Thread: "Melinda's 2x2x2x2 solved"

Hi Bob,

Good stuff!   I'm glad to see that you put your puzzle together and made
progress on it.   Unless we unexpectedly identify a problem, I think
you've got the first valid solution recipe sketched out. Congrats!

I had used the identical approach -- totally identical in all respects
except for how to achieve the basic 4-cycle of the top 2x2x2 upon which
it depends -- and nicknamed it 4tega because it orients a pair of faces
first like the Ortega method does for the 2^3 cube.   However, I was
simply using parity-violating "quarter puzzle twists" to walk through
the solution, which was very unsatisfying and non-kosher, and prevented
me from saying "First!".   You have resurrected quarter puzzle twists in
a kosher way, using the minimalist move set.  Yay!   A kosher 4tega!

Since our approaches were so similar, I could recognize what you're
doing right away, and I endorse it as a valid approach.

When I had worked through it before, I was worried that I would still
occasionally end up with the single-piece-double-twist problem at the
end.  I think I was hallucinating, since it seems clearly impossible to
be in that state with two faces oriented.  However, I would be reassured
if you had a look at the following.

Question:  if you take a single piece of your puzzle and give it a 180
degree twist along any axis, how would you solve the resulting puzzle
state?   I believe that's a correct statement of the basic
single-piece-double-twist problem.   Maybe you'd like to give that a
try.   I had sketched out a tentative solution using quarter-puzzle
twists and 3 sequential puzzle reorientations, based on pure guesswork. 
I should dig it out and try to rehab it.   I'm still working out my
understanding of the 12 orientations of the pieces -- I find that to be
the most fascinating part of this puzzle, theory wise -- and I think a
good step to getting the hang of it will be to fully understand the
solution to the single-piece-double-twist. What's the fewest number of
full puzzle reorientations required to undo a single-piece-double-twist
leaving the rest of the puzzle unchanged?  I think it might be three,
and that the reason will become clear to me after a bit.   But it might
be two, given how confused I still am about how the 12 orientations work.


It would be fun to see a video of you manipulating your puzzle. I'm
curious what feels natural for you.


Some time, a truly dedicated person should make a video of the
following:   generate a MC4D scramble of the 2^4.   Duplicate it on the
physical 2^4 by taking the puzzle apart.   Solve the physical 2^4 while
following along with MC4D with one click per physical puzzle move,
leaving both puzzles solved.


I'm also looking forward to jointly tackling some entries on the list of
open questions about the physical 2^4.  After writing down the list of
open questions, which I have not done.  :)   Do you have any entries for
it?  Directions you'd like to explore?


Cheers
Marc

Hi Marc,

> Good stuff! I'm glad to see that you put your puzzle together and made=
=20
> progress on it. Unless we unexpectedly identify a problem, I think=20
> you've got the first valid solution recipe sketched out. Congrats!

Thanks! Great to see we had the same ideas.

> When I had worked through it before, I was worried that I would still=20
> occasionally end up with the single-piece-double-twist problem at the=20
> end. I think I was hallucinating, since it seems clearly impossible to=20
> be in that state with two faces oriented.=20=20

Right...

> However, I would be reassured=20
> if you had a look at the following.
>=20
> Question: if you take a single piece of your puzzle and give it a 180=20
> degree twist along any axis, how would you solve the resulting puzzle=20
> state? ... What's the fewest number of=20
> full puzzle reorientations required to undo a single-piece-double-twist=20
> leaving the rest of the puzzle unchanged? I think it might be three,=20
> and that the reason will become clear to me after a bit. But it might=20
> be two, given how confused I still am about how the 12 orientations work.

Great question. Three is clearly sufficient, because this state is an insta=
nce of where you can be in step 3a in my solution, and from there three reo=
rientations are required.

I=E2=80=99m about convinced you can=E2=80=99t do it in two, though I don=E2=
=80=99t have a tidy proof yet. You need to show that no matter what you do =
before the first reorientation, that reorientation will leave you in a stat=
e where no opposing color pairs can be solved into opposite faces, because =
of corner twist parity.

