Пример #1
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 def decrypt(self, rsa_blob):
     """Expects numeric argument, not string."""
     this_hash = sha1(str(rsa_blob)).digest()
     for logged_hash in self.log:
         if this_hash == logged_hash:
             return "GO AWAY."
     self.log.append(this_hash)
     return rsa.decrypt_string(rsa_blob, self.priv)
Пример #2
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 def decrypt(self, rsa_blob):
     """Expects numeric argument, not string."""
     this_hash = sha1(str(rsa_blob)).digest()
     for logged_hash in self.log:
         if this_hash == logged_hash:
             return "GO AWAY."
     self.log.append(this_hash)
     return rsa.decrypt_string(rsa_blob, self.priv)
Пример #3
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def run_tests(m):
    """
    Small sanity test suite
    """
    menc = PKCS1_encode(m, k)

    print("1. (un)pad:", PKCS1_decode(menc) == m)

    m1 = rsa.decrypt_string(sk, rsa.encrypt_string(pk, m))
    print("2. rsa w/o pad:", m == m1)

    m2 = PKCS1_decode(rsa.decrypt_string(sk, rsa.encrypt_string(pk, menc)))
    print("3. rsa w/ pad:", m == m2)

    m3 = oracle(rsa.encrypt_string(pk, menc)) == True
    print("4. oracle well-formed:", m3)

    m4 = oracle(rsa.encrypt_string(pk, m)) == False
    print("5. oracle not well-formed", m4)
Пример #4
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def oracle(ciphertext, privkey, bits):
    """bits should equal the max bits of a message, not bit length of
    key.
    """
    plaintext = rsa.decrypt_string(ciphertext, privkey)
    assert bits % 8 == 0
    bytes = bits / 8
    diff = bytes - len(plaintext)
    plaintext = "\x00" * diff + plaintext
    assert len(plaintext) == bytes, len(plaintext)
    return plaintext[0] == "\x00" and plaintext[1] == "\x02"
Пример #5
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def oracle(ciphertext):
    """
    Placeholder for some server which talks RSA PKCS1 v1.5
    It can be used as an oracle, because it tells whether
    the given ciphertext decodes to a valid PKCS1 v1.5 encoding scheme,
    i.e. first 2 bytes of the plaintext == "\x00\x02"
    """
    global queries

    queries += 1
    t = time.perf_counter()
    if queries % 500 == 0:
        print("Query #{} ({} s)".format(queries, round(t - t_start, 3)))

    encoded = rsa.decrypt_string(sk, ciphertext)

    if len(encoded) > k:
        raise Exception("Invalid PKCS1 encoding after decryption!")

    if len(encoded) < k:
        zero_pad = b"\x00" * (k - len(encoded))
        encoded = zero_pad + encoded

    return encoded[0:2] == b"\x00\x02"
Пример #6
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def verify(sig, message, pubkey):
    block = rsa.decrypt_string(sig, pubkey)
    for i in range((bits/8) - len(block)):
        block = "\x00" + block
    return sha1(message).digest() == unpad(block)
Пример #7
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k = 3  # How many times to encrypt the same plaintext, under different
# public keys.

message = 'Hello, world! I am gonna encrypt this thrice; uh oh.'
bits = len(message) * 8 / 2
c = [None] * k
n = [None] * k
for i in range(k):
    U, R = rsa.keypair(bits)
    ciphertext = rsa.encrypt_string(message, U)
    c[i] = ciphertext
    n[i] = U[1]  # the second part of the pubkey
    print "public     " + str(U[1])[:60] + "...."
    print "ciphertext " + str(ciphertext)[:60] + "...."

decrypt = rsa.decrypt_string(ciphertext, R)
print
print "Bob gets this message:", decrypt

#### Eve

# Calculate products of the moduli (pubkeys) EXCEPT pubkey number i.
ms = [None] * k
for i in range(k):
    x = copy.copy(n)
    del x[i]
    ms[i] = reduce(lambda a, b: a * b, x)

# Work thru Chinese Remainder Theorem
result = 0
for i in range(k):
Пример #8
0
k = 3 # How many times to encrypt the same plaintext, under different
      # public keys.

message = 'Hello, world! I am gonna encrypt this thrice; uh oh.'
bits = len(message) * 8 / 2
c = [None]*k
n = [None]*k
for i in range(k):
    U, R = rsa.keypair(bits)
    ciphertext = rsa.encrypt_string(message, U)
    c[i] = ciphertext
    n[i] = U[1] # the second part of the pubkey
    print "public     " + str(U[1])[:60] + "...."
    print "ciphertext " + str(ciphertext)[:60] + "...."

decrypt = rsa.decrypt_string(ciphertext, R)
print
print "Bob gets this message:", decrypt

#### Eve

# Calculate products of the moduli (pubkeys) EXCEPT pubkey number i.
ms = [None]*k
for i in range(k):
    x = copy.copy(n)
    del x[i]
    ms[i] = reduce(lambda a, b: a*b, x)

# Work thru Chinese Remainder Theorem
result = 0
for i in range(k):