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Posted 22 Mar 2006

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# Fast, memory efficient Levenshtein algorithm

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26 Mar 2012CPOL
A version of the Levenshtein algorithm that uses 2*Min(StrLen1,StrLen2) bytes instead of StrLen1*StrLen2 bytes.

## Introduction

The Levenshtein distance is the difference between two strings. I use it in a web crawler application to compare the new and old versions of a web page. If it has changed enough, I update it in my database.

## Description

The original algorithm creates a matrix, where the size is `StrLen1*StrLen2`. If both strings are 1000 chars long, the resulting matrix is 1M elements; if the strings are 10,000 chars, the matrix will be 100M elements. If the elements are integers, it will be 4*100M == 400MB. Ouch!

This version of the algorithm uses only 2*`StrLen` elements, so the latter example would give 2*10,000*4 = 80 KB. The result is that, not only does it use less memory but it's also faster because the memory allocation takes less time. When both strings are about 1K in length, the new version is more than twice as fast.

## Example

The original version would create a matrix[6+1,5+1], my version creates two vectors[6+1] (the yellow elements). In both versions, the order of the strings is irrelevant, that is, it could be matrix[5+1,6+1] and two vectors[5+1].

## The new algorithm

### Steps

Step Description
1 Set n to be the length of s. ("GUMBO")
Set m to be the length of t. ("GAMBOL")
If n = 0, return m and exit.
If m = 0, return n and exit.
Construct two vectors, v0[m+1] and v1[m+1], containing 0..m elements.
2 Initialize v0 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] is not equal to t[j], the cost is 1.
6 Set cell v1[j] equal to the minimum of:
a. The cell immediately above plus 1: v1[j-1] + 1.
b. The cell immediately to the left plus 1: v0[j] + 1.
c. The cell diagonally above and to the left plus the cost: v0[j-1] + cost.
7 After the iteration steps (3, 4, 5, 6) are complete, the distance is found in the cell v1[m].

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

 v0 v1 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

 v0 v1 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

SWAP(v0,v1): If you look in the code you will see that I don't swap the content of the vectors but I refer to them.

Set v1[0] to the column number, e.g. 2.

 v0 v1 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

SWAP(v0,v1).

Set v1[0] to the column number, e.g. 3.

 v0 v1 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

SWAP(v0,v1).

Set v1[0] to the column number, e.g. 4.

 v0 v1 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

SWAP(v0,v1).

Set v1[0] to the column number, e.g. 5.

 v0 v1 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, v1[m] == 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 one insertion = two changes).

## Improvements

If you are sure that your strings will never be longer than 2^16 chars, you could use `ushort` instead of `int`, if the strings are less than 2^8 chars, you could use `byte`. I guess, the algorithm would be even faster if we use unmanaged code, but I have not tried it.

## History

• 2006-03-22
• Version 1.0.
• 2006-03-24
• Detailed description of the algorithm. The code has been rewritten so that it now follows the description. :-)

## Share

 Software Developer (Senior) SEB bank Sweden
I work as a developer in the 'Risk control' department at SEB bank in Stockholm,Sweden and I have been designing software since the early 80's.

