[GAP Forum] some question on sets
Levie Bicua
lee_bkua at yahoo.com
Mon Apr 27 06:35:37 BST 2009
Dear forum,
Is it possible to group the elements of a set into subsets of 6 elements each? Suppose I
want the set T:=[elt1,elt2,elt3, ..., elt42] into S:=[[elt1 to elt6],[elt7 to elt 12], ..., [elt37 to
elt42]].. For the set
T:=[[ (), (2,3), (1,2) ], [ (), (2,3), (1,3) ], [ (), (1,2), (2,3) ],
[ (), (1,2), (1,3) ], [ (), (1,3), (2,3) ],[ (), (1,3), (1,2) ],
[ (2,3), (), (1,2) ], [ (2,3), (), (1,3) ], [ (1,2), (), (2,3) ],
[ (1,2), (), (1,3) ],[ (1,3), (), (2,3) ], [ (1,3), (), (1,2) ],
[ (2,3), (2,3), (1,2) ], [ (2,3), (2,3), (1,3) ], [ (1,2), (1,2), (2,3) ],
[ (1,2), (1,2), (1,3) ], [ (1,3), (1,3), (2,3) ], [ (1,3), (1,3), (1,2) ],
[ (2,3), (1,2), () ], [ (2,3), (1,3), () ], [ (1,2), (2,3), () ],
[ (1,2), (1,3), () ], [ (1,3), (2,3), () ], [ (1,3), (1,2), () ],
[ (2,3), (1,2), (2,3) ], [ (2,3), (1,3), (2,3) ], [ (1,2), (2,3), (1,2) ],
[ (1,2), (1,3), (1,2) ], [ (1,3), (2,3), (1,3) ], [ (1,3), (1,2), (1,3) ],
[ (2,3), (1,2), (1,2) ], [ (2,3), (1,3), (1,3) ], [ (1,2), (2,3), (2,3) ],
[ (1,2), (1,3), (1,3) ], [ (1,3), (2,3), (2,3) ], [ (1,3), (1,2), (1,2) ],
[ (2,3), (1,2), (1,3) ], [ (2,3), (1,3), (1,2) ], [ (1,2), (2,3), (1,3) ],
[ (1,2), (1,3), (2,3) ], [ (1,3), (2,3), (1,2) ], [ (1,3), (1,2), (2,3) ] ];
I used
S:=[T{[1..6]},T{[7..12]},T{[13..18]},T{[19..24]},T{[25..30]},T{[31..36]},T{[37..42]}];
and it returned
S:=[
[[ [ (), (2,3), (1,2) ], [ (), (2,3), (1,3) ], [ (), (1,2), (2,3) ],
[ (), (1,2), (1,3) ], [ (), (1,3), (2,3) ], [ (), (1,3), (1,2) ]],
[[ (2,3), (), (1,2) ], [ (2,3), (), (1,3) ], [ (1,2), (), (2,3) ],
[ (1,2), (), (1,3) ], [ (1,3), (), (2,3) ], [ (1,3), (), (1,2) ]],
[[ (2,3), (2,3), (1,2) ], [ (2,3), (2,3), (1,3) ], [ (1,2), (1,2), (2,3) ],
[ (1,2), (1,2), (1,3) ], [ (1,3), (1,3), (2,3) ], [ (1,3), (1,3), (1,2) ]],
[[ (2,3), (1,2), () ], [ (2,3), (1,3), () ], [ (1,2), (2,3), () ],
[ (1,2), (1,3), () ], [ (1,3), (2,3), () ], [ (1,3), (1,2), () ]],
[[ (2,3), (1,2), (2,3) ], [ (2,3), (1,3), (2,3) ], [ (1,2), (2,3), (1,2) ],
[ (1,2), (1,3), (1,2) ], [ (1,3), (2,3), (1,3) ], [ (1,3), (1,2), (1,3) ]],
[[ (2,3), (1,2), (1,2) ], [ (2,3), (1,3), (1,3) ], [ (1,2), (2,3), (2,3) ],
[ (1,2), (1,3), (1,3) ], [ (1,3), (2,3), (2,3) ], [ (1,3), (1,2), (1,2) ]],
[[ (2,3), (1,2), (1,3) ], [ (2,3), (1,3), (1,2) ], [ (1,2), (2,3), (1,3) ],
[ (1,2), (1,3), (2,3) ], [ (1,3), (2,3), (1,2) ], [ (1,3), (1,2), (2,3) ]] ];
which is exactly what I want, but the method is not that efficient.
I wonder if there is a more efficient way of doing this.
Levi
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