예제 #1
0
    def undigest(self, blocks):
        """undigest(blocks : [string]) : string

        Perform the reverse package transformation on a list of message
        blocks.  Note that the ciphermodule used for both transformations
        must be the same.  blocks is a list of strings of bit length
        equal to the ciphermodule's block_size.
        """

        # better have at least 2 blocks, for the padbytes package and the hash
        # block accumulator
        if len(blocks) < 2:
            raise ValueError("List must be at least length 2.")

        # blocks is a list of strings.  We need to deal with them as long
        # integers
        blocks = list(map(bytes_to_long, blocks))

        # Calculate the well-known key, to which the hash blocks are
        # encrypted, and create the hash cipher.
        K0 = self.__K0digit * self.__key_size
        hcipher = self.__newcipher(K0)
        block_size = self.__ciphermodule.block_size

        # Since we have all the blocks (or this method would have been called
        # prematurely), we can calculate all the hash blocks.
        hashes = []
        for i in range(1, len(blocks)):
            mticki = blocks[i-1] ^ i
            hi = hcipher.encrypt(long_to_bytes(mticki, block_size))
            hashes.append(bytes_to_long(hi))

        # now we can calculate K' (key).  remember the last block contains
        # m's' which we don't include here
        key = blocks[-1] ^ reduce(operator.xor, hashes)

        # and now we can create the cipher object
        mcipher = self.__newcipher(long_to_bytes(key, self.__key_size))

        # And we can now decode the original message blocks
        parts = []
        for i in range(1, len(blocks)):
            cipherblock = mcipher.encrypt(long_to_bytes(i, block_size))
            mi = blocks[i-1] ^ bytes_to_long(cipherblock)
            parts.append(mi)

        # The last message block contains the number of pad bytes appended to
        # the original text string, such that its length was an even multiple
        # of the cipher's block_size.  This number should be small enough that
        # the conversion from long integer to integer should never overflow
        padbytes = int(parts[-1])
        text = b('').join(map(long_to_bytes, parts[:-1]))
        return text[:-padbytes]
예제 #2
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def MGF1(mgfSeed, maskLen, hash):
    """Mask Generation Function, described in B.2.1"""
    T = b("")
    for counter in range(ceil_div(maskLen, hash.digest_size)):
        c = long_to_bytes(counter, 4)
        T = T + hash.new(mgfSeed + c).digest()
    assert (len(T) >= maskLen)
    return T[:maskLen]
예제 #3
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 def _lengthOctets(self, payloadLen):
     """Return a byte string that encodes the given payload length (in
             bytes) in a format suitable for a DER length tag (L).
             """
     if payloadLen > 127:
         encoding = long_to_bytes(payloadLen)
         return bchr(len(encoding) + 128) + encoding
     return bchr(payloadLen)
예제 #4
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def generateQ(randfunc):
    S = randfunc(20)
    hash1 = SHA.new(S).digest()
    hash2 = SHA.new(long_to_bytes(bytes_to_long(S) + 1)).digest()
    q = bignum(0)
    for i in range(0, 20):
        c = bord(hash1[i]) ^ bord(hash2[i])
        if i == 0:
            c = c | 128
        if i == 19:
            c = c | 1
        q = q * 256 + c
    while (not isPrime(q)):
        q = q + 2
    if pow(2, 159) < q < pow(2, 160):
        return S, q
    raise RuntimeError('Bad q value generated')
예제 #5
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파일: RSA.py 프로젝트: BarryYBL/Crypro
    def exportKey(self, format='PEM', passphrase=None, pkcs=1):
        """Export this RSA key.

        :Parameter format: The format to use for wrapping the key.

