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Longest common subsequence

Posted on 2011-03-07
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Last Modified: 2012-05-11
I have a programming assignment that I have to implement the algorithm for solving the Longest Common Subsequence problem.
the program has to be
•prompts the user to enter a pair of strings
•displays the LCS table produced by the algorithm
•displays the longest common subsequence found
•allows the user to repeat the process with a new pair of strings

This is what I have so far... It compilied but get stoped after I enter two strings...
Could you help me out please?


import java.util.*;

public class LCS 
{
    static Scanner console = new Scanner(System.in);
    
    public static void main(String[] args) 
    {
        int i,j;
        String X;  /* String X */
        String Y;    /* String Y */
        System.out.println("Enter the pair of strings");
        X = console.next();
        Y = console.next();
        System.out.println();
        
        int m = X.length();
        int n = Y.length();
        int[][] T = new int[m+1][n+1];
      
        // C[i][0] = 0 for 0 to m 
        for (i = 0; i <= m; i++) 
        {
            T[i][0] = 0;
        }
	
        // C[0][j] = 0 for  j=0 to n 
        for (j = 0; j <= n; j++) 
        {
            T[0][j] = 0;
        }
        
        //FOR i = 1 TO m
        for (i = 1; i <= m; i++) 
        {
            for (j = 1; j <= n; j++) // FOR i = 1 TO n 
            {
                if (X.charAt(i-1) == Y.charAt(j-1)) 
                    T[i][j]=T[i-1][j-1]+1;
                    else 
                    {	
                    	    T[i][j]=java.lang.Math.max(T[i][j-1],T[i-1][j]);
                    	    
                    }
                }
        }
     	  
        // Backtracking 
        String lcs = new String();
        i=m;
        j=n;
        while (i!=0 && j!=0) 
        {
            if (T[i][j] ==1) 
            {    // diagonal 
            	lcs =X.charAt(i-1) + lcs;
                i = i - 1;
                j = j - 1;
            }
            if (T[i][j] == 2) 
            {  // up 
                i = i - 1;
            }
            if (T[i][j] == 3) 
            {  // backword 
                j = j - 1;
            }
        }
     
        /* print out the result */
        System.out.println("String X is " + X);
        System.out.println("String Y is " + Y);
        System.out.println("The length of LCS is " + T[n][m]);
        System.out.println("The LCS is " + lcs);
        System.out.println(T[i][j]);
         
    }
         
}

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Question by:slritmmi
11 Comments
 
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Expert Comment

by:TommySzalapski
Comment Utility
Can you step through the code line by line with a debugger?

If not, put in some debugging output lines like this until you see where it is getting stuck.
System.out.println("I got to line 23, i = " + i + " and j = " + j);
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Accepted Solution

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TommySzalapski earned 500 total points
Comment Utility
I think your problem is in the while loop.
You only check if T[i ][j] is 1, 2, or 3. You need to also check for other things.
Also, if T[i ][j] ==1, you check the next one before restarting the loop.
            if (T[i][j] ==1) 
            {    // diagonal 
            	lcs =X.charAt(i-1) + lcs;
                i = i - 1;
                j = j - 1;
            }
            else if (T[i][j] == 2) 
            {  // up 
                i = i - 1;
            }
            else if (T[i][j] == 3) 
            {  // backword 
                j = j - 1;
            }
            else
            {
              System.out.println("Is this an error?, i = " + i + " and j = " + j + " and T[i][j] = " + T[i][j]);
            }

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Expert Comment

by:mccarl
Comment Utility
Without trying to totally understand your algorithm, from some simple debugging I can see that many of the elements of T[][] are zero after loops at lines 34-46. However, in the last loop at lines 52-68, you are never doing any action if the current element T[ i ][ j ] is zero, and because you never do anything, i and j stay the same, which means the current element will always be the same zero and you will never exit the loop!

