I am using OpenCV and NumPy to apply a Sobel filter to an image in Python, and I expected the output to have smaller dimensions after applying the filter. According to the equation output = input – kernel_size // stride + 1, I thought the output should be smaller than the input by the kernel size (which is 3 in this case). However, when I apply the Sobel filter, the output image dimensions remain the same as the input image.

Here is the code I am using:

import cv2
import numpy as np

input_image = cv2.imread('img3.pgm', cv2.IMREAD_GRAYSCALE)
print(f"length of input image rows : {len(input_image)}")
print(f"length of input image columns: {len(input_image[0])}")

sobel_x = cv2.Sobel(input_image, cv2.CV_64F, 1, 0, ksize=3)  # Sobel in x-direction
sobel_y = cv2.Sobel(input_image, cv2.CV_64F, 0, 1, ksize=3)  # Sobel in y-direction

edge_magnitude = cv2.magnitude(sobel_x, sobel_y)

edge_magnitude_cropped = np.clip(edge_magnitude, 0, 255).astype(np.uint8)
print(f"length of filtered_image row {len(edge_magnitude_cropped)}" )
print(f"length of filtered_image col {len(edge_magnitude_cropped[0])}" )
cv2.imwrite('filtered_image_cv_cropped.pgm', edge_magnitude_cropped)
print("nFiltered image saved as 'filtered_image_cv.pgm'.")

with open('original_image_cv_cropped.txt', 'w') as f:
    for row in input_image:
        f.write(" ".join(str(pixel) for pixel in row) + "n")
print("nOriginal image data saved as 'original_image_cv.txt'.")

with open('filtered_image_cv_cropped.txt', 'w') as f:
    for row in edge_magnitude_cropped:
        f.write(" ".join(str(pixel) for pixel in row) + "n")
print("nFiltered image data saved as 'filtered_image_cv.txt'.")

In the code above, I apply the Sobel filter with a kernel size of 3. The input image is 736×736, and I expected the output to be smaller, as per the formula. But after applying the Sobel filter, the dimensions of the output are the same as the input (736×736). When I manually perform the operation, I get the expected smaller image. Why is this happening with OpenCV?