Message #3707

From: Joel Karlsson <>
Subject: Re: [MC4D] Physical 4D puzzle V2
Date: Sun, 14 May 2017 15:16:45 +0200

Yes, that is correct and in fact, you should divide not only with 24 for
the orientation but also with 16 for the placement if you want to calculate
unique states (since the 2x2x2x2 doesn’t have fixed centerpieces). The
point, however, was that if you don’t take that into account you get a
factor of 24*16=384 (meaning that the puzzle has 384 representations of
every unique state) instead of the factor of 192 which you get when
calculating the states from the virtual puzzle and hence every state of the
virtual puzzle has two representations in the physical puzzle. Yes exactly,
they are indeed the same solved (or other) state and you are correct that
the half rotation (taking off a 2x2 layer and placing it at the other end
of the puzzle) takes you from one representation to the same state with the
other representation. This means that the restacking move (taking off the
front 2x4 layer and placing it behind the other 2x4 layer) can be expressed
with half-rotations and ordinary twists and rotations (which you might have
pointed out already). I think I’ve found six moves including ordinary
twists and a restacking move that is identical to a half-rotation and thus
it’s easy to compose a restacking move with one half-rotation and five
ordinary twists. There might be an error since I’ve only played with the
puzzle in my mind so it would be great if you, Melinda, could confirm this
(the sequence is described later in this email).

To be able to communicate move sequences properly we need notation for
representing twists, rotations, half-rotations, restacking moves and folds.
Feel free to come with other suggestion but you can find mine below. Please
read the following thoroughly (maybe twice) to make sure that you
understand everything since misinterpreted notation could potentially
become a nightmare and feel free to ask questions if there is something
that needs clarification.

*Coordinate system and labelling:*

Let’s introduce a global coordinate system. In whatever state the puzzle is
let the positive x-axis point upwards, the positive y-axis towards you and
the positive z-axis to the right (note that this is a right-hand system).
Now let’s name the 8 faces of the puzzle. The right face is denoted with R,
the left with L, the top U (up), the bottom D (down), the front F, the back
B, the center K (kata) and the last one A (ana). The R and L faces are
either the outer corners of the right and left halves respectively or the
inner corners of these halves (forming octahedra) depending on the
representation of the puzzle. The U, D, F and B faces are either two
diamond shapes (looking something like this: <><>) on the corresponding
side of the puzzle or one whole and two half diamond shapes (><><). The K
face is either an octahedron in the center of the puzzle or the outer
corners of the center 2x2x2 block. Lastly, the A face is either two diamond
shapes, one on the right and one on the left side of the puzzle, or another
shape that’s a little bit hard to describe with just a few words (the white
stickers at 5:10 in the latest video, after the half-rotation but before
the restacking move).

We also need a name for normal 3D-rotations, restacking moves and folds
(note that a half-rotation is a kind of restacking move). Let O be the name
for a rOtation (note that the origin O doesn’t move during a rotation, by
the way, these are 3D rotations of the physical puzzle), let S represent a
reStacking move and V a folding/clamshell move (you can remember this by
thinking of V as a folded line).

Further, let I (capital i) be the identity, preserving the state and
rotation of the puzzle. We cannot use I to indicate what moves should be
performed but it’s still useful as we will see later. Since we also want to
be able to express if a sequence of moves is a rotation, preserving the
state of the puzzle but possibly representing it in a different way, we can
introduce mod(rot) (modulo rotation). So, if a move sequence P satisfies P
mod(rot) = I, that means that the state of the puzzle is the same before
and after P is performed although the rotation and representation of the
puzzle are allowed to change. I do also want to introduce mod(3rot) (modulo
3D-rotation) and P mod(3rot) = I means that if the right 3D-rotation (a
combination of O moves as we will see later) is applied to the puzzle after
P you get the identity I. Moreover, let the standard rotation of the puzzle
be any rotation such that the longer side is parallel with the z-axis, that
is the puzzle forms a 2x2x4 (2 pieces thick in the x-direction, 2 in the
y-direction and 4 in the z-direction), and the K face is an octahedron.

*Rotations and twists: *Now we can move on to name actual moves. The
notation of a move is a combination of a capital letter and a lowercase
letter. O followed by x, y or z is a rotation of the whole puzzle around
the corresponding axis in the mathematical positive direction
(counterclockwise/the way your right-hand fingers curl if you point in the
direction of the axis with your thumb). For example, Ox is a rotation
around the x-axis that turns a 2x2x4 into a 2x4x2. A name of a face (U, D,
F, B, R, L, K or A) followed by x, y or z means: detach the 8 pieces that
have a sticker belonging to the face and then turn those pieces around the
global axis. For example, if the longer side of the puzzle is parallel to
the z-axis (the standard rotation), Rx means: take the right 2x2x2 block
and turn it around the global x-axis in the mathematical positive
direction. Note that what moves are physically possible and allowed is
determined by the rotation of the puzzle (I will come back to this later).
Further note that Ox mod(rot) = Oy mod(rot) = Oz mod(rot) = I, meaning that
the 3D-rotations of the physical puzzle corresponds to a 4D-rotation of the
represented 2x2x2x2.

