ZZ-CT Thread

[JTWong71]

Algorithms: Link

 

[PhillipEspinoza]

So is this what you mean?

Yes, but I didn't once consider that the pseudo pairs could be diagonal, I thought to myself they had to be adjacent along L or R, I thought maybe they could be adjacent along F and B, but now I see they can be any 2 pairs. Wow cool!

"F2L-1 slot, create the pseudo pair and insert with wv?"

That was where I originally started in this thought train, then I thought "what other methods could orient LL pieces while inserting a pair? What would CLS look like? That would be a lot of algs tho for one sub step... but wait, not if both of the L2S corners were oriented... that would narrow down CLS to the 7 OLL cross cases, with 4 different positionings of the LS corner...then TTLL"

EDIT: Wait would it be just 26 because of H case symmetry?

I suppose someone who learns full CLS could just orient pseudo pair and essentially get away with having a truer "Last 3 Edges" step.

BTW going through the posts of the original CT thread I realized Chris kinda had this idea of doing the L2S (in one alg as opposed to my 2 step) but quickly dismissed it realizing that you'd have to have the 2 U-layer corners on the LL in order to have a reasonable set of algs.

Maybe mine is a beginners method to Chris' dream method of doing L2S+LL in one alg?? Surely algs could be generated to orient and permute all LL pieces while inserting the last 2 corners, right?? Woah...

I like the idea of a LL skip method with 100 algs though!

Algorithms: Link

Wow did you just do that? Awesome man thanks!

 

[JTWong71]

Wow did you just do that? Awesome man thanks!

Yep, they're not the best ones of course, I just took the not horrible algorithms that I generated.

 

[PhillipEspinoza]

Revisiting Chris' dream L2S+LL method:

For final block: It's possible to insert pseudo pair into DBR slot, then TSLE the LS leaving you with oriented LL and potential to do the L2S and LL in 1 alg. You would have 4 subsets of cases:

1) diagonal swap UFR and ULB
2) adjacent swap UFR and URB
3) TTLL-1 (the TTLL we know)
4) TTLL-2 (Where DFR belongs in DBR)

How many algs would this be? Anyone?

 

[JTWong71]

Revisiting Chris' dream L2S+LL method:

For final block: It's possible to insert pseudo pair into DBR slot, then TSLE the LS leaving you with oriented LL and potential to do the L2S and LL in 1 alg. You would have 4 subsets of cases:

1) diagonal swap UFR and ULB
2) adjacent swap UFR and URB
3) TTLL-1 (the TTLL we know)
4) TTLL-2 (Where DFR belongs in DBR)

How many algs would this be? Anyone?

I don't understand that very well.
Do you mean we leave have an Oriented Right Block with not necessarily correct Corner Permutation but only Edge Permutation and Corner Orientation, then we just Orient the U-Layer and use one algorithm for the rest?

 

[PhillipEspinoza]

There might be no point in learning this entire method when you could just use ZZ-CT as I imagine a very high alg count and it would indeed be a long hike, like the PCT, even for a seasoned alg thru-hiker like Chris, but the idea is out there.

 

[JTWong71]

There might be no point in learning this entire method when you could just use ZZ-CT as I imagine a very high alg count and it would indeed be a long hike, like the PCT, even for a season alg thru-hiker like Chris, but the idea is out there.

I'm pretty sure I saw Chris Tran talk about that idea somewhere else in the forums, where you do TTLL for 2 F2L Corners.
He said that the problem is that the skip rate is reduced drastically that it doesn't become worth learning.

 

[PhillipEspinoza]

I'm pretty sure I saw Chris Tran talk about that idea somewhere else in the forums, where you do TTLL for 2 F2L Corners.
He said that the problem is that the skip rate is reduced drastically that it doesn't become worth learning.

Yeah he said the problem is the cases he was talking about were where the 2 F2L corners are in the top layer, which is rare, but not when you make false pairs. He was just thinking of inserting the L3E regardless of orientation and permutation of everything else. But yeah the skip rate goes down because LL is skipped entirely lol. I wouldn't learn it but the idea is out there. Nothing is impossible!

BRB I'm going to develop a method where you do both 2x2x3 blocks AND LL in one look LMAO

 

[wir3sandfir3s]

BRB I'm going to develop a method where you do both 2x2x3 blocks AND LL in one look LMAO

Lol dm me I wanna help with this xD

 

[4Chan]

I'm pretty sure I saw Chris Tran talk about that idea somewhere else in the forums, where you do TTLL for 2 F2L Corners.
He said that the problem is that the skip rate is reduced drastically that it doesn't become worth learning.

Yeah, for the two corners in the top, it's around the same amount of algs as ZBLL.
However, the algs aren't nice, and recog for the cases when the corners are diagonal from each other is nearly impossible. That's why I dismissed that method.

 

[PhillipEspinoza]

Yeah I think the most promising sub method of these spinoffs I started coming up with is just making one pseudo pair and using that as your LS which will give you CLS into PLL.

Would it be possible to do COLL with CLS to end up with an edge cycle for LL after LS? This would be how many algs? 6x26-12? 144? Recognition would be on point and LL skip probability would increase to 1/12. But can good algs be generated for something like this?


This is a random but good scramble used CT for solution, I imagine some good FMC people could do good with this but just thought I'd throw another example solve for CT.
(Do an x' rotation before scrambling for this one if you're not CN)
D2 R2 L F B2 D2 R2 U' R B' L F B2 U F2 B' L B' F R' L D2 F B D'

Double X-EO-Line: x y U' F' U2 l U' l r U r' (9/9)
1x2x2: d R U2 R (4/13)
TSLE: (U2) R U' R U' L' U R2 U' L (10/23)
PLL: (G PERM) U R' d' F R2 u R' U R U' R u' R2 U (14/37)

 

[JTWong71]

Yeah I think the most promising sub method of these spinoffs I started coming up with is just making one pseudo pair and using that as your LS which will give you CLS into PLL.

Would it be possible to do COLL with CLS to end up with an edge cycle for LL after LS? This would be how many algs? 6x26-12? 144? Recognition would be on point and LL skip probability would increase to 1/12. But can good algs be generated for something like this?

So would this be done at the 2 Pseudo Pairs state? Or when there's just a single Pseudo Slot, where it's basically CPLS?

 

[PhillipEspinoza]

So would this be done at the 2 Pseudo Pairs state? Or when there's just a single Pseudo Slot, where it's basically CPLS?

CPLS, though I'm not sure which CPLS you're referencing. The one I know doesn't orient the corners, though it would leave a 2gen ZBLL case which would be cool.

 

[JTWong71]

CPLS, though I'm not sure which CPLS you're referencing. The one I know doesn't orient the corners, though it would leave a 2gen ZBLL case which would be cool.

So then I'm now guessing that this doesn't have to do with your Pseudo Slot idea, and that it very different then.

For FMC Solutions, the times that I think is best is when you get a case that 8- Moves for TSLE and or you get a TTLL Skip, or some sort of short algorithm for both steps.

 

[PhillipEspinoza]

So then I'm now guessing that this doesn't have to do with your Pseudo Slot idea, and that it very different then.

For FMC Solutions, the times that I think is best is when you get a case that 8- Moves for TSLE and or you get a TTLL Skip, or some sort of short algorithm for both steps.

Yes it involves making a pseudo pair for the LS. From there you would easily recognize CP as you would for COLL. Sorry for the confusion but for the sake of not spamming this thread we can continue this convo via pm or I will make a separate thread about my pseudo pair idea, because it really has a lot of potential outcomes clearly.

 

[Shiv3r]

I think that you could try to do a speed-heise LS for FMC, with the psuedo-pair idea.
I call it ZZ-Snyder, where you do speed-heise so that you have a L3C last layer. you could force that with some kind of TSLE where you solve the F2L edge and permuting all the U edges, like LPELL but you don't need to create a pair. then you do comms to solve the last 5 corners.(best idea for FMC, especially if using this in comination with ZZ-NR)

 

[genericcuber666]

Unless I'm wrong for new cubers we should say tell has ~90 algs because all is still just a subset. Right?

Is there any sort of 2 side recognition system for ttll?

Once I finish 2 corners mis orientated for tsle should I skip 3 and 4 as I can use basic corner control to never get these cases or are they still worth knowing before ttll?

How are you doing ttll recognition do you do blocks or track edges

 

[ijustreallylikenaruto]

Unless I'm wrong for new cubers we should say tell has ~90 algs because all is still just a subset. Right?

Is there any sort of 2 side recognition system for ttll?

Once I finish 2 corners mis orientated for tsle should I skip 3 and 4 as I can use basic corner control to never get these cases or are they still worth knowing before ttll?

How are you doing ttll recognition do you do blocks or track edges

 

[4Chan]

Whoaaa, two sub-10s in the same average of 5! 11.476 average of 5!
I'm practicing for New York Summer 2016 and I'm doing it on camera!

I mess up a lot because I just reduced tensions on this Prototype Valk3 by a whole 360 turn for each side.
It would've been a great average if I wasn't for the messups, I think maybe 11 average of 12!

Spoiler: Scramble and Times

Statistics for Mon Sep 05 2016 10:38:45 GMT-0400 (Eastern Daylight Time)

Average of 12/12: 12.840
Standard deviation: 2.438
Number of DNFs: 0
Best time: 9.248
Worst time: 20.409

(9.248), 13.432, 11.548, 14.167, 9.448, 14.677, 11.303, (20.409), 11.936, 18.636, 10.892, 12.361 = 12.840 ave12

1. (9.248) F2 U2 R' F2 D2 F2 L B2 U2 L' F2 D F2 U' B' D B L2 F L'
2. 13.432 L2 U R2 D' B2 F2 U' L' D2 L F' L2 R2 F2 L' D2 L' U
3. 11.548 D' L2 D' B2 F2 R2 D' F2 L2 D' B' L' U2 F' U' B' L' D' R F' R'
4. 14.167 B2 U2 L2 F L2 D2 F' L B2 L' D R U2 B L F' R' B2
5. 9.448 R2 D' F2 D' L2 R2 D' F2 D2 L2 B L B2 R2 F' L2 R2 U F' R2
6. 14.677 L' B2 D2 L B2 R U2 R' U2 R2 U F' L' R2 D L2 D2 B' R'
7. 11.303 R2 D2 B2 L2 R U2 L' U2 F U F L' B' L2 F' D U' B2 L2
8. (20.409) R' D2 L' B2 R2 U2 R' D2 U2 B' D' F L' U F' L' R B' F2 L
9. 11.936 U2 R2 B L2 F' D2 F D2 B L D F' D' U L2 B2 D2 F R F'
10. 18.636 D U R2 D' F2 D' B2 F2 D2 B D' F' D2 R U B2 D' L' D
11. 10.892 L' B' R F L F' B R' U2 F L2 U D F2 R2 U F2 D2 R2 U'
12. 12.361 B' R U R' U R2 F2 R F U' L2 B2 U2 R2 D2 B'

I also got a bunch of sub-10 times today on video!
(Since people prefer to see fast things, not slow things LOL)

 

[Rusca]

Hi! I'm new to this forum (I'll introduce myself or whatever later).

I just wanted to share a TSLE alg I just found that isn't on gyroninja's website (and solves a case that has a line of question marks on it, so maybe it helps with filling that sort of database).

It's a rather trivial one, but it seems it went unnoticed so here it is:
On the "edge in slot" subset, the second ccw case (corners point left and down), you can just do the inverse of the alg not in bold from the similar CW case. That is: (U) L U' R U L' U' R'.
I find it pretty ergonomic, and it's a lot shorter than the one in bold.