stuff

training/IntroZ3/tasks/chall1_rot128/README.md

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+This crypto challenge was part of the HTB University CTF we played in 2024
+
+The problem is basically the following (all values are 128 bitvectors):
+given h1 and h2, find r1,r2,r3,r4,s1,s2, such that:
+ROL(s1, r1) ^ ROL(s2, r2) == h1
+ROL(s1, r3) ^ ROL(s2, r4) == h2
+
+z3 doesn't seem to be able to solve this directly in a fast manner (challenge expects solution in less than 2 seconds)
+
+The crypto problem was then "reduced" to the following:
+
+Given any value x with parity 0, find y and c, such that (y ^ (ROL(y,c))) == x
+Whether this even has a (good) solution, would also not seem clear, but sometimes it's worth just asking z3...
+
+Find a solution for this problem with z3 in `z3test.py`
+
+If you want to get an actual flag, copy paste the `compute_yc` function in the exploit.py under "TODO" and run
+`./exploit.py courses.sec.in.tum.de 13371`

training/IntroZ3/tasks/chall1_rot128/exploit.py

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+#!/usr/bin/env python
+
+import pwn
+import sys
+import re
+import binascii
+import random
+
+N = 128
+_ROL_ = lambda x, i : ((x << i) | (x >> (N-i))) & (2**N - 1)
+_ROR_ = lambda x, i : ((x >> i) | (x << (N-i))) & (2**N - 1)
+
+class HashRoll:
+    def __init__(self):
+        self.reset_state()
+
+    def hash_step(self, i):
+        r1, r2 = self.state[2*i], self.state[2*i+1]
+        r = _ROL_(self.state[-2], r1) ^ _ROL_(self.state[-1], r2)
+        print(f'h{i} = {r}')
+        return r
+
+    def update_state(self, state=None):
+        if not state:
+            self.state = [0] * 6
+            self.state[:4] = [random.randint(0, N) for _ in range(4)]
+            self.state[-2:] = [random.randint(0, 2**N) for _ in range(2)]
+        else:
+            self.state = state
+    
+    def reset_state(self):
+        self.update_state()
+
+    def digest(self, buffer):
+        buffer = int.from_bytes(buffer, byteorder='big')
+        m1 = buffer >> N
+        m2 = buffer & (2**N - 1)
+        self.h = b''
+        for i in range(2):
+            self.h += int.to_bytes(self.hash_step(i) ^ (m1 if not i else m2), length=N//8, byteorder='big')
+        return self.h
+
+
+def parity(x):
+    res = 0
+    while x:
+        res ^= x & 1
+        x >>= 1
+    return res
+
+
+def compute_yc(x):
+    import z3
+    s = z3.Solver()
+    y = z3.BitVec("y", N)
+    c = z3.BitVec("c", N)
+    s.add((y ^ z3.RotateLeft(y,c)) == x)
+    s.check()
+    m = s.model()
+
+    return m[y].as_long(), m[c].as_long()
+
+# h = m1 ^ h(0) + m2 ^h(1)
+# h(x) = rol(s1, vx1) ^ rol(s2, vx2)
+
+# given x, y is it possible to find:
+# s1, s2, r1, r2, r3, r4,
+# so that rol(s1, r1) ^ rol(s2, r2) = x
+# and     rol(s1, r3) ^ rol(s2, r4) = y
+# r1 = 0, r2 = 0, r3 = 0
+# x = s1 ^ s2
+# rol(s1, r3) ^ rol(s2, r4) = y
+
+# h0 = rol(s1, r1) ^ rol(s2, r2)
+# s1 = (h0^y)
+# r1, r2 = 0
+# s2 = y
+# r3 = 0, r4 = c
+def solve_round(h1, h2):
+    """
+    Return r1,r2,r3,r4,s1,s2 such that 
+    _ROL_(s1, r1) ^ _ROL(s2, r2) == h1
+    _ROL_(s1, r3) ^ _ROL(s2, r4) == h2
+    """
+    y,c = compute_yc(h1^h2)
+
+    s1 = h1^y
+    s2 = y
+    r1, r2, r3, r4 = 0, 0, 0, c
+    j = 1
+
+
+    # ensure non-zero vals
+    return [r1+j, r2+j, r3+j, (r4+j) % N, _ROR_(s1, j), _ROR_(s2, j)]
+
+r = pwn.remote(sys.argv[1], int(sys.argv[2]))
+
+
+
+
+hashfunc = HashRoll()
+
+
+
+
+ROUNDS = 3
+
+for i in range(ROUNDS):
+    r.recvuntil(b"/3!\n")
+    l = r.recvuntil(b"\n").decode()
+    print(l)
+
+
+    m = re.search(r'H\(([0-9a-f]+)\) = ([0-9a-f]+)', l)
+
+
+    message, h = map(lambda x: int(x, 16), m.groups())
+
+    m1 = message >> N
+    m2 = message & (2**N - 1)
+
+    th1 = h >> N
+    th2 = h & (2**N - 1)
+
+    print(f'H({hex(message)[2:]}) = {hex(h)[2:]}')
+
+    h0 = th1 ^ m1
+    h1 = th2 ^ m2
+
+    state = solve_round(h0, h1)
+    hashfunc.update_state(state)
+    print(hashfunc.digest(message.to_bytes(N, byteorder='big')))
+    print(h.to_bytes(2*N//8, byteorder='big'))
+    print(hashfunc.digest(message.to_bytes(N, byteorder='big')) == h.to_bytes(2*N//8, byteorder='big'))
+    print(state)
+
+    r.recvuntil(b' :: ')
+
+    r.sendline(b','.join(map(lambda x: str(x).encode(), state)))
+
+    print(r.recvline().decode())
+print(r.recvline().decode())
+
+    
+
+    # we want hashstep(0) to be m3 ^ m1
+    # we want hashstep(1) to be m4 ^ m2
+
+

