EDIT: The following descriptions, simulations and algorithms of the method are obsolete now, thanks to the efforts of the community in developing the method. For a better understanding of the method, please check out https://devagio.github.io/Mehta/ or read page 5 onwards on this thread, and join the discord community if interested (invite link is available in the about section of the website and here after page 5).

Algorithmic methods are doing fairly well compared to block building methods, and for good reason. TPS on the new hardware has shot up and it makes sense to make best use of it. A method with a bearable number of algorithms, yet highly algorithmically reliant and with good recognition has the potential to beat a good deal of methods. A means to achieve that is to have 3 algorithmic steps (compared to 2 for CFOP for example); while ensuring that those algorithms do indeed tackle a large set of cases (2LLL CFOP handles ~15k cases).

Orientation steps are almost always recognition friendly as long as they involve one (or two) top colour (like TSLE is decent for recog, OLL is pretty good, etc. but LEOR EO is hard) Permutation steps are good for recognition if a single glance at a cube, and registering 4-5 stickers at most can uniquely identify each case (like 2 sided PLL recognition system if good, ZBLL recognition on the other hand is hard, so is 4x4 PLL).

Cube symmetries can play a huge role in reducing number of algorithms while getting tons of cases solved (There are 72 PLL cases, but symmetry reduces it to 22 (21+1)). These symmetries can be exploited much better once we think about it.

Based on these primers and a bit of work, there seems to be one very promising end to a solve using 3 algorithms:

Suppose a 2x2x3 is solved on DL; and FR and BR edges are solved.

Alg-1: orients 6 corners. This will be a short <R,U> alg; and surprisingly, there are very few cases thanks to the freedom of U layer and the S slice symmetry. Also, R2 setups cut this down much further. There are 23 + (4 x 8) + (2 x 8) = 71 cases, though only 39 algorithms excluding mirrors. If we consider R2 setups, this number goes below 30.

Alg-2: Permutes 6 corners. These will be <R,U,D> algs (<R2,U,D> is enough, but algs may get lengthy); and again surprisingly, there are very few cases thanks to the same freedom and symmetry. There are 5 + (4 x 6) + 18 = 47 cases, though only 29 algorithms excluding mirrors. 2 sided recognition is possible here.

Alg-3: L5E, <M,U> gen, very simple to recognise during the previous alg; though almost 200 cases (60-70 algs excluding mirrors and inverses)

This 3-alg-system, which without any reduction has around 300 algs can reduce the cube state from a possible 6! * 5! * 3^5 * 2^3 = 168M cases! Although, it is possible to influence L5E during belt by doing partial EO (6CO and 6CP doesn't influence EO).

For comparison, CFOP LSLL is a 3-alg-system with 119 algs with 9M cases.

The algorithm quality (length, movesets, recognition and look-ahead ability) is at worst comparable to CFOP LSLL.

How do we achieve " a 2x2x3 is solved on DL; and FR and BR edges are solved"?

Two premises about this.

Firstly, blockbuilding is probably the best thing that can be done with the inspection time in terms of reducing the number of cubestates. A roux block does this slightly better than a petrus block, which is better than a cross; all solving the same number of pieces (counting centres, since roux centres aren't fixed).

Second, there is a reason CFOP is as successful as it is, and that is the symmetry in F2L. We can choose to do whichever F2L pair we please, and we can do these in any order. This is very good for a step that is neither algorithmic, nor block building; because we can simply do whatever we see first. Finally, we can use only the last pair to influence LL.

This gives us a beautiful way to achieve the above mentioned cubestate. A roux block can be done in inspection on bottom, and then each of the 4 edges FR, BR, BL, FL can be placed between corresponding centres in any order convenient; while influencing EO if one wants to reduce the L5E set. And you have your required cubestate! The best part is, this central "belt" doesn't have to align with FB, because at the end of the solve, we can do an AUF with an ADF and will be done.

Step 1: FB (~6 moves)

Step 2: Belt (~9 moves) <R,U,u>

Step 3: 6COLL (~13 moves) <R,U>

Step 4: 6CPLL (~12 moves) <R,U,D>

Step 5: L5E (~15 moves) <M,U>

Avg ~ 55 moves

Promises a higher TPS than CFOP, since FB and cross should be at worst similar TPS, belt is higher TPS than F2L since you don’t have to track 2 pieces or rotate, and a higher percentage of the solve is algorithms; which in all makes a higher TPS objectively certain.

The symmetry exploitation keeps the method as efficient as CFOP despite being much more algorithmic.

