title: Draw random black-and-white forest

text:
# Question
Title: Draw random black-and-white forest
Your task is to write program which will draw 800x600 black-and-white image with something resembling a forest.

Like this (it is dithered photo):

# Rules

* You are disallowed to use any existing images - you should generate image purely algorithmically
* Use only 2 colors - black and white (no grayscale)
* Each time program runs image should be new - random every time
* One tree is not a forest (let say 5 trees minumum)
* Special libraries for drawing trees/forests are disallowed
* Answer with most votes wins

# Answer
> 101 votes
## C: 3863 1144 1023 999 942 927

The original solution saves 2 pnm files per run (one with g appended, before dithering). Because the dithering wasn't beautiful for the first few lines, there is a hack in place to render more lines than needed, and crop during output.

The golfed solution has a simpler dithering and saves only the dithered image. (no warnings with gcc -std=c11 -pedantic -Wall -Wextra)

Example images from 3 original program runs and one run of the golfed version (last image):

Golfed version

```
  #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);}

```

Original version

```
  #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);
  }

```

# Answer
> 43 votes
# Java Jungle

### (954 golfed)

Full of deep, twisting undergrowth, this is a forest not easily traversed.

It's basically a fractal random walk with slowly shrinking, twisty vines. I draw 75 of them, gradually changing from white in the back to black up front. Then I dither the whole thing, shamelessly adapting Averroes' code here for that.

Golfed: <sub>(Just because others decided to)</sub>

```
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 original code:

```
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 };

}

```

One more? Okay! This one has the dithering tuned down a bit, so the blacks in front are much flatter.

Unfortunately, the dither doesn't show the fine details of the vine layers. Here's a greyscale version, just for comparison:

# Answer
> 31 votes
# Javascript + HTML - not golfed

A javascript porting of the algorithm of @Manuel Kansten - it's amazing how good these trees look.

Just to do something different, I draw the image in color, then dither to b/w at the last step.

I don't know why, but my forest is less dark and less frightening respect to Manuel's.

Test with JSfiddle or run the new **Snippet** below. That's NOT fast. Be patient and watch the forest grow.

```
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>
```

# Answer
> 24 votes
# Context Free Art 3 (1133)

CF is a vector graphics rendering language, so I cannot avoid anti-alising. I worked that around by drawing square at the same place several (variable `N`) times. Fog is done by drawing small squares on random places.

```
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 ]
}

```

More renders using different numbers

# Answer
> 23 votes
## IFS with JAVA

This solution uses an Iterated Function System (IFS) to describe one (proto) tree. The IFS is applied 100 times (=forest). Before each tree is painted (planted into the forest) the IFS is altered slightly in place (random walk style). So each tree looks slightly different.

Pictures are from random seeds:

* -824737443
* -1220897877
* -644492215
* 1133984583

No dithering is needed.

```
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;
    }
}

```

# Answer
> 21 votes
## C: 301

This program creates a simple, abstract image in the PGM format. You can open it with GIMP.

```
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);

```

Here is an example run:

# Answer
> 16 votes
I noticed a distinct lack of conifers here, so I hacked something together in Python.

```
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()

```

This was my first Code Golf, it was a lot of fun!

# Answer
> 7 votes
This answer isn't as pretty as I hoped, but it's a stepping stone to a more 3D idea I'm working on, and I really like the idea of actually simulating which trees get resources

```
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);
    }

}

```

---
Tags: popularity-contest, graphical-output, image-processing
---


title: Tuning Strings

text:
# Question
Title: Tuning Strings
## Task

Write a program to determine the note sounded, along with how many cents out of tune, of a string tuned to a given frequency and pressed down at a given point.

For the sake of simplicity, assume that the frequency of the sound produced and the length of the string to the right of where it is pressed are inversely proportional.

Note: this task deals solely in the fundamental tone, and not with overtones/other harmonics.

## Input

Your program is given two pieces of data:

* A string of arbitrary length, representing the string in question. This string will be marked with an X where the string is to be held down.

```
[-----] is a string divided in six sections (five divisions).
[--X--] is a string pressed at the exact center of the string.
[X----] is a string pressed at 1/6 the length of the string. (Length used is 5/6)
[-X--] is a string pressed at 2/5 of the length of the string. (Length used is 3/5)

```

Assume the note is sounded using the part of the string to the right of the `X`.
* A number (not necessarily an integer), signifying the frequency at which the string is tuned. The precision of this number will be at most four digits past the decimal.

It may be assumed that frequencies passed will lie between `10 Hz` and `40000 Hz`.

Input may be passed in a format of your choice. Please specify how input is accepted into your program in your answer.

## Output

Your program must output both the closest note\* in the twelve-tone equal temperament tuning system, and the number of cents away from the closest note that the sound denoted by the string would be (rounded to the nearest cent).

`+n` cents should be used to denote `n` cents sharp/above the note, and `-n` cents for flat/below the note.

The note should be outputted in scientific pitch notation. Assume A4 is tuned to `440Hz`. Use b and # for flat/sharp notes. Note: Either sharp or flat may be used. For the note at `466.16Hz`, either `A#` or `Bb` may be outputted for the note.

Format of output is up to you, as long as the output contains only the two previously specified pieces of information (i.e. printing every single possible output is not allowed).

\*closest note refers to the note that is closest to the sound denoted by the input, measured in the number of cents (therefore, the note that is within `50 cents` of the sound). If the sound is `50 cents` away from two different notes (after rounding), then either of the two notes may be outputted.

## Examples

Your program should work for all cases, not just the following examples.

```
Output             Input Frequency   Input String
A4,  +0  cents     220               [-----X-----]
A5,  +0  cents     220               [--------X--]
D5,  -2  cents     440               [--X--------]
B4,  -49 cents     440               [X----------]
A#4, +19 cents*    314.1592          [X-]
Eb9, +8  cents*    400               [-----------------------X]
Eb11,+8  cents*    100               [--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------X]
D#1, +49 cents*    10                [--X]
A0,  -11 cents     11.7103           [---X--]

```

\*Either sharp or flat could have been outputted.

## Potentially Helpful Links

This is so shortest answer wins.

# Answer
# BBC Basic, 161#

```
  REM get frequency and string. Store length of string in n for later.
  INPUT f,s$
  n=LEN(s$)

  REM store floating-point value of note in semitones above (C0-0.5). Store integer value in n% (offset effectively means it will be rounded not truncated.)
  n=LN(f*(n-1)/(n-INSTR(s$,"X"))/15.8861)/LN(2)*12
  n%=n

  REM format printing to whole numbers only
  @%=2^17

  REM Output note name and octave. Output cents (discrepancy between n and n%, minus the offset of 0.5)
  PRINT MID$("C C#D D#E F F#G G#A A#B ",n%MOD12*2+1,2);INT(n/12)'(n-0.5-n%)*100'

```

Score excludes comments. Not golfed yet.

**Output**

Performs correctly on all test cases except the two long ones. For `Eb9` it seems there is one dash missing from the test case: There are 22 `-` and one `X`, which divides the string into 24 equal pieces. According to my manual calculations, this is 9600Hz, which is 37 cents above a D9. This is exactly what my program outputs. If I add another dash I get Eb9 + 8 cents. Unfortunately BBC Basic cannot handle strings over 255 characters, so the `Eb11` case gives an error.
> 1 votes

# Answer
# C, 179

```
main(n,d){float f;scanf("%*[^X]%nX%*[-]%n]%f",&n,&d,&f);f=log(f*d/(d-n))*17.3123-57.376;n=d=f+.5;n=n%12*7+784;printf("%c%d%c,%+2.0f cents\n",n/12,(d+9)/12,n%12/7*3+32,(f-d)*100);}

```

Receives the ascii picture on a line by itself, and the frequency on a separate line.

A few characters can be dropped by reducing the accuracy of the magic numbers `17.3123` and `57.376`.

Without the golfing, the program looks like this:

