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ZZ Statue of Kitten - a somewhat new ZZ variant

Only a question: Why place anything?

Is it possible (by means of recognition ) to just bring the cube into 2Gen state (is it 7 cases ?) - maybe doing some partila CO stuff by the way

And than happily solve 2-Gen

This would get rid of setups and would bring down the move count, but i dont see how recog would work... (and would still have to solve a F2L slot afterwards).

And would be just 2 algs right? (Y and J)
J-> L' U R U' R' L (6)
Y-> R U2 L' U R' U' L (7)
 
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If you don't solve anything in the LS+LL area, then you would need to solve:
-EO, which is done in ZZ, so nothing to worry about here
-CP, which is only 6 cases or 5 algs.

To get to 2GLS+LL

However, I feel like just doing this is very little, and much more can be done in that step for the amount of recognition needed.
 
The biggest disadvantage I see with this variation is that you're less likely to get lucky with a pre-placed corner than a pre-placed edge, because any corner only has a 1/3 chance of being properly oriented. There's also the 1/3 chance that, if you wait for the last slot to place your corner, you'll have the ugly upside-down orientation.

But! There's few enough cases of CPLS-E and CPLS-C that you could reasonably learn both and get an extra chance to be lucky. Of course, you'd probably want to prioritize the systems for yourself so that you'd have an immediate plan if you came to the last slot without a lucky placement.
 
Move counts:

CPLS+2GLL:
2.6 (Edge setup) + 9 (CPLS) + 14 (2GLL)= 25.6

Reduce to 2gen before last slot:
4.33 (Reduce to 2gen)+ 7.66 (RU-LS) + 14 (2GLL)= 26.00

Silly Way to go to 2GLL:
7.66 (RU-LS) + 6 (Reduce to 2gen) + 14 (2GLL) = 27.66

COLL+EPLL:
7.66 (RU-LS) + 9.78 (COLL)+ 10.75(2gen EPLL) = 28.19

ZZ SK (with the pointing up case):
4.8 (Corner setup) + 10 (CPLS variant) + 14 (2GLL)= 28.8
 
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The move count are mayble sightly off, 2GLL movecount based on your file is about 14.3 and the CPLS i think is lower then 9, so its still the better movecount.

If there was a way to recog reduce to 2gen before last slot would be the better, since is only 2 algs with 4.33 avg move count (0+6+7 /3) and then a 22 moves RU spam.


Edit:

RUL Optimal Algs for ZZ-SK:

Solved: R' U' R' U' R' U R U R (9)
Y: (U') R U R' L U' R U L' U' R' (10)
J right: R U2 L U2 R' U2 R U2 R' L' (10)
J back: (U2) R L U2 R' U2 R U2 L' U2 R' (10)
J left: (U') R2 L U2 R' U' R U2 L' U R2 (10)
J front: R U L U2 R' U' R U2 L' U' R' (11)
 
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Move counts:

Silly Way to go to 2GLL:
7.66 (RU-LS) + 4.66 (Reduce to 2gen) + 14 (2GLL) = 26.32

I know this is the Silly Version but 4.66 is not possibe after the Slot is occupid I think 6,3 is correct (0+7+7+7+7+10/6)

To give Coll a better standing: Stachu compiled a list of "Last 5 edges" algorithmes as part of L2L4 methode but ther would be an option to:

[Corner last Slot (2gen) ] + COLL + L5E : ~3 + 9.78 (COLL)+ ~12 L5E = ~25




@stachu I smile, too


EDIT: @diniz (0+6+7 /3) shouldn' it be (0+6+6+6+6+7 /6)
 
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@diniz (0+6+7 /3) shouldn' it be (0+6+6+6+6+7 /6)

I dont think so, you can just do U, U' or U2 and do the same alg, so they are just 3 different cases (solved, adjacent swap, opposite swap). You are counting adjacent swap multiple times.

Example: Numbering from here: http://www.speedsolving.com/wiki/index.php/Special:AlgDB?mode=view&view=default&puzzle=3&group=COLL

#2 R' U' R U' L U' R' U L' U2' R
#3 U' (R U' L' U R' U' L)
#4 U (R U' L' U R' U' L )
#5 (R U' L' U R' U' L )
#6 U2 (R U' L' U R' U' L )
 
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I dont think so, you can just do U, U' or U2 and do the same alg, so they are just 3 different cases (solved, adjacent swap, opposite swap). You are counting adjacent swap multiple times.

An adjacent swap would come up more often than the other two, so you can't weight them the same.
 
Just for fun I calculated what the weighted movecount for RUL - placing the Last D-Corner in an empty Slot should be:
the .25/.75 comes from AUF

3.25 *4 + 3.75 *4 + 6.75 * 4 + 7 + 7 + 0 = 4.6

What this number doesn't tell is that it is maybe already just placed by some F2L magic.
 
I just had a thought about this. What if you do this instead:

(Assume EO and F2L-1 is complete, and that the last pair isn't 3-4 moves)

1. Separate the pair: Place the corner in the correct slot, and leave its corresponding edge on top
2. Solve the last pair and permute corners on top. There will be many new corner permutation cases to "learn" however there will still be only a few algs. If the FR slot is the last slot, then you will only have to learn 18 algorithms. (6 CPLLs * 3 orientations of the last F2L corner).
3. 2GLL to finish the solve.

If you're ambitious enough, you could learn the second step for each possible last slot, which totals to 72 algorithms.
 
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I think this would work well if you also included the cases with the edge placed in the slot just in case you end up with the edge and corner in the slot already. This would up the algorithm count by 8 (9 minus the solved case) and it would slightly reduce the average move count for setting the pair up.


I think it would also be useful if you made the algorithms end with the pair left in thetop layer so that you can intuitively phase before doing 2GLL.
 
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FRD twisted clockwise:

No swap:
R U R' U' R U2 R' U' R U R'
R U' R' U' R U R' U' R U2 R'
R U' R' U' R U R' U2 R U' R'
Backswap:
L R U' R' U L' R U' R' U' R U R'
Diagonal swap:
R U' R' U R L U' R' U R L' U' R'

FRD twisted anticlockwise:

No swap:
R U R' U2 R U' R' U R U R'
R U' R' U R U2 R' U R U' R'
R U2 R' U R U' R' U R U R'
Back swap:
R U' R' U R U R' L U' R U R' L'
Diagonal swap:
R U R' L U' R U R' L' U' R U R'

Here are some algs. I think I'll take back what I said, these algs seem ok and not very long IMO. So I guess it's worth learning algs for these cases.
 
I just had a thought about this. What if you do this instead:

(Assume EO and F2L-1 is complete, and that the last pair isn't 3-4 moves)

1. Separate the pair: Place the corner in the correct slot, and leave its corresponding edge on top
2. Solve the last pair and permute corners on top. There will be many new corner permutation cases to "learn" however there will still be only a few algs. If the FR slot is the last slot, then you will only have to learn 18 algorithms. (6 CPLLs * 3 orientations of the last F2L corner).
3. 2GLL to finish the solve.

If you're ambitious enough, you could learn the second step for each possible last slot, which totals to 72 algorithms.

I have came up with algorithms for this (by hand, so they're not all optimal), but I also included cases where the edge was in the slot. There are 26 algorithms total with an average move count of 11-12. I would make pictures using VisualCube, but I don't know how to use the arrows to show corner permutation.

Right swap:
L' U R U' L U R' // 7
L U' R' U L' U' R' // 7

Diagonal swap:
L' U2 R U' R U2 L U R U2 R' // 11
No swap:
(R U' R' U) (R U2 R') L' (U' R U R') // 13
(R U2 R') U (R U' R' U) (R U R') // 11

Right swap:
(U') (R U' R' U) (R U R') L (U' R U R') L' // 13

Diagonal swap:
(R U R') L U' (R U R') L' (U' R U R') // 13
L' U R2 U' L (U' R2)3 // 11
No swap:
(R U R' U') (R U2 R') (U' R U R') // 11
(R U' R') U' (R U R' U') (R U2 R') // 11

Right swap:
U' L (R U' R' U) L' (R U' R') (U' R U R') // 13

Diagonal swap:
(R U' R' U) R L (U' R' U R) L' U' R' // 13
No swap:
(U2) (R' U')2 (R U)2 R // 9

Right swap:
(U') R2 U R U' R2 D R' U' R D' // 10

Back swap:
(U') F' (R U R' U') R' F R // 8

Left swap:
(U2) R2 U' L U2 R' U R U2 L' R2 // 10

Front swap:
R' F' R (U R U' R') F // 8
U L R U2 R' U2 R U2 L' U2 R' // 10

Diagonal swap:
R U L U' R' U L' (R U R') // 10
(U') (R U R') L U' R U L' U' R' // 10
No swap:
(R U R' U') (R U R') // 7

Right swap:
L' (R U R' U') L (R U R') // 9

Back swap:
(U') R U2 L U' R' U' L' // 7

Left swap:
(U') R U2 L U' R2 U' L' U' R // 9

Front swap:
(U2) L R U' R' U L' (R U2 R') // 9

Diagonal swap:
(U) R U L U' R' U L' (U' R U R') // 11
No swap:
(R U R') (U R U' R') // 7

Right swap:
(U') L' U R U' R U R U' L R' // 10

Back swap:
L R U' R' U L' U2 R U' R' // 10

Left swap:
R U' R' L' (U R U' R') L // 9
R U2 L R' U' R U R' L' // 9
(U) R' U L U' R2 U' L' U2 R' // 9

Front swap:
(U2) L U' R U L' U2 R' // 7

Diagonal swap:
(U) L' U R2 U' L U R' U2 R' // 9
(U) (R U' R' U) L U' R U L' U' R' // 11
 
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