I created this game and the goal is to find the worst possible solution :
Rules:Solve the previous scramble using a silly method (e.g. only jperms t perms or doing Cfop algs - 1 move)
Every move must have a purpose (Not Solving the cube -1 move and then doing random moves then Solving it)
Awards (For Fun!):If you get a legitimate solve with over 100 moves Bronze over 150 Silver Over 200 Gold rank of Worst solve (you can put it in your Bio if you want)
You don't have to do a reconstruction for each solve just tell us what your method was.
Hope you have fun this is a great way to kill time for those days where you don't want to actually do work to get better you can get worse!!
First scramble D2 F U' R B' U' D F' R' F2 L2 F R2 F' R2 B' U2 F2 R2 F R2 Good luck
Edit: I know how long it may take to do this so if someone gets to it before you feel free to post your solution as well just first person gets to pick the next scramble.
Green front, white top
L F' R' // Daisy
D' F2 D2 R2 D' B2 L2 //Cross
[rotate to green front, yellow top]
U' L' U L // corner 1
[rotate to orange] R' U2 R // corner 2
[rotate to green] R' U R // corner 3
U R U' R' // corner 4
[rotate to orange] U2 L' U L F' L F L' // edge 1
R U' R' F R' F' R // edge 2
[rotate to red] U' R' F R F' R U R' // edge 3
U L' U L [rotate to blue] U2 R' F R F' U2 R' F R F' // edge 4
R U R' U' R' F R F' f R U R' U' f' // Orient yellow edges
U2 F' R F R2 D' R U' R' D R2 F' R' F U F' R F R2 D' R U' R' D R2 F' R' F // Orient yellow corners
U M D' M' U2 M D M' U' M D' M' U' M D M' // PLL
Next: U2 L2 R2 B2 D2 R2 U2 B L2 F R2 U B2 R F2 L U R B' D
Green front, white top
L F' R' // Daisy
D' F2 D2 R2 D' B2 L2 //Cross
[rotate to green front, yellow top]
U' L' U L // corner 1
[rotate to orange] R' U2 R // corner 2
[rotate to green] R' U R // corner 3
U R U' R' // corner 4
[rotate to orange] U2 L' U L F' L F L' // edge 1
R U' R' F R' F' R // edge 2
[rotate to red] U' R' F R F' R U R' // edge 3
U L' U L [rotate to blue] U2 R' F R F' U2 R' F R F' // edge 4
R U R' U' R' F R F' f R U R' U' f' // Orient yellow edges
U2 F' R F R2 D' R U' R' D R2 F' R' F U F' R F R2 D' R U' R' D R2 F' R' F // Orient yellow corners
U M D' M' U2 M D M' U' M D' M' U' M D M' // PLL
Next: U2 L2 R2 B2 D2 R2 U2 B L2 F R2 U B2 R F2 L U R B' D
y2 R U R' F U2 R' U' R' U' R' F R2 F' R // yellow side
y' L' U' L F L' U' L U L F' L2 U L U F2 L' S' L F2 L' S L z2 R U R' U' R' F R2 U' R' U' R U R' F' // red side
z' y' U2 R' B' M2 U M2 U2 M2 U M2 B R U2 x L' U' L F L' U' L U L F' L2 U L U x2 y' R' D' R D R' D' R U' R' D R D' R' D R U // green side
x U2 R' B' M2 U M2 U2 M2 U M2 B R U2 // orange side
z U' L R2 U' R U' R U R U' R' U R U R2 U' R' U2 R L' U // blue side and white side skip
Lucky the white side was solved because I usually get stuck after solving 5 sides.
Next: F R' U' B2 L2 R2 U2 R2 U B2 D L2 F2 U2 F D' R' D' B2 D' R'
White top Green front
R2 U y' L' U' L y U' F' L U L U' L' y R U R' U' R U R' U' L' U L y U M' U2 M y2 R' F R M' U' R' F' R U R U' R' F M R U' R' y' U R' U2 R U2' R' U R y U' y U2 R' F' R U R U' R' F y' R2 U2 R2 U2 R2 M' U2 M U' M' U M // 4 xcross
U' F R U R' U' F' F R U R' U' F' y' R' U2 R U R' U R' D' R U' R' D R U R // oll
z2 R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U'
// Pll
Total move count: 252 moves GOLD STATUS (Tell me if you think the H Perm should count )
Next: F' U2 L2 B2 D2 F R2 U2 F' U2 F2 L2 R' D' L F2 U' L B2 U' B
White top Green front
R2 U y' L' U' L y U' F' L U L U' L' y R U R' U' R U R' U' L' U L y U M' U2 M y2 R' F R M' U' R' F' R U R U' R' F M R U' R' y' U R' U2 R U2' R' U R y U' y U2 R' F' R U R U' R' F y' R2 U2 R2 U2 R2 M' U2 M U' M' U M // 4 xcross
U' F R U R' U' F' F R U R' U' F' y' R' U2 R U R' U R' D' R U' R' D R U R // oll
z2 R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U'
// Pll
Total move count: 252 moves GOLD STATUS (Tell me if you think the H Perm should count )
Next: F' U2 L2 B2 D2 F R2 U2 F' U2 F2 L2 R' D' L F2 U' L B2 U' B
Scramble:
F' U2 L2 B2 D2 F R2 U2 F' U2 F2 L2 R' D' L F2 U' L B2 U' B
I chose a BLD method: old pochmann corners, old-old pochmann edges (only T-perms).
The move count is already 374h, 404q, 372s, 372e, and this is a completely valid and sensible method!
And there was not even a parity there!
/*corners*/
F R' (R U' R' U' R U R' F' R U R' U' R' F R) R F'
R' (R U' R' U' R U R' F' R U R' U' R' F R) R
D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D
R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
D2 R (R U' R' U' R U R' F' R U R' U' R' F R) R' D2
R2 (R U' R' U' R U R' F' R U R' U' R' F R) R2
(R U' R' U' R U R' F' R U R' U' R' F R)
R' D' (R U' R' U' R U R' F' R U R' U' R' F R) D R
/*edges*/
R F R' L' (R U R' U' R' F R2 U' R' U' R U R' F') L R F' R'
d2 L (R U R' U' R' F R2 U' R' U' R U R' F') L' d2
D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2
d' L (R U R' U' R' F R2 U' R' U' R U R' F') L' d
R F' L' R' (R U R' U' R' F R2 U' R' U' R U R' F') R L F R'
L (R U R' U' R' F R2 U' R' U' R U R' F') L'
D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'
l d' l (R U R' U' R' F R2 U' R' U' R U R' F') l' d l'
R S R' (R U R' U' R' F R2 U' R' U' R U R' F') R S' R'
d L (R U R' U' R' F R2 U' R' U' R U R' F') L' d'
l' D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D' l
d2 L' (R U R' U' R' F R2 U' R' U' R U R' F') L d2
However, I was trained on a cube where the R side was stuck and you couldn't turn it. Luckily, you can do R moves through the other sides! I wanted to be as efficient as possible, so I'm not gonna use R' = R R R. Instead, I learned the R, R' and R2 moves to do so in this way:
R = F B U2 D2 B' F' L F B U2 D2 B' F'
R' = F B U2 D2 B' F' L' F B U2 D2 B' F'
R2 = F B U2 D2 B' F' L2 F B U2 D2 B' F'
Applying this to my BLD technic, here is a completely valid, although very old method to solve the cube in a handicapped situation. But hey, every turn is perfectly valid and justified!