BTW, also just an additional point, the t[][] array will always be initialised to be all zeros for you (by Java), no need to do it again at lines 21-31.
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Expert Comment

by:Kendor
Comment Utility
Seems like we had to implement exactly the same thing... (Pattern Recognition 2?)
check out my version of your code...

ps. I am aware that if [option] is not given in the command line that the script fails as it assignes args[1] and args[2]...
public class StringCompare {

	public static void main(String[] args) {
		
		if(args.length > 3 || args.length < 2){
			System.err.println("Check the arguments!");
			System.err.println("usage: java StringCompare [option] x y" +
					"\n options:"+
					"\n 0 -- compute string edit distance between x and y (default)" +
					"\n 1 -- compute longest common subsequence of x and y");
			System.exit(1);
		}
		
		String x = args[1];
		String y = args[2];
		
		int[] c = {2,1,1};	/* Costs: Substitute s,Delete d,Insert i */
		int i,j = 0;
		
		int n = x.length();
		int m = y.length();
		int[][] D = new int[n+1][m+1];;			/* Cost Matrix */
		char[][] pointer = new char[n+1][m+1];	
		
		/* Calculate cost matrix as described in the script*/
		D[0][0] = 0;
		for(i = 1; i<=n; i++){ D[i][0] = D[i-1][0] + c[1]; }
		for(j = 1; j<=m; j++){ D[0][j] = D[0][j-1] + c[2]; }
		
		for(i = 1; i<=n; i++){
			for(j = 1; j<=m; j++){
				
				int m1 = D[i-1][j-1];
				if(x.charAt(i-1) != y.charAt(j-1)){
					m1 = m1 +c[0]; 
				}
				
				int m2 = D[i-1][j] + c[1];
				int m3 = D[i][j-1] + c[2];
				
				D[i][j] = Math.min(Math.min(m1,m2),m3);
				
				/* the bookkeeping part */
				if (m1 == D[i][j]) { 	  pointer[i][j]='s';}
				else if (m2 == D[i][j]) { pointer[i][j]='d';}
				else pointer[i][j]='i';
			}
		}
		
		/* Option 1 selected - do the LCS backtrace */
		if(args[0].charAt(0)=='1'){ 

			/* Backtrace to find LCS */
			StringBuffer theLCS = new StringBuffer();
			
			i=n; j=m; /* start from last */
			while(j>0 && i>0){
				if(x.charAt(i-1) == y.charAt(j-1)) theLCS.insert(0, y.charAt(j-1));
				/* decide on how to go to the next */
				switch (pointer[i][j]) {
				  case 'd': i--;      break; /* delete operation -> go up */
				  case 's': i--; j--; break; /* substitution -> go up and left */
				  case 'i': j--;      break; /* insert operation -> go left */
				  default: break;
				} 
			}

			System.out.println(theLCS.toString());
			
		/* Option 0 selected or default - echo the editdistance of x and y */
		} else {
			System.out.println(D[n][m]);
		}
	}
}

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Expert Comment

by:TommySzalapski
Comment Utility
Kendor, note in the original question that the Asker stated that this was a programming assignment to implement an algorithm.
Your solution no doubt solves the problem but there are two main issues:
1. slritmmi didn't write it so it is academic dishonesty
2. slritmmi won't learn anything if he just uses your code so he'll do poorly on future assignments or fail the tests.

As experts we need to pay attention to see if questions are academic in nature (which slritmmi fortunately told us; a lot of askers don't know to tell us). If it is, we are to guide the asker to the find the solution as much on his own as possible.

Congrats on reaching Master, by the way, and welcome to EE.
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Expert Comment

by:Kendor
Comment Utility
TommySzalapski, you are completely right. By posting the code I didn't mean slritmmi to copy my code nor to directly hand it in as is - but rather asked him/her to "have a look" at a possible solution to the problem. And to be honest - I've been away the last 1.5 months and didn't get the chance to answer any questions (+ there are no question I could answer in "my area"... signal processing) - so I hope to get those "least" points to keep my expert status.
Thanks for reminding me anyways.
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Expert Comment

by:Kendor
Comment Utility
slritmmi is your question resolved? then it would be great from you to close it and assign some points :)
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Expert Comment

by:mlmcc
Comment Utility
I've requested that this question be deleted for the following reason:

This question has been classified as abandoned and is closed as part of the Cleanup Program. See the recommendation for more details.
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Expert Comment

by:TommySzalapski
Comment Utility
http:#a35063128 identified the main critical issues that were causing the errors.
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