The Algorithms logo
The Algorithms
À proposFaire un don

Chinese Remainder Theorem

P
d
p
"""
Chinese Remainder Theorem:
GCD ( Greatest Common Divisor ) or HCF ( Highest Common Factor )

If GCD(a,b) = 1, then for any remainder ra modulo a and any remainder rb modulo b
there exists integer n, such that n = ra (mod a) and n = ra(mod b).  If n1 and n2 are
two such integers, then n1=n2(mod ab)

Algorithm :

1. Use extended euclid algorithm to find x,y such that a*x + b*y = 1
2. Take n = ra*by + rb*ax
"""

from __future__ import annotations


# Extended Euclid
def extended_euclid(a: int, b: int) -> tuple[int, int]:
    """
    >>> extended_euclid(10, 6)
    (-1, 2)

    >>> extended_euclid(7, 5)
    (-2, 3)

    """
    if b == 0:
        return (1, 0)
    (x, y) = extended_euclid(b, a % b)
    k = a // b
    return (y, x - k * y)


# Uses ExtendedEuclid to find inverses
def chinese_remainder_theorem(n1: int, r1: int, n2: int, r2: int) -> int:
    """
    >>> chinese_remainder_theorem(5,1,7,3)
    31

    Explanation : 31 is the smallest number such that
                (i)  When we divide it by 5, we get remainder 1
                (ii) When we divide it by 7, we get remainder 3

    >>> chinese_remainder_theorem(6,1,4,3)
    14

    """
    (x, y) = extended_euclid(n1, n2)
    m = n1 * n2
    n = r2 * x * n1 + r1 * y * n2
    return (n % m + m) % m


# ----------SAME SOLUTION USING InvertModulo instead ExtendedEuclid----------------


# This function find the inverses of a i.e., a^(-1)
def invert_modulo(a: int, n: int) -> int:
    """
    >>> invert_modulo(2, 5)
    3

    >>> invert_modulo(8,7)
    1

    """
    (b, x) = extended_euclid(a, n)
    if b < 0:
        b = (b % n + n) % n
    return b


# Same a above using InvertingModulo
def chinese_remainder_theorem2(n1: int, r1: int, n2: int, r2: int) -> int:
    """
    >>> chinese_remainder_theorem2(5,1,7,3)
    31

    >>> chinese_remainder_theorem2(6,1,4,3)
    14

    """
    x, y = invert_modulo(n1, n2), invert_modulo(n2, n1)
    m = n1 * n2
    n = r2 * x * n1 + r1 * y * n2
    return (n % m + m) % m


if __name__ == "__main__":
    from doctest import testmod

    testmod(name="chinese_remainder_theorem", verbose=True)
    testmod(name="chinese_remainder_theorem2", verbose=True)
    testmod(name="invert_modulo", verbose=True)
    testmod(name="extended_euclid", verbose=True)