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

y2 // inspection

F U2 B' L F B2 // FB (6/6)

u R u R' u R2 u' R' U2 R // Belt (10/16)

U R' U' R U2 R' U' R // 6CO (8/24)

R2 U2 R2 U'D' R2 U R2 U' R2 D R2 D' // 6CP (13/37)

M' U M U2 M' U M' U M2 U' M' U2 M' U2 M2 // L5E (15/52)

U'D' // Adjusting faces (2/54)

(50 ETM)

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

x2 // inspection

B L' F U' B2 R B' // FB (7/7)

U R u' R u2 R2 u' R' U R // Belt (10/17)

R' U' R2 U' R' U R2 U R // 6CO (9/26)

D' R2 D R2 U' R2 U2D' R2 U' R2 U' R2 // 6CP (13/39)

U2 M' U' M U' M' U M2 U2 M U M' U2 M' // L5E (14/53)

(47 ETM)

Through this thread, I hope to improve upon this idea and generate 6CO, 6CP and L5E algs (and perhaps a better recognition system if there is one). Constructive criticism is more than welcome. If you intend to, please let me know if you would be willing to assist me in generating the algorithms. Most importantly, let it be known if you're willing to give this method a try and attempt to beat your CFOP global with it.

I will soon be making a github (similar to YruRU) where I will put all the algs and relevant development discussed on this thread and elsewhere.

Disclaimer: The title isn't meant to offend. It is both a milestone setter, and a hilarious continuation of the theme started in YruRU.

The method has evolved quite a bit thanks to the efforts of numerous people on this forum and otherwise.

Here is the improved version of the method, with move statistics for non-CN solvers:

Step 1: First block (~7 moves) [There are 24 possibilities for a fully CN solver]

Step 2: 3-quarters belt (~7 moves) [There are 4 possible 3QB for each FB, and 6 piece solving orders for each 3QB]

Step 3: Edge orientation+Last edge (~6 moves) [55 cases]

Step 4: 6-Corners' Orientation (~8.5 moves) [71 cases]

Step 5: 6-Corners' Permutation (~10 moves) [47 cases]

Step 6: Last 5 edges' Permutation (~7.5 moves) [16 cases]

Step 0: Various face adjustments (~4 moves) [Between algorithms]

Average move-count: ~50

Total algorithm count: 134 (+55 intuitive)

Spoiler: All Algorithms
Spoiler: EOLE algs

The 56 cases are divided into 8 subsets (8x4 + 6x4). The first four are when the last edge is in UF and the slot is in FR. The subsets are denoted by the orientation of the last edge (in UF) and the edge in FR respectively. Each case in these sets is denoted by the orientation of UL, UB, UR and DR edges respectively. The next two cases are when the last edge is in FR itself, either solved or flipped. The final two cases are when the last edge is in DR.

Note that all the algorithms are entirely intuitive, and should be treated so when beginning. The overall average movecount of these inserts is 5.80 moves.

Spoiler: 6CO algs

The 72 cases (71+1) are divided into 9 subsets, depending on the orientations of DFR and DBR corners. For example, the subsection "RB" is when the DFR corner is oriented towards R and the DBR corner is oriented towards B.

In 3 of the subsections, there are familiar OCLL patterns like T, U, L, O, H, Pi, etc. The other 6 subsets (3+3) have different patterns which I have tried to name based on how they look. Note that there is a symmetry here. For instance, The mirror image of "dog" in one of the subsets is "cat" in the other subset; because dog and cat are similar words. Other such pairs are hand-leg, pigeon-eagle, sickle-hammer, etc. This will become obvious with time.

Under every subset, the average movelength of the main algs is mentioned. The overall average movelength of all of these algorithms is 8.47 STM. There are shorter algorithms for some of these cases, however I have tried to pick the fastest ones. It is interesting to note that in the 5 subsets (40/72 cases) where there is at least one of the D corners oriented, the average alg length is on average 20% less than the remaining 4, so an advanced solver may try to force at least one of the D corners to be oriented for better cases.

Spoiler: 6CP algs

Spoiler: Beginners' version

Spoiler: Beginner CO algs (step 4c)

Or use regular OCLL algs.

Spoiler: Beginner CP algs (step 5b)

Spoiler: Stats

Spoiler: Beginners' Example

Spoiler: Stats Simulation

Ran a 10000 solve simulation of a constrained version of the method.

Constraints:

- only one FB instead of a possible 24 FB

- exactly the same 3 belt edges solved first instead of 4 possible combinations

And then there were the general constraints that there is no influencing of one step on the next (which is kind of the point though).

The best part is, this number actually holds for the method because there isn't much scope of inefficiency. FB is planned in inspection so any intermediate solver is going to be almost optimal, and the last 4 steps are algorithms. The only place where fluid intuition is needed is the first 3 belt edges (3 pieces).

In comparison, theoretical and practical CFOP numbers are nowhere close because of the move inefficiencies even top solvers have during F2L (8 pieces).

**Philosophy and motivation:**Algorithmic methods are doing fairly well compared to block building methods, and for good reason. TPS on the new hardware has shot up and it makes sense to make best use of it. A method with a bearable number of algorithms, yet highly algorithmically reliant and with good recognition has the potential to beat a good deal of methods. A means to achieve that is to have 3 algorithmic steps (compared to 2 for CFOP for example); while ensuring that those algorithms do indeed tackle a large set of cases (2LLL CFOP handles ~15k cases).

Orientation steps are almost always recognition friendly as long as they involve one (or two) top colour (like TSLE is decent for recog, OLL is pretty good, etc. but LEOR EO is hard) Permutation steps are good for recognition if a single glance at a cube, and registering 4-5 stickers at most can uniquely identify each case (like 2 sided PLL recognition system if good, ZBLL recognition on the other hand is hard, so is 4x4 PLL).

