INTRO
I published my first Rubik's Cube post half a year ago.It was useful for me, and useless for anyone else who tried to read it - Mom, Dad, Liên, George, and Jack. These are all very smart people, and the panel spent hours upon hours. The problem was most certainly the post: more specifically, this stuff should not be learned from any reading of any kind, of any post.
My second post was an essay on parity. Nobody but Jack even tried that one.
My third post was a syllabus of the permutation puzzle class I wrote for my kids. It actually worked - my kids understand the content better than any member of the panel, but this is because I played with them; the kids weren't ever reading anything. That's the ticket.
My fourth and final post ventured into how to solve puzzles that don't readily allow for the creation of 3-cycles. Like the second, it remained unexplored by anybody other than Jack. Whataguy. Just yesterday, upon my request, he wrote me an email about the French Revolution.
This week, George and I found ourselves at The Coffee Shark while the kids were in COTA. We hadn't been out and about together for some time, and I thought I'd start "the cube discussion" all over. (Flashback: Look at him - he's such a handsome, smart guy. We're sharing a pecan porter and a chocolate milkshake. Life has just handed me a precious moment.)
But y'all, it was a shitshow. I started laying a foundation, and he interrupted me to ask a question that had a long answer. He wanted to relate what I was explaining to something he already knew, and it wasn't time for that. I should have said, "I don't want to answer that question right now," but instead, I took us down the path to that answer, and the path was too long. So he got mad at how long the path was, and I got mad because we were only going there because he wouldn't focus on what I was saying in the first place.
For this to work, I need to stay on track. This post, like all of them really, is me organizing myself. I won't even ask Jack to read it. I won't even notate inverses nicely, no!! No explaining every detail - that's what my other posts were for. What's about to happen is some barebones shit, and you better bet your ass it requires prerequisites.
NOT A DRILL
CONJUGATES are X Y X'.
Variables can represent a single rotation or any number of rotations.
X and Y are not the same.
Let's call single-rotation-variable conjugates "simple conjugates."
We'll call conjugates in which either X or Y represents multiple rotations "complex conjugates."
For me, even in complex conjugates and commutators, one variable is a single rotation.
"Doubly complex" structures would yield results that are too difficult to track.
Simple conjugates are used all the time in moves that are called "intuitive" in the cubing world.
When this happens, we're usually tracking the movement of ONE piece in particular.
For a Rubik's cube, a simple conjugate alters 8 pieces.
We cannot possibly keep track of that; we simply push them into areas that are still unsolved.
Complex conjugates are often of the form:
1. Set up (X)
2. Commutator (Y)
3. Undoing of the set up (X')
COMMUTATORS are X Y X' Y'.
Again, variables can represent a single rotation or any number of rotations.
Again, X and Y are not the same.
We'll use the descriptors "simple" and "complex" as before.
(In chapter 17 of the first post, I referred to simple commutators as Z/Y commutators.)
Commutators mean nothing until we discuss the intersection of X and Y.
This is the slot or slots shared by X and Y.
What is permuted by a commutator is:1. What is in the intersection of X and Y
2. What X brings to that intersection
3. What Y brings to that intersection
For the most control, we want to alter the fewest pieces.
Thus, we want to make the intersection of X and Y is as small as possible, which is 1.
When we do that, the resulting commutator creates a 3-cycle.
Commutators with bigger intersections yield results that are too complicated.
They can be used, but the results need to be written down into a cheat sheet.
(That's how the Skewb is solved.)
Even in 3-cycles, we're paying most of our attention to ONE piece.
We care tremendously about both its destination position and its destination orientation.
Let's call "destination position" DP and "destination orientation" DO.
The DP of piece ONE determines the DP of all the pieces in the cycle.
If we're cycling the last 3 pieces, we must also check the DO of piece TWO.
Paying attention to THREE pieces is too hard - we're not jugglers.
Luckily, the "laws of the cube" look after that third piece for us.
We'll hand the syllabus over to the puzzles now.
THE IVY SKEWB
Once the corners are set, centers or "ivy leaves" are solved with simple commutators.
Due to the nature of the puzzle, any two different rotations will have an intersection.
And that intersection will always be exactly one slot (no more).