F F B U2 D2 B' F' L' F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' F' F B U2 D2 B' F' L' F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' D' F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L' F B U2 D2 B' F' D F B U2 D2 B' F' L2 F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F' F B U2 D2 B' F' L2 F B U2 D2 B' F' D2 F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L' F B U2 D2 B' F' D2 F B U2 D2 B' F' L2 F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L2 F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L' F B U2 D2 B' F' D' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' D F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' F F B U2 D2 B' F' L' F B U2 D2 B' F' L' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L F B U2 D2 B' F' L F B U2 D2 B' F' F' F B U2 D2 B' F' L' F B U2 D2 B' F' d2 L F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L' d2 D2 L2 F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L2 D2 d' L F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L' d F B U2 D2 B' F' L F B U2 D2 B' F' F' L' F B U2 D2 B' F' L' F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' L F F B U2 D2 B' F' L' F B U2 D2 B' F' L F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L' D L2 F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L2 D' l d' l F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' l' d l' F B U2 D2 B' F' L F B U2 D2 B' F' S F B U2 D2 B' F' L' F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' S' F B U2 D2 B' F' L' F B U2 D2 B' F' d L F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L' d' l' D L2 F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L2 D' l d2 L' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L d2
R' B' D R' F' // Daisy
z r2 F2 x' r' F2 x' F2 x' r' F2 // Cross
z y2 R U R' y' U R U' R' // First corner
y' U2 R U' R' // Second corner
y' U' R U R' U' R U R' U' R U R' U' // Third corner
y' U3 R U' R' // Last corner
y U' U R U' R' y U' L' U L Blue/Red edge
U U R U' R' y U' L' U L // Red/Green edge
U U R U' R' y U' L' U L // Green/Orange edge
y U' U' L' U L y' U R U' R' // Orange/Blue edge
y F R U R' U' F' // Orient yellow cross
U' R U R' U' R U2 R' r' R U R' U' R U2 R' U // Permute yellow cross
y U R U' L' U R' U' L U R U' L' U R' U' L // Permute yellow corners
z2 y R U R' U' R U R' U' D R U R' U' R U R' U' R U R' U' R U R' U' D' Orient yellow corners
Next: F' B' D' L2 D2 R' F U2 R D B2 U' R2 U' L2 D' F2 D' B2 U
z2 R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U' D R U R' U'
Scramble:
F' U2 L2 B2 D2 F R2 U2 F' U2 F2 L2 R' D' L F2 U' L B2 U' B
I chose a BLD method: old pochmann corners, old-old pochmann edges (only T-perms).
The move count is already 374h, 404q, 372s, 372e, and this is a completely valid and sensible method!
And there was not even a parity there!
/*corners*/
F R' (R U' R' U' R U R' F' R U R' U' R' F R) R F'
R' (R U' R' U' R U R' F' R U R' U' R' F R) R
D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D
R2 F (R U' R' U' R U R' F' R U R' U' R' F R) F' R2
D2 R (R U' R' U' R U R' F' R U R' U' R' F R) R' D2
R2 (R U' R' U' R U R' F' R U R' U' R' F R) R2
(R U' R' U' R U R' F' R U R' U' R' F R)
R' D' (R U' R' U' R U R' F' R U R' U' R' F R) D R
/*edges*/
R F R' L' (R U R' U' R' F R2 U' R' U' R U R' F') L R F' R'
d2 L (R U R' U' R' F R2 U' R' U' R U R' F') L' d2
D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2
d' L (R U R' U' R' F R2 U' R' U' R U R' F') L' d
R F' L' R' (R U R' U' R' F R2 U' R' U' R U R' F') R L F R'
L (R U R' U' R' F R2 U' R' U' R U R' F') L'
D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'
l d' l (R U R' U' R' F R2 U' R' U' R U R' F') l' d l'
R S R' (R U R' U' R' F R2 U' R' U' R U R' F') R S' R'
d L (R U R' U' R' F R2 U' R' U' R U R' F') L' d'
l' D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D' l
d2 L' (R U R' U' R' F R2 U' R' U' R U R' F') L d2
However, I was trained on a cube where the R side was stuck and you couldn't turn it. Luckily, you can do R moves through the other sides! I wanted to be as efficient as possible, so I'm not gonna use R' = R R R. Instead, I learned the R, R' and R2 moves to do so in this way:
R = F B U2 D2 B' F' L F B U2 D2 B' F'
R' = F B U2 D2 B' F' L' F B U2 D2 B' F'
R2 = F B U2 D2 B' F' L2 F B U2 D2 B' F'
Applying this to my BLD technic, here is a completely valid, although very old method to solve the cube in a handicapped situation. But hey, every turn is perfectly valid and justified!
F F B U2 D2 B' F' L' F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' F' F B U2 D2 B' F' L' F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' D' F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L' F B U2 D2 B' F' D F B U2 D2 B' F' L2 F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F' F B U2 D2 B' F' L2 F B U2 D2 B' F' D2 F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L' F B U2 D2 B' F' D2 F B U2 D2 B' F' L2 F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L2 F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L' F B U2 D2 B' F' D' F B U2 D2 B' F' L F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L F B U2 D2 B' F' D F B U2 D2 B' F' L F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' F F B U2 D2 B' F' L' F B U2 D2 B' F' L' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L F B U2 D2 B' F' L F B U2 D2 B' F' F' F B U2 D2 B' F' L' F B U2 D2 B' F' d2 L F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L' d2 D2 L2 F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L2 D2 d' L F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L' d F B U2 D2 B' F' L F B U2 D2 B' F' F' L' F B U2 D2 B' F' L' F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' L F F B U2 D2 B' F' L' F B U2 D2 B' F' L F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L' D L2 F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L2 D' l d' l F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' l' d l' F B U2 D2 B' F' L F B U2 D2 B' F' S F B U2 D2 B' F' L' F B U2 D2 B' F' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' F B U2 D2 B' F' L F B U2 D2 B' F' S' F B U2 D2 B' F' L' F B U2 D2 B' F' d L F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L' d' l' D L2 F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L2 D' l d2 L' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' F F B U2 D2 B' F' L2 F B U2 D2 B' F' U' F B U2 D2 B' F' L' F B U2 D2 B' F' U' F B U2 D2 B' F' L F B U2 D2 B' F' U F B U2 D2 B' F' L' F B U2 D2 B' F' F' L d2
This competition got 4 contestants. Thank you for participation. This time the winner is not clear as there are some answers that I should probably consider wrong.
The problems are the following: @TipsterTrickster gave answer to 2A which had F2L -1 solved, so it already had 2x2x2 blocks solved so he should get 0 points for problem 2A. His solution was a 7-moved to a 2x2x2 on FR slot. So I think it wouldn't break the spirit of a competition if he could resubmit, but on the other hand I feel the rules were...
The method is basically to solve a cube learning only 2 algorithms!!
sequence A and B and every solution is just the application of A and B many times (without any extra moves or rotations).
it allows for old-pochman like solve with first permuting all corners and edges and then flipping corners.
for the example scramble it was (2291h, 3451q) before the flipping stage.
I want to go with kubesolver's idea, so kudos to him for the idea.
Let's say I want to go with another Pochmann-style BLD solve. Again, not too fancy! And I don't want to do any wide turns like Dw or such, so I learned to do the T-perm from various angles!
Scramble:
F' B' D' L2 D2 R' F U2 R D B2 U' R2 U' L2 D' F2 D' B2 U
// corners
D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D
F2 (R U' R' U' R U R' F' R U R' U' R' F R) F2
(R U' R' U' R U R' F' R U R' U' R' F R)
D' (R U' R' U' R U R' F' R U R' U' R' F R) D
R' F (R U' R' U' R U R' F' R U R' U' R' F R) F' R
D R (R U' R' U' R U R' F' R U R' U' R' F R) R' D'
R' (R U' R' U' R U R' F' R U R' U' R' F R) R
R2 D' (R U' R' U' R U R' F' R U R' U' R' F R) D R2
// edges
U B (F U F' U' F' L F2 U' F' U' F U F' L') B' U'
L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2
(R U R' U' R' F R2 U' R' U' R U R' F')
U B' (F U F' U' F' L F2 U' F' U' F U F' L') B U'
D' L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D
D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'
L' (R U R' U' R' F R2 U' R' U' R U R' F') L
R F' L2 (R B R' B' R' U R2 B' R' B' R B R' U') L2 F R'
R2 U' L2 (R D R' D' R' B R2 D' R' D' R D R' B') L2 U R2
D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2
U2 R (L U L' U' L' B L2 U' L' U' L U L' B') R' U2
U' F (B U B' U' B' R B2 U' B' U' B U B' R') F' U
Again, no parity, nice and simple solution of 354 moves.
But, I don't like knowing so many algorithms! I want to do it with as few as possible! It turned out, it's enough if you know only two (and their inverse)! And they are super short!
