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fixed bug and added implementation

This commit is contained in:
Martin Thoma 2013-09-01 12:38:44 +02:00
parent 93cd7986b9
commit b0a72dc66c
3 changed files with 68 additions and 1 deletions

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@ -35,7 +35,11 @@
\State $p_1, p_2, \dots, p_n \gets \Call{Factorize}{a}$
\State \Return $\prod_{i=1}^n \Call{CalculateLegendre}{p_i, p}$
\Else \Comment{now: $a \in \mathbb{P}, \sqrt{p-2} \geq a \geq 3$}
\State \Return $(-1) \cdot \Call{CalculateLegendre}{p, a}$
\If{$\frac{p-1}{2} \equiv 0 \mod 2$}
\State \Return $(-1) \cdot \Call{CalculateLegendre}{p, a}$
\Else
\State \Return $(-1) \cdot \Call{CalculateLegendre}{p, a}$
\EndIf
\EndIf
\EndProcedure
\end{algorithmic}

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@ -0,0 +1,63 @@
#!/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)
-1
>>> calculateLegendre(113, 41)
1
"""
if a >= p or a < 0:
return calculateLegendre(a % p, p)
elif a == 0 or a == 1:
return a
elif a == 2:
if a%8 == 1 or a%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:
return calculateLegendre(p, a)
else:
return (-1)*calculateLegendre(p, a)
if __name__ == "__main__":
import doctest
doctest.testmod()