# Roux-breaker? The YruRU method

#### Devagio

##### Member
Inspecting CP in inspection allows us to solve the entire cube 2-gen.

Many cubers, notably Jayden and Bhargav are particularly excited by this idea; and this can potentially beat Roux for One-handed solving.

If you find it game-changing, or otherwise, let me know. I'll be putting this up along with the numbering system on some facebook groups, and on youtube.

On an overall level, it has an average move count of 45-50, and is similar to Roux in execution, except we have 2GLL instead of CMLL, and there is no MU turning necessary.

The idea is simple: Reduce the cube to be solvable using
(First) u, r, U, R
(Then) r, U, R
(And finally) U, R

Scramble: D2 R2 D2 R2 F’ R2 B’ D2 B F’ D F D’ U2 B D’ U B’ L R

R U x’ // 113
It takes 2-3 moves to make a 1x1x3 with white or yellow on one of the sides with 3 pieces (there are 8 such bars possible). This has to be placed in bottom left.

U2 F’ U F // CP
This takes 3-4 moves, and I have developed a numbering system to figure out how to do CP. Since all this is seen in inspection, there is no question of recognition time; and this step is extremely flexible (my numbering system gives 12 solutions that are 4 moves or less) so influencing the next step is very easy.

This is as far as needs to be seen in inspection, however, seeing much farther is possible (for reference, doing this with my numbering system is far easier than making an EO line in inspection)..

Now the cube is reduced to RUru.

U r u R’ U2 R’ u2 // 123

The cube is now reduced to RUr.

The previous step was quite simple and almost always ends with u2, so it is a great opportunity to look ahead.

r U2 r’ U’ R r U’ r’ // EO
R’ U’ r2 // 223

The above two can in advanced solves be combined like EOLR in Roux.

Now, the cube is solvable 2gen, we know how quick that can be.

1/6 of the time we get a CP skip. That’s 4 less moves, two of which are F or D; and allows us to see further into the solve during inspection.

Here are some example solves.

These by no means are perfectly done solves, but they do show the method's general idea; and the high probability of lucky solves adds to it as evident in solve 5.

Scramble 1: R2 U’ B2 R2 D’ F2 D B2 R2 D2 B F2 U L2 R U L’ R’ F2 D

Solution 1: z // inspection
R S // 113
R2 F’ U’ F // CP
u’ R u’ R U R u2 // 123
R2 U R’ r U’ r // EO
R U2 R U r2 // 2-gen reduction

Scramble 2: F U2 B’ U2 L2 R2 F2 R2 F D’ U’ B R D’ R2 D’ L R2 D’ U

Solution 2: x2 // inspection
R’ F // 113
R F’ U’ F // CP
u R u R’ U2 R u2 // 123
R U R U r U’ r’ // EO
R’ U R U’ r2 // 2-gen reduction

Scramble 3: D2 R2 D2 R2 F’ R2 B’ D2 B F’ D F D’ U2 B D’ U B’ L R

Solution 3: x’ // inspection
R F // 113
U2 F’ U F // CP
U r u R’ U2 R’ u2 // 123
r U2 r’ U’ R r U’ r’ // EO
R’ U’ r2 // 2-gen reduction

Scramble 4: D2 F2 R B2 D2 B2 R’ F2 L’ B2 F2 U F D L’ F’ D B’ U L2

Solution 4: z y // inspection
F U’ f’ U F // 113 + CP
U r U r’ R’ U’ R’ u2 // 123
r U’ R2 U r U’ r // EO
U R’ U’ r2 // 2-gen reduction

Scramble 5: D’ B2 U B2 U’ F2 L2 U2 B2 U’ F L2 B U’ R F2 D R’ B2 L2

Solution 5: z2 y // inspection
S2 U’ F2 // 113
CP Skip!
U R’ u’ r’ R2 U R’ u2 // 123
U’ r U R U r U’ r’ // EO (+ block)
U r’ U2 r’ // 2-gen reduction (+ block)
R U2 R U2 R’ U’ R U R’ // F2L
U’ // 2-GLL Skip!

1/6 chance of CP skip, 1/324 chance of LL skip (1/108 if you do phasing, 1/12 if you do WV) Plus extremely high TPS for most of the solve, which would be under 50 moves long.