Cube symmetries can play a huge role in reducing number of algorithms while getting tons of cases solved (There are 72 PLL cases, but symmetry reduces it to 22 (21+1)). These symmetries can be exploited much better once we think about it.

Based on these primers and a bit of work, there seems to be one very promising end to a solve using 3 algorithms:

Suppose a 2x2x3 is solved on DL; and FR and BR edges are solved.

Alg-1: orients 6 corners. This will be a short <R,U> alg; and surprisingly, there are very few cases thanks to the freedom of U layer and the S slice symmetry. Also, R2 setups cut this down much further. There are 23 + (4 x 8) + (2 x 8) = 71 cases, though only 39 algorithms excluding mirrors. If we consider R2 setups, this number goes below 30.

Alg-2: Permutes 6 corners. These will be <R,U,D> algs (<R2,U,D> is enough, but algs may get lengthy); and again surprisingly, there are very few cases thanks to the same freedom and symmetry. There are 5 + (4 x 6) + 18 = 47 cases, though only 29 algorithms excluding mirrors. 2 sided recognition is possible here.

Alg-3: L5E, <M,U> gen, very simple to recognise during the previous alg; though almost 200 cases (60-70 algs excluding mirrors and inverses)

This 3-alg-system, which without any reduction has around 300 algs can reduce the cube state from a possible 6! * 5! * 3^5 * 2^3 = 168M cases! Although, it is possible to influence L5E during belt by doing partial EO (6CO and 6CP doesn't influence EO).

For comparison, CFOP LSLL is a 3-alg-system with 119 algs with 9M cases.

The algorithm quality (length, movesets, recognition and look-ahead ability) is at worst comparable to CFOP LSLL.

**Now to the hard part:**How do we achieve " a 2x2x3 is solved on DL; and FR and BR edges are solved"?

Two premises about this.

Firstly, blockbuilding is probably the best thing that can be done with the inspection time in terms of reducing the number of cubestates. A roux block does this slightly better than a petrus block, which is better than a cross; all solving the same number of pieces (counting centres, since roux centres aren't fixed).

Second, there is a reason CFOP is as successful as it is, and that is the symmetry in F2L. We can choose to do whichever F2L pair we please, and we can do these in any order. This is very good for a step that is neither algorithmic, nor block building; because we can simply do whatever we see first. Finally, we can use only the last pair to influence LL.

This gives us a beautiful way to achieve the above mentioned cubestate. A roux block can be done in inspection on bottom, and then each of the 4 edges FR, BR, BL, FL can be placed between corresponding centres in any order convenient; while influencing EO if one wants to reduce the L5E set. And you have your required cubestate! The best part is, this central "belt" doesn't have to align with FB, because at the end of the solve, we can do an AUF with an ADF and will be done.

**Summing it all up:**Step 1: FB (~6 moves)

Step 2: Belt (~9 moves) <R,U,u>

Step 3: 6COLL (~13 moves) <R,U>

Step 4: 6CPLL (~12 moves) <R,U,D>

Step 5: L5E (~15 moves) <M,U>

Avg ~ 55 moves

Promises a higher TPS than CFOP, since FB and cross should be at worst similar TPS, belt is higher TPS than F2L since you don’t have to track 2 pieces or rotate, and a higher percentage of the solve is algorithms; which in all makes a higher TPS objectively certain.

The symmetry exploitation keeps the method as efficient as CFOP despite being much more algorithmic.

**Examples:**Scramble: B L' U F' L' R2 D' B F' U B U' L R2 U2 F2 B2 L' D2 U2 F' B R2 L B

y2 // inspection

F U2 B' L F B2 // FB (6/6)

u R u R' u R2 u' R' U2 R // Belt (10/16)

U R' U' R U2 R' U' R // 6CO (8/24)

R2 U2 R2 U'D' R2 U R2 U' R2 D R2 D' // 6CP (13/37)

M' U M U2 M' U M' U M2 U' M' U2 M' U2 M2 // L5E (15/52)

U'D' // Adjusting faces (2/54)

(50 ETM)

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

x2 // inspection

B L' F U' B2 R B' // FB (7/7)

U R u' R u2 R2 u' R' U R // Belt (10/17)

R' U' R2 U' R' U R2 U R // 6CO (9/26)

D' R2 D R2 U' R2 U2D' R2 U' R2 U' R2 // 6CP (13/39)

U2 M' U' M U' M' U M2 U2 M U M' U2 M' // L5E (14/53)

(47 ETM)

Through this thread, I hope to improve upon this idea and generate 6CO, 6CP and L5E algs (and perhaps a better recognition system if there is one). Constructive criticism is more than welcome. If you intend to, please let me know if you would be willing to assist me in generating the algorithms. Most importantly, let it be known if you're willing to give this method a try and attempt to beat your CFOP global with it.

I will soon be making a github (similar to YruRU) where I will put all the algs and relevant development discussed on this thread and elsewhere.

Disclaimer: The title isn't meant to offend. It is both a milestone setter, and a hilarious continuation of the theme started in YruRU.

