Backdoor 2017 Imagerev 250
Solved Backdoor 2017 image rev 250
Imagerev 200 Backdoor CTF 2017
Reverse the encrypted file and recover the flag. encrypted.txt encrypt.py
The challenge gives you two files One encrypt.py and enctrypted.txt
from PIL import Image
def bin_return(dec):
return(str(format(dec, 'b')))
def bin_8bit(dec):
return(str(format(dec, '08b')))
def convert_32bit(dec):
return(str(format(dec, '032b')))
def convert_64bit(dec):
return(str(format(dec, '064b')))
def hex_return(dec):
return expand(hex(dec).replace('0x', '').replace('L', ''))
def dec_return_bin(bin_string):
return(int(bin_string, 2))
def dec_return_hex(hex_string):
return(int(hex_string, 16))
def some_LP(l, n):
l1 = []
j = 0
k = n
while k < len(l) + 1:
l1.append(l[j:k])
j = k
k += n
return(l1)
def rotate_right(bit_string, n):
bit_list = list(bit_string)
count = 0
while count <= n - 1:
list_main = list(bit_list)
var_0 = list_main.pop(-1)
list_main = list([var_0] + list_main)
bit_list = list(list_main)
count += 1
return(''.join(list_main))
def shift_right(bit_string, n):
bit_list = list(bit_string)
count = 0
while count <= n - 1:
bit_list.pop(-1)
count += 1
front_append = ['0'] * n
return(''.join(front_append + bit_list))
def addition(input_set):
value = 0
for i in range(len(input_set)):
value += input_set[i]
mod_32 = 4294967296
return(value % mod_32)
def str_xor(s1, s2):
return ''.join([str(int(i) ^ int(j)) for i, j in zip(s1, s2)])
def str_and(s1, s2):
return ''.join([str(int(i) & int(j)) for i, j in zip(s1, s2)])
def str_not(s):
return ''.join([str(int(i) ^ 1) for i in s])
def not_and_and_xor(x, y, z):
return(str_xor(str_and(x, y), str_and(str_not(x), z)))
def and_and_and_xor_xor(x, y, z):
return(str_xor(str_xor(str_and(x, y), str_and(x, z)), str_and(y, z)))
def some_e0(x):
return(str_xor(str_xor(rotate_right(x, 2), rotate_right(x, 13)), rotate_right(x, 22)))
def some_e1(x):
return(str_xor(str_xor(rotate_right(x, 6), rotate_right(x, 11)), rotate_right(x, 25)))
def some_s0(x):
return(str_xor(str_xor(rotate_right(x, 7), rotate_right(x, 18)), shift_right(x, 3)))
def some_s1(x):
return(str_xor(str_xor(rotate_right(x, 17), rotate_right(x, 19)), shift_right(x, 10)))
def expand(s):
return '0' * (8 - len(s)) + s
def get_pixels_list(filename):
im = Image.open(filename)
return list(im.getdata())
def data_encrypted(list_of_pixels):
data = ''
for i in list_of_pixels:
d = ''.join([chr(j) for j in i])
d = encryption(d)
data += ''.join(d)
return data
def message_pad(bit_list):
pad_one = bit_list + '1'
pad_len = len(pad_one)
k = 0
while ((pad_len + k) - 448) % 512 != 0:
k += 1
back_append_0 = '0' * k
back_append_1 = convert_64bit(len(bit_list))
return(pad_one + back_append_0 + back_append_1)
def message_bit_return(string_input):
bit_list = []
for i in range(len(string_input)):
bit_list.append(bin_8bit(ord(string_input[i])))
return(''.join(bit_list))
def message_pre_pro(input_string):
bit_main = message_bit_return(input_string)
return(message_pad(bit_main))
def message_parsing(input_string):
return(some_LP(message_pre_pro(input_string), 32))
def message_schedule(index, w_t):
new_word = convert_32bit(addition([int(some_s1(w_t[index - 2]), 2), int(