I’ll put up the numbering system.
If the method looks promising, let me (and others) know!

The method is called YruRU (Vaay-roo-roo).

EDIT: Here is the website for the method if you’re interested in having a look at the updated version (as of late June 2020).

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#### sqAree

##### Member
Can you explain the CP step a bit more? Because 2-gen redux methods have been around for a while and it's almost always the CP step that stops them from being viable, so we kinda need that explanation.

#### CuberStache

##### Member
Yes, please put the numbering system here! I don't understand the CP step at all but doing it in inspection seems like a great way to eliminate the problems other methods have had with that step (like ZZ-d). I'm super interested in this method and I would consider switching to it for OH if I knew how it worked.

#### Devagio

##### Member
This is going to be an attempt to explain how to do the first step of the method. This is basically one of the first places where I'm posting an explanation out in public than explaining in person or chat, so it may be difficult to understand. I'll try to clarify doubts via a 2-part video that I’ve made. I will also put few examples later.

The first step is the CP-line, where we have to make a 1x1x3 block in bottom left while permuting corners, and this takes 4 - 6 STM on most scrambles. Before explaining how to do it, I will gloss over some bare minimum theory required to understand what we are actually doing.

Just like EO of edges in R and U layers allows them to be solved without the intervention of F and B faces, CP of corners in R and U layers allows them to be solved without the intervention of F, B, L and D faces. What is a solved CP exactly? We do not really have to care to answer that question beyond noting that it is a permutation of the 6 corners in the R and U layers such that it is solvable using R and U; I have already done the work and you just gotta learn some rules.

1) After you solve the two corners in the bottom left, you can solve the CP by "swapping" a pair of the remaining corners. Determining which corners to swap exactly is the hard part, we'll get to that later. We'll first see how to swap them.

2) The "swapping" can be done by using one of the following 3 triggers:
a. F' U' F swaps corners in spots UFL and UFR
b. F' U F or F R’ F’ swap corners in spots UFR and UBR (edited)
c. F R F' swaps corners in spots UBL and UBR
Any or all of these moves may even be wide moves, as per the convenience of solving the DL edge of the CP-line. Also, if the corners you want to swap aren't in these locations, simply get them there using R and U moves (or their wide versions).

Now we get into the numbering system, which will be used to determine which corners to swap. For now, assume that the two corners in bottom left are already solved, and for now say we have yellow on bottom. The following bit is quite complicated, but with practice, it'll become second nature in less than an hour.

I have numbered the white-green-orange corner as 1, the white-orange-blue as 2, white-blue-red as 3 and the white-red-green as 4. The corner that is supposed to be solved in DFR is called 5, and the corner that is to be solved in DBR is called 6.

Now, imagine a thread in the cube going from locations UFL-UBL-UBR-UFR-DFR-DBR. The locations UFL, UBR and DFR are called odd places, and the locations UBL, UFR and DBR are called even places. Also, there are three couples: UFL-UBL, UBR-UFR and DFR-DBR. Trace these on a cube and you will see the obvious patterns.

Now that the naming is out of the way, we can have a look at how to identify which corners to swap to solve CP.

1) Look for corner 5. If it is in an odd location, we will "read" the thread from UFL to DBR. If corner 5 is in an even location, we will read from DBR to UFL (The meaning of reading is obvious later).

2) Now look for corner 6. It can be in 3 possible locations.
a. If corner 6 is in the couple of location of 5, do nothing.
b. If corner 6 is in another couple and the parity of its location is different from the parity of location of 5, then visualise the corners in the third couple swapped, and corner 6 swapped with the corner in the couple of corner 5.
c. If corner 6 is in another couple and the parity of its location is the same as the parity of location of 5, then visualise the third corner in the same parity location swapped with the other corner in the couple of corner 6, and corner 6 swapped with the corner in the couple of corner 5.

3) Now read the thread in the direction as dictated by rule 1, with the considerations of rule 2, ignoring 5 and 6. You will obtain some permutation of the numbers 1, 2, 3 and 4. Starting from 4, cyclically read the numbers (For eg: 3421 changes to 213).