My main algorithms are: a = U F
b = R B L
See? Super simple to learn! YES! Let's use them! Using only a and b I can solve any cube! How, you may ask? Well, every single turn on the cube can be achieved using nothing but a, a', b and b' moves! Here:
U = a'a'a'b'b'ab'a'a'a'bbaabba'b'a'bba'a'a'baaaaaaba'bab'ab'a'a'
F = aaba'ba'b'ab'a'a'a'a'a'a'b'aaab'b'abab'b'a'a'b'b'aaaba'bbaaaa
R = b'a'babba'b'a'b'a'a'ba'b'a'a'bab'abab'a'a'b'abbbbabababab
D = a'a'baaab'b'aba'a'bab'b'abbbab'a'b'aab'b'a'baaaab'b'aabb
B = b'a'ba'a'a'a'bab'a'a'b'b'a'b'aaba'a'bba'bbaaaaba'ba'bba'b'
L = b'a'a'b'aba'a'b'b'b'a'b'a'b'a'b'a'b'b'b'aaaaaba'b'ababa'b'aba'b
Thanks for rokicki and others for this result! See more here:
So, any position of the cube can be written as something like \( g1 g2^2 g1^{-1} g2^3 \), except longer.
Apparently this was known in 1980. Joyner cites this set by "F. Barnes" in a book by Bandelow (which I have yet to see).
I believe this, but I'd like to see a proof by generating each of {B, F, U, D, L, R} as an "alg" in <g1, g2>. (Or at least 5 of them.)
ksolve can't quite handle this. Since 4.3*10^19 is about 65 binary digits...
For example, this is the U turn:
(F' U') (F' U') (F' U') (L' B' R') (L' B' R') (U F) (L' B' R') (F' U') (F' U') (F' U') (R B L) (R B L) (U F) (U F) (R B L) (R B L) (F' U') (L' B' R') (F' U') (R B L) (R B L) (F' U') (F' U') (F' U') (R B L) (U F) (U F) (U F) (U F) (U F) (U F) (R B L) (F' U') (R B L) (U F) (L' B' R') (U F) (L' B' R') (F' U') (F' U')
But I'm a bit lazy, I don't want to generate U' and U2... So I'm gonna use repeated U turns for U2 and U'.
So, let's summarize:
Step 1: I have a nice Pochmann-style solution.
Step 2: I replace every U' turn into U U U turns, and every U2 turns into U U turns (similarly with the other sides).
Step 3: I replace every U turn into a series of my little algorithms of a, a', b, b' (similarly with the other sides).
Step 4: There! I solved the whole cube using only two algorithms with no setup, no conjugates, no cube rotations!
So, how does the full solution look after Step 3? It looks like something like this:
Well, the forum didn't let me post it all in once, because it would have exceed the limit of 200,000 characters. alg cubing and alpha twizzle both struggled to load the solution (but they did eventually), but again, the link is too long (more than 420,000 characters for twizzle), and not even sites like tinyurl let me shorten it...
So, I instead wrote a little Python program for you (see below). Just copy-paste it into an online Python programming page like https://www.programiz.com/python-programming/online-compiler/ then run it, and it generates the result for you. Open alg.cubing.net or https://alpha.twizzle.net/, insert the scramble, insert the solution... Wait for a few seconds...
And voila! A full solution, with a grand total of 73574 turns! That's right, that is more than 73 thousand turns!!!
And every single move has its purpose! Sure, there could be some optimizations I guess to make it shorter... But hey, you asked for silly!
Can I get my 367 gold medals please?
Also, next:
B' L D F' U2 B L B' U R2 F2 U' L2 U' L2 B2 L2 F2 D2 R
Below is the Python code:
Python:
sol = "D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D\n" \
"F2 (R U' R' U' R U R' F' R U R' U' R' F R) F2\n" \
"(R U' R' U' R U R' F' R U R' U' R' F R)\n" \
"D' (R U' R' U' R U R' F' R U R' U' R' F R) D\n" \
"R' F (R U' R' U' R U R' F' R U R' U' R' F R) F' R\n" \
"D R (R U' R' U' R U R' F' R U R' U' R' F R) R' D'\n" \
"R' (R U' R' U' R U R' F' R U R' U' R' F R) R\n" \
"R2 D' (R U' R' U' R U R' F' R U R' U' R' F R) D R2\n" \
"U B (F U F' U' F' L F2 U' F' U' F U F' L') B' U'\n" \
"L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2\n" \
"(R U R' U' R' F R2 U' R' U' R U R' F')\n" \
"U B' (F U F' U' F' L F2 U' F' U' F U F' L') B U'\n" \
"D' L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D\n" \
"D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'\n" \
"L' (R U R' U' R' F R2 U' R' U' R U R' F') L\n" \
"R F' L2 (R B R' B' R' U R2 B' R' B' R B R' U') L2 F R'\n" \
"R2 U' L2 (R D R' D' R' B R2 D' R' D' R D R' B') L2 U R2\n" \
"D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2\n" \
"U2 R (L U L' U' L' B L2 U' L' U' L U L' B') R' U2\n" \
"U' F (B U B' U' B' R B2 U' B' U' B U B' R') F' U\n"
U="a'a'a'b'b'ab'a'a'a'bbaabba'b'a'bba'a'a'baaaaaaba'bab'ab'a'a'"
F="aaba'ba'b'ab'a'a'a'a'a'a'b'aaab'b'abab'b'a'a'b'b'aaaba'bbaaaa"
R="b'a'babba'b'a'b'a'a'ba'b'a'a'bab'abab'a'a'b'abbbbabababab"
D="a'a'baaab'b'aba'a'bab'b'abbbab'a'b'aab'b'a'baaaab'b'aabb"
B="b'a'ba'a'a'a'bab'a'a'b'b'a'b'aaba'a'bba'bbaaaaba'ba'bba'b'"
L="b'a'a'b'aba'a'b'b'b'a'b'a'b'a'b'a'b'b'b'aaaaaba'b'ababa'b'aba'b"
sol = sol.replace("U'", f"{U} {U} {U}")
sol = sol.replace("U2", f"{U} {U}")
sol = sol.replace("U", f"{U}")
sol = sol.replace("F'", f"{F} {F} {F}")
sol = sol.replace("F2", f"{F} {F}")
sol = sol.replace("F", f"{F}")
sol = sol.replace("R'", f"{R} {R} {R}")
sol = sol.replace("R2", f"{R} {R}")
sol = sol.replace("R", f"{R}")
sol = sol.replace("D'", f"{D} {D} {D}")
sol = sol.replace("D2", f"{D} {D}")
sol = sol.replace("D", f"{D}")
sol = sol.replace("B'", f"{B} {B} {B}")
sol = sol.replace("B2", f"{B} {B}")
sol = sol.replace("B", f"{B}")
sol = sol.replace("L'", f"{L} {L} {L}")
sol = sol.replace("L2", f"{L} {L}")
sol = sol.replace("L", f"{L}")
a = "(U F) "
ai = "(F' U') "
b = "(R B L) "
bi = "(L' B' R') "
sol = sol.replace("a'", ai)
sol = sol.replace("a", a)
sol = sol.replace("b'", bi)
sol = sol.replace("b", b)
print(sol)
# uncomment this if want to print to file instead (does not work in browsers):
# with open("sol.txt", "w") as f:
# f.write(sol)
I want to go with kubesolver's idea, so kudos to him for the idea.
Let's say I want to go with another Pochmann-style BLD solve. Again, not too fancy! And I don't want to do any wide turns like Dw or such, so I learned to do the T-perm from various angles!