```
main(n,d)
{
    float f; // 'float' and '%f' better than 'double' and '%lf'

    scanf("%*[^X]%nX%*[-]%n]%f", &n, &d, &f);
    // n is the number of chars before 'X'
    // d is the number of chars before ']'
    // f is the frequency

    // Calculate the tuned frequency
    f = f * d / (d - n);

    // Convert the frequency to logarithmic scale, relative to pitch A0
    f=log(f)*17.3123-57.376;
    // alternatively: f = log2(f / (440 / 16)) * 12;

    // Round to nearest integer
    n=d=f+.5;

    // Calculate the note name ('A', 'B', etc), multipled by 12 for convenience
    n=n%12*7+784;

    printf("%c%d%c,%+2.0f cents\n", // output example: B4 ,-49 cents
        n/12,        // note name
        (d+9)/12,    // octave number
        n%12/7*3+32, // if n%12 is large enough, it's '#' else ' ' (natural)
        (f-d)*100);  // number of cents; stdio rounds it to integer
}

```
> 3 votes

# Answer
# JavaScript (199)

Call it e.g. as `t('[X-]',314.1592)`

```
t=(s,f)=>{l=s.length-1;p='C C# D D# E F F# G G# A B H'.split(' ');n=12*Math.log2(f*l/(l-s.indexOf('X'))/16.3515978);m=n+.5|0;return p[m%12]+(n/12|0)+' '+((n-m)*100+.5|0)}

```

Fixed. (Since I live in europe I used B instead of Bb and H instead of B=)
> 1 votes

# Answer
# Python 3: 175

```
import math
def t(b,s):l=len(s)-1;n=12*math.log2(b*l/(l-s.index("X"))/16.35);m=round(n);return"%s%s%+d"%(("C C# D D# E F F# G G# A A# B".split()*99)[m],m//12,round(100*(n-m)))

```

Ungolfed:

```
import math

c0 = 16.35

def tuning (base_frequency, string):
    return formatted (note_number (frequency (base_frequency, string)))

def formatted (note_number):
    return "{name}{octave:d}{cents:+d}".format (name=note_name (note_number),
                             octave=octave (note_number),
                             cents=cents_out (note_number))

def note_name (note_number):
    return ("C C# D D# E F F# G G# A A# B".split() * 99)[round (note_number)]

def note_number (frequency):
    return 12 * math.log2 (frequency / c0)

def octave (note_number):
    return round (note_number) // 12

def cents_out (note_number):
    return round (100 * (note_number - round (note_number)))

def frequency (base_frequency, string):
    string_length = len (string) - 1
    held_length = string_length - string.index ("X")
    return base_frequency * string_length / held_length

if "__main__" == __name__:

    print ("Testing functions against known values...")
    assert "A4+0"     == tuning (220,      "[-----X-----]")
    assert "A5+0"     == tuning (220,      "[--------X--]")
    assert "D5-2"     == tuning (440,      "[--X--------]")
    assert "B4-49"    == tuning (440,      "[X----------]")
    assert "A#4+19"   == tuning (314.1592, "[X-]")
    assert "D#9+8"    == tuning (400,      "[-----------------------X]")
    assert "D#11+8"   == tuning (100,      "[--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------X]")
    assert "D#1+49"   == tuning (10,       "[--X]")
    assert "A0-11"    == tuning (11.7103,  "[---X--]")
    print ("Tests passed.")

```
> 1 votes

---
Tags: code-golf, music
---


title: Slime: The Territory War

text:
# Question
Title: Slime: The Territory War
You are a glob of slime. Naturally, being slime, you want to ooze over as much area as possible. But there are 3 other slimes who want to do the exact same thing. Who will be the superior slime?

## Description

All slimes will be gathered into one arena. The judges (i.e. the control program) will go through an exhaustive list of all possible 4-slime combinations, place them on the corners of a table, and observe to see which slime oozes out over the most area.

Your slimes can do one of 3 actions each turn: spread out, jump, or merge. Further description on what these mean will be provided in the **Output** section.

## Board / Arena

The arena will be a square board (currently 8x8, but this may change in the future). Here is an example arena of a game in progress:

```
11111222
11111444
11.22444
.1222.4.
333.3244
33333.44
333...44
333....4

```

Slime is represented by the numbers 1 through 4 (players 1 to 4), and empty space is represented by a dot (`.`). Initially, the board starts out as all empty space except for a single unit of player 1's slime in the top left corner, player 2 in the top right, player 3 in the bottom left, and player 4 in the bottom right.

Coordinates are represented by 0-based row and column index, for readability in the code. For example, the coordinates (3, 6) represent the 7th square in the 4th row (in the above example, a `4`). (This makes it easier to access squares: `board[coords.x][coords.y]`.) Here's a visual illustration:

```
(0, 0) (0, 1) (0, 2)
(1, 0) (1, 1) (1, 2)
(2, 0) (2, 1) (2, 2)

```

## Input

Your program's input will be which player you are (1, 2, 3, or 4), a comma (`,`), then the content of the board / arena (with newlines replaced with commas). For example, if you were player 3 in the above scenario, your input would be:

```
3,11111222,11111444,11.22444,.1222.4.,333.3244,33333.44,333...44,333....4

```

## Output

Your program must output 4 integers. The first two are the row and column index respectively of the slime you would like to move, and the next two are the row and column index of where you want to move them.

There are three choices you have on each turn: Spread out, jump, or merge.

* **Spread**

To spread, the target coordinates must be exactly one square away from the slime being moved, and the square at the target coordinates must be empty space. When spreading, a new slime is created at the target coordinates and the old slime is *not* removed. After the new slime is created, all enemy slimes in the 8 squares around this new slime are converted to the player that moved.

For example, with the board in Fig. 1, if player 1 were to output `0 1 1 2`, the result would be the board in Fig. 2.

```
1.         2.
  11.22      11.12
  1..22      1.112
  ..22.      ..11.
  .....      .....

```
* **Jump**

To jump, the target coordinates must be exactly two squares away from the slime being moved, and the square at the target coordinates must be empty space. When jupming, a new slime is created at the target coordinates and the old slime is removed. After the new slime is created, all enemy slimes in the 8 squares around this new slime are converted to the player that moved.

For example, with the board in Fig. 1, if player 1 were to output `0 1 2 3`, the result would be the board in Fig. 2.

```
1.         2.
  11..2      1...2
  1...2      1...1
  ....2      ...11
  ...22      ...11

```
* **Merge**

To merge, the target coordinates must be exactly one square away from the slime being moved, and the square at the target coordinates must be the same player's slime. When merging, the old slime is removed. Then, all empty spaces in the 8 squares around the target slime are converted to the player that moved (not including the old slime being moved).

For example, with the board in Fig. 1, if player 1 were to output `0 1 1 2`, the result would be the board in Fig. 2.

