Python/blockchain/chinese_remainder_theorem.py

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# 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
# Extended Euclid
def extended_euclid(a, b):
"""
>>> 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, r1, n2, r2):
"""
>>> 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, n):
"""
>>> 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, r1, n2, r2):
"""
>>> 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
# import testmod for testing our function
from doctest import testmod
if __name__ == '__main__':
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)