training/IntroZ3/tasks/chall1_rot128/server.py

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+import random, os, signal
+from Crypto.Util.number import long_to_bytes as l2b, bytes_to_long as b2l
+from secret import FLAG
+
+ROUNDS = 3
+USED_STATES = []
+_ROL_ = lambda x, i : ((x << i) | (x >> (N-i))) & (2**N - 1)
+N = 128
+
+def handler(signum, frame):
+    print("\n\nToo slow, don't try to do sneaky things.")
+    exit()
+
+def validate_state(state):
+    if not all(0 < s < 2**N-1 for s in user_state[-2:]) or not all(0 <= s < N for s in user_state[:4]):
+        print('Please, make sure your input satisfies the upper and lower bounds.')
+        return False
+    
+    if sorted(state[:4]) in USED_STATES:
+        print('You cannot reuse the same state')
+        return False
+    
+    if sum(user_state[:4]) < 2:
+        print('We have to deal with some edge cases...')
+        return False
+
+    return True
+
+class HashRoll:
+    def __init__(self):
+        self.reset_state()
+
+    def hash_step(self, i):
+        r1, r2 = self.state[2*i], self.state[2*i+1]
+        return _ROL_(self.state[-2], r1) ^ _ROL_(self.state[-1], r2)
+
+    def update_state(self, state=None):
+        if not state:
+            self.state = [0] * 6
+            self.state[:4] = [random.randint(0, N) for _ in range(4)]
+            self.state[-2:] = [random.randint(0, 2**N) for _ in range(2)]
+        else:
+            self.state = state
+    
+    def reset_state(self):
+        self.update_state()
+
+    def digest(self, buffer):
+        buffer = int.from_bytes(buffer, byteorder='big')
+        m1 = buffer >> N
+        m2 = buffer & (2**N - 1)
+        self.h = b''
+        for i in range(2):
+            self.h += int.to_bytes(self.hash_step(i) ^ (m1 if not i else m2), length=N//8, byteorder='big')
+        return self.h
+
+
+print('Can you test my hash function for second preimage resistance? You get to select the state and I get to choose the message ... Good luck!')
+
+hashfunc = HashRoll()
+
+for _ in range(ROUNDS):
+    print(f'ROUND {_+1}/{ROUNDS}!')
+
+    server_msg = os.urandom(32)
+    hashfunc.reset_state()
+    server_hash = hashfunc.digest(server_msg)
+    print(f'You know H({server_msg.hex()}) = {server_hash.hex()}')
+
+    signal.signal(signal.SIGALRM, handler)
+    signal.alarm(2)
+
+    user_state = input('Send your hash function state (format: a,b,c,d,e,f) :: ').split(',')
+
+    try:
+        user_state = list(map(int, user_state))
+
+        if not validate_state(user_state):
+            print("The state is not valid! Try again.")
+            exit()
+
+        hashfunc.update_state(user_state)
+
+        if hashfunc.digest(server_msg) == server_hash:
+            print(f'Moving on to the next round!')
+            USED_STATES.append(sorted(user_state[:4]))
+        else:
+            print('Not today.')
+            exit()
+    except:
+        print("The hash function's state must be all integers.")
+        exit()
+    finally:
+       signal.alarm(0)
+
+print(f'Uhm... how did you do that? I thought I had cryptanalyzed it enough ... {FLAG}')

training/IntroZ3/tasks/chall1_rot128/z3test.py

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+import random
+import z3
+
+_ROL_ = lambda x, i : ((x << i) | (x >> (N-i))) & (2**N - 1)
+
+
+N = 128
+
+def parity(x):
+    res = 0
+    while x:
+        res ^= x & 1
+        x >>= 1
+    return res
+
+
+while True:
+    x = random.randint(0, 2**128)
+    if parity(x) == 0:
+        break
+
+# Compute y and c, such that y ^ _ROL_(y,c) == x
+# Hint: to implement _ROL_, z3 already has the function z3.RotateLeft
+def compute_yc(x):
+    s = z3.Solver()
+    y = z3.BitVec("y", N)
+    c = z3.BitVec("c", N)
+    s.add((y ^ z3.RotateLeft(y,c)) == x)
+    s.check()
+    m = s.model()
+
+    return m[y].as_long(), m[c].as_long()
+
+
+y, c = compute_yc(x)
+
+print(f'{y=}, {c=}')
+print(f'{x=}, {y ^ _ROL_(y, c)=}')
+print(f'{x==y ^ _ROL_(y, c)=}')