Assumed Knowledge: To complete this puzzle, you must have a basic understanding of the 3x3x3 Rubik's cube, and the CFOP method of solving one. You need not have the algorithms memorized or be able to do them on a cube, but you need to know that there are 21 algorithms to solve all 72 pll cases, and each of the 72 cases can be solved using exactly one of the 21 algorithms (with the exception of the solved state which is already solved). You also need enough understanding to be able to find out what would happen when one of the 21 algorithms is applied to the wrong case (that is, a case where said algorithm will not solve the cube).
The basic problem: Find a series of all 21 PLL Algorithms that, when applied in order to a solved cube, results in a solved cube. Here are some additional rules:
- All 21 algorithms must be used
- No algorithm may be used twice
- 2 different algorithms that solve the same case count as the same algorithm (i.e. the Aa algs [l' U R' D2 R U' R' D2 R2] and [R' F R' B2 R F' R' B2 R2]) are both considered to be the same algorithm for the purpose of this puzzle)
- The cube should be solved at the beginning and end of the series, but not at any point in between. So, beginning with Aa, then rotating the cube and applying Ab, thus solving the cube, is not allowed.
- U moves (auf) and cube rotations (x/y/z moves) for fixing orientation before and after the algorithms are allowed, as long as they're not inserted into the middle of an algorithm.
- Computer scripts are allowed to help you solve this, provided you program your own.
- Existing virtual cubes online are also allowed.
- Bonus: Find a series that never creates the same state twice (Solved + Aa = Ab, Ab + F = F, and F + Z = Ab again. So, the series starting with Aa, F, Z is not allowed unless different U moves are applied so that the Ab state doesn't happen twice).
- Extra Bonus: Use no U turns between algorithms
- A "case" is defined as a state of the cube that can be solved with 0 or more U turns, followed by one of the 21 algorithms, followed by 0 or more U turns.
Spirit of the puzzle: I came up with this when I was trying to create a warm-up/practice series where I can run through all the algorithms quickly to freshen up my muscle memory of each case. It could also be used to drill all 21 cases repeatedly to help commit them to muscle memory without neglecting any lesser-used cases. So, that's the basic idea: A series you could do over and over again that causes you to use every algorithm once, but still wind up with a solved cube.
Example: Since there are many solutions, if you need an example solution, here is one with the single bonus.
The formatting is such that "E => E" would mean I applied an E perm algorithm which produced the E state. "U' Ab => Aa" would mean I did a U' turn followed by an Ab perm algorithm, producing the Aa state...
F => F
Ua => Gd
V => Rb
U Ja => E
Ub => Y
U Ra => Ab
U' Gb => Ra
U Ga => Ub
Jb => T
Z => Aa
Ab => H
Rb => Ga
U2 H => Ja
Gd => Ua
Na => V
U' T => Gc
U2 Aa => Gb
U Y => Jb
U' Gc => Z
E => Nb
Nb => Solved!
And since they can affect the U face state, the algorithms I used are:
Aa => "x R' U R' D2 R U' R' D2 R2 x'"
Ab => "x L U' L D2 L' U L D2 L2 x'"
E => "x' L' U L D' L' U' L D L' U' L D' L' U L D x"
F => "R' U' F' R U R' U' R' F R2 U' R' U' R U R' U R"
Ga => "R2 U R' U R' U' R U' R2 U' D R' U R D'"
Gb => "R' U' R U D' R2 U R' U R U' R U' R2 D"
Gc => "R2 U' R U' R U R' U R2 U D' R U' R' D"
Gd => "R U R' U' D R2 U' R U' R' U R' U R2 D'"
H => "M2 U M2 U2 M2 U M2"
Ja => "x R2 F R F' R U2 R' M U R M' U2 x'"
Jb => "R U R' F' R U R' U' R' F R2 U' R' U'"
Na => "R U R' U R U R' F' R U R' U' R' F R2 U' R' U2 R U' R'"
Nb => "R' U R U' R' F' U' F R U R' F R' F' R U' R"
Ra => "R U' R' U' R U R D R' U' R D' R' U2 R' U'"
Rb => "L' U L U L' U' L' D' L U L' D L U2 L U"
T => "R U R' U' R' F R2 U' R' U' R U R' F'"
Ua => "R U' R U R U R U' R' U' R2"
Ub => "R2 U R U R' U' R' U' R' U R'"
V => "R' U R U' R' F' S' U' R U2 R' U' R U' R' F S R"
Y => "F R U' R' U' R U R' F' R U R' U' R' F R F'"
Z => "M' U M2 U M2 U M' U2 M2"
