The New Method / Substep / Concept Idea Thread

TDM said:
Yeah sorry, that was stupid of me. But I counted 40 for opposite flip:
Ua, Ub: 2+2=4
H, Z: 1+2=3
Aa, Ab: 2+2=4
E, F: 2+2=4
Ga, Gb, Gc, Gd: 2+2+2+2=8
J, L: 2+2=4
Na, Nb: 1+1=2
Ra, Rb: 2+2=4
T, V, Y: 2+2+2=6
4+3+4+4+8+4+2+4+6=39, plus one for solved permutation = 40.
If I'm right, that leaves 156 cases, or 155 that aren't a LL skip. A lot less than the 576 I originally calculated.
Click to expand...

That sounds about right. ZBLL has 177 cases not counting mirrors and inverses and this has slightly less, so it's a reasonable answer.

Thanks for your help. When supercavitation finishes alg generation, I'll be sure to give you credit.
 
I just manually generated every single case (algs will follow ASAP) and came up with 160 cases:

Ua: 8
Ub: 8
H:4
Aa: 8
Ab: 8
Z: 6
E: 6
Ja: 8
Jb: 8
Ra: 8
Rb: 8
Ga: 8
Gb: 8
Gc: 8
Gd: 8
Na: 6
Nb: 6
T: 8
Y: 8
V: 8
F: 8
Solved: 4

Where are the 4 extras?
 
notfeliks said:
That sounds about right. ZBLL has 177 cases not counting mirrors and inverses and this has slightly less, so it's a reasonable answer.

Thanks for your help. When supercavitation finishes alg generation, I'll be sure to give you credit.
Click to expand...
Ty, but you don't have to... I wouldn't say that I've helped much. All I've done is count cases, and even that's taken me quite a few attempts.
supercavitation said:
Where are the 4 extras?
Click to expand...
N perms, like H perms, have a rotational symmetry of 4, and therefore are only counted once for each EO case. The four extras are 2 in each N perm.
 
hey guys, i invented a new method. i dont have a name yet but here are the substeps.

Do first to layers but make it look exactly like a pll case ( if you dont know what i mean just do f2l then x2 with t perm or any other)
Do the last layer untill you get to pll then solve the whole cube in one step.

Pros: easier f2l if you can use ur head. you don't have to look at the bottem when doing doublepll
Cons: maybe hard for beginners to do f2l.

if your confused i'll put an example solve. Also anyone want to help me get the algorithms. i think there are about 300, but like me most people would do tperm f2l so it drops it down to 84 and if you move the d or u layer, its 21
 
I personally don't think this is terribly viable, but I'll post it anyway.
An LS/F2L method + 1LLL (watch this get slammed by Kir)

Step 1:
The goal is to complete the entire F2L, with a solved 2x2 block in the LL.
There are multiple approaches, but the easiest way I found was to create the last pair, attaching a 2x1 to an edge, and inserting the last pair using 3 move regular inserts or a sledge, depending on which one preserves the 2x1. There isn't always a 2x1, but there often are 3 move ones.
If there are no oriented edges, I orient some first, because it is difficult(personally) to orient edges while preserving the pair and the 2x1.

This step was much harder than I thought it would be, because I kept making the 2x2 block and trying to preserve it while making the last pair.

Step 2:
Why make a 2x2 block? To cut down LL cases.
If I calculated correctly, there are 9 corner orientation cases(Solved, La, Lb, Ta, Tb, Ua, Ub, S, AS), 2 edge orientation cases (solved, 2 flip), and 6 PLL cases(solved, V, Aa, Ab, Ja, Jb).
This means there are 9 * 2 * 6 = 108 LL cases.
1/2 are ZBLLs. The other 1/2 need to be generated. I have made some, but they are not pretty: R' U R U2 L' U R' U' L U l R U' R' U; l' U2 l F' l' U2 l r U' R' U M; etc.
Example solve:
D B2 R2 U2 B2 D' B2 U2 F2 U F' R F' R2 B' R' B' U2 R U R' U' R U' R'
L U' R U R' // pair
U L' U L U2 L' U2 // Fix back pair, solve 2x2
R2 F2 R2 U2 B2 L2 B' L2 B' U2 R' F' R //LL
Click
Well, thats it. To explain the corner orientations, there are 2 Ls, where 1 of the 2x2 blocks is solved, but the other isn't. Same with T and U.
 
