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CFOP-Breaker? Mehta method

Devagio

Member
Joined
Apr 21, 2020
Messages
216
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
GGGGR U' R'
GGBBU F R F' U2 R'U R' F R2 F' U2 R'
GBGBU2 F R F' U R'U2 R' F R2 F' U R'
GBBGU' R U2 R2 F R F'U F R' F' R2 U2 R'
BGGBU R F' U2 F R'
BGBGR2 U' F R F' R2
BBGGU2 F R' F' R2 U R'
BBBBU S' U S R U2 R'U R' F R2 F2 U2 F R'
avg5.25
GB
GGGBU2 R F' U F R'
GGBGU' R U R2 F R F'
GBGGF' U2 F R U' R'
BGGGU' R' F R F'
GBBBU2 F R F2 U' F U2 R'
BGBBU F R F2 U2 F R'
BBGBU2 R' F R2 F2 U F R'U2 R F' U F2 R2 F' R
BBBGU S' U' S U2 R' F R F'
avg6.125
BG
GGGBF R F' R'
GGBGF R' F' R
GBGGR U R' F' U' FF U R' U' F' U R
BGGGU' F R' F' R2 U' R'
GBBBU2 R F' U2 F2 R' F'
BGBBU' S' U S R U R'U R U2 F' U F2 R' F'
BBGBU' R' F R2 F2 U' F R'U2 S' U' S U R U' R'
BBBGU S' U S U' F R' F' R
avg5.875
BB
GGGGU' F' U FE R' U' R
GGBBF' U F2 R F' R'U R F R' F2 U' F
GBGBU2 R F R' F2 U2 FF R F2 U2 F U2 R'
GBBGU R' F R F2 U' FF R' F2 U2 F U2 R
BGGBu' F R F'
BGBGF R' F2 U' F U Ru' F U R2 U' R F'
BBGGF' U2 F2 R' F' RU2 R' F R F2 U2 F
BBBBU S' U S u' R' U' R
avg5.625
solved
0Bnil
1BR' F R2 F' R'
2B (adj)R F R2 F' RF U R U' R' F'
2B (opp)F R U R' U' F'
3BS' U SF R F' U2 F R' F'
4BR2 S' U S U2 R2
avg4.17
flipped
0BF R F2 U F U2 R'R U R' F R F' R'
1BF' 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'
3BR' F R F' U' R' F R F'
4BS' U S R U2 R' F' U' F
avg7.17
G
0BR' U' R U R
1BR F R' F'R U F R' F'
2B (adj)R2 F R F' R2R U' F R' F'
2B (opp)R F U R' U' F'
3BF R' F2 U2 F R U2 R
4BF R' F' R2 U2 F R' F'
avg6.00
B
0BF u' R u F2
1BF 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'
3BR F' U' F2 R2 F' R
4Bu' F R' F' U2 R F R2 F' R
avg6.17

