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Petro Leum

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Does anyone use wide or slice turns during the eoline to make it easier or to orient edges?[/QUOTE]

yes, i actually use M and S slices alot, or start from a different oriantation and finish EOLine with a wide turn.
 

Berkmann18

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When I do CFOP, I orient edges with the last slot, Heise style. I think that counts.

For sure it counts as an F2LL "way of solving", so I guess yes.

I belive that's intuitive VHLS. Although I'm not entirely sure.

VHLS is two of the basic F2L cases ([U, R] and R U R') of ZBLS, so if you do go through one of those two F2L cases during the LS than you are using VH otherwise ZB.

Well, no. But I will on occasion do CP, CO, EP, or OCLL and PLL.

Oh, so you would approach it as 3LLL (with the first look skipped off course) or more LbL's-like.
 

2180161

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yeah so an example solve:

Scramble: B2 D U2 R2 D' F2 U' F2 D2 R2 L' B F2 U R D2 L B' U F' D'

F2L-1: z D L D2 L U L D2 F' D' U' R L' U' R' L2 U' L' U' L U L' U' f R' f' U L' U L U' L' U' L

Intuitive ZBLS: d' R' U' R U R' U R U R B' R' B

2-look COLL: L' U2 L U L' U2 R U' L' U R' U R U2 R2 U' R2 U' R2 U2 R

EPLL: F2 U R' L F2 R L' U F2
 
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Millet

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Does anyone use wide or slice turns during the eoline to make it easier or to orient edges?

I frequently use slice turns, mostly M and E turns, to simplify awkward EO cases. Strangely I don't use any S slices though.. Don't know why.

So... I have tried many methods, but the one that I really seem to like is ZZ. So, how should I go about practicing my EOline? I first build the line, then do Eo, but have the line built so it is a D, D', or D2 away from being solved. As a 40 second ZZ solver, is that a decent way to start?

I don't think that doing the line before the EO is a good way to start learning EO-line, as at some point you will want to switch to full EO line, where the line is made after/simultaneously as the EO. To do this I would suggest practicing like this (only move onto the next step when you are comfortable with the current):

1. EO only, planning everything in the inspection (limitless inspection)
2. EO only (Blind), making use of what you've learnt in the previous step.
3. Simpler cases EO-Line, planning everything in the inspection (limitless inspection)
4. Simpler cases EO-Line (Blind)
5. All cases of EO-Line, planning everything in the inspection (limitless inspection)
6. All cases of EO-Line (blind).

I started out somewhat like this, but keep in mind that you are going to fail some times. I failed a lot in the beginning. But just keep at it and you will get better.
 
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aym

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1-LLL using 38 algs (EP/N3C-LS)

I've been working on a way to get a 1-LLL using 38 algs:
an 8-alg LS,
an LL that uses 28-ZBLL lags, and
if you have an OLL skip, use E or H perm.

The idea is instead of phasing, completely solve edge perm (6 algs). In addition, before using one of the algs to place the LS, figure out how to break up corner 3-cycles (2 algs). On the LL, figure out the PLL (4 cases), then the OLL (7 cases if there's no skip) and use one of the 28 ZBLL algs to complete the solve.

I call the method EP/N3C-LS. EP for edge permutation and N3C for No (corner) 3-Cycles.

The EP Part
Look at the non-yellow colors of the UL, FR and UR edges in this order. There are 24 cases. These can be grouped into 6 sets of 4. The 6 sets are:

Case 1 - R U' R' (3h, 3q)
BGR
ORB
GBO
ROG​

Case 2 - U R U2 R' (4h, 5q)
BOG
OGR
GRB
RBO​

Case 3 - R U R' U' R U' R' (7h, 7q)
BRO
OBG
GOR
RGB​

Case 4 - L' U2 R U R' U2 L (7h, 9q)
BOR
OGB
GRO
RBG​

Case 5 - U L' U R U' L R' U' R U' R' (11h, 11q)
BRG
OBR
GOB
RGO​

Case 6 - U' R U' L R' U' R U L' R' (10h, 10q)
BGO
ORG
GBR
ROB​

The first 3 use the same algs using in phasing and they already permute the edges. The last 3 are the ones needed so that all the cases permute the edges.

