General - XOR

XOR Starter

"Given the string label, XOR each character with the integer 13. Convert these integers back to a string and submit the flag as crypto{new_string}."

The following script should do the trick of printing the flag.

#!/usr/bin/env python3

def xor_encrypt(s: str, key: int) -> str:
    """
    Encrypts a string using XOR with a key.
    """
    new_string = ""
    for c in s:
        new_string += chr(ord(c) ^ key)
    return new_string


def main():
    s = "label"
    key = 13
    encrypted_string = xor_encrypt(s, key)
    print("crypto{{{}}}".format(encrypted_string))


if __name__ == "__main__":
    main()

XOR Properties

We get the following outputs where three random keys have been XOR'd together and with the flag:

KEY1 = a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313 
KEY2 ^ KEY1 = 37dcb292030faa90d07eec17e3b1c6d8daf94c35d4c9191a5e1e 
KEY2 ^ KEY3 = c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1 
FLAG ^ KEY1 ^ KEY3 ^ KEY2 = 04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf

The flag is in the last line. It is xored with KEY1, KEY2 and KEY3. Thus, all we need to do is to xor that line with all these keys. Lines 1 and 3 contains those combinations.

That is, KEY1 ^ KEY2 ^ KEY3 ^ FLAG ^ KEY1 ^ KEY3 ^ KEY2 = FLAG

#!/usr/bin/env python3
from pwnlib.util.fiddling import xor

KEY1 = bytes.fromhex('a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313')
KEY2_KEY1 = bytes.fromhex('37dcb292030faa90d07eec17e3b1c6d8daf94c35d4c9191a5e1e')
KEY2_KEY3 = bytes.fromhex('c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1')
FLAG_KEY1_KEY3_KEY2 = bytes.fromhex('04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf')

rm_key1 = xor(KEY1,FLAG_KEY1_KEY3_KEY2)
flag = xor(KEY2_KEY3, rm_key1)
print(flag)

Favourite byte

For the next few challenges, you'll use what you've just learned to solve some more XOR puzzles. I've hidden some data using XOR with a single byte, but that byte is a secret. Don't forget to decode from hex first. 73626960647f6b206821204f21254f7d694f7624662065622127234f726927756d"

We know that the flag starts with 'c' as it follows the format crypto{flag}.

So, to find the one byte key we can xor the first byte of the string with the int value of 'c'.

#!/usr/bin/env python3
from pwnlib.util.fiddling import xor

s = bytes.fromhex('73626960647f6b206821204f21254f7d694f7624662065622127234f726927756d')

key = s[0] ^ ord('c')
print(xor(s, key))

You either know, XOR you don't

"I've encrypted the flag with my secret key, you'll never be able to guess it. 0e0b213f26041e480b26217f27342e175d0e070a3c5b103e2526217f27342e175d0e077e263451150104"

This challenge is similar to the previous one. We start with the fact that we know the flag start like this crypto{.

#!/usr/bin/env python3
from pwnlib.util.fiddling import xor

s = bytes.fromhex('0e0b213f26041e480b26217f27342e175d0e070a3c5b103e2526217f27342e175d0e077e263451150104')

known_plain_txt = 'crypto{'
print(xor(s, known_plain_txt))
# Outputs: b'myXORke+y_Q\x0bHOMe$~seG8bGURN\x04DFWg)a|\x1dTM!an\x7f'
# From this we derive the key
key = 'myXORkey'
print(xor(s, key))