assistant: Here you go:
```
import numpy as np

def squared_pairwise_distances_vectorized(p: np.ndarray, q: np.ndarray) -> np.ndarray:
    """
    Compute the squared pairwise distances between two sets of points p and q.

    Parameters
    ----------
    p : np.ndarray
        The first set of points. Shape is (n, d).
    q : np.ndarray
        The second set of points. Shape is (m, d).

    Returns
    -------
    np.ndarray
        The squared pairwise distances. Shape is (n, m).
    """
    return (p**2).sum(axis=1, keepdims=True) + (q**2).sum(axis=1) - 2 * p @ q.T
    
    
def squared_pairwise_distances_naive(a: np.ndarray, b: np.ndarray) -> np.ndarray:
    """
    This function should do the same as squared_pairwise_distances_vectorized,
    but without using any vectorized operations. It is provided for
    comparison purposes only.
    """
    distances = np.zeros((a.shape[0], b.shape[0]))

    for i in range(a.shape[0]):
    	for j in range(b.shape[0]):
    		distances[i, j] = np.sum((a[i]-b[j])**2)

    return distances
    
def main() -> None:
    p = np.random.rand(7, 2)
    q = np.random.rand(5, 2)

    distances_vectorized = squared_pairwise_distances_vectorized(p, q)
    distances_naive = squared_pairwise_distances_naive(p, q)
    # print(distances_naive)
    assert np.allclose(distances_vectorized, distances_naive)

if __name__ == "__main__":
    main()
```

assistant: You're right, I should have provided the code there. Here's one possible implementation using explicit loops:
```python
for i in range(a.shape[0]):
    for j in range(b.shape[0]):
        distances[i, j] = (a[i, :] - b[j, :])**2).sum()
```

This loops over the rows of a and b and computes the squared L2 distance between each pair of rows

assistant: I have a function `squared_pairwise_distances_vectorized` and I want to test whether it is correct by comparing it to a naive implementation with for-loops. The naive implementation should be easy to understand and not use any NumPy functions. Fill in the part marked with "[...]" below:

import numpy as np

def squared_pairwise_distances_vectorized(p: np.ndarray, q: np.ndarray) -> np.ndarray:
    """
    Compute the squared pairwise distances between two sets of points p and q.

    Parameters
    ----------
    p : np.ndarray
        The first set of points. Shape is (n, d).
    q : np.ndarray
        The second set of points. Shape is (m, d).

    Returns
    -------
    np.ndarray
        The squared pairwise distances. Shape is (n, m).
    """
    return (p**2).sum(axis=1, keepdims=True) + (q**2).sum(axis=1) - 2 * p @ q.T

def squared_pairwise_distances_naive(a: np.ndarray, b: np.ndarray) -> np.ndarray:
    """
    This function should do the same as squared_pairwise_distances_vectorized,
    but without using any vectorized operations. It is provided for
    comparison purposes only.
    """
    distances = np.zeros((a.shape[0], b.shape[0]))

    [...]

    return distances

def main() -> None:
    p = np.random.rand(7, 2)
    q = np.random.rand(5, 2)

    distances_vectorized = squared_pairwise_distances_vectorized(p, q)
    distances_naive = squared_pairwise_distances_naive(p, q)

    assert np.allclose(distances_vectorized, distances_naive)

if __name__ == "__main__":
    main()

Here's a possible implementation of the squared_pairwise_distances_naive function:

python

def squared_pairwise_distances_naive(a: np.ndarray, b: np.ndarray) -> np.ndarray:
    """
    This function should do the same as squared_pairwise_distances_vectorized,
    but without using any vectorized operations. It is provided for
    comparison purposes only.
    """
    distances = np.zeros((a.shape[0], b.shape[0]))

    for i in range(a.shape[0]):
        for j in range(b.shape[0]):
            distances[i, j] = np.sum((a[i, :] - b[j, :])**2)

    return distances


Here's a possible implementation of the squared_pairwise_distances_naive function:

python

def squared_pairwise_distances_naive(a: np.ndarray, b: np.ndarray) -> np.ndarray:
    """
    This function should do the same as squared_pairwise_distances_vectorized,
    but without using any vectorized operations. It is provided for
    comparison purposes only.
    """
    distances = np.zeros((a.shape[0], b.shape[0]))

    for i in range(a.shape[0]):
        for j in range(b.shape[0]):
            distances[i, j] = np.sum((a[i, :] - b[j, :])**2)

    return distances

prompter: Can you answer again, with only one implementation of the squared_pairwise_distances_naive function, and nothing else please?