INTRO
This is my fourth post on permutation puzzles. The others are here:
The Rubik's Cube (1/5)
The 4x4x4 (2/5)
Putting It All Together (5/5)
The first post covers conjugates and commutators and how they pertain to the Rubik's Cube, the second covers the notion of a "parity problem," and the third documents my kids' course. From developing an interest in these puzzles, I've acquired the following collection:
15 Puzzle
Tower Cuboid, Domino Cuboid
Ivy Skewb, Dino Skewb, Skewb
Pocket Cube, Rubik's Cube, Rubik's Revenge, Professor's Cube
Cylinder Shape Mod, Windmill Shape Mod, Fisher Shape Mod, Axis Shape Mod
Pyraminx, Pyramorphix, Mastermorphix
Kilominx, Megaminx
Square 1
20 Stickerless QiYi Brand Permutation Puzzles
Sleepin' in Their Basket
The reasoning behind solving all of these puzzles is related, but the Skewb and the Square 1 (SQ1) seem to require some additional attention. That, and the fact that my faithful friend Jack keeps editing these posts out of the goodness of his heart, is why I'm back. I'll try to make it short.
Skewb, SQ1, Domino Cuboid
CHAPTER 1: THE SKEWB
Chapter 17 of my original Rubik's Cube post reads:
The 3-cycles we've been creating are more complicated commutators that yield simpler outcomes, and Z/Y commutators are simpler commutators that yield more complicated outcomes. In this chapter, we're observing what these simple commutators do, and taking advantage of them to permute our pieces in a way that's convenient for us.(Aside: the Skewb permutation puzzle relies on simple commutators that yield more complicated outcomes only, as it's impossible to construct more complicated commutators that yield simpler outcomes!)
The Skewb has 6 centers and 8 corners. One twist on the Skewb displaces half the puzzle: 3 centers and 4 corners. It's like the Pocket Cube in terms of severe displacement, but without the option to create simple X's and Y's that that have an intersection of 1.
If we can't easily create 3-cycles, we must try a (simple) commutator X*Y*X-1*Y-1 in which both X and Y are two different 120° rotations, and make a note of the results. Such a commutator on the Skewb preserves 6 pieces: 4 corners surrounding a center, that center, and an adjacent center. The remaining 4 centers have been permuted by two 2-cycles, and the remaining 4 corners have been reoriented. Because of the two 2-cycles, doubling the commutator, X*Y*X-1*Y-1*X*Y*X-1*Y-1, is an algorithm that affects corner orientation only!
Thus, the way the Skewb is solved is as follows:
One center and its surrounding four corners are solved by blockbuilding.
The corners must coordinate with one another, but not with other centers.
This automatically places the final four corners.
Let X and Y be 120° rotations that do not disturb the solved center.
The commutator X*Y*X-1*Y-1 is then used to place the remaining centers.
Finally, doubling the commutator, X*Y*X-1*Y-1*X*Y*X-1*Y-1, reorients the final four corners.
The Skewb has to be correctly positioned for this to work, but I'm sure you can figure that part out much more easily than if I write another paragraph about it. Doing it wrong won't permute anything anyway, so you don't have much to lose.
Although the initial reasoning behind this is a little tricky, solving the cube becomes very easy rather quickly. It's a lovable little thing.
CHAPTER 2: SQ1 INTRO & NOTATION
If the Skewb is a lovable little thing, SQ1 is quite the opposite. It's nerdy. And for an official "World Cube Association" event, unpopular. Maybe even annoying. It's so unlike its peers.
The first thing to notice is that it's structured differently. SQ1 is a 3 layer cube in which the middle layer is a 2-piece equator that is very easy to manipulate. Solving the puzzle, then, is really a matter of solving the top and bottom layers. I'll call these U and D.
SQ1 is a shape mod that's first made into a cube. Typically, the yellow and white pieces are placed into their respective sides, and then everything gets permuted around. At some point in this process, the parity state must be addressed. More on that in a mo'.
Because the SQ1 moves so differently, it requires its own notation.
