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Athey!

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Dec 13, 2021
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Here's an idea(OH-only) which I was thinking about for a few weeks:

CP(or 2-generator reduction) is a cool concept, but I feel that most methods that involve it use it in the wrong place, i.e the start of the solve. This would result in a waste of inspection, the inability to be colour neutral and lack of choices. This Mehta variant that I'm proposing has CP later in the solve which allows you to be colour neutral.

Step 1: Mehta-style FB
Step 2: Belt
Step 3: DCAL(80 cases)
Step 4: CPEO(25 cases excluding the cases which are AUFs to other cases and the cases where the edges are oriented)
Step 5(and 6): one of two variants below

Here are the variants for steps 6 and 7:
Variant 1: Solve the rest of the cube in 1 alg(2GLL+1)(408 cases)
Variant 2: CDRLL(or OCLL since there's only 7 cases)+L5EP
Variant 3: JTLE+EPLL

ok, let's be honest here(for full CN):

FB: 6-6.5
Belt: 9-10
DCAL: 9-10
CPEO: 10-12
Total: 34-38

2GLL+1: 14-14.5 according to @Athefre iirc
CDRLL+L5EP: 9.5+10=19.5
JTLE+EPLL: 10+10=20

Total avg movecount= 48-54 moves

Allows for a 2-gen finish while allowing full CN(which most other CP methods don't allow).
Has a movecount comparable to Roux/ZZ-a.
Could be a viable alternative to Roux/ZZ-a if CPEO recog is fixed.

High algcount
FB+belt ergonomics can be bad for some people.
I can't seem to come up with a good recog system for CPEO(If someone comes up with an idea I'd be super grateful to them).
There isn't any algsheet for 2GLL+1 yet afaik.


I'd name it Mehta-CPEO or something ig.
Anyone is welcome to voice their opinions/concerns for this.
Thank you.

I like this, it isnt anything like a method breaker but it just feels good to have a Mehta variant specifically for OH

Also i think the belt could be built better if we do a rotation, kind of cross on left, so you only need R, r and U moves for making the belt
 

IsThatA4x4

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Jul 18, 2021
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UK
WCA
2022RITC01
Just an idea I thought of, not really sure if it is that good.

There are 5 general steps:

1. EO-LOL (EO-Line on left)
Reduces the cube to <R,U,D>

2. Basically just domino reduction, but it could have a better name.
We need to put E-layer pieces in the E-layer and orient the corners. This is what I thought up (from a speedsolving context):

2a. Put 2 oriented corners on D

Then either:

2b. Finish "F2L" by making pairs out of belt edges and corners. For the first pair, it will quite often be inserted with just an R move due to the freedom here.

2c. OCLL

Or:

2b. Insert one pair (again, will quite often be inserted with just an R move)

2c. TSLE

The OCLL path has less algs, but I believe the TSLE path may be faster.

3. DL triplet
Build a triplet of D-layer pieces and put it at DL. Can be done with <R2, r2, U, D> moves

4. CPDR
This step solves the triplet at DR and CP. There are 2 approaches you could take:

Either:

4a. Solve a pair of a corner and an edge at DR (and fix the belt)

4b. Use a 5CP algorithm to solve CP

Or:

4a. Solve the edge at DR

4b. Use a 6CP algorithm to solve CP

The 6CP path is faster, but 5CP has less algs. Both paths use the <R2, r2, U, D> move set. It is important that these preserve the DR edge, so are not mehta algs.

5. L6EP
You could learn full L6EP, or just do it like Roux by solving UL/UR and then the rest.

I am not posting example solves, as I do not have the algs for 5/6CP that preserve the DR edge, so the solves will have an unrealistically high movecount.

Anyway, is this any good?
 