*Inverses and performing a move more than once*

To mark that a move should be performed n times let’s put ^n after it. For
convenience when writing and speaking let ‘ (prime) represent ^-1 (the
inverse) and n represent ^n. The inverse P’ of some permutation P is the
permutation that satisfies P P’ = P’ P = I (the identity). For example
(Rx)’ means: do Rx backwards, which corresponds to rotating the right 2x2x2
block in the mathematical negative direction (clockwise) around the x-axis
and (Rx)2 means: perform Rx twice. However, we can also define powers of
just the lowercase letters, for example, Rx2 = Rx^2 = Rxx = Rx Rx. So
x2=x^2 means: do whatever the capital letter specifies two times with
respect to the x-axis. We can see that the capital letter naturally is
distributed over the two lowercase letters. Rx’ = Rx^-1 means: do whatever
the capital letter specifies but in the other direction than you would have
if the prime wouldn’t have been there (note that thus, x’=x^-1 can be seen
as the negative x-axis). Note that (Rx)^2 = Rx^2 and (Rx)’ = Rx’ which is
true for all twists and rotations but that doesn’t have to be the case for
other types of moves (restacks and folds).

*Restacking moves*

A restacking move is an S followed by either x, y or z. Here the lowercase
letter specifies in what direction to restack. For example, Sy (from the
standard rotation) means: take the front 8 pieces and put them at the back,
whereas Sx means: take the top 8 pieces and put them at the bottom. Note
that Sx is equivalent to taking the bottom 8 pieces and putting them at the
top. However, if we want to be able to make half-rotations we sometimes
need to restack through a plane that doesn’t go through the origin. In the
standard rotation, let Sz be the normal restack (taking the 8 right pieces,
the right 2x2x2 block, and putting them on the left), Sz+ be the restack
where you split the puzzle in the plane further in the positive z-direction
(taking the right 2x2x1 cap of 4 pieces and putting it at the left end of
the puzzle) and Sz- the restack where you split the puzzle in the plane
further in the negative z-direction (taking the left 2x2x1 cap of 4 pieces
and putting it at the right end of the puzzle). If the longer side of the
puzzle is parallel to the x-axis instead, Sx+ would take the top 1x2x2 cap
and put it on the bottom. Note that in the standard rotation Sz mod(rot) =
I. For restacks we see that (Sx)’ = Sx’ = Sx (true for y and z too of
course), that (Sz+)’ = Sz- and that (Sz+)2 = (Sz-)2 = Sz. We can define
Sz’+ to have meaning by thinking of z’ as the negative z-axis and with that
in mind it’s natural to define Sz’+=Sz-. Thus, (Sz)’ = Sz’ and (Sz+)’ =

*Fold moves*

A fold might be a little bit harder to describe in an intuitive way. First,
let’s think about what folds are interesting moves. The folds that cannot
be expressed as rotations and restacks are unfolding the puzzle to a 4x4
and then folding it back along another axis. If we start with the standard
rotation and unfold the puzzle into a 1x4x4 (making it look like a 4x4 from
above) the only folds that will achieve something you can’t do with a
restack mod(rot) is folding it to a 2x4x2 so that the longer side is
parallel with the y-axis after the fold. Thus, there are 8 interesting fold
moves for any given rotation of the puzzle since there are 4 ways to unfold
it to a 4x4 and then 2 ways of folding it back that make the move different
from a restack move mod(rot). Let’s call these 8 folds interesting fold
moves. Note that an interesting fold move always changes which axis the
longer side of the puzzle is parallel with. Further note that both during
unfold and fold all pieces are moved; it would be possible to have 8 of the
pieces fixed during an unfold and folding the other half 180 degrees but I
think that it’s more intuitive that these moves fold both halves 90 degrees
and performing them with 180-degree folds might therefore lead to errors
since the puzzle might get rotated differently. To illustrate a correct
unfold without a puzzle: Put your palms together such that your thumbs
point upward and your fingers forward. Now turn your right hand 90 degrees
clockwise and your left hand 90 degrees counterclockwise such that the
normal to your palms point up, your fingers point forward, your right thumb
to the right and your left thumb to the left. That was what will later be
called a Vx unfold and the folds are simply reversed unfolds. (I might have
used the word “fold” in two different ways but will try to use the term
“fold move” when referring to the move composed by an unfold and a fold
rather than simply calling these moves “folds”.)

