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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
#
#      https://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 https://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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