Note: I had a proper post on reddit. I wanted a neat one here too. I know this is not a groundbreaking method. Whether it's useful to speedsolving or not I just want a proper post of my method here. The original thread wasn't so nice. I just want a proper post of my method here. TL;DR I want a neater post of my method here.

You all know 1LLL right? How many alg is it? 3915 algs just to do LL in one step. CT is a LL skip method with 200 algs. M-CELL is a 96 algorithm method that solves LSLL with 2-algs with I recognition.

All these methods solve LSLL in 2 steps: Last Slot and Last Layer.

What if we did it in one step. 1 alg and 1 recognition. That's crazy it has too many algs. What if I told you can do it an with even less algs than full 1LLL? It's not that much but a decrease is a decrease. My method solves both the last slot and last layer in one step. A true LL skip method and a 5LS method. (5-Looks Solve)

Steps:

1. EOLine or EO 3/4 Cross

Nothing new here.

2. Get to F2L-2

Just blockbuild as usual. Nothing new. ANY slots can be open. It does not matter which ones are as lonh there are 2. The reason I mentioned EO 3/4 Cross above is because I advise against building the FB on that cross edge or else it will add another look on adding the 4th cross piece down to get to F2L-2. But like any rule it can be broken because we all know ergonomics and/or efficiency will always be nice.

3. My Last/3rd Slot

This is to mock how we solve LSLL before. That's why I named it My Last Slot as well.

Choose any open slot to solve and while doing so solve 3 LL edges relative to each other. This step can be intuitive if you know many ways to solve F2L or just 2-look it by making a pair and solve the LL edges using the 24 EP solving algs.

The 1-look version is 791 algs. Because of y2 symmetry there're 3 F2L-2 cases. Diag Slots and the 2 Adj Slots cases. With these there are only 60 F2L cases. 36 of which have the edge in meaning you only need 6 algs for those cases and the rest the usual 24 algs. -1 the solved cases to get 791

I advise just mirroring/inversing/mirror-inversing a alg for 1-slot than to rotation or just learn an alg for that slot if you think it's better. (If you're slot neutral which I advise you do. Treat it like TSLE)

The algs I assume is 2-gen unlike solving CP requires 3-gen sometimes. All of the algs are just F2L algs done differently so the learning curve will be alright.

4. Finale/My Last Layer

Time to solve LL.

Because the EP is solved only the corners can move. As usual 6 CP cases no matter what the F2L case you get. There 7 F2L cases (permutation). Solved, Corner/Edge out, and the 4 cases when they're both at the top layer. So permutation is 6 x 7 or 42. 81 CO cases.

42 x 81 will not spit out the unique # of cases since the solve F2L cases still have symmetry.

81 CO cases is reduced to the unique 24. 3 of which have symmetry and another 3 that has 1/2 symmetry. If symmetrical then CP is reduced to 3 and to 4 if 1/2 symmetrical.

((3 x 3) + (3 x 4) + (18 x 6)) - 1 = 128

The remaining 6 cases can't be reduced further.

6 CP x 6 F2L x 81 CO = 2916

Total # of LSLL = 3044.

Total # of algs = 3835

# of looks of a typical solve:

1 EOLine

2 F2L-2

1 My 3rd Slot

1 Finale

Total # of looks = 5

Example Solve: White Top Green Front

Scramble: F R2 B2 R2 D U R2 D' L2 B2 U B2 L2 R B D L B' L2 B' U2

EO 3/4 Cross: B D R' U2 F

First Square: L U L2

First Block: R' U R U2 L

My Last Slot: U' R' U' R U' R' U' R

LSLL: R2 U2 R' U' R U2 L' U R' U' M'

You all know 1LLL right? How many alg is it? 3915 algs just to do LL in one step. CT is a LL skip method with 200 algs. M-CELL is a 96 algorithm method that solves LSLL with 2-algs with I recognition.

All these methods solve LSLL in 2 steps: Last Slot and Last Layer.

What if we did it in one step. 1 alg and 1 recognition. That's crazy it has too many algs. What if I told you can do it an with even less algs than full 1LLL? It's not that much but a decrease is a decrease. My method solves both the last slot and last layer in one step. A true LL skip method and a 5LS method. (5-Looks Solve)

Steps:

1. EOLine or EO 3/4 Cross

Nothing new here.

2. Get to F2L-2

Just blockbuild as usual. Nothing new. ANY slots can be open. It does not matter which ones are as lonh there are 2. The reason I mentioned EO 3/4 Cross above is because I advise against building the FB on that cross edge or else it will add another look on adding the 4th cross piece down to get to F2L-2. But like any rule it can be broken because we all know ergonomics and/or efficiency will always be nice.

3. My Last/3rd Slot

This is to mock how we solve LSLL before. That's why I named it My Last Slot as well.

Choose any open slot to solve and while doing so solve 3 LL edges relative to each other. This step can be intuitive if you know many ways to solve F2L or just 2-look it by making a pair and solve the LL edges using the 24 EP solving algs.

The 1-look version is 791 algs. Because of y2 symmetry there're 3 F2L-2 cases. Diag Slots and the 2 Adj Slots cases. With these there are only 60 F2L cases. 36 of which have the edge in meaning you only need 6 algs for those cases and the rest the usual 24 algs. -1 the solved cases to get 791

I advise just mirroring/inversing/mirror-inversing a alg for 1-slot than to rotation or just learn an alg for that slot if you think it's better. (If you're slot neutral which I advise you do. Treat it like TSLE)

The algs I assume is 2-gen unlike solving CP requires 3-gen sometimes. All of the algs are just F2L algs done differently so the learning curve will be alright.

4. Finale/My Last Layer

Time to solve LL.

Because the EP is solved only the corners can move. As usual 6 CP cases no matter what the F2L case you get. There 7 F2L cases (permutation). Solved, Corner/Edge out, and the 4 cases when they're both at the top layer. So permutation is 6 x 7 or 42. 81 CO cases.

42 x 81 will not spit out the unique # of cases since the solve F2L cases still have symmetry.

81 CO cases is reduced to the unique 24. 3 of which have symmetry and another 3 that has 1/2 symmetry. If symmetrical then CP is reduced to 3 and to 4 if 1/2 symmetrical.

((3 x 3) + (3 x 4) + (18 x 6)) - 1 = 128

The remaining 6 cases can't be reduced further.

6 CP x 6 F2L x 81 CO = 2916

Total # of LSLL = 3044.

Total # of algs = 3835

# of looks of a typical solve:

1 EOLine

2 F2L-2

1 My 3rd Slot

1 Finale

Total # of looks = 5

Example Solve: White Top Green Front

Scramble: F R2 B2 R2 D U R2 D' L2 B2 U B2 L2 R B D L B' L2 B' U2

EO 3/4 Cross: B D R' U2 F

First Square: L U L2

First Block: R' U R U2 L

My Last Slot: U' R' U' R U' R' U' R

LSLL: R2 U2 R' U' R U2 L' U R' U' M'

Last edited: May 10, 2017