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Subject: Antipodal Edge Position
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.600.5834.0C18D8B1@canrem.com>
Date: 	Wed, 8 Dec 1993 12:33:00 -0500
Organization: CRS Online  (Toronto, Ontario)

>>It's got to be all edges flipped in place.

Oops. Well I figured if all edges flipped was one of the hardest
know cube states that in the case of edges-only it would be the
antipode. I'm now sure (I think) that it is really:

       all edges flipped + 4 X
(with the 4 X on sides F, R, B, L which should match Dan's diagram)

Hmmmm, I don't know if this is a standard form of representation,
but this picture looks like a folded out cube:

               + T +                        + F +
               T   T                        R   L
               + T +                        + B +

        + L +  + F +  + R +          + D +  + D +  + D +
        L   L  F   F  R   R    =>    F   B  R   L  B   F
        + L +  + F +  + R +          + T +  + T +  + T +

               + D +                        + B +
               D   D                        R   L
               + D +                        + F +

               + B +                        + T +
               B   B           --------->   R   L
               + B +           |            + D +
                               |

                             + D +
In my program I would have   L   R  on the screen for the bottom face.
                             + T +

The idea is you are always looking at a cube face head-on (just to
clarify the difference in diagrams).

More quotes for Jerry Bryan:
>The "edges of the 3x3x3 without centers" is a little tougher.  Early
>in the project, I generated a data base for the first few levels
>(six or seven, I think), and I have a "Solver program" that will
>work up to that level.  However, the full "edges of the 3x3x3 without
>centers" is a 4.2 gigabyte file on tape, so it is hard to process.
>Also, the size of the equivalence classes is not in the data base,
>only the level.  I have to calculate the size of each equivalence
>class, and it is an expensive calculation.
>
>I made a pass at the
>file and calculated the number of equivalence classes (took
>125 hours on a mainframe), but I only saved a summary.  I did not
>save the number of equivalence classes for each state.  I found
>the antipodal by looking for level 15, since I knew there was
>only one occurrence, and since the level was in the data base.
>
>I am not yet for sure what they look like, but of the other two states
>with order-24 equivalence classes, one is at level 9 and the other
>is at level 12.  Since the only one at an even level is at level 12,
>I am assuming that will be the one which is Start with the edges all
>flipped.  The one at level 9 will probably be the mirror image of
Start.

I'd still like to see the process for all-edges-flipped (not
caring about the centres or corners). So "level" is the number of moves
required to solve the position? That means edges flipped in place
can be done in 12 qtw.