Zane_C
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This is for anyone wanting a detailed description of TuRBo edges, and also ways to increase efficiency of the method.
UF > UL > UR = R2 U R U R' U' R' U' R' U R'
UF > UR > UL = R U' R U R U R U' R' U' R2
UF > LU > RU = M U M' U2 M U M'
UF > RU > LU = M U' M' U2 M U' M'
UF > LU > UR = U' Rw U R' U' M U R U' R' U
UF > UR > LU = U' R U R' U' M' U R U' Rw' U
UF > RU > UL = U Lw' U' L U M U' L' U L U'
UF > UL > RU = U L' U' L U M' U' L' U Lw U'
T-perm + flip = x' R2 U' R' U R' x F' U' F R U R' U'
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Parity example solves:
Here are two example solves with parity, both using the same scramble.
-Example 1 will be executing the corners first.
-Example 2 will be executing the edges first.
The solving methods will be TuRBo edges and Old Pochmann corners. (No matter how easy the cases are to freestyle)
Bolded stickers represent breaking into a new cycle. (When the buffer has become solved and there are still unsolved pieces).
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Example 1: (corners first)
Scramble in preferred solving orientation: B U' F' U2 B2 F D' F U' F B2 U L D' U2 R L2 B L' R2 D B2 F R2 U'
Corners:
ULB > FUL: - F' D (R U' R' U' R U R' F' R U R' U' R' F R) D' F
ULB > FDR: - R F (R U' R' U' R U R' F' R U R' U' R' F R) F' R'
The buffer is now solved, but this scramble has parity, so ULB and URB need to end up swapped:
ULB > URB: - U (R' U2 R U R' U2' L U' R U L') U'
ULB > DLB: - D F' (R U' R' U' R U R' F' R U R' U' R' F R) F D'
ULB > UFR: - F (R U' R' U' R U R' F' R U R' U' R' F R) F'
ULB > DBR: - R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
ULB > LFD: - D2 R (R U' R' U' R U R' F' R U R' U' R' F R) R' D2
ULB > DBR: - R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
UBL and URB are now swapped. If they were NOT swapped, UB and UL would be swapped instead.
This means your edges would be different from how you memorised them!
Edges:
UF > DF > RU: - D' L2 (U L' U' L U M' U' L' U Lw U') L2 D
UF > DL > RD: - L2 R2 (U L' U' L U M' U' L' U Lw U') R2 L2
UF > DB > UB: - B R' L (M U' M' U2 M U' M') L' R B'
UF > RF > LU: - R (M U' M' U2 M U' M') R'
UF > BR > BL: - L (U' R U R U R U' R' U' R') L'
RB is the last edge target. We must set it up to a PLL case, then swap the two edges and corners simultaneously.
An R-perm would be the easiest to set up:
L' E L (L U2 L' U2 L F' L' U' L U L F L2 U) L' E' L
FL is flipped.
Only an even amount of edges can be flipped. So if we have only memorised one flipped edge, the buffer must also be flipped.
Flip FL and UF however you want, one way would be: L' y' R' U2 R2 U R' U' R' U2 r U R U' r' F
----------------------------------------------------------------
Example 2: (edges first)
Scramble in prefered orientation: B U' F' U2 B2 F D' F U' F B2 U L D' U2 R L2 B L' R2 D B2 F R2 U'
Edges:
UF > DF > RU: - D' L2 (U L' U' L U M' U' L' U Lw U') L2 D
UF > DL > RD: - L2 R2 (U L' U' L U M' U' L' U Lw U') R2 L2
UF > DB > UB: - B R' L (M U' M' U2 M U' M') L' R B'
UF > RF > LU: - R (M U' M' U2 M U' M') R'
UF > BR > BL: - L (U' R U R U R U' R' U' R') L'
RB is the last edge target.
As you know, this solve has parity. And this can be fixed similar to the corners:
The edges need to be finished off with UB and UL swapped. (As it's impossible to solve all the edges while keeping the corners is the same permutation).
We will now solve RB, while also swapping UB and UL, by setting up to a Z-perm.
Dw' R (U R' U' R U' R U R U' R' U R U R2 U' R' U) R' Dw
Flip FL and UF however you want, one way would be: L' y' R' U2 R2 U R' U' R' U2 r U R U' r' F
Corners:
ULB > FUL: - F' D (R U' R' U' R U R' F' R U R' U' R' F R) D' F
ULB > FDR: - R F (R U' R' U' R U R' F' R U R' U' R' F R) F' R'
ULB > URB: - U (R' U2 R U R' U2' L U' R U L') U'
ULB > DLB: - D F' (R U' R' U' R U R' F' R U R' U' R' F R) F D'
ULB > UFR: - F (R U' R' U' R U R' F' R U R' U' R' F R) F'
ULB > URB: - U (R' U2 R U R' U2' L U' R U L') U'
ULB > DBR: - R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
ULB > LFD: - D2 R (R U' R' U' R U R' F' R U R' U' R' F R) R' D2
ULB > DBR: - R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
UF > UL > UR = R2 U R U R' U' R' U' R' U R'
UF > UR > UL = R U' R U R U R U' R' U' R2
UF > LU > RU = M U M' U2 M U M'
UF > RU > LU = M U' M' U2 M U' M'
UF > LU > UR = U' Rw U R' U' M U R U' R' U
UF > UR > LU = U' R U R' U' M' U R U' Rw' U
UF > RU > UL = U Lw' U' L U M U' L' U L U'
UF > UL > RU = U L' U' L U M' U' L' U Lw U'
T-perm + flip = x' R2 U' R' U R' x F' U' F R U R' U'
----------------------------------------------------------------
Parity example solves:
Here are two example solves with parity, both using the same scramble.
