• Welcome to the Speedsolving.com, home of the web's largest puzzle community!
    You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.

    Registration is fast, simple and absolutely free so please, join our community of 40,000+ people from around the world today!

    If you are already a member, simply login to hide this message and begin participating in the community!

CLL+1: An Advanced LL Method (Complete and fully generated!)

Athefre

Member
Joined
Jul 25, 2006
Messages
1,184
CLL+1

COLL+1

Around 10 years ago Thom Barlow (Kirjava) was talking about a new LL method. The idea is to solve CLL and a single edge in one algorithm. The second step would then just be an edge 3-cycle. For cross->F2l this means all that’s left is L3E. For ZZ or Petrus, this means all that’s left is a U-Perm. The problem was that a system hadn’t been fully developed to make it work. In July I thought of how to solve the problem and discussed it on the Roux Discord server. It turned out that TDM had similar ideas a few years ago. We discussed it for a while then I started development.

The way this LL method works is that there are four algs per CLL case when edges aren’t oriented (for when doing cross then F2L for example). The four algs are split into two sets. One set cycles the edges a certain way and the other set cycles them in a different way. Then, within each of these two-alg sets, one alg orients edges a certain way and the other alg orients the edges the opposite way. When all edges are oriented, such as in ZZ or Petrus, there are two algs per CLL case. One alg cycles edges a certain way and the other alg cycles them a different way. In total there are 166 cases for CLL+1 and 83 for COLL+1.

When edges aren’t oriented, you recognize the CLL case and the edge case for that CLL. You will memorize for each case which edge will be moved to the solved position. In the document linked at the top of this post, each case in the images has a number on it. Below those images will be two columns showing “Cycle 1” and “Cycle 2”. Locate your number below one of those columns. Each number shows which edge will be moved to the solved position when one of the algs above the number list is performed. To the left of each algorithm is an image that shows how the alg will flip all of the edges. During a solve, you note the orientation of the edge associated with the number and perform the algorithm that solves that case while also flipping that edge if necessary or keeping its orientation if it is already oriented. When all edges are oriented, you simply locate your number and perform the alg that solves the case. The all edges oriented version is also linked in this post as COLL+1.

This LL method makes for a great alternative. It works for when LL edges are either misoriented or oriented. It isn’t a huge number of algorithms compared to ZBLL and it provides the solver with a very short and easy second step compared to PLL. One major benefit is the high chance of skipping the second step. For Cross->F2L->CLL+1, there are only 10 L3E cases so skips will be pretty frequent. For ZZ, Petrus, and other methods which lead to an EO LL, there are only two L3E cases and those are the simple U-Perms. As for negatives, recognition for the first step takes longer than OLL because you are mostly recognizing the LL case. However, this is made up for by three additional advantages. The first is the short L3E cases. There are fewer long and or bad cases compared to PLL. The worst L3E cases are the 2-flip cases, of which there are two. The rest are either 7 or 9 moves each. The 2-flips can even be easily avoided by just using a different alg and more on that is described in the notes section below. The second advantage over PLL is the fact that L3E skips are more frequent than PLL skips. A final advantage over PLL is that, eventually, the L3E step takes zero recognition time. Because the solver is recognizing the edge case, they will eventually come to know exactly what L3E case will come after performing the CLL+1 alg. This means a perfect transition from CLL+1 to L3E.