You will have one of 6 possible sequences:

123: CP is solved

132: swap either 2-3 or 1-6 or 4-5
213: swap either 1-2 or 3-5 or 4-6 (edited)

231: swap either 1-4 or 2-5 or 3-6
312: swap either 3-4 or 1-5 or 2-6

321: swap either 3-1 or 2-4 or 5-6

You can choose any one of these swaps based on convenience of simultaneously putting the DL edge.

This took weeks to come up with and optimise, I'm fairly certain this is one of the quickest ways to do it; but again minor optimizations may be possible.

You can similarly fix a number system on yellow corners. This allows you to make one of a total of 8 1x1x3 bars. Being colour neutral would be dumb, since the number of bars only increases to 12 possible bars, but the EO will get massively difficult to determine later on.

Of course, we will not have the bottom left corners solved in each scramble, but more than 80% of the times it there will be a 1-move solution, and there always exists a 2 move solution. With under a week of practice, I could consistently plan CP-lines with 2 moves to solve the DL corners in under 15 seconds. Given I can barely plan cross+1 sub-15, I guess this means its really achievable to do this competitively with ease and maybe even plan further.

This is basically all you need to get started with this method given you understand the concept of EO as well, but i will make further posts on some great ideas I have developed that make this much more worth it.

PS. I've named the method YruRU [pronounced vaay-roo-roo], short for Yash's r-u-R-U reduction, which is also a dig at Roux [also pronounced roo], mocking "Why Roux?"

The above explanation is quite hard to understand; it’s great if you do, but I’ll add video links for ease of understanding. Till then, here is an example.

Scramble: L2 F2 D2 B' R2 B' D2 B' U2 L2 R' B D' F2 L' D R2 F L B2

x' y' (putting solved corners in bottom left)
a. 5 is in UFR, an even position, thus we read starting from DRB
b. 6 is in DRB, also an even position. Thus we visualise UBL swapped with DFR and UBR swapped with DRB.
c. Now, skipping the couple that will now have 5 and 6 as we have visualised, we read - DRB-DRF-UBL-UFL, which after the swaps has converted to UBR-UBL-DFR-UFL, which is 2143
d. Removing 4 we have 321, i.e. we have to swap either 1-3 or 2-4 or 5-6.
e. A good way to solve CP while solving DL is to swap 5-6 as follows:
U R2 f' u' F (the final u' is a wide move to facilitate the next step, which will be U R' u' R' u2)
Here, the U R2 was to set up 5 and 6 to swappable positions, and the trigger used was a variation of F' U' F.

Can you explain the CP step a bit more? Because 2-gen redux methods have been around for a while and it's almost always the CP step that stops them from being viable, so we kinda need that explanation.
Done

Yes, please put the numbering system here! I don't understand the CP step at all but doing it in inspection seems like a great way to eliminate the problems other methods have had with that step (like ZZ-d). I'm super interested in this method and I would consider switching to it for OH if I knew how it worked.
Precisely! That was the point of this idea.

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#### CuberStache

##### Member
Precisely! That was the point of this idea.
I hope to put a lot of effort into learning this later today, once I can give if my full attention. The CP recognition system seems to have a steep learning curve. I have to memorize which pieces are which numbers, figure out the weird tracing and swaps, remember what to do once I decide the CP case, and all while the pieces I'm tracing are 1-2 moves away from actually being there... It'll be a while before I have enough memorized to use this in solves but I'll be trying my best. Quarantine sucks but it's the perfect time to learn a completely new method - no comps to worry about ruining results in. Thank you for your contribution to the cubing community and I hope this becomes a popular OH method! Out of curiosity, how fast are you with this method compared to other methods? And you say you can plan CP-Line in inspection; is it nearly perfect or do you make mistakes sometimes?

#### dudefaceguy

##### Member
Interesting - I was just working out how to do last layer edge permutation using wide r and u moves for OH, so this is very timely for me. I'll look into it.

