# Solving without R face turns



## unixpickle (May 25, 2013)

Someone recently posed a question to me: can you solve a 3x3x3 without turning one of the faces? To respond to this, I first tried to figure out how to perform an R' without turning the R face at all. After coming up with a 50 move intuitive solution, I ran a program on an Amazon server overnight to find that B F U2 D2 F' B' L' F B U2 D2 F' B' (13 HTM) is the optimal solution to an R turn without touching the R face (at least, optimal when measured in HTM).

With all this in mind, I now know that it is possible to solve the cube without touching one of the faces. Now I am left wondering if anybody has looked into such a question before. Has anybody found bounds on God's number if one face is omitted from the solution? Has anybody developed a specific solving method that does not require turns of a specific face? This latter question seems more trivial: as a ZZ solver I can imagine solving without the back face if I sacrificed some efficiency in making my EOLine.

I am very interested in what others have to say about solutions which omit a specific face. For any of those who are interested, my initial human solution to an R turn was as follows:
(F2 D2 L D2 F U' F L' U' L) (F2 U2 F2 U2 F2 U2 B2 U2 B2 U2 B2 U2) (L' U L U2 L U' L' U B' U' B U B' U2 B) (U' F' L F L' U2 L' U L U' L' U2 L)

*Update:*
Incase it was not clear from my question: for obvious reasons, I am only interested in solutions which use face turns. This means no lowercase letters or slices.


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## Username (May 25, 2013)

Interesting! I've never even thought about it. I'm gonna try doing it now 

ZZ solve without B/B'/B2 done


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## AndersB (May 25, 2013)

With CFOP it's actually rather easy to not use one layer. If you do not use D you can still solve almost like normally.  If you decide to not use for example R it's a little bit harder, but still pretty easy. To not use U, though, I find is almost impossible!


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## YddEd (May 25, 2013)

Username said:


> Interesting! I've never even thought about it. I'm gonna try doing it now
> 
> ZZ solve without B/B'/B2 done


Ahahaha easy. Unless your algs use B moves. But if not then thats real easy.


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## Genesis (May 25, 2013)

AndersB said:


> With CFOP it's actually rather easy to not use one layer. If you do not use D you can still solve almost like normally.  If you decide to not use for example R it's a little bit harder, but still pretty easy. To not use U, though, I find is almost impossible!



I just used cross on left when I restricted L


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## Cubenovice (May 25, 2013)

unixpickle said:


> B F U2 D2 F' B' L' F B U2 D2 F' B' (13 HTM) is the optimal solution to an R turn



My human solution to R would be l

With this ingenious solution you can solve the complete cube without using L R U D F AND B


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## Stefan (May 25, 2013)

unixpickle said:


> Now I am left wondering if anybody has looked into such a question before.



I believe this has come up a couple times (here or in the old yahoo group or elsewhere). Easiest way is to move the R pieces to another layer and do the turn there. If you haven't noticed, this is also what you found with your program. Btw, Cube Explorer also does the job and finds it in seconds.


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## KongShou (May 25, 2013)

Cubenovice said:


> My human solution to R would be l



l x


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## Username (May 25, 2013)

KongShou said:


> l x



L M x


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## jaap (May 25, 2013)

On my Useful Mathematics page I wrote this:


> It is a strange fact that you only need 5 faces to solve the cube. In other words, {F, B, R, L, U} generates the whole Cube Group. If you know how to solve the cube layer by layer, it is fairly easy to do it without turning the first layer at all. It is therefore fairly obvious that you can solve the position with just the first layer rotated without using any turns of that layer. So one face can be turned by using only the other five. A simple way to prove it is by means of the sequence P = R2L2U2R2B2R2L2F2L2U2, which does the 4-spot. It moves all the pieces from the D layer to the U layer in the same relative positions without actually turning the D layer, so that PUP' has the same effect as D. The same reasoning shows that the whole square group is also generated by just 5 faces.



On my Cube subgroups page I list a few other neat move sequences for this kind of thing:
D = F2R2D2F2U2R2F2 U F2R2U2F2D2R2F2 ( uses U, F2, R2, D2 )
D2 = F2R2L2B2 U2 F2R2L2B2 ( uses F2, B2, R2, L2, U2 )
U2 = FR' FLFL' F2R2 B'RBR F'R ( uses F, B, R, L )


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## ben1996123 (May 25, 2013)

bruce's devils alg for 3x3 doesnt use any B moves


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## Smiles (May 25, 2013)

i used to wonder if that was possible, but then i remembered i use CFOP, therefore the cross didn't require any D moves.
just like in Roux i'd guess it wouldn't be necessary to use any D moves for first block.
and in ZZ instead of B moves just bring the bad edges to F.


