So, I got curious if I could solve from the cube in a cube in a cube in a cube position to the solved state just using its own algorithm. This is the one I was using:

B' 2R2 2L2 U2 2R2 2L2 B F2 R U' R U R2 U R2 F' U F' u l u' f2 d r' u f d2 r2

Where 2R2 and 2L2 mean to move the center slice 180 degrees and u, l, f, d, r, and b mean to move both the outer and center by a quarter turn and so forth.

I did get some interesting cases on the cube, such as one where most of the cube is solved except for 2 opposite 2x2 corners with edges swapped (I think it took 10-12 repeats of the algorithm to get into this state of edges being swapped in 2 opposite 2x2 corners), and I kept going until about the fifteenth repeat.

Then I thought that since I don't even know if I can solve it from the cube in a cube in a cube in a cube position just using that 1 algorithm, I might as well solve it using the reduction method, because I don't know if this property is true for 4x4 cubes.

I know that on the 3x3, you can get from any state to solved, just by using the algorithm that got you into the state. It might be only once like for the Checkerboard or Superflip patterns, or it may be multiple times as in the Cube in a Cube in a Cube, but it's possible. I think this property might have a special name, though if it does, I don't know it. But, is this property true of the 4x4 as well? Was it just a matter of repeating the algorithm again and again after 15 to get back to solved? Or does parity get in the way of doing that?

What I was trying to do was to get from this:

enter image description here

to the solved 4x4 only using this algorithm:

B' 2R2 2L2 U2 2R2 2L2 B F2 R U' R U R2 U R2 F' U F' u l u' f2 d r' u f d2 r2

Is that possible or does this algorithm cause parity that makes it unsolvable with just that 1 algorithm?

  • $\begingroup$ "I know that on the 3x3, you can get from any state to solved" Minor quibble, but you can get from any solvable, or valid--if you will--state to a solved state. There are some permutations on a 3 x 3 x 3 that aren't actually solvable. Like if only 1 corner is rotated, you can't actually get to a solved state. But I'm assuming you meant that given the other context in your question. Hence, why this is a minor quibble :D $\endgroup$
    – Chipster
    Dec 6, 2020 at 5:37
  • $\begingroup$ Welcome to Puzzling! Could you explain what "the cube in a cube in a cube in a cube position" is? We want all questions to be self-contained, and as of right now I have absolutely no idea what you're referring to. $\endgroup$
    – bobble
    Dec 6, 2020 at 5:38
  • $\begingroup$ Well, I did add an image to clarify what the cube in a cube in a cube in a cube position that I was trying to solve from is. And I got that image and algorithm from this site: cubingcheatsheet.com/algs4x_patterns.html under the 3 Cube in a Cube pattern on that page. $\endgroup$
    – Caters
    Dec 6, 2020 at 5:41
  • $\begingroup$ If you start at the solved state, and then repeat an algo (any algo) without errors for long enough (heat death of the universe may become an issue in some cases), you are guaranteed to reach the start state eventually: for any given algo, the resulting state is uniquely defined by the starting state and vice versa, and there are no states where you cannot perform the algo, so since there are a finite number of possible cube states, there will always be a loop, and the starting state will always be part of that loop. (Not sure if this is what you were asking, which is why this is a comment.) $\endgroup$
    – Bass
    Dec 6, 2020 at 6:48
  • $\begingroup$ Apply it 29 more times $\endgroup$ Dec 6, 2020 at 8:15

1 Answer 1


It is always possible to solve a cube state using its own algorithm, regardless of the cube size or the algorithm.

Here's why:

Let's name the starting state "S", and its algorithm (the one that transforms the solved state into S) "A".

enter image description here

There are three important points to notice about cube states and algorithms:

  1. If we know the starting state and the algorithm, the end state is uniquely defined.
  2. If we know the end state and the algorithm, the starting state is uniquely defined.
  3. No matter what the cube state, you can always perform any algorithm on it.

These may seem obvious, but with these, we can prove the bold claim from above.

If we repeatedly perform algorithm A on state S, we get a chain of states. Because of observation 1 above, we know the chain will never branch. Now if we think about what's eventually going to happen to that chain, there are only a couple of options:

  1. At some point, the chain ends. Given observation 3 from above, this won't happen. enter image description here
  2. The chain goes on and on forever, never repeating a state. This is of course impossible; even though the cube has quite a few possible states, the number of possible states is very obviously finite.

enter image description here

  1. At some point, the chain loops without visiting S again. This is also impossible: now the state the chain looped back into has to have two different predecessors, violating observation 2 from above.

enter image description here

  1. At some point, the chain loops back into S. enter image description here

Option 4 is the only possible outcome, so we know that the chain will eventually loop back to S. But we also know (from the definition of A) that you can reach state S from the solved state by performing algorithm A! From observation 2, we know that the solved state is then the only state from which S can be reached, so we have shown that the solved state must be in the chain of states reached by repeating algorithm A on state S.

Since we didn't use the dimensions of the cube or the specifics of the algorithm anywhere in the proof, this applies to all algorithms on all cubes.


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