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$\begingroup$ "I could eliminate the duplicates, but that still requires calculating the full list of nonunique results as an intermediate step, which I expect to be many orders of magnitude longer than the unique results. Is it possible to get the result I'm after without having to cull a humongously longer list first to get there?" $\endgroup$

thecommexokid
– thecommexokid

2020-04-09 06:33:17 +00:00
Commented Apr 9, 2020 at 6:33
$\begingroup$ the second approach does not give the full list of subsets. Take, for example, theSet = {a, a, b, c, c, c, d}. $\endgroup$

kglr
– kglr

2020-04-09 08:33:47 +00:00
Commented Apr 9, 2020 at 8:33
$\begingroup$ Yes, I agree. I think in this case one needs in addition to construct some subsets of list of different elements. $\endgroup$

Acus
– Acus

2020-04-09 08:42:32 +00:00
Commented Apr 9, 2020 at 8:42

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