SKEWB
The skewb is a cube which has been subdivided into regions
differently than the Rubik's cube. First, fix an orientation of
the cube in space, so we may talk about a front face, a back face,
up, down, left, and right. Each of these 6 square faces are subdivided
into 5 facets as follows (where, unfortunately, the square comes out
looking rectangular in the picture below):
________________
 /\ 
 / \ 
 / \ 
 / \ 
 / \  x denotes one of the
 / \  letters f(ront), b(ack),
/ x \ r(ight), l(eft),
\ / u(p), d(own).
 \ / 
 \ / 
 \ / 
 \ / 
 \ / 
______\/______
The 4 corner facets are labeled exactly as in the case of the Rubik's
cube (as the lower case xyz, where x is the label of the face the facet
lives on, y and z the two neighboring faces).
The skewb itself is a cube subdivided as follows: there are 8 corner
pieces which are each in the shape of a tetrahedron. For example,
if you hold a cube in front of you the upper right hand corner of the
front face is the facet of a tetrahedron whose facets are labels
f1, r4, u2.
The moves of the skewb are different from the Rubik's cube as well:
Label the corners as XYZ, where xyz is the notation for any of the
facets belonging to that corner piece. Pick a corner XYZ of the cube and
draw a line L passing through that corner vertex and the opposite corner
vertex ("skewering the cube"). That line defines a 120 degree rotation
in the clockwise direction (viewed from the line looking down onto the
corner you picked). One move of the skewb is defined in terms of this
rotation as follows: Of course a 120 degree rotation of the entire cube
about the line L will preserve the cube but swap some faces and some
vertices. The skewb has a mechanism so that you can actually rotate half
(a "skewed" half) the skewb by 120 degrees about L and leave the other
half completely fixed. This rotation of half the skewb abouut L will
also be denoted XYZ.
We may also label the 5x6 = 30 facets as follows:
++
 20 17 
 16 u 
 19 18 
+++++
 5 2  10 7  25 22  30 27 
 1 l  6 f  21 r  26 b 
 4 3  9 8  24 23  29 28 
+++++
 15 12 
 11 d 
 14 13 
++
Consider the rotation UFR associated to the corner ufr. This move
permutes the facets of the skewb. As a permutation , the disjoint cycle
notation for this move is
UFR = (6 16 21)(7 18 25)(10 17 24)(8 19 22)
Note, in particular UFR does not move the 9facet. The 8 "basic moves" are
given by
FUR = (6 16 21)(7 18 25)(10 17 24)(8 19 22)
RUB = (21 16 26)(22 17 30)(25 20 29)(23 18 27)
BUL = (26 16 1)(27 20 5)(28 17 2)(30 19 4)
LUF = (1 16 6)(2 19 10)(5 18 9)(3 20 7)
FDR = (11 6 21)(25 13 9)(23 15 7)(24 12 8)
BDR = (26 11 21)(29 13 23)(27 12 22)(30 14 24)
FDL = (6 11 1)(9 15 3(10 12 4)(8 14 2)
LDB = (1 11 26)(3 13 27)(4 14 28)(5 15 29).
All other moves are obtained by combining these moves sequentially.
Exercise: Verify that the properties of a permutation puzzle are satisfied
for this puzzle.
nxn RUBIK'S CUBE
The only other Rubik's cube manufactured, as far as I know, is the
5x5 Rubik's cube. Apparently, for n larger than 6 or 7, there are
mechanical problems which cause the manufacture of the nxn cubes to
be overly expensive or perhaps even impossible. For information, at
least theoretically, on the solution of such cubes, the reader is
referred to the article [L] or the postings in the archives of the
"cubelovers" list [CL].
PYRAMINX
The pyraminx is a puzzle in the shape of a tetrahedron (a 4sided
regular platonic solid). Each of the 4 faces is divided into 9 triangular
facets:
/\
/ \
/____\
/\ /\
/ \ / \
/____\/____\
/\ /\ /\
/ \ / \ / \
/____\/____\/____\
There are a total of 4x9 = 36 facets on the pyraminx. They will be labeled
as follows:
/\ /\
/ \ / \
/_1__\ /_36_\
/\ 5 /\ /\ 34 /\
(left) / \ / \ (right) / \ / \ (left)
/_4__\/__6_\ /_35_\/_33_\
/\ 14 /\ 16 /\ /\ 31 /\ 29 /\
/ \ / \ / \ / \ / \ / \
/_13_\/_15_\/_17_\ /_32_\/_30_\/_28_\
front down
/\ /\
/ \ / \
/__3_\ /__2_\
/\ 11 /\ /\ 8 /\
(right) / \ / \ (front) / \ / \ (left)
/_10_\/_12_\ /_7__\/__9_\
/\ 11 /\ 26 /\ /\ 19 /\ 21 /\
/ \ / \ / \ / \ / \ / \
/_23_\/_25_\/_27_\ /_18_\/_20_\/_22_\
left right
We fix an orientation of the tetrahedron in space, so we may speak of
a "front", "right", "left", and "down" face. We label the 4 faces as
f(ront), r(ight), l(eft), d(own).
The tetrahedron itself has been subdivided into subtetrahedrons as
follows: to each face x (so x is either f, r, l, or d) there is an
opposing vertex Vx of the solid. For this face, we slice the solid
along two planes parallel to the face x and lying in between the face
x and the vertex Vx. We want these planes, along with the face x
itself and the vertex Vx to be spaced apart equally. The subtetrahedrons
in the slice of the face itself will be called the "first slice"
associated to the face x, denoted X1, the subtetrahedrons in the "middle"
slice parallel to the face x will be called the "second slice"
associated to that face, denoted X2, and the subtetrahedron
containing the vertex Vx to the face "third slice" associated to
that face, denoted X3.
To each face labeled x, we have a clockwise rotation by 120
degrees of the first slice X1 of the face. We shall denote this
rotation also by X1. This rotation only moves the facets living on the
slice X1. Similarly, we have a clockwise rotation by 120
degrees of the second slice X2 of the face. We shall denote this
rotation also by X2. X3 denotes the clockwise rotation by 120 degrees
of the opposing subtetrahedron containing the vertex Vx. These moves
permute the labels for the 36 facets, hence mmay be regarded as
a permutation of the numbers 1, 2, ..., 36. For example, the
clockwise rotation by 120 degrees (looking at the front face) of the
subtetrahedron opposite to the front face will be denoted F3. The
disjoint cycle notation for this move, regarded as a permutation, is
F3 = (23 22 36).
The "basic" moves are given as follows:
F1 = (2 32 27)(8 31 26)(7 30 12)(19 29 11)
(18 28 3)(1 17 13)(6 15 4)(5 16 14)
F2 = (9 35 25)(21 34 24)(20 33 10)
F3 = (23 22 36)
R1 = (3 36 17)(11 34 16)(10,35,6)
(24 31 5)(23 32 1)(2 22 18)(9 20 7)(8 21 19)
R2 = (12 33 15)(26 29 14)(25 30 4)
R3 = (27 28 13)
L1 = (1 28 22)(5 29 21)(4 33 9)
(14 34 8)(13 36 2)(3 27 23)(11 26 24)(12 25 10)
L2 = (6 30 20)(16 31 19)(15 35 7)
L3 = (17 32 18)
D1 = (13 18 23)(14 19 24)(15 20 25)
(16 21 26)(17 22 27)(28 32 36)(29 31 34)(30 35 33)
D2 = (4 7 10)(5 8 11)(6 9 12)
D3 = (1 2 3)
All other moves are obtained by combining these moves sequentially.
Exercise: Verify that the properties of a permutation puzzle are satisfied
for this puzzle.
MEGAMINX
This puzzle is in the shape of a dodecahedron. A dodecahedron is a
12sided regular platonic solid for which each of the 12 faces is a
pentagon. Here is an ascii picture of a pentagon (which has about a 2%
error):
*