When we're about to create a 3-cycle, we have to first know what is going where.
That's the same thing as checking the DP of ONE piece.
This piece, the first piece we follow, our leader, I call Vixen.
(Dad likes to blame his hatred of my permutation posts solely on this.)
When X brings Vixen to her slot, the direction of the 3-cycle is determined.
The other ivy leaves have no choice but to follow suit.
If Vixen moves clockwise, so do the other two.
To solve the last 3-cycle, any one of the unsolved ivy leaves could be Vixen.
Vixen's identifying element is that with a single rotation, both her DP and DO will be correct.But in this puzzle, all unsolved pieces have one color and one color only.
In other words, there are no orientations to mind.
X must bring any Vixen into her destination, which is the intersection.
Y must ensure that a piece we want to cycle goes into the intersection next.
And that's all she wrote; completing the commutator solves that puzzle.
THE DINO SKEWB
This time, pieces have two colors, so orientation appears to be a new issue.
But it's not!
Due to the nature of the puzzle, pieces cannot be both correctly placed and disoriented.
For further reading, skip down to the discussion under "Skewb," which also has this quirk.
The big difference between the Dino and the Ivy are the rotations.
The Dino has twice as many rotations to choose from, and not all of them overlap.
We begin solving by choosing one piece that we will designate as correct.
All other DPs will be solved relative to that piece.
Then we blockbuild some sort of mass for a while.
As we approach the last unsolved pieces, simple commutators once again come to the rescue.
We must find the two rotations that contain the 3 pieces we want to cycle.
The XY intersection must contain one of them, and the piece that belongs there is Vixen.
Again, X must bring a Vixen into her destination slot, which is the intersection.
(For any 3 pieces, there is now only one correct choice for Vixen, not all 3 as before.)
Again, Y must ensure that an unsolved piece goes into the intersection next.
And as before, the rest of the commutator solves the puzzle.
As is the way of the world, Vixen's DP dictates the direction of the 3-cycle.
But what about DOs?
All 3 pieces will end up reoriented based on the number of times they were rotated.
Vixen will be rotated once.
The piece that replaces Vixen is called Mrs. Claus, and she'll be rotated twice.
The piece that replaces Mrs. Claus is called The Mayor of Southtown, and he'll be rotated once.
Because their DOs always work out, this is something to notice, not something to worry about.
Regarding the final three pieces, in the Ivy, the DPs of all three are on separate cube faces.
In the Dino, two pieces share a cube face, and the third may or may not also share that face.
So in the Dino, all three pieces could be on the same cube face.
ONE TWO THREE
As previously stated, Vixen's DP and DO are always correct after a single rotation.
She's our leader because DO is the only one that can be checked with a single rotation.
Because we're paying most attention to her, we think of the cycle as following her.
(In reality, we can determine the cycle's direction by following any one of the 3.)
Vixen is icy blue because she grows quite ill in the heat.
Mrs. Claus' takes Vixen's original slot, and her DO is the second we check.
She's red because she's Mrs. Claus.
The Mayor of Southtown's identifying element is that we don't check his DO.
He takes Mrs. Claus' original slot.
He's orange because I decided that Southtown is known for its oranges and bright sun.
Like the Florida Orange Bird.
For the Rubik's Cube, and for most permutation puzzles, we're usually focused on two planes.
More generally, we need to think in terms of two opposite places.
Vixen belongs in a cold place, so her slot is typically on a plane I call Snowmiser's.
Claus also belongs in a cold place, so she also typically belongs on Snowmiser's.
Mayor starts off in freezing and miserable in Snowmiser's.
But he belongs in the "opposite" place, which is Heatmiser's.
In other words,
Vixen starts in Snowmiser's and ends in Snowmiser's.
Claus starts in Heatmiser's and ends in Snowmiser's.
Mayor starts in Snowmiser's and ends in Heatmiser's.
THE PYRAMINX
This is a very interesting one - for the first time, DO matters!
First of all, twist the 4 big corner pieces around until they're solved, just like the Ivy.
The 4 little corner pieces just get twisted until they match the big corner pieces.
Their DPs are already in place, just like the DPs of centers on a Rubik's cube.
What's left are 6 edges; this is simply a 6-piece puzzle.
