# 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 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)