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V Achyuthan

Member
Joined
Nov 29, 2020
Messages
310
That gives you some just straight up bad 1LLLs. You might as well just do ZB. 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.
OK ig
 

Athey!

Member
Joined
Dec 13, 2021
Messages
8
Location
Central America
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

Member
Joined
Jul 18, 2021
Messages
149
Location
UK
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

Member
Joined
Nov 29, 2020
Messages
310
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

Member
Joined
Nov 29, 2020
Messages
310
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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Jan 4, 2018
Messages
1,000
WCA
2016TUDO02
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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

Member
Joined
Aug 30, 2020
Messages
736
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.
 

GodCubing

Member
Joined
May 13, 2020
Messages
154
2 Look LSLL method (most of the time) with few algs:

1. Have an oriented LL edge attached to an oriented LL corner (probably already done)
2. Conjugated OLL-CP (331 algs, recognition is the same as 42)
3. L5EP (16 algs)
I like it. Zipper L5EP algs are very fast and having a pair like this is very common. This probably better than zipper not accounting for recognition
 
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