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Konrad’s Overview “How to solve the Bagua Cube”
This is a summary of the TP thread, where I had made an overview of the relevant contributions here.
Names and notation
Names:
Notation:
F B R L U D clockwise 90° turns as on the Rubik’s Cube
F’ B’ R’ L’ U’ D’ counter clockwise 90° turns as on the Rubik’s Cube
F+ B+ R+ L+ U+ D+ clockwise 45° face turns
F- B- R- L- U- D- counter clockwise 45° turns
M slice between the R and L face, clockwise turn viewed from the L face (as in the old WCA notation; in actual WCA notation this is Lw L’)
M’ inverse of M
E slice between the U and D face, clockwise turn viewed from the D face (as in the old WCA notation; in actual WCA notation this is Dw D’)
E’ inverse of E
Synonyms:
Composite edges = edges = 3x3x3 edges
Composite corners (shortcut CC) = envelopes (coined by rline)
Outer edges = edge wings
Outline:
0. Build a normalised cube shape where all outer edges become regular outer edges in a composite edge
1. Reduce 8 composite edges except four that share a single face colour on the solved cube (I’ll use yellow as this last layer, usually on the U face). The sequences TP-B1 and TP-B2 (find details about the two workhorses from Tall Pawn’s method below) are sufficient for this step.
2. Reduce two more composite edges on the U layer using two more commutators
3. Check and fix parity
4. Reduce the last two composite edges using TP-B1, TP-B2a 3-cycle of kite+triangle pairs
Sequences
Sequences
To the left: 2-2 swap of outer edges between two composite edges
TP-B3= TP-B2 F E' F TP-B2
Right picture: 3x3x3 restored
Restored (showing how to setup triangles)
Kite+triangle 3-cycle
Sequence LLL-B1
[[U+ R2 : U+], [R R+ L- : D2]] a [[2:1], [2:1] expands to
U+ R2 U+ R2 U- (R R+) L- D2 L+ (R- R') U+ R2 U- R2 U- (R R+) L- D2 L+ (R- R') (20 turns).
The result performed on a solved cube is
The diagrams to the right show both, the X part and Y part result performed on a solved cube. The overlap of the two is very limited, just a single triangle+kite pair.
So X Y X' Y' provides a pure 3-cycle of three such pairs.
(Please, see the Addendum for full detailed sequence!)
3-cycle
Sequence α (Alpha)
To the left: 3-cycle of piece groups I shall call “large chunks”. A “large junk” consists of 6 pieces adjacent to each other. A “small junk” is the compliment of 3 pieces in a composite edge:
B B+ F- L2 F+ B+ U F’ U’ F B- F- L2 F+ B- B’
The left part of the picture shows the result of such sequence for the U layer. At your right I show it as a restored 3x3x3 Cube. Just 3 large junks are permuted within three composite edges as the arrows indicate.
Written as a conjugate:
[B2:[[B- F-:L2]:[U,F’]]]
B2 is the setup for permuting composite edges on the U layer
Mirrored: [B2:[[B+ F+:R2]:[U',F]]]
Impure 3-cycle of piece groups:
Sequence β (Beta)
R2 U+ R2 D- R2 U+ R2 U- R2 U- D+ R2
Performed on a solved cube it cycles 1 -> 2 -> 3 -> 1
Restored to full cube shape:
The lower two layers stay untouched.
Old Cuboid sequence!
Originally, I had used it a lot. Now, I’m using this just in my parity fix (see below).
Composite Corner 3-cycle:
Sequence γ (Gamma)
U+ R' D' R U' R' D R U+
Originally, I had used this a lot, too. Now, I’m using this just in my parity fix (see below).
Keeping the cube shape helps to avoid errors.
Triangle 4-cycle
Sequence δ (Delta)
3x3x3 turns are very simple setups to do a triangle 4-cycle, like here:
A 45° turn of U (e.g. U+) makes the 4 triangles part of the 3x3x3 centre.
The sequence U+ M E M' U M E' M' U' U-
takes care of all four.
Note: M turns like L (The slice under L) and E like D. (old WCA notation)
Stepwise Solution
0. Build a normalised cube shape where all outer edges become regular outer edges in a composite edge
I do this intuitively. I Use my sequence b in some cases.
1. Reduce 8 composite edges except those on one last face
Currently, I leave four that share a single face colour on the solved cube The sequences TP-B1 and TP-B2 (find details about the two workhorses from Tall Pawn’s method above) are sufficient for this step.
Usually, I build two small junks first + the composite corner. Using TP-B1 I build a large junk and plug this together by TP-B2 with the small junk.
