Re: [xsl] All combinations from a sequence

["Michael Kay mike@xxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

There's a nice algorithm here

https://www.geeksforgeeks.org/power-set/

which abstracts to

for $i in 1 to math:pow(2, count($input))
return combination($i)

where combination($i) includes or excludes each $input[$N] depending on
whether bit $N is set in $i, which you can determine using bin:shift() from
the EXPath binary module.

Michael Kay

> On 30 Sep 2021, at 15:20, Michael MC<ller-Hillebrand mmh@xxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> Good afternoon,
>
> I have a sequence of items and I need all combinations (not permutations) in
all possible lengths.
>
> I saw what I want described as "powerset" in the Python docs:
powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)
>
> In XPath notation and based on strings:
>
> my:powerset(('A','B','C','D'))
>
> This sequence of 4 items should result in a sequence of 16 strings (order
not important) representing all possible combinations: 'ABCD', 'ABC', 'ABD',
'ACD', 'AB', 'AC', 'AD', 'A', 'BCD', 'BC', 'BD', 'B', 'CD', 'C', 'D', ''
>
> Or more general, the result could be an array of sequences.
>
> To get this as a solution in XSLT/XPath I am currently fiddling around with
a recursive function including head() and tail() and count() but I have the
impression I am overcomplicating things.
>
> I am wondering, if this is a use case for fold-left() or if I should rather
think of a filter that drops 0, 1, 2 or 3 items from the sequence. Or is there
a well-known algorithm with a cool name?
>
> Any hints are, as always, very welcome, thanks a lot,
>
> - Michael