A XOR Challenge

Description

Haha! I encrypted a message containing the flag with a 12-byte XOR key. Let’s see if you can crack it. Btw, the phrase “Your flag is” shows up in the plaintext, good luck! Note: The plaintext containing the flag contains some non-printable ascii characters to throw off auto-xor solvers 😉 Flag format: FLAG{some_string}

challenge.zip

(Credits to Sean for the challenge)

Solution

A classic XOR challenge. The trick of this challenge is to make use of XOR’s unique properties. Mainly the property A ⊕ A = 0.

Let’s assume the plaintext containing the flag is A, while the key is B. If we XOR these two together, we get A ⊕ B = C. Then, if we apply the property above, and try to XOR the ciphertext with a known part of a ciphertext, we can try to find the key. Like this, C ⊕ A = (A ⊕ B) ⊕ A = B.

Hence, let’s apply this in this script to solve for the flag.

First, we XOR the known part of the plaintext with the ciphertext.

from binascii import unhexlify
from regex import findall
with open('cipher', "rb") as f:
    ciphertext = unhexlify(f.read().strip())
key = "Your flag is"
key = bytes(key, 'ascii')
keyList = []
x = 0
data = ciphertext
while x < len(ciphertext) - 12:
    res = ""
    try:
        for i in range(len(key)):
            res += chr(data[x + i] ^ key[i % len(key)])
    except Exception as e:
        break
    res = res[-(x % 12):] + res[:-(x % 12)] #padding
    keyList.append(res)
    x = x + 1

The padding function is used to account for the previous x letters of the ciphertext, and wraps-around the key according to the position of the character in each block of 12 characters.

Then, we XOR the list of possible keys with the ciphertext. We can then use the power of Regex to single out the plaintext containing the flag.

data = ciphertext
for k in keyList:
    #k = bytes(k, 'ascii')
    counter = 0
    res = ''
    for j in range (len(ciphertext)):
        res += chr(data[j] ^ ord(k[j % len (k)]))
    if findall('FLAG{.+}', res):
        print(findall('FLAG{.+}', res))
        break

And then we get the flag, FLAG{gg_y0u_kn0w_h0w_t0_und0_x0r_w17h_k3y_l3ngth_4nd_par7ial_pl41n73x7_kn0wl3d63}.

Appendix

This is the challenge author’s code.

#python 2 please
import string
import binascii
def xor(msg, key):
    o = ''
    for i in range(len(msg)):
        o += chr(ord(msg[i]) ^ ord(key[i % len(key)]))
    return o
  
def pad_key(key,keylength):
    key_fixed = ""
    list_key=[]
    list_key[:0]=key
    for j in range(i%keylength):
        tmp = list_key[-1]
        list_key.pop(-1)
        list_key.insert(0,tmp)
    key_fixed = ''.join(str(i) for i in list_key)
    return key_fixed
msg = binascii.unhexlify("")
part_of_msg = "Your flag is"
partial_pt = "FLAG{"
keylength = 12
for i in range(len(msg)):
    print i,
    candidate = xor(msg[i:i+len(part_of_msg)], part_of_msg)
    candidate_padded = pad_key(candidate,keylength)
    possible_pt = xor(msg,candidate_padded)
    if (partial_pt in possible_pt):
        print(possible_pt)

My full code:

from binascii import unhexlify
from regex import findall
with open('cipher', "rb") as f:
    ciphertext = unhexlify(f.read().strip())
key = "Your flag is"
key = bytes(key, 'ascii')
keyList = []
x = 0
data = ciphertext
while x < len(ciphertext) - 12:
    res = ""
    try:
        for i in range(len(key)):
            res += chr(data[x + i] ^ key[i % len(key)])
    except Exception as e:
        break
    res = res[-(x % 12):] + res[:-(x % 12)]
    keyList.append(res)
    x = x + 1
data = ciphertext
for k in keyList:
    #k = bytes(k, 'ascii')
    counter = 0
    res = ''
    for j in range (len(ciphertext)):
        res += chr(data[j] ^ ord(k[j % len (k)]))
    if findall('FLAG{.+}', res):
        print(findall('FLAG{.+}', res))
        break

Arul Kumaran

CTFer