Christopher Mowla
Math 3900
November 18, 2011
The Rubik’s Cube
Rubik’s Cubes of Order n
• A Rubik’s Cube of order n is referred to as
the nxnxn cube, where n n  1, n  .
• A cube with n = an even integer value is
referred to as an even cube.
• A cube with n = an odd integer value is
referred to as an odd cube.
• A cube of size n = 4 and larger are
referred to as big cubes.
• Other commonly solved cube sizes are:
n=2
n=4
n=5
n=6
n=7
n=9
n = 11
Piece Types of the 3x3x3
Corners: They each have 3 stickers.
Edges: They each have 2 stickers.
(Fixed) centers: They each have 1 sticker.
They are fixed on a xyz axis.
Piece Types of the nxnxn
• Unlike the common 3x3x3 Rubik’s Cube
size, there are 4 additional types of pieces
on the nxnxn cube, in general.
• Therefore, there is a total of 7 different
pieces types which can be seen on larger
cube sizes.
• Side Note: We will exclude the 1x1x1
cube size for consistency (it’s an exception
in more than one respect).
Types of Pieces (7)
• Corners
• Edges
– Middle Edges
– Wing Edges
• Centers
– Fixed Centers
– Non-Fixed Centers
• X-center pieces, oblique center pieces, and
+ center pieces.
• For illustration, we will construct cube
sizes up to the 7x7x7 to see all 7 of the
possible piece types.
– The standard 3x3x3 Rubik’s Cube does not
have all of the possible piece types.
– The 7x7x7 Rubik’s Cube has all of the
possible piece types.
• We can construct all cube sizes from the
2x2x2 cube because all cube sizes from
n ù 2 have 8 corners.
Constructing the nxnxn
Odd Cube Size Construction
2x2x2  3x3x3
3x3x3  5x5x5
New piece types on the 5x5x5
• There are three new piece types on the
5x5x5: wing edges, X-center pieces, and
+ center pieces.
Wing Edges
• (They are commonly called wing edges by
the cubing community due to their
symmetry.)
X-Center Pieces
• They are called X-center pieces by the
cubing community because they form an X
about the composite center.
+ Center Pieces
• These are called + center pieces (many refer to
them as T-center pieces as well) because they
form a plus sign about the big cube center.
5x5x5  7x7x7
Wing Edges
• For the 7x7x7, we have a second set of 24
wing edges.
• The term orbit is used to differentiate
different sets of wing edges. It means
where the pieces are able to move.
– On the 7x7x7 cube, we have 2 orbits of wing
edges.
X Center Pieces
+ Center Pieces
New piece type on the 7x7x7
Oblique Center Pieces
• The term oblique is used to describe
these center pieces because they are
neither X-center pieces nor + center
pieces.
Permutations
• A permutation is an arrangement of
objects of the same type in some order.
• Permutations can be decomposed into
cycles.
Definition of a Cycle
• An n-cycle is moving n pieces (2 or more)
of the same type (i.e. edges, corners, or
centers) at the same time so that, when
the algorithm generating the cycle is
repeated exactly n times (and not until
then), the cube will be restored to the
original state it was in.
Examples of N-Cycles
3-Cycles
(Permutations of Middle Edges and Corners on the 3x3x3)
2-Cycles
(Permutations of Wing Edges on the 5x5x5)
4-Cycles
(Permutations of Wing Edges on the 5x5x5)
Combinations of Disjoint Cycles
• Definition:
If an algorithm affects 4 or more pieces of
the same type on a cube (corners, edges,
or centers), all of the pieces affected need
not be part of an n-cycle.
• In other words, n pieces of the same type
affected by an algorithm  an n-cycle of
pieces.
2 2-Cycles
(Permutations of Middle Edges on the 3x3x3)
2 2-Cycles
(Permutations of Wing Edges on the 5x5x5)
2 2-Cycles
(Permutations of Wing Edges on the 5x5x5)
2 3-Cycles
(Permutations of Wing Edges on the 5x5x5)
Calculating the Number of
Permutations (Positions) on the nxnxn
Cube (for n>1)
• By the multiplication rule of probability, the
main idea is to determine the number of
possible permutations for each piece type
and multiply them all together.
More Background Information
Permutation Notation
• Permutations are bijective maps (one to
one and onto) and therefore can be
represented by:
2
3
4
5
6... 
 n   1


