s = -s *,

(6*n) * (6*n-1) * (6*n-2) * (6*n-3) * (6*n-4) * (6*n-5),

/,

((3*n) * (3*n-1) * (3*n-2) * n * n * n * k3_power3)

t = t + s * (k1 + n*k2)

Hi Bertram for the square root you will need to implement *Newtons *quadratic convergence algorithm

And in order to remove the slow division from the inner loop, you'll have to go for *binary splitting*

There's a nice (and readable) explanation at http://www.craig-wood.com/nick/articles/pi-chudnovsky/ showing how to do above in Python - so it's an almost 1:1 to convert to ooRexx.

Still, to get decent performance, you will have to use the gmp library (above article shows that for the Python implmentation). Unfortunately ooRexx doesn't yet have an interface to gmp - one of my really long-time wishes ..

Erich

On Wed, Mar 5, 2014 at 3:52 AM, Bertram Moshier <bertrammoshier@gmail.com> wrote:

Hi,I'm working on a program to compute Pi and noticed two interesting situations.

- Bug #445 "RxMath functions do not honor precision on Windows" is back.
- Something I accept, but can't figure out why and would like to understand.
Bug 445: RxMath functions do not honor precision on Windows

The Pi program I'm writing is using Chudnovsky and as such needs the square root of K3 to the same precision as the final result (say Pi to 100,000 places or even more). Yet, when I call RxCalcSqrt(k3), the number of places is only 11, which is not close enough to provide valid results. I tried it and the result wasn't even close to the known results of Pi at the 100,000 places.

As I read bug 445, Mark closed it as "Committed revision 986." Thus it looks like bug 445 is back in 4.2.0 64-bit.

As for the second item: Something just doesn't make sense, as when I change the values of the variables used in the MATH (same precision) the amount of time it takes to compute is up to 10X longer. The code is the same and precision is the same for both runs. I know you need details to be specific. The "heart" of the loop is:

s = -s *,

(6*n) * (6*n-1) * (6*n-2) * (6*n-3) * (6*n-4) * (6*n-5),

/,

((3*n) * (3*n-1) * (3*n-2) * n * n * n * k3_power3)

t = t + s * (k1 + n*k2)

and what I'm seeing is more time spent here when the values for s, k3_power3, k1, k2, and t are different, but the precisions are the same. They are different because the code supports a fast and normal convergence (14 versus 25). The difference is the fast and normal have different values for the previously mentioned variables, but the same precision. I'm noticing the "fast" is actually 10X slower than the normal. I just don't understand why and thus am thinking there is something "inside" Rexx causing it to go slower, but what?

Well, thanks for listening. :-)

Bertram Moshier

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