 First PrevNext
 My vote of 5 Alper Sünter27-Nov-20 17:53 Alper Sünter 27-Nov-20 17:53
 My changes to this algorithm jjcasalo20-Oct-17 3:46 jjcasalo 20-Oct-17 3:46
 OK, so I did some work since yesterday, and here is what I came up with. The changes I did : 1) Before I start filling up the S vectors, I remove all the identical first letters. For example, for the 2 words CHEMISE and CHEIMSE, the "CHE" part is the same, so I disregard this. And the 2 new words that will be processed (and inserted in the 2 vectors) are "MISE" and "IMSE". Should save a few milliseconds, but when you have millions of rows to compare, they add up... 2) As I wrote previously, I wanted to stop if the current Levenshtein distance was greater than a number. For example, I don't want all the words with a Levenshtein distance greater than 3. The way I do is is that I look at the v1 column (once all the slots have numbers), and take the MIN. This is the current Levenshtein distance. If this current distance is greater than my limit, I exit the function with a value of 99, as I don't need to compute the other columns (as the overall Levenshtein distance can only increase, and never decrease (as you only add up 1s) 3) Optional. If you have to work with accentuated characters, and you want that the Levenshtein distance don't count the difference between, say, "Helene" and "Hélène", then you have to use a function that replaces all the accentuated characters with their "normal" letters. ```Function Levenshtein(ByVal String1 As String, ByVal String2 As String, ByVal LMax As Integer) As Integer Dim cost As Integer ' cost Dim CurrentCost As Integer ' Current Levenshtein distance, for each Column. Dim MinLen As Integer = Math.Min(String1.Length, String2.Length) If (String1.Length > 1000 Or String2.Length > 1000) Then MsgBox("La taille limite des mots est de 1000 lettres", MsgBoxStyle.Exclamation, "Texte trop long") Return 99 End If ' If difference of length between 2 strings is greater than LMax, then return 99 If (Math.Abs(String1.Length - String2.Length) > LMax) Then Return 99 End If String1 = String1.ToUpper String2 = String2.ToUpper ' OPTIONAL for accentuated words String1 = No_accents(String1) String2 = No_accents(String2) ' Step 0 : going fast through the first N identical letters. For i = 0 To MinLen - 1 If (String1(i) = String2(i)) Then If i = MinLen - 1 Then String1 = String1.Substring(i + 1) String2 = String2.Substring(i + 1) Exit For Else Continue For End If End If String1 = String1.Substring(i) String2 = String2.Substring(i) Exit For Next Dim String1Len As Integer = String1.Length ' length of String1 Dim String2Len As Integer = String2.Length ' length of String2 ' Step 1 If (String1Len = 0 Or String2Len = 0) Then Return Math.Max(String1.Length, String2.Length) ' Create the two vectors Dim v0 As Integer() = New Integer(String1Len + 1) {} Dim v1 As Integer() = New Integer(String1Len + 1) {} 'Dim vTmp As Integer() ' Step 2 ' Initialize the first vector For i = 1 To String1Len v0(i) = i Next ' Step 3 ' For each column For i = 1 To String2Len ' Set the 0'th element to the column number v1(0) = i ' Reset CurrentCost to 0 ' Step 4 For j = 1 To String1Len ' Step 5 If (String1(j - 1) = String2(i - 1)) Then cost = 0 Else cost = 1 End If ' Step 6 ' Find minimum v1(j) = Math.Min(v0(j) + 1, Math.Min(v1(j - 1) + 1, v0(j - 1) + cost)) ' We get the current Levenshtein distance, which is the minimum value of the column If j = 1 Then CurrentCost = v1(1) Else CurrentCost = Math.Min(v1(j), CurrentCost) End If Next ' The current Levenshtein distance is greater than the set limit. ' No need to continue, as Levenshtein distance can only grow up If CurrentCost > LMax Then Return 99 ' Swap the vectors Dim vTmp = v0 v0 = v1 v1 = vTmp Next ' Step 7 ' The vectors where swaped one last time at the end of the last loop, ' that is why the result is now in v0 rather than in v1 Return (v0(String1Len)) End Function Function No_accents(ByVal Chaine As String) As String Dim a As String = "ÀÁÂÃÄÅÈÉÊËÌÍÎÏÑÒÓÔÕÖÙÚÛÜÝàáâãäåèéêëìíîïðñòóôõöùúûüýÿ" Dim b As String = "AAAAAAEEEEIIIINOOOOOUUUUYaaaaaaeeeeiiiionooooouuuuyy" For i As Integer = 0 To a.Count - 1 Chaine = Chaine.Replace(a.Substring(i, 1), b.Substring(i, 1)) Next i Return Chaine.ToUpper End Function``` Hope this will be usefull.modified 20-Oct-17 14:37pm.
 What do I do wrong ? (VB.net code) jjcasalo19-Oct-17 11:05 jjcasalo 19-Oct-17 11:05
 Helpful SusannePorte21-Nov-16 22:24 SusannePorte 21-Nov-16 22:24
 Re: Helpful Sten Hjelmqvist22-Nov-16 6:37 Sten Hjelmqvist 22-Nov-16 6:37
 Cleaned up the code John Henckel12-Sep-16 9:45 John Henckel 12-Sep-16 9:45
 Re: Cleaned up the code jjcasalo19-Oct-17 11:06 jjcasalo 19-Oct-17 11:06
 Awesome work! Member 1079435925-Sep-15 14:25 Member 10794359 25-Sep-15 14:25
 Re: Awesome work! Sten Hjelmqvist28-Sep-15 21:49 Sten Hjelmqvist 28-Sep-15 21:49
 Re: Awesome work! Member 107943598-Oct-15 14:33 Member 10794359 8-Oct-15 14:33
 Thank you very much! How can I get the detail change list? Member 1191077915-Aug-15 17:51 Member 11910779 15-Aug-15 17:51
 Results are not accurate Tom Medhurst18-Jun-15 1:35 Tom Medhurst 18-Jun-15 1:35
 Re: Results are not accurate Sten Hjelmqvist5-Jul-15 21:53 Sten Hjelmqvist 5-Jul-15 21:53
 Re: Results are not accurate jfos28-Sep-15 6:39 jfos 28-Sep-15 6:39
 What about the list of changes? UzairSyed10-Jun-15 4:41 UzairSyed 10-Jun-15 4:41
 What about the transpose operation? Member 1164580029-Apr-15 20:45 Member 11645800 29-Apr-15 20:45
 Re: What about the transpose operation? Sten Hjelmqvist1-May-15 2:14 Sten Hjelmqvist 1-May-15 2:14
 Re: What about the transpose operation? james.manning14-May-15 11:44 james.manning 14-May-15 11:44
 Bug xidongzhang25-Nov-14 7:57 xidongzhang 25-Nov-14 7:57
 Re: Bug Sten Hjelmqvist8-Dec-14 23:38 Sten Hjelmqvist 8-Dec-14 23:38
 My vote of 5 Amir Mehrabi-Jorshari13-Nov-13 11:06 Amir Mehrabi-Jorshari 13-Nov-13 11:06
 Perfect for my needs Greg Lindberg5-Nov-13 4:32 Greg Lindberg 5-Nov-13 4:32
 Re: Perfect for my needs Sten Hjelmqvist5-Nov-13 5:01 Sten Hjelmqvist 5-Nov-13 5:01
 My vote of 5 meys_online26-Aug-12 14:38 meys_online 26-Aug-12 14:38
 Thanks - very interesting. SimonDrift20-Jan-12 16:14 SimonDrift 20-Jan-12 16:14
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