Python number.GCD屬性代碼示例

# 需要導入模塊: from Crypto.Util import number [as 別名]
# 或者: from Crypto.Util.number import GCD [as 別名]
def test_getStrongPrime(self):
        """Util.number.getStrongPrime"""
        self.assertRaises(ValueError, number.getStrongPrime, 256)
        self.assertRaises(ValueError, number.getStrongPrime, 513)
        bits = 512
        x = number.getStrongPrime(bits)
        self.assertNotEqual(x % 2, 0)
        self.assertEqual(x > (1L << bits-1)-1, 1)
        self.assertEqual(x < (1L << bits), 1)
        e = 2**16+1
        x = number.getStrongPrime(bits, e)
        self.assertEqual(number.GCD(x-1, e), 1)
        self.assertNotEqual(x % 2, 0)
        self.assertEqual(x > (1L << bits-1)-1, 1)
        self.assertEqual(x < (1L << bits), 1)
        e = 2**16+2
        x = number.getStrongPrime(bits, e)
        self.assertEqual(number.GCD((x-1)>>1, e), 1)
        self.assertNotEqual(x % 2, 0)
        self.assertEqual(x > (1L << bits-1)-1, 1)
        self.assertEqual(x < (1L << bits), 1) 

# 需要導入模塊: from Crypto.Util import number [as 別名]
# 或者: from Crypto.Util.number import GCD [as 別名]
def test_getStrongPrime(self):
        """Util.number.getStrongPrime"""
        self.assertRaises(ValueError, number.getStrongPrime, 256)
        self.assertRaises(ValueError, number.getStrongPrime, 513)
        bits = 512
        x = number.getStrongPrime(bits)
        self.assertNotEqual(x % 2, 0)
        self.assertEqual(x > (1 << bits-1)-1, 1)
        self.assertEqual(x < (1 << bits), 1)
        e = 2**16+1
        x = number.getStrongPrime(bits, e)
        self.assertEqual(number.GCD(x-1, e), 1)
        self.assertNotEqual(x % 2, 0)
        self.assertEqual(x > (1 << bits-1)-1, 1)
        self.assertEqual(x < (1 << bits), 1)
        e = 2**16+2
        x = number.getStrongPrime(bits, e)
        self.assertEqual(number.GCD((x-1)>>1, e), 1)
        self.assertNotEqual(x % 2, 0)
        self.assertEqual(x > (1 << bits-1)-1, 1)
        self.assertEqual(x < (1 << bits), 1) 

# 需要導入模塊: from Crypto.Util import number [as 別名]
# 或者: from Crypto.Util.number import GCD [as 別名]
def _nfold(str, nbytes):
    # Convert str to a string of length nbytes using the RFC 3961 nfold
    # operation.

    # Rotate the bytes in str to the right by nbits bits.
    def rotate_right(str, nbits):
        nbytes, remain = (nbits//8) % len(str), nbits % 8
        return ''.join(chr((ord(str[i-nbytes]) >> remain) |
                           ((ord(str[i-nbytes-1]) << (8-remain)) & 0xff))
                       for i in xrange(len(str)))

    # Add equal-length strings together with end-around carry.
    def add_ones_complement(str1, str2):
        n = len(str1)
        v = [ord(a) + ord(b) for a, b in zip(str1, str2)]
        # Propagate carry bits to the left until there aren't any left.
        while any(x & ~0xff for x in v):
            v = [(v[i-n+1]>>8) + (v[i]&0xff) for i in xrange(n)]
        return ''.join(chr(x) for x in v)

    # Concatenate copies of str to produce the least common multiple
    # of len(str) and nbytes, rotating each copy of str to the right
    # by 13 bits times its list position.  Decompose the concatenation
    # into slices of length nbytes, and add them together as
    # big-endian ones' complement integers.
    slen = len(str)
    lcm = nbytes * slen / gcd(nbytes, slen)
    bigstr = ''.join((rotate_right(str, 13 * i) for i in xrange(lcm / slen)))
    slices = (bigstr[p:p+nbytes] for p in xrange(0, lcm, nbytes))
    return reduce(add_ones_complement, slices) 

# 需要導入模塊: from Crypto.Util import number [as 別名]
# 或者: from Crypto.Util.number import GCD [as 別名]
def recover_rsa_modulus_from_signatures(m1, s1, m2, s2, e=0x10001):
   """
   Calculates the modulus used to produce RSA signatures from
   two known message/signature pairs and the public exponent.

   Since the most common public exponent is 65537, we default
   to that.
   
