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LaTeX-examples/source-code/Pseudocode/Calculate-Legendre/calculate_legendre.py

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2015-11-20 22:36:38 +01:00
#!/usr/bin/env python
# -*- coding: utf-8 -*-
def is_prime(a):
"""
Check if `a` is a prime number.
Parameters
----------
a : int, a >= 2
"""
return all(a % i for i in xrange(2, a))
# http://stackoverflow.com/a/14793082/562769
def factorize(n):
factors = []
p = 2
while True:
# while we can divide by smaller number, do so
while(n % p == 0 and n > 0):
factors.append(p)
n = n / p
p += 1 # p is not necessary prime, but n%p == 0 only for prime numbers
if p > n / p:
break
if n > 1:
factors.append(n)
return factors
def calculate_legendre(a, p):
"""
Calculate the legendre symbol (a, p) with p is prime.
The result is either -1, 0 or 1
>>> calculate_legendre(3, 29)
-1
>>> calculate_legendre(111, 41) # Beispiel aus dem Skript, S. 114
-1
>>> calculate_legendre(113, 41) # Beispiel aus dem Skript, S. 114
1
>>> calculate_legendre(2, 31)
1
>>> calculate_legendre(5, 31)
1
# http://math.stackexchange.com/q/221223/6876
>>> calculate_legendre(150, 1009)
1
# http://math.stackexchange.com/q/221223/6876
>>> calculate_legendre(25, 1009)
1
# http://math.stackexchange.com/q/221223/6876
>>> calculate_legendre(2, 1009)
1
# http://math.stackexchange.com/q/221223/6876
>>> calculate_legendre(3, 1009)
1
"""
if a >= p or a < 0:
return calculate_legendre(a % p, p)
elif a == 0 or a == 1:
return a
elif a == 2:
if p % 8 == 1 or p % 8 == 7:
return 1
else:
return -1
elif a == p-1:
if p % 4 == 1:
return 1
else:
return -1
elif not is_prime(a):
factors = factorize(a)
product = 1
for pi in factors:
product *= calculate_legendre(pi, p)
return product
else:
if ((p-1)/2) % 2 == 0 or ((a-1)/2) % 2 == 0:
return calculate_legendre(p, a)
else:
return (-1)*calculate_legendre(p, a)
if __name__ == "__main__":
import doctest
doctest.testmod()