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JTWong71

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However this is done, I think the best way to do this is to intuitively position the last corner of a LL pair so that it joins with the pair when the last F2L pair is inserted. It may take some time to get used to and fast at, but I think it will be much like F2L in the sense that it will just come naturally eventually. EO may also be able to be solved during the insertion.

I have tried generating a few of these...
Some of the movecounts go up to 16 moves.
This could work out awesome IMO, we just need an efficient way of solving the line.
EO is pretty easy, just solve in the beginning or after a 2x2x3 block.
Here is my idea for the line:
Solve EO and F2L-1.
Solve 2 pairs: the last F2L pair and a LL pair. Both pairs should be able to be aligned so that the R face is free (besides F2L).
Position the corner so that however you insert it, the corner connects with the LL pair forming a line.
I just thought about something else about this last night...
Since there are 17 Last Layer cases, you have a 1/18 chance of getting a Last Layer Skip.
Also, you only need to look at 3 stickers at most to know the permutation cycle for Line Last Layer.
Orientation recognition is really done almost subconsciously for me, so I will only need to look for COLL recognition for 2 Corners. If there is a swap, then there is a J-Perm or an F-Perm, looking at one edge can tell you. If there is no corner swap, there is a U-Perm or pure flip.
 

wir3sandfir3s

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I have tried generating a few of these...
Some of the movecounts go up to 16 moves.

I just thought about something else about this last night...
Since there are 17 Last Layer cases, you have a 1/18 chance of getting a Last Layer Skip.
Also, you only need to look at 3 stickers at most to know the permutation cycle for Line Last Layer.
Orientation recognition is really done almost subconsciously for me, so I will only need to look for COLL recognition for 2 Corners. If there is a swap, then there is a J-Perm or an F-Perm, looking at one edge can tell you. If there is no corner swap, there is a U-Perm or pure flip.
New idea:
The recog would be look for headlights and then find the edge that goes between them, probably. If there are no headlights, then use a different alg. Simple.
At that point, it can either be a les alg intensive 2LLL or a LS method for LL.
 

JTWong71

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New idea:
The recog would be look for headlights and then find the edge that goes between them, probably. If there are no headlights, then use a different alg. Simple.
At that point, it can either be a les alg intensive 2LLL or a LS method for LL.

I don't see where the LS part is said.
The headlights idea is pretty good.
 

Shiv3r

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1 Look Line Last layer! (nicknamed 1LOL LOL LOL This is awesome)
This CFOP/ZZ/Petrus variant was inspired by brainstorming ways of restricting the last layer during F2L so that the amount of cases for a 1-look last layer dropped. This is the result of me and JTWong71's brainstorming.
The Idea is to have a solved 1x2x3 Line, with edges already oriented as in WV.
Some of the benefits is that 1/3 of the time you get a really easy PLL(U-perms, F-perm, J perms), and recognition is actually really good. There is also a 1/18 chance of skipping the last layer altogether, and 8 out of 18 times you get a really fast 2-gen case.

This variant comes in three parts:
-solve F2L-1 and eo
-solve 2 corner edge pairs
CE pair a:U Layer corner-edge pair and insert in slot(part of the solved line)
CE pair b:F2L corner-edge pair placed at BL
-F2L + L(aka LS+L) - 9 cases
-1LLL(1 Look Line Last Layer) - 18 cases(5 PLL, 1 solved)

The algorithms are grouped by the position of the final Line corner.(2/3 of the line is in the F2L slot). The F2L pair is at BL. The orientations are relative to the U layer, not the rest of the Line.

UFR:
Correct Orientation: (U) L' U2 R U' R' U2 L
Clockwise Twist: (U2) R' U2 R U R' U R2 U2 R'
Counterclockwise Twist: R U2 R'

UFL:
Correct Orientation: (U2) R2 D R' U R D' R2 U R U' R'
Clockwise Twist: (U2) R U R' F' U' F U R' F R F'
Counterclockwise Twist:

UBR:
Correct Orientation: (U2) R' F' R U2 R U2 R' F
Clockwise Twist: R U2 R D' R U' R' D R U R
Counterclockwise Twist: R2 D R' U2 R D' R2

corners oriented:
Just do one of your friendly PLL's. It will either be an F-perm, one of the U-perms, or a J-perm.

(Note: LL Line is placed at UL)
2GLL:

T-Set:
Pure Flip: R' U R U2 R' U2 R' U' R U' R' U2 R U2 R
Clockwise U-Perm: (U2) R U2 R U' R2 U' R U R2 U2 R2 U' R U' R
Counterclockwise U-Perm: (U2) R U R' U R U2 R' U' R U2 R' U' R U' R'

U-Set:
Pure Flip: (U) R U2 R' U' R U' R' U2 R' U2 R U R' U R
Clockwise U-Perm: (U2) R' U' R U R' U R U2 R' U R U2 R' U' R
Counterclockwise U-Perm: (U2) R U R' U' R U' R' U2 R U' R' U2 R U R'

Corner Swap:

T-Set:
F-Perm: F D R' U R U' D' R U F R F' U' R' F'
Ja-Perm (Edge Swap: UB + UR): R2 F' R2 U' R' F' U' F R U R' F R
Jb-Perm (Edge Swap: UF + UR): F R D' R2 D R D' R D F' U' R' U R

U-Set:
F-Perm: F U' R U R' F R' F' R2 U' R' F' U' F U F'
Ja-Perm (Edge Swap: UB + UR): (U') R U' R2 F2 D' F' D F' R U R' F R F'
Jb-Perm (Edge Swap: UF + UR): x' U' R U R' F R U' R' U2 F' U' F R' F' R F'
for the first pair, pair up a corner and an edge in the U-layer(make sure they can fit in the slot)
for the second pair, intuitively solve the final CE pair so it preserves the first corner-edge pair(may take some practice). If possible, track the Final Line corner during this step.
 

