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# Chinese Remainder Theorem:
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# GCD ( Greatest Common Divisor ) or HCF ( Highest Common Factor )
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"""
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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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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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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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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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"""
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from typing import Tuple
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# Extended Euclid
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def extended_euclid(a: int, b: int) -> (int, int):
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def extended_euclid(a: int, b: int) -> Tuple[int, int]:
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"""
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>>> extended_euclid(10, 6)
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(-1, 2)
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# Diophantine Equation : Given integers a,b,c ( at least one of a and b != 0), the
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# diophantine equation a*x + b*y = c has a solution (where x and y are integers)
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# iff gcd(a,b) divides c.
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# GCD ( Greatest Common Divisor ) or HCF ( Highest Common Factor )
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from typing import Tuple
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def diophantine(a: int, b: int, c: int) -> (int, int):
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def diophantine(a: int, b: int, c: int) -> Tuple[float, float]:
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"""
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Diophantine Equation : Given integers a,b,c ( at least one of a and b != 0), the
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diophantine equation a*x + b*y = c has a solution (where x and y are integers)
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iff gcd(a,b) divides c.
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GCD ( Greatest Common Divisor ) or HCF ( Highest Common Factor )
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>>> diophantine(10,6,14)
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(-7.0, 14.0)
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return (r * x, r * y)
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# Lemma : if n|ab and gcd(a,n) = 1, then n|b.
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# Finding All solutions of Diophantine Equations:
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# Theorem : Let gcd(a,b) = d, a = d*p, b = d*q. If (x0,y0) is a solution of Diophantine
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# Equation a*x + b*y = c. a*x0 + b*y0 = c, then all the solutions have the form
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# a(x0 + t*q) + b(y0 - t*p) = c, where t is an arbitrary integer.
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# n is the number of solution you want, n = 2 by default
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def diophantine_all_soln(a: int, b: int, c: int, n: int = 2) -> None:
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"""
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Lemma : if n|ab and gcd(a,n) = 1, then n|b.
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Finding All solutions of Diophantine Equations:
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Theorem : Let gcd(a,b) = d, a = d*p, b = d*q. If (x0,y0) is a solution of
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Diophantine Equation a*x + b*y = c. a*x0 + b*y0 = c, then all the
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solutions have the form a(x0 + t*q) + b(y0 - t*p) = c,
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where t is an arbitrary integer.
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n is the number of solution you want, n = 2 by default
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>>> diophantine_all_soln(10, 6, 14)
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-7.0 14.0
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-4.0 9.0
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print(x, y)
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# Euclid's Lemma : d divides a and b, if and only if d divides a-b and b
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# Euclid's Algorithm
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def greatest_common_divisor(a: int, b: int) -> int:
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"""
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Euclid's Lemma : d divides a and b, if and only if d divides a-b and b
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Euclid's Algorithm
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>>> greatest_common_divisor(7,5)
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1
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return b
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# Extended Euclid's Algorithm : If d divides a and b and d = a*x + b*y for integers
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# x and y, then d = gcd(a,b)
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def extended_gcd(a: int, b: int) -> (int, int, int):
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def extended_gcd(a: int, b: int) -> Tuple[int, int, int]:
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"""
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Extended Euclid's Algorithm : If d divides a and b and d = a*x + b*y for integers
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x and y, then d = gcd(a,b)
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>>> extended_gcd(10, 6)
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(2, -1, 2)
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# Modular Division :
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# An efficient algorithm for dividing b by a modulo n.
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# GCD ( Greatest Common Divisor ) or HCF ( Highest Common Factor )
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# Given three integers a, b, and n, such that gcd(a,n)=1 and n>1, the algorithm should
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# return an integer x such that 0≤x≤n−1, and b/a=x(modn) (that is, b=ax(modn)).
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# Theorem:
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# a has a multiplicative inverse modulo n iff gcd(a,n) = 1
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# This find x = b*a^(-1) mod n
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# Uses ExtendedEuclid to find the inverse of a
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from typing import Tuple
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def modular_division(a: int, b: int, n: int) -> int:
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"""
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Modular Division :
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An efficient algorithm for dividing b by a modulo n.
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GCD ( Greatest Common Divisor ) or HCF ( Highest Common Factor )
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Given three integers a, b, and n, such that gcd(a,n)=1 and n>1, the algorithm should
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return an integer x such that 0≤x≤n−1, and b/a=x(modn) (that is, b=ax(modn)).
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Theorem:
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a has a multiplicative inverse modulo n iff gcd(a,n) = 1
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This find x = b*a^(-1) mod n
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Uses ExtendedEuclid to find the inverse of a
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>>> modular_division(4,8,5)
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2
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return x
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# This function find the inverses of a i.e., a^(-1)
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def invert_modulo(a: int, n: int) -> int:
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"""
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This function find the inverses of a i.e., a^(-1)
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>>> invert_modulo(2, 5)
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3
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# ------------------ Finding Modular division using invert_modulo -------------------
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# This function used the above inversion of a to find x = (b*a^(-1))mod n
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def modular_division2(a: int, b: int, n: int) -> int:
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"""
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This function used the above inversion of a to find x = (b*a^(-1))mod n
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>>> modular_division2(4,8,5)
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2
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return x
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# Extended Euclid's Algorithm : If d divides a and b and d = a*x + b*y for integers x
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# and y, then d = gcd(a,b)
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def extended_gcd(a: int, b: int) -> (int, int, int):
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def extended_gcd(a: int, b: int) -> Tuple[int, int, int]:
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"""
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Extended Euclid's Algorithm : If d divides a and b and d = a*x + b*y for integers x
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and y, then d = gcd(a,b)
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>>> extended_gcd(10, 6)
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(2, -1, 2)
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return (d, x, y)
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# Extended Euclid
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def extended_euclid(a: int, b: int) -> (int, int):
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def extended_euclid(a: int, b: int) -> Tuple[int, int]:
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"""
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Extended Euclid
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>>> extended_euclid(10, 6)
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(-1, 2)
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return (y, x - k * y)
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# Euclid's Lemma : d divides a and b, if and only if d divides a-b and b
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# Euclid's Algorithm
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def greatest_common_divisor(a: int, b: int) -> int:
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"""
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Euclid's Lemma : d divides a and b, if and only if d divides a-b and b
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Euclid's Algorithm
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>>> greatest_common_divisor(7,5)
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1
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