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## FinnGamer (May 25, 2013)

Just solved it using no R moves. Wasn't that hard. I did the cross with D and build the Right block with B and F, fixing the cross with M. Just PLL was a bit tricky, used the RUD TPerm mirrored and the UPerm mirrored


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## Smiles (May 25, 2013)

FinnGamer said:


> Just solved it using no R moves. Wasn't that hard. I did the cross with D and build the Right block with B and F, fixing the cross with M. Just PLL was a bit tricky, used the RUD TPerm mirrored and the UPerm mirrored



i actually cube left handed so R-turn less would be easy.
and normally i would think of solving without turning one face as something like "don't turn the red face" so for like T perm you could just do Dw (t perm) Dw'


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## Stefan (May 25, 2013)

FinnGamer said:


> fixing the cross with M



Cheater.


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## Lucas Garron (May 25, 2013)

Chrs Hardwick has an alg in his signature.

I still don't think you can get more obvious than [E' M2 E: R], though.


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## cuBerBruce (May 25, 2013)

Lucas Garron said:


> I still don't think you can get more obvious than [E' M' E: R], though.



From a fixed-centers perspective, however, that alg turns all six faces.


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## Lucas Garron (May 25, 2013)

cuBerBruce said:


> From a fixed-centers perspective, however, that alg turns all six faces.



You're right, that should be an M2.

(Or use the more compact [[E:M2]:R] or [E:M2;R].)


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## antoineccantin (May 25, 2013)

My first try and this is 31 moves: A 9 move commutator leaving 6 corners, then a 9-mover plus an 8-mover and a 9 mover.


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## unixpickle (May 26, 2013)

Stefan said:


> Btw, Cube Explorer also does the job and finds it in seconds.



When you solved this with Cube Explorer, did you use the optimal solver or the two-phase solver? I used an optimal solver because I wanted to be guaranteed of my solution's optimality. If you used a two-phase algorithm, you would have had no way to be certain that your solution was optimal until the phase-one search exceeded 13 moves (of course, you _did_ know that your solution was optimal because I mentioned it in my original post). However, I ran into trouble with my optimal solver because none of my heuristic tables seemed to provide as much support as they usually do during solves. Thus, my optimal solver expanded over 100 billion nodes and needed to be run on an Amazon server. If you did use Cube Explorer's optimal solver and it _did_ work in seconds, the only explanation (which is quite plausible) is that its heuristic tables are much more effective than my simple ones.


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## Mikel (May 26, 2013)

y2 L


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## kcl (May 26, 2013)

A while back I heard something like this about doing a U2 on 2x2 and solving without any 180˚ turns. My solution was U D lol.. My non cheating solution was to do sexy move*6 with a U instead of U' at the end.. As for this one, again I find loopholes haha.. Do a Y rotation, F or F', and then Y'. Loopholes are so nice  

EDIT: yeah Mikel just posted that and I didn't see it..


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## TMOY (May 26, 2013)

Just rotate your cube by 45° around either the y or the z axis. Then you can solve your cube with any method without having to apply R face turns since the R face doesn't exist anymore


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## Stefan (May 26, 2013)

unixpickle said:


> When you solved this with Cube Explorer, did you use the optimal solver or the two-phase solver?



Hmm, good question. I used the "incomplete cubes" solver and both output and runtime make me think it's optimal, but I didn't find anything about it in the documentation, so I'm not sure.



Spoiler: the output



computation started...

Searching depth 7

Searching depth 8

Searching depth 9

Searching depth 10

Searching depth 11

Searching depth 12

Searching depth 13

U D F2 B2 U' D' R' U D F2 B2 U' D' (13f*)
U D F2 B2 U' D' R' U' D' F2 B2 U D (13f*)
U D' F2 B2 U D' R' U D' F2 B2 U D' (13f*)
U D' F2 B2 U D' R' U' D F2 B2 U' D (13f*)


Your 100 billion nodes do seem excessive, given that you only need to go about half way, and there are only about 100 million nodes reachable in seven moves even if you allow all six sides (see cube20.org).

I just tried with ACube searching for optimal sequences:

```
java -cp ACube3.jar ACube f o
760 UF BR UB UL DF FR DB DL UR FL DR BL BUR BRD UBL ULF FRU DFL DLB FDR
```
On my old laptop, it takes 8 seconds to prepare the tables and then 6 seconds for the search.


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## elrog (May 27, 2013)

I've done this before. I don't use one side while I solve my jupiter mf8 cube. After matching up the corners and edges, I solved the cube CFOP style without using the bottom face (the side that the center doesn't turn).


