## Re: [xsl] A super-efficient way to compute the sum of A[i] * B[i] for i=1 to n?

 Subject: Re: [xsl] A super-efficient way to compute the sum of A[i] * B[i] for i=1 to n? From: "Dimitre Novatchev dnovatchev@xxxxxxxxx" Date: Sat, 9 May 2020 17:52:59 -0000
```Hi Roger,

> I need a super-efficient way to compute the sum of A[i] * B[i] for i=1
to n.

In case you have n cores / processors, and the compiler knows to optimize
functions like map(), zipWith() (for-each-pair() in XPath 3.0), then all
processors can each do a corresponding multiplication in parallel in a
single moment (computing cycle).

Then what remains is the summing of the results of these multiplications.
This essentially is the map-reduce technique.

Interestingly enough, while not all reduce operations can be parallelized,
in the case of sum() one can add together n numbers in Log2(n) summing
steps -- very similar to the DVC (DiVide and Conquer) technique, but doing
it parallel -- not sequentially. So, first there would be n/2 additions,
then n/4 additions (of the results of the first step), then n/8 additions,
and so on -- in a total of ceiling(Log2(n)) steps. If for each steps we
have sufficient number of processors (n/2 for the first step, less for the
following steps), then the summing could be performed in Log2(n) computing
cycles.

So, it seems that the whole computation could be performed in something
like ~ ceiling(Log2(n)) + 1 computing cycles.

Or maybe I am being wrong here? :)  Please, correct me.

Cheers,
Dimitre

On Sat, May 9, 2020 at 4:59 AM Costello, Roger L. costello@xxxxxxxxx <
xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:

> Hi Folks,
>
> I need a super-efficient way to compute the sum of A[i] * B[i] for i=1 to
> n.
>
> For example, suppose A is this:
>
> <row>
>     <col>0.9</col>
>     <col>0.3</col>
> </row>
>
> and B is this:
>
> <row>
>     <col>0.2</col>
>     <col>0.8</col>
> </row>
>
> I want to compute:
>
> (0.9 * 0.2) + (0.3 * 0.8)
>
> Here's one way to do it:
>
> sum(for \$i in 1 to count(\$A/col) return number(\$A/col[\$i]) *
> number(\$B/col[\$i]))
>
> I suspect that is not the most efficient approach.
>
> What is the most efficient approach? I will be doing hundreds of thousands
> of these computations, so I want to use the most efficient approach.
>
> /Roger
>
>

--
Cheers,
Dimitre Novatchev
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