-Example 1 will be executing the corners first.
-Example 2 will be executing the edges first.
The solving methods will be TuRBo edges and Old Pochmann corners. (No matter how easy the cases are to freestyle)
Bolded stickers represent breaking into a new cycle. (When the buffer has become solved and there are still unsolved pieces).
----------------------------------------------------------------
Example 1: (corners first)
Scramble in preferred solving orientation: B U' F' U2 B2 F D' F U' F B2 U L D' U2 R L2 B L' R2 D B2 F R2 U'
Corners:
ULB > FUL: - F' D (R U' R' U' R U R' F' R U R' U' R' F R) D' F
ULB > FDR: - R F (R U' R' U' R U R' F' R U R' U' R' F R) F' R'
The buffer is now solved, but this scramble has parity, so ULB and URB need to end up swapped:
ULB > URB: - U (R' U2 R U R' U2' L U' R U L') U'
ULB > DLB: - D F' (R U' R' U' R U R' F' R U R' U' R' F R) F D'
ULB > UFR: - F (R U' R' U' R U R' F' R U R' U' R' F R) F'
ULB > DBR: - R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
ULB > LFD: - D2 R (R U' R' U' R U R' F' R U R' U' R' F R) R' D2
ULB > DBR: - R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
UBL and URB are now swapped. If they were NOT swapped, UB and UL would be swapped instead.
This means your edges would be different from how you memorised them!
Edges:
UF > DF > RU: - D' L2 (U L' U' L U M' U' L' U Lw U') L2 D
UF > DL > RD: - L2 R2 (U L' U' L U M' U' L' U Lw U') R2 L2
UF > DB > UB: - B R' L (M U' M' U2 M U' M') L' R B'
UF > RF > LU: - R (M U' M' U2 M U' M') R'
UF > BR > BL: - L (U' R U R U R U' R' U' R') L'
RB is the last edge target. We must set it up to a PLL case, then swap the two edges and corners simultaneously.
An R-perm would be the easiest to set up:
L' E L (L U2 L' U2 L F' L' U' L U L F L2 U) L' E' L
FL is flipped.
Only an even amount of edges can be flipped. So if we have only memorised one flipped edge, the buffer must also be flipped.
Flip FL and UF however you want, one way would be: L' y' R' U2 R2 U R' U' R' U2 r U R U' r' F
----------------------------------------------------------------
Example 2: (edges first)
Scramble in prefered orientation: B U' F' U2 B2 F D' F U' F B2 U L D' U2 R L2 B L' R2 D B2 F R2 U'
Edges:
UF > DF > RU: - D' L2 (U L' U' L U M' U' L' U Lw U') L2 D
UF > DL > RD: - L2 R2 (U L' U' L U M' U' L' U Lw U') R2 L2
UF > DB > UB: - B R' L (M U' M' U2 M U' M') L' R B'
UF > RF > LU: - R (M U' M' U2 M U' M') R'
UF > BR > BL: - L (U' R U R U R U' R' U' R') L'
RB is the last edge target.
As you know, this solve has parity. And this can be fixed similar to the corners:
The edges need to be finished off with UB and UL swapped. (As it's impossible to solve all the edges while keeping the corners is the same permutation).
We will now solve RB, while also swapping UB and UL, by setting up to a Z-perm.
Dw' R (U R' U' R U' R U R U' R' U R U R2 U' R' U) R' Dw
Flip FL and UF however you want, one way would be: L' y' R' U2 R2 U R' U' R' U2 r U R U' r' F
Corners:
ULB > FUL: - F' D (R U' R' U' R U R' F' R U R' U' R' F R) D' F
ULB > FDR: - R F (R U' R' U' R U R' F' R U R' U' R' F R) F' R'
ULB > URB: - U (R' U2 R U R' U2' L U' R U L') U'
ULB > DLB: - D F' (R U' R' U' R U R' F' R U R' U' R' F R) F D'
ULB > UFR: - F (R U' R' U' R U R' F' R U R' U' R' F R) F'
ULB > URB: - U (R' U2 R U R' U2' L U' R U L') U'
ULB > DBR: - R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
ULB > LFD: - D2 R (R U' R' U' R U R' F' R U R' U' R' F R) R' D2
ULB > DBR: - R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
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