To specify the unfold let’s use V followed by one of x, y, z, x’, y’ and
z’. The lowercase letter describes in which direction to unfold. Vx means
unfold in the direction of the positive x-axis and Vx’ in the direction of
the negative x-axis, if that makes any sense. I will try to explain more
precisely what I mean with the example Vx from the standard rotation (it
might also help to read the last sentences in the previous paragraph
again). So, the puzzle is in the standard rotation and thus have the form
2x2x4 (x-, y- and z-thickness respectively). The first part (the unfolding)
of the move specified with Vx is to unfold the puzzle in the x-direction,
making it a 1x4x4 (note that the thickness in the x-direction is 1 after
the Vx unfold, which is no coincidence). There are two ways to do that;
either the sides of the pieces that are initially touching another piece
(inside of the puzzle in the x,z-plane and your palms in the hand example)
are facing up or down after the unfold. Let Vx be the unfold where these
sides point in the direction of the positive x-axis (up) and Vx’ the other
one where these sides point in the direction of the negative x-axis (down)
after the unfold. Note that if the longer side of the puzzle is parallel to
the z-axis only Vx, Vx’, Vy and Vy’ are possible. Now we need to specify
how to fold the puzzle back to complete the folding move. Given an unfold,
say Vx, there are only two ways to fold that are interesting (not turning
the fold move into a restack mod(rot)) and you have to fold it
perpendicular to the unfold to create an interesting fold move. So, if you
start with the standard rotation and do Vx you have a 1x4x4 that you have
to fold into a 2x4x2. To distinguish the two possibilities, use + or -
after the Vx. Let Vx+ be the unfold Vx followed by the interesting fold
that makes the sides that are initially touching another piece (before the
unfold) touch another piece after the fold move is completed and let Vx- be
the other interesting fold move that starts with the unfold Vx. (Thus,
continuing with the hand example, if you want to do a Vx+ first do the Vx
unfold described in the end of the previous paragraph and then fold your
hands such that your fingers point up, the normal to your palms point
forward, the right palm is touching the right-hand fingers, the left palm
is touching the left-hand fingers, the right thumb is pointing to the right
and the left thumb is pointing to the left). Note that the two halves of
the puzzle always should be folded 90 degrees each and you should never
make a fold or unfold where you fold just one half 180-degrees (if you want
to use my notation, that is). Further note that Vx+ Sx mod(3rot) = Vx- and
that Vx+ Vx+ = I which is equivalent to (Vx+)’ = Vx+ and this is true for
all fold moves (note that after a Vx+ another Vx+ is always possible).

*The 2x2x2x2 in the MC4D software*

The notation above can also be applied to the 2x2x2x2 in the MC4D program.
There, you are not allowed to do S or V moves but instead, you are allowed
to do the [crtl]+[left-click] moves. This can easily be represented with
notation similar to the above. Let’s use C (as in Centering) and one of x,
y and z. For example, Cx would be to rotate the face in the positive
x-direction aka the U face to the center. Thus, Cz’ is simply
[ctrl]+[left-click] on the L face and similarly for the other C moves. The
O, U, D, F, B, R, L, K and A moves are performed in the same way as above
so, for example Rx would be a [left-click] on the top-side of the right
face. In this representation of the puzzle almost all moves are allowed;
all U, D, F, B, R, L, K, O and C moves are possible regardless of rotation
and only A moves (and of course the rightfully forbidden S and V moves) are
impossible regardless of rotation. Note that R and L moves in the software
correspond to the same moves of the physical puzzle but this is not
generally true (I will come back to this later).

*Possible moves (so far) in the standard rotation*

In the standard rotation, the possible/allowed moves with the definitions
above are:

O moves, all of these are always possible in any state and rotation of the
puzzle since they are simply 3D-rotations.

R, L moves, all of these as well since the puzzle has a right and left
2x2x2 block in the standard rotation.

U. D moves, just Ux2 and Dx2 since the top and bottom are 1x2x4 blocks with
less symmetry than a 2x2x2 block.

F, B moves, just Fy2 and By2 since the front and back are 2x1x4 blocks with
less symmetry than a 2x2x2 block.

K moves, all possible since this is a rotation of the center 2x2x2 block.

A moves, only Az moves since this is two 2x2x1 blocks that have to be
rotated together.

S moves, not Sx+, Sx-, Sy+ or Sy- since those are not defined in the
standard rotation.

V moves, not Vz+ or Vz- since the definition doesn’t give these meaning
when the long side of the puzzle is parallel with the z-axis.

Note that for example Fz2 is not allowed since this won’t take you to a
state of the puzzle. To allow more moves we need to extend the definitions
(after the extension in the next paragraph all rotations and twists (O, R,
L, U, D, F, B, K, A) are possible in any rotation and only which S and V
moves are possible depend on the state and rotation of the puzzle).

*Extension of some definitions*

It’s possible to make an extension that allows all O, R, L, U, D, F, B, K
and A moves in any state. I will explain how this can be done in the
standard rotation but it applies analogously to any other rotation where
the K face is an octahedron. First let’s focus on U, D, F and B and because
of the symmetry of the puzzle all of these are analogous so I will only
explain one. The extension that makes all U moves possible (note that the U
face is in the positive x-direction) is as follows: when making an U move
first detach the 8 top pieces which gives you a 1x2x4 block, fold this
block into a 2x2x2 block in the positive x-direction (similar to the later
part of a Vx+ move from the standard rotation) such that the U face form an
octahedron, rotate this 2x2x2 block around the specified axis (for example
around the z-axis if you are doing an Uz move), reverse the fold you just
did creating a 1x2x4 block again and reattach the block. The A moves can be
done very similarly but after you have detached the two 2x2x1 blocks you
don’t fold them but instead you stack them similar to a Sz move, creating a
2x2x2 block with the A face as an octahedron in the middle and then reverse
the process after you have rotated the block as specified (for example
around the negative y-axis if you are doing an Ay’ move). Note that these
extended moves are closely related to the normal moves and for example Ux =
Ry Ly’ Kx Ly Ry’ in the standard rotation and note that (Ry Ly’) mod(rot) =
(Ly Ry’) mod(rot) = I (this applies to the other extended moves as well).