JTWong71

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So, Heise?
Not exactly.
1LLLL is meant mainly for Speedsolving.
With Heise, look ahead and recognition are much more difficult, having to look for 3-5 edges and 2 Corners.
With 1LLLL, you only have to look for 2 Pieces at a time.
2 For the first CE Pair, 2 for the Next CE Pair, and 1 For the Corner.
 

JTWong71

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East Coast of the US
The main difference is that you don't permute all of the edges.
Permuting Edges and 2 Corners in one look isn't the best for Speedsolving in most cases.
So instead of solving 7 Pieces at once in Heise, we are solving 5 Pieces, with substeps in between.
Heise can have substeps, but the Edge Permutation a little hard to do in a Speedsolve while preserving Pairs.
 

Vexatious

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Apr 19, 2016
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Ya I'm completely lost. So you make a 1x2x2 line where? F2l-1 is basically 2x2x3 like Petrus right? Can you make a tutorial or some example solves
 

adimare

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Here's an example solve using ZZ for the F2L-1:

Scramble: z2 B2 L2 B2 D2 R2 D2 R2 U R2 F2 U' R' U' B' R2 B2 R2 U L2 F' U'
F' L2 B R' D2 R' D // EO Line
U2 L2 U2 L2 U R U L U L' // First block
U2 R' U R2 U2 R U' R2 // F2L-1
U' R U R' // CE Pair
U' R U2 R' U' // F2L Pair
U2 R2 D R' U R D' R2 U R U' R' //F2L+L
U2 L' U' L F L' U' L U L F' L2 U L // PLL


It's very similar to Heise's two pair approach. In Heise you solve 2 1x1x2 corner-edge pairs and then solve the edges without breaking them up, solving 5 edges and 2 corners in the process (so you're left with 3 corners to solve). With this method, you solve 2 1x1x2 corner-edge pairs with the restriction that one of them has to be the missing F2L block, then you solve those 2 blocks plus one more corner attached to the 1x1x2 block from the top layer (creating the so-called line, which is just a 1x1x3 block), solving 2 edges and 3 corners in the process (so you're left with 3 edges and 2 corners to solve).

I use the Heise two pair approach a lot when solving for fun and it's my go to approach for FMC, and I can tell you it's not really viable for speedsolving for the same reasons I think this method (at its current state) isn't either: creating the 2 corner-edge pairs quickly is not trivial, if you throw in the restriction that one of the pairs has to be the missing F2L block and that when you're done the other one has to be inserted in its place, you're complicating the last layer so much that the effort isn't worth it.

Just my 2 cents tho, probably worth it to keep looking into this one and try to figure out a way to simplify F2L+L; this is light years away from the nonsensical Hawaiian Kociemba, for instance.
 

Shiv3r

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Here's an example solve using ZZ for the F2L-1:

Scramble: z2 B2 L2 B2 D2 R2 D2 R2 U R2 F2 U' R' U' B' R2 B2 R2 U L2 F' U'
F' L2 B R' D2 R' D // EO Line
U2 L2 U2 L2 U R U L U L' // First block
U2 R' U R2 U2 R U' R2 // F2L-1
U' R U R' // CE Pair
U' R U2 R' U' // F2L Pair
U2 R2 D R' U R D' R2 U R U' R' //F2L+L
U2 L' U' L F L' U' L U L F' L2 U L // PLL


It's very similar to Heise's two pair approach. In Heise you solve 2 1x1x2 corner-edge pairs and then solve the edges without breaking them up, solving 5 edges and 2 corners in the process (so you're left with 3 corners to solve). With this method, you solve 2 1x1x2 corner-edge pairs with the restriction that one of them has to be the missing F2L block, then you solve those 2 blocks plus one more corner attached to the 1x1x2 block from the top layer (creating the so-called line, which is just a 1x1x3 block), solving 2 edges and 3 corners in the process (so you're left with 3 edges and 2 corners to solve).

I use the Heise two pair approach a lot when solving for fun and it's my go to approach for FMC, and I can tell you it's not really viable for speedsolving for the same reasons I think this method (at its current state) isn't either: creating the 2 corner-edge pairs quickly is not trivial, if you throw in the restriction that one of the pairs has to be the missing F2L block and that when you're done the other one has to be inserted in its place, you're complicating the last layer so much that the effort isn't worth it.

Just my 2 cents tho, probably worth it to keep looking into this one and try to figure out a way to simplify F2L+L; this is light years away from the nonsensical Hawaiian Kociemba, for instance.
yes, especially because it restricts the amount of cases for a 1LLL to 18, including solved.

And I can do the 2 pairs fast, its just figuring out hoe to manipulate the pairs so you dont destroy the other one.
 
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