STRATEGIES
So far, I have some strategies for solving the 3x3 Rubik's cube,
the 4x4 Rubik's cube, and the masterball.
3x3 Rubik's cube
Consider the group of transformations of Rubik's cube. If we
number the faces of this cube as follows
+--------------+
| 1 2 3 |
| 4 U 5 |
| 6 7 8 |
+--------------+--------------+--------------+--------------+
| 9 10 11 | 17 18 19 | 25 26 27 | 33 34 35 |
| 12 L 13 | 20 F 21 | 28 R 29 | 36 B 37 |
| 14 15 16 | 22 23 24 | 30 31 32 | 38 39 40 |
+--------------+--------------+--------------+--------------+
| 41 42 43 |
| 44 D 45 |
| 46 47 48 |
+--------------+
then the group is generated by the following generators,
corresponding to the six faces of the cube:
U:= ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19),
L:= ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35),
F:= (17,19,24,22)(18,21,23,20)( 6,25,43,16)( 7,28,42,13)( 8,30,41,11),
R:= (25,27,32,30)(26,29,31,28)( 3,38,43,19)( 5,36,45,21)( 8,33,48,24),
B:= (33,35,40,38)(34,37,39,36)( 3, 9,46,32)( 2,12,47,29)( 1,14,48,27),
D:= (41,43,48,46)(42,45,47,44)(14,22,30,38)(15,23,31,39)(16,24,32,40).
The reader may want to verify this by printing out a hard copy of the
this page and cut+fold+tape the above diagram into a cube.
The size of the group generated by these permutations is
43252003274489856000.
Strategy for solving the cube
Let x^y=y^(-1)*x*y denote conjugation and [x,y]=x^(-1)*y^(-1)*x*y
denote the commutator, for x,y group elements. If x,y,z denote 3 group
elements, let [x,y,z]=x^(-1)*y^(-1)*z^(-1)*x*y*z. If x is a group element
and n>0 is an integer then x^n=x*x*...*x (n times).
The solution strategy is composed of 3 stages:
Stage 1: Solve the top face and top edges. For this the following
moves are useful:
"monotwist":[F,R^(-1)]^2
"monoswap":D^(F^(-1))*(D^2)^F*(D^(-1))^(F^(-1))
"monoflip":(epsilon R)^4, where epsilon is the counterclockwise middle
slice quarter turn
"edgeswap":(U^2)^(R^2*L^2)
Stage 2: Solve the middle edges (and bottom edges as best as
possible). For this the following "clean edge moves" are useful:
R^2*U*F*B^(-1)*R^2*F^(-1)*B*U*R^2 is the top edge 3-cycle (uf,ub,ur),
[U,F^(-1),R]*[U^(-1),B,R^(-1)]^L This flips, but does not permute,
the top edges uf, ub
(R^2U^2)^3 permutes 2 pairs of edges (uf,ub)(fr,br)
(L^2*F^2*B^2*R^2*F^2*B^2)^(D*B^2*F^2) permutes 2 pairs of top
edges (uf,ul)(ur,ub)
Stage 3: Solve the bottom corners (and bottom edges if necessary).
For this the following "clean corner moves" and "clean corner-edge
moves" are useful:
((D^2)^R*(U^2)^B)^2 twists the ufr corner clockwise and the bld
corner counterclockwise
The move ((U^2(D^2)^(F*R^(-1)))^2)^(R^(-1)) has the same
2-corner-twist effect as the one above.
((D^2)^(F*D^(-1)*R)*U^2)^2 permutes 2 pairs of corners
(ufr,ufl)(ubr,ubl)
[(D^(-1))^R,U^(-1)] corner 3-cycle (brd,urb,ulb)
B^(U^(-1)*F)*U^2*U^B*U^2*B^(-1) permutes two top edges and 2 top
corners (ulb,urb)(ub,ur)
These moves were compiled with help from the books [Si]
and [BCG].