Scramble:
F' B' D' L2 D2 R' F U2 R D B2 U' R2 U' L2 D' F2 D' B2 U
// corners
D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D
F2 (R U' R' U' R U R' F' R U R' U' R' F R) F2
(R U' R' U' R U R' F' R U R' U' R' F R)
D' (R U' R' U' R U R' F' R U R' U' R' F R) D
R' F (R U' R' U' R U R' F' R U R' U' R' F R) F' R
D R (R U' R' U' R U R' F' R U R' U' R' F R) R' D'
R' (R U' R' U' R U R' F' R U R' U' R' F R) R
R2 D' (R U' R' U' R U R' F' R U R' U' R' F R) D R2
// edges
U B (F U F' U' F' L F2 U' F' U' F U F' L') B' U'
L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2
(R U R' U' R' F R2 U' R' U' R U R' F')
U B' (F U F' U' F' L F2 U' F' U' F U F' L') B U'
D' L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D
D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'
L' (R U R' U' R' F R2 U' R' U' R U R' F') L
R F' L2 (R B R' B' R' U R2 B' R' B' R B R' U') L2 F R'
R2 U' L2 (R D R' D' R' B R2 D' R' D' R D R' B') L2 U R2
D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2
U2 R (L U L' U' L' B L2 U' L' U' L U L' B') R' U2
U' F (B U B' U' B' R B2 U' B' U' B U B' R') F' U
Again, no parity, nice and simple solution of 354 moves.
But, I don't like knowing so many algorithms! I want to do it with as few as possible! It turned out, it's enough if you know only two (and their inverse)! And they are super short!
My main algorithms are: a = U F
b = R B L
See? Super simple to learn! YES! Let's use them! Using only a and b I can solve any cube! How, you may ask? Well, every single turn on the cube can be achieved using nothing but a, a', b and b' moves! Here:
U = a'a'a'b'b'ab'a'a'a'bbaabba'b'a'bba'a'a'baaaaaaba'bab'ab'a'a'
F = aaba'ba'b'ab'a'a'a'a'a'a'b'aaab'b'abab'b'a'a'b'b'aaaba'bbaaaa
R = b'a'babba'b'a'b'a'a'ba'b'a'a'bab'abab'a'a'b'abbbbabababab
D = a'a'baaab'b'aba'a'bab'b'abbbab'a'b'aab'b'a'baaaab'b'aabb
B = b'a'ba'a'a'a'bab'a'a'b'b'a'b'aaba'a'bba'bbaaaaba'ba'bba'b'
L = b'a'a'b'aba'a'b'b'b'a'b'a'b'a'b'a'b'b'b'aaaaaba'b'ababa'b'aba'b
Thanks for rokicki and others for this result! See more here:
So, any position of the cube can be written as something like \( g1 g2^2 g1^{-1} g2^3 \), except longer.
Apparently this was known in 1980. Joyner cites this set by "F. Barnes" in a book by Bandelow (which I have yet to see).
I believe this, but I'd like to see a proof by generating each of {B, F, U, D, L, R} as an "alg" in <g1, g2>. (Or at least 5 of them.)
ksolve can't quite handle this. Since 4.3*10^19 is about 65 binary digits...
For example, this is the U turn:
(F' U') (F' U') (F' U') (L' B' R') (L' B' R') (U F) (L' B' R') (F' U') (F' U') (F' U') (R B L) (R B L) (U F) (U F) (R B L) (R B L) (F' U') (L' B' R') (F' U') (R B L) (R B L) (F' U') (F' U') (F' U') (R B L) (U F) (U F) (U F) (U F) (U F) (U F) (R B L) (F' U') (R B L) (U F) (L' B' R') (U F) (L' B' R') (F' U') (F' U')
But I'm a bit lazy, I don't want to generate U' and U2... So I'm gonna use repeated U turns for U2 and U'.
So, let's summarize:
Step 1: I have a nice Pochmann-style solution.
Step 2: I replace every U' turn into U U U turns, and every U2 turns into U U turns (similarly with the other sides).
Step 3: I replace every U turn into a series of my little algorithms of a, a', b, b' (similarly with the other sides).
Step 4: There! I solved the whole cube using only two algorithms with no setup, no conjugates, no cube rotations!
So, how does the full solution look after Step 3? It looks like something like this:
Well, the forum didn't let me post it all in once, because it would have exceed the limit of 200,000 characters. alg cubing and alpha twizzle both struggled to load the solution (but they did eventually), but again, the link is too long (more than 420,000 characters for twizzle), and not even sites like tinyurl let me shorten it...
So, I instead wrote a little Python program for you (see below). Just copy-paste it into an online Python programming page like https://www.programiz.com/python-programming/online-compiler/ then run it, and it generates the result for you. Open alg.cubing.net or https://alpha.twizzle.net/, insert the scramble, insert the solution... Wait for a few seconds...
And voila! A full solution, with a grand total of 73574 turns! That's right, that is more than 73 thousand turns!!!
And every single move has its purpose! Sure, there could be some optimizations I guess to make it shorter... But hey, you asked for silly!
Can I get my 367 gold medals please?
Also, next:
B' L D F' U2 B L B' U R2 F2 U' L2 U' L2 B2 L2 F2 D2 R
Below is the Python code:
Python:
sol = "D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D\n" \
"F2 (R U' R' U' R U R' F' R U R' U' R' F R) F2\n" \
"(R U' R' U' R U R' F' R U R' U' R' F R)\n" \
"D' (R U' R' U' R U R' F' R U R' U' R' F R) D\n" \
"R' F (R U' R' U' R U R' F' R U R' U' R' F R) F' R\n" \
"D R (R U' R' U' R U R' F' R U R' U' R' F R) R' D'\n" \
"R' (R U' R' U' R U R' F' R U R' U' R' F R) R\n" \
"R2 D' (R U' R' U' R U R' F' R U R' U' R' F R) D R2\n" \
"U B (F U F' U' F' L F2 U' F' U' F U F' L') B' U'\n" \
"L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2\n" \
"(R U R' U' R' F R2 U' R' U' R U R' F')\n" \
"U B' (F U F' U' F' L F2 U' F' U' F U F' L') B U'\n" \
"D' L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D\n" \
"D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'\n" \
"L' (R U R' U' R' F R2 U' R' U' R U R' F') L\n" \
"R F' L2 (R B R' B' R' U R2 B' R' B' R B R' U') L2 F R'\n" \
"R2 U' L2 (R D R' D' R' B R2 D' R' D' R D R' B') L2 U R2\n" \
"D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2\n" \
"U2 R (L U L' U' L' B L2 U' L' U' L U L' B') R' U2\n" \
"U' F (B U B' U' B' R B2 U' B' U' B U B' R') F' U\n"
U="a'a'a'b'b'ab'a'a'a'bbaabba'b'a'bba'a'a'baaaaaaba'bab'ab'a'a'"
F="aaba'ba'b'ab'a'a'a'a'a'a'b'aaab'b'abab'b'a'a'b'b'aaaba'bbaaaa"
R="b'a'babba'b'a'b'a'a'ba'b'a'a'bab'abab'a'a'b'abbbbabababab"
D="a'a'baaab'b'aba'a'bab'b'abbbab'a'b'aab'b'a'baaaab'b'aabb"
B="b'a'ba'a'a'a'bab'a'a'b'b'a'b'aaba'a'bba'bbaaaaba'ba'bba'b'"
L="b'a'a'b'aba'a'b'b'b'a'b'a'b'a'b'a'b'b'b'aaaaaba'b'ababa'b'aba'b"
sol = sol.replace("U'", f"{U} {U} {U}")
sol = sol.replace("U2", f"{U} {U}")
sol = sol.replace("U", f"{U}")
sol = sol.replace("F'", f"{F} {F} {F}")
sol = sol.replace("F2", f"{F} {F}")
sol = sol.replace("F", f"{F}")
sol = sol.replace("R'", f"{R} {R} {R}")
sol = sol.replace("R2", f"{R} {R}")
sol = sol.replace("R", f"{R}")
sol = sol.replace("D'", f"{D} {D} {D}")
sol = sol.replace("D2", f"{D} {D}")
sol = sol.replace("D", f"{D}")
sol = sol.replace("B'", f"{B} {B} {B}")
sol = sol.replace("B2", f"{B} {B}")
sol = sol.replace("B", f"{B}")
sol = sol.replace("L'", f"{L} {L} {L}")
sol = sol.replace("L2", f"{L} {L}")
sol = sol.replace("L", f"{L}")
a = "(U F) "
ai = "(F' U') "
b = "(R B L) "
bi = "(L' B' R') "
sol = sol.replace("a'", ai)
sol = sol.replace("a", a)
sol = sol.replace("b'", bi)
sol = sol.replace("b", b)
print(sol)
# uncomment this if want to print to file instead (does not work in browsers):
# with open("sol.txt", "w") as f:
# f.write(sol)
When I created this I did not expect the existent that people would go to and how there would be people using deep puzzle theory to create the worst solution but this is really cool and I guess as promised here are your three hundred and SIX SEVEN gold medals:
Also if twizzle net were to do this solution on 6x speed (Which I believe is 6tps) it would take it 3.4062037037 Hours to complete
Okay, okay, okay... I have one more idea. Trust me, the solution won't be that long! But it's kinda silly. But also genius! It's a combination of two!
