On 2014/06/04 07:19:29, Yang wrote:
On 2014/06/03 16:51:30, Raymond Toy wrote:
> On 2014/06/03 07:01:45, Yang wrote:
> > https://codereview.chromium.org/303753002/diff/40001/src/math.js
> > File src/math.js (right):
> >
> >
https://codereview.chromium.org/303753002/diff/40001/src/math.js#newcode262
> > src/math.js:262: }
> > On 2014/06/02 17:26:11, Raymond Toy wrote:
> > > As you mentioned via email, you've removed the 3rd iteration. This
is
really
> > > needed if you want to be able to reduce multiples of pi/2
accurately.
> >
> > That's true. However, the reduction step is not exposed as a library
function.
> > From what I have seen, the third step seems to only affect y1. With a
y0
> really
> > close to y1, it does not change the result of sine or cosine. This is
also
why
> I
> > was asking for a test case where removing this third step would make a
> > difference.
>
> I don't understand what you mean by "y0 really close to y1". What are
you
> saying?
>
>
> tan(Math.PI*45/2) requires the 3rd iteration. ieee754_rem_pio2 returns
> [45, -9.790984586812941e-16, -6.820314736619894e-32]
>
> If you ignore the y1 result, we have
> kernel_tan(-9.790984586812941e-16, 0e0, -1) -> 1021347742030824.2
>
> If you include the y1 result:
> kernel_tan(-9.790984586812941e-16,-6.820314736619894e-32, -1) ->
> 1021347742030824.1
I somehow didn't type what I thought. I meant to say: if y0 is really
close to
0, there does not seem to be any point to invest in the third loop. (I am
aware
that omitting y1 changes the result in some cases. I'm not arguing this).
So in the example here, if I omit the third iteration, I get
[45, -9.790984586812941e-16, -6.820199415561299e-32]
y0 is the same, y1 differs slightly, but the end result is still
1021347742030824.1.
While I understand your desire to reduce the complexity, you are modifying
an
algorithm written by an expert. I think the burden is on you to prove that
by
removing the third iteration you do not change the value of y0.
Also, where is this coming from? In reality, how often will you compute
sin(x)
where x is very near a multiple of pi/2 (where the third iteration is
needed)?
I suspect it occurs more often than we might expect, but also that if you're
doing that, I think you're also computing zillions more values that are not
a
multiple of pi/2.
For example, in 3d-morph, we compute sin((n-1)*pi/15) for n = 0 to 119.
Thus
out of 120 values, we have a multiple of pi just 8 times out of 120. If the
cost
of reduction for multiples of pi/2 AND the computation of sin were reduced
to
exactly zero, you would save about just 6.6% in runtime.
I think there are more important things to look at. We need profile
results. We
need to understand what is really expensive in the reduction, not what we
think
is expensive.
https://codereview.chromium.org/303753002/
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