w_t[index - 7], 2), int(some_s0(w_t[index - 15]), 2), int(w_t[index - 16], 2)]))
return(new_word)
initial = ['6a09e667', 'bb67ae85', '3c6ef372', 'a54ff53a',
'510e527f', '9b05688c', '1f83d9ab', '5be0cd19']
values = ['428a2f98', '71374491', 'b5c0fbcf', 'e9b5dba5', '3956c25b', '59f111f1', '923f82a4', 'ab1c5ed5', 'd807aa98', '12835b01', '243185be', '550c7dc3', '72be5d74', '80deb1fe', '9bdc06a7', 'c19bf174', 'e49b69c1', 'efbe4786', '0fc19dc6', '240ca1cc', '2de92c6f', '4a7484aa', '5cb0a9dc', '76f988da', '983e5152', 'a831c66d', 'b00327c8', 'bf597fc7', 'c6e00bf3', 'd5a79147', '06ca6351', '14292967',
'27b70a85', '2e1b2138', '4d2c6dfc', '53380d13', '650a7354', '766a0abb', '81c2c92e', '92722c85', 'a2bfe8a1', 'a81a664b', 'c24b8b70', 'c76c51a3', 'd192e819', 'd6990624', 'f40e3585', '106aa070', '19a4c116', '1e376c08', '2748774c', '34b0bcb5', '391c0cb3', '4ed8aa4a', '5b9cca4f', '682e6ff3', '748f82ee', '78a5636f', '84c87814', '8cc70208', '90befffa', 'a4506ceb', 'bef9a3f7', 'c67178f2']
def encryption(input_string):
w_t = message_parsing(input_string)
a = convert_32bit(dec_return_hex(initial[0]))
b = convert_32bit(dec_return_hex(initial[1]))
c = convert_32bit(dec_return_hex(initial[2]))
d = convert_32bit(dec_return_hex(initial[3]))
e = convert_32bit(dec_return_hex(initial[4]))
f = convert_32bit(dec_return_hex(initial[5]))
g = convert_32bit(dec_return_hex(initial[6]))
h = convert_32bit(dec_return_hex(initial[7]))
for i in range(0, 64):
if i <= 15:
t_1 = addition([int(h, 2), int(some_e1(e), 2), int(
not_and_and_xor(e, f, g), 2), int(values[i], 16), int(w_t[i], 2)])
t_2 = addition([int(some_e0(a), 2), int(
and_and_and_xor_xor(a, b, c), 2)])
h = g
g = f
f = e
e = addition([int(d, 2), t_1])
d = c
c = b
b = a
a = addition([t_1, t_2])
a = convert_32bit(a)
e = convert_32bit(e)
if i > 15:
w_t.append(message_schedule(i, w_t))
t_1 = addition([int(h, 2), int(some_e1(e), 2), int(
not_and_and_xor(e, f, g), 2), int(values[i], 16), int(w_t[i], 2)])
t_2 = addition([int(some_e0(a), 2), int(
and_and_and_xor_xor(a, b, c), 2)])
h = g
g = f
f = e
e = addition([int(d, 2), t_1])
d = c
c = b
b = a
a = addition([t_1, t_2])
a = convert_32bit(a)
e = convert_32bit(e)
value_0 = addition([dec_return_hex(initial[0]), int(a, 2)])
value_1 = addition([dec_return_hex(initial[1]), int(b, 2)])
value_2 = addition([dec_return_hex(initial[2]), int(c, 2)])
value_3 = addition([dec_return_hex(initial[3]), int(d, 2)])
value_4 = addition([dec_return_hex(initial[4]), int(e, 2)])
value_5 = addition([dec_return_hex(initial[5]), int(f, 2)])
value_6 = addition([dec_return_hex(initial[6]), int(g, 2)])
value_7 = addition([dec_return_hex(initial[7]), int(h, 2)])
value = (hex_return(value_0), hex_return(value_1), hex_return(value_2), hex_return(
value_3), hex_return(value_4), hex_return(value_5), hex_return(value_6), hex_return(value_7))
return(value)
list_pixels = get_pixels_list('./flag.png')
data = data_encrypted(list_pixels)
f = open('./encrypted.txt','w')
f.write(data)
f.close()
encrypted.txt contains the hashed output
709e80c88487a2411e1ee4dfb9f22a861492d20c4765150c0c794abd70f8147c709e80c88487a2 ......
