# Is it possible to generate a solution with only clockwise turns?



## Cuc (Jul 9, 2019)

1. The ideal answer I am looking for is if it's possible to turn each layer only once clockwise at a time, and that the opposite layer can not be turned immediately after.

We are looking for an alg to describe U' in terms of U, D, R, L, F, B, i.e., clockwise quarter turns without repetition and each turn can not be followed by the opposite layer. Note. If we can do that for U', we can also do it for any other layer by transposing the turns.

2. Secondly, perhaps it's easier if we allowed half turns as well. 

3. Thirdly, if opposite turns are allowed, we are not allowing the a pair of opposite turns to be repeated more than twice, otherwise, we'd get U' = D (U D)3. 

Can anybody help?


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## Billabob (Jul 9, 2019)

(R U)104 R = U'


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## Wish Lin (Jul 9, 2019)

Billabob said:


> (R U)104 R = U'


That is the answer, right?


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## HonestBook (Jul 9, 2019)

Wish Lin said:


> That is the answer, right?


It's not. We cannot just substitute this into algorithms. The left side starts with and ends with R, so it cannot work for any algorithm that has an R before or after U'. (e.g. sexy move)


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## Billabob (Jul 9, 2019)

HonestBook said:


> It's not. We cannot just substitute this into algorithms. The left side starts with and ends with R, so it cannot work for any algorithm that has an R before or after U'. (e.g. sexy move)



Good point - In this case you can use (B U)104 B.


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## HonestBook (Jul 9, 2019)

Billabob said:


> Good point - In this case you can use (B U)104 B.


Oh boy, how stupid I was. We have the solution then.


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## shadowslice e (Jul 9, 2019)

You can also generate clockwise only moves solutions using UTM on ksolve++


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## Piotr Grochowski (Jan 2, 2020)

So it would be the 3×3 version of the Irreversible Cube?


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## Cuc (Jan 3, 2020)

Thanks @Billabob, @Wish Lin, @HonestBook, and @shadowslice e.

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shadowslice e said:


> You can also generate clockwise only moves solutions using UTM on ksolve++



What solutions do you find?

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Based on this insightful observation that (R U)105 = I gives (R U)104 R = U' in 209 turns--which I actually knew is true--I found some shorter solutions with the help of this *Order Calculator*:

(F U B U)45 = I, so (F U B U)44 (F U B) = U'. This takes only 179 turns rather than 209.
I bet that there are shorter solutions!
Can you find any?

If, in case 2, half turns are allowed, we can not allow (F2 U F2)4 = I, but we can allow

(F2 U2 F2 U)6 = I, hence (F2 U2 F2 U)5 F2 U F2 = U'. This takes 23 turns.
(R2 U2)3 = I, so (R2 U2)2 R2 U = U'. This takes 6 turns.
If, in case 3, opposite turns are allowed, I found the following solutions in decreasing number of turns:

(D F R U)36 = I, so (D F R U)35 D F R = U'. This takes 143 turns.
(F B F B U)24 = I, so (F B F B U)23 (F B)2 = U'. This takes 119 turns
(B F R U)24 = I, so (B F R U)23 (B F R) = U'. This takes 95 turns.
(R B F U)24 = I, so (R B F U)23 (R B F) = U'. This takes 95 turns.
(D L R U)6 = I, so (D L R U)5 (D L R) = U'. This takes 23 turns.


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## Billabob (Jan 4, 2020)

ksolve gives the following optimal solutions for the scramble U:

Depth 15
D L R D U B F D U L R D U B F
D L R B F L R D U L R B F L R
D B F D U L R D U B F D U L R
D B F L R B F D U B F L R B F
L R D U B F D U L R D U B F D
L R B F L R D U L R B F L R D
B F D U L R D U B F D U L R D
B F L R B F D U B F L R B F D

They all rely on the same idea of repeating (B F R L U D).


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## Billabob (Jan 4, 2020)

Cuc said:


> (R2 U2)3 = I, so (R2 U2)2 R2 U = U'. This takes 6 turns.



This actually doesn't work, it's (R2 U2)5 R2 U which is 12 turns. However the optimal is 8 turns:

Depth 8
U L2 R2 D2 U2 L2 R2 D2
U B2 F2 D2 U2 B2 F2 D2
D2 U L2 R2 D2 U2 L2 R2
D2 U B2 F2 D2 U2 B2 F2
D2 L2 R2 D2 U2 L2 R2 U
D2 B2 F2 D2 U2 B2 F2 U
L2 R2 D2 U2 L2 R2 D2 U
B2 F2 D2 U2 B2 F2 D2 U


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