Levenshtein Distance, in Three Flavors

by Michael Gilleland, Merriam Park Software

The purpose of this short essay is to describe the Levenshtein distance algorithm and show how it can be implemented in three different programming languages.

What is Levenshtein Distance?
Demonstration
The Algorithm
Source Code, in Three Flavors
References
Other Flavors


What is Levenshtein Distance?

Levenshtein distance (LD) is a measure of the similarity between two strings, which we will refer to as the source string (s) and the target string (t). The distance is the number of deletions, insertions, or substitutions required to transform s into t. For example,

The greater the Levenshtein distance, the more different the strings are.

Levenshtein distance is named after the Russian scientist Vladimir Levenshtein, who devised the algorithm in 1965. If you can't spell or pronounce Levenshtein, the metric is also sometimes called edit distance.

The Levenshtein distance algorithm has been used in:


Demonstration

The following simple Java applet allows you to experiment with different strings and compute their Levenshtein distance:


The Algorithm

Steps

Step Description
1 Set n to be the length of s.
Set m to be the length of t.
If n = 0, return m and exit.
If m = 0, return n and exit.
Construct a matrix containing 0..m rows and 0..n columns.
2 Initialize the first row to 0..n.
Initialize the first column to 0..m.
3 Examine each character of s (i from 1 to n).
4 Examine each character of t (j from 1 to m).
5 If s[i] equals t[j], the cost is 0.
If s[i] doesn't equal t[j], the cost is 1.
6 Set cell d[i,j] of the matrix equal to the minimum of:
a. The cell immediately above plus 1: d[i-1,j] + 1.
b. The cell immediately to the left plus 1: d[i,j-1] + 1.
c. The cell diagonally above and to the left plus the cost: d[i-1,j-1] + cost.
7 After the iteration steps (3, 4, 5, 6) are complete, the distance is found in cell d[n,m].

Example

This section shows how the Levenshtein distance is computed when the source string is "GUMBO" and the target string is "GAMBOL".

Steps 1 and 2

    G U M B O
  0 1 2 3 4 5
G 1          
A 2          
M 3          
B 4          
O 5          
L 6          

Steps 3 to 6 When i = 1

    G U M B O
  0 1 2 3 4 5
G 1 0        
A 2 1        
M 3 2        
B 4 3        
O 5 4        
L 6 5        

Steps 3 to 6 When i = 2

    G U M B O
  0 1 2 3 4 5
G 1 0 1      
A 2 1 1      
M 3 2 2      
B 4 3 3      
O 5 4 4      
L 6 5 5      

Steps 3 to 6 When i = 3

    G U M B O
  0 1 2 3 4 5
G 1 0 1 2    
A 2 1 1 2    
M 3 2 2 1    
B 4 3 3 2    
O 5 4 4 3    
L 6 5 5 4    

Steps 3 to 6 When i = 4

    G U M B O
  0 1 2 3 4 5
G 1 0 1 2 3  
A 2 1 1 2 3  
M 3 2 2 1 2  
B 4 3 3 2 1  
O 5 4 4 3 2  
L 6 5 5 4 3  

Steps 3 to 6 When i = 5

    G U M B O
  0 1 2 3 4 5
G 1 0 1 2 3 4
A 2 1 1 2 3 4
M 3 2 2 1 2 3
B 4 3 3 2 1 2
O 5 4 4 3 2 1
L 6 5 5 4 3 2

Step 7

The distance is in the lower right hand corner of the matrix, i.e. 2. This corresponds to our intuitive realization that "GUMBO" can be transformed into "GAMBOL" by substituting "A" for "U" and adding "L" (one substitution and 1 insertion = 2 changes).


Source Code, in Three Flavors

Religious wars often flare up whenever engineers discuss differences between programming languages. A typical assertion is Allen Holub's claim in a JavaWorld article (July 1999): "Visual Basic, for example, isn't in the least bit object-oriented. Neither is Microsoft Foundation Classes (MFC) or most of the other Microsoft technology that claims to be object-oriented."

