Roux Last 5 Corners

trinity.db

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2006BEYE01
Greetings,

I wanted to propose a way to optimize the steps of Roux.

First Block (FB)
Second Block (SB)
CmLL
Last Six Edges (LSE)

The process is intuitive. I'll provide the algorithms and concepts.

Typically, the SB is solved with 2-Gen Slice moves (RrUM). This means that the Permutation of the last four corners is fixed from this algorithm subset. Therefore, if it does not chance, we can recognize it sooner and respond proactively in the solve.

The common theme of B1-C6 (Sune/Anti-Sune CMLLs) and F1-6 is to kick out a corner edge pair from either FR or BR and re-insert it differently.

This made me question: What if you just insert the last corner edge pair intelligently to avoid corner permutation?
Typically, we would insert the last corner edge pair with RU'R' or rUr'. So, all of the alternative insertions will be compared to this standard move count. The goal is to solve the Second Block and Last 5 Corners simultaneously (2B5C).

There are 12 cases for the FR slot. I'll call the extra moves required for 2B5C penalty moves relative to the 3 move requirement to finish the second block. Then, we can evaluate the costs and the benefits.

Extra moves required: (0,5,4,1,4,4;0,5,5,0,1,4) = 33 moves for 12 cases = 33/12 = 2.75* penalty moves on average
*(updated algorithm and move count).
So for 3 moves, you can eliminate 32 cases from CmlL and increase your odds of a CmLL skip significantly.

You will be left with 7 Cmll Cases: B1, C1, D1, E1, F1, G1, H1
  • as well as the lucky CmLL skip A1 Case.
  • We are just applying the Finger Tricks from the B, C, and F CmLLs while the Corner Edge pair is already in the U Layer.
    • FR'F'R
    • F'UF
    • R'FRF'
    • L'UR'U'L

The algorithms above are either conjugated with or substituted for RU'R', RU2R', and rUr'.

Essentially, we break the Second Block Slice 2-Gen rule to reduce the cases for CmLL.



Here is a sample of the algorithm set.


These algorithms Insert the 1x1x2 into the FR and DFR slot.

If you solve the block most efficiently with Roux, you will notice that there are six permutations after solving the 2nd block.

Rather than solving with the
standard RU'R' or rUr', use the alternative to swap the respective corner pairs on the respective planes.

To Setup the 2B5C FR Cases always do RUR' to see the case.

5C (FR Slot)
R U' R' - Pure 1 -- +0 Penalty
R'FRF'U rUr' - (2)B -- +5 Penalty
URUL'UR'U'L - (3)R -- +4 Penalty
U'R' F R F' - (4)F -- +1 Penalty
R'FRF'UFR'F'R - (5)L -- +6 Penalty --> New Algorithm R U2 r' F R' F' R +4 Penalty
UR'FRF'RU2R' - OPP 6 -- +4 Penalty

To Setup the 2B5C' FR Cases always do rU'r' to see the case.
5C' (FR Slot)
rUr' - Pure 1 -- +0 Penalty
rU'r'U2 R'FRF' - (2)B -- +5 Penalty
U2 F'UFU FR'F'R - (3)R -- +5 Penalty
U'F'UF - (4)F -- +0 Penalty
FR'F'R - (5)L -- +1 Penalty
U2 F'UFU rUr' - OPP 6 -- +4 Penalty
----

To Setup the 2B5C BR Cases always do R'U'R to see the case.
5C (BR Slot)
R' U R - Pure 1 -- +0 Penalty
U'R B' R' B - (2)B -- +1 Penalty
U'R'U'LU'RUL' - (3)R -- +4 Penalty
RB'R'BU' r'U'r - (4)F -- +5 Penalty
RB'R'BU'B'RBR' - (5)L -- +6 Penalty New Algorithm R' U2 r B' R B R' +4 Penalty
U'RB'R'BR'U2R - OPP 6 -- +4 Penalty
To Setup the 2B5C' BR Cases always do r'Ur to see the case.
5C' (BR Slot)
r'U'r - Pure 1 -- +0 Penalty
UBU'B' - (2)B -- +0 Penalty
U2 BU'B'U' B'RBR' - (3)R -- +5 Penalty
r'UrU2 RB'R'B - (4)F -- +5 Penalty
B'RBR' - (5)L -- +1 Penalty
U2 BU'B'U' r'U'r - OPP 6 -- +4 Penalty

The desired output is the subset of CmLL Cases A1, B1, C1, D1, E1, F1, G1, H1.
The original inserts (RU'R', R'UR, rUr', r'U'r) solve the pure cases. You would have gotten the desired cases above naturally.
If any of the other permutations are present, you need to adapt this insertion with the respective alternative insertion.

The possible scenarios of the 2B5C comes from the product of three choices/scenarios.

1x1x2 Block Slot (FR/BR) [2]
1x1x2 Block Orientation (Clockwise/Counter-Clockwise) [2]
CmLL Corner Permutation (1-6) [6]

[2]*[2]*[6] = 24 algorithms.
 
Last edited:
Examples from Speedsolving.com 3x3x3 Competition as of 3-31-2023

1680319160053.png



1.
  • z D' R F R' r' F2 x' R2 U R' D' (FB) [10]
  • r M' U' R r U r' U' r' U' R U2 R' (Second 1x2x2) [13]
  • U R' U R U' M' (1x1x2 BR slot paired) [6]
    • r'U'r would finish the SB
    • Notice the C6 permutation
    • Insert with U2 BU'B'U' r'U'r - OPP 6 -- +4 Penalty [8]
      • r'U'r {3}; AUF + R U R' U R' F R F' R U2 R' {12}
  • L U L' U L U2 L' (CmLL C1) [7]
  • U' M U M (4a) [4]
  • U M2 U' M' U2 M' (4b) [6]
  • U' M' U2 M U2 r2 [6]
Move count: [10] + [13] + [6] + [8] + [7] + [4] + [6] + [6] = 60 moves.
 

Attachments

  • 1680319121415.png
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Examples from Speedsolving.com 3x3x3 Competition as of 3-31-2023

View attachment 23076
3.

  • x y' F r U2 R2 U2 B F2 r' F (FB) [9]
  • R' U r2 R' U' B' R' B R' U' R' U M (SB) [13]
    • Notice rUr' yields a B case
  • rU'r'U2 R'FRF' - (2)B -- +5 Penalty [8]
    • CmLL Skip!
  • M' U M U M (4a) [5]
  • U M' U2 M (4b) [4]
  • U' M U2 M U2 R2 [6]
Move Count = 45
 
Examples from Speedsolving.com 3x3x3 Competition as of 3-31-2023

View attachment 23076
4.
  • z U F2 R U r B (FB) [6]
  • r' R' U R U2 R' U' r U r' (SB) [10]
    • U r' U' r would solve resulting in B6
  • U' B U' B' U' r' U' r (2B5C) [8]
  • U' R' U' R U' R' U2 R2 U R' U R U2 R' (CmLL) [14]
  • M2 U M U' (4a) [4]
  • M U2 M U (4b) [4]
  • r2 U2 M' U2 M' [5]
Move Count = 51
 
Examples from Speedsolving.com 3x3x3 Competition as of 3-31-2023

View attachment 23076
5.

  • x' B' D' d' B U R' U2 F2 (FB) [8]
  • U R' U r U R' U' r' U2 M2 (SB) [10]
    • r U' r would solve and yield a Bruno Case.
  • r'UrU2 RB'R'B - (4)F (2B5C) -- +5 Penalty [8]
  • U R U R' U R U2 R' (CmLL) [8]
  • M U M U M' (4a) [5]
  • U' M2 (4b) [2]
  • U r2 M' U2 M U2 M2 [7]
Move Count = 48 Moves
 
Hmm, I am very interested but not able to try yet,
Is it fair to oversimplify this as a 1.5 look CMLL with greater chance of skipping?
 
That is a fair over simplification.

I'm looking at how to influence LSE with A1-H1 next. Ideally this will connect with EOLR.

It's also a continuation of my Second Block Pseudo Block techniques and 4x4x4 parity skip.
 
this is really cool, but i have 3 critiques things.

Firstly, how easy is it to recognize in a speedsolve.

Secondly, the CMLLs are not called A1 or B3 or whatever. you call them by there orientations case (like O, T Pi) then their CP like diag or checkerboard or rowa or columns or something like that.

Finally, do not call it 2b5c (I'll get back to you on its official name, but this is a subset of a larger algset which solves CMLL everytime from last slot(this algset is purely theoretical but has a name))

Oh, and i guess im hyped for EOLR influencing from CP solved CMLLs
 
this is really cool, but i have 3 critiques things.

Firstly, how easy is it to recognize in a speedsolve.

Secondly, the CMLLs are not called A1 or B3 or whatever. you call them by there orientations case (like O, T Pi) then their CP like diag or checkerboard or rowa or columns or something like that.

Finally, do not call it 2b5c (I'll get back to you on its official name, but this is a subset of a larger algset which solves CMLL everytime from last slot(this algset is purely theoretical but has a name))

Oh, and i guess im hyped for EOLR influencing from CP solved CMLLs

nmCmLL will do for recognition.
I'll update the tables for recognition. I recognize the CmLL prior to solving the 5th block. It is easy to recognize. The Second Block inserts are easy.

I am not solving for speed, I am solving for fewest moves at this point and being intentional to use the method. I've gotten several 4a skips through CmLL and a few CmLL skips.

I am just using Roux's old CmLL Step 3 diagrams for cases. I also recognize the cases with nmCmLL and have been doing so for about 15 years. Maybe I am old school? 😂

Pseudo Blocks on 4x4x4 lead me to skip OLL Parity on 4x4x4.
Which lead me to Pseudo Blocks on 3x3x3.
Which lead me to this Second Block + CmLL Hybrid.


By permuting corners with the 2nd Block,
I am evaluating whether I should work on

  • An alg set to solve CmLL + 4a
  • EOLR
 
I have an updated algorithm list.

Screenshot_20230821-111443-382.png


This ties into the case recognition. I used to recognize based on the Permutation that you would get with the default insert (RU'R').


You are able to validate the permutation of the 3 corners in the U Layer relative to the 1x1x2 block. Match the L/R corners.

When you find the pair in the U layer, simply use the respective algorithm. The new move count is 5.33 moves on average.

this is really cool, but i have 3 critiques things.

Firstly, how easy is it to recognize in a speedsolve.

Secondly, the CMLLs are not called A1 or B3 or whatever. you call them by there orientations case (like O, T Pi) then their CP like diag or checkerboard or rowa or columns or something like that.

Finally, do not call it 2b5c (I'll get back to you on its official name, but this is a subset of a larger algset which solves CMLL everytime from last slot(this algset is purely theoretical but has a name))

Oh, and i guess im hyped for EOLR influencing from CP solved CMLLs
 
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