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TEG1 is a useful set, useful both for 2x2 and for 3x3 LMCF. Its advantage is not so much in lower movecount but easier 1-look of the solution in 15 seconds. There are 8 sets, two twists of each of the 4 bottom corners. Before generating it I would calculate the average number of moves to create a TEG1 face. I believe for full EG the average is 3.8? If full EG plus TEG1 results in sub-3 average moves to make a face then it is worth it.

I generated the algs for one set of TEG and a lot were pretty good, but it was the set where the layer can be solved with F R U' R'
 
@Cale S Oh hey do you want to split the workload and you could do the minus set and I do the plus set?

If we start on April 1st, split the work and we both do a set a day we could finish in a month. The only reason I'm willing to do this is cuz I have spring break coming up :) if you're too busy we could do like a set every other day and then we'd still finish in two months.

Sound good? Oh yeah and what program are you using to generate the algs?

Also after doing 100 2×2 test solves, the average to solve a TEG face was 2.69 moves for me. On five different occasions the face was already solved and eleven times it was one move away.

@shadowslice e so..

2×2: abandoning L5C to work on TEG with someone after hearing your feedback. And you're right, not all algs would be sub 10 moves, but if you think about it, the average OLL is about 10 moves. However, VLS (inserting last pair+OLL, kinda similar to this) averages less than 10. Sometimes actually doing more than 2 things at once can reduce the movecount for a step, especially when you insert a pair/corner and solve something on the u-layer.

Moving on to the 3×3 method, I understand your concerns but I still want to pursue this idea. If I'm not going to use L5C algs, I'll take a step back to my original layout for the method, and you guys can tell me how you might execute this:
1. Blockbuild some shape on the D-layer
2. Do an algorithm to bring you to a point where you have three edges on the d-layer solved and all the corners solved (exactly like waterman)
3. Solve 2-3 edges somewhere on the cube, and finally
4. Solve rest of the edges in one algorithm, then permute midges.

I still think with AEFs and some setup TEUL could be cut down to around 500 algorithms all averaging under 10 moves. However, I'm still open to change on the last two step of this method. Oh yeah and when I said "solve the last six edges" I didn't mean solve em all at once, I meant solve the ledges/redges and orient the midges in one alg, then permute the midges in one alg like in LMCF.

Pyraminx: I'm still interested in this idea, but three questions:
1. Why did you abandon this method?
2. How many cases are there?
3. Would a better method be sloving the 1-flip top (~4 moves) and then solving the rest of the pyraminx in one algorithm? Has this been explored before?

Skewb: lol I probably should've realized someone would have thought of this before, but who, and also do you know where I could access the TCLL algs? Did the person succeed in making this method?
And the reason I said "complete last layer" instead of "corners last layer" was because this method on skewb not only solves corners but also edges.

Megaminx: I'll continue to explore 2-4 look last slot+last later options if my proposed methods are too many algorithms.

@Karl Ferber There's a Roux technique that works like this where you solve non-matching blocks, but you do the last two steps as CMLL and LSE like normal, and your method is the same but just the last two steps are different. I'm not sure if your method would be any better than the non-matching blocks technique in Roux, you need to elaborate a bit more on the last two steps (there's no such thing as "FM" edges since you can't have an "edge" in the "middle") but it could be decent.
 
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2.69 moves to create a TEG1 face is amazing, especially with 16% chance of 0-1 move face, that is 1 in 6 being 0-1 moves. Even the best 2x2 solvers like Lucas occasionally cannot 1-look the solve in 15 seconds but with TEG it becomes really easy. Worth generating. The biggest constraint is to allow freedom of the D face. For example some LEG1 are way better than their EG1 algs just because the bar is in a different spot. In those cases I prefer to start with a D move to put the bar in the LEG1 location and use the LEG1 alg instead of the EG1.
 
So using the latest scramble from the 3x3 example solve game I got a 39 move ZZ-Tri solve. I did the same solve with ZZ-CT and it ended up being 46 moves. Here are the two links:

ZZ-Tri:

ZZ-CT:

EDIT: Do any of you have any interest in developing stats on the probability of skipping NLS/TELL? I'm not very skilled with that sort of thing so I definitely need some help with it.
 