#### Devagio

##### Member
I hope to put a lot of effort into learning this later today, once I can give if my full attention. The CP recognition system seems to have a steep learning curve. I have to memorize which pieces are which numbers, figure out the weird tracing and swaps, remember what to do once I decide the CP case, and all while the pieces I'm tracing are 1-2 moves away from actually being there... It'll be a while before I have enough memorized to use this in solves but I'll be trying my best. Quarantine sucks but it's the perfect time to learn a completely new method - no comps to worry about ruining results in. Thank you for your contribution to the cubing community and I hope this becomes a popular OH method! Out of curiosity, how fast are you with this method compared to other methods? And you say you can plan CP-Line in inspection; is it nearly perfect or do you make mistakes sometimes?
I have been developing this idea since quarantine started, so I’ve simply been polishing the entire thing and haven’t practiced with this method. In Inspection I do seem to have a 100% accuracy sub-15, but even if you don’t, you’ll end up with a PLL after your 2GLL, so your solve will basically have like 3-4 seconds added at worst.
The CP in inspection is a very new idea so it’ll certainly take time to understand, but mastering it is quite easy once you get it.
Also, if you find this method a big enough breakthrough, like potentially one of big four, then do help me popularise it. Thanks

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#### CuberStache

##### Member
I have been developing this idea since quarantine started, so I’ve simply been polishing the entire thing and haven’t practiced with this method. In Inspection I do seem to have a 100% accuracy sub-15, but even if you don’t, you’ll end up with a PLL after your 2GLL, so your solve will basically have like 3-4 seconds added at worst.
The CP in inspection is a very new idea so it’ll certainly time time to understand, but mastering it is quite easy once you get it.
Also, if you find this method a big enough breakthrough, like potentially one of big four, then do help me popularise it. Thanks
Understandable that you would want to post the method as soon as you're done with it. I'd love to hear how it works for you in solves after a bit of practice. My initial thoughts are that the EO step needs to be developed more. I struggle to figure out how to use the r U/U' r' trigger to solve EO consistently. It should be more like Roux, where you recognize different cases of where bad edges are and have an "algorithm" of triggers to solve it. Also, EO recognition in the middle of a solve kinda sucks in general but it can be mitigated by the fact that the last move of 2x2x3 is basically always u2, so you can check the U-layer edges easily, and you can ignore the BD one because it's impossible to have one edge flipped. I also have trouble doing the 2x2x3 step since it's so different from any other method, but that should be easy enough to figure out with practice. I'll be posting example solves with this method in the example solve thread, and if I think it has enough potential, I might make a quest thread about it (you're new to the forums but if you search "quest" you will find many threads about people documenting their progress to improve with a certain method; it's a bit of a fad right now). That should help with popularizing it.

#### Devagio

##### Member
Understandable that you would want to post the method as soon as you're done with it. I'd love to hear how it works for you in solves after a bit of practice. My initial thoughts are that the EO step needs to be developed more. I struggle to figure out how to use the r U/U' r' trigger to solve EO consistently. It should be more like Roux, where you recognize different cases of where bad edges are and have an "algorithm" of triggers to solve it. Also, EO recognition in the middle of a solve kinda sucks in general but it can be mitigated by the fact that the last move of 2x2x3 is basically always u2, so you can check the U-layer edges easily, and you can ignore the BD one because it's impossible to have one edge flipped. I also have trouble doing the 2x2x3 step since it's so different from any other method, but that should be easy enough to figure out with practice. I'll be posting example solves with this method in the example solve thread, and if I think it has enough potential, I might make a quest thread about it (you're new to the forums but if you search "quest" you will find many threads about people documenting their progress to improve with a certain method; it's a bit of a fad right now). That should help with popularizing it.
This post is to explain the EOBF step of YruRU, so that given all the information on this group, one could use the method with complete knowledge of all its steps.

After making the CP-line, the line needs to be extended to a 1x2x3 block by attaching 1 centre and 2 edges. This step almost always ends with a u2, providing good opportunity to look ahead to the next step, which EOBF. Here, I have explained EO and BF separately, and the two can be combined with experience and/or algorithms as in EOLR in Roux.

With white or yellow centre on the U face, determine the orientation of the remaining 9 unsolved edges. You may have 0, 2, 4, 6 or 8 unoriented edges.

To solve the case with 4 edges, we have to make a staircase pattern: i.e. place the unoriented edges in DF, RF, UF and UL; while keeping white or yellow centre on top (stuff like r U2 r’ is allowed, though usually not necessary). Now to solve the EO, do (r U r’) or (r U’ r’) or (r U r) or (r U’ r). You can also mirror the staircase from the back and do the same triggers starting with r’ instead of r. The case with 8 unoriented edges can be solved by solving 4 at a time twice.

