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elrog

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I think algorithms for CO and placing the last F2L edge have been made and posted on this forum somewhere. I will search for them in a little bit. Also, this would be more algorithms for this step, but the remaining step would be much less algs than ZBLL.
 

Tao Yu

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From what I can tell, you just want to end the solve with an edges oriented LL without needing to know full ZBF2L.

I think a better way to do this would be to just do 2 look ZBF2L. You could for example, first solve the cross and three F2L pairs, insert the last edge, and then do step 4 from diaper.

You could also pair up the last corner and edge without inserting them, and then insert them using the 16 ZBF2L cases that are applicable.
 
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elrog

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I believe solving the pair (after it is paired up) with EO is called VHF2L and it is a subset of ZBF2L.
I will search for them in a little bit.
I found it. It is called JJLS. JJLS solves the last slot corner with EO and then the last slot edge with CO. You place the corner and then use the second step in it, but disregard the EO and solve the LL in 1 look with PLL+EO.
 
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PLL+EO is COALL (or KLL, MBLL, or whatever else it's been named). I have algs (156 of them) somewhere, but most of them suck. The number is actually smaller than that, though, since, if I remember correctly, it includes both PLL and ELL. It's still a fairly large number of algs.
 

Dane man

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I found it. It is called JJLS. JJLS solves the last slot corner with EO and then the last slot edge with CO. You place the corner and then use the second step in it, but disregard the EO and solve the LL in 1 look with PLL+EO.
That actually sounds like a brilliant idea because it ends with a 21 alg 1LLL (PLL). Though, I think it better to place the corner while doing CO, and the edge while doing EO to reduce the alg count. I found the link too. Nice idea. Someone should set up an alg set for it.
 

elrog

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The alg counts for the JJLS are 65 and 34 respectively with a 21 alg LL. MGLS is 21 and 104 respectively with a 21 alg LL. If you do intuitive F2L, you can solve a slot, then do OLL and PLL for a 57 + 21 alg LL.

What I was suggesting it intuitively placing the last slot corner, then the 34 algs from the second step of JJLS (possibly shortened because of not preserving EO), and finishing with a 156 alg LL.
 

Dane man

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The alg counts for the JJLS are 65 and 34 respectively with a 21 alg LL. MGLS is 21 and 104 respectively with a 21 alg LL. If you do intuitive F2L, you can solve a slot, then do OLL and PLL for a 57 + 21 alg LL.

What I was suggesting it intuitively placing the last slot corner, then the 34 algs from the second step of JJLS (possibly shortened because of not preserving EO), and finishing with a 156 alg LL.
Ah, Didn't know that CLS was 104 cases. That seems like much. Though, the alg counts for a 1LLL are high, they are the price that needs paid for that efficiency.
 

elrog

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You could also do ELS, orient the 5 corners, and then do PLL+C giving you 21+23+72 algorithms. If you phase while doing ELS, you reduce the number of algorithms for PLL+C down to 33, but increase ELS from 21 to 56. This is the lowest alg count LS+LL I know of (not counting ones with intuitive steps).
 

IRNjuggle28

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What would be the alg count for ELL including both parities? 116?

Other than things like bad recognition and high alg count, this seems like it's at least worth some thought for 4x4. Yau/CFCE with the E including solution to either or both parities would guarantee a 2 look last layer.

What about OLL including OLL parity and PLL including PLL parity? Is it as simple as being twice the number of OLLs and PLLs?
 
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TDM

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What would be the alg count for ELL including both parities? 116?

What about OLL including OLL parity and PLL including PLL parity? Is it as simple as being twice the number of OLLs and PLLs?
For the first question, either I've missed something, or it's (4!*2^4)/4 = 96... which doesn't look right at all.
I know there are 22 PLLs with parity and 22 without, but I'm not sure about OLLs. Another calculation which could be wrong: ((4!*2^3)*27)/4 = 1296... now that one looks too high :/

tldr: two calculations that are wrong for some reason I don't know (I TRIED), but I know there are 22 PLLs with and without parity, giving 44 total.
 
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elrog

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29 - Regular ELL
28 - ELL + OLL Parity
38 - ELL + PLL Parity
36 - ELL + OLL Parity + PLL Parity
131 - ELL Total

57 - Regular OLL
54 - OLL + OLL Parity
111 - OLL Total

21 - Regular PLL
24 - PLL + PLL Parity
45 PLL Total

4 - Regular EPLL
6 - EPLL + PLL Parity
10 - EPLL Total
 
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elrog

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Wow, thanks. I have no idea how you got those, but I'll trust that you know what you're doing. :p How about editing alg count for EPLL including PLL parity into that post?
Done.

I found these for ELL + OLL parity by matching each EPLL with each edge orientation pattern and using symmetries to tell how many cases it would require for that specific EPLL. There are 2 cases with EPLL solved, 8 with Ua perm, 8 with Ub perm, 2 with H perm, and 8 with Z perm.

I did the rest in a similar way. OLL + OLL parity was matching each CO with each EO (with OLL parity) and PLL + PLL parity was matching each EPLL (without PLL parity) with either a diagonal or adjacent corner swap.
 

Kirjava

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Cubeshape last example solve

(3,-1) / (0,1) / (-3,6) / (2,1) / (-2,-3) / (2,6) / (0,6) / (-2,-1) / (-4,2) / (-2,6) / (1,2) / (-1,0) / (0,-3) / (-2,5) / (-2,-3) /

edge deisolation:

6,2 / -5,4 / -3,-2 / 0,2 / 1,0

corner orbit resolution (+ reaching selected pairing state)

/ 2,0 / 0,-2 / 4,-2 / 4,0 /

edge pairing (stage needs most improvement)

parity
/ 2,-4 / -2,4 /
easy
4,1 / 1,1 / 1,-1 / -1,-1 /
2-2 cycle
-3,4 / -4,3 / 1,0 / 1,0 /
3 cycle
-1,6 /
1,0 / 5,-1 / 1,0 / -1,0 / 0,1 / -5,0 / -1,0

mini domino esque 2x2x1 pbl

/ -3,0 / 3,-3 / 6,0 / -1,-5
 
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Anouncing a new (2x2) method, STYP!

Hi, today i will be anouncing my new 2x2 method, STYP wich stands for : Skip The Y Perm!
This methode requires 13 algs (14 if you include a Y perm wich u should not really find in a solve since the point is to skip the Y perm).
Steps:

Step1: solve a layer of any color.

Step2a:in this senario you have an oll alg to do but that is the point of this method ( to learn a bit more
algs to skip PLL more often/get rid of those stupid Y perms). You have 5cases where you can do an alg and not get a Yperm, for these 5cases you will learn 1 alg (lets say sune as an expample for this entire explaination) you will use that alg at 5/6 cases you can get at a case. To either skip PLL or give you an adjacent swap fpr PLL.

Step2b: in this step is where the second alg for the same OLL comes over, you will learn the complete oposing alg for the alg you already learnd for the case. Example: if you learn the regular sune wich twists 3 corners and does not swap any pieces, you would lear. The alg that does a y perm+twists the 3 corners, in other words of you would setup the first alg and do the 2nd alg it would give you a Y perm.

Step3: do a adjacent swap to solve the cube or do a U/U'/U2 move to solve the cube.

This is more of a beginner method, i could get to 4.5 seconds with this method ( ofcourse not my main methode). Learning just the 12 OLL algs gives you a 1/3 cance on average skipping PLL.

I hope some people might be intrested in this methode :),
Greeting from Antonie!
 

TDM

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Yeah it could, haven't heard of it thought :/.
Me neither, but I thought the disadvantages of it compared to normal LBL made it not good enough to post. Stuff like longer recognition because you're checking CP and CO at the same time, more algs to learn etc. I don't think this method will be very useful, even to beginners, as Ortega is probably faster.
 

Dane man

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I've had an interesting thought. After introducing the Diaper Method and ZBLBL, an idea came to me. I need to work on some of the algs, but here is the basic idea. The parts in gray are not considered part of the method, though they were the original seed of the idea.

Skipper F2L (SF2L)

1-While performing F2L, one inserts the first two pairs normally.
2-Insert the corner of the third pair, not worrying about it's edge.
3-Insert the fourth pair using Winter Variation (27 algs). (This step is incorrect, and has been replaced with the 3rd and 4th steps below.)
---
3-Insert the fourth pair (edge and corner) without worrying about the orientation of it's corner. (EUOC)
4-Perform CLS (24 algs: CLS: I (and mirror) + OCLL)
--or--
3-Insert the corner of the fourth pair, not worrying about it's edge.
4-Insert the edge of the fourth pair using the second step of JJLS here. (34 algs)

5-Insert the edge of the third pair using ZBLBL algs (Modified to respect corner orientation, 21 algs. I'm still working on making these to see the avg moves.)
6-You are now left with a 1LLL, being just PLL (21 algs).

Comparing to the standard Fridrich(CFOP) OLL/PLL:
avg moves: F2L(6.7*4)+OLL(9.7)+PLL(11.8) = 48.3 HTM
avg moves: F2L(6.7*2)+C3P(~4)+WV(8.07)+ZBLBL(~8.5)+PLL(11.8) = 45.77 HTM (this is incorrect)
avg moves: F2L(6.7*2)+C3P(~4)+EUOC(~5)+CLS(~10.5)+ZBLBL(~8.5)+PLL(11.8) = 53.2 HTM
avg moves: F2L(6.7*2)+C3P(~4)+C4P(~4)+JJLS2(~10)+ZBLBL(~8.5)+PLL(11.8) = 51.7 HTM

The number of moves isn't reduced, though, the algorithm count is reduced a little more. (OLL/PLL = 78 algs; CLS/ZBLBL/PLL = 66 algs; JJLS2/ZBLBL/PLL = 76 algs)

The executional downsides to this method is the added recognition of inserting the corners of the third and fourth pairs to prepare for further algorithms, and the standard look-ahead of the last two F2L pairs becomes increasingly complex (not too much though, it just adds looking for 4 piece orientations per pair). The learning con is that it is quite unorthodox, and a good step out of the way of OLL because it replaces it entirely. Other than that, I see this method as being very effective, holding a lot of potential when it comes to speed cubing and even in exploring methods in FMC.

What you guys think?
 
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