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 <dserp...@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.
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