  • Recognition is about the same as ZBLL. It actually becomes a little easier than that. This is because for each CLL+1 case, one alg solves 9/12 edge permutations and the other alg solves the other 3/12. So all the user is looking for is if the LL case is one of those 3/12 cases or not. If it isn’t, then they know to just use the other alg. This mostly applies to COLL+1. For CLL+1 that doesn’t always work because you need to check the edge that is going to be solved to see if you need to flip or preserve its orientation. Another way to recognize is to determine whether the edges are phased or unphased. This works for COLL+1 if you don’t care about knowing exactly which U-Perm will occur. However, for CLL+1 this recognition is difficult because you have to think during the recognition time about what kind of cycle your alg does to the edges to determine which edge will be moved to the solved position. You need to know which edge is being moved to the solved position so that you can choose the correct alg that either flips or preserves the orientation of that edge. I’m actually not completely sure about the phased/unphased recognition being effective for COLL+1 either. In the analysis that I’ve done, it seems that it doesn’t always work. For some cases, the phased edges alg would be required to be used on an unphased edges state. Also, thinking about it in this phased/unphased way doesn’t reveal what cases each alg is actually able to solve. An alg may be good for when edges are phased, but you wouldn’t know that you could instead use the other alg in the set. This other alg may be one that you greatly prefer.
  • There are patterns within the cases, so that means memorization will be easier than it first appears. Many cases contain the exact same list of numbers and the edges that are moved to the solved position are the same. I based the edge positions around the corners on the ZBLL sets on the wiki. If the edge positions were further made equivalent among all of the CLL cases, there would be even more patterns to make memorization easier.
  • In a set of algs, there is an overlap where both will solve many of the same cases from the list of 12. This means you can choose which alg you want to use. You can use the alg you like more and avoid the one you like less. Another important thing to note is that of course in 2/12 cases, one of the algs will correctly position all of the edges. For CLL+1 this means you can foresee the LL being completely solved and proceed that way. Or, once you are familiar with the cases, you can sometimes use the alg from the other set to force an L3E case that you prefer. Even though there are only a few L3E cases and each are short and good algs, this could provide an additional advantage. You’ll be able to see that using one alg will lead to a 2-flip, so you can use a different alg to avoid the 2-flip and have a more desirable L3E case.
  • You can learn another set of algs or use a different combination to further increase the chance of an L3E skip. You know how your alg cycles the edges and the L3E case it will lead to. For some cases you can use a new alg from a third set to cycle a different way – forcing an L3E skip or an L3E case you prefer.
  • If the corners are already solved, it is best to use ELL. For COLL+1, if the corners are oriented, it is best to use PLL. If using a method that gets to LL without intentionally orienting edges, but all LL edges happen to be oriented, the CLL+1 sets of algs may not contain two algs that preserve the orientation of all edges. So you have a choice of using CLL+1 and leading to L3E or learning another alg set (from COLL+1) to preserve the EO and lead to a U-Perm.
  • The diagonal O case only requires one type of cycle to solve an edge for each of the five possible edge permutations. This means another alg combination isn’t necessary. So for CLL+1, only two algs are needed and for COLL+1 only one alg is required. This works for two of the possible ways of cycling edges (either swapping UB+UL or UB+UR in my document and using my AUF alignment). If the diagonal swap alg cycles edges in ways different from those two, then an alg combination is required.
  • I have attached the Excel files that I created for this method. They contain all of the possible unions of cycles (the possible alg type combinations) that will solve a corner case and an edge. I am providing this to help users generate their own algs. So far I have only a few alg sets generated for each CLL case. The ultimate goal is to have an alg set generated for every cycle. There is also an alternate way of looking at how corners are swapped and edges cycled. An adjacent swap corner case can also be viewed as a cycle of three corners. This would accomplish the same result of solving the corners. Viewing it as a cycle of three corners means that you have to also view the edge cycle in an alternate way. With a corner adjacent swap, you can’t perform a 3-cycle of edges. But if you view that corner case as a 3-cycle of corners, then you can perform a 3-cycle of edges. This alternate way of thinking about cases didn’t prove to be of any use for CLL+1. I’m only providing this explanation in case someone considers this alternate view and questions the benefits. There could be an advantage that I didn’t see, so if anyone wants the analysis of this alternate view I did generate that also and can provide it.
  • You may eventually ask yourself if it’s possible to add a third alg into the union for COLL+1 and solve all edges using just 120 algs. I did update my predictor to test this. It turns out that it is possible…if the edges are in an illegal state. Meaning if you take out two edges and put them back in the wrong positions. So maybe there is a method that could take advantage of this property. Such as having two edges at DF+DB swapped; causing the LL edges to be in that illegal state.
  • Additional Tricks: Can use transformation tricks to further reduce the number of cases, avoid cases you don’t like, or force an L3E skip. R or L moves can position a pre-made pair and change the LL to a different case. M moves can move edges into a more advantageous position. This hasn’t been explored yet, so the beneficial uses need to be documented.

