There are cubes, and then there are cubes that may not always be cubes. One such “Non-Cube” is the Square-1. To learn more about the Square-1, check out the Wikipedia entry. A popular method for solving the Square-1 is the Vandenbergh method (VM). A great resource to understand this method is this page, made by Lars Vandenbergh. If you know the VM , please continue reading. Else, read up on the basic idea of the VM . If you know how to solve cubes other than the 3×3, this should make sense, so please read on.

I was wondering, how do I teach someone to solve the 3×3 if the only cube that they know how to solve is a Square-1? A quick google search came up empty. To answer the question, I created a method for solving the Rubik’s Cube; the method is based on the VM for Square-1. However, I needed to add a few extra steps to make it a valid method for solving 3×3. On a Square-1, the middle layer is always solved, and the pieces cannot change orientation, but this is not the case for 3×3. I needed to add some extra steps to compensate for these added irregularities on the 3×3.

The first step in executing my method is to solve the middle layer. Solving the middle layer is a mostly intuitive step which is done in relatively few moves, compared to the rest of the method. Compared to the VM , this step is one of the “extra” ones, since not only in the VM but on a Square-1 in general, the middle layer is always solved.

The second step is to orient the edges and corners on both of the outer layers. This step is also one of the “extra” steps since orientation cannot be changed on a Square-1. To do this step, you need to use OLL algorithms, as in the CFOP method. Along with all 57 OLL algorithms, there is a strange parity case caused by pieces on both the top and bottom layers not being oriented correctly. I won’t explain the cube theory behind this, but I’ll give you a simplification:

On a 3×3, edges can only be flipped in pairs. This means that in the last layer stage of the CFOP solution, you can only have 0, 2, or 4 edges oriented incorrectly. However, in my solution, you have both the top and bottom layers unsolved so that you can have 0, 2, 4, 6, or 8 edges flipped. When you only have one layer unsolved, there is an even number of edges flipped in that particular layer. If two layers are unsolved, there could be four incorrectly oriented edges across both layers and an odd number of incorrectly oriented edges in each layer, such as 1 incorrectly oriented edge in the top layer and 3 in the bottom layer. It is impossible to flip 3 edges or 1 edge, so the edges have to be redistributed across the layers. This is a parity case, as you cannot solve it with the conventional algorithms.

The third step is to get all the pieces onto their respective layer. It sounds like this step should be done before orienting the pieces, but it does not matter. An advantage to doing this step after orienting the pieces is that if you get parity during the last step, you do not need to worry about switching around the pieces from layer to layer. This shortens the parity algorithm considerably.

As to actually doing the third step, it can be completed with only R2, U, and D moves. It is a reasonably intuitive step, with some restrictions, like those I mentioned previously. You do not need individual R moves as an R2 can bring a piece from the top to the bottom layer and vice versa. Another reason why you do not need to make R moves is that all the edges are correctly oriented. Since all the pieces are correctly oriented, you can bring the pieces onto their correct layer with just R2, U, and D moves.

The fourth and final step is to permute both layers. This step also has a parity case, and the cube theory behind this parity case is similar to the theory of the other case. You see, similarly to how edges can only be flipped in pairs, edges can only be swapped in pairs as well. If I explain more of the cube theory behind the parity case, I’ll be repeating what I said about the other parity case, so I’ll move on now.

The fourth step can be done with PLL algorithms since the final steps can be broken down into doing the last layer on both the top and bottom layers. After you do the parity algorithm, if required, you have a standard PLL case on both the top and bottom layers. Just do the PLL case on the top, then turn the cube over and do the PLL case on the bottom.

How does this method compare to other speedsolving methods, such as CFOP or Roux? After some testing, I found that my method wasn’t very efficient. Unlike CFOP or Roux, where you can always look ahead to the next step, my method requires pauses to turn the cube over and look at the case. These pauses are during steps 2 and 4 and add a bit more time to your solves. Having to bring the pieces onto their correct layer after orientation makes it impossible to look ahead to step 4, and this is only one of many inefficient points of this method.

So, this method is not comparable to speedsolving methods regarding efficiency; however I created this method in one afternoon, and this method was the product of a few random musings, my curiosity has been satisfied.