# The New Method / Substep / Concept Idea Thread

#### mark49152

##### Super Moderator
Staff member
No, it's not. It's just different than F2L.
I find it hard, as do many others, but this is beside the point. If I wanted to learn Roux, I'd learn Roux rather than cluttering up the "ideas" thread.

#### mark49152

##### Super Moderator
Staff member
Just curious as to whether anyone uses the following method.
1. Solve DL and DR with adjacent centres. M slice centres don't matter (2.9 moves).
2. Solve F2L pairs, with free use of the M slice (24.1 moves).
3. Solve DF and DB, and M slice centres (5.2 moves).
4. Finish with OLL/PLL or LL of your choice.

Advantages and reasons I find this appealing:-
1. Step 1 is very short and simple, so planning first pair during inspection is easier (compared to cross).
2. F2L pairs are solved in <URrLlM> with no rotations and no F or B moves.
3. Because the M slice is unrestricted, pair solutions are slightly shorter on average (24.1 versus 27.5 for CFOP).
4. Step 3 is <MU> and based on a few common patterns/triggers, so can be faster TPS than cross, which typically involves several faces.
5. It's easier to look ahead into OLL because the corners aren't changed by step 3.

The only disadvantage I can see is that lookahead is a little harder because you have to be aware of what's in DF and DB while solving pairs.

Regarding move counts, I tried both this and CFOP against the same 12 scrambles, and that's where the estimates above come from (it's only a small sample set, I know). Counts are in STM. Total F2L move count was 32.2 for this method, compared to 33.7 for CFOP. Effectively, cross, which averaged 6.2 moves, was replaced by 8.1 moves for steps 1&3, but that 2-move increase was offset by the more efficient pairs.

Clearly the small decrease in move count isn't enough of an advantage in itself, but move count doesn't appear to be worsened either, so I'm wondering if the other advantages listed above make this worthy of further exploration. I'm considering documenting the F2L cases/solutions.

Thoughts, anyone?

Finally got round to preparing some example solves (only F2L transcribed below, LL is just OLL/PLL).

1. B2 R2 F' L2 B' D2 F' R2 F2 L2 R2 U R' B' F' D B L D L D2

y' x2
B R u // DL/DR
U2 L U' L' l' U' L
U' r' U' R
U M U2 R U' R'
L U L' U L U' L'
U' M2 U2 M // DF/DB = 31 total

2. U' R2 F2 D F2 D L2 U' L2 R2 F' L D' R2 U B2 L' U' F D'

y2 x'
F2 R' // DL/DR
B L' U' L
[L] U L' U' l U L'
r M' U' R'
U' M' U' r' U R
U' M' U M' // DF/DB = 27 total

3. L2 F' U2 B' U2 B2 R2 B' R2 U2 F L' B D' F' D' B D R U2

y2
B D R // DL/DR
U' M U2 R U' R'
M2 U L' M U' L
U l U L'
R' M' U R
U M' U M U M2 // DF/DB = 29 total

Last edited:

#### Costa

##### Member
An interesting last layer alternative?

Instead of going with classic Oll and Pll i thought this:

-Last pair of f2l inserted like VHLS so all of my edges are oriented (http://www.cubewhiz.com/vh.php) 16algs without mirros

-COLL 42 algs

-EPLL which are U,H and Z perms

This alternative will increase the probability of skips A LOT( from 1/216 to 1/27 the oll for example) and wil leave you with a simple fast epll

So yeah my question is: whats better between this or normal oll-pll and WHY?

#### Kirjava

##### Colourful
This is called VH and it sucks.

#### Raviorez

##### Member
This is called VH and it sucks.

Why should it sucks?
For me it seems a pretty good variation

#### TheOneOnTheLeft

##### tOOtl
You still need to know full PLL in case you get an OLL skip.

#### Renslay

##### Member
You still need to know full PLL in case you get an OLL skip.

Technically, no, you don't. J-PLL or Y-PLL as COLL, and then U- Z- or H-PLL as EPLL. That means you didn't have (or forced) a skip. Just sayin'...

