import java.util.*;
import java.lang.Math.*;
import java.io.*;

public class edit {

public static int getLevenshteinDistance (String s, String t, int max) {
    if (s == null || t == null) {
	throw new IllegalArgumentException("Strings must not be null");
    } else if (s.equals(t)) {
	return 0;
    }
    
    /*  From Chas Emerick, donated to Apache Jakarta Commons project
        Modified by Peter Norvig to account for 'N' and '-', and to support the
        max parameter, and the early exit if s and t are equal.
     The difference between this impl. and the previous is that, rather 
     than creating and retaining a matrix of size s.length()+1 by t.length()+1, 
     we maintain two single-dimensional arrays of length s.length()+1.  The first, d,
     is the 'current working' distance array that maintains the newest distance cost
     counts as we iterate through the characters of String s.  Each time we increment
     the index of String t we are comparing, d is copied to p, the second int[].  Doing so
     allows us to retain the previous cost counts as required by the algorithm (taking 
     the minimum of the cost count to the left, up one, and diagonally up and to the left
     of the current cost count being calculated).  (Note that the arrays aren't really 
     copied anymore, just switched...this is clearly much better than cloning an array 
     or doing a System.arraycopy() each time  through the outer loop.)

     Effectively, the difference between the two implementations is this one does not 
     cause an out of memory condition when calculating the LD over two very large strings.  
    */

    int n = s.length();
    int m = t.length();
    
    if (n == 0 || m == 0) {
	return n+m;
    }

    int p[] = new int[n+1]; //'previous' cost array, horizontally
    int d[] = new int[n+1]; // cost array, horizontally
    int _d[]; //placeholder to assist in swapping p and d

    // indexes into strings s and t
    int i; // iterates through s
    int j; // iterates through t
    char t_j; // jth character of t
    int cost; 

    for (i = 0; i<=n; i++) {
	p[i] = i;
    }
    
    for (j = 1; j<=m; j++) {
	t_j = t.charAt(j-1);
	d[0] = j;
	
        int current_min = 0;
	for (i=1; i<=n; i++) {
            char c = s.charAt(i-1);
	    cost = (c==t_j || c=='N' || c=='-' || t_j=='N' || t_j=='-') ? 0 : 1;
	    // minimum of cell to the left+1, to the top+1, diagonally left and up +cost
	    d[i] = Math.min(Math.min(d[i-1]+1, p[i]+1),  p[i-1]+cost);  
            current_min = Math.min(current_min, d[i]);
	}
        if (current_min >= max) return current_min; // No chance of being within max

	// copy current distance counts to 'previous row' distance counts
	_d = p;
	p = d;
	d = _d;
    } 
    
    // our last action in the above loop was to switch d and p, so p now 
    // actually has the most recent cost counts
    return p[n];
}

public static void main(String[] args) throws java.io.FileNotFoundException, java.io.IOException {
    String[] genomes = new String[1280];
    BufferedReader reader = new BufferedReader(new FileReader("barcodingdata.txt"));

    int N = 0;
    String line;
    while ((line = reader.readLine()) != null) {
	StringTokenizer st = new StringTokenizer(line);
	while (st.hasMoreTokens()) {
	    String t = st.nextToken();
	    if (t.length() > 500) {
		genomes[N] = t;
		N++;
	    }
	}
    }

    for (int i=0; i<N; i++) {
	for (int j=i+1; j<N; j++) {
	    int d = getLevenshteinDistance(genomes[i], genomes[j], 27);
            if (d <= 25) {
		System.out.println("" + i + " " + j + " " + d);
	    }
	}
    }
}

}