# How to solve a cube 3x3x3 only with one algorithm



## sm (Nov 10, 2013)

*How to solve a cube 3x3x3 only with one algorithm without additional moves.*
Author of idea – Eugene Benek from Kazakhstan.

For solving we will use a J-permute algorithm: *λ = R U' L U2 R' U R U2 L' R' U*

*Sequence of solving:*

Solving all edges around centers
Solving all corners

My demo-video for this "crazy" method - http://youtu.be/7SopDBFCusI Yes!!! I did it!!! 

*Total, on solving of the cube was required:*
605 / 11 = 55 λ
where:
605 - total number of movements
11 - count of moves in this algorithm (λ)

It is very interesting task! I recommend to try.

*Rules: *It is forbidden to use any additional moves except this algorithm! Also it is forbidden to use the mirror version of this algorithm!

*Tricks with λ-algorithm:*

U = λ y λ y λ
U' = λ y' λ y' λ
U2 = (λ y2 λ y λ y2 λ) or (λ y2 λ y' λ y2 λ)
In my demonstration I didn't use tricks with U, U', U2.
If to emulate algorithms by means of these tricks you need bigger quantity of J-permute. I so think. In this case the algorithm is used as a picklock and the task loses any interest. I use it as a key. 

*PS. Eugene Benek can only with this algorithm solved cubes of any size!!! Too without additional moves. He likes very hard puzzles - a big bandaged cubes and etc.*
One of his topics at the Russian forum - http://twistypuzzles.ru/forum/index.php/topic,305.0.html

*The correction from Eugene for big cubes:*
On the big cubes he uses another version J-permute: *λ = R2 U D' R2 U R2 U' R2 D R2 U' R2 U*
In this algorithm need to replace U/D on u/d - *λ = R2 u d' R2 u R2 u' R2 d R2 u' R2 u* (i.e. instead of external, we twist inside layers) - http://twistypuzzles.ru/forum/index.php/topic,305.msg3296.html#msg3296

For those who used a U-trick, one more task from Evgeny which excludes opportunity to use simply a U-trick with λ.
Modified version of λ-algorithm: *λ = D R2 U' L U2 R' U R U2 L' R' U R' D'*
Video from Eugene about it - http://www.youtube.com/watch?v=B_di2vm682w

This method should be considered as an additional task on the Rubik's cube. For example, as a knight's move on a chessboard. The cube can be used and thus without any physical modifications (bandaging, morphing, etc). Here we limit ourselves to conditions and such tasks sometimes too are interesting and difficult.


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## Dapianokid (Nov 10, 2013)

This algorithm can only solve one particular state of the cube, not counting symmetries.
I solve using only one alg but I apply it following certain guidelines to get around 60-80 move solutions... This move is known as U. I use it from different angles to come up with some nifty solution paths.


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## sm (Nov 10, 2013)

Thanks for the comment!


Dapianokid said:


> This algorithm can only solve one particular state of the cube, not counting symmetries.


Please it is possible in more detail that you mean? You can give example?

As far as I know, this algorithm allows to decide any situation, I checked it on 3x3x3.
I asked Evgeny and he confirms it too: the cube 3x3x3 is possible to solve, from any condition only with λ-algorithm and turns of the cube (x, y, z).


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## legoanimate98 (Nov 10, 2013)

Isn't this just old pochmann with a j-perm?


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## Ollie (Nov 10, 2013)

Reminds me of T-perm only BLD


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## Renslay (Nov 11, 2013)

legoanimate98 said:


> Isn't this just old pochmann with a j-perm?





Ollie said:


> Reminds me of T-perm only BLD



First I thought the same, but here setups are also forbidden (one can use only whole cube rotations).


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## Christopher Mowla (Nov 11, 2013)

This is pretty neat. It's definitely possible to solve all configurations of the 3x3x3 using just cube rotations and a J-perm because sm gave us:


sm said:


> *Tricks with λ-algorithm:*
> 
> U = λ y λ y λ
> U' = λ y' λ y' λ
> U2 = (λ y2 λ y λ y2 λ) or (λ y2 λ y' λ y2 λ)



So you can just solve a 3x3x3 scramble with whatever solution you wish and convert all of the moves in your solution to face turns (if you slice), and then convert all face turns in terms of λ.