The No Corner 3-Cycles Part
1. For cases 1-5 above, look at the DFR and UFR corners:
If they are colorwise-opposite (e.g. blue/red(/yellow), green/orange(/yellow) are diagonally opposite), perform the edge solving algorithm.
If the corners are not colorwise-opposite, look at the UBR and UBL corners and decide which is the one that is colorwise-opposite of the DFR corner.
1a. If it's the UBL, use this algorithm:
 R U' L' U R' U' L U (8h, 8q)
1b. If it's the UBR, use this algorithm:
 U' L' U R U' L U R' (8h, 8q)​
Perform the edge solving algorithm.​

2. For case 6 (BGO, ORG, GBR, ROB), look at the UFR and UBL corners:
If they are colorwise-opposite, perform the edge solving algorithm.
If the corners are not colorwise-opposite, look at the UBR and UBL corners and decide which is the one that is colorwise-opposite of the UFR corner.
2a. If it's the UBR, use this algorithm:
 U' L' U R U' L U R' (8h, 8q)
2b. If the UBR and UBL are the ones that are colorwise-opposite, use this algorithm:
 R U' L' U R' U' L U (8h, 8q)
Perform the edge solving algorithm.​

1-LLL
Since edges are solved and there are no corner 3-cycles, the last layer is greatly simplified. I don't use the standard ZBLL recognition system. AUF so the edges are aligned over their centers, look at the UFR corner, then decide the PLL corner case. There are 4 of them:

  1. Corners are permuted
  2. Corners swap diagonally
  3. Corners swap left to right
  4. Corners swap front to back
Next, look at the OLL case (7 of them or an OLL skip). If OLL skip, then the cube is either solved or needs an E-Perm or H-perm. Of the 7 cases that remain, simple combinatorics says there's only 28 algs at this point. These are all in the ZBLL database (and there's many to choose from). Here's what I use:

1 - Corners Permuted
ZBLL-T (15h, 19q) y' R2 U R U' R U R2 U R2 U' R' U R' U' R2
ZBLL-U (14h, 16q) y R U2' R' U' R U' R' L' U2 L U L' U L
ZBLL-L (15h, 15q) y' R U' L' U R' U' R U R' U' L U R U' R'
ZBLL-H (17h, 23q) y R U R2 U' R2 U' R U2' R U2' R U' R2 U' R2 U R
ZBLL-Pi (15h, 21q) y' R' U2' R U R' U' R2 U2' R2 U' R2 U' R2 U R
ZBLL-S (15h, 19q) y2 L' U L2' U L' U L U2 L U2 L U L' U L2'
ZBLL-AS (15h, 19q) y2 R U' R2 U' R U' R' U2' R' U2' R' U' R U' R2​

2 - Corners Swap Diagonally
ZBLL-T (15h, 20q) y2 R' U2' R2 U' R U' R U' R U' R' U2' R2 U R2
ZBLL-U (14h, 18q) y R' L U2 L2' U R U' L2' U2 L' U' R' U R
ZBLL-L (15h, 16q) R U2' R' U' (R U R' U')2 R U' R'
ZBLL-H (13h, 17q) y L U' R' L2' U R2 U' L2' U R2 L' U' R
ZBLL-Pi (13h, 14q) y' L U' R' U L' U' R2 U' L' U R' U' L
ZBLL-S (11h, 14q) R' U2' R2 U R2 U R U' R U' R'
ZBLL-AS (11h, 14q) L U2 L2' U' L2' U' L' U L' U L​

3 - Corners Swap Left to Right
ZBLL-T (14h, 15q) R L' U' L' U R' U' L U2 R U' L U R'
ZBLL-U (17h, 20q) L U L' U' L2' R' U' R' U L2' U' R U2' L U' L' R
ZBLL-L (16h, 22q) L' R U2' R2 U' R2 U' R' U' L2' U2 R' U R U2' L'
ZBLL-H (13h, 18q) y L U2 L' R' U2 L U2 L' U2 L R U2 L'
ZBLL-Pi (15h, 16q) y' L U2 L' U' L U R' U L' U' R U' L U' L'
ZBLL-S (15q, 15h) R U' L' U R' U' R U' L U R' U' L' U L
ZBLL-AS (15q, 15h) L' U R U' L U L' U R' U' L U R U' R'​

4 - Corners Swap Front to Back
ZBLL-T (18h, 26q) L2' U L' U2 L' U2 L2' R' U L2' U' R U2' L U' L U2 L2'
ZBLL-U (14h, 14q) y2 R U' L' U R' U' L R' U L U' R U L'
ZBLL-L (17h, 24q) L R U2' R2 U L U L2' U L2' U2 L' R2 U2' L' U' R'
ZBLL-H (14h, 14q) R U' L' U R' U' L R U' L' U R' U' L
ZBLL-Pi (15h, 21q) y' L' U L' U2 L U2 R U' L U R' U2 L2 U2 L2
ZBLL-S (15h, 21q) y2 R U R2 U' L' U2 R2 U2' R' U2' R' U2' L U R
ZBLL-AS (15h, 21q) y2 L' U' L2' U R U2' L2' U2 L U2 L U2 R' U' L'​

With cases 3 and 4 if the AUF or cube rotation along the Y axis is a quarter turn, then a left to right swap (case 3) becomes a front to back swap (case 4) and visa versa.

I tried to make my algs match the ZBLL database on speedsolving.com (change leading U turns to y rotations, have the starting position of the U layer match the standards of this database), though the recognition system used here is based on edges and not where the blue/red/yellow corner is located. These algs do work and I checked everyone with alg.cubing.net, however, the AUF's may look a bit weird when looking at it from the perspective of a edge based recognition system. If you find something that doesn't make sense or does not seem to work, let me know and I'll do my best to make the necessary corrections.

Putting It All Together
(Assumptions:
Yellow is the top color.
Blue is the front color.
The blue/red/white CE pair is last.
The blue /red/white pair is to the left it's slot and can be inserted with R U' R'.)​
  1. Figure out what EP case it is based on the UL, FR, and UR edges.
  2. Based on the EP case, figure out what N3C case it is.
  3. Execute the N3c alg.
  4. Execute the EP alg.
  5. AUF so that the U edges line up with the centers.
  6. Figure out the corner PLL. Take special note if it's case 3 (left to right swap) or 4 (front to back swap).
  7. Figure out the corner OLL.
  8. AUF so that the OLL is in the correct starting position.
  9. If the AUF is a quarter turn to get the OLL in position and the the PLL is case 3, use case 4 instead and visa versa.
  10. Execute the ZBLL alg.
Some Thoughts
I think the recognition systems of all the parts EP, N3C and LL are fairly simple and straight forward, though I'm not sure if it is quick and simple enough to be completely practical for speed cubing. I've been trying it for a few weeks now and I seem to be improving, but I need much more practice.

The worst LS case has 19 turns (EP case #5 (11h, 11q) + any N3C case (8h, 8q)). This may turn off some speed cubers. On the bright side the algs are RUL.

I started the Rubik's Cube Club at one of our elementary schools. I taught an LBL method that starts with a white cross on the bottom. Next year, I want to teach a speed cubing method, but trying to each ZZ may prove difficult, not to mention I think throws away much of what they already learned. I'm most likely going to teach CFOP where you orient the edges before creating the last CE pair.

Example Solves
L' R B F U' L2 R' D B F D2 U' R' B R2 B' F D2 U B' D R' D2 L' U'
// ZZ 58h, 64q
x2 // inspection
B U' D' F U' B2 // EOline 8 bad edges
U L R' U' R2 // green/red/white block
L' U' L U' L' U L2' // green/orange/white block
U2 R L' U2 L R' // blue/orange pair
U' R U2 R' // blue/red pair
// EP case #1 (ORB), N3C case 1a (UBL)
R U' L' U R' U' L U // N3C case 1a (UBL)
R U' R' // EP case #1 (ORB)
U // Align edges. PLL case classic E-Perm. OLL skip.
U // AUF for E-Perm
x' U L' U' R U L U' R' U L U' R U L' U' R' x U' // E-Perm

// CFOP with edge orientation 74h, 90q
x2 // inspection
U2 F L2' B2 U' R2 // cross
R' U2 R U' B L' B' L // green/orange pair
U' R L' U R' L F U2 F' // blue/orange pair
y U' R U R' U' R U R' y' // green/red pair
U' F' U F // orient edges FR (yellow/orange) and UR (blue/red)
U2 R2 U R2 U R2 U2 R' U R' // blue/red pair
U L' U R U' L R' U' R U' R' // EP case #5 (GOB), N3C skip
U // Align edges. PLL #1-corners in place. OLL-anti-sune
U // AUF to align OLL-anti-sune
R U' R2 U' R U' R' U2' R' U2' R' U' R U' R2 U2 // Corners in place, anti-sune

U2 L R' D' U2 B2 F L2 U R' F U R F2 D' B2 F' U B' F2 L2 B L' D U2
// ZZ 67h, 86q
x2 // inspection
D2 L2' U F U' F2 // EOLine 4 bad edges
R' U2 R U R // green/red/white block
U2 L' U L U2 L' U2 L U2 L U L U' L' U2 L2 // green/orange/white block
L' U2 L // blue/orange pair
R U R' U2 R U' R' U' //blue/red pair
// EP case #4 (GRO), N3C case 1b (UBR)
U' L' U R U' L U R' // N3C case 1b (UBR)
L' U2 R U R' U2 L U' // EP case #4 (GRO)
U2 // Align edges. PLL #2-corners swap diagonally. OLL-anti-sune
U2 // AUF for OLL-anti-sune
L U2 L2' U' L2' U' L' U L' U L // Corners swap diagonally, anti-sune

// CFOP with edge orientation 65h, 76q
x2 // inspection
D B D2 B' D' F2 // cross
U2 R' L U2 L' R // green/orange pair
F U' F' // blue/orange pair
R U R' U R' U' R // green/red pair (and blue/red pair formed too)
R U R' F' U2 F // orient edges UB (blue/red), UL (yellow/red)
U2 // EP case #6 (ORG), N3C case 2a (UBR)
U' L' U R U' L U R' // N3C case 2a (UBR)
U' R U' L R' U' R U L' R' // EP case #6
U // Align edges. PLL #1-corners in place. OLL-sune.
U' // AUF for OLL-sune
L' U L2' U L' U L U2 L U2 L U L' U L2' U2 // Corners in place, sune

U2 R2 U B2 L2 R2 U2 F2 U' F R2 B2 R2 U' L' D2 U' R F2 L2
// famous solve, Collin Burns 5.25s 42h, 47q

// ZZ 64h, 79q
x2 // inspection
R U R' U' F R D // EOLine, line purposely misaligned.
L' U' R U2 R // orange/white bottom row
L2 U' L // orange/middle edges
D2 // finish orange/white
U2 R' U2' R' U' R2 // green/red/white block
R U2 R' U2 // blue/red pair
// EP case #5 (OBR), N3C case 1b (UBR)
U' L' U R U' L U R' // N3C case 1b (UBR)
U L' U R U' L R' U' R U' R' // EP case #5 (OBR)
// Edges already aligned. PLL #3-corners swap left to right. OLL-L.
U // AUF for OLL-bowtie. PLL #3 changes to PLL #4-corners swap front to back.
L R U2' R2 U L U L2' U L2' U2 L' R2 U2' L' U' R' U' // Corners swap front to back, L.

// CFOP with edge orientation 59h, 65q
x2 // inspection
F' R D L' D2' // cross
R U' R' U' L' U' L // blue/orange pair
U L U' L' // green/orange pair
R' U2' R U2' R' U R // green/red pair.
// Up to this point, everything matches Burn’s solve. He finished with
U F' U F // orient edges
U R U' R' U R U R' U2 // blue/red pair
R U R' U' R U' R' // EP case # 3 (BRO), N3C skip
// Edges already aligned. PLL #3-corners swap left to right. OLL-PI
L U2 L' U' L U R' U L' U' R U' L U' L' U2 // Corners swap left to right, PI
 
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Berkmann18

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Aym, the EP part is excellent and very handy for Keyhole.
I'm confused about the "diagnally opposite" because when I try the algs the corners in questions have the side stickers that are the same instead of being opposites.

For case 6 (BGO, ORG, GBR, ROB), look at the UFR and UBL corners:
If they are diagonally opposite, perform the edge solving algorithm.
If the corners are not diagonally opposite, look at the UBR and UBL corners and decide which is the one that is diagonally opposite of the UFR corner.

Apart from what I said above, this method is great, it might be a bit tough for speed but that's not a problem for me.
It should be DFR right ? Also you forgot a 2c where the DFR edge would be the diagnal opposite.
 

aym

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Aym, the EP part is excellent and very handy for Keyhole.
I'm confused about the "diagnally opposite" because when I try the algs the corners in questions have the side stickers that are the same instead of being opposites.

Apart from what I said above, this method is great, it might be a bit tough for speed but that's not a problem for me.
It should be DFR right ? Also you forgot a 2c where the DFR edge would be the diagnal opposite.


Hi Berkmann18,

I just realized that saying colorwise-opposite is more accurate and probably less ambiguous than saying diagonally opposite.

Case 6 is the one weird case. DFR does not play at all in this case, so it's not likely that a 2c needs to be added. You need to look at UFR, UBL. Yes, these two locations are diagonally opposite each other. But you need to look at the colors on these corners. You are interested in seeing if the corners in these two locations are corners whose colors would be colorwise-opposite of each other on a solved cube. Specifically, if UFR is green/red/yellow, UBL needs to be blue/orange/yellow. If UFR is blue/red/yellow, UBL needs to be green/orange/yellow.

