The endpoints have swapped places, so they have made one transposition.
Now let's say I have a triangle and I rotate the vertices to go from here to here:
The vertices made two transpositions.
For example, first A with B, and then B with C.
Those two transpositions would take us from the initial position to the rotated position:
Count von Count: one, two transpositions, ah-ah-ah.
DC CB
That takes three transpositions. For example, A/B, then B/C, then C/D.
The vertices look an awful lot like corner cubies on a cube, don't they?
I chose vertices because I prefer corner cubies to edge cubies for this visual, but it's also true that rotating 3 edge cubies = 2 transpositions, and rotating 4 edge cubies = 3 transpositions.
It's sometimes helpful to know the parity of how many transpositions it would take to solve a set of center cubies, a set of edge cubies, or a set of corner cubies.
CHAPTER 4: PARITY OF SETS OF PIECES & PARITY OF PUZZLES
Even permutation - a permutation that creates an even number of transpositions
Odd permutation - a permutation that creates an odd number of transpositions
For the 3x3x3, rotating a face by 90° creates a permutation of the edges and a permutation of the corners. In terms of transpositions, the 4-cycle of edges = 3 edge transpositions, and the 4-cycle of corners = 3 corner transpositions, so two simultaneous odd permutations occur.
Every time any move is made on the 3x3x3, the total number of transpositions remains even. The parity of the edge transpositions or the parity of the corner transpositions may be odd, but they'd be odd simultaneously, and their parities would match. When any move on a permutation puzzle creates an even number of total transpositions between all the sets of cubies, that puzzle is said to have even parity. So the parity of the 3x3x3 puzzle is even.
Rotating a face by 90° on the 2x2x2 makes a 4-cycle of corners or 3 corner transpositions, so each move makes an odd permutation, and the 2x2x2 puzzle has odd parity. With each consecutive 90° rotation, the parity state of the corners constantly switches between even and odd.
Rotating a face on the megaminx by 72° makes two 5-cycles. There are 4 edge transpositions and 4 corner transpositions, or two simultaneous even permutations. The megaminx is an even parity puzzle; furthermore, the parity of edges is always even, and the parity of corners is always even.
Let's forget about the corner pieces of a pyraminx since they don't affect anything. A 60° rotation of a face, then, involves a 3-little-edge-cycle and a 3-big-edge-cycle. The 3-little-edge-cycle yields 2 transpositions and the 3-big-edge-cycle yields 2 transpositions. Similar to the megaminx, a move on the pyraminx creates two simultaneous even permutations, so the pyraminx has even parity, and furthermore, all of the edges always have even parity. (I don't know what cubers actually call little edges and big edges because clearly, I'm not a nerd.)
That covers all the permutation puzzles in my Roxenda set except for the 4x4x4.
There are two types of moves you can make on the 4x4x4 - outer rotations and inner rotations. What a pain - it's enough to make you quit right here. But, no - the only reasonable way to tackle a problem like this is to write a public letter to myself, so we'll forge ahead, no matter how rough the terrain. No matter how high the waters. No matter how deep they run.
A 90° outer face rotation makes 3 single-center transpositions, 3 single-edge transpositions, 3 more single-edge transpositions, and 3 corner transpositions, for a total of 12. There are 8 single-edges in an outer face, so you might be tempted to think that an outer face rotation creates 7 single-edge transpositions. However, each rotation does not move an edge cubie to its neighboring slot; if it did, the single-edges would indeed require 7 transpositions. In reality, each rotation creates two different 4-cycles, each of which requires 3 transpositions. So an outer face rotation creates four simultaneous odd permutations, two of which cycle single-edges.
A 90° inner face rotation makes 3 single-center transpositions, 3 more single-center transpositions, and 3 single-edge transpositions for a total of 9. As before, the 8 single-centers create two 4-cycles, each of which requires 3 transpositions. That makes three simultaneous odd permutations, two of which cycle single-centers.
Because an outer face rotation yields an even parity of total single-edge transpositions, outer face rotations affect the parities of single-centers and corners only.
Because an inner face rotation yields an even parity of total single-center transpositions, inner face rotations affect the parity of single-edges only.
The parity of the puzzle (3 + 3 + 3 + 3 + 3 + 3 + 3) is odd.