If the cube is in the half-rotated state, where both the R and L faces are
octahedra, you can extend the definitions very similarly. The only thing
you have to change is how you fold the 2x4 blocks when performing a U, D, F
or B move. Instead of folding the two 2x2 halves of the 2x4 into a 2x2 you
have to fold the end 1x2 block 180 degrees such that the face forms an

These moves might be a little bit harder to perform, to me especially the A
moves seems a bit awkward, so I don’t know if it’s good to use them or not.
However, the A moves are not necessary if you allow Sz in the standard
rotation (which you really should since Sz mod(rot) = I in the standard
rotation) and thus it might not be too bad to use this extended version.
The notation supports both variants so if you don’t want to use these
extended moves that shouldn’t be a problem. Note that, however, for example
Ux (physical puzzle) != Ux (virtual puzzle) where != means “not equal to”
(more about legal/illegal moves later).

*Generalisation of the notation*

Let’s generalise the notation to make it easier to use and to make it work
for any n^4 cube in the MC4D software. Previously, we saw that (Rx)2 = Rx
Rx and if we allow ourselves to rewrite this as (Rx)2 = Rx2 = Rxx = Rx Rx
we see that the capital letter naturally can be distributed over the
lowercase letters. We can make this more general and say that any capital
letter followed by several lowercase letters means the same thing as the
capital letter distributed over the lowercase letters. Like Rxyz = Rx Ry Rz
and here R can be exchanged with any capital letter and xyz can be
exchanged with any sequence of lowercase letters. We can also allow several
capital letters and one lowercase letter, for example RLx and let’s define
this as RLx = Rx Lx so that the lowercase letter can be distributed over
the capital letters. We can also define a capital letter followed by ‘
(prime) like R’x = Rx’ and R’xy = Rx’y’ so the ‘ (prime) is distributed
over the lowercase letters. Note that we don’t define a capital to any
other power than -1 like this since for example R2x = RRx might seem like a
good idea at first but it isn’t very useful since R2 and RR are the same
lengths (and powers greater than two are seldom used) and we will see that
we can define R2 in another way that generalises the notation to all n^4

Okay, let’s define R2 and similar moves now and have in mind what moves we
want to be possible for a n^4 cube. The moves that we cannot achieve with
the notation this far is twisting deeper slices. To match the notation with
the controls of the MC4D software let R2x be the move similar to Rx but
twisting the 2nd layer instead of the top one and similarly for other
capital letters, numbers (up to n) and lowercase letters. Thus, R2x is
performed as Rx but holding down the number 2 key. Just as in the program,
when no number is specified 1 is assumed and you can combine several
numbers like R12x to twist both the first and second layer. This notation
does not apply to rotations (O) folding moves (V) and restacking moves (S)
(I suppose you could redefine the S move using this deeper-slice-notation
and use S1z as Sz+, S2z as Sz and S3z as Sz- but since these moves are only
allowed for the physical 2x2x2x2 I think that the notation with + and – is
better since S followed by a lowercase letter without +/- always means
splitting the cube in a coordinate plane that way, not sure though so input
would be great). The direction of the twist R3x should be the same as Rx
meaning that if Rx takes stickers belonging to K and move them to F, so
should R3x, in accordance with the controls of the MC4D software. Note that
for a 3x3x3x3 it’s true that R3z = Lz whereas R3x = Lx’ (note that R and L
are the faces in the z-directions so because of the symmetry of the cube it
will also be true that for example U3x = Bx whereas U3y=By’).

What about the case with several capital letters and several lowercase
letters, for instance, RLxy? I see two natural definitions of this. Either,
we could have RLxy = Rxy Lxy or we could define it as RLxy = RLx RLy. These
are generally not the same (if you exchange R and L with any allowed
capital letter and similarly for x and y). I don’t know what is best, what
do you think? The situations I find this most useful in are RL’xy to do a
rotation and RLxy as a twist. However, since R and L are opposite faces
their operations commute which imply RL’xy (1st definition) = Rxy L’xy =
Rx Ry Lx’ Ly’ = Rx Lx’ Ry Ly’ = RL’x RL’y = RL’xy (2nd definition) and
similarly for the other case with RLxy. Hopefully, we can find another
useful sequence of moves where this notation can be used with only one of
the definitions and can thereby decide which definition to use. Personally,
I feel like RLxy = RLx RLy is the more intuitive definition but I don’t
have any good argument for this so I’ll leave the question open.

For convenience, it might be good to be able to separate moves like Rxy and
RLx from the basic moves Rx, Oy etc when speaking and writing. Let’s call
the basic moves that only contain one capital letter and one lowercase
letter (possibly a + or –, a ‘ (prime) and/or a number) *simple moves*
(like Rx, L’y, Ux2 and D3y’) and the moves that contain more than one
capital letter or more than one lowercase letter *composed moves*.

