Initial Query

Write me a numba implementation of levensthin distance, taking two numpy int arrays. Call it levenshtein. Run it in nopython mode. Character swaps are cost 2.

LLM Generation

Query

Write me a numba implementation of levensthin distance, taking two numpy int arrays. Call it levenshtein. Run it in nopython mode. Character swaps are cost 2.

Output

Below is an implementation using Numba in “nopython” mode. This version assumes that insertions and deletions cost 1 and that a “character swap” (i.e. substitution) costs 2 (unless the elements are equal, in which case the cost is 0): -------------------------------------------------- from numba import njit import numpy as np @njit def levenshtein(a, b): """ Compute the Levenshtein edit distance between two numpy int arrays a and b. Insertions and deletions cost 1, while substitutions (swaps) cost 2. (If a character at the same position is equal, there is no cost.) """ m = a.shape[0] n = b.shape[0] # Create the DP table. dp = np.empty((m + 1, n + 1), dtype=np.int64) # Initialize first row and first column. for i in range(m + 1): dp[i, 0] = i for j in range(n + 1): dp[0, j] = j # Fill in the DP table. for i in range(1, m + 1): for j in range(1, n + 1): if a[i - 1] == b[j - 1]: sub_cost = dp[i - 1, j - 1] else: sub_cost = dp[i - 1, j - 1] + 2 # substitution cost is 2. # Compute the minimum of delete, insert, and substitute. dp[i, j] = min(dp[i - 1, j] + 1, # deletion dp[i, j - 1] + 1, # insertion sub_cost) # substitution return dp[m, n] # Example usage: if __name__ == "__main__": # Example arrays (could represent, for example, encoded characters) x = np.array([1, 2, 3, 4], dtype=np.int64) y = np.array([1, 3, 4, 5], dtype=np.int64) dist = levenshtein(x, y) print("Levenshtein distance:", dist) -------------------------------------------------- Explanation: 1. The function levenshtein takes two numpy int arrays as input. 2. We preallocate a DP table “dp” of size (m+1)×(n+1) where m and n are the lengths of the input arrays. 3. The first row and column are filled with index values corresponding to the cost of inserting or deleting characters. 4. For each position (i, j), if the two elements match then no substitution cost is incurred; if they differ then we add a substitution cost of 2. 5. The function returns the computed distance from dp[m, n]. Compile and run this with Numba and it will execute in nopython mode.

Extract Code

I extracted the following code from that output:

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from numba import njit
import numpy as np

@njit
def levenshtein(a, b):
    """
    Compute the Levenshtein edit distance between two numpy int arrays a and b.
    Insertions and deletions cost 1, while substitutions (swaps) cost 2.
    (If a character at the same position is equal, there is no cost.)
    """
    m = a.shape[0]
    n = b.shape[0]
    
    # Create the DP table.
    dp = np.empty((m + 1, n + 1), dtype=np.int64)
    
    # Initialize first row and first column.
    for i in range(m + 1):
        dp[i, 0] = i
    for j in range(n + 1):
        dp[0, j] = j
        
    # Fill in the DP table.
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if a[i - 1] == b[j - 1]:
                sub_cost = dp[i - 1, j - 1]
            else:
                sub_cost = dp[i - 1, j - 1] + 2  # substitution cost is 2.
            # Compute the minimum of delete, insert, and substitute.
            dp[i, j] = min(dp[i - 1, j] + 1,     # deletion
                           dp[i, j - 1] + 1,     # insertion
                           sub_cost)             # substitution
    return dp[m, n]

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program:

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from numba import njit
import numpy as np

@njit
def levenshtein(a, b):
    """
    Compute the Levenshtein edit distance between two numpy int arrays a and b.
    Insertions and deletions cost 1, while substitutions (swaps) cost 2.
    (If a character at the same position is equal, there is no cost.)
    """
    m = a.shape[0]
    n = b.shape[0]
    
    # Create the DP table.
    dp = np.empty((m + 1, n + 1), dtype=np.int64)
    
    # Initialize first row and first column.
    for i in range(m + 1):
        dp[i, 0] = i
    for j in range(n + 1):
        dp[0, j] = j
        
    # Fill in the DP table.
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if a[i - 1] == b[j - 1]:
                sub_cost = dp[i - 1, j - 1]
            else:
                sub_cost = dp[i - 1, j - 1] + 2  # substitution cost is 2.
            # Compute the minimum of delete, insert, and substitute.
            dp[i, j] = min(dp[i - 1, j] + 1,     # deletion
                           dp[i, j - 1] + 1,     # insertion
                           sub_cost)             # substitution
    return dp[m, n]




answer = levenshtein(np.array([3, 5, 2, 4, 8, 7]), np.array([3, 4, 2, 4, 8, 7, 9]))
expected = 3
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: All tests passed

Substring Evaluation

Testing if the previous output contains the string All tests passed: True

Check if any of the following conditions are true:

Substring Evaluation

Testing if the previous output contains the string njit: True

Substring Evaluation

Testing if the previous output contains the string nopython: False

Final Answer: True

Check if any of the following conditions are true:

Substring Evaluation

Testing if the previous output contains the string import numba: False

Substring Evaluation

Testing if the previous output contains the string from numba: True

Final Answer: True

Final Answer: True

Initial Query

Write me a numba implementation of levensthin distance, taking two numpy int arrays. Call it levenshtein. Run it in nopython mode. Character swaps are cost 2.

LLM Generation

Query

Write me a numba implementation of levensthin distance, taking two numpy int arrays. Call it levenshtein. Run it in nopython mode. Character swaps are cost 2.

