# rev/Scrambled

We were given a main.py that encrypted the flag in two steps:
1. Each character of the flag is XORed with a single-byte key.
2. The encrypted data is split into chunks of 4 bytes which are then shuffled using a random seed (0-10) and then converted to hex before giving as output.

The given output was:
```bash
1e78197567121966196e757e1f69781e1e1f7e736d6d1f75196e75191b646e196f6465510b0b0b57
```

So in order to reverse the encryption we could write a simple brute script that brutes both the seed and the XOR key to find the correct flag.

```python
import random

def decode_flag(scrambled_result, key, seed):
chunk_size = 4
chunks = [scrambled_result[i:i+chunk_size] for i in range(0, len(scrambled_result), chunk_size)]

random.seed(seed)
shuffled_indices = list(range(len(chunks)))
random.shuffle(shuffled_indices)

unshuffled_chunks = [None] * len(chunks)
for i, chunk in enumerate(chunks):
unshuffled_chunks[shuffled_indices[i]] = chunk

xor_result = [item for chunk in unshuffled_chunks for item in chunk]

flag = "".join([chr(c ^ key) for c in xor_result])
return flag

def main():
scrambled_hex = "1e78197567121966196e757e1f69781e1e1f7e736d6d1f75196e75191b646e196f6465510b0b0b57"

scrambled_result = [int(scrambled_hex[i:i+2], 16) for i in range(0, len(scrambled_hex), 2)]

for seed in range(11):
for key in range(256):
flag = decode_flag(scrambled_result, key, seed)
if all(32 <= ord(c) <= 126 for c in flag):
print(f"Seed: {seed}, Key: {key}, Flag: {flag}")

if __name__ == "__main__":
main()
```

On running this we can brute and hence, reverse the encryption to get our flag.

```bash
ENO{5CR4M83L3D_3GG5_4R3_1ND33D_T45TY!!!}
```