*More about inverses*

This list can obviously be made longer but here are some identities that
are good to know and understand. Note that R, L, U, x, y and z below just
are examples, the following is true in general for non-folds (however, S
moves are fine).

(P1 P2 … Pn)’ = Pn’ … P2’ P1’ (Pi is an arbitrary permutation for
(Rxy)’ = Ry’x’ = R’yx
(RLx)’ = LRx’ = L’R’x
(Sx+z)’ = Sz’ Sx’+ (just as an example with restacking moves, note that
the inverse doesn’t change the + or -)
RLUx’y’z’=R’L’U’xyz (true for both definitions)
(RLxy)’ = LRy’x’ = L’R’yx (true for both definitions)

For V moves we have that: (Vx+)’ = Vx+ != Vx’+ (!= means “not equal to”)

*Some important notes on legal/illegal moves *Although there are a lot of
moves possible with this notation we might not want to use them all. If we
really want a 2x2x2x2 and not something else I think that we should try to
stick to moves that are legal 2x2x2x2 moves as far as possible (note that I
said legal moves and not permutations (a legal permutation can be made up
of one or more legal moves)). Clarification: cycling three of the
edge-pieces of a Rubik’s cube is a legal permutation but not a legal move,
a legal move is a rotation of the cube or a twist of one of the layers. In
this section I will only address simple moves and simply refer to them as
moves (legal composed moves are moves composed by legal simple moves).

I do believe that all moves allowed by my notation are legal permutations
based on their periodicity (they have a period of 2 or 4 and are all even
permutations of the pieces). So, which of them correspond to legal 2x2x2x2
moves? The O moves are obviously legal moves since they are equal to the
identity mod(rot). The same goes for restacking (S) (with or without +/-)
in the direction of the longest side of the puzzle (Sz, Sz+ and Sz- in the
standard rotation) since these are rotations and half-rotations that don’t
change the state of the puzzle. Restacking in the other directions and fold
moves (V) are however not legal moves since they are made of 8 2-cycles and
change the state of the puzzle (note that they, however, are legal
permutations). The rest of the moves (R, L, F, B, U, D, K and A) can be
divided into two sets: (1) the moves where you rotate a 2x2x2 block with an
octahedron inside and (2) the moves where you rotate a 2x2x2 block without
an octahedron inside. A move belonging to (2) is always legal. We can see
this by observing what a Rx does with the pieces in the standard rotation
with just K forming an octahedron. The stickers move in 6 4-cycles and if
the puzzle is solved the U and D faces still *looks* solved after the move.
A move belonging to set (1) is legal either if it’s an 180-degree twist or
if it’s a rotation around the axis parallel with the longest side of the
puzzle (the z-axis in the standard rotation). Quite interestingly these are
exactly the moves that don’t mix up the R and L stickers with the rest in
the standard rotation. I think I know a way to prove that no legal 2x2x2x2
move can mix up these stickers with the rest and this has to do with the
fact that these stickers form an inverted octahedron (with the corners
pointing outward) instead of a normal octahedron (let’s call this
hypothesis * for now). Note that all legal twists (R, L, F, B, U, D, K and
A) of the physical puzzle correspond to the same twist in the MC4D software.

So, what moves should we add to the set of legal moves be able to get to
every state of the 2x2x2x2? I think that we should add the restacking moves
and folding moves since Melinda has already found a pretty short sequence
of those moves to make a rotation that changes which colours are on the R
and L faces. That sequence, starting from the standard rotation, is: Oy
Sx+z Vy+ Ozy’ Vy+ Ozy’ Vy+ = Oy Sx+z (Vy+ Ozy’)3 mod(3rot) = I mod(rot)
(hopefully I got that right). What I have found (which I mentioned
previously) is that (from the standard rotation): RLx Ly2 Sy By2 Ly2 RLx’ =
Sz- and since this is equivalent with Sy = Ly2 RLx’ Sz- RLx Ly2 By2 the
restacking moves that are not legal moves are not very complicated
permutations and therefore I think that we can accept them since they help
us mix up the R and L stickers with other faces. In a sense, the folding
moves are “more illegal” since they cannot be composed by the legal moves
(according to hypothesis *). This is also true for the illegal moves
belonging to set (1) discussed above. However, since the folding moves is
probably easier to perform and is enough to reach every state of the
2x2x2x2 I think that we should use them and not the illegal moves belonging
to (1). Note, once again, that all moves described by the notation are
legal permutations (even the ones that I just a few words ago referred to
as illegal moves) so if you wish you can use all of them and still only
reach legal 2x2x2x2 states. However (in a strict sense) one could argue
that you are not solving the 2x2x2x2 if you use illegal moves. If you only
use illegal moves to compose rotations (that is, create a permutation
including illegal moves that are equal to I mod(rot)) and not actually
using the illegal moves as twists I would classify that as solving a
2x2x2x2. What do you think about this?

