We can see that each square is 10 by 10 pixels. We can recreate the eyes (things on the corners) and try different blurs to find the blurring technique.

After a few trials, we can see that gaussian blur with kernel size 43 is close enough.

Now we can try each square, blur it and see if the result matches with the original image. But because of the kernel size, even 5 square away affects our square, but just a little bit. In fact, trying only 2 square away straight and 1 diagonal is enough (like a diamond shape).


Because we bruteforce in order, we only need to bruteforce 7 squares, which is 2^7 cases (small!).

Lets try bruteforcing the squares, blur the image and see if it matches with the original image.

```python
import cv2 as cv
import numpy as np

original = cv.imread('qr.png')

recovered = np.zeros(original.shape, dtype=np.uint8)
recovered.fill(255)

def paint(x, y, dark, im):
if dark:
im[30+x*10:30+x*10+10, 30+y*10:30+y*10+10] = [0, 0, 0]
else:
im[30+x*10:30+x*10+10, 30+y*10:30+y*10+10] = [255, 255, 255]

def difference(i, j, im):
blurred = cv.GaussianBlur(im, (43, 43), 0)
diff = blurred[30+i*10:30+i*10+10, 30+j*10:30+j*10+10, 0] - original[30+i*10:30+i*10+10, 30+j*10:30+j*10+10, 0]
diff[diff>128] = 255-diff[diff>128]
return diff.sum()

for i in range(0, 37):
for j in range(0, 37):
min_diff = difference(i, j, recovered)
is_paint = False
bruteforce = recovered.copy()

for mask in range(0, 2**7):
for ii, jj, c in zip([0, 0, 0, 1, 1, 1, 2], [0, 1, 2, -1, 0, 1, 0], range(7)):
ii += i
jj += j
if ii < 0 or ii >= 37 or jj < 0 or jj >= 37:
continue

if mask & (1 << (c)):
paint(ii, jj, True, bruteforce)
else:
paint(ii, jj, False, bruteforce)

new_diff = difference(i, j, bruteforce)
if new_diff < min_diff:
min_diff = new_diff
is_paint = mask & 1

paint(i, j, is_paint, recovered)
cv.imshow('recovered', recovered)
cv.waitKey(10)

```

After some time, we have the recovered qr code

Scan it and get the flag: ictf{blurR1ng_is_n0_m4tch_4_u_2ab140c2}