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 $N is
an integer; that is, if sequences are implemented as linked lists rather than
arrays. Saxon will generally use an adaptive implementation where the sequence
is held as an array as soon as you start indexing into it, so the non-recursive
solution will work just fine. Other systems may differ.
Then, if you do use the recursive approach, you run into problems if the
processor doesn't do tail-call optimization. And if that's the case, you need
to consider a divide-and-conquer approach instead.
Michael Kay
Saxonica
mike(_at_)saxonica(_dot_)com
+44 (0) 118 946 5893
On 27 Mar 2015, at 09:36, Leo Studer leo(_dot_)studer(_at_)varioweb(_dot_)ch
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
Dimitre
thanks, this is amazing. With Saxon EE in Oxygen 16.1 I get stack overflow
with 10000 ;-).
Can you compare the time with this solution?
declare namespace my = "my:my";
declare function my:increasing2($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(_at_)gmail(_dot_)com
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> 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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