Around 2011 is when this LL idea was first being talked about. It may have been an idea from Thom Barlow (Kirjava) or he had seen the idea somewhere else. He was sometimes talking about an LL method he was calling CLL+1 and making example solves. The problem was that he didn’t have a system for how to make it work. So he was asking here on the forum if anyone had ideas. I suggested a couple of systems that would work, but they weren’t perfect. The idea eventually kind of went away. A few years later, some users discussed the idea of using two COLLs per case for ZZ and taking advantage of the phased/unphased edges state to force a U-Perm. That was essentially COLL+1. Then TDM had the idea of being able to extend it to an LL with misoriented edges by having one algorithm with a certain edge orientation and another algorithm with the reverted orientation. This made it become the original CLL+1 idea.

In July of this year, unaware of those posts, I had the same ideas myself. Except, instead of the phased/unphased edges idea that others were thinking, I discovered a new kind of system where two or more algorithms can be combined like a Venn diagram. For CLL+1 this means that one algorithm does one type of edge cycle and the other algorithm does a different edge cycle. These two cycles overlap to solve an edge in all cases. The system is called the Cycle Union System and you can learn more about it here. It has many applications beyond CLL+1.

Development File: CLL+1 Unions
Development File: CLL+1 System

Alg sources:
  • CubeExplorer – Most algs were found here.
  • The most popular Roux 2H alg sheet: https://docs.google.com/spreadsheet...1yzTjnqSMS3D3dZcsCIOiJqcI/edit#gid=1701452193
  • Algdb.net
  • SpeedSolving Wiki
  • TDM – The four algs for the adjacent swap O case were provided by TDM. Also, TDM has found good CMLL algs over the years and they are within the CMLL document in the link above. Some of them proved to be helpful in developing CLL+1.
This was a huge project that took close to three months, so it is likely that there are a few mistakes. Possibly even big mistakes. If you notice anything, please let me know.
 
Last edited:

PapaSmurf

Member
Joined
Jan 4, 2018
Messages
1,032
WCA
2016TUDO02
YouTube
Visit Channel
Chish and Fips is completed (sounds way better than CLL+1). But this is really cool (and quite possibly better than a bad OLL->PLL).
damn dis big. how many algs altogether?
It says if you read the first 2 paragraphs.

Having a 1LLL with this is pretty feasible. All it takes is knowing where 2 more edges will end up, leading to a pausless LL. Do you have the average movecount of this too? L3E is ~8, so if Chish is ~12 you have a LL that is extremely efficient.
 

Athefre

Member
Joined
Jul 25, 2006
Messages
1,184
Chish and Fips is completed (sounds way better than CLL+1). But this is really cool (and quite possibly better than a bad OLL->PLL).

It says if you read the first 2 paragraphs.

Having a 1LLL with this is pretty feasible. All it takes is knowing where 2 more edges will end up, leading to a pausless LL. Do you have the average movecount of this too? L3E is ~8, so if Chish is ~12 you have a LL that is extremely efficient.

Very close actually. The average movecount is 13.1. And this is considering the fact that I spent a lot of time choosing longer fast algs instead of prioritizing movecount. This is essentially a 1LLL. Once the user has used the algs for a while they will know what is going to happen to the other three edges after applying a certain alg to a specific case.

Also, I added the number of cases after seeing TheSlykrCubr's post. I just hadn't replied yet.
 

PapaSmurf

Member
Joined
Jan 4, 2018
Messages
1,032
WCA
2016TUDO02
YouTube
Visit Channel
I wonder how difficult CLL+2 would be (in a Zipper context). To simplify everything, you solve the edge that is in FR and the edge that is where FR belongs.

FR solved: 166 cases
FR flipped: 166 cases
FR to UL: 252 cases
FR to LU: 252 cases
FR to UF: 252 cases
FR to FU: 252 cases
FR to UB: 240 cases
FR to BU: 240 cases
FR to UR: 240 cases
FR to RU: 240 cases.

This is a lot (2300). This does assume a few things though and alg count could very possibly be cut down by a fair few (perhaps half). I could've also counted incorrectly. It could be pretty good, but regular Zipper LSLL would work probably equally as well.
 

Athefre

Member
Joined
Jul 25, 2006
Messages
1,184
I wonder how difficult CLL+2 would be (in a Zipper context). To simplify everything, you solve the edge that is in FR and the edge that is where FR belongs.