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
HR' U2 R U2 R U2 R U2 R'R' U' R2 U' R U2 R2 U' R'
PiR U2 R2 U' R2 U' R2 U2 R
SuneR U R2 U' R2 U R
AntisuneR U2 R' U' R U' R'
UF U' R2 U R2 U F'D R D' R2 U R2 D R' D'
TR D' R U2 R' D R'R U R D R' U' R D' R2
LR D' R U' R' D R'
Onil
avg6.63
RR
HR U' R' U2 R' U' R U2 R U' R
PiR U' R' U' R U' R U2 R' U R'
SuneR U' R' D' U' R U2 R' D
AntisuneR' U R D U R' U2 R D'
UR U' R2 U' R U2 R2 U' R
TR U2 R' U' R2 U2 R' U R'R U R U2 R2 U' R U2 R'
LR U R U2 R' U R'
OD' R U' R' D R' U R2 U' RR U R U' R' U R U' R' U R'
avg9.13
FB
HR 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
PiR U R2 U2 R2 U R' U2 R' U' R
SuneD' R U' R' D U' R U2 R'R2 U R' U2 R U' R2 U R U2 R'
AntisuneD R' U R D' U R' U2 RR2 U' R U2 R' U R2 U' R' U2 R
UR' U2 R2 U' R2 U' R2 U2 R'
TR U2 R U2 R U2 R U2 R
LR' U R U R2 U' R2 U R' U2 R
OR2 U' R D' R U2 R' D RR U2 R U' R' U2 R U' R' U2 R'
avg9.75
RD
handL' U R2 U' R2 LR' D' R U' R' D R
asiaR U' R' U' R U2 R'
SD R' U2 R U' R' U' R D'
pR' U R U2 R2 U' R U2 R
dogR U' R' U2 R U' R'
hammerB' R2 U' R2 U BR D U' R' U2 R D' U' R'
clockR' U2 R' U R2 U' R' U2 R
eagleR' U2 R U R2 U' R U2 R
avg7.75
FD
legR D' R U R' D R'
americaD R' U R U R' U2 R D'R2 U' R U2 R' U' R U' R
NR U2 R' U R U R'
bR S' U2 S RR U R' U2 R U R'
catD R' U R U2 R' U R D'R' U' R2 U2 R' U R2 U2 R2 U' R'
sickleB' U' R2 U R2 BR2 U R' U R U2 R' U R'
fanR U' R' U' R U' R' U R U2 R'
pigeonR U2 R' U R U' R' U' R U' R'
avg8.13
DR
legL U' R2 U R2 L'R D R' U' R D' R'
americaR' U R U R' U2 R
ND' R U2 R' U R U R' D
bR U' R' U2 R2 U R' U2 R'D' R U R' U2 R U R' D
catR' U R U2 R' U R
sickleF R2 U R2 U' F'
fanR U2 R U' R2 U R U2 R'
pigeonR U2 R' U' R2 U R' U2 R'
avg7.75
DB
handR' D R' U' R D' R
asiaR2 U2 R' U R' U R U2 R'
SR' U2 R U' R' U' R
pR' S' U2 S R'R' U' R U2 R' U' R
dogD' R U' R' U2 R U' R' DR' U R U R' U' R U R' U2 R
hammerF U R2 U' R2 F'
clockR' U R U R' U R U' R' U2 R
eagleR' U2 R U R' U' R U' R' U' R
avg8.13
FR
handR U' R U' R' U2 R'
asiaR U2 R U' R2 U2 R U R'
SR U' R' U2 R' U R U' R2 U' R'
pR U' R2 U R' U R' U2 RR U2 R' D' U R U R' D
dogR U R U R' U R U' R' U2 R'
hammerR U2 R' U R' U R2 U' RR U2 R' U R2 U' R' U2 R'
clockR' U R U' R U' R U R'R U R U R2 U R U R'
eagleR U2 R' U R2 U2 R2 U2 R U R'
avg9.5
RB
legR' U R' U R U2 R
americaR2 U' R' U R U2 R' U R'
NR' U R U2 R U' R' U R2 U RR' U' R U' R' U' R2 U' R2 U R'
bD' R U' R' D U2 R U' R'D R' U2 R D' U' R' U' R
catR' U' R U R2 U' R2 U R' U' R
sickleR U2 R U R2 U' R U2 R'R' U2 R U' R2 U R U2 R
fanR U' R' U R' U R' U' R
pigeonR' U2 R U' R2 U2 R2 U2 R' U' R
avg9.5
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.

