# Solving a cube using T and U perms



## hawkmp4 (Aug 8, 2010)

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?


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## Zarxrax (Aug 8, 2010)

thats pretty much how lots of people do blindsolving, so I guess its been proved at some point


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## ottothedog (Aug 8, 2010)

it took me a while but i was able to do it starting with two edges flipped


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## Edward (Aug 8, 2010)

Not sure about corners, but for sure you can solve edges this way. And I guess you can use the FRU corner as a buffer to solve corners with T perm.


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## amostay2004 (Aug 8, 2010)

Zarxrax said:


> thats pretty much how lots of people do blindsolving, so I guess its been proved at some point



Uhh no, in blindsolving there're setup moves involved


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## Rinfiyks (Aug 8, 2010)

(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,...


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## hawkmp4 (Aug 8, 2010)

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.


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## Sn0w (Aug 8, 2010)

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


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## Sn0w (Aug 8, 2010)

Ohps. Nvm my above post, to be it proven wrong


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## TheMachanga (Aug 8, 2010)

I managed to do this, but my last "move" involved a set-up move.


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## hawkmp4 (Aug 8, 2010)

Sn0w said:


> 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.


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## theace (Aug 8, 2010)

how do you take care of edges and corners that are flipped or oriented wrong? I don't have a 3x3 right now, so i can't really try this. Lost my cube at college


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## hawkmp4 (Aug 8, 2010)

Using T or U permutations with the U face being something other than the first layer or last layer color changes orientation.


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## Robert-Y (Aug 8, 2010)

Well you would have to permute them, do some set up moves, permute the pieces, and then undo the set up moves I think.


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## riffz (Aug 8, 2010)

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



Rinfiyks said:


> (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,...


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## hawkmp4 (Aug 8, 2010)

Yeah. For example...
(Ua) z (Ua) x y' (Ua) x' z (Ub) z' (Ub) flips two edges, among other things.



riffz said:


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


Yes, but now we're discussing how to actually go about solving a cube like this. Relax.


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## riffz (Aug 8, 2010)

hawkmp4 said:


> Yeah. For example...
> (Ua) z (Ua) x y' (Ua) x' z (Ub) z' (Ub) flips two edges, among other things.
> 
> 
> ...



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


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## TMOY (Aug 9, 2010)

amostay2004 said:


> Zarxrax said:
> 
> 
> > thats pretty much how lots of people do blindsolving, so I guess its been proved at some point
> ...



Not a big problem since you can as well use T and U perms as setups.


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