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The following two transformations are useful in demonstrating that all of the

claimed elements of an equivalence class of positions can be reached. Using

the RLFBUP notation, the transformation RB'R'B'U'BU reverses two of the corner

cubies and leaves all other corner cubies in place (It does, however, shuffle

around the side cubies, and does some random twists to the corners). If the

above transformation is repeated four times in a row, everything is left

exactly fixed, except that three corner cubies are each rotated one-third of a

turn. By then performing the inverse of this operation on two of the three

corners which were turned and a new corner, it is not hard to see that any two

corners can be the only ones moved, and that they are each rotated one-third of

a turn in opposite directions.

If we look at just the corners alone, and ignore in-place rotation, since we

can exchange any adjacent pair, we can obviously get to all permutations of the

corners. A similar argument can be made to show that all the edge cubies can

be arranged arbitrarily (permuted arbitrarily, that is). An easy

transformation rotates three of them (among themselves) on a face, and since we

are also allowed to rotate the face, it is easy to generate a transposition of

any pair.

Unfortunately, the two operations described above are not independent. If we

just look at the blocks and label them ignoring color, a primitive (one-quarter

turn) transformation moves four corners into four corners, and four edge cubies

into four edge cubies. If this is viewed as a member of a permutation group,

it is obviously even (the set being permuted is all the movable cubes). Thus,

at least half of the positions are impossible. If we ignore the corner cubies,

and look at the colors of the edge cubies, every primitive rotation rotates the

four front colors, and the four colors around the outside, again, an even

permutation. Since one of the above used coloring and no corner cubes, and the

other did not, there must be at least a factor of 4 impossible positions. A

much more complicated argument shows the necessity of a factor of three in the

set of impossibles. (Does anyone know of a simple way to see this? I just did

the obvious thing of defining "standard" orientations of every cube in every

corner, and showed that all the primitive transformations caused the total

rotation away from standard to be a multiple of 2 pi.)

Using the transformation which flips any two in place, and the two discussed

above, it is not hard to see that the factor is at most and at least 12.

Boy, It sure is hard to prove things on a computer. On myy notes with

diagrams and all, this is perfectly clear, but I get confused trying to read my

online proof. I hope that someone can make some sense out of it.

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