V Achyuthan

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2022SUND04
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Recently I was thinking about option select during petrus. Here are all the thinks I could think of
APB
223, 2 pairs, EO+insert final cross edge and ZBLL
223, 2 pairs, CxLL, L5E
223, EO, RB, ZBLL
223, EO, CP, RB, 2GLL
B1CEZ (Yes my method. Don't get me wrong here. Could be used if someone doesn't like learning a lot of algs)
223, Just solve F2L, ZBLL
223 + one belt edge, EOLE, DCAL, CDRLL, L5EP
223 + one belt edge, EOLE, DCAL, JTLE, PLL
223 + one belt edge, EOLE, 6CO, 6CP, L5EP
223 + one belt edge, EOLE, 6CO, APDR, PLL

The last 4 are basically just Mehta (because I consider Mehta to be a variant of Petrus)
Can option select be used during petrus? Let me know
 

V Achyuthan

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Ok this might be another variant of petrus
Might be good. Might suck. I don't know
Step 1 - 223
Step 2 - Place the DB and DR edges in FR and DR places in any orientation of permutation
Step 3 - EODRDB. Solve EO and DR and DB edges
Step 4 - insert DFR corner
Step 5 - Solve last pseudo pair
Step 6 - ZBLL
Here are some examples to get an idea
47 STM
46 STM
38 STM
41 STM
43 STM (39 with cancellations)
This method averages 45 moves (same as APB)
Let me know your thoughts on this
 

PapaSmurf

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That's not a variant, it's forcing a specific way of solving RB. Also, you mean DR and BR. Also
Just generally, don't use this thread as a method idea dump. Instead, try to think through the method properly and, if it came to you in about 5 mins, why no one else has mentioned it.
Please, stop just dumping ideas.
 

tsmosher

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Aug 30, 2020
Messages
1,055
More of a LS/LL approach than a method.

Conjugated Last Layer:

1. Get to F2L-1 state.
2. Build any oriented U layer pair.
3. Use oriented U layer pair to conjugate/transform either of the solved slots adjacent to the last slot.
(Conjugation can happen on any face of the cube: F, B, L, or R.)
4. Use any Conjugated Last Layer method to solve the last layer of the cube.
Options include: conjugated C(O)LL > E(P)LL, conjugated NMLL, conjugated ZBLL, etc.
5. AUF; unconjugate the conjugated layer; AUF.
 

guelda

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Apr 2, 2020
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2023WAJN01
Hi everybody!

Not progressing much since quite some time with the Roux method, I guess I need to get back to work and learn some stuff, etc.
For LSE, I learned the EO algs from Kian from the beginning, and spam them, never thinking about good or bad arrow, or even arrow tbh :)
On the excellent "2021 Guide to improve in Roux" which can be found on Reddit, the suggested order of learning is the following: 4c > SBLS > FBLP > CMLL > EOLR

Even if working on block building is important, I'd like to improve on 4th step, ie. start practicing EOLR. I'm still using 2-look CMLL, learning those 42 algs is a bit frightening - and apparently it would be better to start with EOLR, cf. a comment on the mentioned Reddit page:

========================================================
I might be wrong, but my idea is that learning CMLL is not very rewarding. 31 extra algs and recognition would easily take more than a month to learn, and it will be saving around 1s per solve. EOLR would save 1s with a much shorter learning time.
========================================================

By searching the web, I once found Fan Yiqun's page: https://www.fanyiqun.com/
If you click on "Yiqun’s 2-Look", you'll find a page where he talks about his own system for LSE: "Y2L", which should apparently be better than EOLR (translated):

========================================================
If you are a person who is too lazy to learn formulas, then Gilles Ru’s traditional solution is suitable for you, and you can get a good result by using it.
If you are a one-handed player, then Kian Mansor’s plan is suitable for you, one-handed observation should be able to keep up, the number of steps is very short.
If you are a two-handed player, then Fan Yiqun’s Y2L is a very good choice. It has not many steps, the least number of observations, and the observation difficulty is also small. BU recognition method is not required, and dynamic observation is not necessary.
========================================================

I translated the page but it is still not obvious to me how this works... Did any of you tried Y2L and what are your thoughts about it please?

Have a great cubing day!
 

Kaneki Uchiha

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his method has 128 cases
you would be better off getting better at blocks
eolr is mostly done with intuition no one sits and learns 300+ cases
you should learn cmll as soon as possible

I dont think this is worth it
other things you might want to look at are: DFDB recognition for lse and line blocks
 
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