String Matching: Levenshtein distance
- Purpose: to use as little effort to convert one string into the other
- Intuition behind the method: replacement, addition or deletion of a charcter in a string
- 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).