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\title{How to solve the Rubik's Cube}
\author{}
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\begin{document}
\maketitle
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\section{Introduction}
The following is an algorithm for solving the Rubik's Cube. The algorithm is
not my own; rather, it is one I memorized from a book I read (and subsequently
lost) twenty years ago. I don't remember either the author's name or the
title, so I can't attribute it properly.
(Note June 2008: The book was \emph{The Simple Solution to Rubik's Cube}
by James G. Nourse, Bantam, 1981.)
Some of the (very few) group-theoretical observations come from David
Singmaster's (apparently self-published) 1980 pamphlet \emph{Notes on Rubik's
Magic Cube}. Singmaster's pamphlet contains a different algorithm for solving
the cube.
Most of this document describes notation, provides context, etc. The actual
number of moves that must be memorized is quite small. Using this algorithm,
after some practice you can expect solution times of 2-3 minutes.
\emph{John Kerl, Oct. 2003}
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\section{Notation}
\begin{itemize}
\item \textbf{Faces}: Pick a face of the cube; consider this to be the top
face. Of the remaining four side faces, pick one to be the front. In this
orientation, the faces may be labeled U, D, F, B, L and R, for up, down, front,
back, left and right.
\item \textbf{Pieces}: Edge and corner pieces will be named by pairs and
triples of letters, respectively, e.g. $UF$ for the upper front edge and $UFR$
for the upper front right corner.
\item \textbf{Turns}: Denote a clockwise (as viewed from outside the cube)
quarter-turn of a given face by its letter, in boldface. For example, $F$ is
the front face; $\mathbf{F}$ is a clockwise quarter-turn of the front face.
Then a half turn of the face will be denoted by the exponent 2, e.g.
${\mathbf{F}}^2$, and a counterclockwise quarter turn will be denoted by an
apostrophe, e.g ${\mathbf{F}'}$. (Mathematically it would be more appropriate
to use an exponent of $-1$, but the little apostrophes are less obtrusive.)
\item \textbf{Processes}: A process is nothing more than a sequence of moves
applied left to right. A process will be denoted by the string of letters and
their associated exponents, e.g. $\mathbf{UF^2R'}$ means turn the upper face a
quarter-turn clockwise, then turn the front face a half turn, then turn the
right face a quarter-turn counterclockwise. Spaces or dots have no
significance other than to separate mnemonic units.
Also note that $U$ will be $U$ for the entire duration of the solution.
However, $F$ will change as you rotate the cube.
\end{itemize}
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\section{Overview}
The general idea is:
\begin{itemize}
\item The center pieces are fixed via internal posts.
\item Put the top edges in their correct positions and orientations.
\item Put the top corners in their correct positions and orientations.
\item Make sure side centers are correct.
\item Put the four middle-row edge pieces in the correct positions, with
the correct orientations.
\item Put the four bottom corner pieces in the correct positions.
\item Put the four bottom corner pieces in the correct orientations.
\item Put the four bottom edge pieces in the correct positions.
\item Put the four bottom edge pieces in the correct orientations.
\end{itemize}
At each step, you put another 4 pieces into place, while preserving all the
pieces already done. The entire top face is easy to do, since you have so many
degrees of freedom --- so easy, in fact, I won't bother to write it down. The
processes get more and more complicated as you go along, since you have more
solved pieces to keep in place.
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\section{The constructible group vs. the solvable group}
\emph{This section is optional, but do read its last two paragraphs.}
If you take apart the cube and randomly assemble it, you will get a permuation
on the positions and orientations of the edge and corner pieces. Corners can
only go to corners and edges can only go to edges, and centers are fixed.
Since there are 12 edges with 2 orientations each and 8 corners with 3
orientations each, there is a total of
$$
12! 2^{12} \; 8! 3^8 = 2^{29} 3^{15} 5^3 7^2 11 = 519,024,039,293,878,272,000
$$
(519 quintillion) possible constructions, called the \emph{constructible
group}.