To read notation, it is assumed that
the shorter part of the equator when viewed from the front is always on the left.
The equator only allows 180° turns, so the notation / means a 180° turn on the right.
Meanwhile, measuring angles from the center of U or D, each corner piece is 60°.
Each edge piece is 30°.
(Each of 12 corner and edge pieces on a SQ2 are 30°!)
U and D rotations are therefore based on the number of 30° angles.
As is standard, a positive number indicates clockwise, and negative indicates counterclockwise.
Putting this together, 3 = 90° clockwise.
5 = 150° clockwise.
-3 = 90° counterclockwise.
() are used to organize numbers to mean a rotation on U and a rotation on D.
The notation (3,3) means a 90° clockwise U turn and a 90° clockwise D turn.
Remember, clockwise is determined "when looking at the face."
So (3,3) requires that U and D faces to shift in opposite directions.
(1,-1) means a 30° clockwise U turn and a 30° counterclockwise D turn.
U and D would then move in the same direction, imitating a middle slice move.
(1,6) means a 30° clockwise U turn and a 180° D turn.
Putting this together with 180° R turns (/), an example of SQ1 notation is:
(6,0)/(6,0)/(6,0)/(6,0)This algorithm flips the right side of the equator.
For all algorithms, the left side of the equator never moves.
Another example of SQ1 notation is:
/(6,6)/(-1,1)That one exchanges U with D.
If an algorithm affects either U or D only, knowing how to exchange U and D can be very handy. For example, let's say that I want to permute something and my algorithm affects U only, but my issue is in D. I can bring D to U, perform my algorithm, and swap them back. Conjugates :)
CHAPTER 3: SQ1 CUBE SHAPE
Fuss about until all 8 adjacent edges and 2 corners are in U, leaving D with 6 corners. Of course you could do this backwards and put the 6 corners in U instead, but you wouldn't be able to see what you're doing, so I wouldn't recommend it. Tingman calls that shape "the Millennium Falcon," and I think that's a pretty good name for it. After we find the Millennium, we split the 8 edges into two 4's, the two 4's into four 2's, four 2's into four 1's and two 2's, and then finally into all eight 1's. Then it's a cube.
Millennium Falcons in Pretty Colors
Once the SQ1 is in cube shape, if U or D (not both) is offset by exactly 1, then / maintains cube shape. Of course, 3's, -3's, and 6's after (1,0) or (0,-1) won't destroy cube shape either. It's handy to know this when exploring your own algorithms.
For example, the algorithm (1,0)/(-1,0) exchanges half of U with half of D.
Try it and its inverse.
Here's the right equator flip we saw in the last chapter, but without leaving cube shape.
(1,0)/(6,0)/(6,0)/(5,0)
Let's call it the "Equator Flip."
Repeating it will restore the cube.
Here's the U and D flip without leaving cube shape.
(1,0)/(6,6)/(-1,0) For all you Stranger Things fans, let's call it the "Upside Down."
Repeating it will restore the cube.
(1,0)/(-1,-1)/(0,1)
transposes the pair of front and back edges of U and D.
Because those edges gallop around so fast, let's call it the "4 Horsemen."
Repeating it will restore the cube.
swaps a quarter of U with a quarter of D;
It also shifts a quarter of U and D clockwise.
Interesting!
So (1,0) puts us in cube shape position and /(3,0)/ starts permuting quarter squares (quarters).
Playing with that, it appears that
(1,0)/(3,0)/(3,0)/(3,0)/(3,0)/(3,0)/(2,0)
cycles quarters around; the whole thing is an identity algorithm.
CHAPTER 4: SQ1 PARITY PROBLEMS
It's time to talk about parity.
Imagine there are 3 corners on the right side of U and 3 corners on the right side of D. It's true that a / creates 3 2-cycles of corners. A 2-cycle is a transposition, so that's an odd number of transpositions!! The entire point of my second post was to explain that an odd number of transpositions is... unkind. The Rubik's Cube never shows such unkindness.
