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Firstly, are you interested in solving the big cubes blindfolded, yet have trouble seeing the commutators?
http://tinyurl.com/pft5d
This is Chris' thread, it is very useful. Not comfy with using commutators for the wing edges still? Here is my solution to solve the wings, it is very easy, fast and is a stepping stone to using full blown commutators.
What I use is called a helper buffer method. It is almost like Pochmann's 2cycle method for the 3x3x3, however you are cycling 3 edges back and forth.
Buffer>Target>Helper This is done w/ an ABA'B' commutator
Helper>Target>Buffer This is done w/ an BAB'A' commutator
There are three special cases that we will encounter:
Starting a new cycle
Solving the helper edge
Solving the OLL Parity
Let's establish the Helper, Buffer, and Targets (notice plural, makes setups easy, sorta like w/ 2cycling the 3x3x3 blindfolded)
Buffer FRu
Helper UFr
Targets the other 6 edges on the u and r slices.
FDr, DBr, BUr, LFu, BLu, RBu
The algs:
We use these two commutators
A: RU'R'; B:
A: [r]; B: U'RU
let = u, u2, or u'  likewise [r] = r, r2, or r'
You alternate between ABA'B' commutators and BAB'A' commutators.
AB commutators: RU'R'u2RUR'u2; RU'R'u'RUR'u; rU'RUr'U'R'U
BA commutators: uRU'R'u'RUR'; U'RUr2U'R'Ur2 ...
so on and so forth
You do an AB commutator to solve the first wing, then you proceed to solve the piece after that, which was cycled to the Helper's location. Use the proper setup + a BA commutator to solve that piece, you alternate back and forth.
Special Case 1:
Starting a new cycle: What happens, the Buffer is brought into the helper buffer area before the other 23 wings are solve. You must start a new cycle. Simply pick a new location and do the setup to solve that piece. The flow continues as normal, the algorithm used to start the new cycle should be the inverse of the last one used.
Special Case 2:
Solving the Helper: This is what I call integration. You want to solve the helper and the piece paired with it.
Starting with 1, every other piece is the start of a new pair. 1,2 is a pair; 3,4 is a pair; 5,6 is a pair etc.
Odd numbers are solved w/ the AB commutator, Even Numbers are solved w/ the BA commutator. The rule of thumb the pair w/ the helper: The helper is integrated and the other wing is solved in one alg, using the commutator that would solve the target wing in a normal pair.
Solve this pair: The UFr and the LBd
LBd is the second wing in the pair, so a BA commutator would be used. I would solve that pair w/ (Ll)2 u'RU'R'uRUR' (Ll)2
Solve this pair: RDb and the UFr
RDb is the first wing in the pair, so an AB commutator would be used. I would solve that pair w/ (Dd)' rU'RUr'U'R'U (Dd)
Continuing with the solve after integration:
Ok there is nothing special about this integration thing (save for it's 1 alg to solve the pair rather than 2) The net result is you solved another pair, so you continue solving the next pair as you normally would. The first wing of the pair using an AB commutator; The second wing using a BA commutator. I did not realize this until recently, that there is no other significance to integrating the helper.
Special Case 3:
Solving the Orientation Parity: How do you determine if there is a parity? The answer, the last wing is unpaired, in other words, two solve the other 22 wings (everything but the helper and buffer) you'll require an odd number of algorithms. Starting a new cycle counts as an algorithm solving a piece counts as an algorithm, integration counts as two algorithms, because the net effect is solving two pieces. Using a memorization system that pairs locations, such as Letter Pairs or a Person Action system allows you to easily determine if there is a parity or not.
So, you've solved all of the other 22 wings. Now the Helper and Buffer are all that's left to be solved. We don't have an algorithm to directly swap the UFr and FRu. However, with a simple commutator, we can simply use a well known algorithm.
r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2
Can you see the setup + commutator so that you can perform that algorithm with no cube rotations?










