[FFmpeg-devel] [PATCHv2] lavc/cbrt_tablegen: speed up tablegen

[Daniel Serpell]

Hi!,

El Tue, Jan 05, 2016 at 08:08:35AM -0800, Ganesh Ajjanagadde escribio:
> On Tue, Jan 5, 2016 at 7:44 AM, Daniel Serpell <dserpell at gmail.com> wrote:
> > Hi!,
> >
> > El Mon, Jan 04, 2016 at 06:33:59PM -0800, Ganesh Ajjanagadde escribio:
> >> This exploits an approach based on the sieve of Eratosthenes, a popular
> >> method for generating prime numbers.
> >>
> >> Tables are identical to previous ones.
> >>
> >> Tested with FATE with/without --enable-hardcoded-tables.
> >>
> >> Sample benchmark (Haswell, GNU/Linux+gcc):
> >> prev:
> >> 7860100 decicycles in cbrt_tableinit,       1 runs,      0 skips
> >> 7777490 decicycles in cbrt_tableinit,       2 runs,      0 skips
> >> [...]
> >> 7582339 decicycles in cbrt_tableinit,     256 runs,      0 skips
> >> 7563556 decicycles in cbrt_tableinit,     512 runs,      0 skips
> >>
> >> new:
> >> 2099480 decicycles in cbrt_tableinit,       1 runs,      0 skips
> >> 2044470 decicycles in cbrt_tableinit,       2 runs,      0 skips
> >> [...]
> >> 1796544 decicycles in cbrt_tableinit,     256 runs,      0 skips
> >> 1791631 decicycles in cbrt_tableinit,     512 runs,      0 skips
> >>
> >
> > See attached code, function "test1", based on an approximation of:
> >
> >   (i+1)^(1/3) ~= i^(1/3) * ( 1 + 1/(3i) - 1/(9i) + 5/(81i) - .... )
> 
> I assume 1/(3i), 1/(9i^2), etc obtained via a Taylor series at x = 0.
> 

Yes, more specifically (in wxmaxima):

(%i1) at( taylor((i+x)^(1/3)/i^(1/3), x, 0, 6) , x=1 );

               1     1       5       10       22       154
(%o1)     1 + --- - ---- + ----- - ------ + ------ - -------
              3 i      2       3        4        5         6
                    9 i    81 i    243 i    729 i    6561 i


I noticed that if you do a change of variable, the series simplifies:

(%i2) at( taylor((i+x)^(1/3)/i^(1/3), x, 0, 6) , [ x=1, i = (1/3)/j ] );

                        6        5       4      3
                   154 j     22 j    10 j    5 j     2
(%o2)           (- ------) + ----- - ----- + ---- - j  + j + 1
                     9         3       3      3

> >
> > Generated values are the same as original floats (max error in double
> > is < 4*10^-10), it is faster (and I think, simpler) than your version.
> 
> Had thought of these ideas, but did not examine as I was a little
> concerned about accuracy. Thanks, will give it a spin. Or
> alternatively, you can submit a patch since you put it into action.
>

Best if you make the patch, as you can test the speed in your same setup.

> Alternatively, one could directly expand the series for (i+1)^(4/3).

Yes, but the first two coefficients are not 1 anymore, so it needs one
more multiplication, canceling the advantage:

(%i3) at( taylor((i+x)^(4/3)/i^(4/3), x, 0, 6) , [ x=1, i = (4/3)/j ] );

                         6    5       4    3    2
                     11 j    j     5 j    j    j
(%o3)                ----- - --- + ---- - -- + -- + j + 1
                     9216    384   768    48   8


> And it may be possible to tighten the number of terms needed by
> expanding not about x = 0, but x = i to get i+1. Or fancier polynomial
> approximations can be used. Have you tried these?

I think that this is what I'm already doing. As I said, the slowest part
is the first division ( r = (1.0/3.0) / i ), and I can't think of any way
to avoid it.

I altered the last constants to reduce the error, but stopped trying after
getting better than 10^-10 absolute error, that is higher than the
precision of a float.

    Daniel.