CHAPTER 5: THE 3x3x3 REDUCTION
Before we go much further, I ought to mention that the way nearly everybody solves the 4x4x4 is through a method called "reduction." The cube is "reduced" to a 3x3x3-lookalike by first creating six correctly placed 4-block centers, and then creating twelve randomly placed 2-cubie edges, known affectionately as dedges. Let's make up a name for the 4-block centers. "Quenters" is really stupid, but it matches dedges. Once you write about "cubies," you've already lowered the bar so far that there's no use in trying to pick it back up. In fact, you may have already noticed that I replace the word "cubie" with "single-center," "single-edge," or "corner" every now and again.
After the quenters are solved, if the last dedges aren't solved in a nice, neat re-ordering of 3 dedges all at once, they will instead present themselves as 2 unsolved dedges, which is more of a headache. Nobody really talks about it, except to say "here's an algorithm for that."
Single-edge conjugates and commutators can fix that.
In fact, back in 2016 when I went through my first phase of the Rubik's cube, I got very annoyed by the same problem when reducing a 5x5x5. That puzzle, by the way, is much nicer than the 4x4x4. After all six of the 9-cubie nonaters (not a real word) are solved, the next step is to create twelve 3-cubie edges, affectionately known as tredges. I prefer to begin that process by first creating twelve dedges. In turning dedges into tredges, sometimes you can get stuck pairing up the last two.
My sister came up with a solution, but if I were faced with that problem again, I'd turn to single-edge cycles before anything else.
When a 4x4x4 is in a reduced 3x3x3 configuration, the 24 single-centers and the 24 single-edges have morphed into six quenters and twelve dedges, so the pieces are redefined. An outer face rotation, then, makes 0 center transpositions, 3 dedge transpositions, and 3 corner transpositions. An inner face rotation messes up the quenters and dedges, so those moves aren't used. From here, a 3x3x3 solve is attempted.
CHAPTER 6: PARITY PROBLEMS
I also ought to mention that cubers talk about parity all wrong, and they do it on purpose. The terms "parity" and "parity error" are used colloquially to describe any sort of problem in cubing.
"False equivocation" is the situation in which an incorrectly oriented piece is assumed to be correctly oriented, or where two identical pieces have had a transposition. Sometimes the terms "parity" or "parity error" are used to describe the issue of false equivocation. The two are easily confused, because both lead to a seemingly unsolvable state, and both happen invisibly.
So while mathematicians use "parity" simply to describe the evenness or oddness of a number, and cubers use "parity" to describe any sort of problem, I'll use the phrase "parity problem" to mean "problem related to parity states." That seems like a totally reasonable compromise to me.
Why anybody would want puzzles with parity problems or false equivocation issues is beyond me. I feel strongly about this. Like if I were the head of a puzzle factory, I would employ a person to check permutation puzzle designs for such oversights and reject them before they went into production. The 12-color megaminx is an excellent puzzle, and the 6-color megaminx, which doubles 6 colors across the 12 faces, is like its stupid sibling who comes with false equivocation issues.
CHAPTER 7: DISCOVERING POSSIBLE PARITY PROBLEMS
Let's sort everything out, for we
do love organizing. We have the following sets of pieces:
single-centers (24); quenters (6); single-edges (24); dedges (12); and corners (8).
In solving the cube, the parities of all the sets eventually need to match.
An outer face rotation affects these groups in this way:
single-centers - 3 transpositions, odd, does affect parity
quenters - 0 transpositions, even, does not affect parity
single-edges - 6 transpositions, even, does not affect parity
dedges - 3 transpositions, odd, does affect parity
corners - 3 transpositions, odd, does affect parity
An inner face rotation affects these groups in this way:
single-centers - 6 transpositions, even, does not affect parity
quenters - ruins quenters
single-edges - 3 transpositions, odd, does affect parity
dedges - ruins dedges
corners - entirely unaffected
If one set of pieces has odd parity and another has even parity, they are un-matching.
If inner face rotations ruin quenters or dedges, it is preferred to use outer face rotations only.
If two sets of pieces have un-matching parity, and only outer face rotations are available, sometimes the two sets of pieces get locked into un-matching parity states. When that happens, we'll flag it as a "possible parity problem."