Remember how when we were getting into cube shape, we put 6 corners into D? Well, it's also true rotating the 6 corners once, like (0,2), creates a 6-cycle of corners, which requires 5 transpositions. That's an odd number of transpositions again! Unkindness again!
(The SQ2 looks like a tremendous pain, but at least it does not have these problems. A / would create 6 transpositions instead of 3, and a (0,2) would create 10 transpositions instead of 5. But I digress.)
So there are parity problems in SQ1, and two ways we know to wrong the right or right the wrong.
Advanced SQ1 competitors know how to check for parity problems before getting into cube shape. While that's ideal, it requires memorizing a "reference scheme" and making a bunch of calculations concerning 6 different things. I'm not against calculations, as ridiculous as they may be, but I'll be damned if I'm memorizing a "reference scheme."
Other cubers nearly solve the SQ1, find there is parity half the time, curse, and then apply some monstrous algorithm at the very end. That is also entirely unreasonable.
What to do, what to do. Well.
Because I refuse to memorize a reference scheme, the first step would have to be for me to get this thing into cube shape. After that, instead of orienting or permuting pieces, I could start counting all the cycles for the corners and the edges as a way of checking the parity of all transpositions required to solve the cube. Do you remember when I was talking about subgroups under chapter 3 on group theory in my first post? Of course not. That part goes over counting cycles.
Let's say I begin my parity checking process by counting a corner cycle. At random I'd choose a corner, C1, which is in a slot, S1. If C1 is solved, meaning that S1 is C1's correct slot, then we can say C1 is part of a 1-cycle and it requires 0 transpositions to become solved, and that cycle's parity is even.
If C1 is not solved, meaning that S1 is not C1's correct slot, I'd look at C1's correct slot to find a new corner, C2. If C2's correct slot was S1, then C1 and C2 would make a 2-cycle, which requires 1 transposition, and that cycle's parity would be odd.
But if C2's correct slot was not S1, I would look at C2's correct slot to find C3. If C3's correct slot was S1, then C1, C2, and C3 would make a 3 cycle, which requires 2 transpositions, and that cycle's parity would be even.
And so on and so forth.
I'd add up the number of transpositions of all the cycles for corners and all the cycles for edges on U and D to determine the overall parity state. If the cube happened to be completely solved, that number would be 0. If all 8 corners were involved in a single cycle, that would require 7 transpositions, and if all 8 edges were involved in another single cycle, that would require 7 more. In other words, I must count to some number ≥ 0 but ≤ 14.
Once the puzzle is in cube shape and remains in cube shape, the parity of transpositions is locked. To avoid parity problems, we want the cube shape to be locked in an even parity.
If the parity of total transpositions were even, I would simply continue solving the cube. If it were odd, I'd have to fix that by taking my cube back to the Millennium with /(3,3)/(1,2)/(4,2)/. Then I'd (0,2) to swap parity, and restore to cube shape once again.
Altogether, that would look like this:
/(3,3)/(1,2)/(4,2)/
(0,2)
/(-2,4)/(1,2)/(-3,-3)/
But wait! We know of not one, but two ways to right the wrong! I could also toggle parity by taking my cube back only most of the way towards the Millennium, /(3,3)/(1,2)/, noticing an abundance of adjacent corners on both my U and D. Then I could move the 3 adjacent corners on U and and the 3 adjacent corners on D over to the right side with (2,-2), and swap parity with /. Then I'd restore cube shape. That would be kind of showoffy.
Altogether, that would look like this:
/(3,3)/(1,2)/
(2,-2)/
(4,-4)/(1,2)/(-3,-3)/
CHAPTER 5: BUILDING SQ1 ALGORITHMS
By now, you may have wondered why I included the Domino Cuboid in my photo. To solve the Domino Cuboid in terms that made sense to me, I used simple commutator/blockbuilding skills to simultaneously solve the edges, and then I wrote a commutator for the corners. The X in my commutator was R2*U*R2*U-1*R2, and the Y was some form of D. Mainly, my point is that the R2 reminds me a lot of the /. The 180° flip is weirdly constricting, but solving the Domino gives me the courage to explore the SQ1. I recommend trying these puzzles in that order. With the notion that we're going to be using something like R2's all the time, let's start building algorithms.