,
 f  n    f 1 f  2  f  3 f  4  f  5  f  6  ... 
where f 1 , f  2  , ..., f  n  are each different values of n.
Classifying Permutations
• A permutation can be either even or odd.
– An even permutation is a permutation in
which can be solved (achieve the identity) in
an even number of 2-cycle swaps.
• The identity is an even permutation.
– An odd permutation is a permutation in
which can be solved (achieve the identity) in
an odd number of 2-cycle swaps.
• For example, 4-cycles are odd
permutations because they can be split
into three connected 2-cycles.
Example:
1 2 3 4


2
4
1
3


means 12, 24, 43, 31.
• In order to achieve the identity (an even
permutation), where each domain value is
equal to its corresponding co-domain
value,
1 2 3 4 


1
2
3
4


We can perform the following three swaps
(2-cycles)
1 2 3

2 4 1
4 2  1 2

1 2
4  21 1 2 3 4 


3
1
4
2
3


3 4  4 3  1 2 3 4 

  e.
4 3
1 2 3 4 
• On the other hand, two disjoint 4-cycles of
a specific object type is an
even permutation.
For example, the two 4-cycle:
1 2 3 4 5 6 7 8


7
8
1
2
3
4
5
6


can be decomposed to an even number of
2-cycles:
1 71 51 31 8 2 6 2 4
• It turns out that the number of even
permutations is equal to the number of
odd permutations for sets of 2 or more
objects.
– Half of the permutations are even.
– Half of the permutations are odd.
• For n objects of the same type, there are
n! possible permutations.
Rubik’s Cube Interpretation of
Permutations
• Permutations on the nxnxn Rubik’s Cube
mean how many possible positions there
are for each piece type combined with the
rest of the other types of pieces.
A second meaning of the term
“permutation” on Rubik’s Cubes
• As with the mathematical definition of
permutations, the term permutation is also used
to describe the location where each piece is
located.
• The first row of the permutation notation can be
assigned to fixed slots on a cube.
For example, the slots containing the wing
edges in the last layer of a 5x5x5 can
correspond to the permutation notation as
follows:
 1

 f 1
2
3
4
5
6
7
f  2
f  3
f  4
f 5
f 6
f 7
7 6
5
8
3
1
2
4
8 

f 8  
– Each number on the
cube designates the
slot number (the first
row of the 2-line
permutation notation)
and f(1) is the wing
edge piece in slot 1,
f(2) is the wing edge
piece in slot 2, etc.
Orientation
• The term orientation refers to how a
particular piece type on a cube is twisted
in its location.
Orientation of Corners
• Corners can be twisted 3 different ways,
and thus have 3 orientations each
independently.
Orientation of Middle Edges
• Edges can be oriented in two ways:
flipped or not flipped.
• No other piece type can have a different
orientation: their movement is strictly said
to be permutations (movements to
different locations).
• Fixed center pieces rotate in their
locations, but cannot “twist.”
• If a wing edge appears to be flipped, it’s
the other wing edge that is in the same
composite edge.
The Corners
• There are 8 Corners.
8!
• Each corner can be oriented (twisted) in 3
ways independently.
8
8!  3  3  3  3  3  3  3  3  8! 3
Laws of the Cube (1)
• Actually, by what is commonly referred to
as “cube laws,” only 1/3 of the 3^8
orientations of corners are possible.
• If we go back to the illustration of the three
ways a corner can be twisted,
• Suppose the left most image represents the
identity. Then:
– The corner in the left image is turned
clockwise 0 times from the identity.
– The corner in the middle image is turned
clockwise 1 time from the identity.
– The corner in the right image is turned
clockwise 2 times from the identity.
0
1
2
• Once one chooses a point of reference, that is,
one fixes one corner out of the 8 to be in the
identity, then the sum of the values of either +90
degree twists or -90 degree twists (not both)
from the identity must be evenly divisible by 3.
For example, an impossible state is to have the
entire cube solved except for one corner twisted,
since 1 is not evenly divisible by 3.
• Neither can there be two corners twisted clockwise once
since 1+1 = 2, which is not evenly divisible by 3.
Orange
• On the other hand, you could have one corner twisted
clockwise and the other twisted clockwise twice (or anti
clockwise) since 1+2 = 3.
Orange
The Corners
• There are 8 Corners.
8!
• Each corner can be oriented (twisted) in 3
ways independently.
8
8!  3  3  3  3  3  3  3  3  8! 3
• Thus we divide this result by 3 to get
7
8! 3
Even Cube Symmetry
• On even cubes, we can always fix a corner to be
solved (in the identity) and eliminate the number
of permutations it contributes to the whole.
– This is because on even cubes, there are no fixed
centers and therefore there is no unique way to set
the cube on a table, when considering all possible
ways it can be scrambled.
– Since each corner can be placed in 8 different
locations and twisted in 3 different ways, we divide
the total number of corner permutations by
to have a value of zero for odd
 n1 mod 2
24
values of n.
Corners Conclusion
• Therefore, the total number of possible
permutations of corners on the nxnxn
cube, taking into account the cube law and
even cube symmetry is:
8
8! 3
 n1 mod 2
3   3  8
7