4x4 Rubik's cube
Consider the group of transformations of the 4x4 Rubik's cube. If
we number the faces of this cube as follows
+-----------------+
| 49 50 51 52 |
| 61 62 63 64 |
U
| 73 74 75 76 |
| 77 78 79 80 |
+------------------+-----------------+-----------------+-----------------+
| 53 54 55 56 | 1 2 3 4 | 5 6 7 8 | 9 10 11 12 |
| 65 66 67 68 | 13 14 15 16 | 17 18 19 20 | 21 22 23 24 |
L F R B
| 77 78 79 80 | 25 26 27 28 | 29 30 31 32 | 33 34 35 36 |
| 89 90 91 92 | 37 38 39 40 | 41 42 43 44 | 45 46 47 48 |
+------------------+-----------------+-----------------+-----------------+
| 57 58 59 60 |
| 69 70 71 72 |
D
| 81 82 83 84 |
| 93 94 95 96 |
+-----------------+
then the group is generated by the following 12 generators (written in
disjoint cycle notation), corresponding 2 each to the six faces of the
cube:
U1=(49, 52, 88, 85)( 62, 63, 75, 74)( 50,64,87,73)
(51,76,86,61)(5,1,53,9)(6,2,54,10)(7,3,55,11)(8,4,56,12),
U2=(17, 13, 65, 21)( 18, 14, 66, 22)( 19,15,67,23)(20,16,68,24),
L1=(96,48,49,1)(84,36,61,13)(72,24,73,25)(60,12,85,37)
(89,53,56,92)(90,65,55,80)(91,77,54,68)(66,67,79,78),
L2=(59,11,86,38)(71,23,74,26)(83,35,62,14)(95,47,50,2),
F1=(89,5,93,92)(77,17,81,80)(65,29,69,68)(53,41,57,86)
(1,4,40,37)(2,16,39,25)(3,28,38,13)(14,15,27,26),
F2=(73,6,81,91)(74,18,82,79)(75,30,83,67)(76,42,84,55),
R1=(40,88,9,57)(28,76,21,69)(16,64,33,81)(4,52,49,93)
(41,5,8,44)(42,17,7,32)(43,29,6,20)(18,19,31,30),
R2=(39,87,10,58)(27,75,22,70)(15,63,34,82)(3,51,46,94),
B1=(52,53,44,60)(51,65,32,59)(50,77,20,58)(49,89,8,57)
(9,12,48,45)(10,24,47,33)(11,36,46,21)(22,23,35,34),
B2=(54,72,43,64)(66,71,31,63)(78,70,19,62)(90,69,7,61),
D1=(57, 60, 96, 93)( 58, 72, 95, 81)(59, 84, 94, 69)
(70,71,83,82)(45,89,37,41)(46,90,38,42)(47,91,39,43)(48,92,40,44),
D2=(33, 77, 25, 29)( 34, 78, 26, 30)(35, 79, 27, 31)(36,80,28,32),
(To check these are correct, the reader may want to print out a hard
copy of this page and cut-fold-tape the above diagram into a cube.)
Strategies for solving the 4x4 cube
The solution strategy is composed of 4 stages:
Stage 1: Solve the corners. For this use moves for the 3x3 Rubik's
cube.
Stage 2: "Pair" the edges so that the neighboring facets on
neighboring middle edges have the same color as each other. For this the
following "clean edge moves" are useful:
"flipedge":=L2^2*D1^2*U2*F1^3*U2^3*F1*D1^2*L2^2*L1*U1*L1^3*U2^3*L1*U1^3*L1^3
(due to J. Adams [A] who calls it "move 8"). This flips and swaps
the two middle edge facets on the UF boundary. It affects some
centers, but no other edges or corners.
"upedgeswap"=R2*B1^2*D1^2*B1^3*R2^3*B1*D1^2*B1^3*R2*B1^3*R2^2
(due to Thai [T], who calls it an "11 gram"). This move affects some
centers but no corners and only 4 edge facets. It swaps and
flips the right-most UF edge cubie with the right-most UB edge
cubie, sending the U facet of the right-most UF edge cubie to
the B facet of the right-most UB edge cubie.
Stage 3: Solve the edges. For this the "clean edge moves" for
the 3x3 Rubik's cube (see section 1 on the 3x3 Rubik's cube).
Stage 4: Solve the centers. For this, use the following
"clean center moves":
center3cycle:=R1^(-1)*F2*R2^(-1)*F2^(-1)*R1*F2*R2*F2^(-1)
(also called "move 9", due to J. Adams). This move is a 3-cycle
on center facets, affecting no edges, no corners, and no other center
facets. It is the 3-cycle (15,19,18) in the above notation.
Some similar clean center moves:
center1=B1^2*R2^3*F2*R2*B1^2*R2^3*F2^3*R2,
center2=R2^2*B1^2*R2^3*F2*R2*B1^2*R2^3*F2^3*R2^3
These aren't really necessary since the the center3cycle can always
be applied after a suitable set-up move (i.e., in combination with
a suitable conjugation).
The following move is occasionally useful:
"centerswap"=(R2^2*U2^2)^4
This affects only 6 center facets (on the front and back faces) and
no others. It is the product of 2 3-cycles: (15,34,23)(27,14,22)
in the above notation.
These moves were compiled with help from the books [A] and [T].
Rainbow masterball
The solution strategy
Step 1:
The idea is to first get all the middle bands aligned first, so
you get ball corresponding to a matrix of the form
* * * * * * * *
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
* * * * * * * *
Here, * denotes any color. We have labeled the colors on the
masterball as 1, 2, ..., 8 in order of occurrence.
We describe a method, which I call "crab fishing", for
achieving this. (Mathematically, this amounts to performing some
carefully choosen commutators.) Without too much trouble
you can always assume that we have one column aligned. You may
need to flip or rotate the ball a little bit to do this. Call
this aligned column "column 1" and call the color in column 1,
"color 1". We want to get the middle two entries in column 2
aligned. Call the color in the (2,3)-entry "color 2".