Here's where we left off at the end of chapter 3:
(1,0)/(3,0)/(-1,0)
swaps a quarter of U with a quarter of D;
it also shifts a quarter of each U and D clockwise.
This is an algorithm about making quarters, so let's call it the "Washington."
And if we're going to be talking about quarters, let's abbreviate them.
FR - front right quarter
FL - front left quarter
BR - back right quarter
BL - back left quarter
Say we wanted to permute quarters but maintain orientation.
We know
(1,0)/(3,0)/ would get us started.
(3,3)/(3,0)/ would move all the quarters to their original faces.
(5,6) would reorient the cube so we could see what happened.
Altogether,
(1,0)/(3,0)/(3,3)/(3,0)/(5,6)
swaps BR & BL in both U and D.
This is an algorithm about 4 oriented quarters, so let's call it "2 Heads 2 Tails."
Repeating it will restore the cube.
Also, because
(1,0)/(3,0)/(3,3)/(3,0)/(5,6)
swaps BR & BL in both U and D,
I should be able to perform the algorithm, reposition one face only, and repeat the algorithm.
The second iteration will re-solve the face I didn't reposition!
That's just like what I was talking about with the Skewb.
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(0,3) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(5,0)
Aha!
This one swaps FL & BL in D only.
Repeating the whole thing will restore the cube.
This is an algorithm about two oriented quarters only, so let's call it the "Half Dollar."
But actually, "Kennedy" is cooler.
Repeating it will restore the cube.
Related Kennedys:
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(0,-3) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(5,0)
swaps FR & BR in D only.
Repeating it will restore the cube.
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(0,6) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(5,3)
swaps FL & BR in D only.
Repeating it will restore the cube.
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(3,0) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(-2,6)
swaps FR & BR in U only.
Repeating it will restore the cube.
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(-3,0) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(-1,6)
swaps FL & BL in U only.
Repeating it will restore the cube.
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(6,0) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(2,6)
swaps FR & BL in U only.
Repeating it will restore the cube.
Let's go back to Washington.
(1,0)/(3,0)/(-1,0)
Specifically, this moves
BL in U to FR in D
BR in D to FL in U
FL in U to BL in U
FR in D to BR in D
Washington doesn't affect FR & BR in U or FL & BL in D.
That means I should be able to
1. Flip 4 quarters around with Washington,
2. Reposition U and D with (3, 3) so the scrambled quarters are in the back, and
3. Swap adjacent quarters with 2 Heads 2 Tails to get one unsolved quarter on each face.
Let's try it out:
(1,0)/(3,0)/(-1,0) (no slash, just resetting)
(3, 3) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(2,3)
Eureka, it worked!!!
BL in U transposed with BR in D.
Repeating it restores the cube, natural, natural.
This is an algorithm that maintains 3/4 of each face!
Let's add up those quarters and celebrate with some "Charles Shaw."
I was hoping to get to something like this. Originally, I was thinking that from here I'd further isolate a corner or an edge to make a commutator. But as I'm entirely unwilling to leave cube shape even for a moment, all I can permute are quarter squares. Also, so many of my algorithms are quarter square transpositions that leave the rest of the cube intact, which means I've already acquired more powerful tools than I had predicted. So it looks like I need to reevaluate my plan and solve this thing by reduction!
CHAPTER 6: THE DOUBLE KENNEDY
There comes a time in blockbuilding quarters where you can run into something that feels like a parity error due to the limitations of my algorithms. So we need one more tool. You see, our quarter swapping algorithms make a 2-cycle of corners and a 2-cycle of edges. The 4 Horsemen makes one 2-cycle of edges in U, and another in D. But what we can't do yet is make 2 corner or edge transpositions in only one face. That's also known as a 3-cycle, and those are all over my other permutation puzzles. It took me a little while to figure this one out, but I finally got it.
First, let's just explore what (1,0) and (0,-1) quarters are in the first place.