That should be enough proof that this is possible to do, given that we are allowed to use any cube rotations we wish to rotate the equivalencies of U, U' and U2 in terms of λ to be F, R', L2, or whatever moves we use in our solution.


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## Blade8227 (Nov 11, 2013)

I bet I could do it if I was okay with using several hundred moves..


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## sm (Nov 11, 2013)

Thanks to all for comments!



cmowla said:


> That should be enough proof that this is possible to do, given that we are allowed to use any cube rotations we wish to rotate the equivalencies of U, U' and U2 in terms of ? to be F, R', L2, or whatever moves we use in our solution.


Thank you that confirmed it too!



Renslay said:


> First I thought the same, but here setups are also forbidden (one can use only whole cube rotations).


Yes, I used only with λ-algorithm and turns of the cube (x, y, z).



Blade8227 said:


> I bet I could do it if I was okay with using several hundred moves..


I understand you!
The decide to creation of this video was hard to for me.
In the course of decision, I took a few breaks to get some rest I was afraid to make a mistake. It took me about 40 minutes for creation of video.
Eugene does it about 7 minutes without breaks! - http://www.youtube.com/watch?v=GF5xWMXfHps


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## Christopher Mowla (Nov 11, 2013)

This reminds me of another topic I started because in that topic, qqwref and I concluded that you can solve the 4x4x4 cube without using even one inner slice turn (including no cancellations like Rw2 R2 = r2, just solid wide turns and outer face turns). The move set can strictly be <U, Rw> and of course, cube rotations of it.

R2 = ([[U2 Rw U2 Rw2 U2 Rw' U' Rw' Rw2 U2 Rw2 U2 Rw2 U2 Rw U Rw U2 Rw2 U2 Rw' U2, Rw2], Rw] Rw2)3
r2 = R2 + Rw2 (but remember that the move set is <U, Rw>, so we do not have the move R2, but a sequence of Rw's and Us, U2s, etc. + an additional Rw2)

R (in SiGN notation) =
r' (r2 U2 r U2 r2 U' r' U2 r' U r' U' r U r' U' r U2 r2 U2 r2 U2 r U r U' r' U r U2 r' U' r U r' U' r' U2 r2 U2 r2 U2 r' U r U' r' U r U r U2 r U r2 U2 r' U2 r2 U2 r2 U2 r U2 r2 U' r' U2 r' U' r' U' r U r' U' r U2 r2 U2 r2 U2 r U r U' r' U r U2 r' U' r U r' U' r' U2 r2 U2 r2 U2 r' U r U' r' U r U' r U2 r U r2 U2 r' U2 r2 U2 r U2 r2 U2 r U2 r2 U' r' U2 r' U r' U' r U r' U' r U2 r2 U2 r2 U2 r U r U' r' U r U2 r' U' r U r' U' r' U2 r2 U2 r2 U2 r' U r U' r' U r U r U2 r U r2 U2 r' U2 r2 U2 r2 U2 r U2 r2 U' r' U2 r' U' r' U' r U r' U' r U2 r2 U2 r2 U2 r U r U' r' U r U2 r' U' r U r' U' r' U2 r2 U2 r2 U2 r' U r U' r' U r U' r U2 r U r2 U2 r' U2 r)3 r (r U r U r U' r' U2 r2 U2 r U2 r' U r2 U' r2 U' r U2 r' U' r U2 r' U r U2 r' U' r U2 r' U' r U2 r' U r U2 r' U' r U2 r' U' r U2 r' U r U2 r' U' r U2 r' U' r U2 r' U r U2 r U r2 U' r U2 r2 U2 r2 U2 r' U2 r U2 r' U2 r' U2 r2 U2 r2 U2 r' U2 r2 U r' U' r' U' r' U' r U2 r' U r U2 r' U2 r U r' U r U2 r' U2 r U r' U r U2 r' U2 r U r' U r U2 r' U2 r U' r' U' r U r' U2 r U2 r' U' r U' r' U2 r U2 r' U' r U' r' U2 r U2 r' U' r U2 r' U' r U r' U2 r U2 r' U' r U' r' U2 r U2 r' U' r U' r' U2 r U2 r' U' r U2 r')2 (r U r' U r' U r2 U2 r2 U r' U2 r U2 r' U2 r' U' r' U' r U r' U2 r U2)2 r U r U r' U r' U' r2 U r2 U2 r' U2 r U2 r' U2 r' U r' U' r U r' U2 r U

r' = R + Rw'

etc.
...


