Rating:
https://abdullahzamir.github.io/posts/UIUCTF-2024-X-Marked-the-Spot/
For the challenge we were given and encrypted file and the encryption algorithm
```python
from itertools import cycle
flag = b"uiuctf{????????????????????????????????????????}"
# len(flag) = 48
key = b"????????"
# len(key) = 8
ct = bytes(x ^ y for x, y in zip(flag, cycle(key)))
with open("ct", "wb") as ct_file:
ct_file.write(ct)
```
This is simple XOR, and we can retrieve the key using the association property of the XOR operation
as we know some part of the plaintext, which is `uiuctf` and key is ok 8 charaters, to get the full key we only need to bruteforce one character
#### Solve Script:
```python
from itertools import cycle
from pwn import xor
f = open('ct','rb')
flag = f.read()
pt = bytes.fromhex("uiuctf{".encode().hex())
key = xor(flag[:7],pt)
for i in range(0xff):
key = key + bytes([i])
pt = xor(flag,key)
if b'uiuctf{' in pt:
if b'}' in pt[len(pt)-1:]:
print(pt)
key = key[:-1]
```