```
1.         2.
  11..2      1.112
  1.1.2      11112
  ....2      .1112
  ..222      ..222

```

You can also pass, by simply outputting invalid coordinates (ex. `0 0 0 0`).

## Rules and constraints

Additional rules are:

* You may read and write files within your own folder in order to persist data (submissions will be stored in `players/YourBotName/yourBotName.language`), but you may not modify or access anything else outside of it. Internet access is prohibited.
* Your submission may not be coded specifically to help or hurt another submission. (You may have multiple submissions, but they must not specifically interact with each other in any way.)
* Your submission must take no more than 0.1 seconds per turn. If your submission takes 0.105 seconds occasionally, that's fine, but it may not consistently take significantly longer than this time limit. (This is mainly a sanity check to avoid testing taking an overly long time.)
* Your submission must not be an exact duplicate (i.e. use the exact same logic) of another, even if it's in a different language.
* Your submission must be a serious submission. This is opinion-based, but if your submission is clearly not attempting to solve the challenge (ex. if you pass every turn), it will be disqualified.

If your submission breaks any of these rules or doesn't follow the specification, it will be disqualified, removed from `playerlist.txt`, and the game will restart from the beginning. If your submission is disqualified, I will leave a comment on your post explaining why. Otherwise, your submission will be added to the leaderboard. (If you don't see your submission on the leaderboard, have no explanatory comment on your post, and posted your submission before the "Last updated" time below, please tell me! I may have inadvertently overlooked it.)

In your entry, please include:

* A name.
* A shell command to run your program (for example, `java MyBot.java`, `ruby MyBot.rb`, `python3 MyBot.py`, etc.).
    + Note that the input (your player and map) will be appended to this as a command line argument.
    + Programs will be tested on Ubuntu 14.04, so make sure your code can be run (freely) on it.
* A version number, if your code works differently on different versions of your language.
* Your bot's code.
* Instructions on how to compile the code, if necessary.

## Controller code / testing, example bot

The controller code is written in C++, and can be found on Github. Further instructions on how to run and test your code can be found there.

`simplebot.rb`, a very simple bot that spreads or jumps a random slime to a random location each turn, is also posted on Github.

## Scoring and leaderboard

When all squares on the board are filled, the game ends and scores are calculated. The final score of a player is the amount of squares that contain their slime at the end of the game. If 2000 turns have passed (500 for each player) and the game still isn't over, the game will end anyway and the scores will be reported as if the game had ended.

At the end of the tournament, the scores from all games will averaged to calculate each player's final score, which will be posted on the leaderboard. There is no submission deadline; I'll continue to update the leaderboard periodically as new submissions come in.

*4 submissions are required until a real leaderboard appears.*

```
+--------------------------+-----------+--------------------+
| Name                     | Avg Score | Last Updated (UTC) |
+--------------------------+-----------+--------------------+
| GreedySlime              | 47.000    | Jul 22 10:27 PM    |
| Jumper                   | 12.000    | Jul 22 10:27 PM    |
| ShallowBlue              | 5.000     | Jul 22 10:27 PM    |
| Lichen                   | 0.000     | Jul 22 10:27 PM    |
+--------------------------+-----------+--------------------+

```

Last updated: Jul 22 10:27 PM (UTC).

# Answer
> 4 votes
# GreedySlime

Simply makes the move that produces the greatest net gain of slime units.

Note that this is written in **Python 2.x**.

```
def gen_moves(board, pos):
    """Generate valid moves for a given position.

    Return value is a tuple of the form
       (type, from_x, from_y, to_x, to_y)

    The move 'type' is a single character with:
        - 's' = spread
        - 'j' = jump
        - 'm' = merge
    """

    N = len(board)
    x0, y0 = pos
    player = board[x0][y0]

    for i in -2,-1,0,1,2:
        for j in -2,-1,0,1,2:
            if (i == 0 and j == 0):
                continue

            x1, y1 = x0 + i, y0 + j

            if not ((0 <= x1 < N) and (0 <= y1 < N)):
                continue

            c = board[x1][y1]

            if -1 <= i <= 1 and -1 <= j <= 1:
                if c == '.':
                    yield ('s', x0, y0, x1, y1)
                elif c == player:
                    yield ('m', x0, y0, x1, y1)
            else:
                if c == '.':
                    yield ('j', x0, y0, x1, y1)

def eval_move(board, move, initial_net={'s': 1, 'j': 0, 'm': -1}):
    """Evaluates given move in given context.

    - Assumes move is valid.
    - `move` argument is a tuple of the form
       (type, from_x, from_y, to_x, to_y)
    - The move 'type' is a single character with:
        - 's' = spread
        - 'j' = jump
        - 'm' = merge
    """

    N = len(board)
    move_type = move[0]
    x0, y0, x1, y1 = move[1:]
    player = board[x0][y0]

    net = initial_net[move_type]
    for i in -1,0,1:
        for j in -1,0,1:
            if (i == 0 and j == 0):
                continue

            x2, y2 = x1 + i, y1 + j

            if not ((0 <= x2 < N) and (0 <= y2 < N)):
                continue

            c = board[x2][y2]

            if (move_type == 'm' and c == '.') or (move_type != 'm' and c != player and c != '.'):
                net += 1

    return net

def main():
    from sys import argv
    data = argv[1]

    player, board = data.split(',', 1)
    board = map(list, board.split(','))
    N = len(board)

    all_pos_gen = ((a,b) for a in range(N) for b in range(N) if board[a][b] == player)
    all_move_gen = (move for pos in all_pos_gen for move in gen_moves(board, pos))
    move = max(all_move_gen, key=lambda move: eval_move(board, move))

    print move[1], move[2], move[3], move[4]

if __name__ == "__main__":
    main()

```

---

Example run (using the example given in the challenge description, and assuming the code is saved in a file called `slime.py`):

```
$ python slime.py 3,11111222,11111444,11.22444,.1222.4.,333.3244,33333.44,333...44,333....4
4 0 2 2

```

# Answer
> 3 votes
# Shallow Blue

Shallow blue tries to figure out what might happen in the future by doing a~~n exhaustive~~ tree-search of possible moves, unfortunately, he doesn't get any further than his next turn. He then slaps some half-assed score on each possible boardstate after his next turn, calculates a score for each individual branch with an equally ridiculous formula and: voila the ideal move is known!

EDIT: The original code ran waay too slowly so I modified it so that it only takes a random sample of all possible moves. It will try almost all moves when there are little moves possible and a smaller percentage when there are more moves possible.

```
import java.awt.Point;

    public class ShallowBlue {
        private static final int MAX_ROUNDS = 5, PLAYERS = 4;
        static int me = 0;

        public static void main(String[] args) {
            if (args[0] == null) {
                return;
            }

            me = Integer.parseInt(args[0].split(",", 2)[0]);
    String board = args[0].split(",", 2)[1];

    System.out.println(getBestMove(board, me, MAX_ROUNDS - 1));
}

private static String getBestMove(String board, int player, int rounds) {
    String [] boards = new String[24];
    int checkedBoards = 1;
    char playerChar = Integer.toString(player).charAt(0);
    String tempMove = getMove(0, 0, 0, 0);
    String tempBoard = calculateMove(board, tempMove);
    boards[0] = tempBoard;
    String bestMove = tempMove;
    double us = numberOfUs(board, playerChar);
    double skip = (us*2.5/(us*2.5 + 1))/4 + 0.735;
    if (rounds == MAX_ROUNDS - 2) {
        skip = skip*skip;
    }

    float bestScore, worstScore, averageScore, tempScore;
    int scores;

    if (rounds == 0) {
        tempScore = calculateScore(tempBoard, MAX_ROUNDS - rounds - 1);
    } else {
        tempScore = getScore(getBestMove(tempBoard, player%PLAYERS + 1, rounds - 1));
    }

    scores = 1;
    bestScore = tempScore;
    worstScore = tempScore;
    averageScore = tempScore;

    for (int x = 0; x < 8; x++) {
        for (int y = 0; y < 8; y++) {
            if (getCharAt(board, x, y) == playerChar) {
                Point[] possibleMergers = getNeighboringMatches(board, new Point(x, y), playerChar);
                if (possibleMergers[0] != null) {
                    tempMove = getMove(possibleMergers[0].x, possibleMergers[0].y, x, y);
                    tempBoard = calculateMove(board, tempMove);
                    if (addIfUnique(boards, tempBoard, checkedBoards)) {
                        checkedBoards++;
                        if ((rounds != MAX_ROUNDS - 1) && (rounds == 0 || Math.random() < skip)) {
                            tempScore = calculateScore(tempBoard, MAX_ROUNDS - rounds - 1);
                        } else {
                            tempScore = getScore(getBestMove(tempBoard, player%PLAYERS + 1, rounds - 1));
                        }