Let's take a closer look.
Can single-centers and quenters be locked into un-matching parity states?
No. The question is confusing because quenters are made from single-centers.
Technically, inner face rotations ruin quenters, so we'll ignore them for now.
Outer face rotations toggle the parity of single-centers and maintain the parity of quenters.
No parity problem.
Can single-centers and single-edges be locked into un-matching parity states?
No. Inner face rotations toggle the parity of single-edges and maintain the parity of single-centers, and outer face rotations toggle the parity of single-centers and maintain the parity of single-edges.
No parity problem.
Can single-centers and dedges be locked into un-matching parity states?
Yes. Inner face rotations ruin dedges, so we'll ignore them for now.
Outer face rotations affect both parities simultaneously.
Possible parity problem.
Can single-centers and corners be locked into un-matching parity states?
Yes. Inner face rotations affect neither parity.
Outer face rotations affect both parities simultaneously.
Possible parity problem.
Can quenters and single-edges be locked into un-matching parity states?
Yes. Inner face rotations ruin quenters, so we'll ignore them for now.
Outer face rotations affect neither parity.
Possible parity problem.
Can quenters and dedges be locked into un-matching parity states?
No. Inner face rotations ruin quenters, so we'll ignore them for now.
Outer face rotations toggle the parity of dedges and maintain the parity of quenters.
No parity problem.
Can quenters and corners be locked into un-matching parity states?
No. Inner face rotations ruin quenters, so we'll ignore them for now.
Outer face rotations toggle the parity of corners and maintain the parity of quenters.
No parity problem.
Can single-edges and dedges be locked into un-matching parity states?
No. The question is confusing because dedges are made from single-edges.
Technically, inner face rotations ruin dedges, so we'll ignore them for now.
Outer face rotations toggle the parity of dedges and maintain the parity of single-edges.
No parity problem.
Can single-edges and corners be locked into un-matching parity states?
No. Inner face rotations toggle the parity of single-edges and don't affect corners.
Outer face rotations toggle corners and maintain single-edges.
No parity problem.
Can dedges and corners be locked into un-matching parity states?
Yes. Inner face rotations ruin dedges, so we'll ignore them for now.
Outer face rotations affect both parities simultaneously.
Possible parity problem.
CHAPTER 8: WHICH ONES MATTER?
Commutators always use an even number of rotations because every single rotation used in an X or a Y has its inverse coming up somewhere along the way. Making commutators into the center parts of conjugates doesn't change anything, because every new rotation brings along its inverse as well.
If two sets of pieces are locked into un-matching parity states and one of them is solvable with commutators, then the other wouldn't be. That's bad, and that's why we marked those situations as "possible parity problems."
The four possible parity problems we found were between the following sets:
1. single-centers & dedges
2. single-centers & corners
3. quenters & single-edges
4. dedges & corners
Let's take a closer look at the second one, single-centers & corners. Inner face rotations affect neither parity, and outer face rotations affect both parities simultaneously. Single-centers & corners, then, must always match! So they can never reach an un-matching state in which to get locked in the first place! That means we don't have to worry about them, so our new list is:
1. single-centers & dedges
2. quenters & single-edges
3. dedges & corners
Because we know that single-centers & corners always have matching parity, that means the first and the third groups mean the same thing. So we can get rid of one of those as well. Would you rather keep tabs on 24 single-centers or 8 corners? Me too.
1. quenters & single-edges
2. dedges & corners
CHAPTER 9: THE 4x4x4 OLL & PLL PARITY PROBLEMS
The two parity problems in the 4x4x4 are the "orientation of the last layer" (OLL) parity problem, and the "permutation of the last layer" (PLL) parity problem. The OLL parity problem happens when quenters and single-edges are locked into un-matching parity states, and the PLL parity problem happens when dedges and corners are locked into un-matching parity states.
OLL
An OLL parity problem occurs or does not occur the moment the last quenters are solved and placed. When the cube appears to approach a solved state, the OLL parity problem looks like two incorrectly placed single-edges or one flipped dedge when the cube is otherwise solved.
