Difference between revisions of "L5E"

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* f(moves) = f2 (moves) [U] f2 ... This is the same as the F() function above, but with double layer f turns. Note that sometimes the adjusting U turns are not mentioned.
 
* f(moves) = f2 (moves) [U] f2 ... This is the same as the F() function above, but with double layer f turns. Note that sometimes the adjusting U turns are not mentioned.
  
Here are the algorithms for permutation of all 5 edges; make sure to use AUF to place teh FD edge at the position it has in the image.
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Here are the algorithms for permutation of all 5 edges; make sure to use AUF to place the FD edge at the position it has in the image.
  
 
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Revision as of 08:51, 6 January 2010

E15 and E35 are experimental methods that solve the last six or the last eight edges of the U and D layers. First one (in E15) or three (in E35) D-layer edges are solved, and then the last five are finished in one step. Of course, corners and E-layer edges are always preserved.

Description

E15 and E35 use a combination of intuition and algorithms to solve the last six or eight edges. An advanced user can even include ELL algorithms to save turns in some cases. It is similar to the last step of Roux, but here the orientation is done after all of the centers are solved, to make recognition easier.

Both E15 and E35 consist of two steps. In the first step of E15, E1, the BD edge is placed intuitively along with the B and D centers. In the first step of E35, E3, the solver first pairs up the RD and LD edges and places them along with the R centers (or simply does one at a time by placing a center-edge pair and then using an M'UM-type algorithm), and then places BD and the B center as in E1.

In the second step of both methods, E5, there are two main sub-parts. First the edges are oriented using M'UM like in Roux; there are 5 cases for this, and the worst case in Roux (6 edges) is impossible. Then the edges must be permuted; you can place FD during orientation or right after and end with EPLL, or just permute all five edges in one step.

E35 can be useful for Corners First if the E-slice is solved after the corners. It is also useful in any method that starts with columns, such as one which does the F2L pairs first and then finishes with CMLL. E15 is an alternative to the last step of Roux. The E5 part can be used as a standalone method, which can be used (for example) if F2L minus one D edge has been solved and then the LL corners have been finished with CLL or CMLL.

Intermediate system for E5

Intuitive:

This style is not completely intuitive, because it is still necessary to know EPLL. First orient all unoriented edges, then place the FD edge to its correct position and finally end the solve in EPLL. The necessary intuitive steps are:

  • P = M' U2 M
  • O = M' U M
  • O' = M' U' M

Using that plus some U turns you can solve any case, even all ELL case. For example, O U P = (M' U M) U (M' U2 M) is a sample algorithm which solves orientation, then places FD. There is a 1 in 12 chance that EPLL is skipped; otherwise there is one algorithm left.

There is another notation which can make it shorter to write algorithms:

  • F(moves) = F2 (moves) F2

It is important here that the moves inside the parentheses keep the corners solved, so that they will still be correct after this algorithm. An example is the U-perm, F(U P U) = F2 (U M' U2 M U) F2. F(O U') is a nice way to solve orientation and place FD all at once if the FD edge is in the middle of three unoriented edges in LL.

Using algorithms:

The intuitive style lets you find and understand the orientation algorithms yourself. Another approach is to learn the algorithms instead of finding them; the case are:

Basic orientations
FUL Solved.jpg
Intuitive solution

(Possibly optimised) alg

Comment
EO5 2 adjacent.jpg
O U2 O

(M' U M) U2 (M' U M)

Orients UL and UB.
EO5 2 opposite.jpg
O U' O

(M' U M) U' (M' U M)

Orients UF and UB.
EO5 4.jpg
P U2 O'

(M' U2 M) U2 (M' U' M)

Orients UR, UF, UL and UB
EO5 1.jpg
O U O'

(M' U M) U (M' U' M)

Orients FD and UB
EO5 3.jpg
O'

(M' U' M)

Orients FD, UR, UF and UL
All images show white on top (U) and green in front (F).

Sometimes this will also solve the FD edge; the images show where the FD edge must be for the algorithm to solve it, but otherwise you should simply treat the yellow sticker as white. If you learn these positions, it can also be useful to learn the mirror algorithm for the case where FD is on the opposite side from the image. The easiest example is to use O' instead of O for the case with three flipped edges on U. Just like the intuitive step, if FD is not solved use P, and either way finish with EPLL.

