Bruce MacKenzie
Member
Good. That makes sense. After my post I solved all of my 83 cases and I noticed that many of the solutions started with U. It occurred to me that counting cubes that differ only by U turns as the same might be what they were doing. Thanks.To restate what DG said a bit more formally, PLL cases are considered to be identical if you can compose with U moves to get from one to the other, i.e. ∼ X \sim Y iff =U∘∘U X = \mathrm U^a\circ Y\circ \mathrm U^b for some ,∈{0,1,2,3} a, b\in\{0,1,2,3\} . Thus there should be sixteen symmetries, not just the four rotational symmetries (where you restrict it to +=0 a+b=0 ).
83 (actually 84 including the identity state) is the number of equivalence classes the group of 288 up face position permutations may be partitioned into by conjugations with the C4 rotation symmetries. Such equivalence classes are invariant under conjugation with those symmetries.How exactly did you end up with 83?
m<class>m' = <class> for all elements m of the C4 group
Form the symmetry conjugate of a class member and you get another member of the class. You don't get exactly four fold reduction because some elements are symmetric:
m * g * m' = g
e.g. the identity state which is the only member of its symmetry class.
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