> It would be fun to see a video of you manipulating your puzzle. I'm=20
> curious what feels natural for you.

I=E2=80=99m still traveling for Thanksgiving, but will try to put something=
together when I get home.

> I'm also looking forward to jointly tackling some entries on the list of=
=20
> open questions about the physical 2^4. After writing down the list of=20
> open questions, which I have not done. :) Do you have any entries for=
=20
> it? Directions you'd like to explore?

For me the most interesting thing about Melinda=E2=80=99s 2x2x2x2 is that i=
t exists at all. It seems kind of a miraculous accident. I wondered what th=
e equivalent 2d representation of a 2x2x2 would be, and realized that it do=
esn=E2=80=99t exist, because squares do not have 3-fold rotational symmetry=
. We luck out here that the tetrahedral group is a subgroup of the octahedr=
al group. I.e., that you can get the required 4-fold isotropic symmetry in =
a cube. Likewise, you can=E2=80=99t make a similar 3x3x3x3 =E2=80=94 the th=
ree-facet pieces can=E2=80=99t be instantiated as cubes.

So what kinds of 4d puzzles can be implemented this way? Is this it?

Bob

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Aw damn, I was hoping to get this one, but I've not been able to get my
hands on a puzzle yet. I was attempting to bodge one out of paper so I
could attempt this, but it seems I've been bested.

I like the look of your solution though. It reminds me of what I was
thinking of. Well done.

~Luna


On 27 Nov 2017 00:57, "Bob Hearn bob.hearn@gmail.com [4D_Cubing]" <
4D_Cubing@yahoogroups.com> wrote:



Hello MC4Ders,

I saw Melinda=E2=80=99s 2x2x2x2 at a puzzle party last month, where I also =
met
Marc. I knew I had to have one. Melinda sent me the Shapeways link, and I
finally got everything together and got it assembled a few days ago.

I=E2=80=99m happy to say that I=E2=80=99ve solved it! Melinda asked me to d=
escribe my
solution to the list. I should say that I have not solved the virtual
version, or read anything about solutions =E2=80=94 I wanted a pure solving
experience. But that means I may be missing some obvious insights and
standard techniques; apologies if so.

To start, let me establish my terminology. I am new to the list =E2=80=94 I=
=E2=80=99ve
watched Marc=E2=80=99s ROIL video, and Melinda has pointed me towards Joel=
=E2=80=99s posts
on notation from before I joined. But I hope you will forgive me if I use
my own terminology here. The reason is that I orient the puzzle vertically
rather than horizontally. To me this makes sense, since I am generally
focusing on the upper 2x2x2. I refer to the the 8 faces as: upper, lower,
front, back, left, right, inner, and outer. Hopefully the meaning is clear.
In an earlier post I saw Ed refer to the facets of what I call the upper
and lower faces as =E2=80=9Cinverted=E2=80=9D =E2=80=94 I don=E2=80=99t kno=
w whether this is standard, but
I=E2=80=99ll do that too. To be concrete, in this pic, the upper face is pu=
rple,
lower is pink, outer is blue, front is red, and right is green. (Also inner
is white, back is orange and left is yellow =E2=80=94 I used the original
blue-opposite-white coloring scheme.)

https://drive.google.com/file/d/114zgLIhLz5ZnD4UxEo2eWG2yvv06l
KHj/view?usp=3Dsharing

For at least the first few solves I restricted myself to strict moves only,
i.e., 2x2x2x2 operations that correspond directly to MC4D individual turns
or whole-puzzle reorientations. In particular I use upper-face moves
(reorient the upper 2x2x2), lower-face moves (same for lower), and
inner-face moves. Not all inner-face moves are legal, as Marc has explained
in a video: only those that are 90 degrees about the long axis, or 180
degrees about the other two axes. The 90-degree moves I make by rotating
the center 2x2x2 in place, rather than putting it on the end, rotating, and
putting it back, as Marc does in his ROIL video. This is more convenient
for the way I use them. Note that without a whole-puzzle reorientation, the
inverted facets (two faces) can only permute among themselves, as can the
non-inverted facets (six faces).