The OLL parity problem occurs when the parity of the solved quenters does not match the parity of the single-edges. Once the quenters are solved, the parity of the single-edges is locked. Pairing up from single-edges dedges while maintaining solved quenters uses an even number of inner slice moves, so that won't change the parity state of single-edges. If the single-edge parity is odd the moment the quenters are solved, disturbing a quenter to make a dedge will toggle the single-edge parity to even, and re-solving the quenter will toggle the single-edge parity back to odd. The same is true if the single-edge parity is even the moment the quenters are solved; as long as the quenters are split and then re-solved with an even number of inner slice moves, the parity of single-edges will remain even.
The parity of single-edges remains locked, unless the quenters are reconfigured with an odd number of inner face rotations. So that's how you fix the OLL parity problem - breaking the quenters with one inner face rotation and then resolving everything. It's a huge pain, and this is why the 4x4x4 belongs on a rocket.
If the number of inner face rotations used in the scramble plus the number of inner face rotations used in solving the quenters happens to be odd, then the ending result is a cube with solved quenters with single-edges locked into an odd parity state. Because an outer face rotation creates 6 single-edge transpositions, no amount of outer face rotations can affect single-edge parity, which means no amount of outer face rotations can solve the single-edges in relation to their quenters.
PLL
A PLL parity problem occurs or does not occur the moment the last dedges are paired. When the cube appears to approach a solved state, the PLL parity problem looks like two incorrectly permuted but correctly oriented dedges or two incorrectly permuted but correctly oriented corners when the cube is otherwise solved.
The PLL parity problem occurs when the parity of the dedges does not match the parity of the corners. The two affected dedges or corners could be any two dedges or corners; they wouldn't have to be dedges that are directly across from each other like in the picture above. With a PLL parity problem, if the dedges are first solved, the problem will present itself in two corners; and if the corners are first solved, the problem will appear in two dedges.
The moment the last dedges are paired, their parity is locked, so if the parity of the dedges is even and the parity of the corners is odd (or vice versa), there will be PLL parity problem. Inner face rotations would ruin both quenters and dedges, so they aren't used. Outer face rotations create 3 dedge transpositions and 3 corner transpositions, so they can't help to match the parities.
As with the OLL parity problem solve, solving a PLL parity problem requires breaking out of the 3x3x3-lookalike with an odd number of inner face rotations. This time, instead of reconfiguring quenters, the dedges are reconfigured in such a way that their parity matches that of the corners.
In sum,
Single-edges and quenters with unmatching parity = OLL parity problem.
Inner face rotations ruin solved quenters (and paired dedges).
Outer face rotations affect the parity of neither single-edges nor solved quenters.
Corners and dedges with unmatching parity = PLL parity problem.
Inner face rotations ruin solved dedges (and solved quenters).
Outer face rotations affect the parity of both corners and dedges simultaneously.
Vixen switches tracks to the Cupid Carousel and collects a cup of cocoa.
The Cupid Carousel takes Vixen and Mrs. Claus to their respective homes.
Mrs. Claus switches over to the Comet Carousel for her cuppa.
Those 3 moves together were the X in our commutator.
The Mayor takes the Comet Carousel to Christmastown.
That was the Y.
The Mayor could always use a cocoa, so he switches tracks to the Cupid Carousel.
(That's just like our first Christmastown rotation).
The Mayor heads home to Southtown, bringing Vixen back into Christmastown.
Vixen, finding herself at The Station, switches to the Comet Carousel.
And as before, Vixen switches tracks to the Cupid Carousel for a cup of cocoa.
The Cupid Carousel brings Mrs. Claus to Christmastown, pushing Vixen into Southtown.
Mrs. Claus changes tracks to the Comet Carousel to get her cocoa.
(X).
(Y, Vixen and Mrs. Claus clink in Southtown.)
Donner wastes no time switching to the Cupid Carousel to grab her cocoa.
The Cupid Carousel brings Vixen back to Christmastown and Donner back to Vixenland.
Vixen changes tracks to the Comet Carousel, collecting her second cup of cocoa.
(X-1).
Vixen and Mrs. Claus take the Comet Carousel home, and everybody has a good night's rest.
(Y-1, Vixen & Donner clink back in Vixenland.)
The main difference between chapter 13 and chapter 14 is that in chapter 13, the carousels are approaching Christmastown from different directions, starting from Southtown and Vixenland. In chapter 14, they're both traveling the same direction, beginning in Vixenland.