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## Tim Major (Nov 11, 2013)

Didn't someone do an FMC solution with a few hundred moves before the reg changed? Anyway, it isn't new but it is interesting to solve with it, until you know how to do a single turn with a combination of Js.


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## qqwref (Nov 11, 2013)

Using only R U2 R' U' R U2 L' U R' U L (equivalent to your λ, apart from the cube rotation) and rotations, I solved a random scramble in 2:54.36. It's kinda fun. I wonder what God's Alg would look like.


cmowla: Rw' U2 Rw U' Rw' U Rw U' Rw U2 Rw U2 Rw' U2 Rw U2 Rw2 U' Rw2 U Rw U' Rw U' Rw' U2


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## Christopher Mowla (Nov 11, 2013)

qqwref said:


> cmowla: Rw' U2 Rw U' Rw' U Rw U' Rw U2 Rw U2 Rw' U2 Rw U2 Rw2 U' Rw2 U Rw U' Rw U' Rw' U2


I'm confused as to what you meant by this. I was the one who asked Bruce to find this very alg for me using k-solve (after I made my own <U, Rw> alg by hand), and later Kåre found the same alg during his parity algorithm search.

Perhaps you're mistaking my long <U, Rw> parity alg for ones I previously made by hand which were not supercube safe. If you look at the thread I linked to (for which you partipated in), my supercube safe one dedge flip alg also did the move R, for which you proved that it wasn't possible to have a supercube safe one dedge flip algorithm in <U, Rw> which did not also do the move R or R' along with the one dedge flip.

Since my pure dedge flip alg in <U, Rw> flipped a dedge and did the move R, you told me to just add that 26 move non-pure alg to it to flip the dedge back so that all that is left (on the non-supercube) is the move R...

EDIT:
My latest <U,Rw> alg for the non-supercube is "only" 187 btm, not 1152 btm...


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## qqwref (Nov 11, 2013)

Nah, I just mean it's a cool alg, and just having a pure edge flip is enough to show that pretty much anything you want can be done without slice moves. I could use ksolve+ to find an optimal <r,U> R move if you want.

Being able to solve the cube with only double turns and outer layer turns (plus rotations) isn't that unexpected though... what would be cool would be a single alg which we could solve the 4x4x4 (plus rotations ofc). I wonder if there is one.


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## sm (Nov 11, 2013)

Tim Major said:


> Didn't someone do an FMC solution with a few hundred moves before the reg changed? Anyway, it isn't new but it is interesting to solve with it, until you know how to do a single turn with a combination of Js.



Probably, someone already tried it to make. I share an interesting task.
*In my demonstration I didn't use tricks with U, U', U2.*
If to emulate algorithms by means of these tricks you need bigger quantity of J-permute. I so think. In this case the algorithm is used as a picklock and the task loses any interest. I use it as a key. 



qqwref said:


> Using only R U2 R' U' R U2 L' U R' U L (equivalent to your λ, apart from the cube rotation) and rotations, I solved a random scramble in 2:54.36. It's kinda fun.


That's great! Congratulation!

One more option of this algorithm - *R2 U D' R2 U R2 U' R2 D R2 U' R2 U*

For those who used a U-trick, one more task from Evgeny which excludes opportunity to use simply a U-trick with λ.
Modified version of λ-algorithm: *λ = D R2 U' L U2 R' U R U2 L' R' U R' D'*
Video from Eugene about it - http://www.youtube.com/watch?v=B_di2vm682w

This method should be considered as an additional task on the Rubik's cube. For example, as a knight's move on a chessboard. The cube can be used and thus without any physical modifications (bandaging, morphing, etc). Here we limit ourselves to conditions and such tasks sometimes too are interesting and difficult.


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## ~Adam~ (Nov 11, 2013)

I'm sure I read on the forum that someone used only Js during an official FMC.


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## whauk (Nov 11, 2013)

Tim Major said:


> Didn't someone do an FMC solution with a few hundred moves before the reg changed? Anyway, it isn't new but it is interesting to solve with it, until you know how to do a single turn with a combination of Js.





cube-o-holic said:


> I'm sure I read on the forum that someone used only Js during an official FMC.



https://www.worldcubeassociation.org/results/p.php?i=2007DUSS01#333fm


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