So, how about allowing nothing but R turns and cube rotations? So, instead of F, we do y' R y. Instead of U', we do z R' z'. And so on.
Scramble:
B' L D F' U2 B L B' U R2 F2 U' L2 U' L2 B2 L2 F2 D2 R
Let's make a standard CFOP solution, super boring:
x2
F' D L B D' U2 L' R2 // cross
U2 L' U L // F2L-1
R U' R' U2 B U B' // F2L-2
U' R U' R' U2 R U' R' // F2L-3
U2 B' U B U L U' L' // F2L-4
U L' U' L U' L' U L U L F' L' F // OLL
U2 R' U L' U2 R U' R' U2 R L U // PLL
60 turns. Now, if we swap every turn onto a cube rotation + R turn, this is what we get:
x2
y' R' y z' R z y2 R y2 y R y' z' R' z z R2 z' y2 R' y2 R2 // cross
z R2 z' y2 R' y2 z R z' y2 R y2 // F2L-1
R z R' z' R' z R2 z' y R y' z R z' y R' y' // F2L-2
z R' z' R z R' z' R' z R2 z' R z R' z' R' // F2L-3
z R2 z' y R' y' z R z' y R y' z R z' y2 R y2 z R' z' y2 R' y2 // F2L-4
z R z' y2 R' y2 z R' z' y2 R y2 z R' z' y2 R' y2 z R z' y2 R y2 z R z' y2 R y2 y' R' y y2 R' y2 y' R y // OLL
z R2 z' R' z R z' y2 R' y2 z R2 z' R z R' z' R' z R2 z' R y2 R y2 z R z' // PLL
Technically, still 60 turns...
This is obviously silly. But we can think about a little cube solving robot that can do only R turns, and knock the cube over into any orientation.
But that's not why it is genius. The reason why this is genius and actually used in real life, is because this is exactly how to solve the last step of a 4-dimensional Rubik's hypercube (also known as the 3x3x3x3).
You see, when you solve the hypercube with a layer-by-layer approach, the hyper-PLL step is basically solving a regular 3-dimensional cube on the last layer, often left on the so-called inner side (a hypercube has 8 "sides" called cells: front, back, left, right, up, down, inner, outer). You can chose any regular 3D method to solve this step, it will be fine. But there is one catch: to preserve the other hyper-sides, you can only do "cube-rotations" (inner side twists) and "R side turns" (R side twists). This is known as RKT, and it's one of the most fundamental technique when you solve a higher dimensional puzzle.
There are many-many shortcuts; for example you can switch to a different side when the total amount of twists is divisible by 4 (this is known as "paying the RKT debt"), or certain algorithms like the hyper-sexy-move is allowed to be executed from other orientations (because they are RKT-dept-free), but that is beyond the scope of this topic.
The amazing thing is, that the solution above works perfectly fine on a 4D cube if you have this "PLL case" (meaning; your hyper-PLL is the scrambled 3D 3x3 case above)!
Mind. Blown!
Let me show you this: for the sake of simplicity, I add an additional R turn to the scramble (paying the RKT debt). So, our regular 3D cube looks like this:
And the equivalent 4D case would look like this:
Maybe it is more clear if I hide some of the stickers:
Do you see it? Do you see how they are identical?
This is how our 3D cube looks like after cross + two F2L cases (scroll to the end state):
And our hypercube looks like this after we apply the very same moves (in 4D notations: x = IR, y = IU, z = IF, R = OR)
And after we finished with F2L (scroll to the end case):
On the 4D cube:
And after we are finished with the 3D cube (scroll to the end case):
Our 4D cube is nearly done!
The only thing remains on the 4D cube is re-orienting the inner side with x2 (IR2 in 4D notation) and paying the RKT-debt (which is equivalent of doing a 180 center cap rotation on a 3D cube - any algorithm that does that in 3D works on the 4D cube, as long as we restrict ourselves to R moves like the rest of the solution).
Confusing? Yes. Mind-blowing? Definitely YES!
So, is it silly, or genius?!
BTW this also means that the number of hyper-PLL cases is roughly the same as the number of all 3x3x3 cases (43 quintillion; but there are certain RKT-parities and cube orientations that affects the final number). Good luck learning all the PLL cases in higher dimensions!
Next scramble, for future silliness:
U2 L2 D2 R' B2 L' F2 L' D2 L B' F U' R2 U R U' L2 F
I want to go with kubesolver's idea, so kudos to him for the idea.
Let's say I want to go with another Pochmann-style BLD solve. Again, not too fancy! And I don't want to do any wide turns like Dw or such, so I learned to do the T-perm from various angles!
Scramble:
F' B' D' L2 D2 R' F U2 R D B2 U' R2 U' L2 D' F2 D' B2 U
// corners
D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D
F2 (R U' R' U' R U R' F' R U R' U' R' F R) F2
(R U' R' U' R U R' F' R U R' U' R' F R)
D' (R U' R' U' R U R' F' R U R' U' R' F R) D
R' F (R U' R' U' R U R' F' R U R' U' R' F R) F' R
D R (R U' R' U' R U R' F' R U R' U' R' F R) R' D'
R' (R U' R' U' R U R' F' R U R' U' R' F R) R
R2 D' (R U' R' U' R U R' F' R U R' U' R' F R) D R2
// edges
U B (F U F' U' F' L F2 U' F' U' F U F' L') B' U'
L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2
(R U R' U' R' F R2 U' R' U' R U R' F')
U B' (F U F' U' F' L F2 U' F' U' F U F' L') B U'
D' L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D
D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'
L' (R U R' U' R' F R2 U' R' U' R U R' F') L
R F' L2 (R B R' B' R' U R2 B' R' B' R B R' U') L2 F R'
R2 U' L2 (R D R' D' R' B R2 D' R' D' R D R' B') L2 U R2
D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2
U2 R (L U L' U' L' B L2 U' L' U' L U L' B') R' U2
U' F (B U B' U' B' R B2 U' B' U' B U B' R') F' U
Again, no parity, nice and simple solution of 354 moves.
But, I don't like knowing so many algorithms! I want to do it with as few as possible! It turned out, it's enough if you know only two (and their inverse)! And they are super short!
My main algorithms are: a = U F
b = R B L
See? Super simple to learn! YES! Let's use them! Using only a and b I can solve any cube! How, you may ask? Well, every single turn on the cube can be achieved using nothing but a, a', b and b' moves! Here:
U = a'a'a'b'b'ab'a'a'a'bbaabba'b'a'bba'a'a'baaaaaaba'bab'ab'a'a'
F = aaba'ba'b'ab'a'a'a'a'a'a'b'aaab'b'abab'b'a'a'b'b'aaaba'bbaaaa
R = b'a'babba'b'a'b'a'a'ba'b'a'a'bab'abab'a'a'b'abbbbabababab
D = a'a'baaab'b'aba'a'bab'b'abbbab'a'b'aab'b'a'baaaab'b'aabb
B = b'a'ba'a'a'a'bab'a'a'b'b'a'b'aaba'a'bba'bbaaaaba'ba'bba'b'
L = b'a'a'b'aba'a'b'b'b'a'b'a'b'a'b'a'b'b'b'aaaaaba'b'ababa'b'aba'b
Thanks for rokicki and others for this result! See more here:
So, any position of the cube can be written as something like \( g1 g2^2 g1^{-1} g2^3 \), except longer.
Apparently this was known in 1980. Joyner cites this set by "F. Barnes" in a book by Bandelow (which I have yet to see).
