OpenCV bilinear downsampling with non-integer scaling - opencv

Question

I need help in figuring out the algorithm/implementation OpenCV is using for image-downsampling with non-linear scaling factors.
I know the question was already ask a few times, but most answers seem to not match OpenCV's implementation (for instance, this answer is not correct when using OpenCV: https://math.stackexchange.com/questions/48903/2d-array-downsampling-and-upsampling-using-bilinear-interpolation).
Minimal problem formulation:
I want to downsample an image of resolution 4x4 to an image of resolution 3x3 using bilinear interpolation. I am interested in the interpolation coefficients.
Example in python:
img = np.asarray([[ 1, 2, 3, 4],
[ 5, 6, 7, 8],
[ 9, 10, 11, 12],
[13, 14, 15, 16]]).astype(np.float32)
img_resized = cv2.resize(img, (3, 3), 0, 0, cv2.INTER_LINEAR).astype(np.float32)
print(img)
# [[ 1. 2. 3. 4.]
# [ 5. 6. 7. 8.]
# [ 9. 10. 11. 12.]
# [13. 14. 15. 16.]]
print(img_resized)
# [[ 1.8333333 3.1666667 4.5 ]
# [ 7.166667 8.5 9.833333 ]
# [12.5 13.833333 15.166666 ]]
Interpolation coefficients:
After a lot of trial-and-error, I figured out the interpolation coefficients OpenCV is using for this specific case.
For the corner points of the 3x3 image:
1.8333333 = 25/36 * 1 + 5/36 * 2 + 5/36 * 5 + 1/36 * 6
4.5000000 = 25/36 * 4 + 5/36 * 3 + 5/36 * 8 + 1/36 * 7
12.5000000 = 25/36 * 13 + 5/36 * 9 + 5/36 * 14 + 1/36 * 10
15.1666666 = 25/36 * 16 + 5/36 * 15 + 5/36 * 12 + 1/36 * 11
For the middle points of the 3x3 image:
8.5 = 1/4 * 6 + 1/4 * 7 + 1/4 * 10 + 1/4 * 11
For the remaining 4 points of the 3x3 image:
3.1666667 = 5/12 * 2 + 5/12 * 3 + 1/12 * 6 + 1/12 * 7
7.1666667 = 5/12 * 5 + 5/12 * 9 + 1/12 * 6 + 1/12 * 10
9.8333333 = 5/12 * 8 + 5/12 * 12 + 1/12 * 7 + 1/12 * 11
13.833333 = 5/12 * 14 + 5/12 * 15 + 1/12 * 10 + 1/12 * 11
Question:
Can someone please help me make sense of these interpolation coefficients? How are they calculated? I tried to read the source of the cv::resize() function, but it did not help me a lot :S

Answer 1

After playing around with various test cases, I think I know the answer to how OpenCV chooses the sample point locations. As #ChrisLuengo has pointed out in a comment, OpenCV seems to not apply a low-pass filter before downsampling, but uses (bi-)linear interpolation only.
(Possible) Solution:
Let's assume we have a 5x5 image, which pixel positions are represented with the blue circles in the graphic below. We now want to downsample it to a 3x3, or a 4x4 image, and need to find the sample positions of the new downsampled image in the original image grid.
It appears to be that OpenCV uses pixel distance of 1 for the original image grid, and a pixel distance of (OLD_SIZE / NEW_SIZE), thus here 5/3 and 5/4, for the new image grid. Moreover, it aligns both grids at the center point. Thus, OpenCV's deterministic sampling algorithms can be visualized as follows:
Visualization 5x5 to 3x3:
Visualization 5x5 to 4x4:
Sample Code (Python 2.7):
import numpy as np
import cv2
# 1. H_W is the height & width of the original image, using uniform H/W for this example
# resized_H_W is the height & width of the resized image, using uniform H/W for this example
H_W = 5
resized_H_W = 4
# 2. Create original image & Get OpenCV resized image:
img = np.zeros((H_W, H_W)).astype(np.float32)
counter = 1
for i in range(0, H_W):
for j in range(0, H_W):
img[i, j] = counter
counter += 1
img_resized_opencv = cv2.resize(img, (resized_H_W, resized_H_W), 0, 0, cv2.INTER_LINEAR).astype(np.float32)
# 3. Get own resized image:
img_resized_own = np.zeros((resized_H_W, resized_H_W)).astype(np.float32)
for i in range(0, resized_H_W):
for j in range(0, resized_H_W):
sample_x = (1.0 * H_W) / 2.0 - 0.50 + (i - (1.0 * resized_H_W - 1.0) / 2.0) * (1.0 * H_W) / (1.0 * resized_H_W)
sample_y = (1.0 * H_W) / 2.0 - 0.50 + (j - (1.0 * resized_H_W - 1.0) / 2.0) * (1.0 * H_W) / (1.0 * resized_H_W)
pixel_top_left = img[int(np.floor(sample_x)), int(np.floor(sample_y))]
pixel_top_right = img[int(np.floor(sample_x)), int(np.ceil(sample_y))]
pixel_bot_left = img[int(np.ceil(sample_x)), int(np.floor(sample_y))]
pixel_bot_right = img[int(np.ceil(sample_x)), int(np.ceil(sample_y))]
img_resized_own[i, j] = (1.0 - (sample_x - np.floor(sample_x))) * (1.0 - (sample_y - np.floor(sample_y))) * pixel_top_left + \
(1.0 - (sample_x - np.floor(sample_x))) * (sample_y - np.floor(sample_y)) * pixel_top_right + \
(sample_x - np.floor(sample_x)) * (1.0 - (sample_y - np.floor(sample_y))) * pixel_bot_left + \
(sample_x - np.floor(sample_x)) * (sample_y - np.floor(sample_y)) * pixel_bot_right
# 4. Print results:
print "\n"
print "Org. image: \n", img
print "\n"
print "Resized image (OpenCV): \n", img_resized_opencv
print "\n"
print "Resized image (own): \n", img_resized_own
print "\n"
print "MSE between OpenCV <-> Own: ", np.mean(np.square(img_resized_opencv - img_resized_own))
print "\n"
Disclaimer:
This is just my theory that I tested via ~10 test cases. I do not claim that this is 100% true.