New Idea: The Partial Cross Method

[Johannes91]

Once the F2L has been solved with one missing edge, the remaining edge is inserted and the LL edges are oriented in one step

How many algs would that be?

I would guess 16 cases.

Yep, it's 2^4 = 16 when the missing edge is in the LL.

But it can also be in its correct place. If it's solved, there are 4 cases. If it's flipped, there are 2. So the total count is 22 (including solved).

I remember seeing a list of algs (displayed using Java applets) somewhere. But because there are so few cases it would be easy to generate them. [Edit: This is the list I was thinking of: http://web.archive.org/web/20070811005317/http://www-personal.umich.edu/~dlli/NewAlgSet.html.]

 

[fanwuq]

Looks like corner first and layer methods are merging. There's going to be so many Roux variations if I can could this one.
Mine idea is to orient all edges except 2 LL edges (ignore them, doesn't matter weather they are oriented or not). Then use ELL. Only about 10 cases I think excluding reflections and inverses. I know all from TuRBo except ones with flip that act like Z perm. These algs are typically as short as the EPLLs.

 

[Kenneth]

Basicly you need this:

P = M' U2 M

O = M' U M
O' = M' U' M
O2 = P

From that plus some U turns you can create algs to solve any case, example O U P = (M' U M) U (M' U2 M)

The algs for some cases can be shorter if you also add this to the system:

F() = F2 ( moves ) F2

The U turn can be any U, U' U2 or no U turn depending on how U moves inside the parentesis of the "function". F(U P) makes the usual 7 turn U-PLL = F2 (U M' U2 M) U F2. The alg F(U2 O) U P solves the worst ELL (there is also a mirror case)... it was nice, before I used O' U O U O U2 O' to solve it =)

Using this system it gets wery intuitive to create algs that are easy to remember, try it

 

[Kenneth]

3 Cycles:
UF-UR-DF: M U2 M U M’ U2 M’ U’ (8)
UF-DF-UR: U M U2 M U’ M’ U2 M’ (8)
UF-UB-DF: U2 M’ U2 M (4)
UF-DF-UB: M’ U2 M U2 (4)
UF-UL-DF: M U2 M U’ M’ U2 M’ U (8)
UF-DF-UL: U’ M U2 M U M’ U2 M’ (8)

5 Cycles:
UF-UR-UB-UL-DF: M2 U M2 U M U2 M’ (7)
UF-UR-UB-DF-UL: U2 M U2 M' U M2 U M2 U2 (9)
UF-UR-UL-UB-DF: U2 M’ U2 M’ U’ M’ U2 M U’ M2 (10)
UF-UR-UL-DF-UB: M' U2 M U M' U2 M U' (8)
UF-UR-DF-UB-UL: U M U2 M' U M2 U M2 U' (9)
UF-UR-DF-UL-UB: U2 M U2 M' U M2 U M U2 M (10)

UF-UB-UR-UL-DF: U2 M’ U2 M U’ M’ U2 M U’ (9)
UF-UB-UR-DF-UL: M U2 M' U' M2 U' M U2 M U2 (10)
UF-UB-UL-UR-DF: U2 M' U2 M U M' U2 M U (9)
UF-UB-UL-DF-UR: M’ U2 M’ U’ M2 U’ M U2 M’ U2 (10)
UF-UB-DF-UR-UL: U' M' U2 M U M' U2 M (8)
UF-UB-DF-UL-UR: U M’ U2 M U’ M’ U2 M (8)

UF-UL-UR-UB-DF: U2 M' U2 M' U M' U2 M U M2 (10)
UF-UL-UR-DF-UB: M’ U2 M U’ M’ U2 M U (8)
UF-UL-UB-UR-DF: M2 U' M2 U' M U2 M' (7)
UF-UL-UB-DF-UR: U M2 U’ M2 U’ M U2 M’ U’ (9)
UF-UL-DF-UR-UB: U2 M’ U2 M’ U M2 U M U2 M’ (10)
UF-UL-DF-UB-UR: U2 M2 U’ M2 U’ M U2 M’ U2 (9)

UF-DF-UR-UB-UL: M U2 M’ U M2 U M2 (7)
UF-DF-UR-UL-UB: U’ M’ U2 M U’ M’ U2 M U2 (9)
UF-DF-UB-UR-UL: M2 U’ M’ U2 M U’ M U2 M U2 (10)
UF-DF-UB-UL-UR: M2 U M’ U2 M U M U2 M U2 (10)
UF-DF-UL-UR-UB: U M’ U2 M U M’ U2 M U2 (9)
UF-DF-UL-UB-UR: M U2 M’ U’ M2 U’ M2 (7)

2 Cycles:
UF-UR, UB-DF: U2 M U2 M' U' M' U2 M' U M2 U2 (11)
UF-UR, UL-DF: U' M U2 M' U M' U2 M' U' M2 U (11)
UF-UL, UB-DF: U2 M U2 M' U M' U2 M' U' M2 U2 (11)
UF-UL, UR-DF: U M U2 M' U' M' U2 M' U M2 U' (11)
UF-UB, UR-DF: U M U M2 U2 M2 U M' U' (9)
UF-UB, UL-DF: U' M U M2 U2 M2 U M' U (9)
UF-DF, UR-UB: M U2 M' U M' U2 M' U' M2 (9)
UF-DF, UB-UL: M U2 M' U' M' U2 M' U M2 (9)
UF-DF, UR-UL: M U M2 U2 M2 U M' (7)

Jason, you do not need that many algs. If you do AUF then it is only three 3-cycles (of which one is a mirror). three 2-cycles (one mirror) and six 5-cycles (3 are mirrors).

