I'm not going to lie, this looks an awful lot like stadler. If you can find a mostly intuitive approach to LSE I'll be pretty interested though.
Hey man, so I actually did.
I have come to a few situations and I noticed a few things.
I want this to be a thorough post such that it is plausible, looks good on paper, and is applicable so that others can implement and test this approach.
I will share different iterations of the method. For the simple fact that I realized certain behaviors and recurring themes as I was implementing the approach.
My initial thought was to
1. Solve L/R Centers
2. L 1x3x4 Block
3. R 1x3x4 Block
4. Last 4 Centers.
5. LSE (Solving 5 wings is considered an optimal case)
6. Commutators to permute the unsolved wings.
7. Parity Fix.
I usually reduced the cube to a parity where I swapped the UBl and FUl with rU2 rU2F2 rF2 (l'U2lU2) r2. Once you reduced down to parity, you'll do either a quick setup or cube rotation. (You may use the mirror as well).
I then noticed that my approach to Commutators prevented even (PLL) parity. Because each piece was being solved directly. So, therein I found a benefit to utilizing the method.
Ultimately every method has pros and cons. Some methods correlate better to others. I felt that Commutators would carry over from my BLD experience and blend nicely with my Roux solving.
So, you avoid PLL parity however cannot predict odd (OLL) parity mid solve. It was my hope that you could deduce this sooner looking at the "LSE" state of the cube pre-commuators. And I found a simple case that debunked that misconception. Apply this Commutator and you'll see what looks like a "flipped" edge. b' R'uR U2 R'u'R U2 b.
So, next I realized that a 5 cycle could be solved by two 3-cycles. Such that the original (A -> B -> C -> D -> E) could be solved as (A -> B -> C) + (A -> D -> E).
Normally 7 wings are unsolved in the LSE reduction. So, it is a 7 cycle. So, it would take three 3-cycle Commutators to solve that case (in general). So, what if you did a 3 cycle and then a 5 cycle. In blindfolded solving we proposed that the 3 cycle system was plausible because of the congruency with letter pair systems and also it was applicable across centers, wings, corners, and edges. There were a total of 23 x 22 possible cases for wing 3-cycle Commutators for blindfolded solving.
You get 23x22 by taking the number of possible permutations of the first piece. Followed by the possible number permutations of that following piece. If you were to apply 5-cycles to blindfolded solving to create a comprehensive list you would have 23x22x21x20 possible algorothims. It at first seemed daunting.
Remember, we create rules to define subset cases (like PLL, OLL, CMLL, LSE)
So, I proposed then to be able to permute the last 8 wings rather than the last 12. All of them would be on the layer. There were a finite number of cases.
At this point, I am considering an illustrious ZBLL of 4x4 Last wings. I found some cases that were fast.
As I was playing with some 5 cycles, I noticed more patterns. So, I played with doing the 5 cycle once, then twice, inverting it once, and inverting twice as well. I realized that 5 had one solved stae and 4 unsolved states.
And it was interesting because there were nice cancellations and conjugations as well that came into play.
I also noticed that if I was able to "track and permute" a few pieces, I was getting certain patterns as a resultant. Basically a 2 look OLL. With the limited algorithm subset, I realized that simplicity was going to trump complexity.
So, I abandoned a 1-look wing solution. Another side effect to mention was that the algorithm subset actually was not centers safe. Some of the cases would disturb the U/F centers. They would actually only disrupt the Fur and Fdr. My hope was that I would find center safe solutions as I tested further to create the comprehensive list.
So then I played with a few algorithms.
I'll share these with you now.
rUr'U2rUr'
rUr'UrU2r'
(Rw)UR'URU2(Rw)'
And now I am looking at the opportunity to rather than permute the last 12 wings, you now pair the last 12 wings.
I liked the fix of the centers post 3x3 reduction.
Basically, the Roux reduction had a recurring pattern of First 2 blocks, centers. Pair wings in the DB slot. Then Pair wings in the DF slot. Then there were some "shapes" that popped up on the U Layer.
I called them D, L, H, and Z. This is important, because a lot of the post this far has been iterations that led to this development.
D -- 1 single edge pair is paired, the other 3 are scrambled.
L -- 2 adjacent edge pairs are paired, the other 2 are scrambled.
H -- 2 opposite edge pairs are paired, the other 2 are scrambled.
Z -- All 4 edge pairs are scrambled.
Basically look at the D, L, or H case and apply and algorithm and all 4 edge pairs are paired.
For now, apply a favorable algorithm to the Z case and transform it into a D, L, or H, case.
I liked this case nomenclature because you would notice that there are only 2 possible L cases, 2 possible H cases, 4 possible D cases. This case reduction is very similar to a 2 look OLL reduction method.
Next it is interesting because there good cases and bad cases (and by bad, meaning I haven't found a good solution in this infant development stage of the method.)
So, we want to remember the good cases. I shared them earlier.
Here is an H case solution. (Rw)UR'U R'FRF' RU2(Rw)'
So, I am working through the steps and look at how pairing at DF affects the U Layer. Basically, I ask myself do I insert with rUr', rU'r', or rU2r'.
Now, here is where it gets interesting. Apply this algorithm to a solved cube and notice what happens.
rU2r'
...
And I'll follow up later with the rest.
