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I've managed to solve C4 in a way similar to those mentioned, but using a different transform. I solve C3 a little differently than most folk because I use only one (!) transform to do ALL of the edge work. On C3 the result of this transform is (looking at the edges only): a a from b c to ~d b d ~c

Using it and its inverse and a few simple conjugates of it I can do all the

work that is necessary for the edges. For me, at least, it has the twin

advantages of 1) being easy to remember (since it is fairly short and there is

only one of it), and 2) it has such bounded consequences that it is easy to

fix a cube without requiring a lot of planning (in the picture above, nothing

that is not shown changes: no other edges, no corners).

Anyhow, since that little move is a favorite of mine, I tried it on C4. On C3 it comes in two flavors (the move and its inverse, or as it turns out, the right-handed and left-handed versions). On C4 it comes in four versions: the move and its inverse, but each in a `left central slice' and `right central slice' version. Now for the fun part: First off I started ignoring the centers and I noticed that the move (lets call it `M') only moves around edges in a single plane. As I tried to figure out what the damn thing did I discovered that it is a move of order 5!!!. I find it truly hard to plan out what happens when five cubes move around a little orbit, but I'm getting better at predicting it. The result is: from: to: a b a b c d ~d e e f ~f g g h c h

With some pain I have been able to use ONLY M (and conjugates and powers of it,

of course) to get all the edges in place. Then I looked at M a little more to

see what it did with centers. This one is NOT planar, unfortunately, but is

simple enough to be useful: only two sets of centers are affected. If you are

doing M on the top (to get the above edge transform), only the top centers and

the rear centers are affected. And what happens is that there are two disjoint

three-cycles each involving two of the top center cubes and one of the rear

center cubes. thus, you can easily use conjugates of M to move cubes, one or

two at a time, into place on all of the faces around any particular top.

Voila: done!

An amusing thing about M: since its edge permutation is of order 5 and the

center permutation is of order 3, the damn thing is an order 15 move.

/Bernie