Rating:
# XOR
##### Some workers convinced me to invest in their cryptosystem. They said their brand new XOR keystream cipher would revolutionize the crypto world.
##### Luckily they dropped a piece of paper with the source code on it and an encrypted flag. Can you prove they are wrong?!
We have the following ciphertext:
```
48ac3fe745bed053ac14ef2163cc70db58dfe862fe33860330dc22ea589c9f03d922e13365db038626eaee
```
And the following code:
```python
import os
FLAG = b'OFPPT-CTF{...REDACTED...}'
class HoneyComb:
def __init__(self, key):
self.vals = [i for i in key]
def turn(self):
self.vals = [self.vals[-1]] + self.vals[:-1]
def encrypt(self, msg):
keystream = []
while len(keystream) < len(msg):
keystream += self.vals
self.turn()
return bytes([msg[i] ^ keystream[i] for i in range(len(msg))]).hex()
hc = HoneyComb(os.urandom(6))
print(hc.encrypt(FLAG))
```
The key is only 6 bytes long, to match the length of the plaintext it is reused moving the last position to the first in each iteration.
Since we know the start of the plaintext (OFPPT-) finding the key isn't hard.
```python
def find_first_six(enc):
flag_vals = []
for i in range (0, 12, 2):
temp = int(enc[i:i+2],16)
for j in range (255):
if i == 0 and j ^ temp == 79:
flag_vals.append(j)
break
if i == 2 and j ^ temp == 70:
flag_vals.append(j)
break
if i == 4 and j ^ temp == 80:
flag_vals.append(j)
break
if i == 6 and j ^ temp == 80:
flag_vals.append(j)
break
if i == 8 and j ^ temp == 84:
flag_vals.append(j)
break
if i == 10 and j ^ temp == 45:
flag_vals.append(j)
break
return (flag_vals)
```
Know we do the same process as the encryption to decrypt the cipher.
```python
def decrypt(enc, vals):
keystream = []
while len(keystream) < len(enc):
keystream += vals
vals = turn(vals)
return "".join(chr(enc[i] ^ keystream[i]) for i in range(len(enc)))
```
Full code:
```python
def find_first_six(enc):
flag_vals = []
for i in range (0, 12, 2):
temp = int(enc[i:i+2],16)
for j in range (255):
if i == 0 and j ^ temp == 79:
flag_vals.append(j)
break
if i == 2 and j ^ temp == 70:
flag_vals.append(j)
break
if i == 4 and j ^ temp == 80:
flag_vals.append(j)
break
if i == 6 and j ^ temp == 80:
flag_vals.append(j)
break
if i == 8 and j ^ temp == 84:
flag_vals.append(j)
break
if i == 10 and j ^ temp == 45:
flag_vals.append(j)
break
return (flag_vals)
def to_decimal(enc):
dec = []
for i in range(0,len(enc),2):
temp = int(enc[i:i+2],16)
dec.append(temp)
return dec
def turn(vals):
vals = [vals[-1]] + vals[:-1]
return vals
def decrypt(enc, vals):
keystream = []
while len(keystream) < len(enc):
keystream += vals
vals = turn(vals)
return "".join(chr(enc[i] ^ keystream[i]) for i in range(len(enc)))
enc = "48ac3fe745bed053ac14ef2163cc70db58dfe862fe33860330dc22ea589c9f03d922e13365db038626eaee"
vals = find_first_six(enc)
dec = to_decimal(enc)
print(decrypt(dec, vals))
```
```shell
❯ python solve.py
OFPPT-CTF{X0r_w17h_sm4ll_k3y_vuln3r4b1l17y}
```
