Straightforward, and perhaps classic challenge. This takes advantage of the fact that both images are encrypted using the same xor key. We can leak information of the flag by xor the two ciphertext, or the two files given.

There are different snippets of code which can do this, which weirdly is not available online. The following is to do xor on RGB values of pixel by pixel.

daneallen from Cryptohack:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

from PIL import Image

lemur = Image.open("lemur.png")
flag = Image.open("flag.png")

pixels_lemur = lemur.load() # create the pixel map
pixels_flag = flag.load()

for i in range(lemur.size[0]): # for every pixel:
    for j in range(lemur.size[1]):
        # Gather each pixel
        l = pixels_lemur[i,j]
        f = pixels_flag[i,j]

        # XOR each part of the pixel tuple
        r = l[0] ^ f[0]
        g = l[1] ^ f[1]
        b = l[2] ^ f[2]

        # Store the resulatant tuple back into an image
        pixels_flag[i,j] = (r, g, b)

flag.save("lemur_xor_flag.png")

Shorter solution from shoaloak:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

from PIL import Image
from pwn import *

lemur = Image.open("lemur.png")
flag = Image.open("flag.png")

leak_bytes = xor(lemur.tobytes(), flag.tobytes())
leak = Image.frombytes(flag.mode, flag.size, leak_bytes)

leak.save('leak.png')

Last solution using numpy, from Hon:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12

import numpy as np
from PIL import Image

img1 = Image.open("img1.png")
img2 = Image.open("img2.png")

img1np = np.array(img1) * 255
img2np = np.array(img2) * 255

res = np.bitwise_xor(img1np, img2np).astype(np.uint8)

Image.fromarray(res).save('result.png')