189 lines
4.5 KiB
Python
189 lines
4.5 KiB
Python
# -*- coding: utf-8 -*-
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#
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# Copyright 2011 Sybren A. Stüvel <sybren@stuvel.eu>
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#
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# Licensed under the Apache License, Version 2.0 (the "License");
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# you may not use this file except in compliance with the License.
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# You may obtain a copy of the License at
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#
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# https://www.apache.org/licenses/LICENSE-2.0
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#
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# Unless required by applicable law or agreed to in writing, software
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# distributed under the License is distributed on an "AS IS" BASIS,
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# WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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# See the License for the specific language governing permissions and
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# limitations under the License.
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from rsa._compat import zip
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"""Common functionality shared by several modules."""
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class NotRelativePrimeError(ValueError):
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def __init__(self, a, b, d, msg=None):
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super(NotRelativePrimeError, self).__init__(
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msg or "%d and %d are not relatively prime, divider=%i" % (a, b, d))
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self.a = a
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self.b = b
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self.d = d
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def bit_size(num):
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"""
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Number of bits needed to represent a integer excluding any prefix
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0 bits.
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Usage::
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>>> bit_size(1023)
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10
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>>> bit_size(1024)
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11
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>>> bit_size(1025)
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11
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:param num:
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Integer value. If num is 0, returns 0. Only the absolute value of the
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number is considered. Therefore, signed integers will be abs(num)
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before the number's bit length is determined.
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:returns:
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Returns the number of bits in the integer.
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"""
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try:
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return num.bit_length()
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except AttributeError:
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raise TypeError('bit_size(num) only supports integers, not %r' % type(num))
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def byte_size(number):
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"""
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Returns the number of bytes required to hold a specific long number.
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The number of bytes is rounded up.
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Usage::
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>>> byte_size(1 << 1023)
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128
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>>> byte_size((1 << 1024) - 1)
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128
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>>> byte_size(1 << 1024)
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129
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:param number:
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An unsigned integer
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:returns:
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The number of bytes required to hold a specific long number.
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"""
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if number == 0:
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return 1
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return ceil_div(bit_size(number), 8)
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def ceil_div(num, div):
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"""
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Returns the ceiling function of a division between `num` and `div`.
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Usage::
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>>> ceil_div(100, 7)
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15
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>>> ceil_div(100, 10)
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10
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>>> ceil_div(1, 4)
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1
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:param num: Division's numerator, a number
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:param div: Division's divisor, a number
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:return: Rounded up result of the division between the parameters.
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"""
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quanta, mod = divmod(num, div)
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if mod:
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quanta += 1
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return quanta
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def extended_gcd(a, b):
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"""Returns a tuple (r, i, j) such that r = gcd(a, b) = ia + jb
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"""
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# r = gcd(a,b) i = multiplicitive inverse of a mod b
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# or j = multiplicitive inverse of b mod a
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# Neg return values for i or j are made positive mod b or a respectively
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# Iterateive Version is faster and uses much less stack space
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x = 0
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y = 1
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lx = 1
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ly = 0
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oa = a # Remember original a/b to remove
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ob = b # negative values from return results
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while b != 0:
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q = a // b
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(a, b) = (b, a % b)
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(x, lx) = ((lx - (q * x)), x)
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(y, ly) = ((ly - (q * y)), y)
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if lx < 0:
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lx += ob # If neg wrap modulo orignal b
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if ly < 0:
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ly += oa # If neg wrap modulo orignal a
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return a, lx, ly # Return only positive values
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def inverse(x, n):
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"""Returns the inverse of x % n under multiplication, a.k.a x^-1 (mod n)
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>>> inverse(7, 4)
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3
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>>> (inverse(143, 4) * 143) % 4
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1
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"""
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(divider, inv, _) = extended_gcd(x, n)
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if divider != 1:
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raise NotRelativePrimeError(x, n, divider)
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return inv
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def crt(a_values, modulo_values):
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"""Chinese Remainder Theorem.
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Calculates x such that x = a[i] (mod m[i]) for each i.
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:param a_values: the a-values of the above equation
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:param modulo_values: the m-values of the above equation
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:returns: x such that x = a[i] (mod m[i]) for each i
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>>> crt([2, 3], [3, 5])
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8
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>>> crt([2, 3, 2], [3, 5, 7])
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23
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>>> crt([2, 3, 0], [7, 11, 15])
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135
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"""
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m = 1
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x = 0
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for modulo in modulo_values:
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m *= modulo
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for (m_i, a_i) in zip(modulo_values, a_values):
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M_i = m // m_i
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inv = inverse(M_i, m_i)
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x = (x + a_i * M_i * inv) % m
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return x
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if __name__ == '__main__':
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import doctest
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doctest.testmod()
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