Commutator challenge

[irontwig]

I am not sure what you mean here. Are you saying that we already know how to create algorithms to solve 3-cycles which are of the form:
A B A' , where B is not a 3-cycle algorithm?

That can't be true since if (A B A')^3=I then B^3=I since the three A'A cancel out.

 

[Rpotts]

(A B A')*3 = A B*3 A'

 

[Christopher Mowla]

That can't be true since if (A B A')^3=I then B^3=I since the three A'A cancel out.

Yeah, I know but I was just trying to see what he meant. By my interpretation of the wording, this is what appears to be what he meant. So I guess we'll see soon what he meant.

 

[siva.shanmukh]

I am not sure what you mean here. Are you saying that we already know how to create algorithms to solve 3-cycles which are of the form:
A B A' , where B is not a 3-cycle algorithm?

And since that's true (not that I agree that it is or that you are saying that it is), is there a process to write any algorithm that solves 3-cycles into that same form?

Oh.. What I meant is basically that we can solve any 3-cycle algorithm using a commutator. Even if we didn't want to go through the exercise of figuring out A and B, we can do some setup moves, do a known commutator and then undo the setup moves.

To be even more clear, we have found a conjugate version for the U perm as R U' [U R U R U, z' y2] U R'. My question is "Is it possible to write any even move sequence in this form C A B A' B' C' (Assuming this form as conjugate)