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Jun 29, 2019
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My friend taught me this method and I'm not sure if it's original or not and he only taught me the beginner version so I'll try to write as much as I remember of his advanced version

The steps would be
3/4 Cross
F2L minus the edge
CMLL
L5E

The bottom face's center can be on top or bottom depending on the pieces and basically make a line and one white edge and fix the line and edges and then M2 to put it at the bottom or simply do it like a normal cross but only 3 edges

F2L can be done like normal except that face without the edge can be moved to pair as long as no pairs have been inserted in that face
and another trick is the M where if u have an edge that needs to be paired you can do M' to pair and U or U' and M to put the M slice back in it's place and sometimes the edge is not correct so you can put the corner so the other side and in Roux you would do M2 to pair if the pieces were the same but in this method it would be M' U2 M
and the pair would be....paired

CMLL, I don't know how this works but I learnt to do Sune and Anti-Sune and J and A perms to solve the corners but somehow he solves the four corners in one alg
he said it's an alg set u can memorize

L5E I solved like how i would on LSE except I made it into a PLL case but he said I can memorize an alg set for it

He never named this method and could anyone tell if this is original or an existing method
It’s Russo
 

Chris_Cube

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Scramble: B' D' B' D' U2 L D2 B U' B R' U R2 L' B2
Inspection: z2
D2 U' F U' F' // 1x2x2-block
D' // move block to BL
Uw2 U2 Rw U Rw' // 1. edge
Uw U2 R U' R' // 2. edge
Uw U2 Rw U Rw' // 3. edge
Uw U' Rw U Rw' // 4. edge
U2 M' U2 M U2 S' U2 S // Finish Cross (and solve EO, but it is already solved)
R' U' R U R' U' R U' R' U R // F2L 2 (BR)
R U R' U' R U2 R' U R U' R' // F2L 3 (FR)
L' U L U L' U' L U2 L' U L // F2L 4 (FL)
U2 R U R' U R U2 R' // OLL (Sune)
U M2 U' M U2 M' U' M2 // PLL (U-Perm)

I fixed the example solve. There was a problem with F2L 4 and with the OLL. I replaced the E-Moves with Uw-Moves since (for me at least) they are much easier to do. I also pasted it into alg.cubing.net. You can find it here

Why do you solve the E-Layer? Afterwards you do normal F2L which doesn't need solved edges. Actually the F2L-cases with solved edges generally take longer than when the edges are in the U-Layer. You could use Keyhole to improve your method, but I think it doesn't make sense to solve the E-Layer and you should rethink your method.
Yeah ok I understand what you mean. I found it interesting to solve the E-Layer because some last F2L Cases with this situation can skip OLL, so you just have to make PLL. I thought, that it would be a little bit too silly to solve the last F2L after you solve the E-Layer, but if you are fast at rotating the U and R layer this method could improve your F2L solving a little bit. My avg with CFOP is 25 secs and with ESO I got around 30 to 35 secs.
Greetings
 

ProStar

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Ok, back with another idea that probably already exists. It's a combination of CFOP + Roux.

F2B - Solve First 2 Blocks like you normally would do for Roux
CMLL - Same as Roux
EO - Orient Yellow/White edges like normal for Roux, then use M moves to place the remaining cross edges
EPLL - Same as CFOP

Example Solve(excuse my laughably horrific F2B):

z2 U R2 L' U2 L U2 L' U L U2 L U L' R U R' U' M U' M' y R U R' U2 R' U R U' y R U' R' // F2B ( I don't mean any disrespect to Roux users, I'm just awful lol )

R U R' U' R' F R F' // CMLL

U M' U' M U2 M' U' M // EO pt. 1

M' U2 M // EO pt. 2 (I know I could've done U M' U' M U2 M' U M for EO, this is just so it's easier to understand)

U' M2 U M U2 M' U M2 U' // EPLL(+AUF)
 
Joined
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Messages
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Ok, back with another idea that probably already exists. It's a combination of CFOP + Roux.

