Last Layer Options

[elrog]

There are many different possible ways to solve the last layer. I'm interested in knowing these different options because its always nice to have algorithms laying about incase you need them. I'm also curious as to if they would be any good. I won't be discussing anything like L5P here however. The last layer can be divided up into corner orientation (CO), corner permutation (CP), edge orientation (EO), and edge permutation (EP). In these examples, two substeps in parenthesis together means they are done together (example - (OE, PE), this is permutation and orientation of the edges in one step) and L1 is look 1 while L2 is look 2, N/A means it hasn't been named that I know of, and W/M means without mirrors. Given this information, you have the following options for a 2 look last layer:

(CO, EO) (CP, EP) - classic OLL and PLL - L1=48 algs W/M - L1=57 algs - L2=13 algs without mirrors - L2=21 algs - total=78 algs

(CO, CP) (EO, EP) - CLL and ELL - L1=42 algs - L2=19 algs W/M - L2=29 algs - total=71 algs

(CO, EP) (CP, EO) - N/A - L1= 31 algs W/M - L2= 20 algs W/M - total=51 algs

(EO, EP) (CO, EO) - LLEF and L4C - L1=15 algs - L2=84 algs - total=99 algs

(EO, CP) (CO, EP) - N/A and 2GLL - L1=11 algs W/M - L2=49 algs W/M - L2=84 algs - total=62 algs

(CP, EP) (CO, EO) - N/A - L1=21 algs - L2=48 algs W/M - L2=57 algs - total=78 algs

I really don't think that (PC, PE) (OC, OE) would be any good at all because it is just easier to to OLL and PLL, and it would have bad recognition time. I do think that (OC, PE) (PC, OE) and (OE, PC) (PE, OC) could be promising because reduced alg count not only means easier to learn, but also means better recognition time. I would generate algs for these, but I'm already busy generated many other sets of algs I've though of for some various method ideas. Here are the options to solve a 2 look last layer with one substep already solved by solving 2 substeps in the first look:

(EO, CP) (CO) - N/A and Pure CO - L1=20 algs W/M - L2=7algs - total=27 algs

(CP, CO) (EO) - N/A - L1=42 algs - L2=3 algs - total=45 algs

(CO, EO) (CP) - N/A and CPLL- L1=48 algs W/M - L1=57 algs - L2=3 algs - total=60 algs

(CO, EO) (EP) - N/A and EPLL - L1=48 algs W/M - L1=57 algs - L2=3 algs W/M - total=60 algs

(CO, EP) (OE) - N/A - L1=43 algs W/M - L2=3 algs - total=46 algs

(EO, EP) (CO) - ELL and Pure CO - L1=19 algs W/M - L1=29 algs - L2=7 algs - total=36 algs

(EO, CP) (EP) - N/A and EPLL - L1=11 algs W/M - L2=3 algs W/M - total=14 algs

(EO, EP) (CP) - ELL and N/A - L1=19 algs W/M - L1=29 algs - L2=3 algs - total=32 algs

(CP, EP) (EO) - N/A - L1=21 algs - L2=3 algs - total=24 algs

(CP, EP) (CO) - N/A and Pure CO - L1=21 algs - L2= 7 algs - total=28 algs

(CO, EP) (CP) - OCELL and CPLL - L1=32 algs W/M - L1=43 algs - L2=3 algs - total=46 algs

(CO, CP) (EP) - COLL and EPLL - L1=42 algs - L2=3 algs W/M - total=45 algs

These would most likely be used as a buildup to a standard 2L last layer or if someone just doesn't want to learn many algs, they may be learning 2L OLL and PLL. Here are all of the possibilities for a 2 look last layer in which 1 step is already done and are solved by solving 2 substeps in the 2nd look:

(CO) (CP, EP) - OCLL and PLL - L1=7 algs - L2=13 algs W/M - L2=21 algs - total=28 algs

(CP) (CO, EP) - N/A and 2GLL - L1=1 algs W/M - L1=2 algs - L2= 49 algs W/M - L2=84 algs - total=86 algs

