Пример #1
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def recover_key_from_nonce(msg, sig, k, q, hash=sha1):
    """Recovers a DSA private key (`x`) from a message-signature-pair given a known nonce

    Example:
    ```python
    >>> from bop.crypto_constructor import dsa
    >>> my_dsa = dsa()
    >>> msg = b"Hello kind stranger!"
    >>> sig = my_dsa.sign(msg)
    >>> leaked_nonce = sig._k
    >>> my_dsa.x == recover_key_from_nonce(msg, sig, leaked_nonce, my_dsa.q, hash=my_dsa.hash)
    True

    ```

    Arguments:
        msg {bytes or int} -- The message which was signed
        sig {tuple or Signature} -- The signature of the given message in form `(r, s)`
        k {int} -- The leaked nonce
        q {int} -- The public parameter q
    """
    if type(msg) != bytes:
        msg = i2b(msg)

    h = b2i(hash(msg))
    r, s = sig

    return ((s * k - h) * invmod(r, q)) % q
Пример #2
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    def __call__(self, msg):
        plain = self.rsa.decrypt(msg)

        if type(plain) == int:
            plain = i2b(plain)

        # the conversion will hide the first byte if it is zero
        first_byte_is_zero = len(plain) == (self.rsa.key_size // 8 - 1)

        return first_byte_is_zero and plain[0] == 2
Пример #3
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def _dsa_sign(msg, p, q, g, x, k, hash=hash):
    if type(msg) != bytes:
        msg = i2b(msg)

    h = b2i(hash(msg))

    r = pow(g, k, p) % q
    s = (invmod(k, q) * (h + r * x)) % q

    return r, s
Пример #4
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    def decrypt(self, cipher):
        was_bytes = False
        if type(cipher) != int:
            cipher = b2i(cipher)
            was_bytes = True

        m = pow(cipher, self.d, self.n)

        if was_bytes:
            m = i2b(m)

        return m
Пример #5
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    def encrypt(self, plaintext):
        was_bytes = False
        if type(plaintext) != int:
            plaintext = b2i(plaintext)
            was_bytes = True

        c = pow(plaintext, self.e, self.n)

        if was_bytes:
            c = i2b(c)

        return c
Пример #6
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    def verify(self, msg, sig):
        r, s = sig
        if r <= 0 or r >= self.q or s <= 0 or s >= self.q:
            return False

        if type(msg) != bytes:
            msg = i2b(msg)

        h = b2i(self.hash(msg))
        w = invmod(s, self.q)
        u1 = (w * h) % self.q
        u2 = (w * r) % self.q

        v = (pow(self.g, u1, self.p) * pow(self.y, u2, self.p)) % self.q
        return v == r
Пример #7
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    def sign(self, msg):
        if type(msg) != bytes:
            msg = i2b(msg)

        h = b2i(self.hash(msg))
        s = 0
        while s == 0:
            r = 0
            while r == 0:
                if self.generate_nonce is not None:
                    k = self.generate_nonce()
                else:
                    k = secrets.randbelow(self.q - 2) + 2

                r = pow(self.g, k, self.p) % self.q
            s = (invmod(k, self.q) * (h + r * self.x)) % self.q

        return self.Signature(r, s, k)
Пример #8
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Файл: rsa.py Проект: liona24/bop
def bleichenbacher_forge_signature(message,
                                   key_size,
                                   hash=sha1,
                                   protocol=b"DUMMY"):
    r"""Forge a signature RSA (e=3) signature for the given message.

    This method can be used to forge signatures if the checking is incorrectly implemented.

    For this example a dummy PKCS1.5 padding is applied (see argument `protocol`).
    This padding right-aligns the content (i.e. the hash of the message and some
    protocol information). However an incorrect implementation may not check if
    the signature is actually right-aligned.

    You may need a minimum key size for this to work. 1024 bits should be sufficient.

    Example:
    ```python3
    >>> import re
    >>> from bop.hashing import sha1
    >>> from bop.crypto_constructor import rsa
    >>> def check(message, sig, rsa):
    ...     c = rsa.encrypt(sig)
    ...     # note that we are ignoring trailing bytes
    ...     # also note that the leading zero byte gets trimmed
    ...     h = re.match(br"^\x01(\xff)*?\x00DUMMY(.{20})", c, re.DOTALL).group(2)
    ...     return h == sha1(message)
    ...
    >>> some_rsa = rsa(e=3)
    >>> my_msg = b"My very valid message"
    >>> my_sig = bleichenbacher_forge_signature(my_msg, some_rsa.key_size)
    >>> check(my_msg, my_sig, some_rsa)
    True

