Date: Tue, 30 Apr 96 07:42:38 -0700
~~~ ~~~ From: Walt Smith <walts@federal.unisys.com >
~~~ ~~~ Subject: Rubiks Revenge

Rubiks Revenge presents several difficult cases when the last few pieces are
reached.

There has been discussion on this over the years but it does not seem to have
reached closure.

Mark Longridges list of best known solutions,
(http://admin.dis.on.ca:80/~cubeman/cmoves.txt) lists some of these but here
are some improved solutions of my own and additional operators that are
needed to solve it.

If the center four are done on all sides and the edge pieces are paired up,
it can be can be treated like a 3x3x3 Rubik Cube. Several cases come up at
the end that can not occur on a 3x3x3. If the corners are done and the last
few edges are left for last, four cases occur. Here are my methods that are
shorter than any others I have seen.

Case 1. The last three edges can be solved with 3x3x3 techniques.

Case 2. Two edge pairs are swapped. (swaps RF and RB)

```d2  R2  d2  rR2  d2  r2     (7)

This is based on combining the following two sequences:
(d2  R2)2  d2     and    (d2  r2)2
```

Each of these is useful in itself.

This is shorter than Marks Longridges p4.
(Marks p4 is mislabeled Opp. It should be Adj and p5 should be
labeled Opp)

Case 3. A single edge pair is flipped. (flips BL)

```L2
d1
R2  d1  R2  d3  L2  u3  B2  u2  B2  u3
B2  R2  B1  r3  B3  R2  B1  r1  B1     (21)
```

This is shown in four groups because it proceeds in stages. The second move
fixes the parity, the first along with 3rd through 12th fix the faces and the
last group of moves fix the edges.
This is shorter than Marks p3.

Case 4. Both Case 2 and Case 3 exist. The flipped edge pair might be one of
the swapped edge pairs or it might not be. Obviously this can be solved by
using the techniques of Case 2 and 3 applied separately. I have always
thought that it should be possible to find an operator that is shorter than
the sum of these two and possibly shorter that Case 3. I have not done as
much study on this case as the others.

If you want to solve the corners last (avoids Case 3 and 4), you may still
need to solve Case 2 but you may also need to swap two corner pieces. This
can be done by applying Case 2 then fixing the edges then corners. This
will take about 40 turns. The following does it quicker.
(swaps LDB and LDF with the bottom cubie faces remaining bottom faces.)

```R3  D1  L1  D3  R1
D2  L3  D1  L1  D2
L1  U3  r2   F2  r2  fF2  r2  f2  U1  L2    (21)
```

It is listed in three groups to make it easier to memorize as it proceed in
three stages.
This is shorter than the sequence in Mark Jeays on-line solution at