(EP) (CO, CP) - N/A and L4C - L1=2 algs - L2=84 algs - total=86 algs

(EP) (EO, CP) - N/A - L1=2 algs W/M - L2=20 algs W/M - total=22 algs

(CP) (EO, EP) - N/A and ELL - L1=1 algs W/M - L2=19 algs W/M - L2=29 algs - total=30 algs

(EO) (CP, EP) - N/A and PLL - L1=3 algs W/M - L2=13 algs W/M - L2=21 algs - total=24 algs

(EP) (CO, EO) - EPLL and N/A - L1=3 algs W/M - L2=48 algs W/M - L2=57 algs - total=60 algs

(EO) (CO, EP) - N/A and 2GLL - L1=3 algs W/M - L2=49 algs W/M - L2= 84 algs - total=87 algs

(CO) (EO, EP) - N/A and ELL - L1=7 algs - L2=19 algs W/M - L2=29 algs - total=36 algs

(CO) (EO, PC) - N/A - L1=7 algs - L2=20 algs W/M - total=27 algs

(EO) (CP, CO) - N/A and L4C - L1=3 algs - L2=84 algs - total=87 algs

(CP) (CO, EO) - N/A - L1=3 algs W/M - L2=48 algs W/M - L2=57 algs - total=60 algs

These would most likely be used when one substep is solved while solving the final F2L pair. Here are all of the 1 look last layer steps in which 2 substeps are complete:

(CO, EO) - N/A - 48 algs W/M - 57 algs

(CO, CP) - L4C - 84 algs

(CO, EP) - 2GLL - 49 algs W/M - 84 algs

(EO, CP) - N/A - 20 algs W/M

(EO, EP) - ELL - 19 algs W/M - 29 algs

(CP, EP) - PLL - 13 algs W/M - 21 algs

These are the most basic sets of algorithms you would probably use for the last layer while speedsolving. Here are the 1 look last layer options with 1 substep complete:

(CO, EO, CP) - OLLCP - 332 algs

(CO, CP, EP) - ZBLL - 493 algs

(CO, EO, EP) - N/A - 322 algs W/M

(EO, CP, EP) - N/A - 71 algs W/M - 114 algs

I will calculate cases for these whenever I can find time. Here is a 1 look last layer with no substeps solved:

(CO, EO, CP, EP) - 1LLL - 1211 algs W/M or inverses

Doing the whole last layer in 1 look is nearly unfeasable. I've posted this to show just how many options there are not including L5P, L3C, ZBLL, and some others. I also posted this in hopes that somone could look into finding algs for (OC, PE) (PC, OE) and/or (OE, PC) (PE, OC). I am also interested in the 1LLL while a single substep is solved to see if there is a better alternative than ZBLL. Lastly, I would like to discuss doing a type of belt method that would use an interesting last last step.

I've had the idea of a belt method for a long time, but solving a belt really hinders the amount of freedom you have to solve other pieces without messing the belt up. My idea is to orient all of the edges, get all of the bottom edges in the bottom layer, solve 3 of the middle edges while placing the bottom layer corners in the bottom layer, then solve the last middle edge while making sure you don't have corner orientation parity, and finish with some sort of last layer substeps. You could also not solve corner permutation first, and solve that while you place the last edge using ZBF2L algs. I was thinking you could do something like ZBLL, but solve the top 4 corners and permute the bottom 4 edges instead of the top 4 edges. Doing the bottom and top seperately would allow you to AUF reducing the number of cases down to 126 for including CLL. you could use these same algorithms to finish the solve after and x2 rotation. This may require a parity algs such as M2 S2 U M2 S2. There are many different option you could use in getting to the state to use the 126 algorithms, and I'm not sure as to what would be the best route to go.

 

[mark49152]

You've overlooked a few things, like COLL/EPLL, OLLCP, or any combination where you're doing more steps in L1 than L2. Also where phasing, edge control or other techniques are used to reduce the number of cases in the next step. Some of your "n/a" sets already exist e.g. CPEOLL. You don't mention 2GLL. Some of the alg counts don't make sense to me.