A salvo from a different direction is Simson Garfinkels's article in Salon (Jan. 8, 2001) entitled "Java: Slow, ugly and irrelevant", which opens with the unequivocal words "I hate Java".

We prefer to take a neutral stance in these religious wars. As a practical matter, if a problem can be solved in one programming language, you can usually solve it in another as well. A good programmer is able to move from one language to another with relative ease, and learning a completely new language should not present any major difficulties, either. A programming language is a means to an end, not an end in itself.

As a modest illustration of this principle of neutrality, we present source code which implements the Levenshtein distance algorithm in the following programming languages:


Java

public class Distance {

  //****************************
  // Get minimum of three values
  //****************************

  private int Minimum (int a, int b, int c) {
  int mi;

    mi = a;
    if (b < mi) {
      mi = b;
    }
    if (c < mi) {
      mi = c;
    }
    return mi;

  }

  //*****************************
  // Compute Levenshtein distance
  //*****************************

  public int LD (String s, String t) {
  int d[][]; // matrix
  int n; // length of s
  int m; // length of t
  int i; // iterates through s
  int j; // iterates through t
  char s_i; // ith character of s
  char t_j; // jth character of t
  int cost; // cost

    // Step 1

    n = s.length ();
    m = t.length ();
    if (n == 0) {
      return m;
    }
    if (m == 0) {
      return n;
    }
    d = new int[n+1][m+1];

    // Step 2

    for (i = 0; i <= n; i++) {
      d[i][0] = i;
    }

    for (j = 0; j <= m; j++) {
      d[0][j] = j;
    }

    // Step 3

    for (i = 1; i <= n; i++) {

      s_i = s.charAt (i - 1);

      // Step 4

      for (j = 1; j <= m; j++) {

        t_j = t.charAt (j - 1);

        // Step 5

        if (s_i == t_j) {
          cost = 0;
        }
        else {
          cost = 1;
        }

        // Step 6

        d[i][j] = Minimum (d[i-1][j]+1, d[i][j-1]+1, d[i-1][j-1] + cost);

      }

    }

    // Step 7

    return d[n][m];

  }

}

C++

In C++, the size of an array must be a constant, and this code fragment causes an error at compile time:

int sz = 5;
int arr[sz];

This limitation makes the following C++ code slightly more complicated than it would be if the matrix could simply be declared as a two-dimensional array, with a size determined at run-time.

Here is the definition of the class (distance.h):

class Distance
{
  public:
    int LD (char const *s, char const *t);
  private:
    int Minimum (int a, int b, int c);
    int *GetCellPointer (int *pOrigin, int col, int row, int nCols);
    int GetAt (int *pOrigin, int col, int row, int nCols);
    void PutAt (int *pOrigin, int col, int row, int nCols, int x);
}; 

Here is the implementation of the class (distance.cpp):

#include "distance.h"
#include <string.h>
#include <malloc.h>

//****************************
// Get minimum of three values
//****************************

int Distance::Minimum (int a, int b, int c)
{
int mi;

  mi = a;
  if (b < mi) {
    mi = b;
  }
  if (c < mi) {
    mi = c;
  }
  return mi;

}

//**************************************************
// Get a pointer to the specified cell of the matrix
//************************************************** 

int *Distance::GetCellPointer (int *pOrigin, int col, int row, int nCols)
{
  return pOrigin + col + (row * (nCols + 1));
}

//*****************************************************
// Get the contents of the specified cell in the matrix 
//*****************************************************

int Distance::GetAt (int *pOrigin, int col, int row, int nCols)
{
int *pCell;

  pCell = GetCellPointer (pOrigin, col, row, nCols);
  return *pCell;

}

//*******************************************************
// Fill the specified cell in the matrix with the value x
//*******************************************************

void Distance::PutAt (int *pOrigin, int col, int row, int nCols, int x)
{
int *pCell;

  pCell = GetCellPointer (pOrigin, col, row, nCols);
  *pCell = x;

