Dice CTF Quals 2024
Winter [116 pts]
A simple implementation of the Winternitz signature scheme.
nc mc.ax 31001
We’re provided a single file, server.py:
#!/usr/local/bin/python
import os
from hashlib import sha256
class Wots:
def __init__(self, sk, vk):
self.sk = sk
self.vk = vk
@classmethod
def keygen(cls):
sk = [os.urandom(32) for _ in range(32)]
vk = [cls.hash(x, 256) for x in sk]
return cls(sk, vk)
@classmethod
def hash(cls, x, n):
for _ in range(n):
x = sha256(x).digest()
return x
def sign(self, msg):
m = self.hash(msg, 1)
sig = b''.join([self.hash(x, 256 - n) for x, n in zip(self.sk, m)])
return sig
def verify(self, msg, sig):
chunks = [sig[i:i+32] for i in range(0, len(sig), 32)]
m = self.hash(msg, 1)
vk = [self.hash(x, n) for x, n in zip(chunks, m)]
return self.vk == vk
if __name__ == '__main__':
with open('flag.txt') as f:
flag = f.read().strip()
wots = Wots.keygen()
msg1 = bytes.fromhex(input('give me a message (hex): '))
sig1 = wots.sign(msg1)
assert wots.verify(msg1, sig1)
print('here is the signature (hex):', sig1.hex())
msg2 = bytes.fromhex(input('give me a new message (hex): '))
if msg1 == msg2:
print('cheater!')
exit()
sig2 = bytes.fromhex(input('give me the signature (hex): '))
if wots.verify(msg2, sig2):
print(flag)
else:
print('nope')
Given that this problem is about the Winternitz signature scheme, I decided to look it up. The diagram on this site explained pretty much everything about the algorithm succinctly.
Notably, the Winternitz signature scheme is a one-time signature scheme. Meaning, using it again could pose problems. However, looking at the provided source code, it seemed that the service always generated a completely random key upon connection, and we are only given 1 encryption before we are tasked to find our own.
In fact, the implementation in the source file seemed to be a standard Winternitz signature scheme. I was left a bit confused for a while as I tried to look for potential vulnerabilities in the source file or search for them online. Unfortunately, the source code provided nothing apparently obvious to exploit, and all the research papers online seemed to say that Winternitz was very secure.
After a while of searching, I ended up stumbling upon this paper. On page 5, it explains the following:
If the WOTS scheme were used just with the \(l_1\) hash chains representing the message digest \(m\), an adversary could trivially sign any message, where the digest \(r\) consists only of chunks \(r_i\), where \(r_i ≥ m_i\), \(∀i ∈ [0, l_1 − 1]\). This is because the adversary gains information about intermediate hash chain nodes from the original signature. Information that was prior to the signing operation, private. The adversary can simply advance all signature nodes by \(r_i − mi\) to forge a signature.
Basically, this exploit relies on the process of the WOTS scheme, a.k.a. the Winternitz One-Time Signature. Learn how it works before moving on in this writeup! Then the writeup will make a lot more sense.
Let \(n\) represent each byte of the sha256 hash of a message. WOTS hashes each of the 32 private keys \(256-n\) times to obtain the signature, and verifies the signature by hashing each of its 32-byte blocks \(n\) times. Therfore, given message \(m_1\) and \(m_2\), we can essentially forge the signature if \(256-n_1 < 256-n_2\) for every single 32-byte block. Note that here, \(n_1\) represents the byte 1,2,3…32 of \(m_1\)’s sha256 hash, and similarly \(n_2\) does the same with \(m_2\). The comparison occurs between the corresponding bytes, i.e. byte 1 of \(m_1\) and byte 1 of \(m_2\), etc.
This is because, if the aforementioned condition is true, we can simply hash each 32-byte block of the signature that the signature oracle/service provides to us the necessary number of times until it is essentially as if we took each private key block and hashed it \(256-n_2\), for each \(n_2\) of the sha256 hash of \(m_2\).
Here’s an example. Let’s say the first byte of the sha256 hash of \(m_1\) is 137, and the first byte of \(m_2\)’s is 80. The first 32-byte block of the signature of \(m_1\) is obtained by repeatedly hashing the first 32-byte private key \(256-137=119\) times. Meanwhile, in order to obtain the first 32-byte block of \(m_2\)’s signature, we’d have to repeatedly hash the first 32-byte private key \(256-80=176\) times.
Notice how we can essentially just hash the first 32-byte block of \(m_1\)’s signature \(176-119=57\) more times to obtain the first 32-byte block of \(m_2\)’s signature? That’s the key idea. And if \(256-n_1 < 256-n_2\) for each byte of the sha256 hashes of \(m_1\) and \(m_2\), that means we can do this for every single 32-byte block of the signature!