Just curious, but how experienced/fast/knowledgeable are you? And what are the problems you see with existing systems that you are trying to improve on?

 

[Smiles]

I do like that you got it all organized, but you missed some stuff like LS things that reduce LL cases.
plus i found it hard to follow since you wrote PE/OE etc. instead of EP/EO/CP/CO like everyone is familar with.

I also posted this in hopes that somone could look into finding algs for (OC, PE) (PC, OE) and/or (OE, PC) (PE, OC).

COEP is the EP version of COLL. u want EO done for this.
EOCP is CPEOLL but preserving EP makes the algs even worse.
using EOCP as the first step needs CO done to end with EPLL.

basically these have bad recog and bad algs, imo the only things that make sense are what we have already.
old CLL ELL
or OLL PLL for recog purposes (+ great PLL algs)
ideally, for execution speed for 2 look, we want EO already done, and CLL done next if possible.

 

[Escher]

I'm not interested in learning any new 'sets' for LL, but I thought this might be useful for you:

http://www.speedsolving.com/forum/s...PLL-a-2-gen-friendly-alternative-to-COLL-EPLL

 

[elrog]

I did plan on eventually expandind this list to include more last layer options such as one with one substep already solved, then solve 2 more at once, and then the last one such as COLL and EPLL. I will also eventually (maby) add in things like last slot substeps and things like solving 2 top corners and orienting the edges, then finishing the rest, to make this a complete list. I think it could also be beneficial to add links to each of the combinations if algorithms are availiable. I was not aware that others have used a system like mine to name things but in the opposite order. I will go back and change all of my OEs to EOs and ect. Thank you Escher for that page. I will look into that after dinner.

Any comments on my idea to be able AUF during ZBLL? I have been thinking about the best way to get into a position to do it. As of now, I think you should orient the edges and place 3 E layer edges in the E layer, then solve corner orientation parity while inserting the last edge, then seperate out the top/bottom colored pieces while solving the E layer. This should be able to be done in less than 20 moves consistantly and would proabably end up anywhere from 15 to 20 for speedsolving situations.

mark, you've asked how experienced/fast I am. I've been cubing for about a year, and my best time is 27 seconds set with doing CFOP with a 2 look OLL and some blockbuilding influence. I have gotten quicker through constantly playing with the cube, not really trying to get faster. My first solve alone was 2 and 1/2 mins. My second was 2 mins. my 3rd was 1 and 3/4 mins. My 4th was 1 and 1/2 mins. I stopped decreasing my times more when I got around to 1 and 1/2 mins. It took me 3 or 4 days to learn to solve the cube from a not that great teacher who now has 1 and 1/2 min solve as their best time. So far I have figured out every cube I own or have borrowed on my own including the 2x2, 4x4 (I found ways past parity, though it took me a while), curvy copter, megaminx, pyramix, and the mf8 mars and mercury. My greatest feet I beleive was figuring out the mars in 45 minutes on my own. I'm not really that fast, but I do undestand the puzzle and I enjoy solving it. I'm also very enthusiastic about coming up with new ways to solve various puzzles, and would like to create my own puzzles if I had the money to use shapeways.

There is not any particualy thing I'm trying to improve on for the last layer. I would just like to know more about how all of the not yet done or not thought of options would work. This could also be used as a vast list of information for anyone looking to see what options they have for the last layer. The thread could be turned into a sort of one answer one question thread about the last layer, or anything that effects it after the list is completed.

 

[Smiles]

not a bad idea making a thread like this.
on one hand, the amount of options is huge.
on the other hand, there are WAY too many options that are unfeasible, such as:
- EP before EO (bad execution for EO)
- CP before CO (long algs for CO)

and then there are methods that are too unique to categorize, such as:
- Kirjava's "2 good algs" LL system
- LS techniques like phasing
- huge alg sets like LS + OLL

 

[mark49152]

Thanks for answering my question. I've been cubing about 4-5 months and at the outset had a similar interest in exploring all the different methods and steps. I built a similar analysis to yours and planned to put it into a kind of state-transition diagram, although never got round to it. After a while I decided to focus on learning how to do CFOP well rather than exploring methods and steps, but I still read all your posts with interest and might try out some of these ideas once I get bored of CFOP. Thanks for your contributions!