However, if you try to solve a randomly reassembled cube using the following
algorithm (or any other) you will probably fail. It can be shown that the set
of permutations of the cube obtainable by starting at the solved cube and
turning the faces (the \emph{solvable group}) is smaller than the constructible
group by a factor of 12:
\begin{itemize}
\item Permutations of edge and/or corner pieces must be even: Unless your cube
has been taken apart and reassembled incorrectly, you'll never see the cube all
solved except for two interchanged edges, or two interchanged corners. You can
have two pairs of edges interchanged, or three edges cyclically permuted. You
can have two pairs of corners interchanged, or three edges cyclically permuted.
You can have a pair of edges interchanged and a pair of corners interchanged.
\item Flips of edge pieces (in the correct position, but with orientation
changed) must have even parity, i.e. sum to zero mod 2. You won't see the cube
solved except for a flipped edge; flipped edges come in pairs.
\item Spins of corner pieces (in the correct position, but with orientation
changed) must sum to zero mod 3. You won't see the cube solved except for a
spun corner, nor two corners both spun clockwise. If one is spun clockwise,
another will be spun counterclockwise. Or, you may have three corners all spun
clockwise (or counterclockwise).
\end{itemize}
%It can be
%shown that solvable permutations on the positions of edge and corner pieces are
%\emph{even permutations} --- i.e. you can swap four edges or corners, cycle
%three edges or corners, swap two corners and two edges, etc., but you can never
%get just a pair of edges or corners swapped. Also, it can be shown that the
%orientations of the edges (\emph{flips}) maintain parity, e.g. you can't flip
%just one edge, and the orientations of the corners (\emph{spins}) always sum to
%0 mod 3, e.g. you can spin three corners clockwise, or one corner clockwise and
%another counterclockwise, but not just two corners clockwise.
That is, the solvable group has index 12 in the constructible group, and
has order
$$ 12! 2^12 8! 3^8 / 12 = 2^{27} 3^{14} 5^3 7^2 11
= 43,252,003,274,489,856,000 $$
The \emph{sticker group}, the set of patterns obtainable by peeling off the
stickers and replacing them, is astronomically bigger than the constructible
group; your chances of solving such a cube are next to zero. Moral: If a
toddler has been playing with your cube, please disassemble it and reassemble
it first before attempting to solve it by moves on the faces alone.
It may seem odd for me to discuss the things that \emph{can't} happen, but
these facts (for me) serve as mnemonic devices to help remember the choices in
the algorithm below. (Also, if you \emph{do} encounter one of these forbidden
configurations while solving your cube, now you'll know to take it apart and
put it back together into a solvable configuration.)
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\section{Top edges}
First, from a scrambled state, put the upper edges in place. Make sure the
side color of each upper edge piece matches the color of the center piece on
that side --- that is, put each upper edge in the correct position, not merely
somewhere on top of the cube.
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\section{Top corners}
Second, preserving upper edges, put the upper corners in place.
Perhaps obvious but I'll point it out anyway: Make sure the edge and corner
pieces have not only their top color matching the top face of the cube, but
with their side colors matching the corresponding center side pieces. That is,
getting ``all one color'' on top does \emph{not} mean the top is solved.
Rather, once the top is solved you must have three same-color squares along the
top row of each side.
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\section{Middle centers}
Now make sure side centers are correct, e.g. if the top row of the front face
is white, turn the remaining two rows of the cube until the white center is in
front.
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\section{Middle edges}
There are four possibilities for the edge piece that belongs in $FR$:
\begin{enumerate}
\item It is already there, with the correct orientation.
\item It is already there, with the incorrect orientation. Use either one of
the following two processes to put another piece there, which
will push it down to the bottom, then apply the appropriate one of
the following two processes.
\item The piece that belongs in $FR$ is on the bottom row, with its
$R$ color visible from the side of the cube: Turn $D$ to put
this piece onto the $R$ face, then apply:
$$ \mathbf{DFD'F'\cdot D'R'DR} $$
\item The piece that belongs in $FR$ is on the bottom row, with its
$F$ color visible from the side of the cube: Turn $D$ to put
this piece onto the $F$ face, then apply:
$$ \mathbf{D'R'DR \cdot DFD'F'} $$
\end{enumerate}
Do the above for all four middle-row edge pieces. Already, your cube is
starting to look nice --- the top is one solid color, and the four sides are
$2/3$ done.
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\section{Bottom corner positions}
There are three possibilities. Rotate the $D$ face until one of them is true.
\begin{enumerate}
\item All four corners are in the correct position. Proceed to orient them
below.