I found some nice algs for the cases too, I will list all here later today but here is an example:

(y') R U2 R2 U2 R2 U2 R .... solves the easy two cycle

Now it's later and I'm back from my day outside. The rest of the algs:

R2 d M U2 M' d' M' U2 (r' R') ... the tricky 2-cycle, do the last two moves as one. The alg is basicly a U-PLL with a insertion.
L2 d' M U2 M' d M' U2 (l L) ... mirror

M' U2 M ... the "P" case = the easy 3-cycle, to bad it's only a 1:30 chance.

R2 d' M' U2 M' d M' U2 (r' R') ... tricky 3-cycle, basicly the same alg as the 2-cycle but direction is diffrent for d and the first MU2M.
L2 d M' U2 M' d' M' U2 (l L) ... mirror

(M' U2 M) U (M' U2 M) ... H5 PLE (P Last Edges) Two "P"'s and a U = P U P, really fast one and recognition is tow opposite edges in the U-layer, looks just like H from that direction, start from the one not opposite to the FD pice as the UB pice, do P and then move the FD pice to UF = the U turn (or U' in the mirror alg), then do another P.
M' U2 M U' M' U2 M ... mirror

M' U2 M' U M' U2 M U M2 ... U5 PLE, one opposite edge in LL like U-PLL. FD at UF at start. The U or U' (mirror) turn moves the opposite edge initially found in LL to UB. The alg is a P followed by a U-PLL = M2 U M' U2 M U M2
M' U2 M' U' M' U2 M U' M2 ... mirror

M2 U M2 U M U2 M' ... Z5 PLE, as Z-PLL no edges are opposite in LL, move U so it has the side at F that is the same as the pice in the FD position and do a "side swap" = M2 U/U' M2. This moves the case from the F side to the b side and also transformes it to a P cycle
M2 U' M2 U' M U2 M' ... mirror

Preceed all algs with a AUF (if needed) and also end in another AUF (if needed =)

That's the algs I found, some are the same as Jasons.

I think the P case is a lucky case if I do not do anything else to solve the edges. I actully think all times when I solve the last five edges in less than six turns is a lucky case, it may be P, O or O' = [AUF] M' U/U' M [AUF] = the easiest "OLE".

This is a part of my current work. I won't hi-jack this thread anymore when writing about it, I'm planning to start one about my new "floating edge" approach a little later.

 

[Kenneth]

R2 d' M' U2 M' d M' U2 (r' R') ... tricky 3-cycle, basicly the same alg as the 2-cycle but direction is diffrent for d and the first MU2M.
L2 d M' U2 M' d' M' U2 (l L) ... mirror

Here is a much faster one I found yesterday:

(y') r2 U M' U2 M U r2

(y') r2 U' M U2 M' U' r2 ... mirror

 

[Kenneth]

Solving like this gives new possibilitys, I'm currently looking for 2x2x2 style algs to solve CLL, algs that do not preseve the FD edge (almost the same as CMLL). I tried the 5-turn R2 U2 R U2 R2 but it destroys middle layer edges... But you can do it like this:

(x) U M U F2 U' F2 U M' U (x) ... well, it's shorter than 2xSune but not faster =)

But the two algs used for T and L in Ortega 2x2x2 are nice:

r U' r' U' F' U F ... for the L case, also mirror and it makes two CCL's
F' U' F U r U r' ... for the T case, also mirror and it makes two CLL's

The usual algs for these CLL's are also short so you can alter to do edge control during CLL, I use these and many others to do that, usaly the normal alg with a M slice setup, works fine for many CLL/OLL algs. Go for all oriented if the FD pice is oriented initially or the 1 + 3 case (M' U M) if its not.

Method:

1, F2B a la Roux ... also do M-slice + BD edge, it's intuitive and uses at the maximum 5 turns.

2, CLL ... with edge control if possible.

3, orient the edges that are left.

4, permute last edges ... use the algs above.

Or this:

1, F2B
2, Orient corners + solve M-slice and place BD edge .... 7 cases, the rest intuitive or algs for all, it's not many cases if you allow AMS (adjust M slice).
3, Permute corners and orient last edges ... ca 25 cases
4, Permute last edges ... 16 cases

The last method is as you can see easy to learn, it uses less than 50 algs.