L5E

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L5E
[[Image:]]
Information
Proposer(s):
Proposed:
Alt Names: Last five edges
Variants: Sexy*+S(8355)n SexyM Beginner, Commutators (pseudo EF)
Subgroup:
No. Algs:
Avg Moves:
Purpose(s):
Previous state: 5 Corners Missing cube state
Next state: Solved cube state

5 Corners Missing cube state -> L5E step -> Solved cube state


The L5E step is the step between the 5 Corners Missing cube state and the Solved cube state.

L5E, Last five edges, is an experimental method that solves five edges, the four in U and one in the D layer, preserving all other pieces.

ELL is a sub group of L5E, all ELL's may be solved in two steps using this method.

Description

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 using [U] M' U2 M (named 'P') and end with EPLL, or just permute all five edges in one step.

L5E can be useful for Corners First if the E-slice, centres and three FL edges are 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, it is an alternative to the last step of Roux if the centres are placed together with the BD edge at first, also for any LBL style, cross minus one edge at first and then the LL corners have been finished with CLL or CMLL or the same for Fridrich with 2-look OLL, corners first, then L5E orientation, place FD using P if it was not solved and end it all in PLL as usally.

Intermediate system

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 cases 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).

More algs are at the L5EOP page

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

L5EOP is an alternative to L5E, that always solves the FD edge while orienting ('P'lace). Then the last step will always be EPLL.

FD placement + EPLL (L5EP)

Another possibility for improving the intermediate style 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 L5E 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

The most advanced system would be to simply solve 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.

See also