The New Method / Substep / Concept Idea Thread

[One Wheel]

Slightly off topic, but this is the thread that got me thinking about it. If you ignore setup moves the following are possibilities for solving up to 10 oriented edges:
3 edges: 3-cycle
A: cw
B: ccw
4: double swap
5: 5-cycle
A: 1 space cw
B: 2 spaces cw
C: 3 spaces
D: 2 spaces ccw
E: 1 space ccw
6: 2 3-cycles
7: a 7-cycle (7 possibilities, similar to a 5-cycle)
Or
A 3-cycle and a double swap
8: a 5-cycle and a 3-cycle
Or
2 double swaps
9: a 9-cycle (9 possible)
Or
A double swap and a 5-cycle
Or
3 3-cycles
10: 2 5-cycles
Or
2 3-cycles and a double swap
Or
A 3-cycle and a 7-cycle

That's 23 algs and an excellent understanding of commutators to make my method work. And recognition that would make Feliks squirm. Piece of cake. Never more than 3 algs to solve the L10E.

 

[LexCubing]

The lowest we've actually got is 1 look for the entire solve. It's just a lot more difficult.

Spoiler: 1 look 3x3 solves

Yup we already did, I even said SpeedBLD is 1Look.


Can someone calculate how many Unique LSLL cases if the F2L corner is already in, CO is not solved but EO is solved and CP as well. I think I've finally thought of a method:


First is EOS (EOLine Setup)
Training to 1L this will be hard but it will be great. You solve the EOLine while putting the DFR and DBR in their respective places ignoring orientation. Note this step must be EOS not EOLine+S like how EOLine vs EO+Line.

Next you solve the LBD square, if the edges are stuck where is the DR corners are just do an alg. That's just 3 x 12 algs.

Next solve the final left pair with an alg that solves CP and preserves the DR corners if the edge needed is stuck there. Note: DR corners just need need to stay where they are, no need to preserve orientation. This is 21 F2L cases x 6 CP cases + (6 x 6) cases wherein the edge is stuck. Someone help developing a better CP recog.

Next solve the back right square with alg that preserves DFR corner. 3 x 12 cases since the edge can be AUF if it's at the top. (Edge is stuck) There's 9 cases if one edge is stuck, just AUF the free edge. And 9 more cases if both are free thanks to AUF.

Last step is to solve LSLL.

# of Looks:
1 EOS
1 BL Square
1 CP2S
1 BR Square
1 LSLL

5!

 

[AlphaSheep]

A better 5 look method:

1. X-cross
2. F2L Pair
3. F2L Pair
4. ZBLS
5. ZBLL

Or even better:

1. EOLine and plan pair
2. First block
3. Square
4. Finish F2L
5. ZBLL

 

[mDiPalma]

you cited petrus 3x2x2 as 2 looks, when it could actually be only 1 look, which satisfies your 5-look requirement

 

[LexCubing]

you cited petrus 3x2x2 as 2 looks, wq hen it could actually be only 1 look, which satisfies your 5-look requirement

Imagine doing that with a 5LS, effectively making it a 4LS.

Do you know how many LSLL cases are there if the F2L corner is in (its orientation can be any of the 3 states) and EO and CP is solved?

 

[Hazel]

I thought about a Last Slot subset that's a combination of Magic Wondeful (that's how it's spelled, no R) and Tripod, where you insert the Last Slot and solve Last Layer at the same time if there's a solved 2x2x1 block anywhere on the top layer.

Example setup: R2 D R' U R2 D2 F D F2 R F
Solving algorithm: F' R' F2 D' F' D2 R2 U' R D' R2

Anyone know how many algorithms this would be?

 

[Teoidus]

I think this is called tripod

 

[Shiv3r]

I thought about a Last Slot subset that's a combination of Magic Wondeful (that's how it's spelled, no R) and Tripod, where you insert the Last Slot and solve Last Layer at the same time if there's a solved 2x2x1 block anywhere on the top layer.

Example setup: R2 D R' U R2 D2 F D F2 R F
Solving algorithm: F' R' F2 D' F' D2 R2 U' R D' R2

Anyone know how many algorithms this would be?

a lot. look at NLS, that is 12 algs and then like 60 in 2 looks with edges oriented at least 12*60 algs.

 

[Shiv3r]

3 ideas:
CP with TSLE- use a TSLE tree (see below) to get to one of the 3 move cases and finish TSLE and CP with one alg.
http://gyroninja.net/files/tsle.pdf
Improved Hexagonal Francisco
1) CP and Hexagon- possible, but a stretch. My idea is to build the 1x2x3 on D at the back and insert DLF like in 2-gen redux
2a) 3 belt edges
2b) Insert last edge whilst forcing arrow 4-flip case (like BLS, optional)
3) L5C
4) L5E
Francisco style big cube method
I don't know if or how this would work.

Francisco big cube method, there are two:
1.Meyer variant, where after edge pairing you complete the hexagon and then do HF
2.HF-4, a variant by me. Basically you do a Roux block, then insert the DFL corner, then insert 1 wing edge int each F3L slot, insert the last D corner, then finish with K4 F3L, then CLL and L5E(Last 5 edges or Lewis 5 edges)

 

[Shiv3r]

Random probably bad method idea: New setup to Lewis Step 5
1: LR Centers
2: FB
3: D&B centers +DB edge
4: SB (K4/Edge pairing/Blockbuilding) + U&F Centers
5: Lewis Step 5

Lemme try a few solves with this. It may have potential, especially since right now I am playing around with K4 a lot and have fallen in love with F3L.