\item Two corners are in the correct position, and the two which are
interchanged are \emph{not} diagonally opposite one another. Turn the
\emph{entire cube} until the two mispositioned bottom corners are at
$FLD$ and $FRD$, then apply:
$$ \mathbf{R'D'R\cdot FDF'\cdot R'DR \cdot D^2} $$
\item Two corners are in the correct position, and the two which are
interchanged are diagonally opposite one another. Apply the above
process (it doesn't matter which face is front), then turn the cube
$180^\circ$ (so $U$ is still $U$, but $F$ moves to $B$), and apply the
same process again, then rotate $D$ until all four corner pieces are in
their correct positions.
\end{enumerate}
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\section{Bottom corner orientations} There are several possibilities, and
conceivably there are specific processes to handle each. In an attempt to
minimize the number of processes to memorize, though, the author of the book I
read suggested repeated applications of a single process which spins three
corner pieces. You may need to apply this several times. Additionally, there
is a process (from Singmaster) which spins a pair of corners.
\begin{enumerate}
\item To spin $FRD$, $BRD$ and $BLD$ counterclockwise while keeping $FLD$
fixed, apply
$$ \mathbf{R'D'RD'\cdot R'D^2RD^2} $$
\item To spin $FRD$, $BRD$ and $BLD$ clockwise while keeping $FLD$ fixed,
apply
$$ \mathbf{D^2R'D^2R \cdot DR'DR} $$
\item To spin $FLD$ counterclockwise and $BRD$ clockwise, apply:
$$ \mathbf{BL'U^2 \cdot LB'D^2 \cdot BL'U^2 \cdot LB'D^2} $$
\item To spin $FLD$ clockwise and $BRD$ counterclockwise, apply:
$$ \mathbf{D^2BL' \cdot U^2LB' \cdot D^2BL' \cdot U^2LB'} $$
\end{enumerate}
Keep applying the above until you get the bottom corners oriented.
It will take at most a few iterations.
Now your cube is starting to look \emph{really} good. There are at most four
edge pieces jumbled, all on the bottom of the cube.
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\section{Bottom edge positions}
\begin{enumerate}
\item To cycle $RD$, $BD$ and $LD$ clockwise while keeping $FD$
fixed, apply
$$\mathbf{L'RF\cdot LR'D^2 \cdot L'RF \cdot LR'}$$
\item To cycle $RD$, $BD$ and $LD$ counterclockwise while keeping $FD$
fixed, apply
$$\mathbf{L'RF'\cdot LR'D^2 \cdot L'RF' \cdot LR'}$$
\end{enumerate}
If all four edges are out of place, apply either one of the above to
get at least one edge in place. Rotate the cube to put that edge in front.
Then, apply the appropriate one of the above.
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\section{Bottom edge orientations} There are now two flipped edges
directly across from one another, two flipped edges not directly across
from one another, or four flipped edges.
\begin{enumerate}
\item If all four flipped edges are flipped, pick either of the following
processes and do it twice (rotating the entire cube of course
in between iterations, to flip one pair, then the other pair).
\item To flip $FD$ and $BD$, apply
$$
\begin{array}{lll}
\mathbf{L'RF} &\cdot &\mathbf{LR'D}\\
\mathbf{L'RF} &\cdot &\mathbf{LR'D}\\
\mathbf{L'RF^2}&\cdot &\mathbf{LR'D}\\
\mathbf{L'RF} &\cdot &\mathbf{LR'D}\\
\mathbf{L'RF} &\cdot &\mathbf{LR'D^2}\\
\end{array}
$$
\item To flip $LD$ and $BD$, apply
$$
\begin{array}{lll}
\mathbf{L'RF} &\cdot &\mathbf{LR'D'}\\
\mathbf{L'RF'} &\cdot &\mathbf{LR'D'}\\
\mathbf{L'RF^2} &\cdot &\mathbf{LR'}\\
\end{array}
$$
then turn the cube so $R$ becomes $F$ and apply
$$\mathbf{L'RF\cdot LR'D^2 \cdot L'RF \cdot LR'}$$
\end{enumerate}
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\section{You are done!!!}
\end{document}