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From: mitchalsup@aol.com (MitchAlsup1)
Newsgroups: comp.arch
Subject: Re: Continuations
Date: Fri, 19 Jul 2024 16:32:43 +0000
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On Fri, 19 Jul 2024 14:16:01 +0000, Terje Mathisen wrote:

> Thomas Koenig wrote:
>> MitchAlsup1 <mitchalsup@aol.com> schrieb:
>>
>>> What I am talking about is to improve their performance until a
>>> sin() takes about the same number of cycles of FDIV, not 10× more.
>>
>> Maybe time for a little story.
>>
>> Some unspecified time ago, a colleague did CFD calculations which
>> included fluid flow (including turbulence modelling and diffusion)
>> and quite a few chemical reactions together.  So, he evaluated a
>> huge number of Arrhenius equations,
>>
>> k = A * exp(-E_a/(R*T))
>>
>> and because some of the reactions he looked at were highly
>> exothermic or endothermic, he needed tiny relaxation factors (aka
>> small steps).  His calculaiton spent most of the time evaluating
>> the Arrhenius equation above many, many, many, many times.
>>
>> A single calculation took _months_, and he didn't use weak hardware.
>>
>> A fully pipelined evaluation of, let's say, four parallel exp and
>> four parallel fdiv instructions would have reduced his calculation
>> time by orders of magnitude, and allowed him to explore the design
>> space instead of just scratching the surface.
>
> Back when I first looked at Invsqrt(), I did so because an Computation
> Fluid Chemistry researcher from Sweden asked for help speeding up his
> reciprocal calculations (sqrt(1/(dx^2+dy^2+dz^2))), I found that by
> combining the 1/x and the sqrt and doing three of them pipelind together
> (all the water molecules having three atoms), his weeklong simulation
> runs ran in half the time, on both PentiumPro and Alpha hardware.

I, personally, have found many Newton-Raphson iterators that converge
faster using 1/SQRT(x) than using the SQRT(x) equivalent.
>
>>
>> (By the way, if I had found a reasonable way to incorporate the
>> Arrhenius equation into your ISA, I would have done so already :-)
>>
>
> I'm guessing that you would do the exp(-E_a/(R*T)) as
> exp(-E-a)-exp(R*T), since that should give the same result and now you
> could interleave the two exp) calculations, and/or hoist the exp(-E_a)
> term?
>
> Terje