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

#!/usr/bin/env python
# -*- coding: utf-8 -*-

def isPrime(a):
    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(n % p == 0 and n > 0): #while we can divide by smaller number, do so
            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 calculateLegendre(a, p):
	"""
	Calculate the legendre symbol (a, p) with p is prime.
	The result is either -1, 0 or 1

	>>> calculateLegendre(3, 29)
	-1
	>>> calculateLegendre(111, 41) # Beispiel aus dem Skript, S. 114
	-1
	>>> calculateLegendre(113, 41) # Beispiel aus dem Skript, S. 114
	1
	>>> calculateLegendre(2, 31)
	1
	>>> calculateLegendre(5, 31)
	1
	>>> calculateLegendre(150, 1009) # http://math.stackexchange.com/q/221223/6876
	1
	>>> calculateLegendre(25, 1009) # http://math.stackexchange.com/q/221223/6876
	1
	>>> calculateLegendre(2, 1009) # http://math.stackexchange.com/q/221223/6876
	1
	>>> calculateLegendre(3, 1009) # http://math.stackexchange.com/q/221223/6876
	1
	"""
	if a >= p or a < 0:
		return calculateLegendre(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 isPrime(a):
		factors = factorize(a)
		product = 1
		for pi in factors:
			product *= calculateLegendre(pi, p)
		return product
	else:
		if ((p-1)/2)%2==0 or ((a-1)/2)%2==0:
			return calculateLegendre(p, a)
		else:
			return (-1)*calculateLegendre(p, a)

if __name__ == "__main__":
	import doctest
	doctest.testmod()