 

[elrog]

Your welcome. I think I may make this into a chart on the wiki so that it is easier to convey all of the information. I think these charts should include the average move with and without considering proababilities and with using the shortest or most common algorithms. They should also contain the amount of algorithms for the step including and not including mirrors and inverses. There should also be a name given to all of these substeps, or they could be refered to by the way I refered to them. You should also be able to click on the name of the step and get algorithms for it.

I looked at that post for OCELL and found some mistakes. He said the alg number was 40, but I found that he included mirrors for some cases and left out other cases altogether. He sorted the cases out by the corner orientation, while I sorted them out by a solved cross, opposite edge swap, or diagnal edge swap. This is based off the idea that you can always solve atleast 2 of the edges by turning the U layer. I then matched each of these cases with the OLL case for corners and found that this would require 32 algorithms without mirrors and 43 with them. Algorithms for this would be easy to generate because they can be solved with only U and R moves, which I did not realise until I read his post.

Many of these steps such as OLL after everything is permuted, would have all of the same cases as regular OLL, but would use different algorithms, thus shouldn't have the same name or they will get confused such as in all of the various CLLs. This is why I used N/A for many things that would have the same cases as other common steps.

EDIT: I have also found that 2GLL was calculated incorectly. I put the alg number for solving it while being able to AUF. I recalculated taking this into consideration and got 86 algs and 51 without mirrors. On the wiki, the 2GLL page says it has 85 algorithms and the CPLS page says it that 2GLL has 84 algorithms.

mark, when you say you read all my posts, are you talking about this thead specifically?

 

[mark49152]

No, other threads too. Methods and algs are my favourite subjects on the forum, although like I said, I'm focused on mastering basic CFOP first. Any data that can help learn CFOP more effectively is of great interest to me. E.g. I'm currently compiling some data to compare efficiency of 1-look and 2-look OLL case by case, to aid in choosing a learning order for full OLL.

 

[cuBerBruce]

I looked at that post for OCELL and found some mistakes. He said the alg number was 40, but I found that he included mirrors for some cases and left out other cases altogether. He sorted the cases out by the corner orientation, while I sorted them out by a solved cross, opposite edge swap, or diagnal edge swap. This is based off the idea that you can always solve atleast 2 of the edges by turning the U layer. I then matched each of these cases with the OLL case for corners and found that this would require 32 algorithms without mirrors and 43 with them. Algorithms for this would be easy to generate because they can be solved with only U and R moves, which I did not realise until I read his post.

EDIT: I have also found that 2GLL was calculated incorectly. I put the alg number for solving it while being able to AUF. I recalculated taking this into consideration and got 86 algs and 51 without mirrors. On the wiki, the 2GLL page says it has 85 algorithms and the CPLS page says it that 2GLL has 84 algorithms.

I checked some of these using GAP. First, I got 85 cases including solved for 2GLL, so 84 algs is the correct number. With mirrors regarded as equivalent, it reduces to 49 cases (50 including solved).

OLLCP has 332 cases including solved. If you don't count the skip (EPLL), adjacent corner swap PLL, and diagonal corner swap PLL cases, it is 329 cases. This agrees with Robert Yau's page (taking into account cases he notes are equivalent). Counting mirrors as equivalent, the count reduces to 208 including solved, or 205 excluding the all-oriented cases.

OCELL has 43 cases including the solved and other all-oriented cases. So 40 algs is correct if you're excluding the 3 all-oriented cases. With mirrors being considered equivalent, it reduces to 29 cases (including solved), or 26 cases excluding the 3 all-oriented cases.

 

[elrog]

Thanks for the OLLCP number. I had a feeling mine was not correct. For OCELL, I came up with the 43, but I only found 32 without counting mirrors (The 40 algorithms on that page included the solved). Did you by chance not include inverses also? If you did on this one, did you with all the other numbers without mirrors also? I will go back and check my 2GLL cases. I probably made a small error somewhere, knowing that you used a computer to solve if for you, and you tend to know what your talking about.