Example:
5 edges are reduced, already. We leave all edges with yellow for the endgame.
We do the green/white edge next. Triangles do not matter yet. We see some pieces grouped already.
By TP-B2 mirrored we build the white/green composite corner (CC) next.
Next we build a white/green large junk. Green colour on the U face.
After simple setup turns TP-B1 builds it.
Setup for TP-B2 to reduce the edge completely.
Done.
In this style we reduce the remaining two edges of the lower two layers of the 3x3x3.
The beauty of Tall-Pawn’s method is that everything relevant happens in “full daylight” on the U layer. The two workhorses TP-B1 and TP-B2 are very easy to understand.
2. Reduce two more composite edges on the U layer using TP-B1, TP-B2 and/or the 3-cycle α
Example:
The four yellow edges to go.
We build two green/yellow small junks and the green/yellow CC.
TP-B2 will group the two green/yellow small junks
Green / yellow edge is composed.
Green / yellow edge is composed.
The orange/yellow edge shall be the next.
The orange/yellow CC is built.
The two orange kites are paired by TP-B3.
(Since the workspace is now limited, TP-B1 is no longer sufficient.)
(Since the workspace is now limited, TP-B1 is no longer sufficient.)
After TP-B3
After TP-B3
With the two yellow kites at B we do TP-B3 again.
TP-B1 builds the orange/yellow edge.
TP-B1 builds the orange/yellow edge.
This may sound complicated, but practising the use of TP-B1 and TP-B2 (including flipping of edges) will let you recognise that it is easy enough.
Now is the time for an easy parity check
In this specific case a 3-cycle of outer edges is required.
No parity fix necessary!
No parity fix necessary!
3. Check and fix parity.
Any 45° turn on this puzzle flips the parity of the outer edges between “odd” and even”. Such turn is an 8-cycle of outer edges and this causes the parity flipping. (Please, remember that a cycle of an even number cycle of pieces within one piece group means that an odd number of single swaps is needed to put the pieces back to their original position.
E.g. the four elements 1234 are cycled:
1234 -> 2341 (4-cycle)
2341 -> 3241 (single swap of 2 and 3)
3241 -> 1243 (single swap of 3 and 1)
1243 -> 1234 (single swap of 4 and 3)
Because the Ying/Yang symbols on the centre caps show orientation (either diagonal or straight), you can use the little trick to set the centre orientation on your solved cube to a known pattern. (E.g. all straight, pointing from edge to edge).
If you count in step 0 an even number of straight and diagonal oriented centres, this means “no parity”. Otherwise do a single 45° degree turn and put the cube back to its normalised form, carefully watching that the number of 45° turns stays even. Because at step 0 no edges are reduced yet, this is the least effort.
The picture below shows a situation at the end of step 0.
I had counted the diagonally oriented centres looking at the Ying/Yang symbols. I had seen three, made a single 45° turn and now I have got two (orange and yellow). BTW, the easiest way creating a normalised cube is using my sequence b inverse = U2 U+ D-.R2 U+ R2 U- R2 D+ R2 U- R2 (12).
If you consider this little trick somehow as cheating, you can check and fix parity by 22 turns as shown in my diagram below.
4. Reduce the last two composite edges using TP-B1, TP-B2, TP-B3 and LLL-B2 (pure 3-cycle of kite+triangle pairs)
R E’ R TP-B1 R E’ R TP-B1 groups the outer edges of the last two edges.
We put the edge centres to their position via TP-B1 and cycle the kites home using LLL-B1.
In this example, TP-B1, TP-B2 and LLL-B1 were sufficient. No 3-cycle of large junks needed using sequence
5. Triangles using R-B1 (mostly for centre triangles) and (4-cycle) 3x3x3 centre orientation δ
See Addendum d) for Tall Pawn's efficient method!
You can do the centre triangles after step 0. Doing them at the very end is just a no-brainer.
Example after reduction of all less the yellow edges.
Start of this example:
1. Create a similar pattern:
all outer edges yellow on U
Even number of edge centres flipped
2. pair non-yellow kite pairs by TP-B2 and park them using TP-B1 in the F,R,L,B faces (the blue pair is done by a last TP-B2)
3. Group outer edges
here: TP-B3 = TP-B2 R E’ R TP-B2
just have two yellow kite pairs at edge UR!
