Я пытаюсь вычислить кубический корень многосотного числа по модулю P в Python, и терпит неудачу.
Я обнаружил код для алгоритма Тонелли-Шанкса, который якобы легко изменить с квадратных корней на кубические корни, но это ускользает от меня. Я рыскал по веб-библиотекам, математическим библиотекам и нескольким книгам, но безрезультатно. Код был бы замечательным, как и алгоритм, объясненный простым языком.
Вот код Python (2.6?) Для поиска квадратных корней:
def modular_sqrt(a, p):
""" Find a quadratic residue (mod p) of 'a'. p
must be an odd prime.
Solve the congruence of the form:
x^2 = a (mod p)
And returns x. Note that p - x is also a root.
0 is returned is no square root exists for
these a and p.
The Tonelli-Shanks algorithm is used (except
for some simple cases in which the solution
is known from an identity). This algorithm
runs in polynomial time (unless the
generalized Riemann hypothesis is false).
"""
# Simple cases
#
if legendre_symbol(a, p) != 1:
return 0
elif a == 0:
return 0
elif p == 2:
return n
elif p % 4 == 3:
return pow(a, (p + 1) / 4, p)
# Partition p-1 to s * 2^e for an odd s (i.e.
# reduce all the powers of 2 from p-1)
#
s = p - 1
e = 0
while s % 2 == 0:
s /= 2
e += 1
# Find some 'n' with a legendre symbol n|p = -1.
# Shouldn't take long.
#
n = 2
while legendre_symbol(n, p) != -1:
n += 1
# Here be dragons!
# Read the paper "Square roots from 1; 24, 51,
# 10 to Dan Shanks" by Ezra Brown for more
# information
#
# x is a guess of the square root that gets better
# with each iteration.
# b is the "fudge factor" - by how much we're off
# with the guess. The invariant x^2 = ab (mod p)
# is maintained throughout the loop.
# g is used for successive powers of n to update
# both a and b
# r is the exponent - decreases with each update
#
x = pow(a, (s + 1) / 2, p)
b = pow(a, s, p)
g = pow(n, s, p)
r = e
while True:
t = b
m = 0
for m in xrange(r):
if t == 1:
break
t = pow(t, 2, p)
if m == 0:
return x
gs = pow(g, 2 ** (r - m - 1), p)
g = (gs * gs) % p
x = (x * gs) % p
b = (b * g) % p
r = m
def legendre_symbol(a, p):
""" Compute the Legendre symbol a|p using
Euler's criterion. p is a prime, a is
relatively prime to p (if p divides
a, then a|p = 0)
Returns 1 if a has a square root modulo
p, -1 otherwise.
"""
ls = pow(a, (p - 1) / 2, p)
return -1 if ls == p - 1 else ls