__EDIT (19 August):__The method has evolved quite a bit thanks to the efforts of numerous people on this forum and otherwise.

Here is the improved version of the method, with move statistics for non-CN solvers:

Step 1: First block (~7 moves) [There are 24 possibilities for a fully CN solver]

Step 2: 3-quarters belt (~7 moves) [There are 4 possible 3QB for each FB, and 6 piece solving orders for each 3QB]

Step 3: Edge orientation+Last edge (~6 moves) [55 cases]

Step 4: 6-Corners' Orientation (~8.5 moves) [71 cases]

Step 5: 6-Corners' Permutation (~10 moves) [47 cases]

Step 6: Last 5 edges' Permutation (~7.5 moves) [16 cases]

Step 0: Various face adjustments (~4 moves) [Between algorithms]

Average move-count: ~50

Total algorithm count: 134 (+55 intuitive)

GG | ||

GGGG | R U' R' | |

GGBB | U F R F' U2 R' | U R' F R2 F' U2 R' |

GBGB | U2 F R F' U R' | U2 R' F R2 F' U R' |

GBBG | U' R U2 R2 F R F' | U F R' F' R2 U2 R' |

BGGB | U R F' U2 F R' | |

BGBG | R2 U' F R F' R2 | |

BBGG | U2 F R' F' R2 U R' | |

BBBB | U S' U S R U2 R' | U R' F R2 F2 U2 F R' |

avg | 5.25 | |

GB | ||

GGGB | U2 R F' U F R' | |

GGBG | U' R U R2 F R F' | |

GBGG | F' U2 F R U' R' | |

BGGG | U' R' F R F' | |

GBBB | U2 F R F2 U' F U2 R' | |

BGBB | U F R F2 U2 F R' | |

BBGB | U2 R' F R2 F2 U F R' | U2 R F' U F2 R2 F' R |

BBBG | U S' U' S U2 R' F R F' | |

avg | 6.125 | |

BG | ||

GGGB | F R F' R' | |

GGBG | F R' F' R | |

GBGG | R U R' F' U' F | F U R' U' F' U R |

BGGG | U' F R' F' R2 U' R' | |

GBBB | U2 R F' U2 F2 R' F' | |

BGBB | U' S' U S R U R' | U R U2 F' U F2 R' F' |

BBGB | U' R' F R2 F2 U' F R' | U2 S' U' S U R U' R' |

BBBG | U S' U S U' F R' F' R | |

avg | 5.875 | |

BB | ||

GGGG | U' F' U F | E R' U' R |

GGBB | F' U F2 R F' R' | U R F R' F2 U' F |

GBGB | U2 R F R' F2 U2 F | F R F2 U2 F U2 R' |

GBBG | U R' F R F2 U' F | F R' F2 U2 F U2 R |

BGGB | u' F R F' | |

BGBG | F R' F2 U' F U R | u' F U R2 U' R F' |

BBGG | F' U2 F2 R' F' R | U2 R' F R F2 U2 F |

BBBB | U S' U S u' R' U' R | |

avg | 5.625 | |

solved | ||

0B | nil | |

1B | R' F R2 F' R' | |

2B (adj) | R F R2 F' R | F U R U' R' F' |

2B (opp) | F R U R' U' F' | |

3B | S' U S | F R F' U2 F R' F' |

4B | R2 S' U S U2 R2 | |

avg | 4.17 | |

flipped | ||

0B | F R F2 U F U2 R' | R U R' F R F' R' |

1B | F' U2 F R U R' | R U R' F R' F' R |

2B (adj) | F R F2 U' F R' | |

2B (opp) | R F' U F2 R' F' | |

3B | R' F R F' U' R' F R F' | |

4B | S' U S R U2 R' F' U' F | |

avg | 7.17 | |

G | ||

0B | R' U' R U R | |

1B | R F R' F' | R U F R' F' |

2B (adj) | R2 F R F' R2 | R U' F R' F' |

2B (opp) | R F U R' U' F' | |

3B | F R' F2 U2 F R U2 R | |

4B | F R' F' R2 U2 F R' F' | |

avg | 6.00 | |

B | ||

0B | F u' R u F2 | |

1B | F R F' U' R' | |

2B (adj) | u' F R' F' | |

2B (opp) | u' F U R' U' F' | F R F2 U2 F U' R' |

3B | R F' U' F2 R2 F' R | |

4B | u' F R' F' U2 R F R2 F' R | |

avg | 6.17 |

The 56 cases are divided into 8 subsets (8x4 + 6x4). The first four are when the last edge is in UF and the slot is in FR. The subsets are denoted by the orientation of the last edge (in UF) and the edge in FR respectively. Each case in these sets is denoted by the orientation of UL, UB, UR and DR edges respectively. The next two cases are when the last edge is in FR itself, either solved or flipped. The final two cases are when the last edge is in DR.

Note that all the algorithms are entirely intuitive, and should be treated so when beginning. The overall average movecount of these inserts is 5.80 moves.