Last edited:
Phillip1847 said:
I personally don't think this is terribly viable, but I'll post it anyway.
An LS/F2L method + 1LLL (watch this get slammed by Kir)

Step 1: BBLS
The goal is to complete the entire F2L, with a solved 2x2 block in the LL.
There are multiple approaches, but the easiest way I found was to create the last pair, attaching a 2x1 to an edge, and inserting the last pair using 3 move regular inserts or a sledge, depending on which one preserves the 2x1. There isn't always a 2x1, but there often are 3 move ones.
If there are no oriented edges, I orient some first, because it is difficult(personally) to orient edges while preserving the pair and the 2x1.

This step was much harder than I thought it would be, because I kept making the 2x2 block and trying to preserve it while making the last pair.

Step 2: 22LL
Why make a 2x2 block? To cut down LL cases.
If I calculated correctly, there are 9 corner orientation cases(Solved, La, Lb, Ta, Tb, Ua, Ub, S, AS), 2 edge orientation cases (solved, 2 flip), and 6 PLL cases(solved, V, Aa, Ab, Ja, Jb).
This means there are 9 * 2 * 6 = 108 LL cases.
1/2 are ZBLLs. The other 1/2 need to be generated. I have made some, but they are not pretty: R' U R U2 L' U R' U' L U l R U' R' U; l' U2 l F' l' U2 l r U' R' U M; etc.
Example solve:
D B2 R2 U2 B2 D' B2 U2 F2 U F' R F' R2 B' R' B' U2 R U R' U' R U' R'
L U' R U R' // pair
U L' U L U2 L' U2 // Fix back pair, solve 2x2
R2 F2 R2 U2 B2 L2 B' L2 B' U2 R' F' R //LL
Click
Well, thats it. To explain the corner orientations, there are 2 Ls, where 1 of the 2x2 blocks is solved, but the other isn't. Same with T and U.
Click to expand...

This is an interesting idea, but I think it would be better as an opportunistic subset instead (like ZBLL), where you only do the 1LL if you have the 2x2x1 and you know the case.



JKNK said:
hey guys, i invented a new method. i dont have a name yet but here are the substeps.

Do first to layers but make it look exactly like a pll case ( if you dont know what i mean just do f2l then x2 with t perm or any other)
Do the last layer untill you get to pll then solve the whole cube in one step.

Pros: easier f2l if you can use ur head. you don't have to look at the bottem when doing doublepll
Cons: maybe hard for beginners to do f2l.

if your confused i'll put an example solve. Also anyone want to help me get the algorithms. i think there are about 300, but like me most people would do tperm f2l so it drops it down to 84 and if you move the d or u layer, its 21
Click to expand...

I don't see this as being viable. Piece recognition during F2L would be unnecessarily hard, and I would be willing to bet that any single PLL case would be better than an alg that solves two.
 
Phillip1847 said:
I personally don't think this is terribly viable, but I'll post it anyway.
An LS/F2L method + 1LLL (watch this get slammed by Kir)

Step 1: BBLS
The goal is to complete the entire F2L, with a solved 2x2 block in the LL.
There are multiple approaches, but the easiest way I found was to create the last pair, attaching a 2x1 to an edge, and inserting the last pair using 3 move regular inserts or a sledge, depending on which one preserves the 2x1. There isn't always a 2x1, but there often are 3 move ones.
If there are no oriented edges, I orient some first, because it is difficult(personally) to orient edges while preserving the pair and the 2x1.

This step was much harder than I thought it would be, because I kept making the 2x2 block and trying to preserve it while making the last pair.

Step 2: 22LL
Why make a 2x2 block? To cut down LL cases.
If I calculated correctly, there are 9 corner orientation cases(Solved, La, Lb, Ta, Tb, Ua, Ub, S, AS), 2 edge orientation cases (solved, 2 flip), and 6 PLL cases(solved, V, Aa, Ab, Ja, Jb).
This means there are 9 * 2 * 6 = 108 LL cases.
1/2 are ZBLLs. The other 1/2 need to be generated. I have made some, but they are not pretty: R' U R U2 L' U R' U' L U l R U' R' U; l' U2 l F' l' U2 l r U' R' U M; etc.
Example solve:
D B2 R2 U2 B2 D' B2 U2 F2 U F' R F' R2 B' R' B' U2 R U R' U' R U' R'
L U' R U R' // pair
U L' U L U2 L' U2 // Fix back pair, solve 2x2
R2 F2 R2 U2 B2 L2 B' L2 B' U2 R' F' R //LL
Click
Well, thats it. To explain the corner orientations, there are 2 Ls, where 1 of the 2x2 blocks is solved, but the other isn't. Same with T and U.
Click to expand...

