From: Nichael Cramer <nichael@sover.net>
To: Roger Broadie <roger.broadie@iclweb.com>
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: 4x4x4 solution
Message-Id: <Pine.BSI.3.91.971001225256.25718B-100000@granite.sover.net>

Roger Broadie wrote:
> A slice turn produces a 4-cycle on the edges and two 4-cycles on the
> face centres; a face turn produces a 4-cycle on the face centres and
> two 4-cycles on the edges (and a 4-cycle on the corners, which may or
> may not be relevant).
>
> I think it is very relevant. We can set the effects out as follows:
>
> Turn		Piece		Cycle(s)	Parity
> ------        -------         --------        -------
>
> Slice         edge            1x4                odd
>               centre          2x4                even
>
> Face          edge            2x4                even
>               centre          1x4                odd
>               corner          1x4                odd
>
> The consequence is that the parity of the centre pieces depends
> entirely on the number of face turns - any slice turns do not affect
> the parity of these pieces since the changes they introduce will be of
> even parity.  For face turns, the changes to the parity of the corner
> pieces and the centre pieces are the same.  Hence if the corner pieces
> are in place, the centres will be in an even permutation, and that will
> not be changed even if the edge pieces are in an odd permutation, which
> was the essence of Clive McCaig's original question. Nor will that be
> changed by any turn of a central slice to bring them back to an even
> permutation.


As one of the folks who advocated rotating a center slice, let me explain
my (admittedly non-optimal) process for getting out of this fix and perhaps
you can explain where my reasoning is wrong.

1] Imagine a 4X which is completely solved except for two flipped (i.e.
swapped) edge-pieces.

2] For simplicity's sake --and without loss of generality--, assume the 2
flipped/swapped pieces are adjacent and in the top front location.  So the
top of the cube will look like this:

   X X X X
   X X X X
   X X X X
   X 1 2 X

(Here the numbers are meant to indicate only where the cubies are located,
having nothing to do with their colors.)

3] I now rotate one of the center slices (say, the one on the right, i.e.
the one containing the cubie "2") 90dg away from me.

4] The top of the cube now looks like:

   X X 2 X
   X X O X
   X X O X
   X 1 3 X

5] I can now perform the 3-cycle 1->3->2 (i.e. without affecting any of the
rest of the cube).

The top of the cube now looks like:

   X X 3 X
   X X O X
   X X O X
   X 2 1 X

6] In particular, note that "2" and "1" are now in their correct positions
(and, of course, necessarily in their proper "flip" orientation).

7] Moreover, note that I now have exactly three edge cubes in the wrong
place (i.e. "3" from above and the other two edge cubes which were
misplaced during my original 90dg rotation of the center slice).

I can now perform a 3-cycle on these edges pieces (similar to the one used
in step 5 above) again without affecting any of the other locations on the
cube.

8] My cube now has all the edge pieces in their correct location.

9] I now have only to "fix" the 8 central-face cubes which were misplaced
during my initial 90dg twist.  I can now do this is short order.

QED[?]

Nichael Cramer
work: ncramer@bbn.com
home: nichael@sover.net
http://www.sover.net/~nichael/

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