            - *'DER'*. Binary encoding, always unencrypted.
            - *'PEM'*. Textual encoding, done according to `RFC1421`_/`RFC1423`_.
              Unencrypted (default) or encrypted.
            - *'OpenSSH'*. Textual encoding, done according to OpenSSH specification.
              Only suitable for public keys (not private keys).
        :Type format: string

        :Parameter passphrase: In case of PEM, the pass phrase to derive the encryption key from.
        :Type passphrase: string 

        :Parameter pkcs: The PKCS standard to follow for assembling the key.
         You have two choices:

          - with **1**, the public key is embedded into an X.509 `SubjectPublicKeyInfo` DER SEQUENCE.
            The private key is embedded into a `PKCS#1`_ `RSAPrivateKey` DER SEQUENCE.
            This mode is the default.
          - with **8**, the private key is embedded into a `PKCS#8`_ `PrivateKeyInfo` DER SEQUENCE.
            This mode is not available for public keys.

         PKCS standards are not relevant for the *OpenSSH* format.
        :Type pkcs: integer

        :Return: A byte string with the encoded public or private half.
        :Raise ValueError:
            When the format is unknown.

        .. _RFC1421:    http://www.ietf.org/rfc/rfc1421.txt
        .. _RFC1423:    http://www.ietf.org/rfc/rfc1423.txt
        .. _`PKCS#1`:   http://www.ietf.org/rfc/rfc3447.txt
        .. _`PKCS#8`:   http://www.ietf.org/rfc/rfc5208.txt
        """
        if passphrase is not None:
            passphrase = tobytes(passphrase)
        if format == 'OpenSSH':
            eb = long_to_bytes(self.e)
            nb = long_to_bytes(self.n)
            if bord(eb[0]) & 0x80: eb = bchr(0x00) + eb
            if bord(nb[0]) & 0x80: nb = bchr(0x00) + nb
            keyparts = ['ssh-rsa', eb, nb]
            keystring = ''.join(
                [struct.pack(">I", len(kp)) + kp for kp in keyparts])
            return 'ssh-rsa ' + binascii.b2a_base64(keystring)[:-1]

        # DER format is always used, even in case of PEM, which simply
        # encodes it into BASE64.
        der = DerSequence()
        if self.has_private():
            keyType = {1: 'RSA PRIVATE', 8: 'PRIVATE'}[pkcs]
            der[:] = [
                0, self.n, self.e, self.d, self.p, self.q,
                self.d % (self.p - 1), self.d % (self.q - 1),
                inverse(self.q, self.p)
            ]
            if pkcs == 8:
                derkey = der.encode()
                der = DerSequence([0])
                der.append(algorithmIdentifier)
                der.append(DerObject('OCTET STRING', derkey).encode())
        else:
            keyType = "PUBLIC"
            der.append(algorithmIdentifier)
            bitmap = DerObject('BIT STRING')
            derPK = DerSequence([self.n, self.e])
            bitmap.payload = bchr(0x00) + derPK.encode()
            der.append(bitmap.encode())
        if format == 'DER':
            return der.encode()
        if format == 'PEM':
            pem = b("-----BEGIN " + keyType + " KEY-----\n")
            objenc = None
            if passphrase and keyType.endswith('PRIVATE'):
                # We only support 3DES for encryption
                import Crypro.Hash.MD5
                from Crypro.Cipher import DES3
                from Crypro.Protocol.KDF import PBKDF1
                salt = self._randfunc(8)
                key = PBKDF1(passphrase, salt, 16, 1, Crypro.Hash.MD5)
                key += PBKDF1(key + passphrase, salt, 8, 1, Crypro.Hash.MD5)
                objenc = DES3.new(key, Crypro.Cipher.DES3.MODE_CBC, salt)
                pem += b('Proc-Type: 4,ENCRYPTED\n')
                pem += b('DEK-Info: DES-EDE3-CBC,') + binascii.b2a_hex(
                    salt).upper() + b('\n\n')

            binaryKey = der.encode()
            if objenc:
                # Add PKCS#7-like padding
                padding = objenc.block_size - len(
                    binaryKey) % objenc.block_size
                binaryKey = objenc.encrypt(binaryKey + bchr(padding) * padding)

