95 lines
2.3 KiB
Python
95 lines
2.3 KiB
Python
# from http://eli.thegreenplace.net/2009/03/07/computing-modular-square-roots-in-python/
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def modular_sqrt(a, p):
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""" Find a quadratic residue (mod p) of 'a'. p
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must be an odd prime.
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Solve the congruence of the form:
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x^2 = a (mod p)
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And returns x. Note that p - x is also a root.
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0 is returned is no square root exists for
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these a and p.
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The Tonelli-Shanks algorithm is used (except
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for some simple cases in which the solution
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is known from an identity). This algorithm
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runs in polynomial time (unless the
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generalized Riemann hypothesis is false).
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"""
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# Simple cases
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#
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if legendre_symbol(a, p) != 1:
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return 0
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elif a == 0:
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return 0
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elif p == 2:
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return p
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elif p % 4 == 3:
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return pow(a, (p + 1) // 4, p)
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# Partition p-1 to s * 2^e for an odd s (i.e.
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# reduce all the powers of 2 from p-1)
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#
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s = p - 1
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e = 0
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while s % 2 == 0:
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s //= 2
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e += 1
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# Find some 'n' with a legendre symbol n|p = -1.
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# Shouldn't take long.
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#
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n = 2
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while legendre_symbol(n, p) != -1:
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n += 1
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# Here be dragons!
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# Read the paper "Square roots from 1; 24, 51,
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# 10 to Dan Shanks" by Ezra Brown for more
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# information
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#
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# x is a guess of the square root that gets better
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# with each iteration.
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# b is the "fudge factor" - by how much we're off
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# with the guess. The invariant x^2 = ab (mod p)
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# is maintained throughout the loop.
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# g is used for successive powers of n to update
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# both a and b
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# r is the exponent - decreases with each update
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#
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x = pow(a, (s + 1) // 2, p)
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b = pow(a, s, p)
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g = pow(n, s, p)
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r = e
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while True:
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t = b
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m = 0
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for m in range(r):
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if t == 1:
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break
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t = pow(t, 2, p)
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if m == 0:
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return x
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gs = pow(g, 2 ** (r - m - 1), p)
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g = (gs * gs) % p
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x = (x * gs) % p
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b = (b * g) % p
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r = m
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def legendre_symbol(a, p):
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""" Compute the Legendre symbol a|p using
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Euler's criterion. p is a prime, a is
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relatively prime to p (if p divides
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a, then a|p = 0)
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Returns 1 if a has a square root modulo
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p, -1 otherwise.
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"""
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ls = pow(a, (p - 1) // 2, p)
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return -1 if ls == p - 1 else ls
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