From 900c2c8557c08c1cca0dd5951d33122fa67b8228 Mon Sep 17 00:00:00 2001
From: thomasv <thomasv@gitorious>
Date: Wed, 1 Feb 2012 21:02:19 +0100
Subject: [PATCH] square root modulo

---
 client/msqr.py | 94 ++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 94 insertions(+)
 create mode 100644 client/msqr.py

diff --git a/client/msqr.py b/client/msqr.py
new file mode 100644
index 00000000..da3c6fde
--- /dev/null
+++ b/client/msqr.py
@@ -0,0 +1,94 @@
+# from http://eli.thegreenplace.net/2009/03/07/computing-modular-square-roots-in-python/
+
+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 p
+    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