Mapna CTF

Crytography/What next?

Challenge Description:

In this task, we explore the realm of cryptographically secure random generators, where predicting the next output is deemed impossible. Are you ready to test your luck and skill?

Provided Files:

• output.txt

  what_next.py

Solution: The output.txt file provides the values of TMP, KEY, and enc.

The what_next.py is a program that generates a key (KEY), encrypts the flag with this key.

#!/usr/bin/env python3

from random import *
from Crypto.Util.number import *
from flag import flag

def encrypt(msg, KEY):
	m = bytes_to_long(msg)
	c = KEY ^ m
	return c

n = 80
TMP = [getrandbits(256) * _ ** 2 for _ in range(n)]
KEY = sum([getrandbits(256 >> _) for _ in range(8)]) 

enc = encrypt(flag, KEY)

print(f'TMP = {TMP}')
print(f'KEY = {KEY}')
print(f'enc = {enc}')

The encryption function is a simple XOR operation, Therefore, the decryption function is also an XOR operation with the same key.

from Crypto.Util.number import long_to_bytes

def decrypt(enc, KEY):
    m = enc ^ KEY
    msg = long_to_bytes(m)
    return msg

enc = 2290064970177041546889165766737348623235283630135906565145883208626788551598431732
KEY = 23226475334448992634882677537728533150528705952262010830460862502359965393545

flag = decrypt(enc, KEY)
print(f'flag = {flag}')

Flag:

MAPNA{R_U_MT19937_PRNG_Predictor?}