Edit distance (Levenshtein) in NLP - ML Experiment: Train & Evaluate

Experiment - Edit distance (Levenshtein)

Problem:You want to measure how different two words are by counting the minimum number of changes needed to turn one word into the other. This is called the Levenshtein edit distance.

Current Metrics:The current implementation calculates edit distance correctly but is slow for long words, taking over 5 seconds for words longer than 100 characters.

Issue:The current code uses a simple recursive method without optimization, causing slow performance and making it impractical for larger inputs.

Your Task

Improve the edit distance calculation to run efficiently on longer words (100+ characters) while keeping the results accurate.

You must keep the output exactly the same (correct edit distance).

You cannot use external libraries like 'python-Levenshtein'.

You should optimize the existing code or rewrite it using dynamic programming.

Hint 1

Hint 2

Hint 3

Solution

NLP

def levenshtein_distance(s1: str, s2: str) -> int:
    m, n = len(s1), len(s2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]

    for i in range(m + 1):
        dp[i][0] = i
    for j in range(n + 1):
        dp[0][j] = j

    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if s1[i - 1] == s2[j - 1]:
                dp[i][j] = dp[i - 1][j - 1]
            else:
                dp[i][j] = 1 + min(dp[i - 1][j],    # deletion
                                   dp[i][j - 1],    # insertion
                                   dp[i - 1][j - 1]) # substitution
    return dp[m][n]

# Example usage:
word1 = "kitten"
word2 = "sitting"
print(f"Edit distance between '{word1}' and '{word2}':", levenshtein_distance(word1, word2))

Replaced the slow recursive method with a dynamic programming approach using a 2D table.

Initialized the table with base cases for empty strings.

Filled the table iteratively to compute minimum edits step-by-step.

This change drastically improved performance for long strings while keeping accuracy.

Results Interpretation

Before: Recursive method took over 5 seconds for long words and was impractical.

After: Dynamic programming method runs in under 0.1 seconds for the same inputs with exact results.

Using dynamic programming to store intermediate results avoids repeated work, making algorithms like edit distance fast and practical for real-world use.

Bonus Experiment

Try to reduce the memory usage of the dynamic programming solution by using only two rows instead of the full 2D table.

💡 Hint

Notice that to compute the current row, you only need the previous row, so you can overwrite rows to save space.

Practice

1. What does the edit distance (Levenshtein distance) between two words measure?

easy

A. The length difference between two words

B. The minimum number of single-character edits to change one word into the other

C. The number of common letters between two words

D. The number of vowels in both words combined

Solution

1. Step 1: Understand the definition of edit distance

   Edit distance counts how many changes like insertions, deletions, or substitutions are needed to convert one word into another.

2. Step 2: Compare options with the definition

   Only The minimum number of single-character edits to change one word into the other correctly describes this minimum number of single-character edits.

3. Final Answer:

   The minimum number of single-character edits to change one word into the other -> Option B

4. Quick Check:

   Edit distance = minimum edits [OK]

Hint: Edit distance = smallest edits to match words [OK]

Common Mistakes:

• Confusing edit distance with common letters count
• Thinking it measures length difference only
• Mixing up vowels or letter counts

2. Which of the following Python code snippets correctly initializes a 2D table for computing edit distance between strings s1 and s2 of lengths m and n?

easy

A. table = [[0] * (n + 1) for _ in range(m + 1)]

B. table = [[0] * m for _ in range(n)]

C. table = [[0] * (m + 1) for _ in range(n + 1)]

D. table = [[0] * n for _ in range(m)]

Solution

1. Step 1: Recall table size for edit distance

   The table size must be (m+1) rows and (n+1) columns to include empty prefixes of both strings.

2. Step 2: Match code to correct dimensions

   table = [[0] * (n + 1) for _ in range(m + 1)] creates a list with (m+1) rows and each row has (n+1) zeros, which is correct.

3. Final Answer:

   table = [[0] * (n + 1) for _ in range(m + 1)] -> Option A

4. Quick Check:

   Table size = (m+1) x (n+1) [OK]

Hint: Table size = (len(s1)+1) x (len(s2)+1) [OK]

Common Mistakes:

• Swapping m and n in dimensions
• Forgetting to add 1 for empty string prefix
• Using wrong list comprehension order

3. What is the edit distance between the words "kitten" and "sitting"?

medium

A. 2

B. 4

C. 3

D. 5

Solution

1. Step 1: Identify edits from "kitten" to "sitting"

   Changes: substitute 'k' -> 's', substitute 'e' -> 'i', insert 'g' at the end.

2. Step 2: Count total edits

   There are 3 edits: 2 substitutions and 1 insertion.

3. Final Answer:

   3 -> Option C

4. Quick Check:

   Edits counted = 3 [OK]

Hint: Count substitutions + insertions + deletions [OK]

Common Mistakes:

• Counting only substitutions
• Missing the final insertion
• Confusing similar letters as no change

4. Consider this Python code snippet for edit distance calculation. What is the error?

def edit_distance(s1, s2):
    m, n = len(s1), len(s2)
    table = [[0] * (n + 1) for _ in range(m + 1)]
    for i in range(m + 1):
        table[i][0] = i
    for j in range(n + 1):
        table[0][j] = j
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if s1[i] == s2[j]:
                cost = 0
            else:
                cost = 1
            table[i][j] = min(table[i-1][j] + 1, table[i][j-1] + 1, table[i-1][j-1] + cost)
    return table[m][n]

medium

A. The return statement should be table[0][0]

B. The table initialization is incorrect

C. The cost should be 2 when characters differ

D. Indexing s1[i] and s2[j] causes an off-by-one error

Solution

1. Step 1: Check string indexing in loops

   Strings are 0-indexed, but loops start at 1, so s1[i] and s2[j] skip first character and cause index errors.

2. Step 2: Correct indexing

   Use s1[i-1] and s2[j-1] to access correct characters for comparison.

3. Final Answer:

   Indexing s1[i] and s2[j] causes an off-by-one error -> Option D

4. Quick Check:

   Indexing error = s1[i-1], s2[j-1] needed [OK]

Hint: Remember strings start at index 0, loops at 1 need offset [OK]

Common Mistakes:

• Using s1[i] instead of s1[i-1]
• Thinking cost must be 2 for difference
• Returning wrong table cell

5. You want to find the closest word to "flame" from the list ["frame", "flan", "flame", "blame"] using edit distance. Which word will the algorithm select as closest?

hard

A. "flame"

B. "frame"

C. "blame"

D. "flan"

Solution

1. Step 1: Calculate edit distances to each word

   Distances: "flame" to "frame" = 1 (substitute 'l'->'r'), to "flan" = 2 (delete 'm' and 'e'), to "blame" = 1 (substitute 'f'->'b'), to "flame" = 0 (same word).

2. Step 2: Identify minimum distance

   The smallest distance is 0 for "flame" itself, meaning exact match.

3. Final Answer:

   "flame" -> Option A

4. Quick Check:

   Exact match distance = 0 [OK]

Hint: Exact match has zero edit distance [OK]

Common Mistakes:

• Choosing word with one letter difference over exact match
• Ignoring exact matches
• Miscounting substitutions