Writeup for solved challenge in DragonSectorCTF 2020
CRYPTO
Bit Flip I-:
description:
Flip bits and decrypt communication between Bob and Alice.
nc bitflip1.hackable.software 1337
#!/usr/bin/python3
from Crypto.Util.number import bytes_to_long, long_to_bytes
from Crypto.Cipher import AES
import hashlib
import os
import base64
from gmpy2 import is_prime
FLAG = open("flag").read()
FLAG += (16 - (len(FLAG) % 16))*" "
class Rng:
def __init__(self, seed):
self.seed = seed
self.generated = b""
self.num = 0
def more_bytes(self):
self.generated += hashlib.sha256(self.seed).digest()
self.seed = long_to_bytes(bytes_to_long(self.seed) + 1, 32)
self.num += 256
def getbits(self, num=64):
while (self.num < num):
self.more_bytes()
x = bytes_to_long(self.generated)
self.num -= num
self.generated = b""
if self.num > 0:
self.generated = long_to_bytes(x >> num, self.num // 8)
return x & ((1 << num) - 1)
class DiffieHellman:
def gen_prime(self):
prime = self.rng.getbits(512)
iter = 0
while not is_prime(prime):
iter += 1
prime = self.rng.getbits(512)
print("Generated after", iter, "iterations")
return prime
def __init__(self, seed, prime=None):
self.rng = Rng(seed)
if prime is None:
prime = self.gen_prime()
self.prime = prime
self.my_secret = self.rng.getbits()
self.my_number = pow(5, self.my_secret, prime)
self.shared = 1337
def set_other(self, x):
self.shared ^= pow(x, self.my_secret, self.prime)
def pad32(x):
return (b"\x00"*32+x)[-32:]
def xor32(a, b):
return bytes(x^y for x, y in zip(pad32(a), pad32(b)))
def bit_flip(x):
print("bit-flip str:")
flip_str = base64.b64decode(input().strip())
return xor32(flip_str, x)
alice_seed = os.urandom(16)
while 1:
alice = DiffieHellman(bit_flip(alice_seed))
bob = DiffieHellman(os.urandom(16), alice.prime)
alice.set_other(bob.my_number)
print("bob number", bob.my_number)
bob.set_other(alice.my_number)
iv = os.urandom(16)
print(base64.b64encode(iv).decode())
cipher = AES.new(long_to_bytes(alice.shared, 16)[:16], AES.MODE_CBC, IV=iv)
enc_flag = cipher.encrypt(FLAG)
print(base64.b64encode(enc_flag).decode())
Solution:
On analysing the challenge script we can deduce that Diffie Hellman Key exchange was done by Bob with alice multiple times. We only have the control over bit-flip and as the challenge description suggested we have to make the use of it to get the alice seed (because recovering seed for Bob is not easy).
There’s a strange piece of information was given to us in the form of number of iterations used to calculate prime.
Let’s have a look over Rng as how prime was generated: In getbits function, this block was never visited when 512 is sent to self.num:
if self.num > 0:
self.generated = long_to_bytes(x >> num, self.num // 8)
thus allowing self.generated to be clean again for the next call . That’s the flaw in the code which means prime for seed = i and i+2 will be same.
Manually testing the code:
seed = b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'
Generated after 83 iterations
prime = 3217336996812784199323541050098699361781489187527078355681535168764692913032949200158631425936108602790839091441050033248993143847385123136499734649619637
seed = b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x02'
Generated after 82 iterations
prime = 3217336996812784199323541050098699361781489187527078355681535168764692913032949200158631425936108602790839091441050033248993143847385123136499734649619637
So if we flip the second last bit then it’s 0 if the iteration count decreased otherwise 1.
The implementation for rest of the bytes are tricky and in the end we have to brute for the last byte as its either 0 or 1.
Implementation:
Suppose we find upto X’th bit , then we have to guess for .....x10100b (last bit is b and it’s undecided)
we can have iteration count for .....x00000b
and
for .....011111b (by flipping the bits)
as the difference between them is 2 so we will have one less iteration count if the bit was 1 otherwise 0.
Solution script:
In beginning we have to solve POW which is same as asked in GoogleCTF’17.
from pwn import *
from base64 import *
from Crypto.Util.number import *
from Crypto.Cipher import AES
import subprocess
from task import Rng , DiffieHellman
def POW(r):
print("----------Solving POW-----------")
command = r.recv().strip().split(b": ")[-1]
hashed = subprocess.check_output(command,shell=True).strip()
r.sendline(hashed)
print("----------Solved----------------")
def sendloop(r,i: int):
r.recv()
a = b64encode(long_to_bytes(i,32))
r.sendline(a)
iters = int(r.recvline().strip().split(b" ")[2])
bob_number = int(r.recvline().strip().split(b" ")[-1]) # bob number
iv = b64decode(r.recvline().strip()) # IV
ciphertext = b64decode(r.recvline().strip()) # ciphertext
return iters,bob_number,iv,ciphertext
def get_flag(seed,bob_number,iv,ciphertext):
alice = DiffieHellman(long_to_bytes(int(seed,2)))
shared = pow(bob_number,alice.my_secret,alice.prime)
cipher = AES.new(long_to_bytes(shared,16)[:16] , AES.MODE_CBC , IV= iv)
flag = cipher.decrypt(ciphertext)
return flag
r = remote("bitflip1.hackable.software",1337)
POW(r)
bits = ""
for pos in range(1,128):
nums = 0 if bits=='' else int(bits,2)*2
flippedbit = 1<<pos | ((nums//2)^((1<<(pos-1))-1))*2
iters1, iters2 = sendloop(r,nums)[0], sendloop(r,flippedbit)[0]
assert iters2 != 0 # to check they generate the same prime, if not rerun script
bits = "1" + bits if iters1 +1 == iters2 else "0" + bits
iters,bob_number,iv,ciphertext = sendloop(r,0)
#guessing LSB
flag = get_flag(bits+"0",bob_number,iv,ciphertext)
print(flag)
flag = get_flag(bits+"1",bob_number,iv,ciphertext)
print(flag)
r.close()
Running the script gives the result:
/DragonSectorCTF$ python solve.py
[+] Opening connection to bitflip1.hackable.software on port 1337: Done
----------Solving POW-----------
----------Solved----------------
Generated after 286 iterations
b'DrgnS{T1min9_4ttack_f0r_k3y_generation}\n '
Generated after 253 iterations
b'XI\x18T\x1c\xd0m\x81\xc8\x06=\xb81\x93\xa9\x01X\xb3}\xf06\xea\xf2\x95_\x87E\xa2\x14z\x9d\xbd;1\xd1\x01\xd6\xc4))\x1bO\xe7\xf0\xbaxeC'
[*] Closed connection to bitflip1.hackable.software port 1337
Here is our flag: DrgnS{T1min9_4ttack_f0r_k3y_generation}
Voila!
PS: Yeah the count of iteration indicates the time used for generating the prime.