(Ll)'(Bb)' RU'R'uRUR'u' (Bb)(Ll)
I would like to make a disclaimer. I've shown setup moves being done with double slice turns. I feel that double slice turns are safer for avoiding lockups. Since Commutators have no affect on other piece types, a double slice turn is fine.
http://tinyurl.com/pft5d
This is Chris' thread, it is very useful. Not comfy with using commutators for the wing edges still? Here is my solution to solve the wings, it is very easy, fast and is a stepping stone to using full blown commutators.
What I use is called a helper buffer method. It is almost like Pochmann's 2cycle method for the 3x3x3, however you are cycling 3 edges back and forth.
Buffer>Target>Helper This is done w/ an ABA'B' commutator
Helper>Target>Buffer This is done w/ an BAB'A' commutator
There are three special cases that we will encounter:
Starting a new cycle
Solving the helper edge
Solving the OLL Parity
Let's establish the Helper, Buffer, and Targets (notice plural, makes setups easy, sorta like w/ 2cycling the 3x3x3 blindfolded)
Buffer FRu
Helper UFr
Targets the other 6 edges on the u and r slices.
FDr, DBr, BUr, LFu, BLu, RBu
The algs:
We use these two commutators
A: RU'R'; B:
A: [r]; B: U'RU
let = u, u2, or u'  likewise [r] = r, r2, or r'
You alternate between ABA'B' commutators and BAB'A' commutators.
AB commutators: RU'R'u2RUR'u2; RU'R'u'RUR'u; rU'RUr'U'R'U
BA commutators: uRU'R'u'RUR'; U'RUr2U'R'Ur2 ...
so on and so forth
You do an AB commutator to solve the first wing, then you proceed to solve the piece after that, which was cycled to the Helper's location. Use the proper setup + a BA commutator to solve that piece, you alternate back and forth.
Special Case 1:
Starting a new cycle: What happens, the Buffer is brought into the helper buffer area before the other 23 wings are solve. You must start a new cycle. Simply pick a new location and do the setup to solve that piece. The flow continues as normal, the algorithm used to start the new cycle should be the inverse of the last one used.
Special Case 2:
Solving the Helper: This is what I call integration. You want to solve the helper and the piece paired with it.
Starting with 1, every other piece is the start of a new pair. 1,2 is a pair; 3,4 is a pair; 5,6 is a pair etc.
Odd numbers are solved w/ the AB commutator, Even Numbers are solved w/ the BA commutator. The rule of thumb the pair w/ the helper: The helper is integrated and the other wing is solved in one alg, using the commutator that would solve the target wing in a normal pair.
Solve this pair: The UFr and the LBd
LBd is the second wing in the pair, so a BA commutator would be used. I would solve that pair w/ (Ll)2 u'RU'R'uRUR' (Ll)2
Solve this pair: RDb and the UFr
RDb is the first wing in the pair, so an AB commutator would be used. I would solve that pair w/ (Dd)' rU'RUr'U'R'U (Dd)
Continuing with the solve after integration:
Ok there is nothing special about this integration thing (save for it's 1 alg to solve the pair rather than 2) The net result is you solved another pair, so you continue solving the next pair as you normally would. The first wing of the pair using an AB commutator; The second wing using a BA commutator. I did not realize this until recently, that there is no other significance to integrating the helper.
Special Case 3:
Solving the Orientation Parity: How do you determine if there is a parity? The answer, the last wing is unpaired, in other words, two solve the other 22 wings (everything but the helper and buffer) you'll require an odd number of algorithms. Starting a new cycle counts as an algorithm solving a piece counts as an algorithm, integration counts as two algorithms, because the net effect is solving two pieces. Using a memorization system that pairs locations, such as Letter Pairs or a Person Action system allows you to easily determine if there is a parity or not.
So, you've solved all of the other 22 wings. Now the Helper and Buffer are all that's left to be solved. We don't have an algorithm to directly swap the UFr and FRu. However, with a simple commutator, we can simply use a well known algorithm.
r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2
Can you see the setup + commutator so that you can perform that algorithm with no cube rotations?










(Ll)'(Bb)' RU'R'uRUR'u' (Bb)(Ll)
I would like to make a disclaimer. I've shown setup moves being done with double slice turns. I feel that double slice turns are safer for avoiding lockups. Since Commutators have no affect on other piece types, a double slice turn is fine.
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