mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-18 01:00:15 +00:00
95 lines
2.2 KiB
Python
95 lines
2.2 KiB
Python
"""
|
|
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
|
|
|
|
|
|
# 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)
|