The cases with 2 or 6 unoriented edges has to be converted to the case with 4 unoriented edges by doing the triggers as above, except by keeping 1 or 3 unoriented edges in the staircase respectively.
EDIT: another way to solve the case with 2 misoriented edges (or reduce case of 6 misoriented edges) is to do stuff like r U R U r.

The fact that multiple triggers can get the same job done means you can pick the trigger to optimise EO when more than one trigger is required, and to optimise BF when only one trigger is required.

Now you’ll have a cube completely solvable using only R and U moves, and it’s really fun to spam TPS here.

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#### Arc

##### Member
Hello.

This method was invented as the Briggs method in 2015.

Here's an Ao5 from Micki.

I think it's really cool that there's excitement over it again. I've been saying that it is the best OH method for a while now. One advancement that has been made is utilizing the CP recognition from Teoidus's 2GR method to solve CP and the bar (or even the entire FB, though that is more difficult) simultaneously in inspection. A variant called Leor also exists, which forgoes the CP at the beginning of the solve and uses ZBLL to finish. Pyjam has many example solves of Leor on the example solve thread.

It would be really cool if this method could pick up traction again and get attention from bigger names.

#### Devagio

##### Member
Hello.

This method was invented as the Briggs method in 2015.

Here's an Ao5 from Micki.

I think it's really cool that there's excitement over it again. I've been saying that it is the best OH method for a while now. One advancement that has been made is utilizing the CP recognition from Teoidus's 2GR method to solve CP and the bar (or even the entire FB, though that is more difficult) simultaneously in inspection. A variant called Leor also exists, which forgoes the CP at the beginning of the solve and uses ZBLL to finish. Pyjam has many example solves of Leor on the example solve thread.

It would be really cool if this method could pick up traction again and get attention from bigger names.
Yeah, steps 2, 3 and 4 are identical! Really cool.
Though, the approach to CP is entirely different here; and arguably much faster / simplistic.
I’ll definitely have a look at this method to see whether there can be any modifications my method, thanks.

This is going to be an attempt to explain how to do the first step of the method. This is basically one of the first places where I'm posting an explanation out in public than explaining in person or chat, so it may be difficult to understand. I'll try to clarify doubts via a 2-part video that I’ve made. I will also put few examples later.

The first step is the CP-line, where we have to make a 1x1x3 block in bottom left while permuting corners, and this takes 4 - 6 STM on most scrambles. Before explaining how to do it, I will gloss over some bare minimum theory required to understand what we are actually doing.

Just like EO of edges in R and U layers allows them to be solved without the intervention of F and B faces, CP of corners in R and U layers allows them to be solved without the intervention of F, B, L and D faces. What is a solved CP exactly? We do not really have to care to answer that question beyond noting that it is a permutation of the 6 corners in the R and U layers such that it is solvable using R and U; I have already done the work and you just gotta learn some rules.

1) After you solve the two corners in the bottom left, you can solve the CP by "swapping" a pair of the remaining corners. Determining which corners to swap exactly is the hard part, we'll get to that later. We'll first see how to swap them.

2) The "swapping" can be done by using one of the following 3 triggers:
a. F' U' F swaps corners in spots UFL and UFR
b. F' U F swaps corners in spots UFR and UBL
c. F R F' swaps corners in spots UBL and UBR
Any or all of these moves may even be wide moves, as per the convenience of solving the DL edge of the CP-line. Also, if the corners you want to swap aren't in these locations, simply get them there using R and U moves (or their wide versions).

Now we get into the numbering system, which will be used to determine which corners to swap. For now, assume that the two corners in bottom left are already solved, and for now say we have yellow on bottom. The following bit is quite complicated, but with practice, it'll become second nature in less than an hour.

I have numbered the white-green-orange corner as 1, the white-orange-blue as 2, white-blue-red as 3 and the white-red-green as 4. The corner that is supposed to be solved in DFR is called 5, and the corner that is to be solved in DBR is called 6.