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## cannon4747 (May 27, 2013)

first off, I'd just not use D face turns. And something I've tried for fun recently is doing left-handed solves (i'm righty). If i get a T-perm I do L' U' L U L F' L2...etc. and the same for OLL. I'd insert and build F2l pairs on the left as well. It was actually an interesting challenge attempting to flip an alogorithm i know solely by muscle memory. For some cases I just used 2-look lol.


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## unixpickle (May 27, 2013)

I finally optimized my cube solver for restricted movesets. I then ran a couple of statistical analyses on the cube without R turns. Here's what I found:

- maximum moves to orient all edges is 9 (usual maximum is 7, worst case is the only good edges at FR and BR, fix with F L F U' D2 B2 L F B)
- maximum moves to solve a specific cross is 10 moves (usual maximum is 8, for all 190080 crosses)
- maximum moves to solve a specific 2x2x2 block is 9 moves (usual maximum is 8, 7 configurations actually _take_ 9 moves)

Some neat sequences without R turns:
R U R' = U2 F' L' U L U2 F [7 moves]
R2 U2 R2 U2 R2 U2 = U B2 L2 D2 F2 D2 L2 B2 U [9 moves]
superflip = L B F' U' F2 D2 L' B D' U2 F2 D' L' F' L2 F2 L2 D F2 U F2 U2 B2 [23 moves, may not be optimal]


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## cuBerBruce (May 27, 2013)

unixpickle said:


> - maximum moves to orient all edges is 9 (usual maximum is 7, worst case is the only good edges at FR and BR, fix with F L F U' D2 B2 L F B)


If you're not allowing a certain face to be turned, it presumably matters what EO scheme you are assuming, <U,D,R,L,F2,B2>, <U,D,R2,L2,F,B>, or <U2,D2,R,L,F,B>. That is, it presumably matters whether or not the restricted layer is one of the layers that can affect the orientation with a quarter turn.


unixpickle said:


> - maximum moves to solve a specific cross is 10 moves (usual maximum is 8, for all 190080 crosses)


If you're not allowing a certain face to be turned, it presumably matters which face you're making the cross on. For example, if you're not allowing R turns, it probably matters whether cross is on R, L, or U/D/F/B.


unixpickle said:


> - maximum moves to solve a specific 2x2x2 block is 9 moves (usual maximum is 8, 7 configurations actually _take_ 9 moves)


If you're not allowing allowing a certain face to be turned, it presumably matters whether or not the 2x2x2 block includes pieces on that face.


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## unixpickle (May 28, 2013)

I will address your questions with more numbers.



cuBerBruce said:


> If you're not allowing a certain face to be turned, it presumably matters what EO scheme you are assuming, <U,D,R,L,F2,B2>, <U,D,R2,L2,F,B>, or <U2,D2,R,L,F,B>



It certainly does matter. I use the <U,R,L,D,F2,B2> scheme. My original number was based on restricting turns of the R face. If you restrict R, U, D, or L, the maximum moves to orient all edges is 9 with 1 taking 9 moves. Interestingly enough, if you restrict F or B, the maximum is 11 with 7 taking 11 moves.



cuBerBruce said:


> If you're not allowing a certain face to be turned, it presumably matters which face you're making the cross on. For example, if you're not allowing R turns, it probably matters whether cross is on R, L, or U/D/F/B.



Here's what I found while restricting the R face. The F, U, D, and B crosses take 10 moves max with 4 that actually take 10 moves. The right cross takes 10 moves max with 792 that take 10 moves. The left cross, interestingly enough, still takes 8 moves with 1324 crosses which actually require a full 8 moves.



cuBerBruce said:


> If you're not allowing allowing a certain face to be turned, it presumably matters whether or not the 2x2x2 block includes pieces on that face.



Any 2x2x2 block which does not intersect with the R face takes a maximum of 9 moves with 7 configurations that actually require 9 moves. Any 2x2x2 block which intersects with the R face still takes a maximum of 9 moves, but with 2452 configurations that actually require such a number.


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## cuBerBruce (May 28, 2013)

unixpickle said:


> I use the <U,R,L,D,F2,B2> scheme. My original number was based on restricting turns of the R face. If you restrict R, U, D, or L, the maximum moves to orient all edges is 9 with 1 taking 9 moves. Interestingly enough, if you restrict F or B, the maximum is 11 with 7 taking 11 moves.
> 
> 
> 
> ...



I confirm from my own independent breadth-first searches that your results are correct.


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## Herbert Kociemba (May 28, 2013)

Stefan said:


> Hmm, good question. I used the "incomplete cubes" solver and both output and runtime make me think it's optimal, but I didn't find anything about it in the documentation, so I'm not sure.



Yes, the "incomplete solver" uses iterative deepening and the first solution found is guaranteed to be optimal.


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