FR solved: 166 cases
FR flipped: 166 cases
FR to UL: 252 cases
FR to LU: 252 cases
FR to UF: 252 cases
FR to FU: 252 cases
FR to UB: 240 cases
FR to BU: 240 cases
FR to UR: 240 cases
FR to RU: 240 cases.

This is a lot (2300). This does assume a few things though and alg count could very possibly be cut down by a fair few (perhaps half). I could've also counted incorrectly. It could be pretty good, but regular Zipper LSLL would work probably equally as well.

I think it would be much fewer cases than even half of this. Keep watch on the forums because I'll be posting something that solves this problem and many others. I'll post it in a day or two once CLL+1 has had time to be seen.
 

Athefre

Member
Joined
Jul 25, 2006
Messages
1,184
I once asked in the old irc about this in 2011 apparently someone had figured out only ~96 algs we're needed.

Do you remember what kind of system was used? Was this for CLL+1 or COLL+1? It may have been one that required an additional trick. One I suggested back then was to ensure that a pair is pre-made on the LL. Another I suggested was to do M2 to place the edge you want to solve at DB and have the algs cycle DB->UR->UL. Both of those would have around that 96 algs, but they require extra effort and moves.

Using my system that solves this the right way it is possible that the number of algs can be reduced. This is because an orientation trick was used instead of being completely incorporated into the algs. I actually have more written about this in a topic that I'll be posting soon about the system.
 

efattah

Member
Joined
Feb 14, 2016
Messages
612
CLL+1 actually dates back to Waterman in the 1980's. In basic Waterman you solve CLL, but near the end of Waterman's cubing career he began to solve 1 top layer edge at the same time as CLL. In his case, what he did was modify CLL algorithms to switch R to r and L to l and in some cases this doesn't change the movecount but changes the effect on the edges, in other cases you needed to inject an M or M' to compensate for the R->r or L->l. The end result was several variants of each CLL algorithm with different effects on the edges, allowing a single edge to be solved with essentially no extra moves vs. standard CLL.
 

Athefre

Member
Joined
Jul 25, 2006
Messages
1,184
CLL+1 actually dates back to Waterman in the 1980's. In basic Waterman you solve CLL, but near the end of Waterman's cubing career he began to solve 1 top layer edge at the same time as CLL. In his case, what he did was modify CLL algorithms to switch R to r and L to l and in some cases this doesn't change the movecount but changes the effect on the edges, in other cases you needed to inject an M or M' to compensate for the R->r or L->l. The end result was several variants of each CLL algorithm with different effects on the edges, allowing a single edge to be solved with essentially no extra moves vs. standard CLL.

Not sure why I didn't know this. Thanks for the information. I would classify the Waterman CLL+1 as something kind of different from the LL only CLL+1. This is because there are so many more edges involved in the Waterman version and because it would be a more intuitive approach to developing the algs.
 

PapaSmurf

Member
Joined
Jan 4, 2018
Messages
1,032
WCA
2016TUDO02
YouTube
Visit Channel
Why not? You have an alg that is equivalent to OLL then you have an alg that is faster than your average PLL. I get that the recog might be a bit slower, but not by enough to outweigh.
 
Joined
Jun 12, 2019
Messages
54
Why not? You have an alg that is equivalent to OLL then you have an alg that is faster than your average PLL. I get that the recog might be a bit slower, but not by enough to outweigh.
but it is slower enough to outweight because oll/pll recog are extremely fast and the algs are also extremely fast
 

Athefre

Member
Joined
Jul 25, 2006
Messages
1,184
OLL+PLL has been around for ~40 years and has had thousands of people analyze and develop the best algs, recognition, and strategies. CLL+1 has been around for 6 days and has had one person put development effort into it. I think I did a great job finding good algs, but with community involvement even better algs will be found.

Step 1:
OLL vs CLL+1:
+ Recognition: OLL takes less time to recognize.
+ Algs: OLL algs are a little shorter and faster.
+ Skip: Higher chance of a skip with OLL, but even then it is only 1/216. So not a very important measurement.

