Rev Rev Rev Writeup - IERAE CTF 2025

Misc/Rev Rev Rev - Rama

Reverse it.

Initial Thoughts

We are given two files: chal.py, a script that encrypts the flag, and output.txt, the encrypted flag.

The contents of chal.py.

[-246, -131, -204, -199, -159, -203, -201, -207, -199, -159, -204, -158, -155, -205, -211,
-206, -201, -206, -205, -211, -158, -159, -207, -202, -211, -199, -206, -155, -206, -211,
-204, -200, -200, -200, -203, -208, -159, -199, -133, -187, -191, -174, -187, -183]

The contents of output.txt.

Methodology

For this challenge, all we have to do is reverse the chal.py script. First, let’s figure out what each line does.

flag = open('flag.txt', 'r').read()

This line opens the flag and stores it in the flag variable.

x = [ord(c) for c in flag]

This converts each character to its corresponding Unicode point value.

x.reverse()

This reverses the order of the list of point values.

y = [i^0xff for i in x]

This XORs each point value with 0xff, or 255 in decimal.

z = [~i for i in y]

This applies the NOT operator to each value, flipping the bits of each integer.

open('output.txt', 'w').write(str(z))

This writes the result to output.txt, which we are given.

All we have to do is write a script that does the opposite of each of these lines in the opposite order.

Script

This is my full solution script

enc = [-246, -131, -204, -199, -159, -203, -201, -207, -199, -159, -204, -158, -155, -205,
  -211, -206, -201, -206, -205, -211, -158, -159, -207, -202, -211, -199, -206, -155, -206,
  -211, -204, -200, -200, -200, -203, -208, -159, -199, -133, -187, -191, -174, -187, -183]

This reverses open('output.txt', 'w').write(str(z)) by storing the contents of output.txt in a variable.

z = [~i for i in enc]

This reverses z = [~i for i in y], as the bitwise NOT operator is involutive, meaning applying it twice reverts to the original value.

y = [i^0xff for i in z]

This reverses y = [i^0xff for i in x], as the XOR operator, just like the bitwise NOT operator, is involutive.

y.reverse()

This reverses x.reverse(), as reversing a list twice reverts it to the original. (Yet another involutive step!)

x = [chr(c) for c in y]

This reverses x = [ord(c) for c in flag], as chr(c) converts each item in the list from its Unicode point value to its character representation.

print(''.join(x))

Finally, we join the list together, getting our flag.

Solution

The flag is IERAE{9a058884-2e29-61ab-3272-3eb4a9175a94}. This was a pretty simple and fun rev challenge, and I liked how it focused on involutive functions.