BBLS: Sorry, the name is taken [wiki]BBF2L[/wiki]. It's a viable method. I like the idea. Are you saying permuted or just oriented.

22LL is also a thing but with 2 2-cycles. More info on meep's site. The algs are on Michael Gottlieb's site (Tripod page)

So naming is used but it is a nice idea. Kinda like CFOP version of tripod.
 
TheNextFeliks said:
BBLS: Sorry, the name is taken [wiki]BBF2L[/wiki]. It's a viable method. I like the idea. Are you saying permuted or just oriented.

22LL is also a thing but with 2 2-cycles. More info on meep's site. The algs are on Michael Gottlieb's site (Tripod page)

So naming is used but it is a nice idea. Kinda like CFOP version of tripod.
Click to expand...

Hey, I found the LL algs. http://www.ws.binghamton.edu/fridrich/L1/ece.htm
bit outdated, hehe
Can't find anything on meeps site, but I took a look at Gottlieb's site.
Definitely permuted and oriented, if just oriented there would be over 3x as many algs.
Personally, I know almost all of the La subset, and have experimented a decent amount with the last slot part, which I have no idea what to call.

Rarely, there are very difficult cases, where getting the 2x1 is too lengthy.
Similarly, sometimes you will get the back 2x3 solved. This is a time to do an x and apply an alg, as from a different reference point, you have a LL alg.
 
JKNK said:
hey guys, i invented a new method. i dont have a name yet but here are the substeps.

Do first to layers but make it look exactly like a pll case ( if you dont know what i mean just do f2l then x2 with t perm or any other)
Do the last layer untill you get to pll then solve the whole cube in one step.

Pros: easier f2l if you can use ur head. you don't have to look at the bottem when doing doublepll
Cons: maybe hard for beginners to do f2l.

if your confused i'll put an example solve. Also anyone want to help me get the algorithms. i think there are about 300, but like me most people would do tperm f2l so it drops it down to 84 and if you move the d or u layer, its 21
Click to expand...

That sounds a little like Ortega for 3x3.
 
New Method Idea

1. 2x2x3 with FD (Either blockbuilding or cross with 2 pairs)
2. Insert two pairs ignoring orientation (Usually 3 or 4 moves)
3. Solve one pair and orient all edges
4. Orient corners (EJF2L)
5. PLL

Decent method. Algs are alright.
 
GGLL (new idea)

So... this is an idea, and I want to know if it could be faster than CFOP (only the OP bit tho)


Cross - F2L - (ZBLL///OLLCP - EPPL)


after f2l either using ZBLL or OLLCP depending on whether the cross comes up or any other case...
I understand that in order to correctly know ´how to recognize and execute roughly 750 algorithms is insane, but technically speaking, could it be faster than CFOP?

OPINIONS PLEASE?
 
Well, it would be faster to use ZBF2L to guarantee the cross on top and then use ZBLL. Even this would require too many algs. The number of algs you would need to learn to use GGLL is ridiculous, like you said. In my opinion, it isn't worth it. Recognition would be extremely difficult too. Just my opinion.
 
In theory, yes, it could be faster than OLL>PLL, but in practice, the number of algs required would be ridiculously huge that it would be practically impossible to retain them all, and practice them regularly, like you would do with OLL and PLL. Not to mention the recognition - that would take a fair chunk of time in a solve as well.

Also, for such a raw idea as this one, please post here next time: http://www.speedsolving.com/forum/showthread.php?40975-The-New-Method-Substep-Concept-Idea-Thread/
 
Thanks for the replies! :D
I am still quite a noob and I'm just thinking of stuff to do, after learning full pll recently it dawned on me that after I've finished learning all 57 oll's I'll run out of stuff to learn, then it occurred to me that I should learn coll later,
another cool idea is ZBF2L and COLL and EPLL,
Not sure if that would be faster than F2L, OLL and PLL tthough
Opinions? xD
 
I don't know I'm a noob xD
But seriously, would it be faster? Cause if it would then I'll stop learning OLL and switch right over.

I think I was describing VH method..
I will go and decide whether to learn CFOP or VH. Please somebody help me.
DECISIONSSS :((
 
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