The New Method / Substep / Concept Idea Thread

Meyer with Hoya
So I have used Meyer for some time and because I don't like Yau so i tried it with Hoya and it worked. So you have to build your first block on the opposite side of your L2C and you have to build it with S moves wich is wierd but you get used to it. Than you build the first edge of the second block and store it with a F2 and a U2 and then you put the sides with the blocks on the bottom and do edge pairing. If you don't understand me than watch Kians tutorial.
 
carcass said:
Is it really like a 1LLL system or is it more of a ZZ-CT type thing when they say one look last layer? Also, how many algs would be in L5E?
Click to expand...
ZZ-CT is touted as an LL skip method, since you solve different parts of last layer as you insert the last F2L pair.

Athefre said:
I was confused about that also. I'm not sure why the wiki says that it is a 1-look 2-alg LL system. I've been meaning to ask shadowslice about that.
Click to expand...
M CELL is 1LLL because you can recognize L5C and L5E at the same time and if you use comms, will be able to execute the algs back to back without having to take a second look.
 
GenTheThief said:
M CELL is 1LLL because you can recognize L5C and L5E at the same time and if you use comms, will be able to execute the algs back to back without having to take a second look.
Click to expand...

For me the confusing part is calling it 1LLL. LL + two more pieces is no longer a layer. If the method was to recognize CLL and ELL then execute both algs, that would make sense. But in this case M-CELL almost always has a layer and two additional pieces. Maybe the term 1LLL is being stretched in this case to mean "1 Look Finish". If that's the case, maybe a different term should be used.
 
Yep. Kians tutorial is good but you have to ignore the edges part because he makes it more difficult than it has to be. So just understand the concept
 
After Roux F2B for a purely algorithmic approach:
L2E+CO 664 algs COALL 155 algs = 819 algs.
704 L2E+EO ---> ZBLL 493 algs = 1197.
42 CMLL + 27 L2E + 25 ELL = 94 Algorithms. 3 Look
sorry if my math is wrong
what is the fewest algorithms needed for 2 algorithms after F2B?
what I could find is L2E(the last f2l edges) and CO into COALL with 819 algs
 
carcass said:
After Roux F2B for a purely algorithmic approach:
L2E+CO 664 algs COALL 155 algs = 819 algs.
704 L2E+EO ---> ZBLL 493 algs = 1197.
42 CMLL + 27 L2E + 25 ELL = 94 Algorithms. 3 Look
sorry if my math is wrong
what is the fewest algorithms needed for 2 algorithms after F2B?
what I could find is L2E(the last f2l edges) and CO into COALL with 819 algs
Click to expand...

An idea of mine that I often think about: It likely wouldn't be very difficult to place an edge at DBM while building SB. Probably a little EO influencing also. And or EO influence during CMLL. Or the other way around - some EO influence during SB and place an edge during CMLL. This combination leads to CMLL then L5E. With analysis on the best ways to influence EO, the solve would end with either good L5E cases or L5EP. What this creates is a 2 look finish with an overall shorter move-count, few algs, and not much effort.

Actually, one MI3 variant ends with CLL and MU L5E. So if the techniques above are applied to Roux, it turns into the same method.
 
carcass said:
After Roux F2B for a purely algorithmic approach:
L2E+CO 664 algs COALL 155 algs = 819 algs.
704 L2E+EO ---> ZBLL 493 algs = 1197.
42 CMLL + 27 L2E + 25 ELL = 94 Algorithms. 3 Look
sorry if my math is wrong
what is the fewest algorithms needed for 2 algorithms after F2B?
what I could find is L2E(the last f2l edges) and CO into COALL with 819 algs
Click to expand...
How about CMLL+DB into algorithmic L5E for a semi 2 alg L10P?(my guess is somewhere around 200-240 algs since solving the DB edge can be intuitive)
CMLLEO into algorithmic L6EP which I think is a total of >500 algs(CMLLEO is a lot of algs)
CMLL->EOLR->L4E is what top solvers use and it has only 42(±10)algs(It can be considered semi 2 alg since L4E is intuitive)
EODFDB(46 algs) into ZBLL(493 algs) = ZBRoux(539 algs)
(also on a sidenote, what we could do for a 3 alg/semi 4 alg L10P is CMLL+DB->EO->MU L5EP)
 
Here's the best method I could come up with that has mid-solve CP:

1. Roux FB
2. Roux SB missing a corner
3. EODFDB (intuitive) while recognizing the CP case
4. CPLC (26 algs + 26 D offsets, there should be no pause for recog since you can recog during the previous step)
5. 2GLL (84 algs)

Example: R' L2 D2 L' B' U L U2 L2 F R2 B' D2 R2 B' R2 L2 U2 L' F
y2 // inspection
U2 L F' L2 U R r2 F // FB (8/8)
U' r2 R' U R r U' r' // SB-C (8/16)
M U' M U2 M U2 M // EODFDB (7/23)
U2 D' R' D R U R' D' R D // CPLC (10/33)
R U2 R' U' R U' R' U2 R U R' U R U2 R' U2 // 2GLL (16/49)

The EODFDB step gets rid of the long pause necessary for recognizing CP mid-solve.
That being said, Roux is better than this for speedsolving. This is just an interesting novelty method.
 
carcass said:
I have an idea for 2 Look post F2B
OLLCP(331 algs) ---> Solve edge in BR ---> L5E, although I am unsure of how many algs that has.
Click to expand...

After F2B, you can have 0, 2, 4, or 6 bad edges,
When edges need to be rearranged to the top, it gives time for recog
When there are 6 bad edges, OLLCP can flip 4 and leaves 2 always on the bottom
No matter which EO case, its always one look

BD is fast and intuitive with 2~3 move solutions, quickly solving a blind edge

L5E can be done with very few algs using either
1. 1 look 2 alg, inserting the DF edge allows EPLL recog (4 cases)
2. LR EP, LR has only 20 cases and ep has a 33% chance of skip

this method can be 331+ 4 = 335 algs or 331+20 = 351 algs

FB SB OLLCP 6BAD DF AUF EPLL AUF
8 +16 +11 +2 +3 +0.7 +8 +0.75 = around 49 moves
Seems quite efficient, especially when there's good lookahead
 
Last edited:
Wait, this may work. Not, two look, but few moves... probably
FB
dbL square (2x2x1)
Get into a winter variation situation with the pair solved but not inserted
solve last pair and CMLL with 172 algs(I think that would be the right number)
Solve BD and FD with one of 119 algs
solve ELL with one of 25 algs
Overall this would have 316 algs, less than OLLCP. I think 'yall should look into this one.

EDIT: I meant dbR square, thanks @PapaSmurf
 
Last edited:
carcass said:
Wait, this may work. Not, two look, but few moves... probably
FB
dbL square (2x2x1)
Get into a winter variation situation with the pair solved but not inserted
solve last pair and CMLL with 172 algs(I think that would be the right number)
Solve BD and FD with one of 119 algs
solve ELL with one of 25 algs
Overall this would have 316 algs, less than OLLCP. I think 'yall should look into this one.
Click to expand...
Just do Roux, which is the unfortunate truth for a lot of methods. I also think you mean dbR square.
 
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