                        if (tempScore > bestScore) {
                            bestMove = tempMove;
                        }
                        bestScore = Math.max(tempScore, bestScore);
                        worstScore = Math.min(tempScore, worstScore);

                        scores++;
                        averageScore = (averageScore*(scores - 1) + tempScore)/scores;
                    }
                }
            } else if (getCharAt(board, x, y) == '.') {
                Point[] possibleSpreaders = getNeighboringMatches(board, new Point(x, y), playerChar);
                int i = 0;
                while (i < possibleSpreaders.length && possibleSpreaders[i] != null) {
                    tempMove = getMove(possibleSpreaders[i].x, possibleSpreaders[i].y, x, y);
                    tempBoard = calculateMove(board, tempMove);
                    if ((rounds != MAX_ROUNDS - 1) && (rounds == 0 || Math.random() < skip)) {
                        tempScore = calculateScore(tempBoard, MAX_ROUNDS - rounds - 1);
                    } else {
                        tempScore = getScore(getBestMove(tempBoard, player%PLAYERS + 1, rounds - 1));
                    }

                    if (tempScore > bestScore) {
                        bestMove = tempMove;
                    }
                    bestScore = Math.max(tempScore, bestScore);
                    worstScore = Math.min(tempScore, worstScore);

                    scores++;
                    averageScore = (averageScore*(scores - 1) + tempScore)/scores;

                    i++;
                }
                Point[] possibleJumpers = getNextNeighboringMatches(board, new Point(x, y), playerChar);
                i = 0;
                while (i < possibleJumpers.length && possibleJumpers[i] != null) {
                    tempMove = getMove(possibleJumpers[i].x, possibleJumpers[i].y, x, y);
                    tempBoard = calculateMove(board, tempMove);
                    if ((rounds != MAX_ROUNDS - 1) && (rounds == 0 || Math.random() < skip)) {
                        tempScore = calculateScore(tempBoard, MAX_ROUNDS - rounds - 1);
                    } else {
                        tempScore = getScore(getBestMove(tempBoard, player%PLAYERS + 1, rounds - 1));
                    }

                    if (tempScore > bestScore) {
                        bestMove = tempMove;
                    }
                    bestScore = Math.max(tempScore, bestScore);
                    worstScore = Math.min(tempScore, worstScore);

                    scores++;
                    averageScore = (averageScore*(scores - 1) + tempScore)/scores;

                    i++;
                }
            }
        }
    }

    if (rounds == MAX_ROUNDS - 1) {
        return (bestMove);
    } else {
        return getScoreString(bestScore, worstScore, averageScore);
    }
}

private static int numberOfUs(String board, char playerChar) {
    int us = 0;

    for (int i = 0; i < board.length(); i++ ) {
         if (board.charAt(i) == playerChar) {
            us++;
        }
    }

    return us;
}

private static float calculateScore(String board, int roundsPassed) {
    int empties = 0;
    int us = 0;
    int enemy1 = 0;
    int enemy2 = 0;
    int enemy3 = 0;
    for (int i = 0; i < board.length(); i++ ) {
        if (board.charAt(i) == '.') {
            empties++;
        } else if (board.charAt(i) == Integer.toString(me).charAt(0)) {
            us++;
        } else if (board.charAt(i) == Integer.toString(me%PLAYERS + 1).charAt(0)) {
            enemy1++;
        } else if (board.charAt(i) == Integer.toString(me%PLAYERS + 2).charAt(0)) {
            enemy2++;
        } else if (board.charAt(i) == Integer.toString(me%PLAYERS + 3).charAt(0)) {
            enemy3++;
        }
    }

    if (us != 0) {
        us += roundsPassed;
    }

    if (enemy1 != 0) {
        enemy1 = enemy1 + (roundsPassed + 3)%PLAYERS;
    }

    if (enemy2 != 0) {
        enemy2 = enemy2 + (roundsPassed + 2)%PLAYERS;
    }

    if (enemy3 != 0) {
        enemy3 = enemy3 + (roundsPassed + 1)%PLAYERS;
    }

    return us*(empties + 1)/(Math.max(Math.max(enemy1, enemy2), enemy3) + 1);
}

private static float getScore(String scoreString) {
    float bestScore, worstScore, averageScore;
    String[] scores = new String[3];

    scores = scoreString.split(",");
    bestScore = Float.parseFloat(scores[0]);
    worstScore = Float.parseFloat(scores[1]);
    averageScore = Float.parseFloat(scores[2]);

    return (float) Math.sqrt(Math.sqrt(bestScore*averageScore*worstScore*worstScore));
}

private static String getScoreString(float bestScore, float worstScore, float averageScore) {
    return Float.toString(bestScore) + ',' + Float.toString(worstScore) + ',' + Float.toString(averageScore);
}

private static boolean addIfUnique(String[] boards, String board, int checkedBoards) {
    int i = 0;

    while (i < boards.length && boards[i] != null) {
        if (boards[i].equals(board)) {
            return false;
        }
        i++;
    }

    if (i < boards.length) {
        boards[i] = board;
    } else {
        boards[checkedBoards%boards.length] = board;
    }

    return true;
}

private static String calculateMove(String board, String move) {
    int x1 = Integer.parseInt(Character.toString(move.charAt(0)));
    int y1 = Integer.parseInt(Character.toString(move.charAt(2)));
    int x2 = Integer.parseInt(Character.toString(move.charAt(4)));
    int y2 = Integer.parseInt(Character.toString(move.charAt(6)));

    if ((Math.abs(y1 - y2) == 2 || Math.abs(x1 - x2) == 2)
            &&  getCharAt(board, x2, y2) == '.') {
        Point[] enemies = new Point[8];

        enemies = getNeighboringEnemies(board, new Point(x1, y1), Integer.parseInt(Character.toString(getCharAt(board, x1, y1))));

        board = replace(board, enemies, getCharAt(board, x1, y1));
        Point[] middle = {new Point(x1, y1)};
        board = replace(board, middle, '.');
    }

    if ((Math.abs(y1 - y2) == 1 || Math.abs(x1 - x2) == 1)) {
        if (getCharAt(board, x2, y2) == '.' || getCharAt(board, x1, y1) == getCharAt(board, x2, y2)) {
            boolean merge = true;
            if (getCharAt(board, x2, y2) == '.') {
                merge = false;
            }

            Point[] spaces = new Point[8];
            spaces = getNeighboringMatches(board, new Point(x1, y1), '.');
            board = replace(board, spaces, getCharAt(board, x1, y1));

            if (merge) {
                Point[] source = {new Point(x1, y1)};
                board = replace(board, source, '.');
            }
        }
    }

    return board;
}

private static String replace(String board, Point[] targets, char source) {
    int i = 0;

    while (i < targets.length && targets[i] != null) {
        if (targets[i].x == 7 && targets[i].y == 7) {
            board = board.substring(0, getIndexAt(targets[i].x, targets[i].y)) + source;
        } else if (targets[i].x == 0 && targets[i].y == 0) {
            board = source + board.substring(getIndexAt(targets[i].x, targets[i].y) + 1);
        } else {
            board = board.substring(0, getIndexAt(targets[i].x, targets[i].y)) + source + board.substring(getIndexAt(targets[i].x, targets[i].y) + 1);
        }
        i++;
    }

    return board;
}

private static Point[] getNeighboringMatches(String board, Point coord, char match) {
    Point[] matches = new Point[8];

    int i = 0;
    for (int x = coord.x - 1; x <= coord.x + 1; x++) {
        for (int y = coord.y - 1; y <= coord.y + 1; y++) {
            if ((y != coord.y || x != coord.x ) && getCharAt(board, x, y) == match){
                matches[i] = new Point(x, y);
                i++;
            }
        }
    }

    return matches;
}

private static Point[] getNeighboringEnemies(String board, Point coord, int player) {
    Point[] enemies = new Point[8];

