Re: Increasing sequence ?

All said is correct.

I also think that the recursive function should qualify for guaranteed
streamable.

Is this correct?

As for Saxon not dealing in the best way with tail-recursion, I still
expect this issue to be fixed in the future -- would be very useful
and makes perfect sense.

Cheers,
Dimitre

On Fri, Mar 27, 2015 at 4:05 AM, Michael Kay mike@xxxxxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
> What we're seeing here is that the best solution is radically influenced by
> the optimization strategies within the processor.
>
> Dimitre's recursive approach is only worth doing if the processor has
> non-constant performance for an expression of the form sequence[$N] where $N
> is an integer; that is, if sequences are implemented as linked lists rather
> than arrays. Saxon will generally use an adaptive implementation where the
> sequence is held as an array as soon as you start indexing into it, so the
> non-recursive solution will work just fine. Other systems may differ.
>
> Then, if you do use the recursive approach, you run into problems if the
> processor doesn't do tail-call optimization. And if that's the case, you
> need to consider a divide-and-conquer approach instead.
>
> Michael Kay
> Saxonica
> mike@xxxxxxxxxxxx
> +44 (0) 118 946 5893
>
>
>
>
> On 27 Mar 2015, at 09:36, Leo Studer leo.studer@xxxxxxxxxxx
> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> Dimitre
>
> thanks, this is amazing. With Saxon EE in Oxygen 16.1 I get stack overflow
> with 10000 ;-).
> Can you compare the time with this solution?
> declare namespace my = "my:my";
> declare function my:increasing2($seq as xs:double*)as xs:boolean
> {every $v in 1 to (count($seq)-1) satisfies ($seq[$v] lt $seq[$v+1])};
> let $v:=(1 to 1000000) return (my:increasing2($v))
>
> Cheers
> Leo
>
> On 27.03.2015, at 05:24, Dimitre Novatchev dnovatchev@xxxxxxxxx
> <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> Hi Leo,
>
> I ran this with BaseX 7.8.2:
>
> declare namespace my = "my:my";
> declare function my:increasing($seq as xs:double*) as xs:boolean
> {empty($seq[2])
> or
>  $seq[1] lt $seq[2]  and  my:increasing(subsequence($seq, 2))
> };
> let $v:=(1 to 10000)
>  return my:increasing($v)
>
>
> And here is the result (do note this below: - marking as ***tail
> call***: my:increasing(fn:subsequence($seq_0, 2))  )
>
> Total Time: 3.74ms (for 100 000 - long sequence the time was 17.77ms,
> for 1 000 000 - long sequence the time was 207.56ms)
>
>
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