }
        update();
    }
    public void repaint(){
        if(this.getGraphics() != null)
        paint(this.getGraphics());
    }


    public void paint(Graphics g) {
        g.drawImage(buffered_image, 0, 0, this);
    }

    public void update() {  
        repaint();
    }

    public static void main(String[] args) throws IOException {
        JFrame main_frame = new JFrame();
        main_frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

        JPanel top_panel = new JPanel();
        top_panel.setLayout(new BorderLayout());
        Forest s = new Forest();
        top_panel.add(s, BorderLayout.CENTER);
        main_frame.setContentPane(top_panel);

        main_frame.pack();
        main_frame.setVisible(true);
    }

}

6
也许垂直翻转它使其类似于枞树?
2014年
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