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# -*- 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()
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