Step 2:
PLL vs L3E:
+ Recognition: L3E has 0 recognition time. There is a pauseless transition from CLL+1 to L3E after someone uses the method for long enough to learn the cases well. PLL recognition is usually fast, but even at the top level there are sometimes unwanted pauses.
+ Algs: L3E averages 9 moves and has better algs. Sometimes the last move of CLL+1 can even be canceled into the first move of L3E because the solver always knows the case that is coming. PLL is fast, but some algs are close to 20 moves.
+ Skip: 1/12 skip chance for L3E vs 1/72 for PLL. Huge difference. Imagine this at a competition. There are several rounds at a competition. You are probably going to get a second step skip with CLL+1 in at least one solve. Not so guaranteed for PLL.

I have provided all of the possible combinations in the development files in the main post. This will allow users to find even better algs than what one person was able to do. There are so many other possible algs that can be used. There are also recognition tricks that haven't been detailed.
 

xyzzy

Member
Joined
Dec 24, 2015
Messages
2,582
Let me preface this by clarifying that I'm not saying that this is a bad method or anything (it looks promising, actually); I'm just nitpicking bad statistics.

+ Skip: 1/12 skip chance for L3E vs 1/72 for PLL. Huge difference. Imagine this at a competition. There are several rounds at a competition. You are probably going to get a second step skip with CLL+1 in at least one solve. Not so guaranteed for PLL.
You cannot assume that the L3E cases will be equally likely! The true probability of an L3E skip is at most 1/24 (plus epsilon), or only around 1/48 for most CLL cases with the basic method (use the "Cycle 1" alg when it'll lead to L3E, the "Cycle 2" alg otherwise).

(The "plus epsilon" is to account for symmetric OLLCP cases, where the probability of an L3E skip can indeed be as high as 1/12 or 1/6. I'll be ignoring this henceforth, since symmetric OLLCP cases are rare.)

You do get a probability of 1/6 L3E skip with COLL+1, but even then, the probability distribution of the U perms (considered together with the angle) is not uniform. Perhaps Ua will be more common, perhaps Ub will be more common, perhaps having a left bar will be more common (RUS U perms!), etc.

Consider a different CLL+1 system, where after recognising the CLL case and AUFing, we choose one of eight algs that will solve whatever edge is currently at the UF position. For example, for the Sune CLL case (R U R' U R U2 R'), we might choose these eight algs:

(U') R' U2 R U R' U R / r U r' U R U' R' r U' r'
R U R' U R U2 R' / F U R' F R F' R U' R' F'
R U R' U R2 U R U R2 U' R' U' R2 / F' U R U R2 u' R2 u R U2 F
(U') R' U' R U R U R' U' R' U R U R U' R' / r U R' U R U2 r'

(I didn't try particularly hard to optimise the alg choices.)

In this CLL+1 system, the permutation and the orientation of the three edges we don't look at are completely undetermined, and so we really have a 1/3 chance of EP skip and 1/4 chance of EO skip, for a total 1/12 chance of L3E skip.

In the CLL+1 system proposed in this thread, we're not solving a specific edge, which means that we can no longer safely assume that the remaining three edges are completely undetermined. Indeed, there's an even simpler way of seeing that the L3E skip rate can't possibly be as high as 1/12: an L3E skip is equivalent to doing 1LLL, but if you're learning only 4 algs out of each CLL subset of (4!/2) × 2^3 = 96 algs, how can you be doing 1LLL more than 4/96 = 1/24 of the time?

---
(throwing in a reply to the bottom post by Athefre because it's not really on-topic)

I received the numbers from a couple of people on Discord. They both gave the same number, so I trusted the calculation.
Probability is weird like that: one could come up with many different probabilities for a well-defined event, all with seemingly reasonable and intuitive justifications, and yet only one of the answers can be correct. Outside of cubing, the Monty Hall problem is the classic example (yes, you should switch); within cubing, case probabilities in steps like LSLL is the natural example (the probability of an LS skip is less than 1/150 under reasonable assumptions).

Probability is hard, but getting it right also seems to be pretty important when trying to develop methods, especially methods that involve new algs that you haven't gotten into muscle memory. In lieu of actual empirical data of how well a new method performs, having some objective metric like average move count to serve as a proxy would be nice. And even then, empirical data is hard to come by when we're looking at marginal differences between methods (e.g. methods that differ by only a quarter of a move on average); the smaller the difference, the more work you have to do to be certain of it. (And it's not linear! You have to do quadratically more work!)