    for (int i = 1; i <= PLAYERS; i++){
        enemies = mergeArr(enemies, getNeighboringMatches(board, coord, Integer.toString((player + i - 1)%PLAYERS + 1).charAt(0)));
    }

    return enemies;
}

private static Point[] getNextNeighboringMatches(String board, Point coord, char match) {
    Point[] matches = new Point[16];

    int i = 0;
    for (int x = coord.x - 2; x <= coord.x + 2; x++) {
        for (int y = coord.y - 2; y <= coord.y + 2; y++) {
            if ((Math.abs(y - coord.y) == 2 || Math.abs(x - coord.x) == 2) && getCharAt(board, x, y) == match){
                matches[i] = new Point(x, y);
                i++;
            }
        }
    }

    return matches;
}

private static char getCharAt(String board, int x, int y) {

    if (x >= 0 && x < 8 && y >= 0 && y < 8) {
        return board.charAt(9*x + y);
    } else {
        return '\0';
    }
}

private static int getIndexAt(int x, int y) {
    return 9*x + y;
}

private static Point[] mergeArr(Point[] arr1, Point[] arr2) {
    int i = 0;
    int j = 0;

    while (i < arr1.length && arr1[i] != null) {
        i++;
    }

    while (j < arr2.length && arr2[j] != null) {
        arr1[i + j] = arr2[j];
        j++;
    }

    return arr1;
}

private static String getMove(int x1, int y1, int x2, int y2) {
    return Integer.toString(x1) + " " + Integer.toString(y1) + " " + Integer.toString(x2) + " " + Integer.toString(y2);
    }
}

```

# Answer
> 2 votes
# Jumper

Likes to jump, even more so towards the middle.

Will pass if no slimes can jump.

**C++**, Should compile simply with `g++ jumper.cpp -o jumper`

```
#include <math.h>
#include <algorithm>
#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#define maxn(x, y) ((x) > (y) ? (x) : (y))
#define absn(x) ((x) < 0 ? -(x) : (x))
class Board {
 public:
    Board(std::string input_string);
    void Move();
 private:
    void ParseBoardState(std::string console_string);
    int Slimes(int cell);
    void GetXY(int cell, int& r, int& c);
    bool CanJumpFromHere(int cell, int& jump_to_cell, int& rad);
    int CalcRadius(int cell);
    bool CheckJumpDist(int x, int y);

    int player_num_;
    std::size_t board_dim_;
    std::size_t sq_;
    std::vector< std::vector<int> > slimes_;
};
Board::Board(std::string input_string)
    : player_num_(0),
      board_dim_(0),
      slimes_() {
    board_dim_ = std::count(input_string.begin(), input_string.end(), ',');
    sq_ = board_dim_ * board_dim_;
    std::istringstream temp(input_string.substr(0,1));
    temp >> player_num_;
    ParseBoardState(input_string);
}
void Board::ParseBoardState(std::string console_string) {
    int place = 0;
    for (std::size_t row = 0; row < board_dim_; ++row ) {
        slimes_.push_back(std::vector<int>());
        place = console_string.find(",",place+1);
        std::string temp2 = console_string.substr(place+1, 8);
        for (std::size_t col = 0; col < board_dim_; ++col ) {
            int sl = 0;
            std::istringstream bint(temp2.substr(col,1));
            bint >> sl;
            slimes_[row].push_back(sl);
        }
    }
}
int Board::Slimes(int cell) {
    int r = 0;
    int c = 0;
    GetXY(cell, r, c);
    return  slimes_[r][c];
}
void Board::GetXY(int cell, int& r, int& c) {
    for (std::size_t row = 0; row < board_dim_; ++row ) {
        for (std::size_t col = 0; col < board_dim_ ; ++col ) {
            if ( (row * board_dim_ + col) == cell) {
                r = row;
                c = col;
            }
        }
    }
}
void Board::Move() {

    // go through each cell:
    int index = 0;
    int jump_to_cell = 0;
    int rad = 0;
    int min_rad = 1000;
    int best_jump_to = -1;
    int best_jump_from = -1;
    for (int c = 0; c < sq_; ++c) {
        if (Slimes(c) == player_num_) {
            if (CanJumpFromHere(c, jump_to_cell , rad)) {
                if (rad < min_rad) {
                    best_jump_from = c;
                    best_jump_to = jump_to_cell;
                    min_rad = rad;
                }
                index += 1;
            }
        }
    }

    int ret_row = 0;
    int ret_col = 0;

    if (index == 0) {
        // can't jump so dont bother:
        std::cout << "0 0 0 0" << std::endl;
    } else {
        GetXY(best_jump_from, ret_row, ret_col);
        std::cout << ret_row << " " << ret_col  << " ";
        GetXY(best_jump_to, ret_row, ret_col);
        std::cout << ret_row << " " << ret_col << std::endl;
    }
}
bool Board::CanJumpFromHere(int cell, int& ret_jump_to_cell, int & ret_rad) {
    int r = 0;
    int c = 0;
    int rad = 10000;
    int jump_to_cell = 0;
    int rad_min_for_this_cell = 10000;
    GetXY(cell, r, c);
    bool jumpable = false;
    for (int row_test = -2; row_test < 3; ++row_test) {
        for (int col_test = -2; col_test < 3; ++col_test) {
            if ( (r + row_test) > 0 &
                 (r + row_test) < board_dim_ &&
                 (c + col_test) > 0 &&
                 (c + col_test) < board_dim_ &&
                 (CheckJumpDist(col_test, row_test)) &&
                 (slimes_[r+row_test][c+col_test] == 0)) {

                jumpable = true;
                jump_to_cell = (r + row_test) * board_dim_ + c + col_test;
                rad = CalcRadius(jump_to_cell);

                if (rad < rad_min_for_this_cell) {
                    ret_rad = rad;
                    ret_jump_to_cell = jump_to_cell;
                    rad_min_for_this_cell = ret_rad;
                }
            }
        }
    }
    return jumpable;
}
bool Board::CheckJumpDist(int x, int y) {
    int maxDelta = maxn(absn(x), absn(y));
    if (maxDelta <= 0 || maxDelta > 2) {
        return false;
    } else {
        return true;
    }
}
int Board::CalcRadius(int cell) {
    int r = 0;
    int c = 0;
    GetXY(cell, r, c);
    // unnecessary accuracy considering how bad this bot is:
    float mid = static_cast<float>(board_dim_) / 2;
    float rad = sqrt((r - mid) * (r - mid) + (c-mid)*(c-mid));
    int ret = static_cast<int>(rad + 0.5);
    return ret;
}
int main(int argc, char* argv[]) {
    if (argc != 2) {
        return 0;
    } else {
        std::string input_string(argv[1]);
        Board board(input_string);
        board.Move();
    }
    return 0;
}

```

I stole your move verification sorry. Also I gave up on correct coding practices just after I started, so don't look. However, it seems to run on any size of board.

# Answer
> 2 votes
**DeathSlime**:

**Description**: Tries to hunt the weakest enemy and destroy them. Repeat.

**How to run**: ruby DeathSlime.rb

**Ruby Version**: 2.1.2

```
#!/usr/bin/env ruby
class PlayerPosition;
  attr_accessor :x, :y;
  def initialize(x, y) @x = x; @y = y; end
  def distance(pos) Math.sqrt((pos.x - @x)**2 + (pos.y - @y)**2); end
end

class Board
  attr_reader :player, :empty_positions
  def initialize(player_id, game_state_string)
    @player_positions = {}
    @empty_positions = []

    @enemies = []
    @player = Player.new

    row = 0
    col = 0
    game_state_string.chars.each do |tile|
      row += 1 and col = 0 and next if tile == ','
      @empty_positions << PlayerPosition.new(col, row) and col += 1 and next if tile == '.'