These algorithms were generated and compiled by @trangium
DF Loc.DB Loc.CaseAUFAlgAUFAlt AlgAUFAlt Alg
56SolvedD'
AdjacentR U R' F' R U R' U' R' F R2 U' R' D'
DiagonalR U' R' U R U' D' R2 U R D R U' D' R2F R U' R' U' R U R' F' R U R' U' R' F R F' D'
65SolvedF 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'
AdjacentR U R' D' R U' D2 R' U' R D2 R'R2 U R2 U2 R2 D' R2 U R2 U' R2
DiagonalR U' R' U R' U2 R2 U R U R' U' R2 D'
26Front OppR2 U2 R2 U' R2 U' R2 D'U2R2 U R2 U R2 U2 R2 D'
Right OppU'D R2 U2 R2 U R2 U R2 D2UD R2 U' R2 U' R2 U2 R2 D2
Double OppU2x 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 BarD' R2 U R2 U' R2 D R2 D'D' R U R' U R2 D R D' R
Right BarUR2 D' R2 U R2 U' R2UD' R' D R' D' R2 U R U' R'
Double BarR2 U R2 U' R2 D R2 U' R2 U R2 D2R2 U R2 F R U R U' R' F' R U2 R' U2 R' D'
62Front OppR2 U2 F U F' R2 F U' F' D'R2 U' R2 D R2 U' R2 U R2 D' R2 D'
Right OppU'R' F' R U R2 U' R' F R U' R2 D'
Double OppU2R2 U' D R2 U' R2 U R2 D2U2R2 F2 U' F2 D R2 D2
Front BarU'R2 U' R2 D'
Right BarR2 U' R2 U' R2 D R2 U' R2 U R2 D2U2R' U2 R D R D' R D R U2 R D2U'D' R U R' U' R U2 R D R' U2 R2 D' R
Double BarU'x R' D' R U2 R' D R U2 F' x'R U2 R2 U R D R' U' R D2 R U2 R'U2R2 U' R2 D' R2 U R2 U R2 U2 R2
52Front OppD' R2 U2 R2 U' R2 U' R2U2D' R2 U R2 U R2 U2 R2 U2
Right OppU'R2 U2 R2 U R2 U R2 D'UR2 U' R2 U' R2 U2 R2 D'
Double OppU2D' 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 BarU2R2 D R2 U' R2 U R2 D2U2D R D' R D R2 U' R' U R D2
Right BarU'D R2 U' R2 U R2 D' R2 D'U'D R' U' R U R2 D' R' D R' D2
Double BarU'R2 U' R2 U R2 D' R2 U R2 U' R2R U' R U' R' U R' U R' F' U R' U' R F D'
25Front OppU2R 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 OppU'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 OppUR2 U D' R2 U R2 U' R2UR2 B2 U B2 D' R2
Front BarUR U2 R' D' R' D R' D' R' U2 R'U'R2 U R2 U R2 D' R2 U R2 U' R2
Right BarR2 U R2 D'
Double Barx 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'
231=2, 3 AdjU'R2 U2 R2 D'
1=2, 3 OppR2 U' R2 U2 R2 U' D' R2 U R2 U' R2D' R U R2 D R' D' R U2 R2 U R2 U R
2=3, 1 AdjU'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 OppR2 U R2 U R2 D R2 U' R2 U R2U'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 oppD' R U' R' U2 D R2 U2 R2 D' R U R'UF U R U' R' F' R2 F R U R' U' F' R2 D'
All diff, 2 and 3 oppU2R2 U R2 U' R2 U' D' R2 U R2 U' R2U2D' R' D R2 U R U2 R' U' R2 D' R' U' R2
321=2, 3 AdjU2D' R U' R U R' D R D' R' U R2 U2 R2
1=2, 3 OppU2D' R2 U R2 U D R2 U2 D' R2U2D' R2 U R' U D R2 U' D' R U' R2F' U R' U' R U R' U' R U R' U' R F D'
2=3, 1 AdjR2 U2 D' R2 U R2 U' R2
2=3, 1 OppF2 D' r2 D r U' r F2 r' U rUR2 U R2 U' R2 U2 D' R2 U R2 U' R2
All diff, 1 and 2 oppR2 U R2 U' R2 U R2 D'U2R2 U' R2 U R2 U' R2 D'
All diff, 2 and 3 oppD' R2 U R2 D R2 U' D' R2
131=2, 3 AdjR2 U R2 U' R2 U2 R2 D'R2 U' R2 U2 R2 U' R2 D'
1=2, 3 OppUR2 U D R2 U' R2 U R2 D2
2=3, 1 AdjU2R2 U' R2 D' R2 D R2 U' R2 D' R2U2R2 U' R' D' R D R2 U' R' U R' U' D' R2U'R U' R F2 R' U R' U' R2 F2 U2 R2 D'
2=3, 1 OppUR2 U R2 D' R2 U R2 U R2 U2 R2
All diff, 1 and 2 oppUR D R' D' R2 U R D R2 U' D' R D'U2R 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 oppU2R2 U' D' R2 U R2 U' R2
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

