Update libs

libs/rsa/common.py

@@ -0,0 +1,185 @@
+# -*- coding: utf-8 -*-
+#
+#  Copyright 2011 Sybren A. Stüvel <sybren@stuvel.eu>
+#
+#  Licensed under the Apache License, Version 2.0 (the "License");
+#  you may not use this file except in compliance with the License.
+#  You may obtain a copy of the License at
+#
+#      http://www.apache.org/licenses/LICENSE-2.0
+#
+#  Unless required by applicable law or agreed to in writing, software
+#  distributed under the License is distributed on an "AS IS" BASIS,
+#  WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
+#  See the License for the specific language governing permissions and
+#  limitations under the License.
+
+'''Common functionality shared by several modules.'''
+
+
+def bit_size(num):
+    '''
+    Number of bits needed to represent a integer excluding any prefix
+    0 bits.
+
+    As per definition from http://wiki.python.org/moin/BitManipulation and
+    to match the behavior of the Python 3 API.
+
+    Usage::
+    
+        >>> bit_size(1023)
+        10
+        >>> bit_size(1024)
+        11
+        >>> bit_size(1025)
+        11
+
+    :param num:
+        Integer value. If num is 0, returns 0. Only the absolute value of the
+        number is considered. Therefore, signed integers will be abs(num)
+        before the number's bit length is determined.
+    :returns:
+        Returns the number of bits in the integer.
+    '''
+    if num == 0:
+        return 0
+    if num < 0:
+        num = -num
+
+    # Make sure this is an int and not a float.
+    num & 1
+
+    hex_num = "%x" % num
+    return ((len(hex_num) - 1) * 4) + {
+        '0':0, '1':1, '2':2, '3':2,
+        '4':3, '5':3, '6':3, '7':3,
+        '8':4, '9':4, 'a':4, 'b':4,
+        'c':4, 'd':4, 'e':4, 'f':4,
+     }[hex_num[0]]
+
+
+def _bit_size(number):
+    '''
+    Returns the number of bits required to hold a specific long number.
+    '''
+    if number < 0:
+        raise ValueError('Only nonnegative numbers possible: %s' % number)
+
+    if number == 0:
+        return 0
+    
+    # This works, even with very large numbers. When using math.log(number, 2),
+    # you'll get rounding errors and it'll fail.
+    bits = 0
+    while number:
+        bits += 1
+        number >>= 1
+
+    return bits
+
+
+def byte_size(number):
+    '''
+    Returns the number of bytes required to hold a specific long number.
+    
+    The number of bytes is rounded up.
+
+    Usage::
+
+        >>> byte_size(1 << 1023)
+        128
+        >>> byte_size((1 << 1024) - 1)
+        128
+        >>> byte_size(1 << 1024)
+        129
+
+    :param number:
+        An unsigned integer
+    :returns:
+        The number of bytes required to hold a specific long number.
+    '''
+    quanta, mod = divmod(bit_size(number), 8)
+    if mod or number == 0:
+        quanta += 1
+    return quanta
+    #return int(math.ceil(bit_size(number) / 8.0))
+
+
+def extended_gcd(a, b):
+    '''Returns a tuple (r, i, j) such that r = gcd(a, b) = ia + jb
+    '''
+    # r = gcd(a,b) i = multiplicitive inverse of a mod b
+    #      or      j = multiplicitive inverse of b mod a
+    # Neg return values for i or j are made positive mod b or a respectively
+    # Iterateive Version is faster and uses much less stack space
+    x = 0
+    y = 1
+    lx = 1
+    ly = 0
+    oa = a                             #Remember original a/b to remove 
+    ob = b                             #negative values from return results
+    while b != 0:
+        q = a // b
+        (a, b)  = (b, a % b)
+        (x, lx) = ((lx - (q * x)),x)
+        (y, ly) = ((ly - (q * y)),y)
+    if (lx < 0): lx += ob              #If neg wrap modulo orignal b
+    if (ly < 0): ly += oa              #If neg wrap modulo orignal a
+    return (a, lx, ly)                 #Return only positive values
+
+
+def inverse(x, n):
+    '''Returns x^-1 (mod n)
+
+    >>> inverse(7, 4)
+    3
+    >>> (inverse(143, 4) * 143) % 4
+    1
+    '''
+
+    (divider, inv, _) = extended_gcd(x, n)
+
+    if divider != 1:
+        raise ValueError("x (%d) and n (%d) are not relatively prime" % (x, n))
+
+    return inv
+
+
+def crt(a_values, modulo_values):
+    '''Chinese Remainder Theorem.
+
+    Calculates x such that x = a[i] (mod m[i]) for each i.
+
+    :param a_values: the a-values of the above equation
+    :param modulo_values: the m-values of the above equation
+    :returns: x such that x = a[i] (mod m[i]) for each i
+    
+
+    >>> crt([2, 3], [3, 5])
+    8
+
+    >>> crt([2, 3, 2], [3, 5, 7])
+    23
+
+    >>> crt([2, 3, 0], [7, 11, 15])
+    135
+    '''
+
+    m = 1
+    x = 0 
+
+    for modulo in modulo_values:
+        m *= modulo
+
+    for (m_i, a_i) in zip(modulo_values, a_values):
+        M_i = m // m_i
+        inv = inverse(M_i, m_i)
+
+        x = (x + a_i * M_i * inv) % m
+
+    return x
+
+if __name__ == '__main__':
+    import doctest
+    doctest.testmod()
+