      @player_positions[tile] ||= []
      @player_positions[tile] << PlayerPosition.new(col, row)
      col += 1
    end

    @player_positions.each do |id, positions|
      @enemies << Player.new(id, positions) if id != player_id
      @player = Player.new(id, positions) if id == player_id
    end
  end

  def border_space(player_positions, possible_border, allowance = 1)
    near = []
    possible_border.each do |border|
      is_near = false
      player_positions.each {|pos| is_near = true and break if pos.distance(border) <= allowance}
      near << border if is_near
    end
    near
  end

  def closest_to(player_positions, enemy_positions)
    player_closest_block = nil
    shortest_distance = 1000
    enemy_closest_block = nil
    player_positions.each do |player|
      enemy_positions.each do |enemy|
        if player.distance(enemy) < shortest_distance
          shortest_distance = player.distance(enemy)
          enemy_closest_block = enemy
          player_closest_block = player
        end
      end
    end
    return player_closest_block, enemy_closest_block
  end

  def empty_space_near(player_positions, allowance = 1); border_space(player_positions, @empty_positions, allowance); end
  def weakest_enemy; @enemies.select{|enemy| !enemy.dead? }.sort {|x,y| x.strength <=> y.strength}.first; end
end

class Player
  attr_reader :positions
  def initialize(id = -1, positions = []); @id = id; @positions = positions; end
  def dead?; @positions.length == 0; end
  def strength; @positions.length; end
  def can_hurt?(enemy)
    is_close_enough = false
    self.positions.each do |my_pos|
      enemy.positions.each {|enemy_pos| is_close_enough = true and break if my_pos.distance(enemy_pos) <= 2 }
    end
    is_close_enough
  end
end

class DeathSlime

  def initialize(arg_string)
    game_state = arg_string[2..-1]
    player_id = arg_string[0]
    @board = Board.new(player_id, game_state)
  end

  def attack
    if @board.weakest_enemy
      try_to_spread_to_weakest || try_to_jump_to_weakest || try_to_merge_to_weakest || try_to_move_to_weakest
    else
      try_to_move if @empty_positions.length > 0
    end
  end

  def try_to_spread_to_weakest
    mine = @board.empty_space_near(@board.player.positions, 1)
    theirs = @board.empty_space_near(@board.weakest_enemy.positions, 1)
    target_space = mine.detect{|space| theirs.include?(space) }
    return move(@board.closest_to(@board.player.positions, [target_space]).first, target_space) if target_space
    false
  end

  def try_to_jump_to_weakest
    mine = @board.empty_space_near(@board.player.positions, 2)
    theirs = @board.empty_space_near(@board.weakest_enemy.positions, 1)
    target_space = mine.detect{|space| theirs.include?(space) }
    return move(@board.closest_to(@board.player.positions, [target_space]).first, target_space) if target_space
    false
  end

  def try_to_merge_to_weakest
    definite_border = nil
    definite_merge = nil
    possible_border = @board.border_space(@board.weakest_enemy.positions, @board.player.positions)
    possible_border.each do |border|
      possible_merges = @board.border_space([ border ], @board.player.positions.select{|space| space != border })
      definite_merge = possible_merges.first and definite_border = border and break if possible_merges.length > 0
    end
    return move(definite_merge, definite_border) if definite_border && definite_merge
    false
  end

  def try_to_move_to_weakest
    player_closest, enemy_closest = @board.closest_to(@board.player.positions, @board.weakest_enemy.positions)
    spreading_distance = @board.empty_space_near([player_closest], 1)
    jumping_distance = @board.empty_space_near([player_closest], 2)
    theirs = @board.empty_space_near(@board.player.positions, 2)

    spreading_space = spreading_distance.detect{|space| theirs.include?(space) }
    return move(@board.closest_to(@board.player.positions, [spreading_space]).first, spreading_space) if spreading_space

    jumping_space = jumping_distance.detect{|space| theirs.include?(space) }
    return move(@board.closest_to(@board.player.positions, [jumping_space]).first, jumping_space) if jumping_space

    return move(@board.closest_to(@board.player.positions, [spreading_distance]).first, spreading_distance) if spreading_distance.length > 0
    return move(@board.closest_to(@board.player.positions, [jumping_distance]).first, jumping_distance) if jumping_distance.length > 0

    #merge randomly
    closest_enemy = @board.closest_to(@board.player.positions, @board.weakest_enemy.positions).first
    return move(@board.closest_to(@board.player.positions.select{|space| space != closest_enemy }, [closest_enemy]).first, closest_enemy)
  end

  def try_to_move
    spreading_distance = @board.empty_space_near(board.player.positions, 1)
    jumping_distance = @board.empty_space_near(board.player.positions, 2)

    return move(@board.closest_to(@board.player.positions, [spreading_distance]).first, spreading_distance) if spreading_distance.length > 0
    return move(@board.closest_to(@board.player.positions, [jumping_distance]).first, jumping_distance) if jumping_distance.length > 0
  end

  def move(start_block, end_block)
    STDOUT.write "#{start_block.x} #{start_block.y} #{end_block.x} #{end_block.y}"
    true
  end
end

slime_of_death = DeathSlime.new(ARGV[0])
slime_of_death.attack

```

# Answer
> 1 votes
## Lichen

This is a bot written in R. It needs to be triggered using `Rscript Lichen.R`.

```
input <- strsplit(commandArgs(TRUE),split=",")[[1]]
me <- input[1]
arena <- do.call(rbind,strsplit(input[-1],""))
n <- sum(arena==me)
where <- which(arena==me,arr.ind=TRUE)
closest <- function(a,b){
    x <- abs(outer(a[,1],b[,1],`-`))
    y <- abs(outer(a[,2],b[,2],`-`))
    matrix(which(x<2&y<2,arr.ind=TRUE),ncol=2)
    }
if(n==0){ #No slime on the board
    out <- "0 0 0 0"
    }else if(n==1){ #One slime on the board
        x <- where[1]+c(1,-1)
        y <- where[2]+c(1,-1)
        out <- paste(where[1]-1,where[2]-1,x[x%in%2:(nrow(arena)-1)]-1,y[y%in%2:(nrow(arena)-1)]-1,sep=" ")
    }else{
        area <- apply(which(arena==me,arr.ind=TRUE),2,range,na.rm=TRUE)
        empty <- matrix(which(arena==".",arr.ind=TRUE),ncol=2)
        opponents <- c("1","2","3","4")[c("1","2","3","4")!=me]
        for(i in seq_along(opponents)){
            if(i==1){
                other <- which(arena==opponents[i],arr.ind=TRUE)
                }else{other <- rbind(other,which(arena==opponents[i],arr.ind=TRUE))}
            }
        fillable <- matrix(empty[empty[,1]%in%area[1,1]:area[2,1]&empty[,2]%in%area[1,2]:area[2,2],],ncol=2)
        enemies <- matrix(other[other[,1]%in%area[1,1]:area[2,1]&other[,2]%in%area[1,2]:area[2,2],],ncol=2)
        if(length(unique(where[,2]))==1 | length(unique(where[,2]))==1){ #Slimes form a line
            W <- closest(where,empty)
            if(nrow(W)){
                out <- paste(c(where[W[1,1],]-1,empty[W[1,2],]-1),collapse=" ")
            }else{out <- "0 0 0 0"}
        }else if(length(enemies)&length(fillable)){ #There are enemies and empty spaces in habitable area
            w <- closest(enemies, fillable)
            if(nrow(w)){
                X <- abs(where[,1]-fillable[w[1,2],1])
                Y <- abs(where[,2]-fillable[w[1,2],2])
                W <- which(X<2&Y<2)
                out <- paste(c(where[W[1],]-1,fillable[w[1,2],]-1),collapse=" ")
            }else{out <- "0 0 0 0"}
        }else if(length(fillable)){ #There are empty spaces in habitable area
            w <- closest(fillable,where)
            out <- paste(c(where[w[1,2],]-1,fillable[w[1,1],]-1),collapse=" ")
        }else{
            x <- area[!area[,1]%in%c(1,nrow(arena)),1]
            y <- area[!area[,2]%in%c(1,ncol(arena)),2]
            if(sum(arena[x+(-1:1),y+(-1:1)]==".")>1){
                w <- where[where[,1]%in%(x+c(1,-1))&where[,2]%in%(y+c(1,-1)),]
                out <- paste(w[1]-1,w[2]-1,x-1,y-1,sep=" ")
            }else{
                W <- closest(where, empty)
                if(nrow(W)){
                    out <- paste(c(where[W[1,1],]-1,empty[W[1,2],]-1),collapse=" ")
                }else{out <- "0 0 0 0"}
            }
        }
    }
cat(out)

```

The intended algorithm is that it tries to cover a rectangular area (filling the blank using `spread`). When the rectangle is complete, it `merges` the two slimes at one of its corner (the one the furthest from the corner of the board) to expand the "habitable" area, then fill that newly defined rectangle, etc. It does not use `jump`.

