Thread: Chain Code (Freeman)

What have you tried? Where are you stuck?

i didn't try anything yet, because i not have any idea about chain code in java. i need someone to give me idea on doing this.

You mean you didn't even try a web search? We have no clue where you are stuck - the algorithm, reading images, java in general...the question is vague enough to have so many different answers you are not helping yourself receive the help you wish.

i have run this program,

import javax.imageio.ImageIO;
import java.io.*;
import java.io.IOException;
import java.awt.image.BufferedImage;
import java.util.Vector;
 
public class Imagersteg {
 
 
    static Vector<Integer> checkx = new Vector<Integer>();
    static Vector<Integer> checky = new Vector<Integer>();
    static Vector<Integer> direct = new Vector<Integer>();
 
  public static void main( String[] args ) throws IOException{
 
    BufferedImage image = null;
 
    File imagefile = new File("C\\Users\\INTAN\\sign.bmp");
    image = ImageIO.read(imagefile);
 
    int w=image.getWidth();
    int h=image.getHeight();
 
    int x, y;
    x = y = 0;
 
    boolean flag = false;
    for(int k=0;k<h;k++)
    {
      for(int j=0;j<w;j++)
      {
        int rgbValue= image.getRGB( j, k) + 0xFFFFFF + 1;
 
        if(rgbValue == 0)
        {
          x = j;
          y = k;
 
          flag = true;
 
          break;
        }
      }
 
      if(flag)
        break;
    }
 
    System.out.println("Success");
    System.out.println( x + " " + y);
 
    int[] lx = new int[8];
    int[] ly = new int[8];
    lx[0] = 0;
    lx[1] = 1;
    lx[2] = 1;
    lx[3] = 1;
    lx[4] = 0;
    lx[5] = -1;
    lx[6] = -1;
    lx[7] = -1;
    ly[0] = -1;
    ly[1] = -1;
    ly[2] = 0;
    ly[3] = 1;
    ly[4] = 1;
    ly[5] = 1;
    ly[6] = 0;
    ly[7] = -1;
    int l = 0;
    int a = 0;
    int b = 0;
    int beforex = 0;
    int beforey = 0;
 
 
    boolean flagz = false;
    for(int j = 0; j < 500; j++)
    {
      if(flagz)
        break;
      for (int i = 0; i < 8; i++)
      {
    	if(check(x + lx[i],y + ly[i]) && !((x + lx[i] == beforex) && (y + ly[i] == beforey)))
        //if(!((x + lx[i] == beforex) && (y + ly[i] == beforey)))
        {	
          if((image.getRGB(x + lx[i], y + ly[i])  == -16777216))
          {
            l = i;
 
            a = x + lx[i];
            b = y + ly[i];
 
            checkx.add(x);
            checky.add(y);
            beforex = x;
            beforey = y;
 
            x = a;
            y = b;
 
            break;
          }
          checkx.add(x);
          checky.add(y);
        }
 
        if(i == 7)
          flagz = true;
      }
 
      System.out.println(x + " " + y + " = " + l);
      direct.add(l);
 
    }
 
 
    Integer[] chain= new Integer[direct.size()];
	direct.copyInto(chain);
	for(int i = 0; i < chain.length; i++)
	{
		System.out.println( "chaincode = " + chain[i]);
	}
 
	int count = 0;
 
	for(int i = 0; (i + 2) < chain.length; i= i+2)  //direction in array
	{
		for(int j = 0; j < 8; j++)  //direction 0 to 7
		{
			if(chain[i] == j)  //if 2nd pixel = i 
			{
				if(chain[i+1] == j+1 || chain[i+1] == j-1) //if (3rd pixel = i+1) || (3rd pixel = i-1)
				{
					count++;
				}
			}
		}
	}
	System.out.println("total input =" + count);
 
  }
 
public static boolean check(int x, int y){
 
 
	Integer[] xed= new Integer[checkx.size()];
	Integer[] yed= new Integer[checky.size()];
	checkx.copyInto(xed);
	checky.copyInto(yed);
	for( int i = 0; i < xed.length; i++)
	{
		//System.out.println(x + " . " + xed[i] + " . "  + y + " . "  + yed[i]);
		if((x == xed[i]) && (y == yed[i])){
			//System.out.println(x + " . " + xed[i] + " . "  + y + " . "  + yed[i]);
			return false;
		}
	}
	return true;
 
 
}
}

But i have got error, why?blabla.jpg

Originally Posted by fzdabc

hello,have you finished this project now? Can you show me your source code ? I also got stuck in this problem these days.

That isn't how this works. You're essentially saying, "You finished your homework, and I haven't, so can I see yours?". Make your own attempt, then post a specific question along with an SSCCE if you get stuck.