4. All kite pairs built, all outer edges grouped
5. Use TP-B1 to put kite pairs to their position, build CC and plug them into edges. In the case below use
(U+ R’ L’ D2 L R U- R E’ R)x6 (60) to flip the two edge centres at UR and UL
b) Sequence LLL-B1 in detail
c) Explanation of the name Composite Corners
By the following picture I want to explain how I came up with the name "Composite Corner"
If you view the Bagua as a puzzle with 12 piece groups I call Composite Corners (CC) + 8 regular corners and you use 45° degree turns to permute CC with other CC and regular corners, only, you can scramble it very differently from a regular 3x3x3. Still, you never disrupt pairs of outer edges and kite pairs or the centre triangles or the edge triangles. If you permute the puzzle in such way, it can easily be solved by putting the CC back between the outer edges where they belong. It may be not a bad exercise to begin with.
Tall Pawn
Post subject: Re: How to solve the Dayan Bagua Cube
Posted: Tue Nov 29, 2016 3:35 pm
I'm down to one sequence for the small triangles now and It's super efficient. F+ M S M' F-
I do all the centers first with the exception of one then I use the triangles on that remaining one to place the edge group ones. Cycling back and forth between 2 centers. At the end I will be left with either everything reduced or a case that requires the use of already reduced edges of the same color.
Doing it this way I can place 4 at a time throughout most of the reduction with very easy setup moves. The only exception is when identical colors are involved.
_________________
Got my fastest time ever with this scramble! LBFDUF2UD2RF2UFU'R'F2D'BRLB2R
Joined: Wed Mar 05, 2014 7:22 pm
Location: Pennsylvania
Tall Pawn wrote:
...
I'd argue the safeness as far as the centers are concerned at least if the method I use is followed. After one iteration of this sequence, a cube rotation of z2, and your setup moves to place more edge groups on the F face to be reduced doing that sequence again will fix the center that was messed up during the first one. The sequence itself isn't center safe but the method is.....
Interesting idea!
It may need some training to use this as an effective method.
At first glance a lot of triangles are permuted (12). The following picture wants to explain, how the triangles travel around:
I start with a fully reduced cube. I have numbered the blue centre triangles clockwise by 1 2 3 4, the orange centre triangles by 5 6 7 8 and the four edge triangles on the F face a b c d. You may need to click onto my picture to enlarge the little numbers.
(In old WCA notation M turns like L and S like F.)
The last two diagrams at the bottom show how the start configuration is changed when I do a full commutator [F+, M S M'].
To make this a method the "z2 + do it again" hint is essential.
EDIT: I want to add this comment:
This is a good example of innovative thinking. Tall Pawn hasn't stopped after hundred of solves to look for improvements.
He came up with a simple [1:[1:1]] conjugate that seems to permute too much at once to be useful, but he could embed this into a very effective method.
Simplicity = pure beauty, as with his workhorses TP-B1 and TP-B2!!!
TP post December 3rd, 2016:
About my last experiences with triangles:
Tall Pawn wrote:
I'm down to one sequence for the small triangles now and It's super efficient. F+ M S M' F-
I do all the centers first with the exception of one then I use the triangles on that remaining one to place the edge group ones. Cycling back and forth between 2 centers. At the end I will be left with either everything reduced or a case that requires the use of already reduced edges of the same color.
Doing it this way I can place 4 at a time throughout most of the reduction with very easy setup moves. The only exception is when identical colors are involved.
I have used this method several times and made an experiment to compare it with the method I'm using in a pCubes example.
I started with 20 unsolved edge triangles and used Tall Pawn's method here (I start with a first setup):
Here is a video that starts with the same pattern. The setups are a bit different, though.
I started from the same configuration again (the start setup is different, because the method is different) using my own method:
If you count the number of turns contained in the "core" sequences (8 times 5 turns in F+ M S M' F- z2); 4 times 9 turns in U+ M E M' U M E' M' (U' U-) and once 7 turns for a 2-2 swap), you see 40 turns in total on the first picture and 43 in total on the second.
At the same time I needed 8 times setups of edges on the F face (including the start setup) in the first case and 5 times setups on the U face in the second case.
I have not counted all setup turns, but I'm pretty confident that the total count of turns is a bit in favour of my own method.
(Probably, somebody more used to Tall Pawn's method could have done the setups cleverer.)
Personally, I find setups of edges on a single face easier and I do'nt have to care about anything else (another face centre), at all.
Note my little innovation
in my last sequence: U+ M E' M2 E' M U- (7 turns) does a 2-2 swap of edge centres.
So, 7 turns instead of 13 ((U+ L' R' U2 R L U' L' R' U2 R L (U' U-)) or 9 ( U+ M E M' U2 M E' M' (U2 U-))
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