F2B - Solve First 2 Blocks like you normally would do for Roux
CMLL - Same as Roux
EO - Orient Yellow/White edges like normal for Roux, then use M moves to place the remaining cross edges
EPLL - Same as CFOP

Example Solve(excuse my laughably horrific F2B):

z2 U R2 L' U2 L U2 L' U L U2 L U L' R U R' U' M U' M' y R U R' U2 R' U R U' y R U' R' // F2B ( I don't mean any disrespect to Roux users, I'm just awful lol )
R U R' U' R' F R F' // CMLL
U M' U' M U2 M' U' M // EO pt. 1
M' U2 M // EO pt. 2 (I know I could've done U M' U' M U2 M' U M for EO, this is just so it's easier to understand)
U' M2 U M U2 M' U M2 U' // EPLL(+AUF)
This has been proposed by a lot of people, but no EO and before the EO. This is usually not very efficient, but this version might be better than placing L and R and then L4E. I can do some comparison solves for refrence.
 
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Ok, back with another idea that probably already exists. It's a combination of CFOP + Roux.

F2B - Solve First 2 Blocks like you normally would do for Roux
CMLL - Same as Roux
EO - Orient Yellow/White edges like normal for Roux, then use M moves to place the remaining cross edges
EPLL - Same as CFOP

Example Solve(excuse my laughably horrific F2B):

z2 U R2 L' U2 L U2 L' U L U2 L U L' R U R' U' M U' M' y R U R' U2 R' U R U' y R U' R' // F2B ( I don't mean any disrespect to Roux users, I'm just awful lol )
R U R' U' R' F R F' // CMLL
U M' U' M U2 M' U' M // EO pt. 1
M' U2 M // EO pt. 2 (I know I could've done U M' U' M U2 M' U M for EO, this is just so it's easier to understand)
U' M2 U M U2 M' U M2 U' // EPLL(+AUF)
thats how i solve roux
 

PapaSmurf

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For the ESO method: just start off by building a 2x2x2. It averages 6 moves if you get good.
Instead of solving the whole E slice, just solve the FL and BR edges. That will only be another 4 ish moves.
EO+cross is an interesting step. I'd guess about 8-10 moves?
2 keyhole F2L pairs which would be maybe 10 moves.
An F2L pair, around 8 moves.
ZBLL, 16 moves (inc. AUF).
All in all, an interesting method, rotationless and good ergonomics, althougn mixed. Probably a low-ish movecount of <50 moves. Would be cool to see if it's good, but would need ZBLL (493 algs) to unlock its proper potential.

Overall, a pretty lucky solve, but it feels like on of those methods that are lucky just because they are. Could have potential.
 

ProStar

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(I asked this another time but don't remember where so I don't know what the answer was)

How many algs would it take to to the last slot while orienting and permuting corners? Obviously the set would be enormous, but it could be like OLS; you learn only certain subsets(like WV or SV for OLS)
 
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I have this idea. It's probably very bad, but here it is :
1. Solve the entire cube ignoring the orientation of every piece.
2. Orient the E-Slice (1 algorithm)
3. Apply pure OLL on one side. Try orienting the side as much as you can, if finish with one twisted corner or edge, don't worry. If you have both, make sure that two bad pieces are next to each other after pure OLL.
4. Do the same thing on the other side.
5. If you have twisted corners, edges, or both, align them so they are on the same layer, then rotate the cube and apply another pure OLL so they are both solved simultaniously.
 
Joined
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I have this idea. It's probably very bad, but here it is :
1. Solve the entire cube ignoring the orientation of every piece.
2. Orient the E-Slice (1 algorithm)
3. Apply pure OLL on one side. Try orienting the side as much as you can, if finish with one twisted corner or edge, don't worry. If you have both, make sure that two bad pieces are next to each other after pure OLL.
4. Do the same thing on the other side.
5. If you have twisted corners, edges, or both, align them so they are on the same layer, then rotate the cube and apply another pure OLL so they are both solved simultaniously.
Seems pretty bad Lol. No offense.
 

ProStar

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So, it's like Petrus, except instead of EO you do CO. Then, when F2L is finished, you have all the corners oriented. Then you can do one alg and solve the rest of the cube. The LL set would have less algs than ZBLL, cause corners instead of edges would be oriented. I can't figure out how to do CO properly though.
 

Etotheipi

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So, it's like Petrus, except instead of EO you do CO. Then, when F2L is finished, you have all the corners oriented. Then you can do one alg and solve the rest of the cube. The LL set would have less algs than ZBLL, cause corners instead of edges would be oriented. I can't figure out how to do CO properly though.
CO is a pain XD. ask WoowyBaby about how to solve it, he's an FMCer who uses domino reduction sometimes, he'd know.
 
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