Rating:
# Farm - cryptoCTF 2021
- Category: Crypto
- Points: 41
- Solves: 149
- Solved by: drw0if
## Description
Explore the Farm very carefully!
## Solution
In this challenge we were given two text files:
- farm.py: a sagemath script that generates a key and encrypts the flag
- enc.txt: the encrypted flag yelds by the previous script
First thing first: we are dealing with `Galois finite field` since the scripts creates this global object:
```python
F = list(GF(64))
```
Let's see the key generation:
```python
def keygen(l):
key = [F[randint(1, 63)] for _ in range(l)]
key = math.prod(key) # Optimization the key length :D
return key
```
At the end of the function we are given a single value in `GF(64)` but this field has only 64 values so we can brute-force the key.
We have to inverse the encryption function and we can go:
```python
for c in enc:
x = ALPHABET.index(c) # c = ALHPABET[x]
y = F[x] # x = F.index(y)
z = farmtomap(y/pkey) # y = pkey * maptofarm(z)
m += z
```
At the end the flag was:
```
CCTF{EnCrYp7I0n_4nD_5u8STitUtIn9_iN_Fi3Ld!}
```