            # Each BASE64 line can take up to 64 characters (=48 bytes of data)
            chunks = [
                binascii.b2a_base64(binaryKey[i:i + 48])
                for i in range(0, len(binaryKey), 48)
            ]
            pem += b('').join(chunks)
            pem += b("-----END " + keyType + " KEY-----")
            return pem
        return ValueError(
            "Unknown key format '%s'. Cannot export the RSA key." % format)
예제 #6
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 def encode(self):
     """Return a complete INTEGER DER element, fully encoded as a TLV."""
     self.payload = long_to_bytes(self.value)
     if bord(self.payload[0]) > 127:
         self.payload = bchr(0x00) + self.payload
     return DerObject.encode(self)
예제 #7
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    def digest(self, text):
        """digest(text:string) : [string]

        Perform the All-or-Nothing package transform on the given
        string.  Output is a list of message blocks describing the
        transformed text, where each block is a string of bit length equal
        to the ciphermodule's block_size.
        """

        # generate a random session key and K0, the key used to encrypt the
        # hash blocks.  Rivest calls this a fixed, publically-known encryption
        # key, but says nothing about the security implications of this key or
        # how to choose it.
        key = self._inventkey(self.__key_size)
        K0 = self.__K0digit * self.__key_size

        # we need two cipher objects here, one that is used to encrypt the
        # message blocks and one that is used to encrypt the hashes.  The
        # former uses the randomly generated key, while the latter uses the
        # well-known key.
        mcipher = self.__newcipher(key)
        hcipher = self.__newcipher(K0)

        # Pad the text so that its length is a multiple of the cipher's
        # block_size.  Pad with trailing spaces, which will be eliminated in
        # the undigest() step.
        block_size = self.__ciphermodule.block_size
        padbytes = block_size - (len(text) % block_size)
        text = text + b(' ') * padbytes

        # Run through the algorithm:
        # s: number of message blocks (size of text / block_size)
        # input sequence: m1, m2, ... ms
        # random key K' (`key' in the code)
        # Compute output sequence: m'1, m'2, ... m's' for s' = s + 1
        # Let m'i = mi ^ E(K', i) for i = 1, 2, 3, ..., s
        # Let m's' = K' ^ h1 ^ h2 ^ ... hs
        # where hi = E(K0, m'i ^ i) for i = 1, 2, ... s
        #
        # The one complication I add is that the last message block is hard
        # coded to the number of padbytes added, so that these can be stripped
        # during the undigest() step
        s = divmod(len(text), block_size)[0]
        blocks = []
        hashes = []
        for i in range(1, s+1):
            start = (i-1) * block_size
            end = start + block_size
            mi = text[start:end]
            assert len(mi) == block_size
            cipherblock = mcipher.encrypt(long_to_bytes(i, block_size))
            mticki = bytes_to_long(mi) ^ bytes_to_long(cipherblock)
            blocks.append(mticki)
            # calculate the hash block for this block
            hi = hcipher.encrypt(long_to_bytes(mticki ^ i, block_size))
            hashes.append(bytes_to_long(hi))

        # Add the padbytes length as a message block
        i = i + 1
        cipherblock = mcipher.encrypt(long_to_bytes(i, block_size))
        mticki = padbytes ^ bytes_to_long(cipherblock)
        blocks.append(mticki)

        # calculate this block's hash
        hi = hcipher.encrypt(long_to_bytes(mticki ^ i, block_size))
        hashes.append(bytes_to_long(hi))

        # Now calculate the last message block of the sequence 1..s'.  This
        # will contain the random session key XOR'd with all the hash blocks,
        # so that for undigest(), once all the hash blocks are calculated, the
        # session key can be trivially extracted.  Calculating all the hash
        # blocks requires that all the message blocks be received, thus the
        # All-or-Nothing algorithm succeeds.
        mtick_stick = bytes_to_long(key) ^ reduce(operator.xor, hashes)
        blocks.append(mtick_stick)

        # we convert the blocks to strings since in Python, byte sequences are
        # always represented as strings.  This is more consistent with the
        # model that encryption and hash algorithms always operate on strings.
        return [long_to_bytes(i,self.__ciphermodule.block_size) for i in blocks]