When using algorithms that begin with (1,0) and not (0,-1),
Quarters are edges then corners when moving clockwise.
Think of corners as having 2 colors, not 3.
The first corner color is yellow or white.
The second edge-matching color is on each corner's counterclockwise side.
The third color is completely ignored.
We'll call these (1,0) quarters.
On the other hand, when using algorithms that begin with (0,-1) and not (1,0),
Quarters are corners then edges when moving clockwise.
The first corner color is yellow or white.
The second edge-matching color is on each corner's clockwise side.
We'll call these (0,-1) quarters.
Because our algorithms were written for (1,0) quarters, we mostly use those.
Luckily, extending our understanding to (0,-1) quarters is not hard at all!
If we momentarily think in (0,-1) quarters and then return to (1,0) quarters, we can move edges around but leave corners untouched! But we never have to stop thinking in quarters! We just have to think in two kinds of quarters with overlapping corners!
A (0,-1) Kennedy looks like this:
(0,-1)/(3,0)/(3,3)/(3,0)/(6,-5) (no slash, just resetting)
(0,3) (rotating)
(0,-1)/(3,0)/(3,3)/(3,0)/(6,1)
That maintains U and swaps FL and BL (0,-1) quarters in D.
To turn that into a edge 3-cycle, follow up with the corresponding (1,0) Kennedy!
We started with a (0,-1) Kennedy DL, so in this case, we'll continue with a (1,0) Kennedy DL:
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(0,3) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(5,0)
Look! It's a clockwise 3-cycle of D front, back, and left edges! Thrilling!!
I'll call the 3-edge cycle a DoubleKennedy.
Specifically, that one was the DoubleKennedyDL, and altogether, it looks like this:
(0,-1)/(3,0)/(3,3)/(3,0)/(6,-5) (no slash, just resetting)
(0,3) (rotating)
(0,-1)/(3,0)/(3,3)/(3,0)/(6,1)
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(0,3) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(5,0)
Cycles D front, back, and left edges clockwise
DoubleKennedy DR:
(0,-1)/(3,0)/(3,3)/(3,0)/(6,-5) (no slash, just resetting)
(0,-3) (rotating)
(0,-1)/(3,0)/(3,3)/(3,0)/(6,1)
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(0,-3) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(5,0)
Cycles D front, back, and right edges clockwise
Counterclockwise DoubleKennedys, on the other hand, begin with (1,0) Kennedys:
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(0,3) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(5,0)
(0,-1)/(3,0)/(3,3)/(3,0)/(6,-5) (no slash, just resetting)
(0,3) (rotating)
(0,-1)/(3,0)/(3,3)/(3,0)/(6,1)
Cycles D front, back, and left edges counterclockwise
(1,0)/(3,0)/(3,3)/(3,0)/(5,6) (no slash, just resetting)
(0,-3) (rotating)
(1,0)/(3,0)/(3,3)/(3,0)/(5,0)
(0,-1)/(3,0)/(3,3)/(3,0)/(6,-5) (no slash, just resetting)
(0,-3) (rotating)
(0,-1)/(3,0)/(3,3)/(3,0)/(6,1)
Cycles D front, back, and right edges counterclockwise
You get the idea. You could cycle edges in U by changing (0,3) to (3,0), etc.
I won't muddy up our toolbox with all of these because you don't actually need to understand it for my method. You just need to be able to start an algorithm with (0,-1), like maybe one time. Of course, now that you know how to make 3-cycles of edges, you can solve a SQ1 a great variety of ways.