So, look at UFR and note the colors of the corner in that location. For example let's say it's the green/red/yellow corner. Next, look at UBL and see if the blue/orange/yellow corner is back there. If so, then running the EP alg will slot the CE pair, solve the edges and make sure the corners are not in a 3-cycle.

If the UBL corner has either green/orange/yellow or blue/red/yellow, then the EP alg will solve the edges, but there will be a corner 3-cycle going in the last layer. To break the 3-cycle, first look at UBR. If blue/orange/yellow is in UBR, blue/orange/yellow needs to move to UFR and UFR (green/red/yellow) needs to move to UBL. Use 2a to do this, U' L' U R U' L U R'.

Now, the really odd part of case 6. If the two corners that are colorwise-opposites are in UBL and UBR, then these need to move to UFR and UBL respectively. To continue on with this example, if green/red/yellow is in UFR, blue/red/yellow is in UBR and green/orange/yellow is in UBL, it won't work to use 2a to move the corners around. It puts green/red/yellow in UBL, blue/red/yellow in UFR and green/orange/yellow in UBR. Thus, 2a leaves UFR and UBL in a state where the corners are not colorwise-opposites. But 2b (R U' L' U R' U' L U) pushes the corners anti-clockwise putting green/red/yellow in UBR, blue/red/yellow in UBL and green/orange/yellow in UFR. Now, the corners in UFR and UBL are colorwise-opposites. Running the EP alg for case 6 will do what we want it to do.

So why break 3-cycles? A big chunk of the ZBLL database solves 3-cycles. Some of the algs are amazingly short. Others are a bit long. I did some quick analysis and without symetries the number of cases for solving LL 3-cycles is 112. Accounting for symetries, I think there are 48. Knowing 2 algs and when to apply them saves a ton of ZBLL memorization. BTW, the algs used here are nothing more than Sune and Anti-Sune variations that 3-cycle corners but leaves edge permutation in tact.
 

Berkmann18

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Hi Berkmann18,

I just realized that saying colorwise-opposite is more accurate and probably less ambiguous than saying diagonally opposite.

Case 6 is the one weird case. DFR does not play at all in this case, so it's not likely that a 2c needs to be added. You need to look at UFR, UBL. Yes, these two locations are diagonally opposite each other. But you need to look at the colors on these corners. You are interested in seeing if the corners in these two locations are corners whose colors would be colorwise-opposite of each other on a solved cube. Specifically, if UFR is green/red/yellow, UBL needs to be blue/orange/yellow. If UFR is blue/red/yellow, UBL needs to be green/orange/yellow.

So, look at UFR and note the colors of the corner in that location. For example let's say it's the green/red/yellow corner. Next, look at UBL and see if the blue/orange/yellow corner is back there. If so, then running the EP alg will slot the CE pair, solve the edges and make sure the corners are not in a 3-cycle.

If the UBL corner has either green/orange/yellow or blue/red/yellow, then the EP alg will solve the edges, but there will be a corner 3-cycle going in the last layer. To break the 3-cycle, first look at UBR. If blue/orange/yellow is in UBR, blue/orange/yellow needs to move to UFR and UFR (green/red/yellow) needs to move to UBL. Use 2a to do this, U' L' U R U' L U R'.

Now, the really odd part of case 6. If the two corners that are colorwise-opposites are in UBL and UBR, then these need to move to UFR and UBL respectively. To continue on with this example, if green/red/yellow is in UFR, blue/red/yellow is in UBR and green/orange/yellow is in UBL, it won't work to use 2a to move the corners around. It puts green/red/yellow in UBL, blue/red/yellow in UFR and green/orange/yellow in UBR. Thus, 2a leaves UFR and UBL in a state where the corners are not colorwise-opposites. But 2b (R U' L' U R' U' L U) pushes the corners anti-clockwise putting green/red/yellow in UBR, blue/red/yellow in UBL and green/orange/yellow in UFR. Now, the corners in UFR and UBL are colorwise-opposites. Running the EP alg for case 6 will do what we want it to do.

So why break 3-cycles? A big chunk of the ZBLL database solves 3-cycles. Some of the algs are amazingly short. Others are a bit long. I did some quick analysis and without symetries the number of cases for solving LL 3-cycles is 112. Accounting for symetries, I think there are 48. Knowing 2 algs and when to apply them saves a ton of ZBLL memorization. BTW, the algs used here are nothing more than Sune and Anti-Sune variations that 3-cycle corners but leaves edge permutation in tact.

