Ok, so I think I have a potential answer. I will put it out as a conjecture, and it can be either proven or disproven.
Hardwick conjecture:
The longest possible algorithm for any effect in the 2-gen 3x3x3 puzzle obtained by turning only R's and U's and without repeating any cube state is 73,483,199 turns. Such an algorithm would be the equivalent of either one quarter turn or one half turn on the puzzle. Any other permutation effect on the cube must be of equal or shorter length than this one.
This is done by constructing a Hamiltonian cycle out of all possible 73,483,200 2-gen [R,U] 3x3x3 states. IF such a hamiltonian cycle exists, then imagine it is constructed of all 73,483,200 states of the 2-gen puzzle. Call this Hamiltonian cycle the 2-gen Devil's algorithm.
Number the states the 2gen Devil's algorithm passes through as:
\( a_1, a_2, a_3, ... , a_{73,483,200} \)
where each \( a_i \) is a unique cube state achieved in the 2-gen permutation group. All \( a_i \) states taken together form a partition of the 2-generator cube group. Also, define \( a_1 \) to be the solved cube.
Now let's number the turns in the Devil's algorithm itself as:
\( s_1, s_2, s_3, ... , s_{73,483,200} \)
where \( s_n \) denotes the particular turn that takes the cube from state \( a_{n} \) to state \( a_{n+1} \)
Now assume that the particular algorithm you want to achieve is state \( a_{k+1} \)
So our currently numbered Devil's algorithm can be viewed as:
\( s_1, s_2, s_3, ... , s_k, s_{k+1}, ... , s_{73,483,200} \)
Now to construct the algorithm that has the effect of taking the solved state and turning it into state \( a_{k+1} \) construct the following algorithm:
\( s_1, s_2, s_3, ... , s_{k-1}, s_{k} \)
This sequence is k turns in length. To construct another version of this algorithm you can use:
\( s_{73,483,200}', s_{73,483,199}', ... , s_{k+2}', s_{k+1}' \)
where \( s_n' \) denotes the inverse of turn \( s_n \) and is the turn that takes the cube from state \( a_{n+1} \) to state \( a_n \). This constructed sequence is \( 73,483,200 - k \) turns in length.
Finding the longest possible alg to achieve a particular U permutation can be constructed by finding an optimal turn length 2gen 3 cycle for that U perm. Call that state \( a_{k+1} \) and construct the longer version of the permutation using the method above, which will be the permutation you wished to achieve.
This leads me to:
Hardwick conjecture number 2:
A longest possible algorithm for any desired permutation \( P \) is of length:
\( 73,483,200 - O_{P} \) turns.
\( O_{P} \) denotes the length of an optimal algorithm to achieve the permutation \( P \)
This algorithm is constructed by creating a new Devil's Algorithm \( r_n \) such that:
\( r_1, r_2, ... , r_{O_P} \) is the optimal length algorithm that takes the solved state and transforms it into the permutation \( P \). Now construct the following algorithm out of this new Devil's algorithm \( r_n \) IF such a new Devil's algorithm exists:
\( r_{73,483,200}', r_{73,483,199}', ... , r_{O_P+2}', r_{O_P+1}' \)
This algorithm will be a longest possible algorithm to achieve the permutation \( P_k \) and this assertion is Hardwick conjecture number 2.
Chris
lol not that I believe that but I'd still like to see some evidence in either direction