8! 3
 n1 mod 2
24
Middle Edges (Odd Cubes Only)
• There are 12 middle edges.
12!
• Each edge can be flipped in two ways
independently.
12
12! 2
Laws of the Cube (2)
• Similar to the reduction of the possible
orientations of the corners, by the “cube
laws,” only 1/2 of the 2^12 orientations of
middle edges are possible.
• Only an even number of middle edges can
be flipped.
Middle Edges (Odd Cubes Only)
• There are 12 middle edges.
12!
• Each edge can be flipped in two ways
independently.
12
12! 2
• By the “cube law,” we divide this number
by 2 to have
12
12! 2
2
11
 12! 2
Laws of the Cube (3)
• It also turns out that the middle edges are
in an odd permutation if and only if the
corners are in an odd permutation.
• Recall that odd permutations make up half
of all permutations of 3 or more objects.
• Since half of the permutations of two
different piece types are dependent on
each other, only half of the permutations of
one of them is allowed on the cube.
Middle Edges Final
• Since middle edges only are in odd cubes
(and thus the presence of both the corners
and middle edges only occurs on odd
cubes to cause the ½ reduction), we divide
what we had previously by 2 and raise it
n mod 2 to have a value of zero for even
cubes.
11 n mod 2
 12! 2

 2







10 n mod 2
 12! 2
Wing Edges
• Whether even or odd cube sizes, big cube
sizes (larger than the 3x3x3) have
n  2
 2  sets (orbits) of 24 wing edges.
• For example, the 6x6x6 has
6  2 4
 2    2    2  2 orbits of wing edges.
• The 7x7x7 has
7  2 5
 2    2    2.5  2
orbits of wing edges.
n  2
 2 
• Since there are
orbits of wing
edges on the nxnxn cube, containing 24
wing edges each, the total number of
permutations of the wing edges is:
 n 2 
 2 
24!
.
X-Center Pieces, + Center pieces,
and Oblique Center Pieces.
• Similar to wing edges, big cubes have a
set number of orbits of all non-fixed center
pieces.
• To determine the number of orbits for the
nxnxn, let’s observe a large big cube
center.
• Suppose we are looking head on at the
white face of a 13x13x13. Let’s shade in
the fixed center piece (not part of the nonfixed center piece calculation), the
corners, and all orbits of wing edges.
• Now, imagine we were going to divide this
composite center into 4 pieces.
• If this was an even cube, then we could divide
the composite center into 4 equal squares, but
as an odd cube, this is the best we can do with
an extra slice row and column in the center.
• The reason we are dividing the center into 4
equal parts it for us to imagine that we are
turning the face (outer most slice) closest to us.
Turning the face in either direction will send all of
the purple pieces in each quadrant to the other 3
quadrants after three rotations.
• Recall that the term “orbit” refers to where
pieces can move to.
• Therefore, since all 4 quadrants can move
to each other, we need only consider one
quadrant to count the number of orbits of
non-fixed center pieces.
• Clearly for even cube sizes, the number of
non-fixed center orbits is the amount of
squares in each quadrant:
2
 n2

 .
 2 
• For odd cubes, notice
that the white cross
also consists of 4
symmetrically
equivalent blocks. If
we rotate the face 3
times, all of these
n3
blocks will
go to each
2
other.
• (These blocks make
up the orbits of the +
center pieces.)
• Therefore for odd cubes, the total number
of non-fixed center orbits is:
 n  3  n  3  n  1 n  3