 

[Hazel]

I think this is called tripod

I'm pretty sure Tripod is where it's just the last layer (with the 2x2x1 block), this is turning it into a LSLL subset

 

[Hazel]

a lot. look at NLS, that is 12 algs and then like 60 in 2 looks with edges oriented at least 12*60 algs.

What is NLS? I can't find it on the Speedsolving wiki or anywhere else on Google.

 

[Shiv3r]

What is NLS? I can't find it on the Speedsolving wiki or anywhere else on Google.

Ask Neuro. its also called ZZ-tripod, invented a few months ago by Neuro.

 

[Hazel]

Ask Neuro. its also called ZZ-tripod, invented a few months ago by Neuro.

It sounds different from I'm proposing. ZZ-tripod preserves the 2x2x1 block while inserting the F2L pair, then uses an alg to solve Last Layer, but what I'm saying is solving Last Layer (with the 2x2x1) WHILE inserting the pair. I wouldn't think it would be too many algs because excluding the F2L pair, there's only 5 scrambled pieces.

 

[Shiv3r]

It sounds different from I'm proposing. ZZ-tripod preserves the 2x2x1 block while inserting the F2L pair, then uses an alg to solve Last Layer, but what I'm saying is solving Last Layer (with the 2x2x1) WHILE inserting the pair. I wouldn't think it would be too many algs because excluding the F2L pair, there's only 5 scrambled pieces.

with just the LL, there are 85 or so algs for tripod. It may be the same for the LS idea, but I doubt it. so the minimum number is 85 algs.

 

[Neuro]

@Aerma I should probably make an official thread for ZZ-Tripod, I've gotten quite a few questions on it. But basically you do ZZ until F2L-1, and form a 2x2x1 on the top, forming the tripod. From there, you use one of 35 algs to solve LS while preserving top block (NLS) and then use one of 50 ZBLL's (subset called TELL) and it's mostly made for people wanting to learn ZBLL. 80 algs where 5 of the TELL's are PLL's, and NLS uses a lot of common triggers/algs similar to TSLE. Here's an example solve, and here's the algsheet. Looking to get better algs and hopefully generate stats on probabilities of skipping steps.

 

[Neuro]

Lemme try a few solves with this. It may have potential, especially since right now I am playing around with K4 a lot and have fallen in love with F3L.

Looking back on it that proposal was just as I predicted: bad. What I'd probably do is L/R centers, FB, centers+DB, then F3L or otherwise to solve the rest of the F2L so you can use Lewis step 5. However, I'd say that unless you really want to do this variant, you'd probably be better off doing a Yau/Hoya variant where you leave out FD and then do F3L while still getting to Lewis Step 5.

 

[Shiv3r]

Looking back on it that proposal was just as I predicted: bad. What I'd probably do is L/R centers, FB, centers+DB, then F3L or otherwise to solve the rest of the F2L so you can use Lewis step 5. However, I'd say that unless you really want to do this variant, you'd probably be better off doing a Yau/Hoya variant where you leave out FD and then do F3L while still getting to Lewis Step 5.

becuz Lewis step 5 is fun. and the F3L thing you are talking about is HF-4, but that's K4 with leaving one edge out.

 

[xyzzy]

here's the algsheet. Looking to get better algs

Some alternative algs (not necessarily better):

C3: (U2) L' U R' U' (L R) U2 R' U' R
(just the inverse of the one you have in the spreadsheet at a different angle)

E4: R U R' L' U2 R U' R' U' R U' M' x'

G11: (U) R (U D') R U R' D R2 U' R U R2 U2 R'
(actually a conjugated J perm)

G12: (U) x' U' R' F' R2 U' R' U R' F R U x
(it's a V perm!)

 

[Thermex]

Hey I just finished the DFR- set for TEG today! The link to the PDF is at the bottom of this post.
I'm pretty lazy/busy so there's no pictures or anything, just the names of the cases and the algorithms. The algorithms are in the same order as the cases in cyotheking's TCLL- doc, so open that if you want to see the case that corresponds with the alg: http://www.cyotheking.com/tcll-1/
I'm pretty sure there are no mistakes in any of the algs, but if you find any, tell me (I kinda doubt anybody's gonna actually go through all of them though). anyway the algs turned out really well for my second or third time generating algorithms, I'm pretty sure the cases averaged around 8.6 moves or something, but about a fourth of them had a y' rotation in them because a lot of cases could be solved quicker that way. There were also a lot of 2-gen algs (~15% of them were 2-gen), and only 2 or 3 algs included L moves and were 4-gen. I found all of the cases to be pretty fast and easy, although there are a couple of algs in the gun set that really suck, maybe someone could help with those. Otherwise I can EASILY sub 2 all of these and pretty easily sub 1.5 most of them. Only 3 more TEG sets left and TEG- is done Also when there's an "a" alg and a "b" alg side-by-side, they're two decently equal algs that both work for one case.

@Shiv3r I've been experimenting with different ways of doing Fransisco on big cubes and many of them seem to have potential. This is my favorite idea so far for 5-7:

1. Centers
2. Solve a 1×2×3 roux-style block on the d-layer, leaving the FD edge unsolved
3. Finish pairing 7 more edges (all edges but two u-layer edges)
4. Add the DFL corner to your 1×2×3 block, creating a "hexagon"
4. Solve three e-layer edges with HF F2L
5. Do two algorithms that solve the rest of the pieces (everything except for the two unpaired edges)
6. Solve the last two edges with some alg/commutator. This step might suck/have way too many algorithms, and in that case you could just pair all the remaining edges in step 2 and solve step 5 like normal HF.