Message #3709
From: Melinda Green <melinda@superliminal.com>
Subject: Re: [MC4D] Physical 4D puzzle V2
Date: Thu, 18 May 2017 18:33:05 -0700
Hello Joel,
Thanks for drilling into this puzzle. Finding good ways to discuss and think about moves and representations will be key. I’ll comment on some details in-line.
On 5/14/2017 6:16 AM, Joel Karlsson joelkarlsson97@gmail.com [4D_Cubing] wrote:
>
> 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).
Yes, I made that claim in the video but didn’t show it because I have yet to record such a sequence. I’ve only stumbled through it a few times. I talked about it at 5:53 <https://www.youtube.com/watch?v=zqftZ8kJKLo&t=5m53s> though I mistakenly called it a twist, when I should have called it a sequence.
>
> 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).
You mean "RLx Ly2 Sy By2 Ly2 RLx’ = Sz-"? Yes, that works. There do seem to be easier ways to do that beginning with an ordinary rolling rotation. I don’t see those in your notation, but the equivalent using a pair of twists would be Rx Lx’ Sx Vy if I got that right.
>
> 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).
I see the utility of a global coordinate system, but this one seems rather non-standard. I suggest that X be to the right, and Y up since these are near-universal standards. Z can be in or out. I have no opinion. If there is any convention in the twisty puzzle community, I’d go with that.
Note also that the wiki may be a good place to document and iterate on terminology, descriptions and diagrams. Ray added a "notation" section to the 3^4 page here <http://wiki.superliminal.com/wiki/3%5E4>, and I know that one other member was thinking of collecting a set of moves on another wiki page.
>
> 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).
>
I think it’s more correct to say that the K face is either an octahedron at the origin (A<>K<>A) or in the center of one of the main halves, with the A face inside the other half(>A<>K<). This was what I was getting at in my previous message. You do later talk about octahedral faces being in either the center or the two main halves, so this is just terminology. But about "the outer corners of the center 2x2x2 block", this cannot be the A or K face as you’ve labeled them. You’ve been calling these the L/R faces, but the left-right distinction disappears in the half-rotated state, so maybe "left" and "right" aren’t the best names. To me, they are always the "outside" faces, regardless. You can distinguish them as the left and right outside faces in one representation, or as the center and end outside faces in the other. (Or perhaps "end" versus "ind" if we want to be cute.)
I’m also a little torn about naming the interior faces ana and kata, not because of the names themselves which I like, but because the mysterious faces to me are the outermost ones you’re calling R and L. It only requires a simple rotation to move faces in and out of the interior (octahedral) positions, but it’s much more difficult to move another axis into the L/R/outer direction.
So maybe the directions can be
* Up-Down
* Front-Back
* Ind-End
* Ana-Kata
I’m not in love with it and will be happy with anything that works. Thoughts anyone?
>
> 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).
>
I think it’s fine to call the clamshell move a fold or denote it as V. I just wouldn’t consider it to be a basic move since it’s a simple composite of 3 basic twists as shown here <https://www.youtube.com/watch?v=Asx653BGDWA&t=19m40s>. In general, I think there are so many useful composite moves that we need to be able to easily make them up ad hoc with substitutions like *Let **↓ = **Rx Lx’*. These are really macro moves which can be nested. That said, it’s a particularly useful move so it’s probably worth describing in some formal way like you do in detail below.
>
>
> 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.
>
There are several, distinct types of rotations, none of which change the state of the puzzle, and I think we need a way to be unambiguous about them. The types I see are
- Simple reorientation of the physical puzzle in the hand, no magnets involved. IE your ‘O’ moves and maybe analogous to mouse-dragging in MC4D?
- Rolling of one 2x2x2 half against the other. Maybe analogous to ctrl-click?
- Half-rotations. Maybe analogous to a "face first" view? (ctrl-click on a 2-color piece of the 3^4 with the setting "Ctrl-Click Rotates: by Cubie")
- Whole-puzzle reorientations that move an arbitrary axis into the "outer" 2 faces. No MC4D analog.
> *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+)’ = 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).
Assuming you complete a folding move in the same representation (<><> or ><><), then there are only two interesting choices. That’s because it doesn’t matter which end of a chosen cutting plane you open it from, the end result will be the same. That also means that any two consecutive clamshell moves along the same cutting plane will undo each other. It further suggests that any interesting sequence of clamshell moves must alternate between the two possible long cut directions, meaning there is no choice involved. 12 clamshell moves will cycle back to the initial state.
There is one other weird folding move where you open it in one direction and then fold the two halves back-to-back in a different direction. If you simply kept folding along the initial hinge, you’d simply have a restacking. When completed the other way, it’s equivalent to a restacking plus a clamshell, so I don’t think it’s useful though it is somewhat interesting.