* \

\
* *

/
*

 /

*
We call two faces "neighboring" if they share an edge. There are 20
vertices and 30 edges on a dodecahedron.
Each of the puzzle faces has been subdivided into 11 facets by slicing
each edge with a cut which is both parallel to that edge and not far
from the edge (say onefifth the way to the opposite vertex). A very
crude ascii picture is as follows:
* *
* a *
* \ / *
* j * b * the 11 facets
* i\ * * / * are labeled a, b, ..., k
* * k * c * for typographical
* /h* *d\ * convenience
*  * * *  *
* g\ f/ e *
* * * * *
There are a total of 11x12 = 132 facets on the puzzle.
Each face of the solid is parallel to a face on the opposite side. Fix a
face of the dodecahedron and consider a plane parallel to that face slicing
through the solid and about onefifth the way to the opposite face. There
are 12 such slices. Two such slices associated to two neighboring edges will
intersect inside the dodecahedron at a 120 degree angle but two such slices
associated to two nonneighboring edges will not intersect inside the
dodecahedron (though they will intersect outside the solid of course).
We slice up the solid dodecahedron in this way. This creates a smaller
dodecahedron in the center and several other irregular smaller pieces.
For each such slice associated to a given face x there is a "basic"
corresponding move of the megaminx X given by clockwise rotating the
slice of the megaminx by 120 degree, leaving the rest of the dodecahedron
invariant. Such a move effects 26 facets of the megaminx and leaves the
remaining 106 facets completely fixed.
Label the 12 faces of the solid as f1, f2, ..., f12 in some fixed way.
Imagine that the dodecahedron is placed in 3space in such a way that
one side on the xyplane and is centered along the positive zaxis
so that one of the vertices of the top face is at the xyzcoordinate
(r,0,s), where r is the radius of the inscribed circle for the pentagon
and s is the distance from the "up" face to the "down" face of the
dodecahedron.
Exercise: Suppose r=1. Find s. (This is fairly hard  see the chapter on
Platonic solids for some ideas.)
The up face we label as f1. The others may be labeled
according to the following graph, where faces are represented by
vertices and two vertices are connected by an edge if the corresponding
faces are neighboring.
________________________________
/ _________________________ \
/ / \ \
  f7 ______ \ \
 / \ \  
f1 is "up"  f6 f2 _____ \  
\ / \ / \ \  / /
f11 f1  f3  f8  f12 _ _/
f12 is "down" \ / \ \ /  \
f5 f4  f9 _________/  
/\ /  
 f10 __/ 
\__________________________/
A more symmetric way to order the faces of the dodecahedron is as follows
(see [B], exercise 18.35):
f1 u f12 d
f2 u0 f11 d1
f3 u1 f10 d0
f4 u2 f9 d4
f5 u3 f8 d3
f6 u4 f7 d2
One property of this labeling is explained in the following
Exercise: Suppose that the permutation (0 1 2 3 4) of the numbers 0, 1, 2,
3, 4 acts on the labels u0, ..., u4 and d0, ..., d4 in the obvious way.
Show that this permutation of the faces corresponds to a rotation of
the dodecahedron.
Notice that, like the cube, each vertex is uniquely determined by
specifying the three faces it has in common. We use the notation x.y.z
for the vertex of the dodecahedron which lies on the three faces x, y
and z. Note that the order is irrelevent: x.y.z denotes the same vertex as
y.x.z or z.y.x.
The facets of the megaminx may be specified as with the Rubik's cube:
a corner facet may be specified as [x.y.z], where x is the face the facet
lives on and y, z are the two neighboring faces of the facet. An edge facet
may be specified by [x.y], where x is the face the facet lives on and y is
the two neighboring face of the facet. The center facet of f1 will simply be
denoted by [f1]. We will call this label the "intrinsic label".
We may label the facets of the up face f1 as follows:
f1 facet symbol numerical label intrinsic label