```
.....   .....   .....   .....   .....   ..333
.....   .333.   3333.   3333.   3333.   33333
333..   3333.   3333.   3333.   3333.   33.33
333..   3.33.   3.33.   3333.   3333.   3333.
333..   333..   333..   333..   3333.   3333.

```

If an enemy is in the habitable area and there is an empty space as well in the area, it fills the empty space next to it. If the slime that should be merged into when expanding the habitable area is surrounded by enemies, then one slime will just `spread` instead of this one merging.

# Answer
> 0 votes
## CornerSlime

This slime has a notion of corners, or at least did when I first wrote it in C#, I'm not really sure any more.

Written in C++, presumably will compile fine with gcc and next to no arguments; hopefully I've not used anything MSVC specific by accident.

Tested exclusively against itself on a modified server (no fancy new C++ compiler where I am) so I have no clue how it will perform, hopefully it won't be disqualified for being too slow. There is currently no randomness in this bot, but I may add some at a later date.

This is C# ported to C++ (due to speed concerns) by someone who doesn't really know C++, and is horrendous. It starts by building an array of Cells, which it then fills with all sorts of unhelpful information about the cells around it (number of my cells, number of my slimes, that sort of thing). Then it uses this information to decide if it needs to look more closely at the information it used to create that information, and then potentially uses said information to produce a meaningful output.

```
#include <iostream>

#define min(a,b) a>b?b:a;
#define max(a,b) a>b?a:b;

#define null 0 // fun times

struct Cell
{
public:
    int t;
    int x, y;
    int counts1[5];
    int counts2[5];
    int ecount1;
    int ecount2;
    bool safe1;
    bool safe2;

    bool canspread;
    bool canjump;
    bool canmerge;

    bool spreadable;
    bool jumpable;
    bool mergeable;

    Cell()
    {
        for (int i = 0; i < 5; i++)
        {
            counts2[i]=counts1[i]=0;
        }
        ecount1=ecount2=0;
        safe1=safe2=mergeable=jumpable=spreadable=canmerge=canjump=canspread=false;
    }

    Cell(int tN, int xN, int yN) // not sure why I can't call () constructor here
    {
        for (int i = 0; i < 5; i++)
        {
            counts2[i]=counts1[i]=0;
        }
        ecount1=ecount2=0;
        safe1=safe2=mergeable=jumpable=spreadable=canmerge=canjump=canspread=false;

        t = tN;
        x = xN;
        y = yN;
    }

    void findOptions(int moi)
    {
        if (t == 0)
        {
            if (counts1[moi] > 0)
                spreadable = true;
            if (counts2[moi] > 0)
                jumpable = true;
        }
        else if (t == moi)
        {
            if (counts1[moi] > 0)
                mergeable = canmerge = true;
            if (counts1[0] > 0)
                canspread = true;
            if (counts2[0] > 0)
                canjump = true;
        }
    }
};

const int dim = 8;
const int hdim = 4;

int moi;
int chezMoi;

int target;
int chezTarget;

Cell cells[dim][dim];

int cornerCounts[4][5];
int totalCounts[5];

// ring ness - why why why

// end ring ness

int tlx;
int tly;
int thx;
int thy;

int alx;
int aly;
int ahx;
int ahy;

int rj;
int rstate;

void ring(int x, int y, int dist)
{
    tlx=x-dist;
    tly=y-dist;
    thx=x+dist;
    thy=y+dist;

    alx=max(0, tlx);
    aly=max(0, tly);
    ahx=min(dim-1, thx);
    ahy=min(dim-1, thy);

    rstate = 0;
}

bool nextR(Cell** outc)
{
    if (rstate == 1)
    {
        goto state1;
    }
    if (rstate == 2)
    {
        goto state2;
    }
    if (rstate == 3)
    {
        goto state3;
    }
    if (rstate == 4)
    {
        goto state4;
    }

    if (alx == tlx)
    {
        rj = aly - 1;
        rstate = 1;
    }
state1:
    if (alx == tlx)
    {
        if (++rj <= ahy)
        {
            *outc = (cells[alx]+rj);
            return true;
        }
        alx++;
    }

    if (ahx == thx)
    {
        rj = aly - 1;
        rstate = 2;
    }
state2:
    if (ahx == thx)
    {
        if (++rj <= ahy)
        {
            *outc = (cells[ahx]+rj);
            return true;
        }
        ahx--;
    }

    if (aly == tly)
    {
        rj = alx - 1;
        rstate = 3;
    }
state3:
    if (aly == tly)
    {
        if (++rj <= ahx)
        {
            *outc = (cells[rj]+aly);
            return true;
        }
    }

    if (ahy == thy)
    {
        rj = alx - 1;
        rstate = 4;
    }
state4:
    if (ahy == thy)
    {
        if (++rj <= ahx)
        {
            *outc = (cells[rj]+ahy);
            return true;
        }
    }

    return null;
}

int cox;
int coy;

int ci;
int cj;

void corner(int idx)
{
    cox = (idx / 2) * hdim;
    coy = (idx % 2) * hdim;

    ci = 0;
    cj = -1;
}

bool nextC(Cell** outc)
{
    for (;ci < hdim;ci++)
    {
        for (;++cj < hdim;)
        {
            *outc = (cells[ci+cox]+cj+coy);
            return true;
        }
        cj = -1;
    }

    return false;
}

void cornerCount(int idx, int* c)
{
    int ox = (idx / 2) * hdim;
    int oy = (idx % 2) * hdim;

    for (int i = 0; i < hdim; i++)
    {
        for (int j = 0; j < hdim; j++)
        {
            c[cells[i+ox][j+oy].t]++;
        }
    }
}

void ringCount(int x, int y, int dist, int* c)
{
    int tlx=x-dist;
    int tly=y-dist;
    int thx=x+dist;
    int thy=y+dist;

    int alx=max(0, tlx);
    int aly=max(0, tly);
    int ahx=min(dim-1, thx);
    int ahy=min(dim-1, thy);

    if (alx == tlx)
    {
        for (int j = aly; j <= ahy; j++)
            c[cells[alx][j].t]++;
        alx++;
    }
    if (ahx == thx)
    {
        for (int j = aly; j <= ahy; j++)
            c[cells[ahx][j].t]++;
        ahx--;
    }
    if (aly == tly)
    {
        for (int i = alx; i <= ahx; i++)
            c[cells[i][aly].t]++;
    }
    if (ahy == thy)
    {
        for (int i = alx; i <= ahx; i++)
            c[cells[i][ahy].t]++;
    }
}

int trans(char c)
{
    return c<48?0:c-48;
}

std::string res(Cell* ca, Cell* cb)
{
    char buff[100]; // shhh
    sprintf_s(buff, "%d %d %d %d\n", ca->x, ca->y, cb->x, cb->y);
    return std::string(buff);
}

std::string go(char* inp)
{
    moi = trans(inp[0]);

    int a = 2;

    for (int i = 0; i < dim; i++)
    {
        for (int j = 0; j < dim; j++)
        {
            cells[i][j] = Cell(trans(inp[a]), i, j);
            a++;
        }
        a++;
    }