An essential technique I use is manipulating the top 2x2x2 as an
independent 2x2x2 magic cube (ignoring its inverted facets), almost freely,
as follows: whichever face of the upper cube you want to turn, reorient the
upper face so that the desired face is on the bottom, adjacent to the lower
cube. Then, make an inner-face move. You=E2=80=99ve just turned a face of t=
he upper
2x2x2, as well as of the lower 2x2x2. But if you leave the lower 2x2x2
alone, all that is happening is that its top face is turning back and forth
=E2=80=94 you are not scrambling it. So by sandwiching each upper-cube move=
you
want to make between reorientations, you can execute any 2x2x2 sequence you
want, while almost leaving the lower 2x2x2 alone.

OK, so on to the solution. First, pick a pair of opposite colors. I prefer
pink and purple, for reasons that will become clear. Then, these are the
steps:

1. Get all of the pink and purple facets out of the front and back faces.
This is easy to do by manipulating the top 2x2x2 to clear two opposite
faces of pink and purple, then putting one of those cleared faces against
the lower face, then switching upper and lower, doing the same for the new
upper cube, then reorienting each to put the cleared faces into the front
and back slots.

2. Perform Melinda's whole-puzzle reorientation. This puts the front and
back faces into the upper and lower positions. Which means that now, all
the purple and pink facets are in non-inverted positions, where they can be
manipulated.

3. Solve purple into the front face, and pink into the back face. This can
be done by manipulating the upper cube to put purple on top, then the lower
cube to put purple on bottom, then combining those faces onto the upper
cube by using a 180-degree inner-face move, then getting all the pinks in
the right spots on the lower cube. All that matters here is facets: you
don=E2=80=99t care whether the pink/purple pieces are in their proper relat=
ive
slots. Again, when manipulating one cube, put it in the upper position,
with the lower cube oriented so that you will not mess anything up with the
inner moves.

There is a possible complication here. In fact unless you are lucky (1/3 of
the time), this will be the most complicated part of the solution.

3a. It may be that when solving the pinks into opposite 2x2x2 faces, one of
them will be out of place: from a 2x2x2 perspective, one corner will be
twisted. Like this:

https://drive.google.com/file/d/1SiA4eOXsJbNpS8t-IpVTwo9TQiikgkV_/view

Now, you may think, this should not be possible: corner twist parity is
conserved on a 2x2x2 =E2=80=94 you can=E2=80=99t twist just one corner. So =
what gives?
Well, here there are no =E2=80=9Ccorners=E2=80=9D per se, with only three o=
rientations.
Each piece actually has 12 possible orientations. There is still overall
orientation parity conservation. When we say a 2x2x2 corner is =E2=80=9Ctwi=
sted=E2=80=9D
here, we mean that it=E2=80=99s rotated 120 degrees about an inverted facet=
from
where we would like it. But this could be matched by another piece which is
rotated the opposite way about a non-inverted facet =E2=80=94 say, a pink o=
ne. Once
you see this, it is in principle simple to fix this situation.

The complication is that you must do this using a conjugate sequence
involving a whole-puzzle reorientation. First, perform R=E2=80=99 on the up=
per cube
(via inner-face moves wrapped between upper-face reorientations). Now, you
would like to twist the bottom-front cubies of the top 2x2x2, about the
pink facet on the left, clockwise, and about the red facet on the right,
counterclockwise. Then, when you undo the R=E2=80=99, the wayward pink face=
t will
be in the right place. But in order to perform this transform you first
have to do a whole-puzzle reorientation. Using Melinda=E2=80=99s sequence, =
this
moves the front face to the lower face. Then, you can use a standard 2x2x2
double-corner twister to fix the corners =E2=80=94 the pink and red facets =
you want
to twist about are now in inverted position. This will rotate one face of
the other 2x2x2 back and forth, but your typical transform here has an
equal number of clockwise and counterclockwise moves, so has no net effect
on the other cube. Now, undo the whole-puzzle reorientation, undo the R=E2=
=80=99,
and voila!