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.


HR' U2 R U2 R U2 R U2 R'R' U' R2 U' R U2 R2 U' R'
PiR U2 R2 U' R2 U' R2 U2 R
SuneR U R2 U' R2 U R
AntisuneR U2 R' U' R U' R'
UF U' R2 U R2 U F'D R D' R2 U R2 D R' D'
TR D' R U2 R' D R'R U R D R' U' R D' R2
LR D' R U' R' D R'
Or use regular OCLL algs.

Location of DFR cornerCaseMain algAlt alg
DFRsolvedD'
DFRopposite on rightR U R' F' R U R' U' R' F R2 U' R' D'
DFRdiagonalR U' R' U R U' D' R2 U R D R U' D' R2F R U' R' U' R U R' F' R U R' U' R' F R F' D'
UBLopposite in frontR2 U2 R2 U' R2 U' R2 D'U2 R2 U R2 U R2 U2 R2 D'
UBLopposite on rightU'D R2 U2 R2 U R2 U R2 D2UD R2 U' R2 U' R2 U2 R2 D2
UBLdiagonalU2 x' R' U R U' R' U R U' R' U R U' F' xR' D R U' R D' R' U2 R D R' U R' D' R D'
UBLheadlights in frontD' R2 U R2 U' R2 D R2 D'D' R U R' U R2 D R D' R
UBLheadlights on rightU R2 D' R2 U R2 U' R2UD' R' D R' D' R2 U R U' R'
UBLsolvedR2 U R2 U' R2 D R2 U' R2 U R2 D2R2 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

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.

1597778730185.png



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).
 
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TheSlykrCubr

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Belt can be called ribbon. Ribbon sounds a bit like rivers. Rivers of babylon is a song. Ribbon of Babylon (ROB for short) is a funny name. Please call it the ROB method and I would be happy.

I think it would be easier if you were to create a set of short algorithms that orient and permute 2-3 each time, and repeat. It would be easier to learn, generate, and (if very short,) could be faster then 2 separate algorithm sets. As far as i know, there isn't any other (good) method that does this.
 
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Devagio

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i strongly doubt belt will have higher tps than F2L with that many wide U moves. I think its better to just a 223 then solve the belt edges
So the main aim is to set up the cube so that a 223 and the remaining 2 belt edges are solved; because 6CO, 6CP and L5E is the entire point.
I completely agree that if it is possible for someone to plan a 223 in inspection (fast people should be able to do that) then this is the way to go.
However, I do not think TPS would drop due to below F2L TPS due to u moves, because there isn’t any need to purposely slow down for lookahead.
Even if it does slow down to F2L TPS or lower, I guess 8-9 moves of slower turning would still be as fast as 2 F2L pairs. If FB is as fast as cross, 3rd pair and LSLL will certainly be slower than 6CO, 6CP, L5E; so things should balance out for the methods to at least be comparable.