 
2
4
 2 
2
• However, we can also just take the floor
function of the even cube number to obtain
an equivalent value:
 n  2  2 

 
 2  
• Note that there are 4 of each non-fixed center
piece type (since we divided the composite
center into 4 equal parts) in every one of the 6
faces of the cube, or 4(6) = 24 non-fixed center
pieces in each orbit.
• This gives the total number of possible
permutations of non-fixed center pieces on odd
and even cubes to be:
 n  2  2 

 
 2  


24!
.
Non-Fixed Centers Final
• Lastly, just as there
were 4 pieces of each
type in each face.
– Each face is a different
color.
– Non-fixed center
pieces are
indistinguishable.
– Therefore, we must
divide by:
 n  2  2 

 
2

   
 1
 6
 4! 
• This gives us a final
number of:
 n  2  2 

 
24!   2  
.
6 


 4! 
The Formula
• Multiplying the total number of
permutations for the corners, middle
edges, wing edges, and non-fixed center
pieces all together, we achieve the formula
for the number of positions of the nxnxn
Rubik’s Cube.
F  n 
7
8! 3
 n 1 mod 2
24
12! 2 
10 n mod 2
 n2 
 2 
24!
 n  2  2 

 
24!   2  
6 


 4! 
Examples
•
•
•
•
•
•
n =2: 3,674,160
n = 3: 43, 252, 003, 274, 489, 856, 000.
n = 4: Approximately 7.40 x 10 ^45.
n = 7: Approximately 1.95 x 10^160.
n = 11: Approximately 1.09 x 10^425.
n = 100: Approximately 2.35 x 10^ 38, 415
Supercubes
• Probably everyone has either seen or
heard of the Sudoku 3x3x3 cube:
Supercubes
• Notice that the
numbers of the
center pieces are
all turned to match
the direction of the
numbers on the
corners and middle
edges perfectly.
Definition
• An nxnxn supercube is similar to the
nxnxn cube, except that all center piece
types, both the fixed center pieces and the
3 non-fixed center pieces types (X-center
pieces, + center pieces, and oblique
center pieces) are distinguishable.
• In other words, the regular 6-colored
nxnxn cube and the nxnxn supercube
have the same number of permutations for
the “Cage” portion of the cube.
• Since the 2x2x2 does not even have
centers, the supercube sizes begin with
n = 3.
• Supercubes have more permutations than
regular 6-colored cubes.
Calculating the Number of
Positions of the nxnxn supercube.
• We merely need to multiply the formula for
the regular nxnxn by a factor determined
by the additional number of ways each
color of center pieces can swap with each
other (in any permutation from the formula
for the regular 6-colored nxnxn cube).
Fixed Centers (Odd Cubes Only)
• There are 6 fixed centers on every odd
cube.
• Each center can be rotated in 4 directions
in its location.
6 n mod 2
• This gives
total permutations.
4
 
Laws of the Cube (4)
• Just as the middle edges are in an odd
permutation if and only if the corners are in
an odd permutation, it carries on to fixed
centers as well.
• That is, middle edges, corners, and fixed
center pieces are all dependent on each
other. If one of them has an odd
permutation (or even permutation), so do
the other two.
Fixed Centers (Odd Cubes Only)
• There are 6 fixed
centers on every odd
cube.
• Each center can be
rotated in 4 directions
in its location.
n mod 2
6
• This gives 4
 
total permutations.
• By the “cube law,” we
divide this number by
2, because only half
of the fixed center
permutations are
allowed:
4

 2

6




n mod 2
Non-Fixed Center Pieces
• From calculating the
formula for the regular
nxnxn, we found that
there are:
 n  2  2 

 
 2  
orbits of non-fixed center
pieces.
• In each orbit, there were
24/6 = 4 center pieces of
each color.
• Now that center
pieces ARE
distinguishable, we
have an additional 4!
permutations of each
color. There are 6
colors, which gives us
 4!
6
• We raise this to the number of orbits of
non-fixed centers, since that is the number
of different sets we are multiplying
together:
 n2 2 


6  2  
4! 

 
 n2 2 
6
 
 2  


 4!
Laws of the Cube (5)
• Now that non-fixed center pieces are
unique to their positions,
– An odd permutation of wing edges exists if
and only if an odd permutation exists in the +
center pieces and oblique center pieces.
– An odd permutation of corners exists if and
only if there is an odd permutation all of the Xcenter pieces  n  2 2   n  2   1  n  2 2 
 

and at least
 
 

2
2
2 2

  
  
+ center and oblique center pieces.
 
