[Cfp-interest 2871] Re: CFP review of TS-4 and TS-5 revisions

[Jim Thomas]

> On Aug 26, 2023, at 11:31 AM, Jim Thomas <jaswthomas at sbcglobal.net> wrote:
> 
>> I rewrote the algorithm to be clearer, and to fix the typo you noticed:
>> 
>> Let the algorithm be
>> 
>>                               # infinite result is a.h +a.t +b.h +b.t
>> 
>>       {u.h;u.t} = augadd(a.h, b.h)
>>                               # infinite result is u.h +u.t +a.t +b.t
>> 
>>       {v.h;v.t} = augadd(a.t, b.t)
>>                               # infinite result is u.h +u.t +v.h +v.t
>> 
>>       {w.h;w.t} = augadd(u.t, v.t)
>>                               # could be regular add since w.t is discarded
>>                               # infinite result is u.h +v.h +w.h +w.t
>> 
>>       {y.h;y.t} = augadd(v.h, w.h)
>>                               # could be regular add since y.t is discarded
>>                               # infinite result is u.h +y.h +y.t +w.t
>> 
>>       {z.h;z.t} = augadd(u.h, y.h)
>>                               # infinite result is z.h +z.t +y.t +w.t
>> 
>> 
>> Now {z.h;z.t} is a pretty good approximation to the intended result,
>> commutative and without conditional branches,
>> with error y.t +w.t.
>> Two of the additions are marked as "could be regular add" rather than
>> augadd - the algorithm above gives a name to w.t and y.t for the
>> didactic purpose of the error formula.

Ugh. I was looking at an earlier algorithm, where u = a.h +b.t, etc. Disregard my previous comments below.

>> 
>> I claim that it's commutative by symmetry inspection - switch a and b,
>> then u and v are the same,
> 
> u and v switch
> 
>> so w is the same,
> 
> yes
> 
>> so  y is the same,
> 
> y is v.h + w.h, which needn’t be the same as u.h + w.h
> 
>> so z is the 
>> same.   

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