#### jorgeskm

##### Member
Idea: Last pair + Last layer

Hi! I Spanish and my English isn´t good.
The fist step is oriented all the corners of last layer while you put the last pair. For this, there are four main cases.

All F2L's finish with this pair.

For each case there are 27 cases of orientation corners. But a lot of are mirror cases. I have generated cases of first pair:

L' U2 R U' R' U2 L (7)
R U' R' (3)
L R' F R F' L' (6)
R D R' U' R D' R2 (7)
y' R2 F R F' R (5)
R U' R' U2 R U' R' U2 R U R' (11)
R U R' U' R U' R' (7)
R U' R' U R U2 R' (7)
F' R U2 R' U2 R' F R (8)
R' F R2 U R' U' R U R' U' F' (11)
R' F R2 U R' U' F' (7)
R' F R F' R' U' R U' R' U2 R (11)
R U' y R U R' U' F' (7)
y R' F R U' R' F' R (7)
R' F R F' (4)
Lw R U' R' U F' Lw' (7)
R U2 R' (3)
R' F' R U2 R U2 R' F (8)
Lw U F' U' Lw' (5)
R U' R' U' R U R' U R U2 R' (11)
R U' R2 U2 R U R' U R (9)
F' U' F U' R U' R' (7)
R U' B U' B' R' (6)
(R U' R' U)x2 R U2 R' (11)
R U2 R2 U2 R U R' U R (9)
R U' R2 U' R U' R' U2 R (9)
(R U R' U')x2 R U' R' (11)

The algs are very good. And the average moves are in this cases 7,18 (if I haven´t mistake).
I suposse the other 3 cases (choosing good cases) will have average moves sub 8. At least the average moves of the four cases should be sub8.

Last layer

Now, we have to solve the last layer with one alg. There are 156 cases. But, 21 are PLL's, other is the last layer solved, other algorithm of BLD, some ELL's.
Removing these, 12x cases have to learn. They are a lot of cases, but wait for the comparison with Fridrich.

The best of thes 156 cases is that the recognition is very easy. All the corners are oriented, and in 94 of 156 cases, 2 edges adjacent will be oriented.
Imagine that you recognize this case for R-F.

You have a lot of information with this. And this OLL has a probability of 72/156. If we add the 22 cases (all corners oriented and all edges oriented):
72+22=94.

The others are easy to recognize too:

Last layer CFOP vs This Method
CFOP:
Put the pair: 3 movs.
OLL: 10 movs.
PLL: 12,5 movs.
This are good algs not optimal algs.

In total, 25,5 movs.

This method:
Put the pair: 8 movs.
The rest: ? movs.
If the rest are less than 14 movs, the total would be sub 22.

I don´t know how many average movements would be the final, but I think that less than 14 moves can be (search good algs)

3,5 movs equals (aproximately) 0,5 seg in Feliks, Mats, Alexander,...

And this aren´t many algs as ZBLL, RLS (or VLS),...

#### TDM

##### Member
long post
This has already been thought of. Also, not every pair finishes as you said; for example, I sometimes finish my F2L with D R U' R' D' or l F' R U' R' U l'.

#### jorgeskm

##### Member
This has already been thought of
I just saw this explanation. He/She explains the idea simply, without say number of algorithms,...
But I explain some things, adventajes, comparison with CFOP,...

Also, not every pair finishes as you said; for example, I sometimes finish my F2L with D R U' R' D' or l F' R U' R' U l'.
These are special cases and most people don't use this.

#### TDM

##### Member
I just saw this explanation. He/She explains the idea simply, without say number of algorithms,...
But I explain some things, adventajes, comparison with CFOP,...
A lot of that has been discussed in later posts, not just the original suggestion. Also I have seen this idea discussed multiple times, and everything has probably been covered already.
These are special cases and most people don't use this.
I know, especially the second, but those are just examples. People use lots of different algs after learning F2L intuitively to speed up their F2L (although some disagree, arguing it makes lookahead harder, but my second alg has the same effect as F' U F U' R U' R' and the first isn't complicated).

