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92 lines
2.1 KiB
Python
92 lines
2.1 KiB
Python
# Chinese Remainder Theorem:
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# GCD ( Greatest Common Divisor ) or HCF ( Highest Common Factor )
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# If GCD(a,b) = 1, then for any remainder ra modulo a and any remainder rb modulo b
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# there exists integer n, such that n = ra (mod a) and n = ra(mod b). If n1 and n2 are
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# two such integers, then n1=n2(mod ab)
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# Algorithm :
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# 1. Use extended euclid algorithm to find x,y such that a*x + b*y = 1
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# 2. Take n = ra*by + rb*ax
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# Extended Euclid
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def extended_euclid(a, b):
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"""
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>>> extended_euclid(10, 6)
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(-1, 2)
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>>> extended_euclid(7, 5)
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(-2, 3)
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"""
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if b == 0:
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return (1, 0)
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(x, y) = extended_euclid(b, a % b)
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k = a // b
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return (y, x - k * y)
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# Uses ExtendedEuclid to find inverses
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def chinese_remainder_theorem(n1, r1, n2, r2):
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"""
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>>> chinese_remainder_theorem(5,1,7,3)
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31
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Explanation : 31 is the smallest number such that
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(i) When we divide it by 5, we get remainder 1
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(ii) When we divide it by 7, we get remainder 3
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>>> chinese_remainder_theorem(6,1,4,3)
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14
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"""
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(x, y) = extended_euclid(n1, n2)
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m = n1 * n2
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n = r2 * x * n1 + r1 * y * n2
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return (n % m + m) % m
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# ----------SAME SOLUTION USING InvertModulo instead ExtendedEuclid----------------
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# This function find the inverses of a i.e., a^(-1)
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def invert_modulo(a, n):
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"""
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>>> invert_modulo(2, 5)
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3
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>>> invert_modulo(8,7)
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1
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"""
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(b, x) = extended_euclid(a, n)
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if b < 0:
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b = (b % n + n) % n
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return b
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# Same a above using InvertingModulo
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def chinese_remainder_theorem2(n1, r1, n2, r2):
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"""
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>>> chinese_remainder_theorem2(5,1,7,3)
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31
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>>> chinese_remainder_theorem2(6,1,4,3)
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14
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"""
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x, y = invert_modulo(n1, n2), invert_modulo(n2, n1)
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m = n1 * n2
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n = r2 * x * n1 + r1 * y * n2
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return (n % m + m) % m
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if __name__ == "__main__":
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from doctest import testmod
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testmod(name="chinese_remainder_theorem", verbose=True)
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testmod(name="chinese_remainder_theorem2", verbose=True)
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testmod(name="invert_modulo", verbose=True)
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testmod(name="extended_euclid", verbose=True)
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