Solving a cube using T and U perms

hawkmp4

Veteran
Joined
May 17, 2008
Messages
1,395
So, I was messing around last night and ended up scrambling a cube normally and solved it only using T and U permutations, without setup moves. Intuition tells me that with T and U permutations only, any position of the cube can be reached.
My question is, how would I go about proving this? It's a pretty simple task to flip two edges, do a pure 3-cycle of corners, etc., but I'm not sure how to go about showing that any position can be reached. Any thoughts?
 
Again, I am pretty certain it's possible. I'm looking for a way to prove that it's possible. Blindsolving, like someone else said, involves setup moves, so that's not really relevant.

EDIT: @Rinfiyks- Thanks! That's perfect. Didn't think about trying to do that.
 
Possible, but you would first have to setup a position that is correctly oriented, then permut piece by piece.
Also, wouldn't it take long to cycle corners,cause you would side affect edges, and fix them per 2 turns, which means like term x6
 
Possible, but you would first have to setup a position that is correctly oriented, then permut piece by piece.
Also, wouldn't it take long to cycle corners,cause you would side affect edges, and fix them per 2 turns, which means like term x6

I solve corners first using T perms. I place the first 3 first layer corners in the correct permutation and orientation. Then I use the last first layer corner position as a sort of buffer, shooting down LL corners and then back up in the correct orientation using T perms from different angles. Then I permute the LL corners correctly.
From that point on, it's a pretty simple matter of U perms to place edges in the correct position and orientation.

The corners first approach allows me to avoid odd edge permutation parity and to avoid having to fix edges after each T perm. After the corners are solved, since the U perms are pure edge cycles, there's no issue with destroying what I've already solved.
 
Using T or U permutations with the U face being something other than the first layer or last layer color changes orientation.
 
Yeah. For example...
(Ua) z (Ua) x y' (Ua) x' z (Ub) z' (Ub) flips two edges, among other things.

I don't know why you guys are still discussing this. The answer was already stated by Rinfiyks:

(T-perm y)*3 Ua-perm y' Ua-perm
That all cancels down to U.

If you can do U with T and U perms, you can do U2, U', and repeat on all faces, blablabla,...
Yes, but now we're discussing how to actually go about solving a cube like this. Relax.
 
Yeah. For example...
(Ua) z (Ua) x y' (Ua) x' z (Ub) z' (Ub) flips two edges, among other things.

I don't know why you guys are still discussing this. The answer was already stated by Rinfiyks:

(T-perm y)*3 Ua-perm y' Ua-perm
That all cancels down to U.

If you can do U with T and U perms, you can do U2, U', and repeat on all faces, blablabla,...
Yes, but now we're discussing how to actually go about solving a cube like this. Relax.

I didn't mean those posts. I realize my post might have come across as rude, which wasn't my intentions.
 
Back
Top