*What moves to use? *Here’s a short list of the simple moves that I think
should be used for the physical 2x2x2x2. Note that this is just my thoughts
and you may use the notation to describe any move that it can describe if
you wish to. The following list assumes that the puzzle is in the standard
rotation but is analogous for other representations where the K face is an

O, all since they are I mod(rot),
R, L, all since they are legal (note Rx (physical puzzle) = Rx (virtual
U, D, only x2 since these are the only legal easy-to-perform moves,
F, B, only y2 since these are the only legal easy-to-perform moves,
K, A, only z, z’ and z2 since these are the only legal easy-to-perform
S, at least z, z+ and z- since these are equal to I mod(rot),
S, possibly x and y since these help us perform rotations and is easy to
compose (not necessary to reach all states and not legal though),
V, all 8 allowed by the rotation of the puzzle (at least one is necessary
to reach all states and if you allow one the others are easy to achieve

If you start with the standard rotation and then perform Sz+ the following
applies instead (this applies analogously to any other rotation where the R
and L faces are octahedra).

O, all since they are I mod(rot),
R, L, only z, z’ and z2 since these are the only legal easy-to-perform
U, D, only x2 since these are the only legal easy-to-perform moves,
F, B, only y2 since these are the only legal easy-to-perform moves,
K, A, at least z, z’ and z2, possibly all (since they are legal) although
some might be hard to perform.
S, V, same as above.

Note that (from the standard rotation): Ux2 Sz+ Ux2 Sz- != I mod(rot) (!=
for not equal) which implies that the Ux2 move when R and L are octahedra
is different from the Ux2 move when K is an octahedron. (Actually, the
sequence above is equal to Uy2).

Regarding virtual n^4 cubes I think that all O, C, R, L, U, D, F, B, K and
A moves should be used since they are all legal and really the only thing
you need (left-clicking on an edge or corner piece in the computer program
can be described quite easily with the notation, for example, Kzy2 is
left-clicking on the top-front edge piece on the K face).

I hope this was possible to follow and understand. Feel free to ask
questions about the notation if you find anything ambiguous.

Best regards,
Joel Karlsson

Den 4 maj 2017 12:01 fm skrev "Melinda Green
[4D_Cubing]" <>:

Thanks for the correction. A couple of things: First, when assembling one
piece at a time, I’d say there is only 1 way to place the first piece, not

  1. Otherwise you’d have to say that the 1x1x1 puzzle has 24 states. I
    understand that this may be conventional, but to me, that just sounds silly.

Second, I have the feeling that the difference between the "two
representations" you describe is simply one of those half-rotations I
showed in the video. In the normal solved state there is only one complete
octahedron in the very center, and in the half-rotated state there is one
in the middle of each half of the "inverted" form. I consider them to be
the same solved state.


On 5/3/2017 2:39 PM, Joel Karlsson [4D_Cubing]

Horrible typo… It seems like I made some typos in my email regarding the
state count. It should of course be 16!12^16/(6*192) and NOT
12!16^12/(6*192). However, I did calculate the correct number when
comparing with previous results so the actual derivation was correct.

Something of interest is that the physical pieces can be assembled in
16!24*12^15 ways since there are 16 pieces, the first one can be oriented
in 24 ways and the remaining can be oriented in 12 ways (since a corner
with 3 colours never touch a corner with just one colour). Dividing with 6
to get a single orbit still gives a factor 2*192 higher than the actual
count rather than 192. This shows that every state in the MC4D
representation has 2 representations in the physical puzzle. These two
representations must be the previously discussed, that the two halves
either have the same color on the outermost corners or the innermost
(forming an octahedron) when the puzzle is solved and thus both are
complete representations of the 2x2x2x2.

Best regards,
Joel Karlsson

Den 30 apr. 2017 10:51 em skrev "Joel Karlsson" <>:

I am no expert on group theory, so to better understand what twists are
legal I read through the part of Kamack and Keane’s *The Rubik Tesseract *about
orienting the corners. Since all even permutations are allowed the easiest
way to check if a twist is legal might be to:

  1. Check that the twist is an even permutation, that is: the same twist can
    be done by performing an even number of piece swaps (2-cycles).
  2. Check the periodicity of the twist. If A^k=I (A^k meaning performing the
    twist k times and I (the identity) representing the permutation of doing
    nothing) and k is not divisible by 3 the twist A definitely doesn’t violate
    the restriction of the orientations since kx mod 3 = 0 and k mod 3 != 0
    implies x mod 3 = 0 meaning that the change of the total orientation x for
    the twist A mod 3 is 0 (which precisely is the restriction of legal twists;
    that they must preserve the orientation mod 3).

For instance, this implies that the restacking moves are legal 2x2x2x2
moves since both are composed of 8 2-cycles and both can be performed twice
(note that 2 is not divisible by 3) to obtain the identity.

Note that 1 and 2 are sufficient to check if a twist is legal but only 1 is
necessary; there can indeed exist a twist violating 2 that still is legal
and in that case, I believe that we might have to study the orientation
changes for that specific twist in more detail. However, if a twist can be
composed by other legal twists it is, of course, legal as well.