Now, imagine a thread in the cube going from locations UFL-UBL-UBR-UFR-DFR-DBR. The locations UFL, UBR and DFR are called odd places, and the locations UBL, UFR and DBR are called even places. Also, there are three couples: UFL-UBL, UBR-UFR and DFR-DBR. Trace these on a cube and you will see the obvious patterns.

Now that the naming is out of the way, we can have a look at how to identify which corners to swap to solve CP.

1) Look for corner 5. If it is in an odd location, we will "read" the thread from UFL to DBR. If corner 5 is in an even location, we will read from DBR to UFL (The meaning of reading is obvious later).

2) Now look for corner 6. It can be in 3 possible locations.
a. If corner 6 is in the couple of location of 5, do nothing.
b. If corner 6 is in another couple and the parity of its location is different from the parity of location of 5, then visualise the corners in the third couple swapped, and corner 6 swapped with the corner in the couple of corner 5.
c. If corner 6 is in another couple and the parity of its location is the same as the parity of location of 5, then visualise the third corner in the same parity location swapped with the other corner in the couple of corner 6, and corner 6 swapped with the corner in the couple of corner 5.

3) Now read the thread in the direction as dictated by rule 1, with the considerations of rule 2, ignoring 5 and 6. You will obtain some permutation of the numbers 1, 2, 3 and 4. Starting from 4, cyclically read the numbers (For eg: 3421 changes to 213).

You will have one of 6 possible sequences:

123: CP is solved

132: swap either 2-3 or 1-6 or 4-5
213: swap either 1-2 or 3-6 or 4-5

231: swap either 1-4 or 2-5 or 3-6
312: swap either 3-4 or 1-5 or 2-6

321: swap either 3-1 or 2-4 or 5-6

You can choose any one of these swaps based on convenience of simultaneously putting the DL edge.

This took weeks to come up with and optimise, I'm fairly certain this is one of the quickest ways to do it; but again minor optimizations may be possible.

You can similarly fix a number system on yellow corners. This allows you to make one of a total of 8 1x1x3 bars. Being colour neutral would be dumb, since the number of bars only increases to 12 possible bars, but the EO will get massively difficult to determine later on.

Of course, we will not have the bottom left corners solved in each scramble, but more than 80% of the times it there will be a 1-move solution, and there always exists a 2 move solution. With under a week of practice, I could consistently plan CP-lines with 2 moves to solve the DL corners in under 15 seconds. Given I can barely plan cross+1 sub-15, I guess this means its really achievable to do this competitively with ease and maybe even plan further.

This is basically all you need to get started with this method given you understand the concept of EO as well, but i will make further posts on some great ideas I have developed that make this much more worth it.

PS. I've named the method YruRU [pronounced vaay-roo-roo], short for Yash's r-u-R-U reduction, which is also a dig at Roux [also pronounced roo], mocking "Why Roux?"
Here are the video tutorial links. They’re not sufficient by themselves and reading the above post is necessary to make sense of what is said. The video explains the idea very thoroughly though.
Part 1:
Part 2:

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#### CuberStache

##### Member
The "swapping" can be done by using one of the following 3 triggers:
a. F' U' F swaps corners in spots UFL and UFR
b. F' U F swaps corners in spots UFR and UBL UBR
c. F R F' swaps corners in spots UBL and UBR
According to the video, the above correction should be made. This is assuming the video is correct, which I don't know for sure, but it seems like it would be easier to point at the correct pieces than to name them

#### y235

##### Member
Here's an alternative CP recognition idea (not fully developped) - suppose the DL corners are solved. We'll divide the other six corners to 3 pairs - the two corners with D-layer color, and the four corners with the U-layer color are divided to opposite pairs - so if we solved the White-Green corners, our pairs are White-Blue-Orange+White-Blue-Red, Yellow-Green-Red+Yellow-Blue-Orange, Yellow-Blue-Red+Yellow-Green-Orange.
Now, recognition goes like this:
1. Look at the two DR corners, let us call them F and B (F is the front one, of course).
2. Find each of their pairs -
case i. F and B are themselves a pair,
case ii. We have two corresponding corners in the U-layer, call them F and B as well.
3. In case i, the other 2 pairs are in the U-layer and are either both positioned adjacently (so e.g. UFR-UFL is a pair and UBR-UBL is a pair), or diagonally.
In case ii., the F and B corners in the U-layer can be in one of only 3 relative positions - adjacent and F is before B going clockwise, adjacent and B is before F going clockwise, or not adjacent.

