From: Jim Mahoney
To: ERCO@compuserve.com
Cc: CUBE-LOVERS@ai.mit.edu
Subject: Re: 4x4x4 solution -- [Digest v23 #159]
>Your problem is one of parity. You have two edges cubies swapped
>(this swap is visible) and two face center (centre) cubies of the
>same color swapped (this swap is invisible). You have to have an
Edwin> I've had this problem as well.... If I understand you
Edwin> correctly, this problem simply doesn't occur anymore as
Edwin> soon as you number (or mark in any other way) the center
Edwin> pieces which a) makes solving the cube a bit more difficult
Edwin> b) makes sure that you'll always get back to the original
Edwin> configuration of center pieces.
This isn't quite true, at least not on the 4x4x4.
While it is true that parity is the question at hand, and also that on
the 4x4x4 cube a quarter of a central slice performs an odd permutation
on the edges which is otherwise "invisible", it is *not* true that
marking the centers will help. The reason is that a quarter turn on a
center slice of the 4x4x4 performs a cyclic rearrangement of 4 edges -
an odd permutation - while at the same time rearranges *two* sets of 4
central pieces - an even permutation of the centers. Thus parity does
not prohibit swapping two edges while leaving the centers untouched.
Moreover, in fact there are move sequences which will exchange two
edges without disturbing the position of any other piece, corner or
center - though I don't have any on hand which are short. If there's
interest, though, I can produce a move sequence to exchange two 4x4x4
edges while leaving all corners and centers in their original
positions.
A cross-section looks like this. A quarter turn cycles the four E's,
the four C1's, and the four C2's. This is an odd permutation of the
E's but an even permutation of the C's. (All the C's are corners, and
can be put into each other's positions with a combination of face and
center turns.)
E C1 C2 E
C2 C1
C1 C2
E C2 C1 E
A full account of parity and possible 4x4x4 moves gives
4x4x4
type , how many , parity after: 1/4 face turn , 1/4 center turn
- ---------------------------------------------------------------------
corners | 8 | odd | even (untouched)
edges | 24=2x(12 edges) | even (8 move) | odd
centers | 24=4x(6 faces) | odd | even (8 move)
Thus to solve a 4x4x4 cube you must have made both (1) an even
total number of moves on the faces (to restore the corners and centers to
even parity), as well as (2) an even total number of moves on the center
slices, to restore the edges to even parity.
The parity constraints on the 5x5x5 are a bit different. In that case
there are two types of edges (the one in the middle of an edge vs the
ones next to the corners) and three types of centers. Each has its
own parity change under each different slice. A bit of playing around
shows that any central slice move which cycles 4 edges must also cycle
several kinds of centers. At least one of those center cycles is odd.
Therefore on the 5x5x5 you cannot exchange a pair of edges without
also exchanging two centers somewhere. So marking where the centers
go will help on the 5x5x5.
Regards,
Jim Mahoney mahoney@marlboro.edu
Physics & Astronomy
Marlboro College, Marlboro, VT 05344
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End of Cube-Lovers Digest
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