But yes; the goal for an aspiring world record holder should be to inspect a 223 and track the remaining edges in inspection. FB+belt is a way of achieving the same thing with more tricks possible; it doesn’t define the method and the choice is upto the user.
Belt can be called ribbon. Ribbon sounds a bit like rivers. Rivers of babylon is a song. Ribbon of babylon is a funny name. Please call it this and i would be happy
Lol no. Firstly, it doesn’t matter how the FB+belt is done, the point of the method is 6CO+6CP+L5E. At the end of the day, blockbuilding FB+belt will the the supreme thing to do.
Plus, I guess I’m gonna keep the name simple this time to avoid various confusions.

I think it would be easier if you were to create a set of short algorithms that orient and permute 2-3 each time, and repeat. It would be easier to learn, generate, and (if very short,) could be faster then 2 separate algorithm sets. As far as i know, there isn't any other (good) method that does this.
Totally disagree. Apart from horrible recognition, there will be too much decision making. Plus, there’s a little theory I built up to “derive” 6CO+6CP+L5E. I’ll post it on this thread in a while.
 

TheSlykrCubr

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Ok, I have an idea which will probably get shot down straight away, but when making the 2x2x3 block, leave out the edge parallel to the one that gets solved in L5E so that you have a lower-case n shape. Then you can do L6E, which has waaaaaay more information out there on how to get good. Also gives it a satanic vibe, which is good since it's the cfop-breaker. But it wouldn't work with your way of deriving, so it would be more work for you to get it right.

Also, wouldn't allowing wide moves in the algs make them faster to execute/ shorter in general?

If you don't like either of these ideas, i'll shut up for a while
 
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Devagio

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Ok, I have an idea which will probably get shot down straight away, but when making the 2x2x3 block, leave out the edge parallel to the one that gets solved in L5E so that you have a lower-case n shape. Then you can do L6E, which has waaaaaay more information out there on how to get good. Also gives it a satanic vibe, which is good since it's the cfop-breaker. But it wouldn't work with your way of deriving, so it would be more work for you to get it right.

Also, wouldn't allowing wide moves in the algs make them faster to execute/ shorter in general?

If you don't like either of these ideas, i'll shut up for a while
No no, keep them coming; you never know what might be ground breaking.

I did think of leaving out that edge to get L6E. The issue with that is, you never know what edge will end up in that spot. If it is a useful edge, it’ll be a big hassle removing it.
Secondly, this doesn’t give any major boost in speed, while L5E is much faster than L6E since it is one-lookable.

Wide turns do make things shorter and rotationless, but they make the cube grip awkward which slows down turning. I do not think it is a major issue here; but sometimes it may be.
 
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No no, keep them coming; you never know what might be ground breaking.

I did think of leaving out that edge to get L6E. The issue with that is, you never know what edge will end up in that spot. If it is a useful edge, it’ll be a big hassle removing it.
Secondly, this doesn’t give any major boost in speed, while L5E is much faster than L6E since it is one-lookable.
People basically do L6E so pauselessly and so fast that they basically one looked it, the way you see LSE is not several sub-steps, but rather one long step.
 

TheSlykrCubr

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No no, keep them coming; you never know what might be ground breaking.

I did think of leaving out that edge to get L6E. The issue with that is, you never know what edge will end up in that spot. If it is a useful edge, it’ll be a big hassle removing it.
Secondly, this doesn’t give any major boost in speed, while L5E is much faster than L6E since it is one-lookable.
From Speedsolving.com's Wiki page for L5E

"The most advanced system would be to simply solve in one step. Nobody has generated the algorithms for this yet, but it may not be a good idea because of the large number of cases and bad recognition. For the worst ones the algorithm will probably just be the same as in the 2-step solution, so the benefit will be small at best."
 

