[next] [prev] [up] Date: Mon, 04 Aug 80 22:04:00 -0700 (PDT)
[next] [prev] [up] From: Tom Davis <Davis@OFFICE-3 >
~~~ [prev] [up] Subject: Cube Permutations

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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