So we have 5 possible cases overall. One also needs to know the parity of the corner permutation, and then we have a total of 10 possible cases. Each one can be solved with a 3-4 move alg (like R F' U' F).

So, overall, recognition goes like this - First take note of the parity. Then look at the DFR and DBR corners. If they "match", see whether U-layer matching corners are adjacent or opposite. If they don't match, find the relative position of their pairs in the U-layer. Do the right fix out of 10 possible algs.

There is one small caveat - the relative position is determined up to AUF, that is, doing a U turn doesn't change the relative position of the pieces in the U-layer. However, it does change the parity of the corner permutation. So one has to take that into account as well.

Also, Devagio's method is more flexible - you can choose which pair to swap out of three possible ones. However, I think that this recognition might be faster (counting parity is quick, finding the pairs is also quick).

The idea behind this is that in a state that can be solved 2-gen, those 3 pairs must be in exactly one of 5 configurations: (taken from https://www.jaapsch.net/puzzles/pgl25.htm)

That is, if we're in case V for example, then one pair must be in UBL-UBR, another one in UFR-DFR, and the last one in DBR-UFL. So we find the locations of two pairs, and then do the right moves to bring us to one of those 5 desired situations. However, there might be a swap of 2 corners in the same pair - this is something we have to take account for.

#### CuberStache

##### Member
213: swap either 1-2 or 3-6 or 4-5
After agonizing over what I could possibly be doing wrong, I think this line is incorrect. I think it should be 1-2 or 3-5 or 4-6

See this scramble for proof, or correct me if I'm wrong

R' B' D F U' R B U' R' L2 U2 F2 D' L2 D F2 L2 D F2 U2 F2

x' R2 F2 // Line, tracing should be 213

#### Arc

##### Member
This is the recognition method I mentioned. There is a learning curve to it but with some practice it becomes very fast. The big advantage of this is avoiding solving DFL and then unsolving it to solve CP. It also gives better ergonomics with only one F, and should let you see further in inspection once you are proficient.

Yeah, steps 2, 3 and 4 are identical! Really cool.
Though, the approach to CP is entirely different here; and arguably much faster / simplistic.
I’ll definitely have a look at this method to see whether there can be any modifications my method, thanks.
This method just splits the first step into three pieces, in essence it is an "intermediate version" of Briggs. Shadowslice also gives suggested substeps of 1x1x2 -> CP + 1x1x3 -> FB, which is identical to your steps except the redundant solving of DFL. And yes indeed, Shadowslice's original CP system isn't good, see above for a much better one. This is what I use and also what Micki used in the video.

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#### Devagio

##### Member
According to the video, the above correction should be made. This is assuming the video is correct, which I don't know for sure, but it seems like it would be easier to point at the correct pieces than to name them
Yes, this correction should be made, the video is correct.

After agonizing over what I could possibly be doing wrong, I think this line is incorrect. I think it should be 1-2 or 3-5 or 4-6

See this scramble for proof, or correct me if I'm wrong

R' B' D F U' R B U' R' L2 U2 F2 D' L2 D F2 L2 D F2 U2 F2

x' R2 F2 // Line, tracing should be 213
Again, correct. I made an error.
By the way, a better CP-line here probably would be
x U’ F’ U’ F’ f2 (which is 2gen too)
This is more practical because it is much easier in inspection to check out CP when the bottom left corners are solved.

Here's an alternative CP recognition idea (not fully developped) - suppose the DL corners are solved. We'll divide the other six corners to 3 pairs - the two corners with D-layer color, and the four corners with the U-layer color are divided to opposite pairs - so if we solved the White-Green corners, our pairs are White-Blue-Orange+White-Blue-Red, Yellow-Green-Red+Yellow-Blue-Orange, Yellow-Blue-Red+Yellow-Green-Orange.
Now, recognition goes like this:
1. Look at the two DR corners, let us call them F and B (F is the front one, of course).
2. Find each of their pairs -
case i. F and B are themselves a pair,
case ii. We have two corresponding corners in the U-layer, call them F and B as well.
3. In case i, the other 2 pairs are in the U-layer and are either both positioned adjacently (so e.g. UFR-UFL is a pair and UBR-UBL is a pair), or diagonally.
In case ii., the F and B corners in the U-layer can be in one of only 3 relative positions - adjacent and F is before B going clockwise, adjacent and B is before F going clockwise, or not adjacent.