Best regards,

2017-04-29 1:04 GMT+02:00 Melinda Green
[4D_Cubing] <>:

> First off, thanks everyone for the helpful and encouraging feedback!
> Thanks Joel for showing us that there are 6 orbits in the 2^4 and for your
> rederivation of the state count. And thanks Matt and Roice for pointing out
> the importance of the inverted views. It looks so strange in that
> configuration that I always want to get back to a normal view as quickly as
> possible, but it does seem equally valid, and as you’ve shown, it can be
> helpful for more than just finding short sequences.
> I don’t understand Matt’s "pinwheel" configuration, but I will point out
> that all that is needed to create your twin interior octahedra is a single
> half-rotation like I showed in the video at 5:29
> <>. The two main
> halves do end up being mirror images of each other on the visible outside
> like he described. Whether it’s the pinwheel or the half-rotated version
> that’s correct, I’m not sure that it’s a bummer that the solved state is
> not at all obvious, so long as we can operate it in my original
> configuration and ignore the fact that the outer faces touch. That would
> just mean that the "correct" view is evidence that that the more
> understandable view is legitimate.
> I’m going to try to make a snapable V3 which should allow the pieces to be
> more easily taken apart and reassembled into other forms. Shapeways does
> offer a single, clear translucent plastic that they call "Frosted Detail",
> and another called "Transparent Acrylic", but I don’t think that any sort
> of transparent stickers will help us, especially since this thing is chock
> full of magnets. The easiest way to let you see into the two hemispheres
> would be to simply truncate the pointy tips of the stickers. That already
> happens a little bit due to the way I’ve rounded the edges. Here is a
> close-up <> of a half-rotation
> in which you can see that the inner yellow and white faces are solved. Your
> suggestion of little mapping dots on the corners also works, but just
> opening the existing window further would work more directly.
> -Melinda
> On 4/28/2017 2:15 PM, Roice Nelson [4D_Cubing] wrote:
> I agree with Don’s arguments about adjacent sticker colors needing to be
> next to each other. I think this can be turned into an accurate 2^4 with
> coloring changes, so I agree with Joel too :)
> To help me think about it, I started adding a new projection option for
> spherical puzzles to MagicTile, which takes the two hemispheres of a puzzle
> and maps them to two disks with identified boundaries connected at a point,
> just like a physical "global chess
> <>" game I have.
> Melinda’s puzzle is a lot like this up a dimension, so think about two
> disjoint balls, each representing a hemisphere of the 2^4, each a "subcube"
> of Melinda’s puzzle. The two boundaries of the balls are identified with
> each other and as you roll one around, the other half rolls around so that
> identified points connect up. We need to have the same restriction on
> Melinda’s puzzle.
> In the pristine state then, I think it’d be nice to have an internal
> (hidden), solid colored octahedron on each half. The other 6 faces should
> all have equal colors split between each hemisphere, 4 stickers on each
> half. You should be able to reorient the two subcubes to make a half
> octahedron of any color on each subcube. I just saw Matt’s email and
> picture, and it looks like we were going down the same thought path. I
> think with recoloring (mirroring some of the current piece colorings)
> though, the windmill’s can be avoided (?)
> […] After staring/thinking a bit more, the coloring Matt came up with is
> right-on if you want to put a solid color at the center of each
> hemisphere. His comment about the "mirrored" pieces on each side helped me
> understand better. 3 of the stickers are mirrored and the 4th is the
> hidden color (different on each side for a given pair of "mirrored"
> pieces). All faces behave identically as well, as they should. It’s a
> little bit of a bummer that it doesn’t look very pristine in the pristine
> state, but it does look like it should work as a 2^4.
> I wonder if there might be some adjustments to be made when shapeways
> allows printing translucent as a color :)
> […] Sorry for all the streaming, but I wanted to share one more
> thought. I now completely agree with Joel/Matt about it behaving as a 2^4,
> even with the original coloring. You just need to consider the corner
> colors of the two subcubes (pink/purple near the end of the video) as being
> a window into the interior of the piece. The other colors match up as
> desired. (Sorry if folks already understood this after their emails and
> I’m just catching up!)
> In fact, you could alter the coloring of the pieces slightly so that the
> behavior was similar with the inverted coloring. At the corners where 3
> colors meet on each piece, you could put a little circle of color of the
> opposite 4th color. In Matt’s windmill coloring then, you’d be able to see
> all four colors of a piece, like you can with some of the pieces on
> Melinda’s original coloring. And again you’d consider the color circles a
> window to the interior that did not require the same matching constraints
> between the subcubes.
> I’m looking forward to having one of these :)
> Happy Friday everyone,
> Roice
> On Fri, Apr 28, 2017 at 1:14 AM, Joel Karlsson
> [4D_Cubing] <> wrote:
>> Seems like there was a slight misunderstanding. I meant that you need to
>> be able to twist one of the faces and in MC4D the most natural choice is
>> the center face. In your physical puzzle you can achieve this type of twist
>> by twisting the two subcubes although this is indeed a twist of the
>> subcubes themselves and not the center face, however, this is still the
>> same type of twist just around another face.
>> If the magnets are that allowing the 2x2x2x2 is obviously a subgroup of
>> this puzzle. Hopefully the restrictions will be quite natural and only some