Thanks for the help.

 

[cuBerBruce]

Did you by chance not include inverses also?

Inverses are applicable for 2GLL. The number of cases I get (excluding the solved case) are:

2GLL, counting mirrors as equivalent: 49
2GLL, counting inverses as equivalent: 48
2GLL, counting mirrors and inverses as equivalent: 34

OLLCP does not have unique inverse cases for each case. Hence, using inverses is not applicable to OLLCP.

Likewise, OCELL cases do not have unique inverses. However, if we only consider 2-gen OCELL algs, then there are only two inverses applicable for each case, and the two inverses differ by an "X-perm" (the standard H-PLL with U2 for the AUF). If we allow X-perm as a setup move, then we can get well-defined antisymmetry classes. The problem with this reduction is that often you might have to perform an X-perm prior to performing the alg (or its inverse). Thus, for each "case," you need to learn two additional algs (combining x-perm with the alg and inverse alg), or else you often have to use an "X-perm" setup (in effect, adding an extra step to the method).

2-gen OCELL, allowing inverses, and positions differing by X-perm as the same: 21
2-gen OCELL, allowing mirrors, inverses, and positions differing by X-perm as the same: 18

Since you may need to in effect learn up to 18*3 = 54 algs, it may be better to simply learn the 34 2GLL cases. I note that 2GLL and 2-gen OCELL are pretty much the same thing, except that 2-gen OCELL has only half the effective number of configurations. If an alg solves a 2GLL case, then that alg along with that alg composed with X-perm will both solve the same 2-gen OCELL case.

 

[elrog]

While solving 2GLL, I realize that you may need to learn the ammount of algorithms without mirrors if you are going to solve it with only U and R moves. If you don't include mirrors cases, you may have to do L and U moves for some cases instead. Is this correct?

 

[qqwref]

You can mirror 2gen algs.

e.g. R2 U R U R' U' R' U' R' U R' <-> R2 U' R' U' R U R U R U' R

 

[elrog]

Oh right.... You cant if you had an F2L slot unsolved though, could you?

I ask this because I've been thinking about different things to solve while doing the last F2L pair. One of my ideas is to solve a 1x2x2 block in the top layer at the same time as the last F2L slot. I reduced the amount of cases in various ways such as having edges already permuted, getting both or aleast the corner F2L piece(s) in the top layer, and conrtolling orientation of the corner in the slot.

 

[cuBerBruce]

For F2L slots, or for last slot + last layer, you can use diagonal mirroring, F becomes R', L' becomes B, etc. if it's the FR slot. But you lose rotational symmetry when considering a last slot in addition to the last layer, though you can still use pre-AUF and post-AUF to more or less make up for that.

 

[elrog]

I'm working on making charts for this. I'm organizing them by how many substeps it solves, and what it preserves. I will leave some of them out such as solving last layer edge orientation while preserving corner orientation, because this would use all of the same algorithms as solving last layer edge orientation while preserving corner orientation and permutation, and it wouldn't be able to be shortened by preserving less. I will not post these for quite some time, because I'm trying to complete them first. They will contain, what the step solves and preserves, its name, alg number, mirror number, inverse number, number of algs that are both a mirror and inverse of another alg, average moves, and average moves wheighted by probability. I will not come up with algorithms to calculate the average moves of some cases that are just too inefficient such as orientation of edges and corners whilst preserving permutation of edges and corners. I won't be able to fit a link into the charts, so I'll provide links at the bottom for substeps that people commonly use, and possibly some that I make algs for. Making a chart to include ways to affect the top layer while solving the last F2L pair, and solving portions of the top layer not divisible into my 4 substeps will be put into charts aswell, but I will inlude only the practical ones that people have come up with and use such as MGLS and L5P.

I will have trouble generating algs for anything that involves permuting a piece before it is oriented, because the solver that I use can't be set to ignore orientation. If there is one that can, I'd like to know about it. A generater that can count opposite colored sides as the same would also be nice to know about, though not particularly for this.