So all we need to do now is find a pair of messages such that their sha256 hashes satisfy this condition, where \(m_1\) is greater than \(m_2\) for every corresponding byte. We can actually very simply brute force this. Essentially, I tried to greedily select a message where the minimum byte of the sha256 hash was the maximum possible, and another message where the maximum byte of the sha256 hash was the minimum possible. At each step, if either message changed, I would check if it satisfied the aforementioned condition for all 32-byte blocks, i.e. the sha256 hash with the greater bytes was greater than the sha256 hash with smaller bytes for every single corresponding byte pair. It’s not a perfect greedy algorithm – I’d love to know if there is actually a better algorithm to do this – but it worked out in the end! Here was my implementation of it:
from hashlib import sha256
largest_min_value_of_maxlist = 0
smallest_max_value_of_minlist = 256
minlist = []
maxlist = []
min_msg = ''
max_msg = ''
for i in range(10000000):
h = sha256(str(i).encode()).digest()
mx = max(list(h))
mn = min(list(h))
smallest_max_value_of_minlist = min(smallest_max_value_of_minlist, mx)
if smallest_max_value_of_minlist == mx:
# print("Min:", minsum, list(h), end='\n\n')
minlist = list(h)
min_msg = str(i).encode()
if maxlist == []: continue
s = ''
for i in range(32):
s += ('1' if minlist[i] < maxlist[i] else '0')
print(s)
if s == '1'*32: break
largest_min_value_of_maxlist = max(largest_min_value_of_maxlist, mn)
if largest_min_value_of_maxlist == mn:
# print("Max:", maxsum, list(h), end='\n\n')
maxlist = list(h)
max_msg = str(i).encode()
s = ''
for i in range(32):
s += '1' if minlist[i] < maxlist[i] else '0'
print(s)
if s == '1'*32: break
for i in range(32):
print(1 if minlist[i] < maxlist[i] else 0, end='')
print()
print(min_msg, minlist)
print(max_msg, maxlist)
print(smallest_max_value_of_minlist, largest_min_value_of_maxlist)
This ended up returning the following messages and corresponding hash bytes:
msg1 = b'1973198'
msg2 = b'752802'
msg1_hash = [144, 246, 206, 112, 109, 246, 159, 156, 133, 235, 192, 200, 226, 105, 136, 87, 207, 242, 207, 117, 130, 177, 187, 166, 152, 86, 219, 116, 247, 147, 129, 94]
msg2_hash = [72, 126, 124, 84, 46, 60, 148, 99, 45, 75, 40, 120, 38, 95, 7, 82, 57, 29, 32, 90, 91, 73, 27, 117, 92, 32, 134, 90, 77, 77, 92, 80]
Notice how every byte in the msg1_hash is greater than its corresponding byte in the msg2_hash? That’s exactly what we need. Now, we can very easily create a short script to interact with the remote service to execute the exploit!
from hashlib import sha256
from pwn import *
import binascii
msg1 = binascii.hexlify(b'1973198')
msg2 = binascii.hexlify(b'752802')
msg1_hash = [144, 246, 206, 112, 109, 246, 159, 156, 133, 235, 192, 200, 226, 105, 136, 87, 207, 242, 207, 117, 130, 177, 187, 166, 152, 86, 219, 116, 247, 147, 129, 94]
msg2_hash = [72, 126, 124, 84, 46, 60, 148, 99, 45, 75, 40, 120, 38, 95, 7, 82, 57, 29, 32, 90, 91, 73, 27, 117, 92, 32, 134, 90, 77, 77, 92, 80]
diffs = [msg1_hash[i] - msg2_hash[i] for i in range(32)]
print(diffs)
def hash(x, n):
for _ in range(n):
x = sha256(x).digest()
return x
p = remote('mc.ax', 31001)
p.sendlineafter(b': ', msg1)
p.recvuntil(b': ')
sig1 = p.recvline().decode('ascii')[:-1]
sig1 = binascii.unhexlify(sig1)
chunks = [sig1[i:i+32] for i in range(0, len(sig1), 32)]
vk = [hash(x, n) for x, n in zip(chunks, diffs)]
for i in range(len(vk)):
vk[i] = binascii.hexlify(vk[i]).decode('ascii')
sig2 = "".join(vk)
p.sendlineafter(b': ', msg2)
p.sendlineafter(b': ', sig2.encode())
print(p.recvline().decode("ascii"))
And with that, we run the script and get the flag!
dice{according_to_geeksforgeeks}
Thoughts: Quite a nice crypto challenge with a cleverly hidden vulnerability in the problem. Perhaps I even could’ve figure it out myself after some time. I still think it was a great solve on my part regardless!