We want to get color 2 in the (2,2)-entry. The remaining large
color 2 tile is the "crab" we will "fish" for. Hold the ball in
front of you in such a way that column 2 is slightly to the left
of center and column 3 is slightly to the right of center. There
are 4 facets in the right upper middle band, 4 facets in the
left upper middle band, 4 facets in the right lower middle
band and 4 facets in the left lower middle band. A flip about
the center on the right half (i.e., perform f2) exchanges these.
We may assume that color 2 is on one of the four facets in the
right lower middle band. (If it isn't you need to apply f2
first). Now perform r2^{-1}*f2^{-1}*r2*f2: first perform r2^{-1}
(this is "baiting the hook"), then f2^{-1} ("putting the hook
in the water"), then r2 ("setting the hook"), and finally
f2 ("reeling in the hook"). You may or may not have color 2
in the (2,2) place like you want but the color 1 stripe is
intact. If necessary, try again. After at most 4 tries you'll
be successful.
Step 2: Repeat this "crab fishing" strategy to get color 2 in
the (1,2) position (using r1^{-1}*f2^{-1}*r1*f2 in place of
r2^{-1}*f2^{-1}*r2*f2). Now, by turning the ball over if necessary,
repeat this idea to get color 2 in the (4,2) position. Now you
have two "aligned" stripes on your ball - color 1 in column 1
and color 2 in column 2. We say, in this case, that columns 1
and 2 have been "solved".
Step 3: Repeat this for columns 3 and 4.
Step 4: Use the moves in the "catalog" below to finish the
puzzle.
A catalog of rainbow moves
Column moves:
We number the columns as 1,...,8. We will use a signed cycle
notation to denote an action of a move on the columns of the
masterball. For example, (a) a move which switches the 1st and 3rd
column but flips both of them over will be denoted by (1,3)_,
(b) a move which sends the 4th column to the 6th column, the
6th column to the 5th column, and switches the 2nd and 3rd column
but flips both of them over will be denoted by (2,3)_(6,5,4),
move cycle
f1 (1,4)_(2,3)_
f2 (2,5)_(3,4)_
f3 (3,6)_(4,5)_
f4 (4,7)_(5,6)_
f5 (5,8)_(6,7)_
f6 (1,6)_(7,8)_
f7 (2,7)_(1,8)_
f8 (3,8)_(1,2)_
f1*f2*f1 (1,2)_(3,5)
f1*f2*f1*f2 (5,4,3,2,1)_
f1*f3*f1 (1,5)(2,6)
f2*f3*f2 (2,3)_(6,5,4)
f1*f4*f1 (1,7)(5,6)
f1*f5*f1 (5,8)_(6,7)_
f1*f8*f1 (2,8)(3,4)_
f8*f1*f8 (1,8)_(4,3,2)
f2*f1*f2 (1,3)(4,5)_
f3*f1*f3 (1,5)(2,6)
f3*f6*f3 (1,3)(4,8)_
f8*f1*f2 (1,4)_(2,3,8,5)_
Some products of 2-cycles on the facets:
These are all based on an idea of Andrew Southern. The polar2swap
and equator2swap were obtained by trying variations of some of
Andrew's moves on a MAPLE implementation of the masterball.
We number the facets in the i-th column, north-to-south, as i1, i2,
i3, i4 (where i=1,2,...,8).
move cycle
r1*f4*r1^(-1)*r4*f4*r4^(-1) (44,84)(41,81)
f1*r1*f4*r1^(-1)*r4*f4*r4^(-1)*f1 (14,84)(11,81)
polar2swap36 (11,14)(31,61)
polar2swap18 (61,64)(11,81)
equator2swap36 (12,13)(32,62)
equator2swap18 (62,63)(12,82)
where
polar2swap36=
f1*r3^(-1)*r4^(-1)*f1*f2*r1*r4^(-1)*f2*r4^4*f2*
r1^(-1)*r4*f2*r4^4*f1*r3*r4*f1
(if you replace r3 by r2 both times in this move you get the same
effect),
polar2swap18=
f1*r3^(-1)*r4^(-1)*f3*f4*r1*r4^(-1)*f4*r4^4*f4*
r1^(-1)*r4*f4*r4^4*f3*r3*r4*f1
equator2swap36=
f1*r4^(-1)*r3^(-1)*f1*f2*r2*r3^(-1)*f2*r3^4*f2*
r2^(-1)*r3*f2*r3^4*f1*r4*r3*f1
equator2swap18=
f1*r4^(-1)*r3^(-1)*f3*f4*r2*r3^(-1)*f4*r3^4*f4*
r2^(-1)*r3*f4*r3^4*f3*r4*r3*f1
Moves in this section were obtained with the help of Andrew
Southern.