Tags: reverse-engineering
Rating: 4.7
The challenge gives us a read.pyc file, so a python compiled file.
When running it, we see a beautiful animation reading "READ" and a scrolling text under it.
There is also an inpt field we shall put our reversed flag into.<br>
<br>
I used a great tool called uncompyle6 for decompiling the binary.
The result is good, but most function and variable are named gibberish like "ubbaaalubba" so I tidied it up a bit.
I noticed this here: helpful_key = 'you-may-need-this-key-1337' in the code and noted it down.
After some more staring at the code I was attracted to one special function: (vars renamed for readibility)
def actions(user_input): #triggers to check if inputted string is right
data_list = [73, 13, 19, 88, 88, 2, 77, 26, 95, 85, 11, 23, 114, 2, 93, 54, 71, 67, 90, 8, 77, 26, 0, 3, 93, 68]
result = ''
for i in range(len(data_list)):
if user_input[i] != chr(data_list[i] ^ ord(helpful_key[i])):
return 'bbblalaabalaabbblala'
b2a = ''
a2b = [122, 86, 75, 75, 92, 90, 77, 24, 24, 24, 25, 106, 76, 91, 84, 80, 77, 25, 77, 81, 92, 25, 92, 87, 77, 80, 75, 92, 25, 74, 77, 75, 80, 87, 94, 25, 88, 74, 25, 95, 85, 88, 94]
for bbb in a2b:
b2a += chr(bbb ^ 57)
else:
return b2a
The above code takes an input (use_input) and then loops over data_list and helpful_key, xor's them and checks if the result represented as a character is the same as input string.
73 13 19 88 88 2 77 26 95 85 11 23 114 2 93 54 71 67 90 8 77 26 0 3 93 68
y o u - m a y - n e e d - t h i s - k e y - 1 3 3 7
That means if we Xor them ourselves, we get the key required!
My small python script for doing so:
for i in range(len(data_list)):
print(chr(data_list[i]^ord(helpful_key[i])),end="")
The output:
0bfu5c4710ns_v5_4n1m4710ns
I at first didn't notice, that there was actual text written but this here is actually our flag, we just have to wrap it into darkCON{XXX}
So the flag is: darkCON{0bfu5c4710ns_v5_4n1m4710ns}
We can enter it into the read.pyc application and it indeed tells us that it is right!
