Dimitre,
I haven't the sleight-of-hand with writing this lingo, but wouldn't it be
worthwhile to try and check the ascending
property by *halving *the sequence and looking at each part separately?
When splitting, the last item of the first half must be less than the first
of the second, and then call recursively for each of the halves. I don't
know whether it would be faster, but it should reduce the risk of stack
overflow.
Cheers
Wolfgang
On 27 March 2015 at 10:37, 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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