The New Method / Substep / Concept Idea Thread

[Teoidus]

I remember 2x2x3 being something insane, like 9 or 10

 

[Shiv3r]

another random Idea for trying to not get OLL parity: (so far may only work for Yau or Meyer, not Hoya)
-When solving the 4th cross dedge, use 2-at-a-time pairing to solve another edge at the same time. This means that after the first 3-2 in 3-2-3 edge pairing there will be either a skip or 2 edges unsolved. if you track the amount of misoriented edges while edgepairing, you can decide which alg to use for the L2E to not get parity.

 

[Teoidus]

Any one want to try and flesh out Gilles Roux's 4x4 method?

1x3x4 on LR
CLL
solve ULUR dedges
"orient" remaining wings
permute wings
L4C

 

[Shiv3r]

Any one want to try and flesh out Gilles Roux's 4x4 method?

1x3x4 on LR
CLL
solve ULUR dedges
"orient" remaining wings
permute wings
L4C

to y'all not in the roux method speedsolvers:
this was a secret until after gilles roux joined the roux method speed solvers group a week or two ago. after posting a 3x3 average on the weekly competition, he asked what methods were approved for 4x4, and then gave an outline of the method he uses, which is posted above almost verbatim.

I want to help flesh it out.

I like the first 2(3) steps, they're basically the first 2 steps of the Lewis Method. As far as I can tell this is basically Lewis method with a sandwich finish.
How about we skip orientation of wings, and solve the M dedges 2/3 at a time with comms? that means we only have 2 looks for dedges.
and for centers, I think that comms are a little inefficient, I think that @rachmaninovian has a post on a faster L4C with more algs. I'll dig it up and then link it here.
EDIT: here's the sandwich L4C thing I was thinking of.
However, I am wondering what other strategies we can come up with.

 

[Neuro]

I remember 2x2x3 being something insane, like 9 or 10

That's about what I expected after doing tests on HARCS. But how difficult is recognizing multiple 2x2x3 blocks in inspection? Seems like full neutrality would be pretty intense but if you were to restrict it to 8 like in Roux possibly less then it could be viable 100% of the time.

 

[Teoidus]

consistently 1looking a 2x2x3 in inspection petrus-style is pretty hard. Though, I suggested a few months ago that one could onelook FB + track DFDB and centers (essentially 1looking a 2x2x3), which gives ~12-14 stm solutions

 

[Shiv3r]

consistently 1looking a 2x2x3 in inspection petrus-style is pretty hard. Though, I suggested a few months ago that one could onelook FB + track DFDB and centers (essentially 1looking a 2x2x3), which gives ~12-14 stm solutions

this I dubbed Pootris.

Also, I want to help flesh out the 4x4 method @Teoidus, look at above post

 

[Neuro]

So the 2x2x3 is actually the most efficient (no surprise there) but Roux is actually a very close second, >.03 away from 2x2x2 moves per piece
1: Roux- ~8, 5 pieces// 8/5= 1.6
2: 2x2x3- ~11, 7 pieces// 11/7= ~1.57
3: CFOP- ~7, 4 pieces// 7/4= 1.75

EO Line takes like 8 moves max but the solves can be super efficient. Hard to generate stats like those previously listed though.

EDIT: This was stupid. If you can solve optimally they should all be about the same for moves per piece. The point where it really changes are steps 2-x due to influencing future portions of the solve and getting efficient subsets. Still, I think that Roux FB may be one of the best ways to start a solve simply because of ease recognition and ability to look ahead into another step (2x2x2, SB, anything really) while not compromising movecount.

 

[shadowslice e]

So the 2x2x3 is actually the most efficient (no surprise there) but Roux is actually a very close second, >.03 away from 2x2x2 moves per piece
1: Roux- ~8, 5 pieces// 8/5= 1.6
2: 2x2x3- ~11, 7 pieces// 11/7= ~1.57
3: CFOP- ~7, 4 pieces// 7/4= 1.75

EO Line takes like 8 moves max but the solves can be super efficient. Hard to generate stats like those previously listed though.