   Parameters:
   m1 - (string) The first message
   s1 - (string) The signature of the first message
      as an unencoded string
   m2 - (string) The second message
   s2 - (string) The signature of the second message
   e - (int) The exponent to use

   Returns the modulus as an integer, or False upon failure.
   """
   m1 = string_to_long(m1)
   s1 = string_to_long(s1)
   m2 = string_to_long(m2)
   s2 = string_to_long(s2)
   gcd_result = number.GCD( s1 ** e - m1, s2 ** e - m2 )

   if gcd_result < s1 or gcd_result < s2:
      # The modulus can never be smaller than our signature.
      # If this happens, we have been fed bad data.
      return False

   else:
      return int(gcd_result) 

# 需要導入模塊: from Crypto.Util import number [as 別名]
# 或者: from Crypto.Util.number import GCD [as 別名]
def crt(list_a, list_m):
    """
    Reference: https://crypto.stanford.edu/pbc/notes/numbertheory/crt.html
    Returns the output after computing Chinese Remainder Theorem on

    x = a_1 mod m_1
    x = a_2 mod m_2
    ...
    x = a_n mod m_n

    input parameter list_a = [a_1, a_2, ..., a_n]
    input parameter list_m = [m_1, m_2, ..., m_n]

    Returns -1 if the operation is unsuccessful due to some exceptions
    """
    try:
        assert len(list_a) == len(list_m)
    except:
        print "[+] Length of list_a should be equal to length of list_m"
        return -1
    for i in range(len(list_m)):
        for j in range(len(list_m)):
            if GCD(list_m[i], list_m[j])!= 1 and i!=j:
                print "[+] Moduli should be pairwise co-prime"
                return -1
    M = 1
    for i in list_m:
        M *= i
    list_b = [M/i for i in list_m]
    assert len(list_b) == len(list_m)
    try:
        list_b_inv = [int(gmpy2.invert(list_b[i], list_m[i])) for i in range(len(list_m))]
    except:
        print "[+] Encountered an unusual error while calculating inverse using gmpy2.invert()"
        return -1
    x = 0
    for i in range(len(list_m)):
        x += list_a[i]*list_b[i]*list_b_inv[i]
    return x % M 

# 需要導入模塊: from Crypto.Util import number [as 別名]
# 或者: from Crypto.Util.number import GCD [as 別名]
def neg_pow(a, b, n):
        assert b < 0
        assert GCD(a, n) == 1
        res = int(gmpy2.invert(a, n))
        res = pow(res, b*(-1), n)
        return res 

# 需要導入模塊: from Crypto.Util import number [as 別名]
# 或者: from Crypto.Util.number import GCD [as 別名]
def rsa_construct(n, e, d=None, p=None, q=None, u=None):
    """Construct an RSAKey object"""
    assert isinstance(n, long)
    assert isinstance(e, long)
    assert isinstance(d, (long, type(None)))
    assert isinstance(p, (long, type(None)))
    assert isinstance(q, (long, type(None)))
    assert isinstance(u, (long, type(None)))
    obj = _RSAKey()
    obj.n = n
    obj.e = e
    if d is None:
        return obj
    obj.d = d
    if p is not None and q is not None:
        obj.p = p
        obj.q = q
    else:
        # Compute factors p and q from the private exponent d.
        # We assume that n has no more than two factors.
        # See 8.2.2(i) in Handbook of Applied Cryptography.
        ktot = d*e-1
        # The quantity d*e-1 is a multiple of phi(n), even,
        # and can be represented as t*2^s.
        t = ktot
        while t%2==0:
            t=divmod(t,2)[0]
        # Cycle through all multiplicative inverses in Zn.
        # The algorithm is non-deterministic, but there is a 50% chance
        # any candidate a leads to successful factoring.
        # See "Digitalized Signatures and Public Key Functions as Intractable
        # as Factorization", M. Rabin, 1979
        spotted = 0
        a = 2
        while not spotted and a<100:
            k = t
            # Cycle through all values a^{t*2^i}=a^k
            while k<ktot:
                cand = pow(a,k,n)
                # Check if a^k is a non-trivial root of unity (mod n)
                if cand!=1 and cand!=(n-1) and pow(cand,2,n)==1:
                    # We have found a number such that (cand-1)(cand+1)=0 (mod n).
                    # Either of the terms divides n.
                    obj.p = GCD(cand+1,n)
                    spotted = 1
                    break
                k = k*2
            # This value was not any good... let's try another!
            a = a+2
        if not spotted:
            raise ValueError("Unable to compute factors p and q from exponent d.")
        # Found !
        assert ((n % obj.p)==0)
        obj.q = divmod(n,obj.p)[0]
    if u is not None:
        obj.u = u
    else:
        obj.u = inverse(obj.p, obj.q)
    return obj