DD | ||

H | R' U2 R U2 R U2 R U2 R' | R' U' R2 U' R U2 R2 U' R' |

Pi | R U2 R2 U' R2 U' R2 U2 R | |

Sune | R U R2 U' R2 U R | |

Antisune | R U2 R' U' R U' R' | |

U | F U' R2 U R2 U F' | D R D' R2 U R2 D R' D' |

T | R D' R U2 R' D R' | R U R D R' U' R D' R2 |

L | R D' R U' R' D R' | |

O | nil | |

avg | 6.63 | |

RR | ||

H | R U' R' U2 R' U' R U2 R U' R | |

Pi | R U' R' U' R U' R U2 R' U R' | |

Sune | R U' R' D' U' R U2 R' D | |

Antisune | R' U R D U R' U2 R D' | |

U | R U' R2 U' R U2 R2 U' R | |

T | R U2 R' U' R2 U2 R' U R' | R U R U2 R2 U' R U2 R' |

L | R U R U2 R' U R' | |

O | D' R U' R' D R' U R2 U' R | R U R U' R' U R U' R' U R' |

avg | 9.13 | |

FB | ||

H | R U R' U' R2 U R2 U' R2 U2 R' U' R' | R' U' R U R2 U' R2 U R2 U2 R U R |

Pi | R U R2 U2 R2 U R' U2 R' U' R | |

Sune | D' R U' R' D U' R U2 R' | R2 U R' U2 R U' R2 U R U2 R' |

Antisune | D R' U R D' U R' U2 R | R2 U' R U2 R' U R2 U' R' U2 R |

U | R' U2 R2 U' R2 U' R2 U2 R' | |

T | R U2 R U2 R U2 R U2 R | |

L | R' U R U R2 U' R2 U R' U2 R | |

O | R2 U' R D' R U2 R' D R | R U2 R U' R' U2 R U' R' U2 R' |

avg | 9.75 | |

RD | ||

hand | L' U R2 U' R2 L | R' D' R U' R' D R |

asia | R U' R' U' R U2 R' | |

S | D R' U2 R U' R' U' R D' | |

p | R' U R U2 R2 U' R U2 R | |

dog | R U' R' U2 R U' R' | |

hammer | B' R2 U' R2 U B | R D U' R' U2 R D' U' R' |

clock | R' U2 R' U R2 U' R' U2 R | |

eagle | R' U2 R U R2 U' R U2 R | |

avg | 7.75 | |

FD | ||

leg | R D' R U R' D R' | |

america | D R' U R U R' U2 R D' | R2 U' R U2 R' U' R U' R |

N | R U2 R' U R U R' | |

b | R S' U2 S R | R U R' U2 R U R' |

cat | D R' U R U2 R' U R D' | R' U' R2 U2 R' U R2 U2 R2 U' R' |

sickle | B' U' R2 U R2 B | R2 U R' U R U2 R' U R' |

fan | R U' R' U' R U' R' U R U2 R' | |

pigeon | R U2 R' U R U' R' U' R U' R' | |

avg | 8.13 | |

DR | ||

leg | L U' R2 U R2 L' | R D R' U' R D' R' |

america | R' U R U R' U2 R | |

N | D' R U2 R' U R U R' D | |

b | R U' R' U2 R2 U R' U2 R' | D' R U R' U2 R U R' D |

cat | R' U R U2 R' U R | |

sickle | F R2 U R2 U' F' | |

fan | R U2 R U' R2 U R U2 R' | |

pigeon | R U2 R' U' R2 U R' U2 R' | |

avg | 7.75 | |

DB | ||

hand | R' D R' U' R D' R | |

asia | R2 U2 R' U R' U R U2 R' | |

S | R' U2 R U' R' U' R | |

p | R' S' U2 S R' | R' U' R U2 R' U' R |

dog | D' R U' R' U2 R U' R' D | R' U R U R' U' R U R' U2 R |

hammer | F U R2 U' R2 F' | |

clock | R' U R U R' U R U' R' U2 R | |

eagle | R' U2 R U R' U' R U' R' U' R | |

avg | 8.13 | |

FR | ||

hand | R U' R U' R' U2 R' | |

asia | R U2 R U' R2 U2 R U R' | |

S | R U' R' U2 R' U R U' R2 U' R' | |

p | R U' R2 U R' U R' U2 R | R U2 R' D' U R U R' D |

dog | R U R U R' U R U' R' U2 R' | |

hammer | R U2 R' U R' U R2 U' R | R U2 R' U R2 U' R' U2 R' |

clock | R' U R U' R U' R U R' | R U R U R2 U R U R' |

eagle | R U2 R' U R2 U2 R2 U2 R U R' | |

avg | 9.5 | |

RB | ||

leg | R' U R' U R U2 R | |

america | R2 U' R' U R U2 R' U R' | |

N | R' U R U2 R U' R' U R2 U R | R' U' R U' R' U' R2 U' R2 U R' |

b | D' R U' R' D U2 R U' R' | D R' U2 R D' U' R' U' R |

cat | R' U' R U R2 U' R2 U R' U' R | |

sickle | R U2 R U R2 U' R U2 R' | R' U2 R U' R2 U R U2 R |

fan | R U' R' U R' U R' U' R | |

pigeon | R' U2 R U' R2 U2 R2 U2 R' U' R | |

avg | 9.5 |

In 3 of the subsections, there are familiar OCLL patterns like T, U, L, O, H, Pi, etc. The other 6 subsets (3+3) have different patterns which I have tried to name based on how they look. Note that there is a symmetry here. For instance, The mirror image of "dog" in one of the subsets is "cat" in the other subset; because dog and cat are similar words. Other such pairs are hand-leg, pigeon-eagle, sickle-hammer, etc. This will become obvious with time.