    ```

    Arguments:
        message {bytes} -- The message to sign
        key_size {int} -- The size of the public key parameter N in bits

    Keyword Arguments:
        hash {callable} -- The hash algorithm to use (default: {sha1})
        protocol {bytes} -- The bytes to use which specify meta data. In reality this would be some kind of ASN.1 scheme but for this toy example it does not really matter. (default: {b"DUMMY"})

    Returns:
        bytes -- The signature of the message
    """
    # we fake this. In reality a little bit of effort has to be made in order to specify the
    # hash algorithm used, size of the hash etc.
    h = hash(message)
    payload = b"\x00" + protocol + h
    payload_len = len(payload) * 8

    # this is more or less a heuristic, place digest at about 2/3
    position = key_size // 8 // 3 * 16

    n = (1 << payload_len) - b2i(payload)
    c = (1 << (key_size - 15)) - n * (1 << position)

    root = cubic_root2(c)
    if len(root) == 1:
        # we are lucky
        root = root[0]
    else:
        root = root[-1]

    return i2b(root)
Пример #9
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Файл: rsa.py Проект: liona24/bop
def decrypt_pkcs_padding_leak(oracle, msg, e, n):
    """Perform an adaptive chosen ciphertext against a weak RSA implementation leaking PKCS1v5 padding information.

    This attack was discovered by Daniel Bleichenbacher and is well described in the original [paper](http://archiv.infsec.ethz.ch/education/fs08/secsem/bleichenbacher98.pdf).

    Side note:\
    This attack is generally faster if `msg` is already properly padded.
    Depending on the keysize the expected running time may be somewhere between a few seconds (key size <= 512 bits) or somewhere around
    one minute for a 1024 public modulus.

    I once tested using a 2048 bit key which ran about an hour..\
    I am pretty sure this implementation is far from optimal, though I am not qualified to judge where I went wrong.


    Example:
    ```python3
    >>> from bop.oracles.padding import PaddingRSAOracle as Oracle
    >>> import bop.utils as utils
    >>> # We will choose smaller params to keep the running time down a little bit
    >>> p, q = utils.gen_rsa_key_params(256)
    >>> o = Oracle(p, q)
    >>> e, n = o.public_key()
    >>> plain = b"Hey! This message is save!"
    >>> # Optional: Pad the message:
    >>> # plain = utils.pad_pkcs1v5(plain, n)
    >>> msg = o.encrypt(plain)
    >>> decrypt_pkcs_padding_leak(o, msg, e, n)
    b'Hey! This message is save!'

    ```


    Args:
        oracle (callable): An oracle which leaks whether the decrypted message has a valid PKCS1v5 padding or not
            (i. e. the first byte is equal to 0 and the second byte is equal to 2)
        msg (bytes or int): The message to decrypt
        e (int): The public exponent
        n (int): The public modulus

    Returns:
        bytes or int: The decrypted message. Depending on the input type the output type is matched.
    """

    was_bytes = False
    if type(msg) != int:
        msg = b2i(msg)
        was_bytes = True

    k = bit_length_exp2(n)
    assert k > 16

    B = 1 << (k - 16)

    s0 = 1
    c0 = msg

    first = True

    M = {(2 * B, 3 * B - 1)}
    if not oracle(msg):
        # Blinding
        # ensure we have a valid padding to work with
        while True:
            s0 = secrets.randbelow(n - 1) + 1
            c0 = (msg * pow(s0, e, n)) % n
            if oracle(c0):
                break

    while True:
        if first:
            # Step 2a
            s = n // (3 * B)
            while oracle((c0 * pow(s, e, n)) % n) is False:
                s += 1
            first = False
        elif len(M) > 1:
            # Step 2b
            s_ = s + 1
            while not oracle((c0 * pow(s_, e, n)) % n):
                s_ += 1
            s = s_
        else:
            # Step 2c
            a, b = next(iter(M))
            found = False
            r = 2 * (b * s - 2 * B) // n
            while not found:
                s_ = (2 * B + r * n) // b

                s_max = (3 * B + r * n) // a

                while s_ <= s_max:
                    if oracle((c0 * pow(s_, e, n)) % n):
                        found = True
                        break
                    s_ += 1

                r += 1

            s = s_

        # Step 3
        M_ = set()
        for a, b in M:
            r_low = (a * s - 3 * B + 1) // n
            r_high = (b * s - 2 * B) // n
            for r in range(r_low, r_high + 1):
                # note the + s - 1 in order to ensure rounding to the next integer
                low = max(a, (2 * B + r * n + s - 1) // s)
                high = min(b, (3 * B - 1 + r * n) // s)

                if low <= high and (low, high) not in M_:
                    M_.add((low, high))