Just like our corner 3-cycle made use of two opposite planes, Snowmiser's place and Heatmiser's place, our center 3-cycle makes use of two carousel tracks. And just like two cubies start in Snowmiser's and the third starts in Heatmiser's for a corner 3-cycle, the center 3-cycle is no different: two cubies will always start on one carousel and the third will always start on the other.
CHAPTER 15: REVIEW
THE TOOLBOX
Have you collected all 12?
1. The Corner 3-Cycle (
review)
I think of this as the key - the very most important commutator.
Once you understand it, you can understand all commutators.
Also, you can solve the 3x3x3 Rubik's cube with nothing else!
2. The 2 Corner Twist (
review)
An orienting commutator is most handy when you'd like to permute without a care in the world.
Pulling a corner out, reorienting it, and placing it back is a little tricky.
Once that sinks in, however, understanding the inverse is nothing.
Our Corner 3-Cycle uses a conjugate in the X of the commutator; this uses two conjugates.
in which M is the first move in the 3-move X
It's the edge version of the corner 3-cycle!
in which M is the second move in the 3-move X
Still very much like the corner 3-cycle.
in which M is the 1-move Y
Not too hard after the first two edge 3-cycles, but a little less related.
This is the first commutator where the 1-move Y is an inner face rotation.
That means the 3-move X will appear to mess up more of the cube than it did before!
in which 3 edges are on the same M
This edge 3-cycle gets first prize for being the shortest and simplest.
The inverse requires very little effort to think through.
Reorients centers.
in which 4 edges are on the same M
An extension of the previous, maybe even easier with all 180° rotations.
Also reorients centers.
This is the other orienting commutator, also most handy.
Pulling an edge out, reorienting it, and placing it back is simpler than twisting a corner.
As with the 2 Corner Twist, the X in the commutator uses two conjugates.
Both reorienting commutator inverses are easier to envision than 3-cycle inverses.
9. The Oriented 3-Cycles (
review)
Without pencil and paper, I find these kind of annoying and hard.
10. The Center 3-Cycle (
review)
in 3 different planes
Understanding both the corner 3-cycle and also the last edge 3-cycle are helpful prerequisites.
Even though there are 3-move X's and 1-move Y's like usual, the rhythms are duples.
11. The Center 3-Cycle (review) in 2 different planes
An important extension of the first, this 3-cycle has two cubies in Vixenland.
12. Conjugates in General (
review)
Conjugates are my favorite!
The C's in conjugates move pieces into optimal positions for commutators.
Their inverses restore the cube. It really is that simple and that amazing.
My my my, we've built quite a lovely toolbox, haven't we? Any commutator can be put into the middle of a conjugate or inverted (or both), which means our tools are very flexible!
CHAPTER 16: THE MEGAMINX 3-CYCLES
In case you also bought the Roxenda speed cube set for the affordable price of $22.49, you might notice that everything we've talked about translates pretty well to all the puzzles except for the megaminx edges, which are a little tricky. At first I couldn't figure out how to cycle them, but
Jamie Mulholland could, and he explained that even though there are no M moves on a megaminx, M's can be imitated by rotating two outer planes around edges.
The megaminx no longer has five related edge cycles; there's just one and its inverse, but it's fairly powerful. It's similar to the first commutator in chapter 12 from
my original post. The main thing to remember is that only outer planes are at our disposal, and to create a commutator, we must create a plane in which nothing but an edge has changed. Let's call that plane Snowmiser's.
When our story begins, edge Vixen and edge Mayor will be hanging out at Snowmiser's. They do that a lot, much to the Mayor's dismay. Mrs. Claus could be in a variety of places, but not just any place. We'll simplify the situation for the purpose of our story and get more into her real specifics in just a mo'. (Once you hear Julie Andrews say, "mo'" in Thoroughly Modern Millie, it just sort of creeps into your life every now and again.)
When we begin, Vixen and the Mayor are in Snowmiser's. Mrs. Claus is not.
Also, Vixen and Mrs. Claus both have a Snowmiser tile. The Mayor does not.