Under every subset, the average movelength of the main algs is mentioned. The overall average movelength of all of these algorithms is 8.47 STM. There are shorter algorithms for some of these cases, however I have tried to pick the fastest ones. It is interesting to note that in the 5 subsets (40/72 cases) where there is at least one of the D corners oriented, the average alg length is on average 20% less than the remaining 4, so an advanced solver may try to force at least one of the D corners to be oriented for better cases.

These algorithms were generated and compiled by @trangium

DF and DB location numbers follow the YruRU numbering scheme.

For the bottom cases, the Case column refers to 3 stickers on the right. For the two cases above that, the case numbers refer to 3 stickers on the front

DF Loc. | DB Loc. | Case | AUF | Alg | AUF | Alt Alg | AUF | Alt Alg |

5 | 6 | Solved | D' | |||||

Adjacent | R U R' F' R U R' U' R' F R2 U' R' D' | |||||||

Diagonal | R U' R' U R U' D' R2 U R D R U' D' R2 | F R U' R' U' R U R' F' R U R' U' R' F R F' D' | ||||||

6 | 5 | Solved | F R U R U' R' F' R U' R U R2 U' R D' | R' U R' F' R U R' U' R' F R2 U' R' U' R2 D' | ||||

Adjacent | R U R' D' R U' D2 R' U' R D2 R' | R2 U R2 U2 R2 D' R2 U R2 U' R2 | ||||||

Diagonal | R U' R' U R' U2 R2 U R U R' U' R2 D' | |||||||

2 | 6 | Front Opp | R2 U2 R2 U' R2 U' R2 D' | U2 | R2 U R2 U R2 U2 R2 D' | |||

Right Opp | U' | D R2 U2 R2 U R2 U R2 D2 | U | D R2 U' R2 U' R2 U2 R2 D2 | ||||

Double Opp | U2 | x R' U R U' R' U R U' R' U R U' F' x' | R' D R U' R D' R' U2 R D R' U R' D' R D' | R2 U R2 D R2 D' R2 U' R2 D R2 D2 | ||||

Front Bar | D' R2 U R2 U' R2 D R2 D' | D' R U R' U R2 D R D' R | ||||||

Right Bar | U | R2 D' R2 U R2 U' R2 | U | D' R' D R' D' R2 U R U' R' | ||||

Double Bar | R2 U R2 U' R2 D R2 U' R2 U R2 D2 | R2 U R2 F R U R U' R' F' R U2 R' U2 R' D' | ||||||

6 | 2 | Front Opp | R2 U2 F U F' R2 F U' F' D' | R2 U' R2 D R2 U' R2 U R2 D' R2 D' | ||||

Right Opp | U' | R' F' R U R2 U' R' F R U' R2 D' | ||||||

Double Opp | U2 | R2 U' D R2 U' R2 U R2 D2 | U2 | R2 F2 U' F2 D R2 D2 | ||||

Front Bar | U' | R2 U' R2 D' | ||||||

Right Bar | R2 U' R2 U' R2 D R2 U' R2 U R2 D2 | U2 | R' U2 R D R D' R D R U2 R D2 | U' | D' R U R' U' R U2 R D R' U2 R2 D' R | |||

Double Bar | U' | x R' D' R U2 R' D R U2 F' x' | R U2 R2 U R D R' U' R D2 R U2 R' | U2 | R2 U' R2 D' R2 U R2 U R2 U2 R2 | |||

5 | 2 | Front Opp | D' R2 U2 R2 U' R2 U' R2 | U2 | D' R2 U R2 U R2 U2 R2 U2 | |||

Right Opp | U' | R2 U2 R2 U R2 U R2 D' | U | R2 U' R2 U' R2 U2 R2 D' | ||||

Double Opp | U2 | D' x R' U R U' R' U R U' R' U R U' x' | U' | R D' R' U R' D R U2 R' D' R U' R D R' D' | U' | R2 U' R2 D' R2 D R2 U R2 D' R2 | ||

Front Bar | U2 | R2 D R2 U' R2 U R2 D2 | U2 | D R D' R D R2 U' R' U R D2 | ||||

Right Bar | U' | D R2 U' R2 U R2 D' R2 D' | U' | D R' U' R U R2 D' R' D R' D2 | ||||