To solve the last 3-cycle, we must check the DOs of both Vixen and Claus.
There are three different commutators we can consider.
The first is a simple commutator.
The second is a complex commutator in which X is a simple conjugate.
The third is a complex commutator in which Y is a simple conjugate.
Let's talk about the simple commutator.
As before, we can find two rotations that contain the 3 pieces we want to cycle.
The intersection contains one of them, and the piece that belongs there is Vixen.
There will be 0-2 Vixen candidates.
1. If there are 0, then simple commutators are out.
2. If there is 1, the simple commutator will only work if Claus' DO works out.
It very well may not!
In this case, Claus and Mayor would be left with correct DPs but incorrect DOs.
3. If there are 2, only one of them will yield the correct DO for Claus, and therefore Mayor.
Assuming a simple commutator is out of the question, we need to try something new.
As mentioned in the previous chapter, we need to think in terms of two opposite places.
In the case of the Pyraminx, one of these places is a plane.
The other is everything else.
Vixen and Mayor are on the plane, which is Snowmiser's.
Claus isn't.
Vixen is in the wrong position.
However, she is already on the correct plane.
Furthermore, she's oriented so her DO will be correct after a rotation of that plane.
That plane rotation moving Vixen to her slot will either be X or Y.
The other one will be a simple conjugate to bring Claus down onto the plane.
There are actually two simple conjugates that will accomplish this.
However, only one of the two will leave Claus with the right DO.
Putting the conjugate in either X or Y will not affect the DPs of anything.
Assuming the correct simple conjugate has been chosen, it won't effect the DOs either.
So which one should you use?
I wrote all my previous posts putting the conjugate in X.
This was a pretty arbitrary decision; I wanted to be consistent with Ryan Heise.
Do both, just for no reason, but remember that either way, Vixen always moves first.
She can either rotate on her plane immediately, or get swapped for Claus in the opposite place.
There are two ways to get swapped for Claus, and Claus' DO determines the right one.
In an inverse of a complex commutator, the conjugate switches from X to Y or vice versa.
In other words, they often look like one of these:
(X1 X2 X1') Y (X1 X2' X1') Y'
X (Y1 Y2 Y1') X' (Y1 Y2' Y1')
Thinking in "both for no reason" is similar to thinking in inverses, but switching Vixen's identity.
Either way, the 3 DPs cycle in accordance to Vixen's lead.
But what happens with DOs?
Vixen always maintains orientation with regard to the plane of rotation that comprises X or Y.
We've been calling that Snowmiser's.
Our Mayor is always rotated once, and Claus twice.
When pieces have one or two faces, this doesn't look like much of anything.
So while the Pyraminx or Rubik's Cube edges orient predictably, it's difficult to tell who's who.
However, Rubik's Cube corners each have 3 sides, and that makes them easier to understand.
Let's assume that we're cycling corners with no 180° rotations on the Rubik's Cube.
As mentioned, Vixen maintains her orientation with regard to Snowmiser's.
That means her Snowmiser tile is on the Snowmiser plane.
Mayor maintains orientation with regard to some other plane.
That plane is neither Heatmiser's or Snowmiser's, but one of his tiles matches that plane.
Claus, having been rotated twice, maintains orientation with regard to no planes!
None of her tiles match the plane they're a part of.
INTERMISSION
THE POCKET CUBE
The Pocket Cube is like the Pyraminx but without the simple commutator option.
This is because any two rotations have an overlap of more than 1!
Also, 6 of its 8 pieces can appear to have solved DPs, and we can't create 2-cycles.
In this case, we must treat some solved piece as if it is unsolved.
Chapter 18 of the first post clears that up.
Remember when I said it we could get 3 correct DPs with 2 wrong DOs in the Pyraminx?
In other words, Vixen had the correct DO, but Claus and Mayor didn't?
Reorienting them is the same as the 2 corner twist, which is a handy tool for the Pocket Cube.
That's a commutator where either X or Y is has two simple conjugates back to back.
The first takes Jingle out of the intersection and the second returns him with a different DO.
Of course, Jangle will also be reoriented by the end of the commutator.
Like before, it doesn't matter whether you put the conjugates in X or Y.
