Feryll
Veteran
- Joined
- Dec 31, 2009
- Messages
- 557
After some thinking and research, I have come up with a new way to (hopefully) speedsolve with. Maybe 'new' isn't the best way to phrase it, as it is a big extension off of the very popular Fridrich method. Basically, you use algorithms as you insert f2l pairs to achieve two things at once, sort of similar to MGLS. This method should be an easy read and idea to intermediate and expert cubers.
1. Solve the cross + 2 adjacent f2l pairs - Basically just that. You build the cross and first f2l pair normally, and then solve one more the isn't diagonal from the first solved pair.
2. Build the 3rd f2l pair and place it above it's slot - Basically, have the 3rd pair U R U' R' away from it's solved position. One more thing is that, with the bottom, or single color, of the f2l pair facing you, the empty slot should be on that same face. Example: F2 R' F2 R would be a properly made one. R' F R U' F' Would not (The UF edge and URF corner being the pair). If you make an improper one, you would have to learn twice as many algorithms for the next step. It should also be on the right side, or you'll have twice as many algs for the next step (but at least they'd be mirrors
)
3. Perform an algorithm to orient all edges and place the f2l pair - If I'm correct, there should be 14 algs to learn, being around 7 moves (I haven't taken a real average yet) They are relatively fingertricky, as they use mostly R, U, F moves to do it near optimally. If you don't want to learn the algorithms, this can be done intuitively, but slower (unless you're extremely good at it).
4. Make the last f2l pair, plus an extra corner - Basically, have an extra corner on the f2l pair ready to be inserted. Be sure not to make any F, F', B or B' moves, as that disturbs edge orientation. Example: L' U2 L U2 is an extended f2l pair to be inserted.
5. Insert 1x1x3 block, and solve corners - This can be done with a one-look or two look. I suggest 2 look, as the one look has 81 algorithms (8-9 moves) and the 2 look only has 9. If you know about commutators, you can orient the corners, while permuting them, but we need not worry about permutation in the first look of the 2 look. You can research commutators if you don't already know about them. Then there are 9 permuting algorithms.
6. Permute the last four edges - Remember how we oriented edges in step 3, and solved the corners in step 5? Well, that only leaves a yummy edge permutation
! This will be an average of 10 moves, depending on how long your 2-gen algorithms are, and a big chance to get a skip on this step.
This means that this method solves the cube in 27 algorithms (Or just 13 if you do edge orientation intuitively), or 99 if you choose to go 1-look on step 5 (So I can say my full optimal method requires less than 100 algorithms!), with an average of 30-35 algorithm moves. A variation would be to go Petrus and replace steps 2 and 3 with a block version of the algorithm, not sure how it would work out.
EXAMPLE: Scramble - D2 F' D' R2 B F D2 F' L2 D L' U D' R' U' F2 U2 L' D' R2 L D2 B2 L' U
Step 1 - L' D B' U B' R2 L U' L' U L U L2 U2 L B' U2 B2 U B2 U B
Step 2 - U' L F' L' U2 F
Step 3 - U R B U' B' R'
Step 4 - B U B' U B U2 B' (This is a generated optimal solution; it will be hard to perfect this step)
Step 5a - U R' D R U' R' D' R (Commutator)
Step 5b - U R' U L' U' R U' L U
Step 6 - M' U' M'2 U' M'2 U' M' U2 M'2 U
Please give me your ideas and comments on this
1. Solve the cross + 2 adjacent f2l pairs - Basically just that. You build the cross and first f2l pair normally, and then solve one more the isn't diagonal from the first solved pair.
2. Build the 3rd f2l pair and place it above it's slot - Basically, have the 3rd pair U R U' R' away from it's solved position. One more thing is that, with the bottom, or single color, of the f2l pair facing you, the empty slot should be on that same face. Example: F2 R' F2 R would be a properly made one. R' F R U' F' Would not (The UF edge and URF corner being the pair). If you make an improper one, you would have to learn twice as many algorithms for the next step. It should also be on the right side, or you'll have twice as many algs for the next step (but at least they'd be mirrors
)3. Perform an algorithm to orient all edges and place the f2l pair - If I'm correct, there should be 14 algs to learn, being around 7 moves (I haven't taken a real average yet) They are relatively fingertricky, as they use mostly R, U, F moves to do it near optimally. If you don't want to learn the algorithms, this can be done intuitively, but slower (unless you're extremely good at it).
4. Make the last f2l pair, plus an extra corner - Basically, have an extra corner on the f2l pair ready to be inserted. Be sure not to make any F, F', B or B' moves, as that disturbs edge orientation. Example: L' U2 L U2 is an extended f2l pair to be inserted.
5. Insert 1x1x3 block, and solve corners - This can be done with a one-look or two look. I suggest 2 look, as the one look has 81 algorithms (8-9 moves) and the 2 look only has 9. If you know about commutators, you can orient the corners, while permuting them, but we need not worry about permutation in the first look of the 2 look. You can research commutators if you don't already know about them. Then there are 9 permuting algorithms.
6. Permute the last four edges - Remember how we oriented edges in step 3, and solved the corners in step 5? Well, that only leaves a yummy edge permutation
! This will be an average of 10 moves, depending on how long your 2-gen algorithms are, and a big chance to get a skip on this step.This means that this method solves the cube in 27 algorithms (Or just 13 if you do edge orientation intuitively), or 99 if you choose to go 1-look on step 5 (So I can say my full optimal method requires less than 100 algorithms!), with an average of 30-35 algorithm moves. A variation would be to go Petrus and replace steps 2 and 3 with a block version of the algorithm, not sure how it would work out.
EXAMPLE: Scramble - D2 F' D' R2 B F D2 F' L2 D L' U D' R' U' F2 U2 L' D' R2 L D2 B2 L' U
Step 1 - L' D B' U B' R2 L U' L' U L U L2 U2 L B' U2 B2 U B2 U B
Step 2 - U' L F' L' U2 F
Step 3 - U R B U' B' R'
Step 4 - B U B' U B U2 B' (This is a generated optimal solution; it will be hard to perfect this step)
Step 5a - U R' D R U' R' D' R (Commutator)
Step 5b - U R' U L' U' R U' L U
Step 6 - M' U' M'2 U' M'2 U' M' U2 M'2 U
Please give me your ideas and comments on this
Last edited:
