Solving without R face turns

[unixpickle]

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.

 

[Username]

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

ZZ solve without B/B'/B2 done

 

[AndersB]

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!

 

[YddEd]

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.

 

[Genesis]

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

 

[Cubenovice]

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

 

[Stefan]

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.

 

[KongShou]

My human solution to R would be l

l x

 

[Username]

l x

L M x

 

[jaap]

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 )

 

[ben1996123]

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

 

[Smiles]

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.

 

[FinnGamer]

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

 

[Smiles]

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'

 

[Stefan]

fixing the cross with M

Cheater.

 

[Lucas Garron]

Chrs Hardwick has an alg in his signature.

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

 

[cuBerBruce]

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.

 

[Lucas Garron]

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].)

 

[antoineccantin]

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.

 

[unixpickle]

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.