Bagua
Konrad’s Overview “How to solve the Bagua Cube”
This is a summary of the TP thread, where I had made on overview of the relevant contributions here.
Names and notation
Names:
Synonyms:
Composite edges = edges = 3x3x3 edges
Composite corners (shortcut CC) = envelopes (coined by rline)
Outer edges = edge wings
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
&xnbsp;
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
TP-B1 (Tall Pawn’s Bagua workhorse 1)
U+ R’ L’ D2 R L U-
Shortcut [U+:[L’ R’:D2]]
TP-B2 (Tall Pawn’s Bagua workhorse 2)
U+ L2 R2 U2 R2 L2 U-
Shortcut [U+:[L2 R2:U2]]
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):
Rline’s triangle sequence
R-B1
U+ L’ U- F2 L’ U L F2 U+ L U- =
[[U+:L’]:[F2 L’: U]]
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
&xnbsp;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 a
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
Performed on a solved cube it cycles 1 -> 2 -> 3 -> 1
Restored to full cube shape:
Impure 3-cycle of piece groups:
Sequence b
R2 U+ R2 D- R2 U+ R2 U- R2 U- D+ R2
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 g
The lower two layers stay untouched.
Triangle 4-cycle
Sequence d
3x3x3 turns are very simple setups to do a triangle 4-cycle, like here:
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.
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&xnbsp; 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 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 a
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 – by accident the orange/yellow outer edges are grouped as well - and build the orange/yellow CC, too.
We put the orange kite in the B face via TP-B1 into another edge and flip it to the U face. We pair the orange kites (using TP-B2 R E’ R TP-B2) plug them to the orange face of the orange/yellow edge (TP-B1) outer and plug in the orange yellow CC by TP-B1.
This may sound complicated, but practising the use of TP-B1 and TP-B2 (including flipping of edges) will let you recognize that it is easy enough.
Now is the time for an easy parity check
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 you 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 normalized 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.
Parity can be handled at step 0, if you set the centre orientation (i.e. the direction of the Ying/Yang symbol to a known state on the solved puzzle, e.g. an even number of diagonals. This means that you can check this state by counting the diagonals towards the end of step 0 and do an additional 45° turn, if necessary.
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
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5. Triangles using R-B1 (mostly for centre triangles) and (4-cycle) 3x3x3 centre orientation d
Example after reduction of all less the yellow edges.
Start of this example:
1. Create a similar pattern:
all outer edges yellow on U
Two edge centres flipped
2. pair non-yellow kite pairs by TP-B1 and park them in the F,R,L,B faces (the blue pair is done by a last TP-B1)
&xnbsp;
3. Group outer edges
here: TP-B3 = TP-B1 R E’ R TP-B1
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.