removed trailing whitespace and fixed indentation in history widget

lib/msqr.py

@@ -3,14 +3,14 @@
 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
@@ -27,7 +27,7 @@ def modular_sqrt(a, p):
         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)
     #
@@ -36,20 +36,20 @@ def modular_sqrt(a, p):
     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
@@ -63,7 +63,7 @@ def modular_sqrt(a, p):
     b = pow(a, s, p)
     g = pow(n, s, p)
     r = e
-    
+
     while True:
         t = b
         m = 0
@@ -71,22 +71,22 @@ def modular_sqrt(a, p):
             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.
     """