Edits: In fact it is a set partition problem.
I have the following subsets:
s:={"a","b","c","d","e","f"}:
I want to divide the set into 2 groups such that each group has exactly 3 elements. We want to generate all solutions. Note that we treat ({"a", "b", "c"}|{"d", "e", "f"}) and ({"d", "e", "f"}|{"a", "b", "c"}) as the same solution.
So first I used the choose function to generate all 3-subsets and then filtered for duplicate solutions.
with(combinat):
M:=[op(choose({"a","b","c","d","e","f"},3))]:
l:=[ListTools:-Categorize((x,y)-> is(x =`minus`(s, y)),M)]:
seq(l[i][1],i=1..10)
{"a", "b", "c"}, {"a", "b", "d"}, {"a", "b", "e"}, {"a", "b", "f"}, {"a", "c", "d"}, {"a", "c", "e"}, {"a", "c", "f"}, {"a", "d", "e"}, {"a", "d", "f"}, {"a", "e", "f"}
But I feel like the solution is a bit inefficient. In fact, we only need half of all 3-subsets. I wonder if we can only keep one representative member of the same solutions when generating the 3-subsets.
gengcuts.mw
In fact, the problem can be generalized as follows: given a list of n elements, divide it into m groups and generate all solutions. Similarly, the same solution is generated only once.