The other one is the plane rotation that holds both of them.
It's which conjugate you perform first that determines the final DOs.
THE RUBIK'S CUBE
The centers of the Rubik's are set.
The corners are just like the Pocket Cube.
What's new are the edges, which move on M slices.
All edge commutators use one and only one M slice.
We'll assume X and Y represent outer plane rotations for the purposes of this discussion.
We'll let M represent an M slice rotation.
There are two kinds of edge commutators: true 3-cycles and untrue 3-cycles.
Let's talk about the true ones first.
Remember, complex commutators often look like one of these:
(X1 X2 X1') Y (X1 X2' X1') Y'
X (Y1 Y2 Y1') X' (Y1 Y2' Y1')
Vixen must first rotate on an M slice or a plane.
1. Let's assume she first rotates on an M slice.
In this case, all our M slices are already used up!
That M slice rotation will either be all of X, which looks like this:
M (Y1 Y2 Y1') M' (Y1 Y2' Y1')
Or it will be the first move of the simple conjugate X1 X2 X1', which looks like this:
(M X2 M') Y (M X2' M') Y'
2. Now let's assume she rotates on a plane.
We still have an M slice to play, and here are all our choices:
X (M Y2 M') X' (M Y2' M')
(X1 M X1') Y (X1 M' X1') Y'
X (Y1 M Y1') X' (Y1 M' Y1')
(X1 X2 X1') M (X1 X2' X1') M'
The untrue commutators appear to create 3-cycles, but the XY intersection has 3 instead of 1.
They secretly reorient centers - an invisible side effect on a standard Rubik's Cube!
The 3-cycle wannabes look like one of these these:
M (Y180) M' (Y180)
(X180) M (X180) M'
In related news, double 2-cycle wannabes look like one of these:
(M180) (Y180) (M180) (Y180)
(X180) (M180) (X180) (M180)
So how do you choose between these 10 options?
First, make sure Vixen is set up with her DP and DO.
Then eliminate anything that yields an incorrect DO for Claus.
If you get stuck, use a set up/complex conjugate.
RUBIK'S REVENGE & THE PROFESSOR'S CUBE
We know how most of this works already.What's new is the notion of parity problems, which are discussed in my second post.
Solving with blockbuilding and reduction can seem to lock puzzles in an unsolvable state.
But that's not totally unfamiliar; we saw 6 "solved pieces" in the Pocket Cube.
Learning why this happens is the only key to fixing it.
We needed corner commutators for the Pocket Cube.
We needed edge commutators for the Rubik's Cube.
And now, we need center commutators for these.
Vixen, Claus, and Mayor rotate on planes that are like M slices.
But we have to call them something new because there are more than one per dimension.
Thus, we can think in terms of outer rotations and inner rotations.
Let's call that O's and I's.
To get to one of our center pieces, we need some kind of I movement.
(I1 I2 I1') is a conjugate that changes more than one piece in any rotation.
(I O I') is also a conjugate that will change more than one piece in any rotation.
But (O I O') is a conjugate that will change one piece in a parallel inner slice!
So (O I O') has to be the X or the Y, and the "parallel inner slice" is the other.
Our center commutator will look like one of these:
(O I1 O') I2 (O I1' O') I2'
I1 (O I2 O') I1' (O I2' O')
Our Vixen, Claus, and Mayor can be in two or three different planes.
If I1 and I2 move in opposite directions, the cycled pieces will span three planes.
If I1 and I2 move in the same direction, however, the cycled pieces will span only two planes.
Vixen will begin in one, while the other two will begin adjacent in the second.
From a solved state, this makes the 3-cycle appear to be a 2-cycle.
Remember, Vixen is our leader, followed by Claus, followed by Mayor.
This is a 3-cycle.
Regarding DOs, Vixen always maintains her orientation.
She's always set up to travel on an I; if she's not, a set up is required.
The other two rotate once in accordance to what O or O' did to them.
I go into more detail about this in chapters 13 and 14 of my second post.
I do call the Mayor "Donner" in two planes due to that piece's DP being adjacent to Vixen.
And you thought I had no heart.
At the very end of chapter 19 in my first post, I direct the reader to Ryan Heise's pair 3-cycles.