I understood, thanks. I will start using it and see if I can combine it with Keyhole and a TSEOLL technic to see how handy it could be.
 

aym

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That's a pretty cool method! :)

It's also pretty good transition to a method that uses more ZBLLs!

Thanks!

Yes, you can do all sorts of things with this method to get to full ZBLL. It also guides the way to get there.

EP + ZBLL set 1 + PLL A, E, H = 2-LLL, something similar to ZZ reduction (16 algs)
EP + N3C + ZBLL sets 1-4 + PLL E, H = 1-LLL (38 algs)
EP + ZBLL sets 1-4 + ZBLL 3-cycle algs + PLL A, E, H = 1-LLL (149 algs)
Phasing + ZBLL subset = 1-LLL (167 algs)
ZBLL!

I understood, thanks. I will start using it and see if I can combine it with Keyhole and a TSEOLL technic to see how handy it could be.

You're welcome and thank you for helping me make things clearer.

I apologize for the noob question, but can you enlighten me on TSEOLL? I haven't heard of it.
 
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MM99

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Hey could someone link me to the cls cases that are useful in ZZ? thanks in adavance
 

Berkmann18

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Hey could someone link me to the cls cases that are useful in ZZ? thanks in adavance

You should take a look at asmallkitten's channel and check the CLS videos or otherwise just try to learn like one case for each subsets.

You're welcome and thank you for helping me make things clearer.

I apologize for the noob question, but can you enlighten me on TSEOLL? I haven't heard of it.

You're welcome too :).
TSEOLL stands for Third Slot Edge Orientation of Last Layer so it's like doing ZB/VH with the 3rd pair instead of the last one. So it would leave the cube to a EP+N3C-LS / OLS (Orientation Last Slot) +PLL / LSCLL (Last Slot Corner of Last Layer) +ELL (Edge of LL) / Keyhole LS (Edge Perm. Last Slot) +Keyhole LL (pure CLL) stage
 
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aym

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You're welcome too :).
TSEOLL stands for Third Slot Edge Orientation of Last Layer so it's like doing ZB/VH with the 3rd pair instead of the last one. So it would leave the cube to a EP+N3C-LS / OLS (Orientation Last Slot) +PLL / LSCLL (Last Slot Corner of Last Layer) +ELL (Edge of LL) / Keyhole LS (Edge Perm. Last Slot) +Keyhole LL (pure CLL) stage

Great info! TSEOLL sounds quite useful. I'm teach elementary school students and I want to get the returning students into speed cubing, and thinking on teaching CFOP. Would love to teach them ZZ, but I think EOLine will just confuse them. The problem is I didn't know of a good way to get them in EP+N3C-LS. This could be it. Thanks!

BTW I like your use of "+" instead of "/" in the name; since "+" implies the concatenation of the two techniques instead of alternatives. I may have to change the name slightly. My other though was to change N3C to NC3C. It's a bit more precise, still a bit cryptic, but also sounds like a Star Wars droid. :)
 

Berkmann18

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Great info! TSEOLL sounds quite useful. I'm teach elementary school students and I want to get the returning students into speed cubing, and thinking on teaching CFOP. Would love to teach them ZZ, but I think EOLine will just confuse them. The problem is I didn't know of a good way to get them in EP+N3C-LS. This could be it. Thanks!

BTW I like your use of "+" instead of "/" in the name; since "+" implies the concatenation of the two techniques instead of alternatives. I may have to change the name slightly. My other though was to change N3C to NC3C. It's a bit more precise, still a bit cryptic, but also sounds like a Star Wars droid. :)

If you want to teach ZZ, you should go through Roux or Petrus' EO detection system first as it will help on getting one of the ZZ EO "rules" down which will make the EO detection on ZZ easier. You should also go through the 2-look approach of the EOline.

For teaching your method, I also don't know which suitable way could be handy :(

I actually use those characters that way and I find myself using | too for alternative (coder's habit).
For the name you might be changing to EpNc3CLS or perhaps EN3CLS or even AymLS.
 

TDM

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If you want to teach ZZ, you should go through Roux or Petrus' EO detection system first as it will help on getting one of the ZZ EO "rules" down which will make the EO detection on ZZ easier. You should also go through the 2-look approach of the EOline.
Roux's EO isn't very similar to ZZ at all. You might as well just teach ZZ EO. Also teach it four (or two) edges at a time, rather than 2-look. The simpler you make it at first, the easier it will be.
 
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