I believe this, but I'd like to see a proof by generating each of {B, F, U, D, L, R} as an "alg" in <g1, g2>. (Or at least 5 of them.)
ksolve can't quite handle this. Since 4.3*10^19 is about 65 binary digits...
For example, this is the U turn:
(F' U') (F' U') (F' U') (L' B' R') (L' B' R') (U F) (L' B' R') (F' U') (F' U') (F' U') (R B L) (R B L) (U F) (U F) (R B L) (R B L) (F' U') (L' B' R') (F' U') (R B L) (R B L) (F' U') (F' U') (F' U') (R B L) (U F) (U F) (U F) (U F) (U F) (U F) (R B L) (F' U') (R B L) (U F) (L' B' R') (U F) (L' B' R') (F' U') (F' U')
But I'm a bit lazy, I don't want to generate U' and U2... So I'm gonna use repeated U turns for U2 and U'.
So, let's summarize:
Step 1: I have a nice Pochmann-style solution.
Step 2: I replace every U' turn into U U U turns, and every U2 turns into U U turns (similarly with the other sides).
Step 3: I replace every U turn into a series of my little algorithms of a, a', b, b' (similarly with the other sides).
Step 4: There! I solved the whole cube using only two algorithms with no setup, no conjugates, no cube rotations!
So, how does the full solution look after Step 3? It looks like something like this:
Well, the forum didn't let me post it all in once, because it would have exceed the limit of 200,000 characters. alg cubing and alpha twizzle both struggled to load the solution (but they did eventually), but again, the link is too long (more than 420,000 characters for twizzle), and not even sites like tinyurl let me shorten it...
So, I instead wrote a little Python program for you (see below). Just copy-paste it into an online Python programming page like https://www.programiz.com/python-programming/online-compiler/ then run it, and it generates the result for you. Open alg.cubing.net or https://alpha.twizzle.net/, insert the scramble, insert the solution... Wait for a few seconds...
And voila! A full solution, with a grand total of 73574 turns! That's right, that is more than 73 thousand turns!!!
And every single move has its purpose! Sure, there could be some optimizations I guess to make it shorter... But hey, you asked for silly!
Can I get my 367 gold medals please?
Also, next:
B' L D F' U2 B L B' U R2 F2 U' L2 U' L2 B2 L2 F2 D2 R
Below is the Python code:
Python:
sol = "D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D\n" \
"F2 (R U' R' U' R U R' F' R U R' U' R' F R) F2\n" \
"(R U' R' U' R U R' F' R U R' U' R' F R)\n" \
"D' (R U' R' U' R U R' F' R U R' U' R' F R) D\n" \
"R' F (R U' R' U' R U R' F' R U R' U' R' F R) F' R\n" \
"D R (R U' R' U' R U R' F' R U R' U' R' F R) R' D'\n" \
"R' (R U' R' U' R U R' F' R U R' U' R' F R) R\n" \
"R2 D' (R U' R' U' R U R' F' R U R' U' R' F R) D R2\n" \
"U B (F U F' U' F' L F2 U' F' U' F U F' L') B' U'\n" \
"L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2\n" \
"(R U R' U' R' F R2 U' R' U' R U R' F')\n" \
"U B' (F U F' U' F' L F2 U' F' U' F U F' L') B U'\n" \
"D' L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D\n" \
"D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'\n" \
"L' (R U R' U' R' F R2 U' R' U' R U R' F') L\n" \
"R F' L2 (R B R' B' R' U R2 B' R' B' R B R' U') L2 F R'\n" \
"R2 U' L2 (R D R' D' R' B R2 D' R' D' R D R' B') L2 U R2\n" \
"D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2\n" \
"U2 R (L U L' U' L' B L2 U' L' U' L U L' B') R' U2\n" \
"U' F (B U B' U' B' R B2 U' B' U' B U B' R') F' U\n"
U="a'a'a'b'b'ab'a'a'a'bbaabba'b'a'bba'a'a'baaaaaaba'bab'ab'a'a'"
F="aaba'ba'b'ab'a'a'a'a'a'a'b'aaab'b'abab'b'a'a'b'b'aaaba'bbaaaa"
R="b'a'babba'b'a'b'a'a'ba'b'a'a'bab'abab'a'a'b'abbbbabababab"
D="a'a'baaab'b'aba'a'bab'b'abbbab'a'b'aab'b'a'baaaab'b'aabb"
B="b'a'ba'a'a'a'bab'a'a'b'b'a'b'aaba'a'bba'bbaaaaba'ba'bba'b'"
L="b'a'a'b'aba'a'b'b'b'a'b'a'b'a'b'a'b'b'b'aaaaaba'b'ababa'b'aba'b"
sol = sol.replace("U'", f"{U} {U} {U}")
sol = sol.replace("U2", f"{U} {U}")
sol = sol.replace("U", f"{U}")
sol = sol.replace("F'", f"{F} {F} {F}")
sol = sol.replace("F2", f"{F} {F}")
sol = sol.replace("F", f"{F}")
sol = sol.replace("R'", f"{R} {R} {R}")
sol = sol.replace("R2", f"{R} {R}")
sol = sol.replace("R", f"{R}")
sol = sol.replace("D'", f"{D} {D} {D}")
sol = sol.replace("D2", f"{D} {D}")
sol = sol.replace("D", f"{D}")
sol = sol.replace("B'", f"{B} {B} {B}")
sol = sol.replace("B2", f"{B} {B}")
sol = sol.replace("B", f"{B}")
sol = sol.replace("L'", f"{L} {L} {L}")
sol = sol.replace("L2", f"{L} {L}")
sol = sol.replace("L", f"{L}")
a = "(U F) "
ai = "(F' U') "
b = "(R B L) "
bi = "(L' B' R') "
sol = sol.replace("a'", ai)
sol = sol.replace("a", a)
sol = sol.replace("b'", bi)
sol = sol.replace("b", b)
print(sol)
# uncomment this if want to print to file instead (does not work in browsers):
# with open("sol.txt", "w") as f:
# f.write(sol)
Not really. It's just a really over-complicated Old Pochmann method (a method developed for solving the cube blindfolded, ergo you solve the cube piece by piece, with minimal changes after each step).
The over-complication comes from the fact that the Cube's group can be generated with only two set of movements. That means, ANY scramble can be solved/generated by using nothing but "a" and "b" (and their inverses) in some crazy, random-looking order. This would be pretty hard to prove normally, you have to go through ALL possible cube scrambles and provide them the series of a/b moves that solves them... OR... It is sufficient to show that a single U move, a single R move, etc. CAN be produced by a series of a/b moves. Which I did (but I take no credit for that discovery!)
And I simply used these to substitute every single turn in my solution (which was already lengthy on purpose) with these series of a/b moves.
This is the most inefficient way to prove that that particular scramble can be solved with nothing but a/b moves... BUT it's easy to generate, and of course, this technique can be trivially used for ANY scramble. An optimal solution of a/b series would be way, way, waaaaay shorter - but much harder to find.
Cage method involves solving the outer pieces and then working inwards a layer at a time. You can start with a cage method solution and use the same substitution of course. I just chose Old Pochmann, because that's one of the longest way to solve a cube (but still in a way that makes sense) in my repertoire.
I want to go with kubesolver's idea, so kudos to him for the idea.
Let's say I want to go with another Pochmann-style BLD solve. Again, not too fancy! And I don't want to do any wide turns like Dw or such, so I learned to do the T-perm from various angles!
Scramble:
F' B' D' L2 D2 R' F U2 R D B2 U' R2 U' L2 D' F2 D' B2 U
// corners
D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D
F2 (R U' R' U' R U R' F' R U R' U' R' F R) F2
(R U' R' U' R U R' F' R U R' U' R' F R)
D' (R U' R' U' R U R' F' R U R' U' R' F R) D
R' F (R U' R' U' R U R' F' R U R' U' R' F R) F' R
D R (R U' R' U' R U R' F' R U R' U' R' F R) R' D'
R' (R U' R' U' R U R' F' R U R' U' R' F R) R
R2 D' (R U' R' U' R U R' F' R U R' U' R' F R) D R2
// edges
U B (F U F' U' F' L F2 U' F' U' F U F' L') B' U'
L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2
(R U R' U' R' F R2 U' R' U' R U R' F')
U B' (F U F' U' F' L F2 U' F' U' F U F' L') B U'
D' L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D
D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'
L' (R U R' U' R' F R2 U' R' U' R U R' F') L
R F' L2 (R B R' B' R' U R2 B' R' B' R B R' U') L2 F R'
R2 U' L2 (R D R' D' R' B R2 D' R' D' R D R' B') L2 U R2
D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2
U2 R (L U L' U' L' B L2 U' L' U' L U L' B') R' U2
U' F (B U B' U' B' R B2 U' B' U' B U B' R') F' U
Again, no parity, nice and simple solution of 354 moves.