So we have 5 possible cases overall. One also needs to know the parity of the corner permutation, and then we have a total of 10 possible cases. Each one can be solved with a 3-4 move alg (like R F' U' F).

So, overall, recognition goes like this - First take note of the parity. Then look at the DFR and DBR corners. If they "match", see whether U-layer matching corners are adjacent or opposite. If they don't match, find the relative position of their pairs in the U-layer. Do the right fix out of 10 possible algs.

There is one small caveat - the relative position is determined up to AUF, that is, doing a U turn doesn't change the relative position of the pieces in the U-layer. However, it does change the parity of the corner permutation. So one has to take that into account as well.

Also, Devagio's method is more flexible - you can choose which pair to swap out of three possible ones. However, I think that this recognition might be faster (counting parity is quick, finding the pairs is also quick).

The idea behind this is that in a state that can be solved 2-gen, those 3 pairs must be in exactly one of 5 configurations: (taken from https://www.jaapsch.net/puzzles/pgl25.htm)
That is, if we're in case V for example, then one pair must be in UBL-UBR, another one in UFR-DFR, and the last one in DBR-UFL. So we find the locations of two pairs, and then do the right moves to bring us to one of those 5 desired situations. However, there might be a swap of 2 corners in the same pair - this is something we have to take account for.
This seems like quite a decent idea once developed, definitely a little quicker than my version; though the lack of flexibility would probably mean adding a lot many more moves to solve the 1x1x3 line alongside CP.
I’d definitely suggest trying to expand this idea and see where it goes.

This is the recognition method I mentioned. There is a learning curve to it but with some practice it becomes very fast. The big advantage of this is avoiding solving DFL and then unsolving it to solve CP. It also gives better ergonomics with only one F, and should let you see further in inspection once you are proficient.

This method just splits the first step into three pieces, in essence it is an "intermediate version" of Briggs. Shadowslice also gives suggested substeps of 1x1x2 -> CP + 1x1x3 -> FB, which is identical to your steps except the redundant solving of DFL. And yes indeed, Shadowslice's original CP system isn't good, see above for a much better one. This is what I use and also what Micki used in the video.
This seems way too impractical tbh. I may be wrong, but this sure is quite a bit harder than YruRU’s recog. Moreover, EO isn’t necessary this early in the solve.

#### mukerflap

##### Member
Roux already reduces cube to RUMr in second block, without doing CP in inspection. F moves are not bad. I cant imagine Uw and Rw being ssuch a good moveset. For example, do r u r' u' 6 times and compare it to R U R' U' 6 times. wide U and R moves mean you have such a small grip on the cube, its not very stable. the MU part of roux is where i actually turn the fastest
this guy is consistently turning 6tps plus with 0 pauses

#### Devagio

##### Member
Roux already reduces cube to RUMr in second block, without doing CP in inspection. F moves are not bad. I cant imagine Uw and Rw being ssuch a good moveset. For example, do r u r' u' 6 times and compare it to R U R' U' 6 times. wide U and R moves mean you have such a small grip on the cube, its not very stable. the MU part of roux is where i actually turn the fastest
this guy is consistently turning 6tps plus with 0 pauses
Roux requires CMLL to do CP, and the cube is not reduced to rRUM in second block. Also, you’ll almost never have to do r and u together if you’ve seen the method. Plus, MU is not a problem but RU is better.

#### tasguitar7

##### Member
This is very interesting. It is certainly similar to Briggs but not obviously (to me at least at the moment) identical and it is basically impossible to have any new method not share some similarities with existing methods. I will look into it more for the next few days and come back with my thoughts

#### mukerflap

##### Member
Roux requires CMLL to do CP, and the cube is not reduced to rRUM in second block. Also, you’ll almost never have to do r and u together if you’ve seen the method. Plus, MU is not a problem but RU is better.
how is the cube not reduced to rRUM in sb? and CMLL is really easy