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Re: [xsl] All combinations from a sequence

2021-09-30 09:37:33
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 Müller-Hillebrand mmh(_at_)docufy(_dot_)de 
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> 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


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