• Despite the complexity of these relationships,
the total number of permutations shared
between the non-fixed center pieces and the
corners and wing edges is just:
2
 n  2  2 

 
 2  


• Therefore, we divide by this amount:
1
 n 2 2 

 
 2  


2
, which means we only consider half of the
permutations of each orbit of non-fixed centers.
The Formula for the Supercube
• We multiply the formula for the regular 6colored nxnxn by all of the factors just
discussed to achieve the formula for the
supercube:
 46 
S n  F n  
 2 
 
4
S n  F n 
 2

6




n mod 2
n mod 2
 n  2  2 


6  2  
4! 

 

 
2
 n  2  2 


6   2  
4! 




 2 


 n  2  2 

 
1   2  
The Factor Increase (Examples)
• The number of permutations of the regular
nxnxn are increased by:
• n = 2: 1
• n = 3: 2,048.
• n = 4: 95, 551, 488.
• n = 7: Approx. 1.56 x 10 ^ 51.
• n = 11: Approx. 8.24 x 10 ^ 162.
• n = 100: Approx. 3.55 x 10 ^ 19, 160.
Relationship Between Wing Edges
and Non-Fixed Center Pieces.
• We need to find a formula to represent the
total number of non-fixed center orbits in
an odd permutation for a given number of
orbits of wings in an odd permutation.
• We will use a 13x13x13.
• As mentioned earlier, a 4-cycle is an odd
permutation and two 4-cycles is an even
permutation.
• In fact, if we do a quarter turn to an inner
layer slice of the cube, we can model it.
• These are the X-center pieces
• If we reflect this now to the top, clearly (n-2)-2 center
orbits in an odd permutation now in the yellow column
(just focus on the yellow column, not its corresponding
yellow row).
• Reflecting this to the top, clearly we have
(n-2)-2-2 center orbits left with an odd
permutation in both slices. That is, 2(n-2-2(2)).
• We now have exactly (13-2)-2-2-2=5 NF
center orbits in an odd permutation in
each wing edge orbit we chose.
• Clearly there is a pattern and if we were to
choose arbitrary wing edge orbits, the
number of non-fixed center orbits in an
odd permutation with the wings would be
the same for any combination of the same
number of wing edge orbits.
• The formula is:
w  n  2  2w   w  n  2   2w
2
Example
• On the 1,000 x 1,000 x 1,000 cube, if w =
65 orbits of wings have an odd
permutation, then
65 1, 000  2   2  65   56, 420
2
orbits of non-fixed center orbits (consisting
of + and oblique centers) has an odd
permutation.
The Maximum Formula
• It’s useful for us to know the maximum
number of non-fixed center orbits that can
be in an odd permutation for a given size
n.
Finding a Maximum Formula By the
Use of Calculus
• The formula w  n  2   2w
variable function because
2
is NOT a 2
n  2
1 w  
 In

 2 
Thus, the maximum value for a given n is
the maximum value achieved by at least
one w from that domain.
• We can differentiate with respect to w,
since that is the changing domain:
 
2
w  n  2   2w  n  2  4w

w 
• To verify that we will obtain the maximum
value, we differentiate once more.

 n  2  4w  4  0 w  I n
w
• Setting the first derivative
equal to zero to find the
critical point, we have:
• By comparing the formula
to the true maximum
values, we just need to
n2
n  2  4w  0  w 
take the floor of this.
4
• Substituting w into
w  n  2   2w
, we get
1n2


2 2 
2
2
 1  n  2 2 
 
 
 2  2  
• We can see a 3-D
graph representation
of the formula
w  n  2   2w
2
very well by rewriting as
.
2
 w   n   2   2  w
n
W
• The graph of the maximum formula
 1   n   2 2 
M  n     
 
2  2  


w

 w n  2
  2 w
2
n
• Instead of resorting to the floor function,
we can determine whether or not there is a
unique value of w to yield the maximum
value for cubes of size n.
• Comparing the formula derived from
calculus with the actual maximum values,
1n2


2 2 
2
1) If n is odd, then we have 1/8 more than necessary.

