Anuar
Member
If you don't know what OPA is, you should watch this video first.
Also, watch this from 7:20 to 8:34. I use the same letter scheme and orientation shown in the video. My buffer is I, but you can choose any buffer you want. It's just the place where you begin the first cycle.
One more thing you need to know is how to differentiate between good and bad wings. This is very similar to ZZ edge orientation. The good wings are A B C D J L R T U V W X. The bad wings are E F G H I K M N O P Q S.
With this method, instead of tracing wings to solve them, you want to trace them to the place next to the other wing with the same colors. For example: A goes next to Q, B goes next to M, C goes next to I, and so on.
You also have to take good and bad edges and the location they are going to into consideration. We're going to assign a value for each case, and that value will be added to the parity value for each traced wing. The parity value always starts at 0.
A good wing going to a good location is 1.
A bad wing going to a bad location is 1.
A good wing going to a bad location is 0.
A bad wing going to a good location is 0.
For example: A (good) going to F (bad) is 0; K (bad) going to E (bad) is 1.
There are three special cases that might happen:
1. If there is a solved edge, you check its orientation. If it's a good edge, you don't need to anything. If it's a bad edge, add 1 to the parity value.
2. If you pick up the cube and realize that your buffer is solved in relation to the wing next to it, just rotate the cube to another orientation where the buffer isn't solved.
3. If you finish the first cycle but there are some unsolved edges you didn't trace, that's a cycle break. Choose any of those edges and start a new cycle there. For every cycle break, add 1 to the parity value.
Finally, let's apply that to the solve. In 4x4, the parity doesn't change after you've solved two adjacent centers. If your parity value is even, you must do an odd number of wide moves until two adjacent centers are solved. If your parity value is odd, you must do an even number of wide moves until two adjacent centers are solved. To see techniques to solve the first centers, watch the first video I mentioned. Since most good methods build 2 opposite centers first, you usually have to count the wide moves until you solve the first three centers.
I've been using this method for about a week, and my inspection is almost sub-15. For me, it's a lot faster than 4BLD tracing. If you have any questions, I'll be happy to answer.
Also, watch this from 7:20 to 8:34. I use the same letter scheme and orientation shown in the video. My buffer is I, but you can choose any buffer you want. It's just the place where you begin the first cycle.
One more thing you need to know is how to differentiate between good and bad wings. This is very similar to ZZ edge orientation. The good wings are A B C D J L R T U V W X. The bad wings are E F G H I K M N O P Q S.
With this method, instead of tracing wings to solve them, you want to trace them to the place next to the other wing with the same colors. For example: A goes next to Q, B goes next to M, C goes next to I, and so on.
You also have to take good and bad edges and the location they are going to into consideration. We're going to assign a value for each case, and that value will be added to the parity value for each traced wing. The parity value always starts at 0.
A good wing going to a good location is 1.
A bad wing going to a bad location is 1.
A good wing going to a bad location is 0.
A bad wing going to a good location is 0.
For example: A (good) going to F (bad) is 0; K (bad) going to E (bad) is 1.
There are three special cases that might happen:
1. If there is a solved edge, you check its orientation. If it's a good edge, you don't need to anything. If it's a bad edge, add 1 to the parity value.
2. If you pick up the cube and realize that your buffer is solved in relation to the wing next to it, just rotate the cube to another orientation where the buffer isn't solved.
3. If you finish the first cycle but there are some unsolved edges you didn't trace, that's a cycle break. Choose any of those edges and start a new cycle there. For every cycle break, add 1 to the parity value.
Finally, let's apply that to the solve. In 4x4, the parity doesn't change after you've solved two adjacent centers. If your parity value is even, you must do an odd number of wide moves until two adjacent centers are solved. If your parity value is odd, you must do an even number of wide moves until two adjacent centers are solved. To see techniques to solve the first centers, watch the first video I mentioned. Since most good methods build 2 opposite centers first, you usually have to count the wide moves until you solve the first three centers.
Scramble:
U2 L' F2 B R F2 D' R2 D2 B U2 F U2 D2 L2 F' B' U2 F2 R' F Fw2 Uw2 B R' Uw2 R' D2 R2 Uw2 U2 B F' D B' D2 B' Uw B2 Fw Uw D2 Rw Uw' Rw' Uw' F2
The wing on I is S (bad), and it goes to N (bad). Parity value = 1
The wing on N is T (good), and it goes to U (good). Parity value = 2
The wing on U is B (good), and it goes to H (bad). Parity value = 2
The wing on H is P (bad), and it goes to S (bad). Parity value = 3
The wing on S is D (good), and it goes to G (bad). Parity value = 3
The wing on G is X (good), and it goes to M (bad). Parity value = 3
The wing on M is F (bad), and it goes to Q (bad). Parity value = 4
The wing on Q is Q (bad), and it goes to E (bad). Parity value = 5
The wing on E is K (bad), and it goes to O (bad). Parity value = 6
The wing on O is V (good), and it goes to P (bad). Parity value = 6
The wing on P is C (good), and it goes to L (good). Parity value = 7
The wing on L is H (bad), and it goes to I (bad). Parity value = 8
That ends the first cycle, because H is going back to where the cycle begun (I). All edges have been verified, so there's no new cycle.