CHAPTER 7: SOLVING THE SQ1
7.1: TOOLBOX OF DERIVABLE ALGORITHMS
Dolla'Dolla': (1,0)/(-1,0) (Chapter 3)
Equator Flip: (1,0)/(6,0)/(6,0)/(5,0) (Chapter 3)
Upside Down: (1,0)/(6,6)/(-1,0) (Chapter 3)
4 Horsemen: (1,0)/(-1,-1)/(0,1) (Chapter 3)
Washington: (1,0)/(3,0)/(-1,0) (Chapter 5)
2 Heads 2 Tails: (1,0)/(3,0)/(3,3)/(3,0)/(5,6) (Chapter 5)
Kennedy DR: (1,0)/(3,0)/(3,3)/(3,0)/(5,6)(0,-3)(1,0)/(3,0)/(3,3)/(3,0)/(5,0) (Chapter 5)
2H2T(0,-3)2H2T
Kennedy DL: (1,0)/(3,0)/(3,3)/(3,0)/(5,6)(0,3)(1,0)/(3,0)/(3,3)/(3,0)/(5,0) (Chapter 5)
2H2T(0,3)2H2T
Kennedy D Opposite: (1,0)/(3,0)/(3,3)/(3,0)/(5,6)(0,6)(1,0)/(3,0)/(3,3)/(3,0)/(5,3) (Chapter 5)
2H2T(0,6)2H2T
Kennedy UL: (1,0)/(3,0)/(3,3)/(3,0)/(5,6)(-3,0)(1,0)/(3,0)/(3,3)/(3,0)/(-1,6) (Chapter 5)
2H2T(-3,0)2H2T
Kennedy UR: (1,0)/(3,0)/(3,3)/(3,0)/(5,6)(3,0)(1,0)/(3,0)/(3,3)/(3,0)/(-1,6) (Chapter 5)
2H2T(3,0)2H2T
Kennedy U Opposite: (1,0)/(3,0)/(3,3)/(3,0)/(5,6)(6,0)(1,0)/(3,0)/(3,3)/(3,0)/(2,6) (Chapter 5)
2H2T(6,0)2H2T
Charles Shaw: (1,0)/(3,0)/(-1/0)(3,3)(1,0)/(3,0)/(3,3)/(3,0)/(2,3) (Chapter 5)
WASH(3,3)2H2T
Parity Rotation: /(3,3)/(1,2)/(4,2)/(0,2)/(-2,4)/(1,2)/(-3,-3)/ (Chapter 4)
Parity Flip: /(3,3)/(1,2)/(2,-2)/(4,-4)/(1,2)/(-3,-3)/ (Chapter 4)
7.2: THE KRATZKE METHOD
1. Get to the Millennium Falcon - Tingman's video is excellent for this.
a. Group pairs of edges together.
b. There are 3 ways to add the final pair to the cluster of 6.
I. They can be added to a side of the cluster of 6.
II. They can be separated into a line and added to straddle the cluster of 6.
III. If the last 2 edges are stuck in an L-shape instead,
Replace an outer pair in the cluster of 6 with the L.
This creates a line of edges, a cluster of 4, and an isolated pair.
Make a cluster of 6, and then add the line.
c. The notion of opposite edges in a line is important; let's call these edge-lines.
2. Do the carbon dating thing to get a cube.
3. Check for odd parity.
a. You can rotate pieces around as much as you like before counting cycles.
b. Once you start counting cycles, however, do not alter the cube.
c. While unnecessary, it's easier count cycles if the pieces are oriented.
That means all the yellow pieces are in U and all the white pieces are in D.
d. To do this, first put yellow corners into U with Dolla'Dolla'.
Whenever I specify an algorithm like that, I always mean:
You can use it as many times as needed, and
You have the freedom to use U and D rotations to your heart's content.
Then use the 4 Horsemen to get one or both yellow edge-lines into U.
It's possible that this leaves an L-shape of edges in white.
In this case, permute white corners to form yellow and white edge-lines.
Whenever I say "permute," feel free to use any of our derivable algorithms.
Finally, use the 4 Horsemen to finish orienting edges.
e. Calculate the parity of transpositions needed to solve all pieces.
4. If the parity is odd, use one of the parity fix algorithms.
5. Blockbuild exactly 4 quarters and hide them from the 4 Horsemen. Then build the remaining 4 quarters with the 4 Horsemen.
You must be hiding 4 (1,0) quarters or 4 (0,-1) quarters; don't mix and match.
a. Odd numbers of built quarters are a huge pain; avoid them if at all possible.
b. (1,0) quarters can be hidden in FR and BL of U and also FL and BR of D.