@efattah I did another 50 just to make I didn't get super super lucky, my results were

43 solves were either 2 or 3 moves, and overall I got a pretty good total average (I didn't count. As for very lucky/unlucky solves:
One solve was 4 moves, one solve was a skip and I had five 1-movers (surprising) one of those one movers was completely rediculous, it was one move from a checkerboard and I solved it using R' F2 R2 U2 lol

On a side note if this Cale S guy doesn't respond or declines my offer, you want to work on the method together? Like I said if I do minus and you do plus, we can fully finish the method in a month if we each generate a a set a day and two months if we each do a set every other day.
 
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Hey guys, @TDM and I were discussing an idea earlier today which would basically be alternative CMLL sets for annoying LS cases. Essentially they solve CMLL + fix one twisted corner in the SB (since these tend to be the worst last pair cases), which lets you do something like r U r' instead of R U R' U' R U R' U' r U r'.

To be able to fix FR slot, you need 2 * 42 more algs (so this is basically the same number of algs as EG).

To be able to handle any orientation of the SB corners (which would allow for some pretty efficient pseudo-SBs but might yield bad algs), you need 9 * 42 = 378 algs.

Thoughts/help generating these algs?
 
So I've read a lot about NLS and how it is a great alternative to ZZ-CT that requires fewer algs and a different recognition method. However, I can't seem to find the original post for NLS. Can someone give me a link, or post number + thread, to the write-up NLS?
 
about TEG: yesterday I tried 50 scrambles and got a 2.41 avg50 for TEG face movecount, worst case was 3 moves

I don't think I'll generate the algs because I would never use them and I haven't been cubing as much lately, there are so many cases
 
@Cale S that's fine, I understand there are a LOT of cases. Also you probably got a lower average than me because I'm a pretty inefficient solver.

@Teoidus I think I know what you mean. I don't think making all 372 is a good idea though, too many algorithms to save like 1-2 moves per solve, and most of the algs probably wouldn't be very ergonomic.
I'd be happy to generate the ones that only have one unoriented corner though, as we could just take TCLL sets and modify them so that they preserve the LR edges/e slice. How about starting April 1st (thats when spring break starts for me) I generate the plus set and you do the minus set, and if we each do two sets (ex. I do the gun and spaceship, next day I do hammer and pinwheel, etc.) a day then we'll be finished by next sunday. If you're too busy, you can extend the time it takes for us to generate these algs if you want, I have a lot of free time though because it's spring break. (Keep in mind you'll have to generate two different minus sets, one for when the twisted corner is in the FRONT and you need to preserved DL and DR, and one for when the corner is in the BACK and you need to preserve DF and DB).

@efattah so yeah do you want to generate the TEG algs together? You could do the plus set and I could do the minus set. If we each did one set a day (~5.5 algs per day), we would finish in a month, and if we each did a set every other day (~3 algs per day) we'd finish in two months. What would be a better schedule for you? (if you're able to do this of course, if not it's not a huge deal)
 
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I just came up with FNPOLL on the 100 signs you have been cubing too long thread and thought i should post my ideas here
so firstly as there are 2 different types of n perm it means that you only need to learn 1834 cases unlike the 3668 for 1 look last layer and seeing as everybody must like N-Perms???? and recognition should be the same as 1LLL
 
So I've read a lot about NLS and how it is a great alternative to ZZ-CT that requires fewer algs and a different recognition method. However, I can't seem to find the original post for NLS. Can someone give me a link, or post number + thread, to the write-up NLS?
I plan to make a tutorial for ZZ-Tri later this week so watch out for that. In the meantime, here's a link to the algsheet. So basically you get to F2L-1 and build a 1x2x2 block in UBR (making the tripod) and NLS is used to insert LS while preserving top block. From there you get an extremely small subset of ZBLL.
 
