In the previous article we looked in to what ‘**Levenshtein distance**‘ is and an example use case. We saw that when the word we are transforming has small number of letters it is easy to find the minimum edit distance, but as the number of letters in the word increases it is difficult to be sure that we have found the ‘minimum’. So we will be discussing a method that will make sure that we end up with the value for minimum edit distance.

We will take the following example.

five | four |

From one glance we can see that we need minimum 3 operations to transform five to four (3 replace operations). Lets see how this new method would give that answer. First we have to make a grid like follows. Columns will have the word before transformation and the rows will have the word after transformation. The ‘**ε**‘ (epsilon) denotes an empty string.

ε |
f | i | v | e | |

ε |
|||||

f | |||||

o | |||||

u | |||||

r |

We start from the top and we fill the cells. We read the grid from top to left. For example, to fill the first cell we have to think ‘how many operations does it take to convert an empty string (**ε**) to an empty string (**ε**) ?’ The answer is zero. So we put 0 there.

ε |
f | i | v | e | |

ε |
0 | ||||

f | |||||

o | |||||

u | |||||

r |

Next, to fill the second cell we have to think ‘how many operations does it take to convert a ‘f’ to an empty string (**ε**) ?’ The answer is 1 (delete ‘f’). So we put 1 there.

ε |
f | i | v | e | |

ε |
0 | 1 | |||

f | |||||

o | |||||

u | |||||

r |

Next to fill the third cell we have to think ‘how many operations does it take to convert ‘fi’ to an empty string (**ε**) ?’ The answer is 2 (delete ‘f’, delete ‘i’). So we put 2 there.

ε |
f | i | v | e | |

ε |
0 | 1 | 2 | ||

f | |||||

o | |||||

u | |||||

r |

We can fill the first row in a similar manner and then we get the following.

ε |
f | i | v | e | |

ε |
0 | 1 | 2 | 3 | 4 |

f | |||||

o | |||||

u | |||||

r |

Next to fill the first cell in the second row we have to think ‘how many operations does it take to convert an empty string ‘**ε**‘ to ‘f’ ?’ The answer is 1 (add ‘f’). So we put 1 there.

ε |
f | i | v | e | |

ε |
0 | 1 | 2 | 3 | 4 |

f | 1 | ||||

o | |||||

u | |||||

r |

Next to fill the second cell in the second row we have to think ‘how many operations does it take to convert ‘f’ to ‘f’ ?’ The answer is 0. So we put 0 there.

ε |
f | i | v | e | |

ε |
0 | 1 | 2 | 3 | 4 |

f | 1 | 0 | |||

o | |||||

u | |||||

r |

similarly we can keep on going filling.

ε |
f | i | v | e | |

ε |
0 | 1 | 2 | 3 | 4 |

f | 1 | 0 | 1 | 2 | 3 |

o | 2 | 1 | 1 | x | |

u | |||||

r |

One more example, so how do we fill the above ‘x’ ? We have to think ‘how many operations does it take to convert ‘fiv’ to ‘fo’ ?’ The answer is 2 (replace ‘i’ with ‘o’. delete ‘v’). So we put 2 there.

ε |
f | i | v | e | |

ε |
0 | 1 | 2 | 3 | 4 |

f | 1 | 0 | 1 | 2 | 3 |

o | 2 | 1 | 1 | 2 | |

u | |||||

r |

Finally we end up with something like this.

ε |
f | i | v | e | |

ε |
0 | 1 | 2 | 3 | 4 |

f | 1 | 0 | 1 | 2 | 3 |

o | 2 | 1 | 1 | 2 | 3 |

u | 3 | 2 | 2 | 2 | 3 |

r | 4 | 3 | 3 | 3 | 3 |

The ‘Minimum Edit Distance’ (**Levenshtein distance**) is the value of the last cell.

So we get the same answer 3. For the example I took a small word, but when dealing with longer words this grid method is much useful in making sure we got the minimum number of operations.

Okay so we now know the grid method, but it takes time right ? What if there is a cheat method? a short cut to fill this table ? There is. We will look in to it in the next part, and we will finish up with the python code for this calculation in the article after that.

Exploring NLP 05 – Minimum Edit Distance 3/4

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