Newsgroups: rec.puzzles
From: hoey@AIC.NRL.Navy.Mil (Dan Hoey)
Date: Mon, 4 Jul 1994 01:15:29 GMT
Subject: Re: Rubik Nightmare (a.k.a. Rubik's Revenge)

Two days ago I wrote of the problem of exchanging the following pairs
of edge pieces on one face of a 4^3 Rubiks cube.

>  c e e c              c e * c
>  e f f e              e f f e
>  e f f *              e f f e
>  c * e c              c * e c
>  adjacent             opposite

I've put together processes for these now.  For people unfamiliar with
my notation:

0.  Moves consist of turning a 1x4x4 slab of the cube by ninety
    degrees with respect to the rest of the cube.
1.  Clockwise turns of the outer slabs are labeled by the first
    letters of the words Back, Front, Top, Down, Left, and Right.
2.  Clockwise turns of the adjacent inner slabs are suffixed "i".
3.  Anticlockwise turns (and inverses generally) are suffixed "'".
4.  Half turns are suffixed "^2", repeated operations generally
    are suffixed "^n".

Let us suppose the pictures refer to the F face in an upright
position.  The processes can be done as follows:

> 1.  Put both edge pieces on the same inner slab of the cube.

For the adjacent case, RT; for the opposite case FRF'T.  This puts
both in the Li slab, at the DF and TF positions.

> 2.  Turn the slab 90 degrees.

This is Li.

> 3.  Use commutators to undo the 90-degree turn of the face centers
>     of the slab.

As I mentioned, there is a shortcut for this: (Li' T^2)^6.  Actually,
the shortcut is a five-cycle of pairs of face centers, but it doesn't
matter unless like-colored centers are distinguished.

> 4.  Use commutators to perform the permutation of the edge pieces of
>     the slab to make step 5 do the right thing.

I use T'R'T Li T'RT Li T'R'T Li^2 T'RT.

> 5.  Do the inverse of step 1.

That's T'R' or T'FR'F'.

I've chosen conjugates of some of these processes so there is
cancellation to form the 29-qt process

  RT' (Li' T^2)^4 Li' TR'T Li T'RT Li T'R'T Li^2 T'

for the adjacent case.  The 35-qt process

  FRF'T' (Li' T^2)^4 Li' TR'T Li T'RT Li T'R'T Li^2 T' RFR'F'

for the opposite case is not so felicitous; I'm sure there's a better
way for this (call it an ObPuzzle).  If I hadn't been willing to allow
permuting like-colored face centers in step 3, a cyclic permutation of
the adjacent case would work.  What goes around, comes around.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil