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xn = 1 << log(n + 1, 4)

Tom's method is slightly better than this, but they both take O(lg(lg(n)) steps through the loop instead of O(lg(n)). That's important if n has hundreds or millions of bits, especially since in those cases the individual steps can take quite a while.

The following takes about (lg(lg(n)) – 5) times through the loop, and those 5 fewer steps probably matter when 2**64 < n < 2**2**20, say.

if n <= 9e307:

xn = int(sqrt(n))

else:

p = log(n, 4)

xn = int(2.0 ** (p % 1 + 256)) << (int(p) – 256)

By the way the original routine needs a fix for when n == 0.

Thanks, y'all.

]]>def isPythagoreanTriplet(a, b, c): if ((a**2) + (b**2)) == (c**2): return 1 else: return 0

and then did:

for a in range(1, 500): for b in range(1, 500): for c in range(1, 500): if isPythagoreanTriplet(a, b, c) and (a+b+c) == 1000: print "A = %d, B = %d, C = %d," % (a,b,c) return a*b*c]]>

xn = (x >> x.bit_length() // 2) + 1

]]>