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Subject:

(Sorry.) The right answer should be:

The state of the cube is not:

X|O|X|X|X X|A|X|C|X X|X|X|X|X X|X|X|X|X X|X|X|X|X But rather: X|X|X|X|X X|X|X|X|X X|X|X|X|X X|X|X|O|X X|X|X|B|XWhere cubie "C" just "looks" like it's in the right place.

You need an operator that rotates A->B->C->A. (Left as an exercise; hints

available on request.)This will very likely leave an inconvenient number of edges flipped. For

the answer to _this_ problem, see my last post. ;)

I think you must be wrong here (but would love to be proved wrong--I'm no

mathematician so group theory is very much beyond me).

Proof #1:

We hold the cube with the red face on top, and the yellow face in

front (colors obviously don't matter, but I find it easier to discuss using

them). We will assign a parity to the edge cubies, being defined by holding

the cube such that the red face of the cubie is on top and the yellow in

front. If the cubie is on the left as we look at it in this position it is

parity 0, on the right it is parity 1.

There are only two operations available which affect the cubie we

are interested in: rotating the front face 90deg; and rotating the slice the

cubie is in 90deg. It is easy to see that neither of these moves alters the

parity (assume the cubie's frame of reference, and think of rotating the

rest of the cube around it--it is clear that it will not end up on the other

side).

Therefore the move C->A in the above is impossible.

Proof #2:

Take apart the order 4 cube (my falls apart depressingly easilly)

and try to reassemble it with the two edges exchanged. It will not fit, as

they are mirror images of each other.

Note that you get an apparant parity reversal by flipping the

cubies, but this does not actually move anything. In other words, no amount

of flipping and moving will allow you to end up moving A->B->C->A. That's

why I solve edges first.

Ronnie