    // count corners and totals
    for (int i = 0; i < 4; i++)
    {
        cornerCount(i, cornerCounts[i]);
        for (int j = 0; j < 5; j++)
        {
            totalCounts[j] += cornerCounts[i][j];
        }
    }

    // count and find cell options
    for (int i = 0; i < dim; i++)
    {
        for (int j = 0; j < dim; j++)
        {
            Cell* c = cells[i]+j;

            ringCount(i, j, 1, c->counts1);
            ringCount(i, j, 2, c->counts2);

            // safeness
            for (int r = 1; r < 5; r++)
            {
                if (r != moi)
                {
                    c->ecount1 += c->counts1[r];
                    c->ecount2 += c->counts2[r];
                }
            }
            c->safe1 = c->ecount1 == 0 && c->counts1[0] == 0; // surrounded by moi
            c->safe2 = c->ecount1 == 0 && c->ecount2 == 0; // no enemies in sight

            // that funcion which does stuff
            c->findOptions(moi);
        }
    }

    // find chezMoi
    chezMoi = moi-1; // might work, can't be bothered to work it out
    for (int i = 1; i < 4; i++)
    {
        if (cornerCounts[i][moi] > cornerCounts[chezMoi][moi])
            chezMoi = i;
    }

    int best = 0;

    if (cornerCounts[chezMoi][moi] < hdim * hdim)
    {
        // fill our corner
        best = 0;
        Cell* ac = null;
        Cell* bc = null;

        corner(chezMoi);
        Cell* c;
        while (nextC(&c))
        {
            if (c->spreadable && c->ecount1 + 1 > best)
            {
                ring(c->x, c->y, 1);
                Cell* oc;
                while (nextR(&oc))
                {
                    if (oc->canspread)
                    {
                        best = c->ecount1 + 1;
                        ac = oc;
                        bc = c;
                        break;
                    }
                }
            }
            if (c->mergeable && c->counts1[0] - 1 > best)
            {
                ring(c->x, c->y, 1);
                Cell* oc;
                while (nextR(&oc))
                {
                    if (oc->safe2 && oc->canmerge && c->counts1[0] > 0)
                    {
                        best = c->counts1[0] - 1;
                        ac = oc;
                        bc = c;
                        break;
                    }
                }
            }
        }

        if (bc != null)
        {
            return res(ac, bc);
        }
    }

    // pick target (why?)
    target = -1;
    best = 0;
    for (int i = 0; i < 4; i++)
    {
        if (i == moi)
            continue;
        int cur = totalCounts[i];
        if (target == -1 || cur > best)
        {
            target = i;
            best = cur;
        }
    }

    if (target != -1)
    {
        for (int i = 0; i < 4; i++)
        {
            if (i == chezMoi)
                continue;
            int cur = cornerCounts[i][target];
            if (chezTarget == -1 || cur > best)
            {
                chezTarget = i;
                best = cur;
            }
        }

        // attack chosen target (not sure it does this anymore...)
        best = 0;
        Cell* ac = null;
        Cell* bc = null;

        for (int i = 0; i < dim; i++)
        {
            for (int j = 0; j < dim; j++)
            {
                Cell* c = cells[i]+j;

                if (c->spreadable && c->ecount1 + 1 > best)
                {
                    ring(c->x, c->y, 1);
                    Cell* oc;
                    while (nextR(&oc))
                    {
                        if (oc->canspread)
                        {
                            best = c->ecount1 + 1;
                            ac = oc;
                            bc = c;
                            break;
                        }
                    }
                }
                if (c->jumpable && c->ecount1 - 1 > best)
                {
                    ring(c->x, c->y, 2);
                    Cell* oc;
                    while (nextR(&oc))
                    {
                        if (oc->safe2 && oc->canjump)
                        {
                            best = c->ecount1 - 1;
                            ac = oc;
                            bc = c;
                            break;
                        }
                    }
                }
            }
        }

        if (bc != null)
        {
            return res(ac, bc);
        }
    }

    return "0 0 0 0\n";
}

int main(int argc, char* args[])
{
    printf(go(args[1]).c_str());
    return 0;
}

```

---
Tags: king-of-the-hill
---


# Question
Title: Standardise leading zeroes in an input string
Write a program in the shortest number of **bytes** possible that will parse any string given to it by input, and output that string with any and all numbers padded with leading zeroes to match the largest number's length.

For example:

Input:

```
This 104 is an -8 example of 4.518 a string 50.

```

The generated output should become:

```
This 104 is an -008 example of 004.518 a string 050.

```

Note that any digits after the period are not considered part of the "length" of a number. Numbers are considered any sequence of digits with either 0 or 1 periods in the sequence. Numbers will be delimited with either the string boundary, spaces, commas, or newlines. They can also be followed by a period, but only if the period is then followed by a delimiting character. They can also be preceded with a '-' to indicated negatives. So something like this:

```
The strings 20.d, 5.3ft and &450^ are not numbers, but 450.2 is.

```

Should output the following:

```
The strings 20.d, 5.3ft and &450^ are not numbers, but 450.2 is.

```

That is to say, no modifications.

String input will be no more than 200 characters, if your program has an upper bound for some reason.

The winning answer will be the answer in the shortest number of bytes in **seven days** from the posting of this question.

### Test cases

Input:

```
2 40 2

```

Output:

```
02 40 02

```

Explanation: both substrings `2` are bounded on one side by a string boundary and on the other side by .

---

Input:

```
E.g. 2,2.,2.2,.2,.2., 2 2. 2.2 .2 .2. 2d 2.d 2.2d .2d .2.d 40

```

Output:

```
E.g. 02,02.,02.2,00.2,00.2., 02 02. 02.2 00.2 00.2. 2d 2.d 2.2d .2d .2.d 40

```

Explanation: in the first two groups the first four numbers are followed by a delimiter (`,` or ) and the final one is followed by a period then a delimiter; in the third group, each sequence is followed by the non-delimiter `d`.

# Answer
## Ruby, 111 bytes

I hope this catches all the details of the spec

```
puts $*[0].gsub(r=/(?<![^,\s])(-?)(\d+)(?=\.?\d*(?![^,\s]))/){$1+$2.rjust(i.scan(r).map{|m|m[1].size}.max,'0')}

```

It reads the input as the first command line argument and prints the result to STDOUT.
> 4 votes

# Answer
# Cobra - 282

```
class P
    def main
        r,s,l,n=RegularExpressions.Regex(r'^[\n,]?-?(\d+).?\d*[\n,]?$'),Console.readLine.split(' '),0,0
        for m in 2,for y,x in s.numbered,if r.isMatch(x),while (n-=n-r.match(x).groups[1].length)<if(n>l,l-=l-n,l),x=s[y]=if(x[0]=='-','-0'+x[1:],'0'+x)
        print s.join(' ')

```

For those of you wishing to test this without installing Cobra, it's rather easily converted into equivalent C# code.
> 2 votes

# Answer
## JavaScript, 201 chars

```
s=prompt();j=k=l=0,t={}
for(i in s){s[i]==+s[i]&&s[i]!=" "?j++:(t[i-j]=j,j=0);if(j>k)k=j;}if(j!=0)t[+i+1-j]=j
for(i in s){if(t[i])while(k-t[i]){s=s.slice(0,+i+l)+"0"+s.slice(+i+l);t[i]++;l++;}}alert(s)

```
> 0 votes

---
Tags: code-golf
---