To be fair, the mistake made here (that L3E skip is 1/12) doesn't substantially affect the calculus, and it's why I just stated up front that this was statistical nitpicking.
 
Last edited:

Athefre

Member
Joined
Jul 25, 2006
Messages
1,184
Let me preface this by clarifying that I'm not saying that this is a bad method or anything (it looks promising, actually); I'm just nitpicking bad statistics.


You cannot assume that the L3E cases will be equally likely! The true probability of an L3E skip is at most 1/24 (plus epsilon), or only around 1/48 for most CLL cases with the basic method (use the "Cycle 1" alg when it'll lead to L3E, the "Cycle 2" alg otherwise).

(The "plus epsilon" is to account for symmetric OLLCP cases, where the probability of an L3E skip can indeed be as high as 1/12 or 1/6. I'll be ignoring this henceforth, since symmetric OLLCP cases are rare.)

You do get a probability of 1/6 L3E skip with COLL+1, but even then, the probability distribution of the U perms (considered together with the angle) is not uniform. Perhaps Ua will be more common, perhaps Ub will be more common, perhaps having a left bar will be more common (RUS U perms!), etc.

Consider a different CLL+1 system, where after recognising the CLL case and AUFing, we choose one of eight algs that will solve whatever edge is currently at the UF position. For example, for the Sune CLL case (R U R' U R U2 R'), we might choose these eight algs:

(U') R' U2 R U R' U R / r U r' U R U' R' r U' r'
R U R' U R U2 R' / F U R' F R F' R U' R' F'
R U R' U R2 U R U R2 U' R' U' R2 / F' U R U R2 u' R2 u R U2 F
(U') R' U' R U R U R' U' R' U R U R U' R' / r U R' U R U2 r'

(I didn't try particularly hard to optimise the alg choices.)

In this CLL+1 system, the permutation and the orientation of the three edges we don't look at are completely undetermined, and so we really have a 1/3 chance of EP skip and 1/4 chance of EO skip, for a total 1/12 chance of L3E skip.

In the CLL+1 system proposed in this thread, we're not solving a specific edge, which means that we can no longer safely assume that the remaining three edges are completely undetermined. Indeed, there's an even simpler way of seeing that the L3E skip rate can't possibly be as high as 1/12: an L3E skip is equivalent to doing 1LLL, but if you're learning only 4 algs out of each CLL subset of (4!/2) × 2^3 = 96 algs, how can you be doing 1LLL more than 4/96 = 1/24 of the time?

Thanks for this. In this community, I'm a full-time method developer and I think I've discovered and developed some great things. However, a huge weakness of mine has always been group theory. It really is something that I should know. It has slowed down development time for some of my projects. It has also kept me from being able to prove advantages. I think CLL+1 is the only time that I've ever provided numbers like this. I received the numbers from a couple of people on Discord. They both gave the same number, so I trusted the calculation. But I guess maybe I should have just left it out of the posts.

I really appreciate the amount of thought and effort that you put into your reply. Thank you.
 

trangium

Member
Joined
Jul 24, 2019
Messages
117
WCA
2019TRAN10
How would one go about recognizing CLL+1 after recognizing the CLL case? Like what exactly would I look for on the scramble B2 R F2 R B2 L' B2 R D2 R B' L B2 R F2 R' B? Would the recognition be significantly harder if CLL+1 was done entirely with cycle unions and without the orientation trick? Depending on how easy the recognition is, I might start learning this set.

Also a note on anyone who wants to contribute to the algs: cycle unions are not needed to generate algs if the orientation trick is used. To generate the algs for a specific CLL case, find two algs for that CLL case where applying the inverse of one algorithm then solving the corners with the other algorithm leads to an ELL case with exactly one solved or flipped edge, then find the two algs that are a 4-flip away from the original two. For example, for the Sune pure twist case, I found R U R' U R U2 R' and r' R2 U R' U r U2 r' U M'. Applying R U2 R' U' R U' R' then r' R2 U R' U r U2 r' U M' leads to an ELL case with exactly one flipped edge, so we know this pair works. Then I found F R U R' U' F' R U R' F' U2 F R U' R' and R' U2 R2 U R2 U R U' R U' R' which are a 4-flip away from the first two cases.
 
Top