Parity value = 8, so you must do an odd number of wide moves to solve two adjacent centers. This is one way you could do it (building the opposite centers first):
z y'
F' B2 Rw U Rw B' Rw
y' U' Rw U2 Rw'
z x U D' Rw U' F' Rw
U2 L' F2 B R F2 D' R2 D2 B U2 F U2 D2 L2 F' B' U2 F2 R' F Fw2 Uw2 B R' Uw2 R' D2 R2 Uw2 U2 B F' D B' D2 B' Uw B2 Fw Uw D2 Rw Uw' Rw' Uw' F2
The wing on I is S (bad), and it goes to N (bad). Parity value = 1
The wing on N is T (good), and it goes to U (good). Parity value = 2
The wing on U is B (good), and it goes to H (bad). Parity value = 2
The wing on H is P (bad), and it goes to S (bad). Parity value = 3
The wing on S is D (good), and it goes to G (bad). Parity value = 3
The wing on G is X (good), and it goes to M (bad). Parity value = 3
The wing on M is F (bad), and it goes to Q (bad). Parity value = 4
The wing on Q is Q (bad), and it goes to E (bad). Parity value = 5
The wing on E is K (bad), and it goes to O (bad). Parity value = 6
The wing on O is V (good), and it goes to P (bad). Parity value = 6
The wing on P is C (good), and it goes to L (good). Parity value = 7
The wing on L is H (bad), and it goes to I (bad). Parity value = 8
That ends the first cycle, because H is going back to where the cycle begun (I). All edges have been verified, so there's no new cycle.
Parity value = 8, so you must do an odd number of wide moves to solve two adjacent centers. This is one way you could do it (building the opposite centers first):
z y'
F' B2 Rw U Rw B' Rw
y' U' Rw U2 Rw'
z x U D' Rw U' F' Rw
Scramble:
R2 U B2 R2 D2 R2 U' B2 R2 F' D' L D' F D2 L' D2 L2 B' Uw2 F L Fw2 B R2 Uw2 B' L B R2 Uw B2 D' R2 L Uw' L Rw Fw Rw' Fw Uw2 D'
In this scramble, my buffer (I) is solved in relation to the wing next to it. I'm going to do a y rotation before I start tracing, so that my buffer isn't solved.
There is one solved misoriented edge. Parity value = 1.
The wing on I is T (good), and it goes to O (bad). Parity value = 1
The wing on O is R (good), and it goes to I (bad). Parity value = 1
That ends the first cycle, so a new cycle begins on A. Parity value = 2
The wing on A is K (bad), and it goes to S (bad). Parity value = 3
The wing on S is A (good), and it goes to M (bad). Parity value = 3
The wing on M is L (good), and it goes to R (good). Parity value = 4
The wing on R is D (good), and it goes to X (good). Parity value = 5
The wing on X is C (good), and it goes to J (good). Parity value = 6
The wing on J is W (good), and it goes to L (good). Parity value = 7
The wing on L is P (bad), and it goes to U (good). Parity value = 7
The wing on U is B (good), and it goes to T (good). Parity value = 8
The wing on T is V (good), and it goes to A (good). Parity value = 9
That ends the second cycle. All edges have been verified, so there's no new cycle.
Parity value = 9, so you must do an even number of wide moves to solve two adjacent centers. This is one way you could do it (building the opposite centers first):
F' U2 Lw F Lw
y F2 Rw U Rw'
y' z F' U Rw2 U' Rw U2 Rw'
R2 U B2 R2 D2 R2 U' B2 R2 F' D' L D' F D2 L' D2 L2 B' Uw2 F L Fw2 B R2 Uw2 B' L B R2 Uw B2 D' R2 L Uw' L Rw Fw Rw' Fw Uw2 D'
In this scramble, my buffer (I) is solved in relation to the wing next to it. I'm going to do a y rotation before I start tracing, so that my buffer isn't solved.
There is one solved misoriented edge. Parity value = 1.
The wing on I is T (good), and it goes to O (bad). Parity value = 1
The wing on O is R (good), and it goes to I (bad). Parity value = 1
That ends the first cycle, so a new cycle begins on A. Parity value = 2
The wing on A is K (bad), and it goes to S (bad). Parity value = 3
The wing on S is A (good), and it goes to M (bad). Parity value = 3
The wing on M is L (good), and it goes to R (good). Parity value = 4
The wing on R is D (good), and it goes to X (good). Parity value = 5
The wing on X is C (good), and it goes to J (good). Parity value = 6
The wing on J is W (good), and it goes to L (good). Parity value = 7
The wing on L is P (bad), and it goes to U (good). Parity value = 7
The wing on U is B (good), and it goes to T (good). Parity value = 8
The wing on T is V (good), and it goes to A (good). Parity value = 9
That ends the second cycle. All edges have been verified, so there's no new cycle.
Parity value = 9, so you must do an even number of wide moves to solve two adjacent centers. This is one way you could do it (building the opposite centers first):
F' U2 Lw F Lw
y F2 Rw U Rw'
y' z F' U Rw2 U' Rw U2 Rw'
I've been using this method for about a week, and my inspection is almost sub-15. For me, it's a lot faster than 4BLD tracing. If you have any questions, I'll be happy to answer.