The same idea can be used to permute more than 3 center pieces for cubes of large dimensions.
Note that (O I O') creates one XY intersection for multiple parallel I's in large dimension cubes.
Any number of these I's can be used in Y.
Rest assured that this is still a 3-cycle, but Vixen, Claus, and Mayor are all a bit bigger.
Or you could think of it as a few simultaneous 3-cycles.
What's more is that the various I's don't even need to be adjacent.
Parallel I's left out of the commutator will simply function like the rest of the cube.
Also, as long as O moves Vixen to a new I, O can move in either direction.
This is really interesting stuff, and the best video out there on it is by Duke of Cubes.
THE CYLINDER SHAPE MOD,
THE WINDMILL, FISHER, & AXIS SHAPE MODS,
THE MASTERMOPHIX & PYRAMORPHIX,
& THE MIRROR & GHOST SHAPE MODS
Pish posh, there's not a lot to say here.
The Cylinder is easier than the Rubik's Cube.
That's because each edge and each corner have 3 identical pieces.
The rest are harder because they're supercubes.
Shape mods are disconcerting, certainly, but the trick is to recognize corners, edges, and centers.
Once those are defined, you just muddle your way through like any solve.
The Mastermorphix and the Pyramorphix are 3x3x3 cube puzzles that look like pyramids.
What a mind trip!
I do not own a Mirror Cube because it wouldn't match my set.
I do not own a Ghost Cube because it only becomes a 3x3x3 shape mod when skewed.
Yes, that sounds difficult, but more importantly, that sounds annoying!
With no frame of reference other than a second Ghost Cube, I'd consider it a waste of space.
CUBOIDS
I only have a couple of the little ones: the 2x3x3 (Domino) and the 2x2x3 (Tower).
Fix the shape mod aspect of them, if necessary.
Write an algorithm, see what it can do, and work backwards, blockbuilding around that.
Specifically, I wrote a commutator that creates a corner 3-cycle, so I solve edges first.
There are 8 edges on the Domino and 4 on the Tower, but they're not hard.
The Tower edges need no explanation.
For the Domino, I solve two adjacent edge pairs, and then sort out the rest.
Then it's corner time.
"But Lan, we have to deal with 180° turns!"
True; and therefore, we cannot create 3-part X's like we do for the Rubik's Cube.
Instead, we need 5-part X's!
Place either the Domino or the Tower so that white or yellow is U.
Then alter one corner of one face by removing two corners and putting one of them back.
For example, (R2 U R2 U' R2) alters exactly one corner in D.
So that's my X, and some sort of D rotation is therefore my Y.
For the commutator (R2 U R2 U' R2) D (R2 U R2 U' R2) D', the intersection is DFR.
That means Vixen begins there.
She always maintains orientation, and lands in the corner D brought to the intersection.
In other words, Vixen travels from DFR to DFL.
The Mayor was (miserable) in DFL, so he moves from DFL to UFR.
Our second member of the trio, Claus, then travels from UFR to DFR to complete the cycle.
THE KILOMINX & MEGAMINX
I love both of these puzzles.
They're the best for setting up complex conjugates with ridiculously long X's!
The solved Professor's Cube is great at teaching blockbuilding tool.
That's because there are so many solved pieces, it's easy to pay attention to what's unsolved.
The same is true of the Kilominx.
It requires the same skillset as the Pocket Cube exactly, but it's even prettier and more fun.
The difference between the Kilominx and the Megaminx are edges.
But the Megaminx edges have no M slices!!
That's a little scary, but remember how to create 3-cycles - we need an XY intersection of 1.
Is it possible to swap out a Megaminx edge for another using outer rotations only?
Absolutely. Jamie Mullholland was the brains behind chapter 16 of my second post.
THE SKEWB & SQUARE 1
I wrote my fourth post exclusively on these two, but the point was that sometimes it's impractical to think in 3-cycles. But what if we really want to? Let's delve a little deeper into the Skewb.
The Skewb is actually a Pyraminx in disguise. In fact, it was originally named the "Pyraminx Cube."