Double Bar | U' | R2 U' R2 U R2 D' R2 U R2 U' R2 | R U' R U' R' U R' U R' F' U R' U' R F D' | |||||

2 | 5 | Front Opp | U2 | R U R2 U R' D R' U' R D2 R U' R' | U' | R2 U R2 D' R2 D R2 U R2 D' R2 | ||

Right Opp | U' | R2 U F U F' R2 F U' F' D' | R D' R' D R' U D' R2 U' R D R' D' R2 | |||||

Double Opp | U | R2 U D' R2 U R2 U' R2 | U | R2 B2 U B2 D' R2 | ||||

Front Bar | U | R U2 R' D' R' D R' D' R' U2 R' | U' | R2 U R2 U R2 D' R2 U R2 U' R2 | ||||

Right Bar | R2 U R2 D' | |||||||

Double Bar | x R D R' U2 R D' R' U2 B' x' | R U R' U R U' R' D' R U R' U' R U' R' | ||||||

2 | 3 | 1=2, 3 Adj | U' | R2 U2 R2 D' | ||||

1=2, 3 Opp | R2 U' R2 U2 R2 U' D' R2 U R2 U' R2 | D' R U R2 D R' D' R U2 R2 U R2 U R | ||||||

2=3, 1 Adj | U' | R F' R2 F R U R' F' R2 F R U R2 D' | D R2 U' R2 U R2 D' R2 U' R2 U2 R2 D' | R' U' R F' R2 F U R' F' R' F U R2 D' | ||||

2=3, 1 Opp | R2 U R2 U R2 D R2 U' R2 U R2 | U' | R2 U R U' R U R' U R2 U D' R U' R' | R D' R U R' D R' U' R' D R' U R D' R D' | ||||

All diff, 1 and 2 opp | D' R U' R' U2 D R2 U2 R2 D' R U R' | U | F U R U' R' F' R2 F R U R' U' F' R2 D' | |||||

All diff, 2 and 3 opp | U2 | R2 U R2 U' R2 U' D' R2 U R2 U' R2 | U2 | D' R' D R2 U R U2 R' U' R2 D' R' U' R2 | ||||

3 | 2 | 1=2, 3 Adj | U2 | D' R U' R U R' D R D' R' U R2 U2 R2 | ||||

1=2, 3 Opp | U2 | D' R2 U R2 U D R2 U2 D' R2 | U2 | D' R2 U R' U D R2 U' D' R U' R2 | F' U R' U' R U R' U' R U R' U' R F D' | |||

2=3, 1 Adj | R2 U2 D' R2 U R2 U' R2 | |||||||

2=3, 1 Opp | F2 D' r2 D r U' r F2 r' U r | U | R2 U R2 U' R2 U2 D' R2 U R2 U' R2 | |||||

All diff, 1 and 2 opp | R2 U R2 U' R2 U R2 D' | U2 | R2 U' R2 U R2 U' R2 D' | |||||

All diff, 2 and 3 opp | D' R2 U R2 D R2 U' D' R2 | |||||||

1 | 3 | 1=2, 3 Adj | R2 U R2 U' R2 U2 R2 D' | R2 U' R2 U2 R2 U' R2 D' | ||||

1=2, 3 Opp | U | R2 U D R2 U' R2 U R2 D2 | ||||||

2=3, 1 Adj | U2 | R2 U' R2 D' R2 D R2 U' R2 D' R2 | U2 | R2 U' R' D' R D R2 U' R' U R' U' D' R2 | U' | R U' R F2 R' U R' U' R2 F2 U2 R2 D' | ||

2=3, 1 Opp | U | R2 U R2 D' R2 U R2 U R2 U2 R2 | ||||||

All diff, 1 and 2 opp | U | R D R' D' R2 U R D R2 U' D' R D' | U2 | R U' R F' R2 F U R' F' R' F D' | x R2 U R' U' R U R' U' R U R' U' R' F' x' | |||

All diff, 2 and 3 opp | U2 | R2 U' D' R2 U R2 U' R2 |

For the bottom cases, the Case column refers to 3 stickers on the right. For the two cases above that, the case numbers refer to 3 stickers on the front

**Step 1:**Make a 1x2x3 in DL (i.e. solve DL, DF, DB edges and DFL, DBL corners. E slice centres are not necessarily solved). If this is hard, make a 1x2x2 in DLF or DLB during inspectiom, and then extend it with a pair. I suggest to try and be at least partially color neutral here. Roux FB tutorials should certainly help.

**Step 2:**Intuitively solve 3 belt edges using <R, U, u> moveset. <F, f> moves are okay too. Remember, you do not need to insert edges with triggers since the R layer is free. Whenever possible, try and solve 2 edges at once, this should become natural with practice.

**Step 3:**Solve the last belt edge while intuitively doing EO. CFOP edge control tutorials should help. Some useful triggers to start with are F<U>F', R'FRF', S'US, R2UR2, etc. Keep this intuitive for now, you can then start to slowly optimise your solutions using the EOLE algorithms.

**Step 4a:**If only 0 or 1 of the 6 remaining corners are oriented (you’ll skip this 66% of the time), use either Sune or Antisune to ensure at least 2 of the 6 corners are oriented.