But, I don't like knowing so many algorithms! I want to do it with as few as possible! It turned out, it's enough if you know only two (and their inverse)! And they are super short!
My main algorithms are: a = U F
b = R B L
See? Super simple to learn! YES! Let's use them! Using only a and b I can solve any cube! How, you may ask? Well, every single turn on the cube can be achieved using nothing but a, a', b and b' moves! Here:
U = a'a'a'b'b'ab'a'a'a'bbaabba'b'a'bba'a'a'baaaaaaba'bab'ab'a'a'
F = aaba'ba'b'ab'a'a'a'a'a'a'b'aaab'b'abab'b'a'a'b'b'aaaba'bbaaaa
R = b'a'babba'b'a'b'a'a'ba'b'a'a'bab'abab'a'a'b'abbbbabababab
D = a'a'baaab'b'aba'a'bab'b'abbbab'a'b'aab'b'a'baaaab'b'aabb
B = b'a'ba'a'a'a'bab'a'a'b'b'a'b'aaba'a'bba'bbaaaaba'ba'bba'b'
L = b'a'a'b'aba'a'b'b'b'a'b'a'b'a'b'a'b'b'b'aaaaaba'b'ababa'b'aba'b
Thanks for rokicki and others for this result! See more here:
So, any position of the cube can be written as something like \( g1 g2^2 g1^{-1} g2^3 \), except longer.
Apparently this was known in 1980. Joyner cites this set by "F. Barnes" in a book by Bandelow (which I have yet to see).
I believe this, but I'd like to see a proof by generating each of {B, F, U, D, L, R} as an "alg" in <g1, g2>. (Or at least 5 of them.)
ksolve can't quite handle this. Since 4.3*10^19 is about 65 binary digits...
For example, this is the U turn:
(F' U') (F' U') (F' U') (L' B' R') (L' B' R') (U F) (L' B' R') (F' U') (F' U') (F' U') (R B L) (R B L) (U F) (U F) (R B L) (R B L) (F' U') (L' B' R') (F' U') (R B L) (R B L) (F' U') (F' U') (F' U') (R B L) (U F) (U F) (U F) (U F) (U F) (U F) (R B L) (F' U') (R B L) (U F) (L' B' R') (U F) (L' B' R') (F' U') (F' U')
But I'm a bit lazy, I don't want to generate U' and U2... So I'm gonna use repeated U turns for U2 and U'.
So, let's summarize:
Step 1: I have a nice Pochmann-style solution.
Step 2: I replace every U' turn into U U U turns, and every U2 turns into U U turns (similarly with the other sides).
Step 3: I replace every U turn into a series of my little algorithms of a, a', b, b' (similarly with the other sides).
Step 4: There! I solved the whole cube using only two algorithms with no setup, no conjugates, no cube rotations!
So, how does the full solution look after Step 3? It looks like something like this:
Well, the forum didn't let me post it all in once, because it would have exceed the limit of 200,000 characters. alg cubing and alpha twizzle both struggled to load the solution (but they did eventually), but again, the link is too long (more than 420,000 characters for twizzle), and not even sites like tinyurl let me shorten it...
So, I instead wrote a little Python program for you (see below). Just copy-paste it into an online Python programming page like https://www.programiz.com/python-programming/online-compiler/ then run it, and it generates the result for you. Open alg.cubing.net or https://alpha.twizzle.net/, insert the scramble, insert the solution... Wait for a few seconds...
And voila! A full solution, with a grand total of 73574 turns! That's right, that is more than 73 thousand turns!!!
And every single move has its purpose! Sure, there could be some optimizations I guess to make it shorter... But hey, you asked for silly!
Can I get my 367 gold medals please?
Also, next:
B' L D F' U2 B L B' U R2 F2 U' L2 U' L2 B2 L2 F2 D2 R
Below is the Python code:
Python:
sol = "D' R (R U' R' U' R U R' F' R U R' U' R' F R) R' D\n" \
"F2 (R U' R' U' R U R' F' R U R' U' R' F R) F2\n" \
"(R U' R' U' R U R' F' R U R' U' R' F R)\n" \
"D' (R U' R' U' R U R' F' R U R' U' R' F R) D\n" \
"R' F (R U' R' U' R U R' F' R U R' U' R' F R) F' R\n" \
"D R (R U' R' U' R U R' F' R U R' U' R' F R) R' D'\n" \
"R' (R U' R' U' R U R' F' R U R' U' R' F R) R\n" \
"R2 D' (R U' R' U' R U R' F' R U R' U' R' F R) D R2\n" \
"U B (F U F' U' F' L F2 U' F' U' F U F' L') B' U'\n" \
"L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2\n" \
"(R U R' U' R' F R2 U' R' U' R U R' F')\n" \
"U B' (F U F' U' F' L F2 U' F' U' F U F' L') B U'\n" \
"D' L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D\n" \
"D L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D'\n" \
"L' (R U R' U' R' F R2 U' R' U' R U R' F') L\n" \
"R F' L2 (R B R' B' R' U R2 B' R' B' R B R' U') L2 F R'\n" \
"R2 U' L2 (R D R' D' R' B R2 D' R' D' R D R' B') L2 U R2\n" \
"D2 L2 (R U R' U' R' F R2 U' R' U' R U R' F') L2 D2\n" \
"U2 R (L U L' U' L' B L2 U' L' U' L U L' B') R' U2\n" \
"U' F (B U B' U' B' R B2 U' B' U' B U B' R') F' U\n"
U="a'a'a'b'b'ab'a'a'a'bbaabba'b'a'bba'a'a'baaaaaaba'bab'ab'a'a'"
F="aaba'ba'b'ab'a'a'a'a'a'a'b'aaab'b'abab'b'a'a'b'b'aaaba'bbaaaa"
R="b'a'babba'b'a'b'a'a'ba'b'a'a'bab'abab'a'a'b'abbbbabababab"
D="a'a'baaab'b'aba'a'bab'b'abbbab'a'b'aab'b'a'baaaab'b'aabb"
B="b'a'ba'a'a'a'bab'a'a'b'b'a'b'aaba'a'bba'bbaaaaba'ba'bba'b'"
L="b'a'a'b'aba'a'b'b'b'a'b'a'b'a'b'a'b'b'b'aaaaaba'b'ababa'b'aba'b"
sol = sol.replace("U'", f"{U} {U} {U}")
sol = sol.replace("U2", f"{U} {U}")
sol = sol.replace("U", f"{U}")
sol = sol.replace("F'", f"{F} {F} {F}")
sol = sol.replace("F2", f"{F} {F}")
sol = sol.replace("F", f"{F}")
sol = sol.replace("R'", f"{R} {R} {R}")
sol = sol.replace("R2", f"{R} {R}")
sol = sol.replace("R", f"{R}")
sol = sol.replace("D'", f"{D} {D} {D}")
sol = sol.replace("D2", f"{D} {D}")
sol = sol.replace("D", f"{D}")
sol = sol.replace("B'", f"{B} {B} {B}")
sol = sol.replace("B2", f"{B} {B}")
sol = sol.replace("B", f"{B}")
sol = sol.replace("L'", f"{L} {L} {L}")
sol = sol.replace("L2", f"{L} {L}")
sol = sol.replace("L", f"{L}")
a = "(U F) "
ai = "(F' U') "
b = "(R B L) "
bi = "(L' B' R') "
sol = sol.replace("a'", ai)
sol = sol.replace("a", a)
sol = sol.replace("b'", bi)
sol = sol.replace("b", b)
print(sol)
# uncomment this if want to print to file instead (does not work in browsers):
# with open("sol.txt", "w") as f:
# f.write(sol)
First let me start with a normal cfop solution:
B' D R2 L2 F L2 F' U2 x2 F' U' F2 U2 F' U' B U B' U L' U2 L U2 L' U' L Fw R U R' U' Fw' U F R U R' U' F' L' U' L U' L' U2 L U'
Let me take out the rotation and use IF to get the optimal insertions:
B' D R2 L2 F L2 F' U2 B' D' B2 D2 B' D' F D F' D2 R D' L' D R' D L D2 L' D' L F D L D' L' F' D B R2 U R' D R U' R2 D' B' D'
A very nice elegant 47 move solution, this is where the madness begins. (Also thanks to @Renslay for this alg.)
