Message #3846

From: Luna Peña <scarecrowfish@gmail.com>
Subject: Re: [MC4D] Melinda’s 2x2x2x2 solved
Date: Mon, 27 Nov 2017 20:36:02 +0000

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’s 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’m happy to say that I’ve solved it! Melinda asked me to describe my
solution to the list. I should say that I have not solved the virtual
version, or read anything about solutions — 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 — I’ve
watched Marc’s ROIL video, and Melinda has pointed me towards Joel’s 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 “inverted” — I don’t know whether this is standard, but
I’ll do that too. To be concrete, in this pic, the upper face is purple,
lower is pink, outer is blue, front is red, and right is green. (Also inner
is white, back is orange and left is yellow — I used the original
blue-opposite-white coloring scheme.)

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

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’ve just turned a face of the 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
— 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:

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.
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.
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’t care whether the pink/purple pieces are in their proper relative
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 — you can’t twist just one corner. So what gives?
Well, here there are no “corners” per se, with only three orientations.
Each piece actually has 12 possible orientations. There is still overall
orientation parity conservation. When we say a 2x2x2 corner is “twisted”
here, we mean that it’s 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 — say, a pink one. 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’ on the upper 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’, the wayward pink facet will
be in the right place. But in order to perform this transform you first
have to do a whole-puzzle reorientation. Using Melinda’s sequence, this
moves the front face to the lower face. Then, you can use a standard 2x2x2
double-corner twister to fix the corners — 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’,
and voila!

Perform Melinda’s whole-puzzle reorientation. This makes the upper face
pink, and the lower face purple.
Now, solve the top and bottom 2x2x2s independently, using standard 2x2x2
techniques. Because we have the original six Rubik’s cube colors here, 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=sharing

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’s or Joel’s 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’s it in a nutshell. Now, once you have done this a few times, 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… is it OK to just take off the top cube and
manipulate it independently, if you know your transform has net 0 clockwise
and counterclockwise moves? I think that’s a matter of taste. In the 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