mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-11-30 16:31:08 +00:00
9316e7c014
* flake8 --max-line-length=88
* fixup! Format Python code with psf/black push
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
153 lines
3.2 KiB
Python
153 lines
3.2 KiB
Python
# Modular Division :
|
||
# An efficient algorithm for dividing b by a modulo n.
|
||
|
||
# GCD ( Greatest Common Divisor ) or HCF ( Highest Common Factor )
|
||
|
||
# Given three integers a, b, and n, such that gcd(a,n)=1 and n>1, the algorithm should
|
||
# return an integer x such that 0≤x≤n−1, and b/a=x(modn) (that is, b=ax(modn)).
|
||
|
||
# Theorem:
|
||
# a has a multiplicative inverse modulo n iff gcd(a,n) = 1
|
||
|
||
|
||
# This find x = b*a^(-1) mod n
|
||
# Uses ExtendedEuclid to find the inverse of a
|
||
|
||
|
||
def modular_division(a, b, n):
|
||
"""
|
||
>>> modular_division(4,8,5)
|
||
2
|
||
|
||
>>> modular_division(3,8,5)
|
||
1
|
||
|
||
>>> modular_division(4, 11, 5)
|
||
4
|
||
|
||
"""
|
||
assert n > 1 and a > 0 and greatest_common_divisor(a, n) == 1
|
||
(d, t, s) = extended_gcd(n, a) # Implemented below
|
||
x = (b * s) % n
|
||
return x
|
||
|
||
|
||
# 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) # Implemented below
|
||
if b < 0:
|
||
b = (b % n + n) % n
|
||
return b
|
||
|
||
|
||
# ------------------ Finding Modular division using invert_modulo -------------------
|
||
|
||
# This function used the above inversion of a to find x = (b*a^(-1))mod n
|
||
def modular_division2(a, b, n):
|
||
"""
|
||
>>> modular_division2(4,8,5)
|
||
2
|
||
|
||
>>> modular_division2(3,8,5)
|
||
1
|
||
|
||
>>> modular_division2(4, 11, 5)
|
||
4
|
||
|
||
"""
|
||
s = invert_modulo(a, n)
|
||
x = (b * s) % n
|
||
return x
|
||
|
||
|
||
# Extended Euclid's Algorithm : If d divides a and b and d = a*x + b*y for integers x
|
||
# and y, then d = gcd(a,b)
|
||
|
||
|
||
def extended_gcd(a, b):
|
||
"""
|
||
>>> extended_gcd(10, 6)
|
||
(2, -1, 2)
|
||
|
||
>>> extended_gcd(7, 5)
|
||
(1, -2, 3)
|
||
|
||
** extended_gcd function is used when d = gcd(a,b) is required in output
|
||
|
||
"""
|
||
assert a >= 0 and b >= 0
|
||
|
||
if b == 0:
|
||
d, x, y = a, 1, 0
|
||
else:
|
||
(d, p, q) = extended_gcd(b, a % b)
|
||
x = q
|
||
y = p - q * (a // b)
|
||
|
||
assert a % d == 0 and b % d == 0
|
||
assert d == a * x + b * y
|
||
|
||
return (d, x, y)
|
||
|
||
|
||
# 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)
|
||
|
||
|
||
# Euclid's Lemma : d divides a and b, if and only if d divides a-b and b
|
||
# Euclid's Algorithm
|
||
|
||
|
||
def greatest_common_divisor(a, b):
|
||
"""
|
||
>>> greatest_common_divisor(7,5)
|
||
1
|
||
|
||
Note : In number theory, two integers a and b are said to be relatively prime,
|
||
mutually prime, or co-prime if the only positive integer (factor) that divides
|
||
both of them is 1 i.e., gcd(a,b) = 1.
|
||
|
||
>>> greatest_common_divisor(121, 11)
|
||
11
|
||
|
||
"""
|
||
if a < b:
|
||
a, b = b, a
|
||
|
||
while a % b != 0:
|
||
a, b = b, a % b
|
||
|
||
return b
|
||
|
||
|
||
if __name__ == "__main__":
|
||
from doctest import testmod
|
||
|
||
testmod(name="modular_division", verbose=True)
|
||
testmod(name="modular_division2", verbose=True)
|
||
testmod(name="invert_modulo", verbose=True)
|
||
testmod(name="extended_gcd", verbose=True)
|
||
testmod(name="extended_euclid", verbose=True)
|
||
testmod(name="greatest_common_divisor", verbose=True)
|