**Step 4b:**Use an R2<U>R2 trigger to put two of the oriented corners in DFR and DBR.

**Step 4c:**You now have 7 possible cases for CO, use an alg.

**Step 5a:**Use an R2<U>R2 trigger to solve the DBR corner.

**Step 5b:**You now have 8 possible cases for CP, use an alg.

**Step 6a:**Insert the D edge to the bottom layer using M'U2M.

**Step 6b:**You now have 4 possible EP cases, use an alg.

H | R' U2 R U2 R U2 R U2 R' | R' U' R2 U' R U2 R2 U' R' |

Pi | R U2 R2 U' R2 U' R2 U2 R | |

Sune | R U R2 U' R2 U R | |

Antisune | R U2 R' U' R U' R' | |

U | F U' R2 U R2 U F' | D R D' R2 U R2 D R' D' |

T | R D' R U2 R' D R' | R U R D R' U' R D' R2 |

L | R D' R U' R' D R' |

Location of DFR corner | Case | Main alg | Alt alg |

DFR | solved | D' | |

DFR | opposite on right | R U R' F' R U R' U' R' F R2 U' R' D' | |

DFR | diagonal | R U' R' U R U' D' R2 U R D R U' D' R2 | F R U' R' U' R U R' F' R U R' U' R' F R F' D' |

UBL | opposite in front | R2 U2 R2 U' R2 U' R2 D' | U2 R2 U R2 U R2 U2 R2 D' |

UBL | opposite on right | U'D R2 U2 R2 U R2 U R2 D2 | UD R2 U' R2 U' R2 U2 R2 D2 |

UBL | diagonal | U2 x' R' U R U' R' U R U' R' U R U' F' x | R' D R U' R D' R' U2 R D R' U R' D' R D' |

UBL | headlights in front | D' R2 U R2 U' R2 D R2 D' | D' R U R' U R2 D R D' R |

UBL | headlights on right | U R2 D' R2 U R2 U' R2 | UD' R' D R' D' R2 U R U' R' |

UBL | solved | R2 U R2 U' R2 D R2 U' R2 U R2 D2 | R2 U R2 F R U R U' R' F' R U2 R' U2 R' D' |

Total number of algs: 7+8+4 = 19

Expected movecounts:

Step 3: 10 moves

Step 4: 13 moves

Step 5: 14 moves

Step 6: 13 moves

Total from step 3 to end: 40 moves

Total: 60-65 moves for a beginner

Source: HARCS

Expected movecounts:

Step 3: 10 moves

Step 4: 13 moves

Step 5: 14 moves

Step 6: 13 moves

Total from step 3 to end: 40 moves

Total: 60-65 moves for a beginner

Source: HARCS

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

L2 D R' // 1a (making a 1x2x2)

R' U2 R D L' y' // 1b (extending to 1x2x3)

U2 R' // 2a (belt edge 1)

u2 U R // 2b (belt edge 2)

u' R // 2c (belt edge 3)

F' U F R U2 R' // 3 (intuitive EOLE)

R' U' R U' R' U2 R // 4a (orienting at least 2 corners)

R2 U2 R2 // 4b (putting in the 2 corners)

R U2 R2 U' R2 U' R2 U2 R // 4c (orienting L4C)

R2 U' R2 // 5a (permuting DBR corner)

U R2 D' R2 U R2 U' R2 // 5b (permuting other corners)

U2 M' U2 M // 6a (permuting D layer edge)

M2 U M2 U M' U2 M2 U2 M' // 6b (EPLL)

This solve had no skips. Without cancellations, movecount = 3+5+2+3+2+6+7+3+9+3+8+4+9 = 64 moves.

L2 D R' // 1a (making a 1x2x2)

R' U2 R D L' y' // 1b (extending to 1x2x3)

U2 R' // 2a (belt edge 1)

u2 U R // 2b (belt edge 2)

u' R // 2c (belt edge 3)

F' U F R U2 R' // 3 (intuitive EOLE)

R' U' R U' R' U2 R // 4a (orienting at least 2 corners)

R2 U2 R2 // 4b (putting in the 2 corners)

R U2 R2 U' R2 U' R2 U2 R // 4c (orienting L4C)

R2 U' R2 // 5a (permuting DBR corner)

U R2 D' R2 U R2 U' R2 // 5b (permuting other corners)

U2 M' U2 M // 6a (permuting D layer edge)

M2 U M2 U M' U2 M2 U2 M' // 6b (EPLL)

This solve had no skips. Without cancellations, movecount = 3+5+2+3+2+6+7+3+9+3+8+4+9 = 64 moves.

Ran a 10000 solve simulation of a constrained version of the method.

Constraints:

- only one FB instead of a possible 24 FB

- exactly the same 3 belt edges solved first instead of 4 possible combinations

And then there were the general constraints that there is no influencing of one step on the next (which is kind of the point though).

The best part is, this number actually holds for the method because there isn't much scope of inefficiency. FB is planned in inspection so any intermediate solver is going to be almost optimal, and the last 4 steps are algorithms. The only place where fluid intuition is needed is the first 3 belt edges (3 pieces).

In comparison, theoretical and practical CFOP numbers are nowhere close because of the move inefficiencies even top solvers have during F2L (8 pieces).

Last edited: Dec 30, 2020