But in principle you only have to produce the numbers 0..2^(N-1) in binary
form and filter (index) your set by the 1's (present), 0 absent..
In fact, if Martin needs to filter out some subsets, then filtering out
integers and only generating subsets for the integers that remain, might
even be faster.
groundset := [1,2,3,4]
subset(li, n) == [x for x in li for i in 0..#li-1 | bit?(n, i)]
sort((x,y)+->#x<#y or #x=#y and x<y,_
[subset(groundset, n) for n in 0..2^(#groundset)-1])
The question is whether sorting the subsets is needed.
Ralf
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