>
> 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.
I think the only legal 90 degree twists of the K face are those about the long axis. I believe this is what Christopher Locke was saying in this message <https://groups.yahoo.com/neo/groups/4D_Cubing/conversations/messages/3645>. To see why there is no straightforward way to perform other 90 degree twists, you only need to perform a 90 degree twist on an outer (L/R) face and then reorient the whole puzzle along a different outer axis. If the original twist was not about the new long axis, then there is clearly no straightforward way to undo that twist.
>
> 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.
I noticed something like this the other day but realized that it only seems to work for rotations along the long dimension (z in your example). These are already easily accomplished by a simple rotation to put the face in question on the end caps, followed by a double end-cap twist.
This is as far as I’m going to comment for the moment because the information gets very dense and I’ve been mulling and picking over your message for several days already. In short, I really like your attempt to provide a complete system of notation for discussing this puzzle and will be curious to hear your thoughts on my comments so far. I hope others will chime in too.
One final thought is that a real "acid test" of any notation system for this puzzle will be attempt to translate some algorithms from MC4D. I would most like to see a sequence that flips a single piece, like the second 4-color series on this page <http://superliminal.com/cube/solution/pages/four_color.htm> of Roice’s solution, or his pair of twirled corners at the end of this page <http://superliminal.com/cube/solution/pages/series_hints.htm>. One trick will be to minimize the number of whole-puzzle reorientations needed, but really any sequence that works will be great evidence that the puzzles are equivalent. I suspect that this sort of exercise will never be practical because it will require too many reorientations, and that entirely new methods will be needed to actually solve this puzzle.
Best,
-Melinda
> 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 octahedron.
>
> 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 cubes.
>
> 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 2^nd 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(1^st definition) = Rxy L’xy = Rx Ry Lx’ Ly’ = Rx Lx’ Ry Ly’ = RL’x RL’y = RL’xy (2^nd 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 i=1,2,…n)
> (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 octahedron.
>
> O, all since they are I mod(rot),
> R, L, all since they are legal (note Rx (physical puzzle) = Rx (virtual 2x2x2x2)),
> 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 moves,
> 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 anyway).
>
> 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 moves,
> 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 melinda@superliminal.com <mailto:melinda@superliminal.com> [4D_Cubing]" <4D_Cubing@yahoogroups.com <mailto:4D_Cubing@yahoogroups.com>>:
>
>
> 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 24. 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.
>
> -Melinda
>
>
> On 5/3/2017 2:39 PM, Joel Karlsson joelkarlsson97@gmail.com <mailto:joelkarlsson97@gmail.com> [4D_Cubing] wrote:
>> 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" <joelkarlsson97@gmail.com <mailto:joelkarlsson97@gmail.com>>:
>>
>> 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,
>> Joel
>>
>> 2017-04-29 1:04 GMT+02:00 Melinda Green melinda@superliminal.com <mailto:melinda@superliminal.com> [4D_Cubing] <4D_Cubing@yahoogroups.com <mailto:4D_Cubing@yahoogroups.com>>:
>>
>>
>> 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 <https://www.youtube.com/watch?v=zqftZ8kJKLo&t=5m29s>. 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 <http://superliminal.com/cube/inverted1.jpg> 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 roice3@gmail.com <mailto:roice3@gmail.com> [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 <http://www.pa-network.com/global-chess/indexf.html>" 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 joelkarlsson97@gmail.com <mailto:joelkarlsson97@gmail.com> [4D_Cubing] <4D_Cubing@yahoogroups.com <mailto:4D_Cubing@yahoogroups.com>> 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 melinda@superliminal.com <mailto:melinda@superliminal.com> [4D_Cubing]" <4D_Cubing@yahoogroups.com <mailto:4D_Cubing@yahoogroups.com>>:
>>>
>>> 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 joelkarlsson97@gmail.com <mailto:joelkarlsson97@gmail.com> [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 melinda@superliminal.com <mailto:melinda@superliminal.com> [4D_Cubing] <4D_Cubing@yahoogroups.com <mailto:4D_Cubing@yahoogroups.com>>:
>>>>
>>>> Dear Cubists,
>>>>
>>>> I’ve finished version 2 of my physical puzzle and uploaded a video of it here:
>>>> https://www.youtube.com/watch?v=zqftZ8kJKLo <https://www.youtube.com/watch?v=zqftZ8kJKLo>
>>>> 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: http://superliminal.com/cube/dessert_cube.jpg <http://superliminal.com/cube/dessert_cube.jpg> I don’t know how practical a solution this is but it sure looks delicious! Welcome Marc!
>>>>
>>>> -Melinda
>>>>
>>>>
>>>
>>>
>>>
>>>
>>>
>>
>>
>>