"pow" (power) function

[David M. Cooke]

"Russ" <uymqlp502 at sneakemail.com> writes:

> Ben Cartwright wrote:
>> Russ wrote:
>
>> > Does "pow(x,2)" simply square x, or does it first compute logarithms
>> > (as would be necessary if the exponent were not an integer)?
>>
>>
>> The former, using binary exponentiation (quite fast), assuming x is an
>> int or long.
>>
>> If x is a float, Python coerces the 2 to 2.0, and CPython's float_pow()
>> function is called.  This function calls libm's pow(), which in turn
>> uses logarithms.
>
> I just did a little time test (which I should have done *before* my
> original post!), and 2.0**2 seems to be about twice as fast as
> pow(2.0,2). That seems consistent with your claim above.
>
> I'm a bit surprised that pow() would use logarithms even if the
> exponent is an integer. I suppose that just checking for an integer
> exponent could blow away the gain that would be achieved by avoiding
> logarithms. On the other hand, I would think that using logarithms
> could introduce a tiny error (e.g., pow(2.0,2) = 3.9999999996 <- made
> up result) that wouldn't occur with multiplication.

It depends on the libm implementation of pow() whether logarithms are
used for integer exponents. I'm looking at glibc's (the libc used on
Linux) implementation for Intel processors, and it does optimize
integers. That routine is written in assembly language, btw.

>> > Does "x**0.5" use the same algorithm as "sqrt(x)", or does it use some
>> > other (perhaps less efficient) algorithm based on logarithms?
>>
>> The latter, and that algorithm is libm's pow().  Except for a few
>> special cases that Python handles, all floating point exponentation is
>> left to libm.  Checking to see if the exponent is 0.5 is not one of
>> those special cases.
>
> I just did another little time test comparing 2.0**0.5 with sqrt(2.0).
> Surprisingly, 2.0**0.5 seems to take around a third less time.
>
> None of these differences are really significant unless one is doing
> super-heavy-duty number crunching, of course, but I was just curious.
> Thanks for the information.

And if you are, you'd likely be doing it on more than one number, in
which case you'd probably want to use numpy. We've optimized x**n so
that it does handle n=0.5 and integers specially; it makes more sense
to do this for an array of numbers where you can do the special
manipulation of the exponent, and then apply that to all the numbers
in the array at once.

-- 
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|David M. Cooke
|cookedm(at)physics(dot)mcmaster(dot)ca