Levenshtein Distance

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Revision as of 04:55, 14 August 2010 by 211.2.129.92 (Talk)

Compute the levenshtein distance between Strings - see http://en.wikipedia.org/wiki/Levenshtein_distance

Implementations

Java

<highlightsyntax language="java122"> public static int getLevenshteinDistance (String s, String t) {

 if (s == null || t == null) {
   throw new IllegalArgumentException("Strings must not be null");
 }
 /*
   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(); // length of s
 int m = t.length(); // length of t
 if (n == 0) {
   return m;
 } else if (m == 0) {
   return n;
 }
 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; // 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;
    for (i=1; i<=n; i++) {
       cost = s.charAt(i-1)==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);  
    }
    // 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];

} </highlightsyntax>

Ruby

def levenshtein_distance(s, t)
  m = s.size
  n = t.size
  d = Array.new(m+1) { Array.new(n+1) }
  for i in 0..m
    d[i][0] = i
  end
  for j in 0..n
    d[0][j] = j
  end
  for j in 0...n
    for i in 0...m
      if s[i,1] == t[j,1]
        d[i+1][j+1] = d[i][j]
      else
        d[i+1][j+1] = [d[i  ][j+1] + 1, # deletion
                       d[i+1][j  ] + 1, # insertion
                       d[i  ][j  ] + 1  # substitution
                      ].min
      end
    end
  end
  d[m][n]
end

s = "sitting"
t = "kitten"
p levenshtein_distance(s, t)    #=> 3