Well basically if you're going by optimal, the greater the number of pieces, the greater the efficiency for basically all alg set which is why we do things like 1LELL rather than 2LPLL.

 

[Teoidus]

this I dubbed Pootris.

Also, I want to help flesh out the 4x4 method @Teoidus, look at above post

Okay, I'm sorry but this name is unacceptable. Please stop using it. Pootris, Gay redux, SuNe... really? These sound like names toddlers would pick...

As for the 4x4 method, I wonder if there is a way to solve wings + some of L4C at the same time?

 

[Shiv3r]

*states all my names that are obviously jokes*
As for the 4x4 method, I wonder if there is a way to solve wings + some of L4C at the same time?

I don't think the algs would be very nice. check out the thread I posted up above, though. I think we can have L4C be like 4.5-3 look.

 

[crafto22]

EDIT: This was stupid. If you can solve optimally they should all be about the same for moves per piece. The point where it really changes are steps 2-x due to influencing future portions of the solve and getting efficient subsets. Still, I think that Roux FB may be one of the best ways to start a solve simply because of ease recognition and ability to look ahead into another step (2x2x2, SB, anything really) while not compromising movecount.

I don't believe that is correct due to the fact that corners have more possible orientations and therefore more ways to be unsolved, if that makes sense.

 

[Teoidus]

Honestly what you will probably find with something like this is that the step that solves the most pieces at once always "wins" in terms of efficiency. This is because human methods are quite bad in terms of move-optimality (god's number for the <M,U> and <R,U> subgroups is like 20 STM--you are, in a sense, not any "closer" to solved in any of these groups vs in a random scramble)

 

[crafto22]

Has this been proposed before?

1. 3 cross edges, last one is any misoriented U-layer edge (5 moves)
2. F2L-1 (20 moves)
3. Last pair whilst ensuring 3 U-layer edges are unoriented (~7 moves using 41 "new" algs, or ~10 moves with 2 "new" algs)
4. COLL (10 moves)
5. L5E with the easiest EO case, probably possible to one-look almost every time (~10 moves)
Total would be 52 moves with 41 new algs, which is better than regular CFOP, all while making cross easier and use of insepction time more efficient.

Basic algs for inserting the last pair:
Connected pair, edges need to be flipped: y R' F' R U2 M' U' M
Disconnected pair, edges need to be flipped: R' D' r U r' D' R2 U R'

 

[Hazel]

I was actually thinking of having a 2×1×1 bar that formed half of a layer, but the second option you mentioned might work. It's pretty much the same amount of work to create half a layer as opposed to half a face, plus the last step is half as many algs and more ergonomic if you have half a layer. And yes, the orientation step does NOT force the d-face to be solved, as PBL algs are kinda meh and it would be a LOT easier if you solve those two pieces during permutation.

@Aerma I don't think the method you came up with was really viable, but some of the ideas you came up with actually could be really good in WaterRoux: (keep in mind for my example here the three solved edges are DB, DR and DL and we are trying to solve L9E)

1. Solve the DF edge and seperate the edges into their correct layers.
2. Solve the four u-layer edges whilst orienting the midges.
3. Do a Z rotation and finally permute the midges.

This idea seems decent and is most likely less algorithms than Three U-layer edges then L6E, but some of the recogniton might suck and the algs might not be very ergonomic. If this idea seems to have any potential I'd be willing to work on it with you.

Yeah, sounds good! Do you think it's okay enough to pursue, or should I go back to square 1 (pun not intended)?

 

[Shiv3r]

Has this been proposed before?

1. 3 cross edges, last one is any misoriented U-layer edge (5 moves)
2. F2L-1 (20 moves)
3. Last pair whilst ensuring 3 U-layer edges are unoriented (~7 moves using 41 "new" algs, or ~10 moves with 2 "new" algs)
4. COLL (10 moves)
5. L5E with the easiest EO case, probably possible to one-look almost every time (~10 moves)
Total would be 52 moves with 41 new algs, which is better than regular CFOP, all while making cross easier and use of insepction time more efficient.

