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Date: Sat, 07 Jun 1997 16:14:04 -0400
To: cube-lovers@ai.mit.edu
From: karen angelli
Subject: 5x5x5 practical Q's
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>From Peter Reitan, not Karen Angelli:
The 'parity' problem Mark Pilloff discussed in his 5x5x5 practical Q's
posting is a common frustrator when it comes to solving either the 4x4x4 or
5x5x5. If you've ever read any of the on-line descriptions of Rubik's
Revenge solutions in various web pages, some of them suggest mixing the
cube back up and starting over, hoping to get the correct orientation 50%
of the time. However, there is help available.
I happen to have a nifty algorithm that solves your problem. However, I am
not fluent in cube-algebra notation, so I will try to explain it as well as
possible. This algorithm is equally applicable to either the 4x4x4 or 5x5x5.
When I solve either cube, I solve opposite sides, corners, edges and center
pieces first. My second goal is to correctly orient the edge cubies. On
the 5x5x5, the center edge cubies is equivalent to solving the 3x3x3
edgies. The 4x4x4 edgies is the same as solving the 5x5x5 non-center
edgies. I have, and it sounds like Mark Pilloff has, moves that will pair
up the appropriate edgies together. As he notes, half of the time, one
pair of the edgies are flipped into the wrong orientation. To flip the
wayward edgie pair (w-edgies) back into line, I use the following method,
which I 'discovered' through brute force trial and error, and my faith in
cube symmetry.
Hold the cube so that the offending w-edgies are oriented horizontally, on
the top of the cube, and at the back of the cube. Each individual edgie
cublet of the offending w-edgie pair belongs to its own vertical slice -
one to the left of center running vertically around the cube (L-slice) and
the slice to the right of center (R-slice) (these are the interior vertical
slices, not the right or left face slices). The move involves only 180
twists of the right Face (T-Face) and 90 twists of the R-Slice, either
toward you or away from you.
1. a. 180 T-Face
b. 90 R-Slice away from you
c. 180 T-Face
d. 90 R-Slice toward you
e. Rotate the entire cube toward you 90(move the top front edge to the
bottom front edge.)
2. Repeat 1.
3. Repeat 1.
4. Repeat 1, steps 1-3.
The parity problem is now more or less solved, but you have a couple more
steps to put the edgies back to their final resting place.
You will notice that two things have happened - 1. the center slices are
now rotated 90 degrees from the orientation of the right and left face
slices (you can fix that simply), and 2. the edgie pair whose parity
orientation you have corrected has now swapped places with one of the other
edge pairs, (the edge pair that started on the front top edge). This can
be quickly corrected with the following algorithm (which you probably
already know):
Orient the cube so that the swapped edge pairs are on the top face, and one
of the pairs is at the front top edge and the other pair is at the back top
edge.
1. a. 180 T-Face
b. 180 R-Slice
2. repeat 1
3. repeat 1
4. a. 180 T-Face
Since this method of arranging the outer, interior edge cubies on the 5x5x5
does not preserve the center edge cubies, I wait solve the center edge
cubies afterwards. This is done in the same way as the 3x3x3. Although
correct orientation of the center cube edgies can corrupt the two opposite
faces that I solved first, as a matter of style, I always like to have the
two opposite sides solved first. That way, I solve a greater percentage of
the cube using intuitive moves, rather than with the more constrained
end-game moves where you need to preserve a greater portion of the cube
with each new solution.
Having written these moves down, I am struck by their similarity to each
other, and their similarity to the ARA'R' method Wei-Hua Huang describes
(with apologies to God) as the most elegant algorithm of all. The core of
both of the algorithms I describe can be represented with ARA'R'.
1. a. 180 T-Face TF
b. 90 R-Slice away RS
c. 180 T-Face TF'
d. 90 R-Slice toward RS'
e. Rotate cube 90 90
1. a. 180 T-Face TF
b. 180 R-Slice RS
2. repeat 1 TF'
RS'
In addition, the first algorithm involves three iterations of one
algorithm, and then an incomplete core algorithm. The second algorithm,
similarly, involves three repetitions of a core algorithm, followed by an
unfinished core algorithm.
Step 1
Repeat 1
Repeat 1
Repeat most of 1
Not being a mathematician, I don't know what else to say. Perhaps I'll
think about it a bit longer. Any comments?
'e-ya later,
Pete