As usual, the plan is to replace edge Vixen (who is at Snowmiser's) with edge Claus (who is not) with the X in our commutator. Then we'll rotate Snowmiser's to select the Mayor for the 1-move Y. The inverse of X will bring Vixen back to Snowmiser's while removing the Mayor, and the inverse of Y will rotate Snowmiser's back to its original state.
In more detail, we must prepare for takeoff.
1. Edge 1 Position & Orientation
Because Vixen is an edge, she's in two planes.
The color of one of her 2 tiles must match the color of the rest of that face.
That face is Snowmiser's.
The other face we'll call Vixenland.
2. Edge 2 Position
The miserable Mayor must be at Snowmiser's with Vixen.
3. Edge 3 Position
For now, Mrs. Claus will be positioned in Vixenland across from Vixen.
Because our planes have 5 sides, she'll be 2 or 3 Vixenland rotations away from Vixen.
4. Edge 3 Orientation
For now, Mrs. Claus will be oriented so that her Snowmiser tile is not in Vixenland.
Here we go.
As it turns out, there's only one way to begin our X, and that's to take the two corners adjacent to Vixen and move them together, away from Mrs. Claus, in the same direction. (That means one of the planes will be moving clockwise and one will be moving counterclockwise, just like opposite planes on a cube when they're moving in the same direction.)
Our next step is to replace Vixen with Mrs. Claus. In our simplified story, we'll rotate Vixenland 2 or 3 times until we've made the switch. Notice that we have steered clear of those moving corner planes!
To restore Snowmiser's, both of Vixen's original corners come back to where they were, snuggling Mrs. Claus instead of Vixen.
That was our 2+1+2 move X. It's just a 3-move X that involves a weird two-part corner move.
Our 1-move Y is a Snowmiser rotation that brings the Mayor to where Claus just arrived.
Taking the inverse of X and then the inverse of Y of course completes the megaminx edge 3-cycle and otherwise restores the dodecahedron that we may have newly discovered feelings of attachment towards.
Okay, once that makes some sense, let's revisit Mrs. Claus' possibilities. We already know that Vixen's position and orientation defines both planes Snowmiser's and Vixenland. We also know that Vixen and the Mayor will be on Snowmiser's.
3. Edge 3 Position
Mrs. Claus' position and orientation are very complicated.
Once the first double corner move has been executed, we cannot rotate Snowmiser's.
Nor can we rotate any of Snowmiser's 5 original corners.
That means we cannot rotate seven out of twelve planes.
But... that means we can move the other five planes!
The five planes that don't bother Snowmiser's corners are:
Vixenland
The two planes across Vixenland opposite Snowmiser's
(Sort of like a two-land Heatmiser's)
The only other plane joining the previous two planes
Moving around the dodecahedron in the same direction, the next plane
Mrs. Claus can be any of the eighteen edges on those five planes.
(That's what I meant when I said this commutator was "fairly powerful.")
4. Edge 3 Orientation
However Mrs. Claus moves across those 5 planes, her second to last move
must have her arrive in Vixenland with her Snowmiser tile outside of Vixenland.
Her final move will be a Vixenland rotation transporting her to Snowmiser's.
When she arrives to Snowmiser's, her snowmiser tile will, of course, match that face.
Wasn't that wild?
Now that the edges are old hat, we can address the corners. The megaminx corner 3-cycle is like the first commutator in our toolbox, so this part is easy.
Similar to the Mrs. Claus specs in the megaminx edge 3-cycle, the long part of the corner 3-cycle commutator (the middle of our X) includes a choice of fourteen corners across seven planes of rotation, while the short part (the 1-move Y) only offers a selection of four Snowmiser corners.
In either case, it makes just as much sense to put a shorter commutator into the middle of a conjugate, but there are times in life when we do crazy things just for the fun of it. I'm pretty sure that's what they mean when they say, "Carpe diem."
CHAPTER 17: THE MEGAMINX ORIENTED COMMUTATORS
Now that we know how to permute edges on a megaminx, we can figure out a 2-edge flip. It starts off the same way as the megaminx edge 3-cycle: Jingle's two adjacent corners move in the same direction, and Snowmiser's is defined. Jingle rotates twice (144°, not 72°) on the non-Snowmiser plane to avoid the seven untouchables. Two more adjacent plane rotations will take care of the reorienting, and 144° in the same direction he's been travelling will bring him back to Snowmiser's but upside down. His two original adjacent corners return to their places, and that's the end of X.