4. Perform Melinda's whole-puzzle reorientation. This makes the upper face
pink, and the lower face purple.

5. Now, solve the top and bottom 2x2x2s independently, using standard 2x2x2
techniques. Because we have the original six Rubik=E2=80=99s cube colors he=
re, you
even get the 2x2x2 color scheme you are familiar with. (If a single corner
is twisted while solving the first cube, fix it by using a 180-degree
inner-face move to mix the cubes, and using a double-corner twister.) To be
strict you have to again sandwich each 2x2x2 move between top-cube
reorientations. This means that when one cube is solved, you now actually
care what happens to its top layer when you put it on the bottom, and solve
the other one. How do you know it will wind up in the right place?

Well, unless you have made a mistake (which happened to me the first two
attempts), you cannot wind up with the top layer of the bottom cube rotated
90 degree. That is a 4-cycle, which has odd parity, and every 2x2x2x2 move
has even parity. However, it is possible this layer will wind up rotated
180 degrees, like this:

https://drive.google.com/file/d/1a9YLXTAbPS-DHxGSm9MbtjceuhTNFJ2u/view?
usp=3Dsharing

5a. Which leads us to the the second possible complication. Here you get
lucky 50% of the time. But if you do wind up here, there is a fairly
straightforward transform built from commutators and conjugate sequences
that fixes it, and does not require a whole-puzzle reorientation.

The details I think I will leave as an exercise for the reader, for now. I
am happy to try to transcribe the sequence into Marc=E2=80=99s or Joel=E2=
=80=99s notation
if there is interest, or make a video. But as a hint, try to find a
sequence that switches the ufr and ubr cubies on the top 2x2x2, while
scrambling the bottom 2x2x2.

--------------

So, that=E2=80=99s it in a nutshell. Now, once you have done this a few tim=
es, you
may get tired of the tediousness of performing all the 2x2x2 moves as
upper-face reorientations wrapped around inner-face moves. Especially
because most 2x2x2 sequences you will want to perform will leave the lower
cube unchanged anyway. So=E2=80=A6 is it OK to just take off the top cube a=
nd
manipulate it independently, if you know your transform has net 0 clockwise
and counterclockwise moves? I think that=E2=80=99s a matter of taste. In th=
e end
the consequence of all those inner-face moves will be simply to leave the
top layer of the bottom 2x2x2 rotated 180 degrees, or not.

Bob Hearn



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First off, huge congratulations, Bob! I've been surprised by how difficult =
this puzzle turned out to be since the MC4D version is not considered to be=
terribly difficult. I knew you could do it, and I'm very happy to see a fi=
rst solution. I'm looking forward to seeing optimizations, other solutions,=
and eventually speed solving contests.

On 11/27/2017 12:26 PM, Bob Hearn bob.hearn@gmail.com [4D_Cubing] wrote:
> [...]
> For me the most interesting thing about Melinda=E2=80=99s 2x2x2x2 is that=
it exists at all. It seems kind of a miraculous accident. I wondered what =
the equivalent 2d representation of a 2x2x2 would be, and realized that it =
doesn=E2=80=99t exist, because squares do not have 3-fold rotational symmet=
ry. We luck out here that the tetrahedral group is a subgroup of the octahe=
dral group. I.e., that you can get the required 4-fold isotropic symmetry i=
n a cube. Likewise, you can=E2=80=99t make a similar 3x3x3x3 =E2=80=94 the =
three-facet pieces can=E2=80=99t be instantiated as cubes.

I'm as surprised as anyone. A lot of things had to fall into place for this=
to happen. Even after coming up with a workable design it was still years =
before I had a mechanism that could implement it. Perhaps the final piece w=
as finding a usable whole puzzle reorientation. Yes, the real accident is t=
he relationship between the cube and tetrahedron in 3D. I expect this will =
be the cause of endless objections when people assume I'm making a huge err=
or in thinking that my 3D cubes have some shape relationship with 4D cubies=
which they definitely do not.

> So what kinds of 4d puzzles can be implemented this way? Is this it?

You mean by using cubes? That's a new and interesting question. I have no i=
dea!

-Melinda