# Crypto - Monkfish

## Challenge

```python
#!/usr/bin/sage
import sys
print("I caught a monkfish in the sea! ")
sys.stdout.flush()
from hashlib import sha256
from Crypto.Util.number import bytes_to_long
from random import SystemRandom
import ast
def verif_pok(v, F, pi):
    com = pi[0]
    resp = pi[1]
    verif = pi[2]
    a = list(FF)[sha256(bytes([list(FF).index(i[0]) for i in list(com) + list(v) + list(verif)])).digest()[0] % len(list(FF))]
    out1 = apply(F, resp)
    out2 = com + (a * a) * v - a * verif
    return out1 == out2
s = random_matrix(FF, n, 1)
m0 = matrix(FF, m, 1, [0]*100)
m1 = matrix(FF, n, 1, [0]*100)
m2 = v
pi = (m0, m1, m2)
res = verif_pok(v, F, pi)
```

I omit the whole question as we can use this method to solve the question in a small number of attempts. A more general and proper solution is given in the [next part](/ctf/corctf-2024/crypto-anglerfish.md).

We give `m0,m1,m2` as input and we need the proof of knowledge to be verified.

Suppose we give `resp` to be 0, then `out1` becomes 0. Then we need to get `out2` to be 0.  We do this by setting `com` to zero and `verif` to v. However, this method only works if the value of a becomes 0 or 1, which has a 40% chance based on the value of v


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