2) If n is even and evenly divisible by 4,
 then we have 1/2 more than necessary.

• We can use trig functions instead.
1  n  2  1 2  n  1
2  n 
2  n 

  sin    cos   cos  
2 2  8
 2  2
 4 
 2 
2
• But also note that:
1  n  2   n  2
n  4n  4


 
2 2 
8
8
2
2
2
n 
n  4n  3  cos 
  n  1 n  3 1
2  n 
2




 cos 

8
8
8
2


• We no longer need to subtract 1/8 for odd cubes
because we multiplied 1/8 by cosine.
2
2
 n 1 n  3  1 cos2  n   1 cos2  n  cos2  n 
8
8


2

 2


4




2



 n  1 n  3 , if n is odd,

8

n  1 n  3 3


 M n  
 , if n is divisible by 4,
8
8

  n  1 n  3 1
 , if n is even, but not divisible by 4.

8
8

Solutions for Even Cubes Divisible
by 4
n  1 n  3 k 


2
solve  w  n  2   w 
 , w  for k  3
8
8 

• Using the quadratic formula, we get:
n
n
w1   1, w2 
2
2
• Both solutions work! Therefore w is not
unique for this cube size class.
Solutions for Even Cubes Not
Divisible by 4
n  1 n  3 k 


2
solve  w  n  2   w 
 , w  for k  1
8
8 

• Using the quadratic formula, we get:
n2
w1  w2 
(double root)
4
• This is equal to the critical point we found, which
is not surprising since the original formula
worked for this cube class from the start.
Solutions for Odd Cubes
n  1 n  3 k 


2
solve  w  n  2   w 
 , w  for k  0
8
8 

• Using the quadratic formula,
n 1
n3
w1 
, w2 
4
4
– w1 is true for every other odd integer starting
with n = 5.
– w2 is true for every other odd integer starting
with n = 7.
Issues
• We do not have a single representation for
the value of w that yields the maximum
value.
• If we approach this problem using the
following 4th order non-homogeneous
recursion relation
ak 3  ak 7  2  k  4 , k  8
with the initial conditions:
a1  0, a2  2, a3  4, and a4  6
• With ingenuity, we can find the solution:
n
 4 
2

 n
n
2
 n  4i   2    n  2    2   
4
4






i 1
• Comparing with the original formula w  n  2   w2
n
• Clearly w  2  
4
, and we can
represent w equal to that for all size cubes.
However, we note that n evenly divisible by 4
has two solutions: w  n  1, n .
2
2
Relationship between the non-fixed
centers and the Corners
• Since we found the formula
 1  n  2 2 
M  n   
 
 2  2  
using calculus, we can resume the
following slide:
Laws of the Cube (5)
• Now that non-fixed center pieces are
unique to their positions,
– An odd permutation of wing edges exists if
and only if an odd permutation exists in the +
center pieces and oblique center pieces.
– An odd permutation of corners exists if and
only if there is an odd permutation all of the Xcenter pieces  n  2 2   n  2   1  n  2 2 
 

and at least
 
 

2
2
2 2

  
  
+ center and oblique center pieces.
 
• As we first were deriving w  n  2   2w ,
we learned that the X-center pieces were
in a 2 4-cycle when inner layer slices were
turned. Hence we should be able to
understand the first bullet completely.
• The second bullet is merely stating that Xcenter pieces are in an odd permutation if
the corners are in an odd permutation.
• We can see this using a similar image as
before, just extrapolating the permutation
to the outer layer.
2
• If a quarter turn of the outer most layer is
done in the same direction as on the
13x13x13 example, then the corners are in
following 4-cycle (odd permutation).
• In addition, the X-center pieces are in a 4cycle (odd permutation) too.
• The only item left to interpret in the second bullet
is:
 n  2  2   n  2   1  n  2  2 
 

 
 

 2    2   2  2  
• The total non-fixed center pieces (besides Xcenter orbits which is equal to  n  2 
 2 
can be no less than the total number of nonfixed center orbits minus the maximum number
of center orbits which can be in an odd
permutation with wings, M(n).
References
• Cube Laws
http://www.ryanheise.com/cube/cube_laws
.html
• Number of Permutations
http://en.wikipedia.org/wiki/Rubik%27s_cu
be