Snowmiser's is rotated to select Jangle.
Jangle's corners rotate the same way Jingle's did, Jangle does the Jingle dance but backwards (I believe that's called the "Elgnij"), and his corners return to their original places to complete the commutator.
The megaminx 2 corner twist is so similar to the cube 2-corner twist that it's hardly worth mention.
For all reorienting commutators,
First, the plane that holds both to-be-reoriented pieces is identified.
In my last post, I called this plane Snowmiser's.
The X in all the reorienting commutators does 3 things:
1. It removes one piece from Snowmiser's
2. Reorients that piece, and
3. Returns that piece to Snowmiser's without otherwise disturbing Snowmiser's.
The Y rotates Snowmiser's to select the second piece that will be reoriented.
The rest of the commutator re-solves everything.
If you know how to build a cross, place corners, and place second-layer edges with double Z-commutators (covered in chapters 16 and 17 of
my original post), then you can create simple conjugates around those skills to solve all but one face on the megaminx. Or you could solve all but two, leaving a "column" open to re-orient the last edges intuitively (as described in chapter 19 of my original post). In either case, once you're at that point, a few conjugates with chapters 16 and 17 of this post will finish the solve.
Chapter 15 of my original post walks you through creating conjugates for single layers. Basically, it just tells you to move your pieces off one layer so that they're ready for takeoff, perform your commutator, and move everything back to where it was with the inverse of your first sequence of moves.
It looks like I've finally written out how to solve all the cubes and the megaminx from my Roxenda speed cube set. There's no way you'd get to this point without being able to solve the pyraminx, but if you have, you've officially gotten away with graduating Rubik's Cube 202 totally unqualified. That would actually be so astounding that I'd be forced to offer my deepest heartfelt congratulations.
CHAPTER 18: MUSINGS ON INVERSES
It should come as no surprise that sometimes I need a piece of paper to help keep track of what I'm doing, or to jot down the C's in my conjugates. If you find yourself jotting something down, and you think there might be a simpler way to approach a problem, you can always try thinking of cycling your Vixen backwards, which means you'd be thinking in terms of inverses.
Speaking of inverses, the stories in chapters 13 and 14 from this post are based on the inverses of the center commutators that are normally presented, so if you learn center commutators from someone else and they're different from mine, that's why. My first post had 3-move X's and 1-move Y's, and I require compelling reasons to be inconsistent.
Let's take a moment to review 3-cycle commutators with longer X's and 1-move Y's, along with their inverses, which are commutators with 1-move X's and longer Y's. It's nice to be a stranger to neither. To begin, we'll need a piece (let's call it V) which is correctly oriented to travel to its slot (let's call that S) through a given plane of rotation (let's call that P). So on the 4x4x4, P could be U, D, R, L, F, B, MU, MR, MF, or even MD, ML, or MB. (Cubers have a different notation for this, but then you have to memorize their arbitrary directions.)
It's rather clumsy to use one variable for a piece, one variable for a position, and one variable for an entire plane, but there it is. That's to say nothing of the fact that two other variables, X and Y, are being used to represent a sequence of rotations. Please bear with me.
V must be able to travel to S by moving P some number of times (P, P-1, or P2), which means both V and S are contained in plane P.
During the longer X, V is moved out of P without otherwise disturbing P. Then, the 1-move Y rotates P so that S occupies V's original position. The longer X inverse puts V into S, and the 1-move Y inverse returns S to where it belongs.
For a 1-move X, P is rotated so that V moves to S's spot immediately, also moving everything else in P with it. The longer Y removes V from P without otherwise disturbing P. The 1-move X inverse returns P back to where it belongs, and the longer Y inverse puts V into S.
Naturally, you have to keep track of what exactly is replacing Vixen, as well as its orientation, if you're wanting to solve more than one piece at a time, so those details give you the specifics of what exactly to do in the longer part of the commutator and how to rotate the 1-move part.
