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Solving the final layer as 270 special cases

These pages are a rough description of what I believe to be the ultimate way of speed solving the final layer with oriented edges. You can get there through the Petrus method, or other means.

Background

Counting symmetries but not inversions as the same case there are in total 270 different positions for this case. The perfect solutions for all these are well known, thanks to Bernhard Helmstetter, and average 12.08 moves.

Why not the "perfect method"?

It is tempting to assume that the best way to speed solve this would be to learn all these algorithms. But I seriously doubt it, for the following reasons:

Why not the usual fast method?

Traditionally, last layer methods are done in 2 or more steps, where each step is independent of the other. After each algorithm you have to check the new position to determine what algorithm from the next step to apply to it. Each such check is called a "look", so the best of these methods are known as "two look" systems. The "perfect method described above would be this idea taken to its extreme as a "one look" system.

The most successful such system is the Friedrich 2 look system, as described here. The first "look" (OLL) orients all pieces, and requires memorization of 57 algs. The second (PLL) places all pieces, using 21 algs. The average length of each step is around 10 moves [anyone know the exact numbers?], resulting in a total move count of around 20 for this approach.

The one look, two algorithm, "perfect enough" system

The system described here can be seen as taking the best and most practical aspects of the two systems rejected above. It is a one look system, that requires learning around 40 algorithms (most of which a serious cuber already knows), and averages 14-15 moves.

The idea is to learn only a reasonably small number of algorithms - certainly less than the 78 needed for Friedrich, and for each of the 270 distinct positions, find the fastest combination of two of those algs for each of the 270 positions, and memorize those. Or to be precise, if you learn 40 algs, you will use a pair of algorithms for 230 positions, and only one for the positions the 40 solve.

Learning all those is still a fairly big memorization undertaking, but much smaller than memorizing 270 individual algorithms. My gut feeling number is that it's 10% of the effort. And for "muscle memory", you only need to keep 40 known fast algs up to speed, which is much easier than 270.

I think that the 2-3 more moves used compared to the perfect 12.08 is more than compensated by these factors in terms of execution speed.

The Algorithms

I started out trying to find good combinations manually, but I soon realized this was a job much better suited for computers. So I wrote a program that can take a set of algorithms, generate every combination of two of them, and see which of the 270 positions it solved. Then you can pick the best solution for each position.

I started by putting in all edge orientation preserving algs of 11 or less moves, as listed by Helmstetter (The "BH54" codes refers to position 54 in the Bernhard Helmstetter list on his 5+6+7 page.). I then tried to determine which ones was least useful. The color coding below is an attempt to show which are the core algs you need to learn (green), the more advanced one that are useful (yellow), and the even more advanced (orange). It needs a lot more work, and is not all that reliable, but it's a start.

A4: Allan
B5: Sune
B6: AntiSune
Bx12: Niklas
Cx11: Clix
Dx1: AntiBH54
Dx12: AntiBH58
Ex1: BH54
Ex12: BH58
Ey11: Evelyn
G4: Bruno
B9: Benny
B10: AntiBenny
Cx5: BH84
Cz5: Czeslaw
Dz5: Denzel
EE4: Eve
Ey6: AntiBH84
A2: Arne
BB2: Bob
BB3: AntiBob
By1: Buffy
Bz1: AntiBuffy
Bq1: BH90
Bq3: AntiBH90
Cx7: BH89
Ey5: AntiBH89
F4: AntiDblSune
F5: DblSune

These are pretty much useless according to the computer, but I've always used them and I'm not about to stop now.

B2: Rune
B8: Su-Sune-ne
B11: AntiSu-Sune-ne
C6: AntiSu-Sune-ne
A3: Bert
CC5: BH141

---------- Renamed?? -----------------

Ax5: Axel
Cy7: BH44