(F' U') (F' U') (F' U') (L' B' R') (L' B' R') (U F) (L' B' R') (F' U') (F' U') (F' U') (R B L) (R B L) (U F) (U F) (R B L) (R B L) (F' U') (L' B' R') (F' U') (R B L) (R B L) (F' U') (F' U') (F' U') (R B L) (U F) (U F) (U F) (U F) (U F) (U F) (R B L) (F' U') (R B L) (U F) (L' B' R') (U F) (L' B' R') (F' U') (F' U')
Also let this alg = a
If every turn can be turned into a U turn, then this is where the fun begins.
B' = x' a a a x
B = x' a x
B2 = x' a a x
L = z a z'
L' = z a a a z'
L2 = z a a z'
R = z' a z
R' = z' a a a z
R2 = z' a a z
F = x a x'
F' = x a a a x'
F2 = x a a x'
D = x2 a x2
D' = x2 a a a x2
D2 = x2 a a x2
So first changing this solution into that we get a solution that is about 10k moves long.
However this is just the beginning, assuming the 95x multiple every time, we could do this again with the new solution to get a 950,000 move solution and once more to get a 90.25 million move solution
and so it continues
8.5 billion
814 billlion
77 trillion
7 quintillion
and 698 quintillion, eventually (I believe) cycling through ever single possible Rubik's cube combination.
Do I win? lol
First let me start with a normal cfop solution:
B' D R2 L2 F L2 F' U2 x2 F' U' F2 U2 F' U' B U B' U L' U2 L U2 L' U' L Fw R U R' U' Fw' U F R U R' U' F' L' U' L U' L' U2 L U'
Let me take out the rotation and use IF to get the optimal insertions:
B' D R2 L2 F L2 F' U2 B' D' B2 D2 B' D' F D F' D2 R D' L' D R' D L D2 L' D' L F D L D' L' F' D B R2 U R' D R U' R2 D' B' D'
A very nice elegant 47 move solution, this is where the madness begins. (Also thanks to @Renslay for this alg.)
(F' U') (F' U') (F' U') (L' B' R') (L' B' R') (U F) (L' B' R') (F' U') (F' U') (F' U') (R B L) (R B L) (U F) (U F) (R B L) (R B L) (F' U') (L' B' R') (F' U') (R B L) (R B L) (F' U') (F' U') (F' U') (R B L) (U F) (U F) (U F) (U F) (U F) (U F) (R B L) (F' U') (R B L) (U F) (L' B' R') (U F) (L' B' R') (F' U') (F' U')
Also let this alg = a
If every turn can be turned into a U turn, then this is where the fun begins.
B' = x' a a a x
B = x' a x
B2 = x' a a x
L = z a z'
L' = z a a a z'
L2 = z a a z'
R = z' a z
R' = z' a a a z
R2 = z' a a z
F = x a x'
F' = x a a a x'
F2 = x a a x'
D = x2 a x2
D' = x2 a a a x2
D2 = x2 a a x2
So first changing this solution into that we get a solution that is about 10k moves long.
However this is just the beginning, assuming the 95x multiple every time, we could do this again with the new solution to get a 950,000 move solution and once more to get a 90.25 million move solution
and so it continues
8.5 billion
814 billlion
77 trillion
7 quintillion
and 698 quintillion, eventually (I believe) cycling through ever single possible Rubik's cube combination.
Do I win? lol
First let me start with a normal cfop solution:
B' D R2 L2 F L2 F' U2 x2 F' U' F2 U2 F' U' B U B' U L' U2 L U2 L' U' L Fw R U R' U' Fw' U F R U R' U' F' L' U' L U' L' U2 L U'
Let me take out the rotation and use IF to get the optimal insertions:
B' D R2 L2 F L2 F' U2 B' D' B2 D2 B' D' F D F' D2 R D' L' D R' D L D2 L' D' L F D L D' L' F' D B R2 U R' D R U' R2 D' B' D'
A very nice elegant 47 move solution, this is where the madness begins. (Also thanks to @Renslay for this alg.)
(F' U') (F' U') (F' U') (L' B' R') (L' B' R') (U F) (L' B' R') (F' U') (F' U') (F' U') (R B L) (R B L) (U F) (U F) (R B L) (R B L) (F' U') (L' B' R') (F' U') (R B L) (R B L) (F' U') (F' U') (F' U') (R B L) (U F) (U F) (U F) (U F) (U F) (U F) (R B L) (F' U') (R B L) (U F) (L' B' R') (U F) (L' B' R') (F' U') (F' U')
Also let this alg = a
If every turn can be turned into a U turn, then this is where the fun begins.
B' = x' a a a x
B = x' a x
B2 = x' a a x
L = z a z'
L' = z a a a z'
L2 = z a a z'
R = z' a z
R' = z' a a a z
R2 = z' a a z
F = x a x'
F' = x a a a x'
F2 = x a a x'
D = x2 a x2
D' = x2 a a a x2
D2 = x2 a a x2
So first changing this solution into that we get a solution that is about 10k moves long.
However this is just the beginning, assuming the 95x multiple every time, we could do this again with the new solution to get a 950,000 move solution and once more to get a 90.25 million move solution
and so it continues
8.5 billion
814 billlion
77 trillion
7 quintillion
and 698 quintillion, eventually (I believe) cycling through ever single possible Rubik's cube combination.
Do I win? lol
I don't think you would necessarily cycle through all cube states... But cycle through some of them, as much as you wish with this recursive approach.
To infinity and beyond!
First let me start with a normal cfop solution:
B' D R2 L2 F L2 F' U2 x2 F' U' F2 U2 F' U' B U B' U L' U2 L U2 L' U' L Fw R U R' U' Fw' U F R U R' U' F' L' U' L U' L' U2 L U'
Let me take out the rotation and use IF to get the optimal insertions:
B' D R2 L2 F L2 F' U2 B' D' B2 D2 B' D' F D F' D2 R D' L' D R' D L D2 L' D' L F D L D' L' F' D B R2 U R' D R U' R2 D' B' D'
A very nice elegant 47 move solution, this is where the madness begins. (Also thanks to @Renslay for this alg.)
(F' U') (F' U') (F' U') (L' B' R') (L' B' R') (U F) (L' B' R') (F' U') (F' U') (F' U') (R B L) (R B L) (U F) (U F) (R B L) (R B L) (F' U') (L' B' R') (F' U') (R B L) (R B L) (F' U') (F' U') (F' U') (R B L) (U F) (U F) (U F) (U F) (U F) (U F) (R B L) (F' U') (R B L) (U F) (L' B' R') (U F) (L' B' R') (F' U') (F' U')
Also let this alg = a
If every turn can be turned into a U turn, then this is where the fun begins.
B' = x' a a a x
B = x' a x
B2 = x' a a x
L = z a z'
L' = z a a a z'
L2 = z a a z'
R = z' a z
R' = z' a a a z
R2 = z' a a z
F = x a x'
F' = x a a a x'
F2 = x a a x'
D = x2 a x2
D' = x2 a a a x2
D2 = x2 a a x2
So first changing this solution into that we get a solution that is about 10k moves long.
However this is just the beginning, assuming the 95x multiple every time, we could do this again with the new solution to get a 950,000 move solution and once more to get a 90.25 million move solution
and so it continues
8.5 billion
814 billlion
77 trillion
7 quintillion
and 698 quintillion, eventually (I believe) cycling through ever single possible Rubik's cube combination.
Do I win? lol
:
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