Output

Below is one solution. In this version, insertion and deletion have a cost of 1, and replacing (i.e. swapping a mismatched character) has a cost of 2. The function takes two numpy integer arrays and computes the Levenshtein distance in nopython mode using Numba. ------------------------------------------------ from numba import njit import numpy as np @njit def levenshtein(a, b): n = a.shape[0] m = b.shape[0] # Create a DP table with dimensions (n+1) x (m+1) dp = np.empty((n + 1, m + 1), dtype=np.int64) # Base cases: distance from empty sequence to a sequence of length i or j for i in range(n + 1): dp[i, 0] = i for j in range(m + 1): dp[0, j] = j # DP computation: # For each position, we take the minimum cost from: # - Deleting an element (cost 1), # - Inserting an element (cost 1), or # - Replacing (swapping) a mismatched element (cost 2, or 0 if they match) for i in range(1, n + 1): for j in range(1, m + 1): if a[i - 1] == b[j - 1]: cost = 0 else: cost = 2 deletion = dp[i - 1, j] + 1 insertion = dp[i, j - 1] + 1 substitution = dp[i - 1, j - 1] + cost dp[i, j] = deletion if deletion < insertion else insertion if substitution < dp[i, j]: dp[i, j] = substitution return dp[n, m] # Example usage: if __name__ == "__main__": # Let's compare two sequences represented as numpy int arrays. # For example, a corresponds to [1, 3, 5, 7] and b corresponds to [1, 4, 5, 9] a = np.array([1, 3, 5, 7], dtype=np.int64) b = np.array([1, 4, 5, 9], dtype=np.int64) print("Levenshtein distance:", levenshtein(a, b)) ------------------------------------------------ Explanation: 1. We use the decorator @njit to compile the function in nopython mode. 2. The DP table dp is created with shape (n+1, m+1). Each cell dp[i,j] holds the minimal cost to convert the first i elements of a into the first j elements of b. 3. Base cases: converting an empty array to a length‑j array costs j (via j insertions) and vice‑versa. 4. In the main loop, if the current elements match (i.e. a[i-1]==b[j-1]), no cost is incurred. Otherwise, substituting one for the other costs 2. 5. Finally, we return the value in dp[n, m] which is the Levenshtein distance. This implementation meets the requirements and runs fully in nopython mode with Numba.

Extract Code

I extracted the following code from that output:

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from numba import njit
import numpy as np

@njit
def levenshtein(a, b):
    n = a.shape[0]
    m = b.shape[0]
    
    # Create a DP table with dimensions (n+1) x (m+1)
    dp = np.empty((n + 1, m + 1), dtype=np.int64)
    
    # Base cases: distance from empty sequence to a sequence of length i or j
    for i in range(n + 1):
        dp[i, 0] = i
    for j in range(m + 1):
        dp[0, j] = j

    # DP computation:
    # For each position, we take the minimum cost from:
    # - Deleting an element (cost 1),
    # - Inserting an element (cost 1), or 
    # - Replacing (swapping) a mismatched element (cost 2, or 0 if they match)
    for i in range(1, n + 1):
        for j in range(1, m + 1):
            if a[i - 1] == b[j - 1]:
                cost = 0
            else:
                cost = 2

            deletion = dp[i - 1, j] + 1
            insertion = dp[i, j - 1] + 1
            substitution = dp[i - 1, j - 1] + cost
            dp[i, j] = deletion if deletion < insertion else insertion
            if substitution < dp[i, j]:
                dp[i, j] = substitution

    return dp[n, m]

Check if all of the following conditions are true:

Run Code Interpreter

Running the following program:

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from numba import njit
import numpy as np

@njit
def levenshtein(a, b):
    n = a.shape[0]
    m = b.shape[0]
    
    # Create a DP table with dimensions (n+1) x (m+1)
    dp = np.empty((n + 1, m + 1), dtype=np.int64)
    
    # Base cases: distance from empty sequence to a sequence of length i or j
    for i in range(n + 1):
        dp[i, 0] = i
    for j in range(m + 1):
        dp[0, j] = j

    # DP computation:
    # For each position, we take the minimum cost from:
    # - Deleting an element (cost 1),
    # - Inserting an element (cost 1), or 
    # - Replacing (swapping) a mismatched element (cost 2, or 0 if they match)
    for i in range(1, n + 1):
        for j in range(1, m + 1):
            if a[i - 1] == b[j - 1]:
                cost = 0
            else:
                cost = 2

            deletion = dp[i - 1, j] + 1
            insertion = dp[i, j - 1] + 1
            substitution = dp[i - 1, j - 1] + cost
            dp[i, j] = deletion if deletion < insertion else insertion
            if substitution < dp[i, j]:
                dp[i, j] = substitution

    return dp[n, m]




answer = levenshtein(np.array([3, 5, 2, 4, 8, 7]), np.array([3, 4, 2, 4, 8, 7, 9]))
expected = 3
assert answer == expected, f'Wrong answer; got {answer} instead of {expected}'
print('All tests passed')
And got the output: All tests passed

Substring Evaluation

Testing if the previous output contains the string All tests passed: True

Check if any of the following conditions are true:

Substring Evaluation

Testing if the previous output contains the string njit: True

Substring Evaluation

Testing if the previous output contains the string nopython: False

Final Answer: True

Check if any of the following conditions are true:

Substring Evaluation

Testing if the previous output contains the string import numba: False

Substring Evaluation

Testing if the previous output contains the string from numba: True

Final Answer: True

Final Answer: True