Okay, so I've been playing around with TCMLL and here are some examples to show how beneficial it can be:

39 STM TCMLL vs 41 Roux

40 STM TCMLL vs 48 Roux

44 STM TCMLL vs 49 Roux

42 STM TCMLL vs 44 Roux

Note: I didn't select times where TCMLL happened to get better movecounts than solving the pair and proceeding normally. These are every instance in which I thought (over the course of ~12 solves) TCMLL would be nice to have, and every time TCMLL yielded movecounts anywhere from 2-8 moves better than proceeding normally.
 
I got all TCMLL algs on one doc. Most of the algs aren't that great but here it is. Just comment if you want to contribute and I'll add you.
 
No idea but it'd probably be ridiculously high and recog wouldn't be good either. What you might be able to do though is orient everything while inserting the edge and get a really small TTLL subset.
 
I've come up with a possible 3x3 method, but it's probably been done before. I haven't come up with a name yet.

Step 1: Solve 3/4 of a cross
Step 2: F2L (still excluding last cross edge)
Step 3: CELL (Corners of the Last Layer without regards to the last layer Edges, 42 cases, algs not generated)
Step 4: LE-OLL (Insert last cross edge while finishing OLL: Last Edge + OLL, 14 cases. Algs: http://bit.ly/2mBdtAV)
Step 5: Edge PLL

Example solve (random scramble from csTimer):
D' F2 R U' L' U B L' D' F U F2 R2 D2 F2 D R2 B2 U' B2 U'

Step 1: x2 U2 L F'
Step 2:
- F2L 1: R2
- F2L 2: U L U2 L' U y L U L'
- F2L 3: Lw' U2 Lw Dw R U2 R' L U' L'
- F2L 4: R U R'
Step 3: U' F R U R' U' F'
Step 4: Dw' M' U2 M (easy case)
Step 5: U' R2 U R U R' U' R' U' R' U R'

48 moves (minus rotations), but this scramble probably could have been solved with much less because I'm not the best at F2L.

Method pros:
- Super easy crosses are pretty common
- Cross to F2L transition is a bit easier than CFOP
- LE-OLL cases are pretty easy to memorize and execute, and there are only 14 of them
- There's a bit more freedom with F2L I think
- Easy to switch from CFOP
- You can very commonly cancel moves while transitioning from LE-OLL to EPLL

Method cons:
- CELL may not be worth it to learn (again, algs haven't been made)
- I find that I rotate more than in CFOP
- I'm not sure how lookahead compares to CFOP

After doing an Average of 50 with this method, I got a 16.02 Mo3, 18.46 Ao5, 19.77 Ao12, 21.82 Ao50, and a 13.20 single! I think this may have a bit of potential.

If anybody remembers when I posted this, I figured out how to combine the last 2 steps into one! For LE-OLL, instead of completely solving OLL, you do an alg to make it so the Last Edge just needs a U' M' U M to be solved, along with the rest of the OLL (most of the LE-OLL cases end with this anyway, so the algs would be shorter and faster!) Then you memorize 11 cases to always be able to cancel into EPLL while solving the last edge! So it wouldn't really be learning new algs, just knowing which case corresponds to which cancel. You would recognize the cases by looking at the top 3 stickers on R, The UB edge, and the FUL/FUR corner.

Here's an example:
setup: R2 U R U R' U' R' U' R' U R' M' U' M U

You would recognize that the headlights on R are the same color as the UB edge, and that the RU edge and FUR edge are opposites.

Alg: U' M' U M' U M U2 M' U M2 (Insert canceling into counterclockwise EPLL)

So I think if you memorize all 11 algs (22 if you want to learn the mirrors for the U M' U' M insert also), then it could be a bit faster than the standard version of this method.

I've also realized that you can solve everything up to LE-OLL in many different ways, including 3/4 cross + F2L, Roux + CMLL + BD edge, some sort of Petrus variant, etc.

I've decided to call this method M3 (because my initials, MM, plus "method" is 3 M's).
 
@TDM just PM me your email or comment on the doc and I'll add you. I'm sure that the algs can be better but all of the cases are there.
 
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