"""
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 typing import Tuple
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)
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
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
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)