a 1 [f1.f6.f2]
b 2 [f1.f2]
c 3 [f1.f2.f3]
d 4 [f1.f3]
e 5 [f1.f3.f4]
f 6 [f1.f4]
g 7 [f1.f4.f5]
h 8 [f1.f5]
i 9 [f1.f5.f6]
j 10 [f1.f6]
k 11 [f1]
For the next face (the f2 face), we label the facets in such a way that
the abc edge of f1 joins the ghi edge of f2:
f2 facet symbol numerical label intrinsic label

a 12 [f2.f6.f7]
b 13 [f2.f7]
c 14 [f2.f7.f8]
d 15 [f2.f8]
e 16 [f2.f8.f3]
f 17 [f2.f3]
g 18 [f2.f3.f1]
h 19 [f2.f1]
i 20 [f2.f1.f6]
j 21 [f2.f6]
k 22 [f2]
In general, we can label the remaining facets in such a way that
the "basic" moves are, as permutations, given by:
F1 = (1 3 5 7 9)(2 4 6 8 10)(20 31 42 53 64)x
x(19 30 41 52 63)(18 29 40 51 62)
F2 = (12 14 16 18 20)(13 15 17 19 21)(1 60 73 84 31)x
x(3 62 75 86 23)(2 61 74 85 32)
F3 = (23 25 27 29 31)(24 26 28 30 32)(82 95 42 3 16)x
x(83 96 43 4 17)(84 97 34 5 18)
F4 = (34 36 38 40 42)(35 37 39 41 43)(27 93 106 53 5)x
x(28 94 107 54 6)(29 95 108 45 7)
F5 = (45 47 49 51 53)(46 48 50 52 54)(38 104 117 64 7)x
x(39 105 118 65 8)(40 106 119 56 9)
F6 = (56 58 60 62 64)(57 59 61 63 65)(49 115 75 20 9)x
x(50 116 76 21 10)(51 117 67 12 1)
F7 = (67 69 71 73 75)(68 70 72 74 76)(58 113 126 86 12)x
x(59 114 127 7 13)(60 115 128 78 14)
F8 = (78 80 82 84 86)(79 81 83 85 87)(71 124 97 23 14)x
x(72 125 98 24 15)(73 126 89 25 16)
F9 = (89 91 93 95 97)(90 92 94 96 98)(80 122 108 34 25)x
x(81 123 109 35 26)(82 124 100 36 27)
F10 = (100 102 104 106 108)(101 103 105 107 109)x
x(91 130 119 45 36)(92 131 120 46 37)x
x(93 122 111 47 38)
F11 = (111 113 115 117 119)(112 114 116 118 120)x
x(102 128 67 56 47)(103 129 68 57 48)x
x(104 130 69 58 49)
F12 = (122 124 126 128 130)(123 125 127 129 131)x
x(100 89 78 69 111)(101 90 79 70 112)x
x(102 91 80 71 113)
OTHER PERMUTATION PUZZLES
I have left out several puzzles: "topspin" ( planar puzzle),
"alexander's star" (a stellated icosahedron), "the "impossiball"
(a spherically shaped icosahedron), "mickey's challenge" (a spherically
shaped cuboctahedron), Christoph's jewel (a cuboctahedron), and a
"Rubicized truncated octahedron" (which I don't know the real name
for). The curious reader may want to see, for example, chapter 5
of [B] or the article [GT].