Message #3702

From: Joel Karlsson <>
Subject: Re: [MC4D] Physical 4D puzzle V2
Date: Wed, 03 May 2017 23:39:29 +0200

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 wats 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