>> "strange" moves would be illegal. Regarding the "families of states" (aka
>> orbits), the 2x2x2x2 has 6 orbits. As I mentioned earlier all allowed
>> twists preserves the parity of the pieces, meaning that only half of the
>> permutations you can achieve by disassembling and reassembling can be
>> reached through legal moves. Because of some geometrical properties of the
>> 2x2x2x2 and its twists, which would take some time to discuss in detail
>> here, the orientation of the stickers mod 3 are preserved, meaning that the
>> last corner only can be oriented in one third of the number of orientations
>> for the other corners. This gives a total number of orbits of 2x3=6. To
>> check this result let’s use this information to calculate all the possible
>> states of the 2x2x2x2; if there were no restrictions we would have 16! for
>> permuting the pieces (16 pieces) and 12^16 for orienting them (12
>> orientations for each corner). If we now take into account that there are 6
>> equally sized orbits this gets us to 12!16^12/6. However, we should also
>> note that the orientation of the puzzle as a hole is not set by some kind
>> of centerpieces and thus we need to devide with the number of orientations
>> of a 4D cube if we want all our states to be separated with twists and not
>> only rotations of the hole thing. The number of ways to orient a 4D cube in
>> space (only allowing rotations and not mirroring) is 8x6x4=192 giving a
>> total of 12!16^12/(6*192) states which is indeed the same number that
>> for example David Smith arrived at during his calculations. Therefore,
>> when determining whether or not a twist on your puzzle is legal or not it
>> is sufficient and necessary to confirm that the twist is an even
>> permutation of the pieces and preserves the orientation of stickers mod 3.
>> Best regards,
>> Joel
>> Den 28 apr. 2017 3:02 fm skrev "Melinda Green
>> [4D_Cubing]" <>:
>> The new arrangement of magnets allows every valid orientation of pieces.
>> The only invalid ones are those where the diagonal lines cutting each
>> cube’s face cross each other rather than coincide. In other words, you can
>> assemble the puzzle in all ways that preserve the overall diamond/harlequin
>> pattern. Just about every move you can think of on the whole puzzle is
>> valid though there are definitely invalid moves that the magnets allow. The
>> most obvious invalid move is twisting of a single end cap.
>> I think your description of the center face is not correct though. Twists
>> of the outer faces cause twists "through" the center face, not "of" that
>> face. Twists of the outer faces are twists of those faces themselves
>> because they are the ones not changing, just like the center and outer
>> faces of MC4D when you twist the center face. The only direct twist of the
>> center face that this puzzle allows is a 90 degree twist about the outer
>> axis. That happens when you simultaneously twist both end caps in the same
>> direction.
>> Yes, it’s quite straightforward reorienting the whole puzzle to put any
>> of the four axes on the outside. This is a very nice improvement over the
>> first version and should make it much easier to solve. You may be right
>> that we just need to find the right way to think about the outside faces.
>> I’ll leave it to the math geniuses on the list to figure that out.
>> -Melinda
>> On 4/27/2017 10:31 AM, Joel Karlsson
>> [4D_Cubing] wrote:
>> Hi Melinda,
>> I do not agree with the criticism regarding the white and yellow stickers
>> touching each other, this could simply be an effect of the different
>> representations of the puzzle. To really figure out if this indeed is a
>> representation of a 2x2x2x2 we need to look at the possible moves (twists
>> and rotations) and figure out the equivalent moves in the MC4D software.
>> From the MC4D software, it’s easy to understand that the only moves
>> required are free twists of one of the faces (that is, only twisting the
>> center face in the standard perspective projection in MC4D) and 4D
>> rotations swapping which face is in the center (ctrl-clicking in MC4D). The
>> first is possible in your physical puzzle by rotating the white and yellow
>> subcubes (from here on I use subcube to refer to the two halves of the
>> puzzle and the colours of the subcubes to refer to the "outer colours").
>> The second is possible if it’s possible to reach a solved state with any
>> two colours on the subcubes that still allow you to perform the previously
>> mentioned twists. This seems to be the case from your demonstration and is
>> indeed true if the magnets allow the simple twists regardless of the
>> colours of the subcubes. Thus, it is possible to let your puzzle be a
>> representation of a 2x2x2x2, however, it might require that some moves that
>> the magnets allow aren’t used.
>> Best regards,
>> Joel
>> 2017-04-27 3:09 GMT+02:00 Melinda Green
>> [4D_Cubing] <>:
>>> Dear Cubists,
>>> I’ve finished version 2 of my physical puzzle and uploaded a video of it
>>> here:
>>> Again, please don’t share these videos outside this group as their
>>> purpose is just to get your feedback. I’ll eventually replace them with a
>>> public video.
>>> Here is an extra math puzzle that I bet you folks can answer: How many
>>> families of states does this puzzle have? In other words, if disassembled
>>> and reassembled in any random configuration the magnets allow, what are the
>>> odds that it can be solved? This has practical implications if all such
>>> configurations are solvable because it would provide a very easy way to
>>> fully scramble the puzzle.
>>> And finally, a bit of fun: A relatively new friend of mine and new list
>>> member, Marc Ringuette, got excited enough to make his own version. He
>>> built it from EPP foam and colored tape, and used honey instead of magnets
>>> to hold it together. Check it out here:
>>> essert_cube.jpg I don’t know how practical a solution this is but it
>>> sure looks delicious! Welcome Marc!
>>> -Melinda