}

//*****************************
// Compute Levenshtein distance
//*****************************

int Distance::LD (char const *s, char const *t)
{
int *d; // pointer to matrix
int n; // length of s
int m; // length of t
int i; // iterates through s
int j; // iterates through t
char s_i; // ith character of s
char t_j; // jth character of t
int cost; // cost
int result; // result
int cell; // contents of target cell
int above; // contents of cell immediately above
int left; // contents of cell immediately to left
int diag; // contents of cell immediately above and to left
int sz; // number of cells in matrix

  // Step 1	

  n = strlen (s);
  m = strlen (t);
  if (n == 0) {
    return m;
  }
  if (m == 0) {
    return n;
  }
  sz = (n+1) * (m+1) * sizeof (int);
  d = (int *) malloc (sz);

  // Step 2

  for (i = 0; i <= n; i++) {
    PutAt (d, i, 0, n, i);
  }

  for (j = 0; j <= m; j++) {
    PutAt (d, 0, j, n, j);
  }

  // Step 3

  for (i = 1; i <= n; i++) {

    s_i = s[i-1];

    // Step 4

    for (j = 1; j <= m; j++) {

      t_j = t[j-1];

      // Step 5

      if (s_i == t_j) {
        cost = 0;
      }
      else {
        cost = 1;
      }

      // Step 6 

      above = GetAt (d,i-1,j, n);
      left = GetAt (d,i, j-1, n);
      diag = GetAt (d, i-1,j-1, n);
      cell = Minimum (above + 1, left + 1, diag + cost);
      PutAt (d, i, j, n, cell);
    }
  }

  // Step 7

  result = GetAt (d, n, m, n);
  free (d);
  return result;
	
}

Visual Basic

'*******************************
'*** Get minimum of three values
'*******************************

Private Function Minimum(ByVal a As Integer, _
                         ByVal b As Integer, _
                         ByVal c As Integer) As Integer
Dim mi As Integer
                          
  mi = a
  If b < mi Then
    mi = b
  End If
  If c < mi Then
    mi = c
  End If
  
  Minimum = mi
                          
End Function

'********************************
'*** Compute Levenshtein Distance
'********************************

Public Function LD(ByVal s As String, ByVal t As String) As Integer
Dim d() As Integer ' matrix
Dim m As Integer ' length of t
Dim n As Integer ' length of s
Dim i As Integer ' iterates through s
Dim j As Integer ' iterates through t
Dim s_i As String ' ith character of s
Dim t_j As String ' jth character of t
Dim cost As Integer ' cost
  
  ' Step 1
  
  n = Len(s)
  m = Len(t)
  If n = 0 Then
    LD = m
    Exit Function
  End If 
  If m = 0 Then
    LD = n
    Exit Function
  End If 
  ReDim d(0 To n, 0 To m) As Integer
  
  ' Step 2
  
  For i = 0 To n
    d(i, 0) = i
  Next i
  
  For j = 0 To m
    d(0, j) = j
  Next j

  ' Step 3

  For i = 1 To n
    
    s_i = Mid$(s, i, 1)
    
    ' Step 4
    
    For j = 1 To m
      
      t_j = Mid$(t, j, 1)
      
      ' Step 5
      
      If s_i = t_j Then
        cost = 0
      Else
        cost = 1
      End If
      
      ' Step 6
      
      d(i, j) = Minimum(d(i - 1, j) + 1, d(i, j - 1) + 1, d(i - 1, j - 1) + cost)
    
    Next j
    
  Next i
  
  ' Step 7
  
  LD = d(n, m)
  Erase d

End Function

References

Other discussions of Levenshtein distance may be found at:

Other Flavors

The following people have kindly consented to make their implementations of the Levenshtein Distance Algorithm in various languages available here:

Other implementations outside these pages include: