Prove It Without A Cube

[Cube Is Life]

Is there a way to prove that two sequences of moves do the same thing without using a cube? D L2 B2 L2 R2 F2 R2 U2 R2 D' U' R' B F L' D R2 and U R B F L D R2 do the same thing but there is not a clear relationship between the moves used.

 

[Michael Womack]

Yes and no. For the yes part do it with a cube simulator and for the no part there is no way around it without using any type of cube if it's real or virtual.

 

[Lucas Garron]

Depends on your definition of "using a cube".

The easiest is a simulation: D L2 B2 L2 R2 F2 R2 U2 R2 D' U' R' B F L' D R2 vs. U R B F L D R2

It's possible to reduce your question using syntactic manipulation:

D L2 B2 L2 R2 F2 R2 U2 R2 D' U' R' B F L2 = U R B F

You could observe that L2 B2 L2 R2 F2 R2 is just a double 2-swap, but I don't see an easy way to prove that these have the same permutation without actually simulating/calculating the effect.

 

[Cube Is Life]

So there is no way to just add a move before U R B F L R2 and then do moves later to undo the move you added in without using a cube?

 

[Lucas Garron]

So there is no way to just add a move before U R B F L R2 and then do moves later to undo the move you added in without using a cube?

Depends. That's very vague. You'd probably get more useful responses if you explained what you're *actually* trying to do.

Are you trying to develop an alg? A new solution? Are you trying to do solves with certain restrictions? Are you trying to solve a brainteaser? Curious about cube identities? Are you trying to understand something you've seen elsewhere?

 

[kunparekh18]

R2 L2 R2 L2 does the same thing as U2 D2 U2 D2

 

[Michael Womack]

R2 L2 R2 L2 does the same thing as U2 D2 U2 D2

Well duh, but how are those algs related to the OPs Question?

 

[Cube Is Life]

Depends. That's very vague. You'd probably get more useful responses if you explained what you're *actually* trying to do.

I am trying to reduce the number of moves I do in a FMC by shortening it like how D L2 B2 L2 R2 F2 R2 U2 R2 D' U' R' B F L' D R2 can be shortenend to U R B F L D R2

 

[Chree]

I'm getting a "God's Algorithm" vibe here.

So I think the answer is either (A) No or (B) Not yet.

 

[Lucas Garron]

I am trying to reduce the number of moves I do in a FMC by shortening it like how D L2 B2 L2 R2 F2 R2 U2 R2 D' U' R' B F L' D R2 can be shortenend to U R B F L D R2

Ah. Sounds sensible on the face of it, but it's about as good an idea as trying to reduce a 4x4x4 to a 2x2x2.


If there are any obvious shortcuts like the example you gave, it probably means you were doing something wrong in the first place, and did something very convoluted instead of something obvious.
Usually, if there *are* short savings to be had, your best bet for finding them is to look for corner commutator insertions that cancel a lot of moves., or to try a whole bunch of promising alternatives throughout your solution.

However, if you've written down an FMC solution by hand, I'm willing to bet that any substring up to around 18 moves is not going to be replaceable by anything significantly shorter. You could look through the FMC thread to get some actual data, but I suspect that once you get to human solutions under 30 moves for FMC even a computer wouldn't help much – unless you're willing to replace basically your entire solution.

 

[Stefan]

Not what he wanted, but here's a little program that checks algorithm equivalence, using my favorite way to simulate:
https://gist.github.com/pochmann/a70e314bbeb954b13ef5

 

[Herbert Kociemba]

I think, the problem to show that two sequences are the same straightforwardly leads to the problem to find a presentation for the Rubik's cube group

http://en.wikipedia.org/wiki/Presentation_of_a_group

What we call "canonical sequences" is for example the group defined by the presentation

<U,R,F,D,L,B|U^4=R^4=F^4=D^4=L^4=B^4=U D U^3 D^3=R L R^3 L^3=F B F^3 B^3=1>

With the presentation given above we can for example formally show that U^2 D^2=D^2 U^2:

Because of U D U^3 D^3=1 we have U D U^3 D^3 D=D and because of D^4=1 we get U D U^3 = D.
Right multiplication with U gives U D = D U.

Now we have U^2 D^2 = U (U D) D = U (D U) D = (U D) (U D) = (D U) (D U) = D (U D) U = D (D U) U = D^2 U^2 q.e.d.

Of course we can *not* show with the presentation given above that for example D2 L2 R2 D2 U2 L2 R2 U2 = 1. So the presentation given above surely does not give the cube group. But we could add the relation D2 L2 R2 D2 U2 L2 R2 U2 = 1 to get a "better" group. The problem to find a presentation for the cube group is still unsolved.