Basic algs for inserting the last pair:
Connected pair, edges need to be flipped: y R' F' R U2 M' U' M
Disconnected pair, edges need to be flipped: R' D' r U r' D' R2 U R'

it looks like Hawaiian Kociemna, but with less algs than the original(hundreds for HKOLL+HKPLL). It's an HK or 3CFCE variant.(look up 3CFCE on the wiki)
Also, looks like you're just using VHLS 4-flip cases to force a 3-flip on U. I use the first alg regularly, especially on 6x6+ because that's really the only place I main CFOP.
cons I can see:
-not sure how good recognition for onelooking the last step would be.
-the cross with 1 U layer edge misoriented is about the same as cross movecount and looking-in-inspection-wise, I think.

I will try a few solves with this method(using L5EOP and the few of the L5EPLL's I know). maybe I'll post some alg.cubing example solves, I have played with 3CFCE-style methods before and they're pretty nice.

 

[Shiv3r]

Has this been proposed before?

1. 3 cross edges, last one is any misoriented U-layer edge (5 moves)
2. F2L-1 (20 moves)
3. Last pair whilst ensuring 3 U-layer edges are unoriented (~7 moves using 41 "new" algs, or ~10 moves with 2 "new" algs)
4. COLL (10 moves)
5. L5E with the easiest EO case, probably possible to one-look almost every time (~10 moves)
Total would be 52 moves with 41 new algs, which is better than regular CFOP, all while making cross easier and use of insepction time more efficient.

Basic algs for inserting the last pair:
Connected pair, edges need to be flipped: y R' F' R U2 M' U' M
Disconnected pair, edges need to be flipped: R' D' r U r' D' R2 U R'

here is an example solve
it's super inefficient because it only saved one move in cross, and it made the F2L actually less efficient, even if you don't include the flipping with the LS.

EDIT: Here is the cross you could do with 1 extra move it's a 5 HTM xcross. after finding this xcross, decided to try a speedsolve with the scramble, and I got a sub-14 solve(I am like sub-25 with CFOP usually, it's not my main method). let me try to reconstruct what I did.

E2: Here is a full reconstruction of what I did. it's 47 htm/stm(same either way)

 

[Rpotts]

We do not just reduce the number of cases, zbll is not really 1lll as it requires a pre-1lll setup to solve in one look meaning last slot takes longer, if someone knew full 1lll that would still be around 4000 algs

ZBLL is really 1 look LL assuming you have edges oriented when you finish F2L. Whether you do that during EOLine or freely during F2L, or some sort of EOLS is up to you.

Also Jabari knows a really large subset of LL cases and LSLL cases, so it may be more possible than we once thought. Who knows.

 

[Neuro]

So here's a solve of a Roux/ZZ hybrid I've been playing with. Scramble is from the example solve thread.

L B2 U2 R2 B2 R2 F' L2 F' U2 L2 B U' L2 R D2 R' F' U' F2 R

R2 L F' R U' F B U D2 R2 U D L2//EOFB + LR (13/13)
U R2 U2 R2 U' R2 U R U' R'//SB (10/23)
U L' U' L U' L' U2 L//COLL (8/31)
U M2 U M' U2 M' U2 M2//4C (8/39)

Got pretty lucky with LR. I think that making FB is kinda awkward though and F2L edges can get trapped in DFDB. Anyone think it can be made decent?

 

[Shiv3r]

So here's a solve of a Roux/ZZ hybrid I've been playing with. Scramble is from the example solve thread.

L B2 U2 R2 B2 R2 F' L2 F' U2 L2 B U' L2 R D2 R' F' U' F2 R

R2 L F' R U' F B U D2 R2 U D L2//EOFB + LR (13/13)
U R2 U2 R2 U' R2 U R U' R'//SB (10/23)
U L' U' L U' L' U2 L//COLL (8/31)
U M2 U M' U2 M' U2 M2//4C (8/39)

Got pretty lucky with LR. I think that making FB is kinda awkward though and F2L edges can get trapped in DFDB. Anyone think it can be made decent?

at least it isn't a CFOP/roux hybrid, but I think if you want to do something like that you're better off doing ZZ-LSE or zz-4c or whatever it's called.