On Oct 7, 5:19 pm, [EMAIL PROTECTED] wrote:
> but I want to make clear that I think that (0).numbits()==-1
> is the natural solution. At least for all square-and-multiply-like
> algorithms needed in [...]
Can you clarify this? Why is -1 the natural solution? I
can see a case for 0, for -infinity (whatever that means),
or for raising an exception, but I can't see why -1 would
be at all useful or natural.
Here's a straightforward (mildly inefficient)
implementation of the left-to-right square-and-multiply algorithm
for powering. It works just fine with nbits(0) == 0.
def nbits(n):
return 0 if n == 0 else len(bin(abs(n)))-2
def LtoRpow(a, n):
acc = 1
for i in reversed(xrange(nbits(n))):
acc *= acc
if n & (1<<i):
acc *= a
return acc
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