First I tried to Reverse the Code and it was boring , Got some ideas about the script that it takes the tuple of the pixel values and convert it to a hash of 64 length
So I tought of Brute forcing the entire space , which is 256x256x256
The only Problem is that it takes time, it takes 1sec in my pc to calculate 10 hashes and the whole will take 2 days ,which is not possible
f = open('./encrypted.txt', 'r'),
text = f.read()
enc = [text[i:i + 64] for i in range(0, len(text), 64)]
l1 = open('./list.txt','w')
for i in enc:
l1.write(i+"\n")
cat list.txt | sort -n | uniq -c | sort -n -r
5821 709e80c88487a2411e1ee4dfb9f22a861492d20c4765150c0c794abd70f8147c
306 ac205167ca956b408a925c3854fdd82ffa43672263ae7dba5a68b29d9a81fa56
291 2ec847d8a31a988b3117a5095dae74f490448223f035ec7eddef6768b91a9028
188 8ae40a3583aef6697d2c2eff57eb915ed0bda54aaa92812ad97982743ac06f37
90 ab5ab0fedc83e5a1a1871c427eccbcd3cf0fc1bb74a82a552adfd9b4e57f391b
85 2ac9a6746aca543af8dff39894cfe8173afba21eb01c6fae33d52947222855ef
79 f1b901847390b0ed7e374e7c1e464ec17b46a427c487a5ad6cbd2906405083d5
73 5ae7e6a42304dc6e4176210b83c43024f99a0bce9a870c3b6d2c95fc8ebfb74c
62 b9e8d0a22760b87553c0b9c55ae93058bf8d4389c87765488cea1637e94bd9b6
59 a30cb1d8569c5c141b2ade1caf57038b2be46c9bc4939c8f702a0ff4fcecfd77
58 91737e71235959a56c524997e18d6d14d6ddd714ed2a450a24f765255a2733ee
53 700af1feb55ab0613bdbc466815643743156af4e869120244eb05ca72c45002c
50 0aad7da77d2ed59c396c99a74e49f3a4524dcdbcb5163251b1433d640247aeb4
47 7b108f7c5c6f1507c4ffe2275dd9b8e25a71d175a5a9d3e19aeec3f27d82caf1
42 204164d223b35aabb54ea32b1d14d8bb5a8df56f7c81f3304987fa4193426729
38 c4289629b08bc4d61411aaa6d6d4a0c3c5f8c1e848e282976e29b6bed5aeedc7
24 5ae0d5195906bfc4f70167cf171ae4d08e7376aa246977acf172187d5d384f10
I spilt the encrypted.txt into fragments of length 64 and analysing the files reveals that there are only 17 uniq values
So I grabbed the most used used RGB colors from here
color_list = [(176, 23, 31),
(220, 20, 60),
(255, 182, 193),
(255, 174, 185),
(238, 162, 173),
(205, 140, 149),
(139, 95, 101),
(255, 192, 203),
(255, 181, 197),
...
...
...
]
for i in range(0, len(color_list), 1):
if data_encrypted(color_list[i:i + 1]) in col_hash:
print(color_list[i:i + 1], data_encrypted(color_list[i:i + 1]))
([(255, 255, 255)], '5ae7e6a42304dc6e4176210b83c43024f99a0bce9a870c3b6d2c95fc8ebfb74c')
([(128, 128, 128)], '8ae40a3583aef6697d2c2eff57eb915ed0bda54aaa92812ad97982743ac06f37')
([(0, 0, 0)], '709e80c88487a2411e1ee4dfb9f22a861492d20c4765150c0c794abd70f8147c')
([(207, 207, 207)], '7b108f7c5c6f1507c4ffe2275dd9b8e25a71d175a5a9d3e19aeec3f27d82caf1')
([(191, 191, 191)], 'ac205167ca956b408a925c3854fdd82ffa43672263ae7dba5a68b29d9a81fa56')
([(143, 143, 143)], 'b9e8d0a22760b87553c0b9c55ae93058bf8d4389c87765488cea1637e94bd9b6')
([(112, 112, 112)], 'c4289629b08bc4d61411aaa6d6d4a0c3c5f8c1e848e282976e29b6bed5aeedc7')
([(64, 64, 64)], '2ec847d8a31a988b3117a5095dae74f490448223f035ec7eddef6768b91a9028')
([(48, 48, 48)], '2ac9a6746aca543af8dff39894cfe8173afba21eb01c6fae33d52947222855ef')
The output shows that the pixels have all the values equal , Let’s test all the values from 0 - 256
for i in range(0, 256):
if data_encrypted([(i, i, i)]) in col_hash:
print(i, data_encrypted([(i, i, i)]))
Result :
