Dear Forum,
I have been thinking about the problem Brandon is asking.
Here is a quick computation:
samecolors := [[1, 5, 12, 16, 17, 23],[3, 7, 10, 14, 19, 21],
[2, 4, 6, 8, 9, 11, 13, 15, 18, 20, 22, 24]];
# this is not stricly necessary, but helps me understanding the problem
lst:=[];
for i in [1..Length(samecolors)] do
for j in samecolors[i] do lst[j]:=i; od; od;
TU := (1, 2, 3, 4)(5, 7, 9, 11)(6, 8, 10, 12);
RU := (1, 2, 3, 4)(5, 7, 9, 11)(6, 8, 10, 12)
(13, 15, 17, 19)(14, 16, 18, 20)(21, 22, 23, 24);
RL := (5, 6, 14, 13)(1, 7, 22, 20)(2, 15, 21, 12)
(4, 8, 23, 19)(3, 16, 24, 11)(10, 9, 17, 18);
G:=Group([TU,RU,RL]); # order 88179840
S:=Group([RU,RL]); # S_4
H:=Stabiliser(G,lst,Permuted); # order 4320
orb:=Orbit(G,lst,Permuted);; # length 20412
orbs:=OrbitsDomain(S,orb,Permuted);; # 894
reps:=List(orbs,Minimum);;
len:=List(orbs,Length);;
Collected(len); # [ [ 4, 3 ], [ 8, 3 ], [ 12, 78 ], [ 24, 810 ] ]
This says that 810 of the 894 S-orbits are indeed regular, but there
are quite a few shorter ones. Usually, one would like to see how the
latter look like... (In the meantime, I have checked privately with
Brandon, that 894 indeed is the correct answer.)
Anyway, this runs in less than a second on my machine. So, although it
involves computing the full orbit of the puzzle group, and the suborbits
under the rotational symmetry group, it does not seem to make much sense
to invoke Burnside's Lemma to just find the number of S-orbits.
Of course this picture might change, when it comes to different problems...
Best wishes, Jürgen
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