On a Pyraminx, it's easy to locate the 4 pyramidical points. Interestingly enough, they remain stationary relative to one another because any rotation simply reorients the entire puzzle. The points look more like corners than centers because they're pyramids, but they function more like centers because they don't permute relative to one another. So let's just be reasonable, and for the purposes of this discussion, call them pyramids.
The Skewb is a puzzle with 6 squares and 8 pyramids, but there's a catch: 4 of the 8 pyramids are stationary. The other 4 permute and are considered to be "floating."
We can think of these 8 pyramids in terms of 2 teams of 4: Team North Pole and Team Florida Bird.
Team North Pole contains the northpole pyramid (red, white, blue) and all the pyramids that can be reached by traveling from the northpole pyramid across square diagonals. If you disassemble the Skewb, you'll find this team attached to the central rotating mechanism. Because Team North Pole does not permute, its 4 pyramids are always correctly positioned but not necessarily corrected oriented.
(This should remind you of the "corners" of the Ivy Skewb.)
The other team is Team Florida Bird, to which the floridabird pyramid (orange, yellow, green) and all the pyramids that can be reached by traveling from the floridabird pyramid across square diagonals belong.
Note that the northpole pyramid is located across the cube diagonal from the floridabird pyramid.
The 4 members of Team Florida Bird and all 6 of the squares permute.
Interestingly, every rotating face on the Skewb has a central pyramid from one team and 3 circumferential pyramids from the other. This never changes, so we can think of the two teams as operating on different tracks.
So what happens when we try to create a 3-cycle of pyramids?
Our first step would be to write an X - any X however long and clunky - such that all but one circumferential pyramid on a given plane gets replaced.
By choosing Y to be the rotation of that given plane, the commutator will create a pyramid 3-cycle.
But we can't write that X!!
If we try to replace any circumferential pyramid that belongs to Team North Pole, we cannot, for they are stationary and immovable. And if we try to alter only one circumferential pyramid on Team Florida Bird, we will be trying to make a single transposition in an even parity puzzle. Remember, two of our Team Florida Bird pyramids are already correctly placed in the circumference, which only leaves two pyramids left to switch, which is 1 transposition.
Any Skewb rotation creates a 3-cycle of pyramids (2 transpositions), a 3-cycle of squares (2 transpositions of squares), and a reorientation of the central pyramid to the rotation (0 transpositions).
Although we cannot write an X that exchanges one circumferential pyramid for a different one, it is possible to write an X that reorients a circumferential pyramid. In this case, the resulting commutator creates a 2-pyramid twist instead of a 3-cycle. Not helpful, but it makes sense.
Okay. What happens when we try to create a 3-cycle of squares?
Well, we can, silly as it may be. I could write a big clunky algorithm through trial and error that breaks up one face and restores it with the exception of one square, but it's easier for me to type and you to read if I don't. The simple commutator we already have generates this situation eventually anyway!
With any two overlapping rotations X and Y, one iteration of (X Y X1 Y1) will create two 2-cycles of squares and re-orient 4 correctly placed pyramids. Notice that any number of iterations will not displace pyramids.
Two iterations will restore squares and reorient the 4 pyramids again; they will remain disoriented. Three iterations will restore pyramids and repeat the two 2-cycles of squares. If two iterations restores squares and three iterations restores pyramids, then 6 restores the cube in its entirety.
Anyway, three iterations is what we want for this purpose - notice that at the point of three iterations, two squares, which have been entirely unaffected this whole time, are still solved. Thus, we have a face with all solved corners and 2 of 3 solved centers. The rotation of this plane is the Y in our square 3-cycle.
An inefficient square 3-cycle, but a 3-cycle on the Skewb nonetheless.
Regarding the SQ1, that little thing loves to generate double 2-cycles all on its own, so it makes sense to work with what it does best. However, you really can generate SQ1 3-cycles with my cuboid 5-part X's! In fact, you can think of a SQ1 with solved quarters as a pocket cube that has a 180° constriction on R, which would function like an imaginary 2x2x2 cuboid.
CONCLUSION
Don't forget how to checkerboard, which is an extra little trick to keep in your back pocket, should the need to demonstrate the nature of shape mods arise. For the next time I share a pecan porter and a chocolate shake with that handsome hunk of a man of mine, it looks like I will have practiced my pick up line.