Devagio

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From Speedsolving.com's Wiki page for L5E

"The most advanced system would be to simply solve in one step. Nobody has generated the algorithms for this yet, but it may not be a good idea because of the large number of cases and bad recognition. For the worst ones the algorithm will probably just be the same as in the 2-step solution, so the benefit will be small at best."
The L5E page was made in 2008, it may have relics of cubing advice that no longer holds. The world record time is literally one third of the world record time back then.
A major thing in cubing is, even if two solutions are same, a one-looked solution is much much better than a two-look solution because of the massive TPS people have today.
Also, EO can be influenced during belt, which will give only the great cases (i.e. ones with 0 or 4 bad edges).
 

mukerflap

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No no, keep them coming; you never know what might be ground breaking.

I did think of leaving out that edge to get L6E. The issue with that is, you never know what edge will end up in that spot. If it is a useful edge, it’ll be a big hassle removing it.
Secondly, this doesn’t give any major boost in speed, while L5E is much faster than L6E since it is one-lookable.

Wide turns do make things shorter and rotationless, but they make the cube grip awkward which slows down turning. I do not think it is a major issue here; but sometimes it may be.
yeah lse is pretty much 0 look since you predict how to do it in CMLL

So the main aim is to set up the cube so that a 223 and the remaining 2 belt edges are solved; because 6CO, 6CP and L5E is the entire point.
I completely agree that if it is possible for someone to plan a 223 in inspection (fast people should be able to do that) then this is the way to go.
However, I do not think TPS would drop due to below F2L TPS due to u moves, because there isn’t any need to purposely slow down for lookahead.
Even if it does slow down to F2L TPS or lower, I guess 8-9 moves of slower turning would still be as fast as 2 F2L pairs. If FB is as fast as cross, 3rd pair and LSLL will certainly be slower than 6CO, 6CP, L5E; so things should balance out for the methods to at least be comparable.

But yes; the goal for an aspiring world record holder should be to inspect a 223 and track the remaining edges in inspection. FB+belt is a way of achieving the same thing with more tricks possible; it doesn’t define the method and the choice is upto the user.

Lol no. Firstly, it doesn’t matter how the FB+belt is done, the point of the method is 6CO+6CP+L5E. At the end of the day, blockbuilding FB+belt will the the supreme thing to do.
Plus, I guess I’m gonna keep the name simple this time to avoid various confusions.


Totally disagree. Apart from horrible recognition, there will be too much decision making. Plus, there’s a little theory I built up to “derive” 6CO+6CP+L5E. I’ll post it on this thread in a while.
cross is faster than FB because there arent B moves
 

Devagio

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yeah lse is pretty much 0 look since you predict how to do it in CMLL
Just, no
So if you do the eo before l5e to leave you with 4 or 0 bad edges, why not just use EOLL and EPLL or even just ELL if you're good at recognition?
Influence EO, not do it. Big difference. Influencing EO barely takes any time if at all.
 

Devagio

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you can predict lse in eo mukerflap is correct on that
Yep, you can predict L6EP in EO; not LSE in CMLL as claimed.

And sure it may be pauseless, but I'm willing to bet that L5E can be executed faster since you're braindead throughout the thing, instead of predicting the second half during the first half.

In any case, influenced L5E, which is practically what will show up most of the time is much faster than the average L6E (influencing EO during CMLL is much harder than doing so during belt).
 

TheSlykrCubr

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In any case, influenced L5E, which is practically what will show up most of the time is much faster than the average L6E (influencing EO during CMLL is much harder than doing so during belt).
What if you used L5EOP? You wouldn't have to do the EO influencing, but you'd have to do EPLL instead, which has very short algorithms, and a 1/12 chance of a skip
 

Devagio

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What if you used L5EOP? You wouldn't have to do the EO influencing, but you'd have to do EPLL instead, which has very short algorithms, and a 1/12 chance of a skip
It's an extra alg at the end of the day. Will add to movecount, will add a pause for EP recog, etc. Sure, it will be a decent way to transition to full l5e, but that's about it. If it is possible to learn the entire set, why not do it; it's not like there are 1000 algs in L5E, its less than 70 excluding mirrors and inverses, and even less if we do setups. And all of them 2-gen, so easy to learn as well.
 
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