(0,-1) quarters can ben hidden in the opposite quadrants.
Built edges simply need to be perpendicular to the 4 Horsemen transpositions.
c. Make U into yellow corners with white edges.
(You can choose to divide pieces any way you like; this is just how I do it.)
(Ex: white corners with yellow edges; hot color corners with cool color edges)
Arrange U with the method I described before the parity count in 3d.
d. Permute quarters in D until the 4 Horsemen can build 2 opposite yellow quarters in U.
e. Build the 2 yellow quarters and hide them from the 4 Horsemen with a 90° U rotation.
f. If you're very lucky,
I. You didn't accidentally build any white quarters, AND
II. You're set up so that the 4 Horsemen can build 2 white quarters in D.
In this case, build those 2 white quarters!
Then use the 4 Horsemen to finish building all 8 quarters.
g. Otherwise, permute quarters in D and build the 2 remaining yellow quarters in U.
We must now evaluate D.
I. If D has 0 built quarters, excellent!! We can hide them from the 4 Horsemen.
Begin with Dolla'Dolla', follow with 2 Heads 2 Tails, and hide.
Use the 4 Horsemen to build all four white quarters.
II. If D has 1 built quarter, we must remedy this with a (0,-1) Kennedy.
We don't care about quarter placement, so we don't need a DoubleKennedy.
(A (0,3) rotation cycles D front, back, and left edges clockwise.)
(A (0,-3) rotation cycles D front, back, and right edges clockwise.)
You have two options.
α. Unbuild all white quarters with a (0,-1) Kennedy and go to the previous step.
β. Solve U completely with (1,0) quarter algorithms first.
Then use your (0,-1) Kennedy to help make all 8 (0,-1) quarters!
That commits you to (0,-1) algorithms through the end. Fancy.
III. If D has 2 built quarters, you counted wrong and you have parity.
IV. D can't possibly have 3 built quarters; what would be in the 4th?
V. If D has 4 built quarters, you already have all 8! Go go go!
7. Permute your 8 quarters to finish the solve!!
CHAPTER 8: CHECKERBOARDING & SOURCES
To make a solved standard Rubik's Cube into one with a checkerboard pattern on each side is very easy. Take each of the 3 middle slices, and in any order, rotate them 180°. That's all.
Checkerboarding doesn't entirely work with puzzles that have an even number of pieces along at least one edge, but some people try. Just like the book Will It Waffle, here are my opinions on Will It Checkerboard.
15 Puzzle: No. What does this even mean?
Tower & Domino Cuboids: Only on some sides; still worth it
Ivy Skewb, Dino Skewb, Skewb: Not at all
Rubik's Cube, Professor's Cube: Classic checkerboard!
Pocket Cube, Rubik's Revenge: I don't like this
Cylinder Shape Mod: Yes, kind of sweet
Windmill Shape Mod: YES; this one is named for its checkerboard pattern
Fisher Shape Mod: Yes, it's wacky
Axis Shape Mod: Yes, yikes, the wackiest
Pyraminx, Pyramorphix: No
Mastermorphix: Looks like bowling shoes
Kilominx, Megaminx: Just gets scrambled really
Square 1: So glad you asked
We should all know how to checkerboard a SQ1!
4 Horsemen: (1,0)/(-1,-1)/(0,1) (no slash, just resetting)
(3,3) (rotating)
(1,0)/(-1,-1)/(0,1) (no slash, just resetting)
(-3,-3) (rotating)
Altogether, (1,0)/(-1,-1)/(0,1)(3,3)(1,0)/(-1,-1)/(-3,-2)
There must be a million people on earth who would gladly read every word of this great post, besides you and me. I would like to meet one of those people.
ReplyDeleteThank you so much Jack!! The internet tells me that there are over 8 billion people on Earth, so I agree with you. Let's look for one together, and when we find one, we win! I'll start with my dad and if he doesn't work out, I'll keep looking. I have a cousin named Emily I could try next.
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