        M = M_

        # Step 4
        if len(M) == 1:
            a, b = next(iter(M))
            if a == b:
                plain = (a * invmod(s0, n)) % n
                if was_bytes:
                    return i2b(plain)
                return plain
Пример #10
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Файл: rsa.py Проект: liona24/bop
def decrypt_parity_leak(oracle, msg, e, n):
    """Performs a RSA-parity attack given an oracle which reports whether a decrypted plaintext is even or odd

    Example:
    ```python
    >>> from bop.oracles.parity import RsaParityOracle as Oracle
    >>> o = Oracle()
    >>> msg = o.encrypt(b"Hello Bob!")
    >>> e, n = o.public_key()
    >>> decrypt_parity_leak(o, msg, e, n)
    b'Hello Bob!'

    ```

    Args:
        oracle (callable): A function which decrypts ciphers which were encrypted using the given public key and reports whether the resulting plain text is even (`True`) or odd (`False`)
        msg (bytes or int): The encrypted message to decrypt
        e (int): The public key exponent e
        n (int): The public key modul n

    Raises:
        RuntimeError: If the plain text cannot be recovered. This is propably caused by rounding errors

    Returns:
        bytes or int: The decrypted message. Depending on the type given (`msg`) the resulting type matches.
    """
    was_bytes = False
    if type(msg) != int:
        was_bytes = True
        msg = b2i(msg)

    i = 2
    upper = n
    lower = 0

    # we will use this to carry the rounding error, though this is VERY vague
    # the desired plain text is usually off by only 1 anyways
    rest = 1

    while i <= n:
        # encrypt i
        f = pow(i, e, n)

        if oracle(f * msg):
            # even
            upper = (upper + lower) // 2
            rest += (upper + lower) & 1
        else:
            # odd
            lower = (upper + lower) // 2

        i <<= 1

    for plain in range(lower, upper + rest):
        if pow(plain, e, n) == msg:
            if was_bytes:
                return i2b(plain)
            else:
                return plain

    raise RuntimeError("Could not find plain text. Something went wrong :(")
Пример #11
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def recover_key_from_duplicate_nonce(gen, public_parameters, hash=sha1):
    """Attempt to recover a DSA private key (`x`) from a stream of messages by searching for a duplicate usage of a nonce.

    Note that this is a pretty naive implementation running in O(n^2) where n is the number of messages (signatures) checked.

    Example:
    ```python
    >>> from bop.crypto_constructor import dsa
    >>> import secrets
    >>>
    >>> # Setup a weak implementation
    >>> my_dsa = dsa()
    >>> def weak_nonce_generator():
    ...     return secrets.choice(range(2, 11))
    ...
    >>> my_dsa.generate_nonce = weak_nonce_generator
    >>> def message_src():
    ...     while True:
    ...         msg = secrets.token_bytes(16)
    ...         yield msg, my_dsa.sign(msg)
    ...
    >>> x, leaked_nonce = recover_key_from_duplicate_nonce(message_src(), my_dsa.public_parameters, hash=my_dsa.hash)
    >>> my_dsa.x == x
    True

    ```

    Arguments:
        gen {generator} -- A generator yielding message-signature-pairs `(msg, signature)`.
            It will be consumed until a duplicate nonce is found or it is empty.
        public_parameters {tuple} -- The public parameters of the DSA algorithm used

    Keyword Arguments:
        hash {callable} -- The hash function to use (default: {sha1})

    Returns:
        (int, int) -- (x, leaked_nonce), i.e. the private key and the leaked nonce. Or (None, None) if no duplicate was found.
    """

    p, q, g, y = public_parameters

    first_msg, first_sig = next(gen)
    if type(first_msg) != bytes:
        first_msg = i2b(first_msg)
    bases = [(first_msg, first_sig)]

    # we do a simple exhaustive search for a duplicated nonce
    for msg1, sig1 in gen:
        if type(msg1) != bytes:
            msg1 = i2b(msg1)
        h1 = b2i(hash(msg1))
        r1, s1 = sig1

        for msg2, sig2 in bases:
            h2 = b2i(hash(msg2))
            r2, s2 = sig2
            # Assume k was equal
            k = ((h2 - h1) * invmod(s2 - s1, q)) % q

            # Check if our assumption holds
            x = recover_key_from_nonce(msg1, sig1, k, q, hash=hash)

            if _dsa_sign(msg1, p, q, g, x, k, hash=hash) == (r1, s1):
                return x, k

        bases.append((msg1, sig1))

    return None, None
Пример #12
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 def __str__(self):
     return binascii.hexlify(i2b(self.r)) + "." + binascii.hexlify(
         i2b(self.s))