Semi-advanced system for E5

Orientation + FD placement

In many cases you can orient the edges and solve FD at the same time without using any more moves. However, for most cases the length of the algorithm is as long as the two-step process, so learning a separate algorithm would not help. Some good cases are:

  • algorithm ... possibly optimized moves ... pieces to orient, location of FD
  • M U F(U' M') ... M d R2 d' M' d R2 ... orients RU and FU, FD at UF.
  • M U' F(U M') ... M d' L2 d M' d' L2 ... orients LU and FU, FD at UF.
  • O U' O' ... (M' U M) U' (M' U' M) ... orients FU and BU, FD at UB
  • M U 3x(M' U) M2 ... orients UF, UL, UB and DF without moving any pieces.

More algorithms will be added later.

FD placement + EPLL (EP5)

Another possibility for improving Intermediate E5 is to permutate the last five edges in one step. This only requires 16 algorithms in total, so counting the 5 orientations creates a system that solves E5 in two short steps. Even if you know this, it is still useful to place FD without extra moves if possible, since EPLL recognition is very fast.

We will introduce the following function:

  • f(moves) = f2 (moves) [U] f2 ... This is the same as the F() function above, but with double layer f turns. Note that sometimes the adjusting U turns are not mentioned.

Here are the algorithms for permutation of all 5 edges; make sure to use AUF to place the FD edge at the position it has in the image.

Double two cycle permutations
FUL Solved.jpg
Intuitive solution

(Possibly optimised) alg

Comment
EP5 22 A.jpg
F(H-PLL)

(y') R U2 R2 U2 R2 U2 R

Swaps UF<->DF and UR<->UL
EP5 22 B1.jpg
F(Z-PLL)

(y') r2 U' 2x(M E2) U r2

Swaps UR<->DF and UL<->UB
EP5 22 B2.jpg
F(Z-PLL')

(y') r2 U 2x(M E2) U' r2

Swaps UL<->DF and UR<->UB
EP5 22 H.jpg
H-PLL

Ra U2 Ra' (y) Ra' U2 Ra

Swaps UR<->UL and UF<->UB
EP5 22 Z.jpg
Z-PLL

M2 U' 2x(M E2) U M2

Swaps UR<->UB and UL<->UF
3-cycle permutations
EP5 3 A.jpg
P

(M' U2 M)

Cycles UF->DF->UB
EP5 3 B1.jpg
f(d' P)

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

Cycles UL->UB->DF
EP5 3 B2.jpg
f(d P)

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

Cycles UR->UB->DF
EP5 3 U1.jpg
U-PLL

F2 U (M' U2 M) U F2

Cycles UF->UL->UR
EP5 3 U2.jpg
U-PLL'

F2 U' (M' U2 M) U' F2

Cycles UF->UR->UL
5-cycle permutations
EP5 5 H1.jpg
P U P

(M' U2 M) U (M' U2 M)

Cycles UF->UL->UR->DF->UB
EP5 5 H2.jpg
P U' P

(M' U2 M) U' (M' U2 M)

Cycles UF->UR->UL->DF->UB
EP5 5 Z1.jpg
P Z-PLL

2x(M2 U) (M U2 M')

Cycles UF->UR->UB->UL->FD
EP5 5 Z2.jpg
P Z-PLL'

2x(M2 U') (M U2 M')

Cycles UF->UL->UB->UR->DF
EP5 5 U1.jpg
P U-PLL

(M' U2 M') U' (M' U2 M) U' M2

Cycles UF->DF->UB->UL->UR
EP5 5 U2.jpg
P U-PLL'

(M' U2 M') U (M' U2 M) U M2

Cycles UF->DF->UB->UR->UL
These images have white on top (U) and green in front (F). Darker pieces change positions.

Advanced system for E5

The most advanced system would be to simply solve E5 in one step. Nobody has generated the algorithms for this yet, but it may not be a good idea because of the large number of cases and bad recognition. For the worst ones the algorithm will probably just be the same as in the 2-step solution, so the benefit will be small at best.

One case where this is useful (and doable) is to solve cases where at least one of the edges is solved at the start. Using setup turns and ELL this can be done in one step, although the recognition and setup moves might mean that this is not worth doing.

Examples

E15 example

Scramble:

1.

L2 R2 U2 L2 R2 F2 U' L2 R2 D2 L R' B' U2 F2 L R'

E31 Scramble 01.jpg

Note that the centers are solved here, but they will not always be.

Intermediate:

  • E1: U2 M U2 M'
  • EO5: U2 O
  • PFD: U' P
  • EPLL: H-PLL

Do: U2 M U2 M' ... U2 (M' U M) ... U (M' U2 M) ... Ra U2 Ra' (y) Ra' U2 Ra

2-step E5:

  • E1: U2 M U2 M'
  • EO5: U2 O
  • EP5: U' F(H-PLL)

Do: U2 M U2 M' ... U2 (M' U M) ... (y') U' R U2 R2 U2 R2 U2 R

See also