#### caters

##### Member
OLL, PLL to solve whole cube

Is it possible to do OLL and PLL of white layer and then OLL and PLL of each middle side to solve the whole cube so that the yellow side is solved at the same time the middle sides are solved?

If so than that means these as advantages:
less algorithms
get to sub-20 and sub-15 and sub-10 faster than with normal CFOP because you practice a particular algorithm at least once and these OLL and PLL algorithms are not the main factor into being slow in CFOP

You do the same process for 5 sides instead of 1 which means for someone who is very good at OLL and PLL 5x time in OLL and PLL in normal CFOP

However in the long run it is more advantageous than regular CFOP and probably would get you at those speeds in LBL much faster.

#### GuRoux

##### Member
Is it possible to do OLL and PLL of white layer and then OLL and PLL of each middle side to solve the whole cube so that the yellow side is solved at the same time the middle sides are solved?

If so than that means these as advantages:
less algorithms
get to sub-20 and sub-15 and sub-10 faster than with normal CFOP because you practice a particular algorithm at least once and these OLL and PLL algorithms are not the main factor into being slow in CFOP

You do the same process for 5 sides instead of 1 which means for someone who is very good at OLL and PLL 5x time in OLL and PLL in normal CFOP

However in the long run it is more advantageous than regular CFOP and probably would get you at those speeds in LBL much faster.

i'm not really understanding what you are proposing, can you show an example solve.

#### caters

##### Member
Well I have not had any success so far because I have not yet found a way to do OLL and PLL of the 4 middle sides without messing up the white but still getting the same outcome.

#### Bindedsa

##### Member
Is it possible to do OLL and PLL of white layer and then OLL and PLL of each middle side to solve the whole cube so that the yellow side is solved at the same time the middle sides are solved?

If so than that means these as advantages:
less algorithms
get to sub-20 and sub-15 and sub-10 faster than with normal CFOP because you practice a particular algorithm at least once and these OLL and PLL algorithms are not the main factor into being slow in CFOP

You do the same process for 5 sides instead of 1 which means for someone who is very good at OLL and PLL 5x time in OLL and PLL in normal CFOP

However in the long run it is more advantageous than regular CFOP and probably would get you at those speeds in LBL much faster.
This would not work, the reason we are able to do LL algorithms is because we solved the other two layers and all of the last layer pieces are in the last layer.

#### TheOneOnTheLeft

##### tOOtl
Is it possible to do OLL and PLL of white layer and then OLL and PLL of each middle side to solve the whole cube so that the yellow side is solved at the same time the middle sides are solved?

This seems to be closest to L2L, which requires more algorithms than CFOP, if I remember correctly. Look up L2L4 or L2Lk.

#### TheNextFeliks

##### Member
This seems to be closest to L2L, which requires more algorithms than CFOP, if I remember correctly. Look up L2L4 or L2Lk.

It really is different but I can see why you would say its similar.

#### TheOneOnTheLeft

##### tOOtl
It really is different but I can see why you would say its similar.

I realise they're by no means the same method, but it's the most similar method that's feasible - that I knew of (in terms solving one side, solving another side separately and having the rest of the cube done once you've finished).

#### caters

##### Member
The OLL and PLL of white and middle sides is supposed to work like this:

Get all white pieces in top layer

Orient and permute them

Get all green pieces on green side

Orient and permute them(orientation messes up white, permutation fixes white)

Do the same for the other 3 middle sides

Yellow and every other color is solved with less algorithms than CFOP and you solve much faster.

#### TheOneOnTheLeft

##### tOOtl
I understood what you meant, but as Bindedsa suggested, there are more possibilities for OLL and PLL if the rest of the cube isn't solved. Also, according to speedpicker's analysis, OLL/PLL is about 38% of the average CFOP solve, so doing OLL/PLL 3 or more times in a solve would generally be slower than a full CFOP solve, not accounting for the extra time taken to put all the white pieces into the white layer, the green pieces into the green layer, and so on.