(0, '709e80c88487a2411e1ee4dfb9f22a861492d20c4765150c0c794abd70f8147c')
(16, 'ab5ab0fedc83e5a1a1871c427eccbcd3cf0fc1bb74a82a552adfd9b4e57f391b')
(32, '0aad7da77d2ed59c396c99a74e49f3a4524dcdbcb5163251b1433d640247aeb4')
(48, '2ac9a6746aca543af8dff39894cfe8173afba21eb01c6fae33d52947222855ef')
(64, '2ec847d8a31a988b3117a5095dae74f490448223f035ec7eddef6768b91a9028')
(80, '204164d223b35aabb54ea32b1d14d8bb5a8df56f7c81f3304987fa4193426729')
(96, 'f1b901847390b0ed7e374e7c1e464ec17b46a427c487a5ad6cbd2906405083d5')
(112, 'c4289629b08bc4d61411aaa6d6d4a0c3c5f8c1e848e282976e29b6bed5aeedc7')
(128, '8ae40a3583aef6697d2c2eff57eb915ed0bda54aaa92812ad97982743ac06f37')
(143, 'b9e8d0a22760b87553c0b9c55ae93058bf8d4389c87765488cea1637e94bd9b6')
(159, '91737e71235959a56c524997e18d6d14d6ddd714ed2a450a24f765255a2733ee')
(175, '700af1feb55ab0613bdbc466815643743156af4e869120244eb05ca72c45002c')
(191, 'ac205167ca956b408a925c3854fdd82ffa43672263ae7dba5a68b29d9a81fa56')
(207, '7b108f7c5c6f1507c4ffe2275dd9b8e25a71d175a5a9d3e19aeec3f27d82caf1')
(223, 'a30cb1d8569c5c141b2ade1caf57038b2be46c9bc4939c8f702a0ff4fcecfd77')
(239, '5ae0d5195906bfc4f70167cf171ae4d08e7376aa246977acf172187d5d384f10')
(255, '5ae7e6a42304dc6e4176210b83c43024f99a0bce9a870c3b6d2c95fc8ebfb74c')
Now we got all the pixels corresponding to the hashes in the encrypted text
Now Decryption , But they have not given the image resolution , there are 7371 pixels now the image resolution will be factor of this number
We try them all !!
key_dic = {'709e80c88487a2411e1ee4dfb9f22a861492d20c4765150c0c794abd70f8147c': 0,
'ab5ab0fedc83e5a1a1871c427eccbcd3cf0fc1bb74a82a552adfd9b4e57f391b': 1,
'0aad7da77d2ed59c396c99a74e49f3a4524dcdbcb5163251b1433d640247aeb4': 3,
'2ac9a6746aca543af8dff39894cfe8173afba21eb01c6fae33d52947222855ef': 4,
'2ec847d8a31a988b3117a5095dae74f490448223f035ec7eddef6768b91a9028': 6,
'204164d223b35aabb54ea32b1d14d8bb5a8df56f7c81f3304987fa4193426729': 8,
'f1b901847390b0ed7e374e7c1e464ec17b46a427c487a5ad6cbd2906405083d5': 9,
'c4289629b08bc4d61411aaa6d6d4a0c3c5f8c1e848e282976e29b6bed5aeedc7': 11,
'8ae40a3583aef6697d2c2eff57eb915ed0bda54aaa92812ad97982743ac06f37': 12,
'b9e8d0a22760b87553c0b9c55ae93058bf8d4389c87765488cea1637e94bd9b6': 14,
'91737e71235959a56c524997e18d6d14d6ddd714ed2a450a24f765255a2733ee': 15,
'700af1feb55ab0613bdbc466815643743156af4e869120244eb05ca72c45002c': 17,
'ac205167ca956b408a925c3854fdd82ffa43672263ae7dba5a68b29d9a81fa56': 19,
'7b108f7c5c6f1507c4ffe2275dd9b8e25a71d175a5a9d3e19aeec3f27d82caf1': 20,
'a30cb1d8569c5c141b2ade1caf57038b2be46c9bc4939c8f702a0ff4fcecfd77': 22,
'5ae0d5195906bfc4f70167cf171ae4d08e7376aa246977acf172187d5d384f10': 23,
'5ae7e6a42304dc6e4176210b83c43024f99a0bce9a870c3b6d2c95fc8ebfb74c': 25,
}
pixels = []
num = 0
print(len(enc))
for i in enc:
num = key_dic[i]
pixels.append((num, num, num))
pos_cor = [(1, 7371),
(3, 2457),
(7, 1053),
(9, 819),
(13, 567),
(21, 351),
(27, 273),
(39, 189),
(63, 117),
(81, 91)]
for j in pos_cor:
img = Image.new('RGB', j, 'white')
pix = img.load()
cordinates = []
for i in range(0, len(pixels), j[1]):
cordinates.append(pixels[i:i + j[1]])
for i in range(0, j[0]):
for k, n in zip(cordinates[i], range(0, j[1])):
pix[i, n] = k[0:3]
img.save("image" + str(j[0]) + ".png")
image21.png contains the flag