The flag is fqtbjfub4uj_0_d00151a52523e510f3e50521814141c. The attached file may be useful.
Category: Reversing
Let's see what's in main.py
:
def xor(s1,s2):
return ''.join(chr(ord(a) ^ ord(b)) for a, b in zip(s1, s2))
def encrypt(a):
some_text = a[::2]
randnum = 14
text_length = len(some_text)
endtext = ""
for i in range(1, text_length + 1):
weirdtext = some_text[i - 1]
if weirdtext >= "a" and weirdtext <= "z":
weirdtext = chr(ord(weirdtext) + randnum)
if weirdtext > "z":
weirdtext = chr(ord(weirdtext) - 26)
endtext += weirdtext
randtext = a[1::2]
xored = xor("aaaaaaaaaaaaaaa", randtext)
hex_xored = xored.encode("utf-8").hex()
return endtext + hex_xored
def decrypt(msg):
raise NotImplementedError
def main():
opt = input("Would you like to [E]ncrypt or [D]ecrypt? ")
if opt[:1].lower() == "e":
msg = input("Enter message to encrypt: ")
print(f"Encrypted message: {encrypt(msg)}")
elif opt[:1].lower() == "d":
msg = input("Enter message to decrypt: ")
print(f"Decrypted message: {decrypt(msg)}")
if __name__ == "__main__":
main()
Alright, we just have to implement decrypt()
to undo the transformations we see in encrypt()
. Let's see what happens when we encrypt a message.
kali@kali:~/Downloads$ python3 main.py
Would you like to [E]ncrypt or [D]ecrypt? e
Enter message to encrypt: helloworld
Encrypted message: vzccz040d161305
kali@kali:~/Downloads$ python3 main.py
Would you like to [E]ncrypt or [D]ecrypt? e
Enter message to encrypt: aaaaaaaa
Encrypted message: oooo00000000
kali@kali:~/Downloads$ python3 main.py
Would you like to [E]ncrypt or [D]ecrypt? e
Enter message to encrypt: bbbb
Encrypted message: pp0303
encrypt()
takes the provided input string and splits it into 2 strings:
-
some_text
with the even letters (starting from 0), which gets passed into a loop that applies ROT-14 on the lowercase letters. -
randtext
with the odd letters (starting from 1), which gets XOR'd with a static string and converted to a string of hex values.
After those steps are done, it returns the concatenation of the 2 resulting strings.
It would have been nice if there was a delimiter of some sort to separate the 2 strings. Oh well, we just need the flag, so I'll add the space in manually. Implement the decrypt()
function to reverse the transformations described above, with a space separating the 2 strings.
import binascii
def xor(s1,s2):
return ''.join(chr(ord(a) ^ ord(b)) for a, b in zip(s1, s2))
def unxor(s1,s2):
return ''.join(chr(ord(a) ^ b) for a, b in zip(s1, s2))
def encrypt(a):
some_text = a[::2]
randnum = 14
text_length = len(some_text)
endtext = ""
for i in range(1, text_length + 1):
weirdtext = some_text[i - 1]
if weirdtext >= "a" and weirdtext <= "z":
weirdtext = chr(ord(weirdtext) + randnum)
if weirdtext > "z":
weirdtext = chr(ord(weirdtext) - 26)
endtext += weirdtext
randtext = a[1::2]
xored = xor("aaaaaaaaaaaaaaa", randtext)
hex_xored = xored.encode("utf-8").hex()
return endtext + ' ' + hex_xored
def decrypt(msg):
# let's just throw a space in there manually, that seems easier.
endtext, hex_xored = msg.strip().rstrip().split(" ")
randnum = 14
some_text = ""
for i in range(1, len(endtext) + 1):
weirdtext = endtext[i - 1]
if weirdtext >= "a" and weirdtext <= "z":
weirdtext = chr(ord(weirdtext) - randnum)
if weirdtext < "a":
weirdtext = chr(ord(weirdtext) + 26)
some_text += weirdtext
xored = binascii.unhexlify(hex_xored)
randtext = unxor("aaaaaaaaaaaaaaa", xored)
cleartext = "\n"
for a, b in zip(some_text, randtext):
cleartext += a + b
return cleartext
def main():
opt = input("Would you like to [E]ncrypt or [D]ecrypt? ")
if opt[:1].lower() == "e":
msg = input("Enter message to encrypt: ")
print(f"Encrypted message: {encrypt(msg)}")
elif opt[:1].lower() == "d":
msg = input("Enter message to decrypt: ")
print(f"Decrypted message: {decrypt(msg)}")
if __name__ == "__main__":
main()
Test some known input:
kali@kali:~/Downloads$ python3 solve.py
Would you like to [E]ncrypt or [D]ecrypt? e
Enter message to encrypt: helloworld
Encrypted message: vzccz 040d161305
kali@kali:~/Downloads$ python3 solve.py
Would you like to [E]ncrypt or [D]ecrypt? d
Enter message to decrypt: vzccz 040d161305
Decrypted message:
helloworld
Now we just have to run the flag from the description through it, with the extra space:
kali@kali:~/Downloads$ perl -e 'print "d\n" . "fqtbjfub4uj_0_d 00151a52523e510f3e50521814141c\n"' | python3 solve.py
Would you like to [E]ncrypt or [D]ecrypt? Enter message to decrypt: Decrypted message:
ractf{n3v3r_g0nn4_g1v3_y0u_up}
The flag is:
ractf{n3v3r_g0nn4_g1v3_y0u_up}