CHAPTER 19: THE SUPERCUBE
Cubes that require specifically oriented single-centers are called "supercubes." Single-centers always rotate around the center point of the four-square quenter area; that's just how outer and inner rotations move them. If you'd like to turn your 4x4x4 cube into a supercube, first solve it, and then put 16 little stickers, all with the same orientation, on each face. That will do the trick, but if you'd like to turn your 4x4x4 into a supercube with some class, make sure to orient opposite sides so that they're both pointing in one direction together, and make sure no corner is adjoining two planes that have orientations 0 or 180° apart. Supercubes don't get classier than that.
If, on the other hand, you'd like to be the proud owner of the least classy supercube on the block, you could take a sharpie and draw a really ugly picture on each face. That would work too.
For the 4x4x4 supercube, nothing we know about the corners or single-edges changes, but the way our center 3-cycles orient our single-centers is suddenly useful.
Whether our cubies traipse across 2 or 3 different lands, the following observations remain:
Vixen never gets re-oriented.
(Of course she doesn't; that's what defines her as our Vixen.)
Claus, Mayor, and Donner always rotate towards the center line between the carousels.
In terms of inverses, Vixen would first travel to a new land instead of switching tracks.
Vixen still never gets re-oriented.
Claus, Mayor, and Donner still always rotate towards the center line between the carousels.
What changes is the direction of the 3-cycle.
Vixen will start moving the 3-cycle in a clockwise or counter-clockwise direction.
Naturally, everyone else will follow suit.
Friendly reminder:
The last two edge commutators in the toolbox (#6 & #7) do reorient centers, so they should be used on supercubes with discretion!
CHAPTER 20: CONCLUDING REMARKS & SOURCES
Most people who don't want to memorize algorithms run into a parity problem, re-scramble the cube, and just hope for the best. A step up from that would be understanding what causes the parity problems, strategically "scrambling" the cube with an inner face rotation, and re-solving. Re-solving with conjugates and commutators specifically makes sense because commutators will maintain parity for any set of pieces. Being of the form C*X*Y*X-1*Y-1*C-1, they guarantee an even number of every type of move.
Solving the 4x4x4 with the centers last makes the most sense to me. It's called the "cage method" because the solved pieces look like a cage. (I've listed a source that goes into the details of how to do this most efficiently below, but I've paid it no attention myself.) My reasoning is that you won't run into a PLL parity problem if you're not married to dedges, and if you run into the OLL parity problem, which could happen because your quenters have a parity state whether or not they're solved, then an inner face rotation would disrupt up to 4 incorrectly solved single-edges instead of up to 4 incorrectly solved single-edges and 8 incorrectly solved single-centers.
Because an OLL parity problem appears to require a 2-cycle, using an inner face rotation to fix it is related to chapter 18 of
my original post. If you'll recall, we discovered that if a 3x3x3 cube appears to require a 2-cycle of edges and a 2-cycle of corners, rotating a face may "un-solve" some pieces, but in return, our newly unsolved pieces could now be prepared for 3-cycles.
By the by, as previously hinted at in chapter 11 of this post, that's also everything you need to know about solving the 2x2x2.
But back to the cage method - because corners don't move on inner face rotations, I'd start with those. After that, there's no real need to avoid dedges should they arise: they just oughtn't be solved all at once. Of course, all three of the edge commutators and the 2-edge flip from my original post work on dedges. And once that cage is complete, you can cocoa your way out of there like a boss, with nary a par'y in sight.
And yes, I can solve the 4x4x4 faster and more easily through reduction and backtracking than with dozens of commutators, but for the times I pride myself on taking a journey with no hidden detours, I solve this thing a little bit faster than a squirrel could.
Center commutators and single-edge commutators might be a little tedious, but you won't be making a deal with the devil 75% of the time. And as always, if you're not ready for take off, you can always use conjugates so that you are (no rules just right). With center commutators, sufficient C's are often just simple outer face (U D R L F B) rotations.
If you'd like to look into this a little more on your own, I should warn you that parity is most often explained by people who don't know what they're talking about. In the last post, I emphasized that
Ryan Heise's website was my favorite source, but that was my favorite source amongst many correct sources. No such luxury this time. Be careful out there, and start with these.
These are good sources.
SOURCES
"Let's talk about parity!"
"It's fascinating."
"You don't mind this, do you?"
"... We don't mind at all... Thank you, Santa."
(My 4x4x4, headed at warp speed towards a big ball of fire)
Entire Series:
