- Joined
- Feb 17, 2019

- Messages
- 14

- Likes
- 11

**Intuitive 4x4 method and tutorial**

I'm an intuitive solver, and I haven't been able to find any good resources for intuitive 4x4 solving, so I made this method and tutorial for others like myself. My hope is that it will be useful for folks who are looking for a completely intuitive 4x4 method that doesn't use any written or memorized algorithms. I think it should eventually be in the "How-tos and Guides" subforum, but I wanted to get your opinions on the method here first. It's based on the Sandwich method.

The most interesting parts are step 4, which is similar to parts of OBLBL, step 5, which I have not found anywhere else so far, and step 6, which I think employs a technique used by advanced Sandwich solvers, but which I haven't seen described anywhere in detail yet. I also like the fact that step 4 sets up good cases for very efficient 4-move center commutators in the last step.

I'm definitely not a speedcuber, so this method is not designed for speed - my major goals are elegance and efficiency. It's somewhat inspired by the Heise method, which I use to solve 3x3 cubes.

As far as move efficiency goes, I think this method has good potential. I have only been cubing for about a month, so I am not very good, but I can get to around 120 moves regularly. I think this will get lower once I get better and more familiar with the method. My personal best is 105 moves, and that includes many sub-optimal cases, so sub-100 is definitely possible. My worst solve full of horrible cases and poor decisions was 137 moves. The average of my last 9 solves is 122, and the average of my last 5 solves is 112 so I'm getting better. Even if this isn't amazingly efficient, at least it isn't terribly inefficient.

Here's the tutorial:

**Solving the 4x4 and Avoiding Parity Problems Intuitively**

This is the method that I use to solve the 4x4 intuitively without using any memorized algorithms, and to easily avoid parity problems. It is based on the Sandwich method.

I assume a good knowledge of blockbuilding and commutators. For a good video on commutators in general and 4x4 commutators in particular, I suggest this video:

I use these 7 steps:

**1.**Solve two opposite centers.

**2.**Solve a 1x3x3 block and a 1x3x4 block (similar to Roux) on the opposite face.

**3a.**Solve two more corners such that the last 3 corners are out of place, then solve the last 3 corners using a commutator.

**3b.**Solve the last edge pair of the 1x3x3 block to extend it to a 1x3x4 block.

**4.**Extend one of the 1x3x4 blocks into a 2x3x4 block by solving one middle slice excluding the top layer.

**5.**Use the unsolved slice to solve exactly 7 of the remaining 10 edge pieces.

**6.**Solve the remaining 3 edge pieces and some center pieces with a commutator.

**7.**Solve the remaining center pieces with 2 or 3 commutators.

Here is a photo album which includes a brief description of each step:

That's it! You should be able to do these steps intuitively without any more instruction (except maybe the parity explanation in step 3a). The remainder of the guide is just clarification and some tips for better efficiency.

My goals in making this method were the following:

1. Completely intuitive - no memorized or written algorithms.

2. Ability to solve parity problems at their source, elegantly and intuitively.

3. Move efficiency.

4. Fun, which includes a varied set of problems and solving techniques during the course of the solve.

**Step 1: Opposite centers**(~10 moves)

This is fairly straightforward. You can sometimes solve edges while solving centers if the edges are already attached to their matching center pieces. But, solving edges is very efficient in the next step so it doesn't make sense to spend more than 3 extra moves to solve an edge pair during this first step.

**Step 2: Blockbuilding**(~25-30 moves)

This step is also straightforward. The goal is to end up with one 1x3x4 block around one of your solved centers, and a 1x3x3 block around the other. I like to solve two 1x3x3 blocks first, then solve the last corner/edge pair to make a 1x3x4 block last. This gives me some more freedom compared to solving a 1x3x4 block first.

The two blocks don't actually have to have matching colors, but subsequent steps are a lot easier if they do. If you really want to save moves, try making unmatched blocks for an extra challenge.

**Step 3: Corners**(~12-17 moves)

First the short version: solve 2 of the 5 unsolved corners such that the last 3 unsolved corners are all incorrectly permuted (i.e. they are out of place, not just twisted in place), then solve these last 3 corners with a commutator. If only 2 corners are incorrectly permuted, they cannot be solved with a commutator because they have odd parity. In this situation, you can just make a quarter turn of the unsolved face to change the parity from odd to even, then solve with commutators.

Okay, now the long version.

This is where we have our first brush with parity problems. If you know the Heise method for the 3x3, you will know that it forces the last 3 corners to have even parity by solving all edges first. This makes it possible to solve the last 3 corners with a single commutator. With the 4x4, this technique does not work because corner parity and edge parity are independent. If we wait to solve the corners until the end of the solve, we can end up with odd corner parity, which is commonly known as a form of PLL parity. Fixing this at the end of the solve can be done intuitively, but it requires discarding and re-solving large portions of the puzzle. So, we are instead going to solve corners now when it is easy to fix parity problems.

First, since we are solving intuitively, we have to understand what parity is. In group theory, parity refers to the number of swaps that it takes to sort a particular permutation. If you can sort a permutation using an odd number of swaps, the permutation has odd parity, and if you can sort a permutation using an even number of swaps, the permutation has even parity. For some really interesting discussions of parity, I recommend the following resources:

Ryan Heise's parity explanation for the 3x3 https://www.ryanheise.com/cube/parity.html

The Parity of Permutations and the Futurama Theorem

A long parity essay about the 4x4

https://hlavolam.maweb.eu/parity-problem

Commutators are the secret weapon of the intuitive solver, but they cannot solve odd parity permutations, because they must always do an even number of swaps. So, we have to make sure that our last three corners have even parity before we can solve them with a commutator. Every quarter-turn of any face cycles corners between odd and even parity, so it is actually very easy to transition between odd and even parity.

The easy way to solve corners using commutators is to keep spamming commutators, and if you end up with 2 or 3 unsolved corners that can't be solved by commutators and conjugates, do a quarter turn of the unsolved face and then keep spamming commutators. However, this is neither elegant nor efficient. In order to be efficient, we need to know whether parity is odd or even before we start to solve the corners. We can do this by counting the number of swaps it will take to permute all of the corners. If you do blind solving, you already know how to do this. For our purposes here, we can use a simplified version. After a bit of practice, this can be done in a few seconds.

First, pick any incorrectly permuted corner (i.e. corners that are in the wrong place - corners twisted in place count as "solved" because we are only considering permutation, not orientation). Find its solved position; this is one swap. Now take note of the corner that is currently in this position and find its solved position; this is a second swap. Continue doing this until you come back to the original piece; this is one cycle. When counting swaps, don't count the starting piece itself, either at the beginning or the end of a cycle.

All of the corners may be in the same cycle, or they may be arranged in multiple separate cycles. Whatever the arrangement, the number of swaps it takes to solve a particular group of pieces will always equal the number of incorrectly permuted pieces minus the number of cycles. To express this as a formula, if p represents the number of pieces, c represents the number of cycles, and s represents the number of swaps needed to solve these pieces, then s=p-c. So in order to determine parity, we really only need to count the number of cycles, then subtract that number from the number of unsolved pieces. We can look at an easy example using four letters:

B C D A

Let's trace the swaps needed to put them in alphabetical order, starting with A. A belongs in B's spot, B belongs in C's spot, C belongs in D's spot, and D belongs in A's spot. All four letters are in the same cycle, and three swaps will sort all of the letters into their correct places. This arrangement has odd parity: 4 letters minus 1 cycle equals 3 swaps. Now look at this example:

C D A B

Here, the letters are arranged in two separate cycles, and we can sort them using only two swaps. This arrangement has even parity: 4 letters minus 2 cycles equals 2 swaps. Note that this example is the same as the previous example, except that the letters have all been rotated one spot to the left. This should help to illustrate how a quarter-turn switches pieces between odd and even parity.

We can apply exactly the same analysis to our corners quite easily. Count the number of corners that are out of place (don't count corners twisted in place), subtract the number of cycles it takes to permute them, and that number will tell you whether the corners have odd or even parity.

After we determine parity, our next goal is to solve two corners such that 3 corners remain unsolved with even parity. We can do this by solving 2 of our 5 unsolved corners using either an odd or an even number of quarter turns. 4-move commutators are very useful here. Remember that a 180 degree turn like U2 counts as two quarter turns. Once you have 3 corners unsolved and even parity, solve the last 3 corners with a single commutator, and this step is complete. (If you end up with all unsolved corners twisted in place, this is an even parity situation as well, it is just less efficient.)

First, since we are solving intuitively, we have to understand what parity is. In group theory, parity refers to the number of swaps that it takes to sort a particular permutation. If you can sort a permutation using an odd number of swaps, the permutation has odd parity, and if you can sort a permutation using an even number of swaps, the permutation has even parity. For some really interesting discussions of parity, I recommend the following resources:

Ryan Heise's parity explanation for the 3x3 https://www.ryanheise.com/cube/parity.html

The Parity of Permutations and the Futurama Theorem

A long parity essay about the 4x4

https://hlavolam.maweb.eu/parity-problem

Commutators are the secret weapon of the intuitive solver, but they cannot solve odd parity permutations, because they must always do an even number of swaps. So, we have to make sure that our last three corners have even parity before we can solve them with a commutator. Every quarter-turn of any face cycles corners between odd and even parity, so it is actually very easy to transition between odd and even parity.

The easy way to solve corners using commutators is to keep spamming commutators, and if you end up with 2 or 3 unsolved corners that can't be solved by commutators and conjugates, do a quarter turn of the unsolved face and then keep spamming commutators. However, this is neither elegant nor efficient. In order to be efficient, we need to know whether parity is odd or even before we start to solve the corners. We can do this by counting the number of swaps it will take to permute all of the corners. If you do blind solving, you already know how to do this. For our purposes here, we can use a simplified version. After a bit of practice, this can be done in a few seconds.

First, pick any incorrectly permuted corner (i.e. corners that are in the wrong place - corners twisted in place count as "solved" because we are only considering permutation, not orientation). Find its solved position; this is one swap. Now take note of the corner that is currently in this position and find its solved position; this is a second swap. Continue doing this until you come back to the original piece; this is one cycle. When counting swaps, don't count the starting piece itself, either at the beginning or the end of a cycle.

All of the corners may be in the same cycle, or they may be arranged in multiple separate cycles. Whatever the arrangement, the number of swaps it takes to solve a particular group of pieces will always equal the number of incorrectly permuted pieces minus the number of cycles. To express this as a formula, if p represents the number of pieces, c represents the number of cycles, and s represents the number of swaps needed to solve these pieces, then s=p-c. So in order to determine parity, we really only need to count the number of cycles, then subtract that number from the number of unsolved pieces. We can look at an easy example using four letters:

B C D A

Let's trace the swaps needed to put them in alphabetical order, starting with A. A belongs in B's spot, B belongs in C's spot, C belongs in D's spot, and D belongs in A's spot. All four letters are in the same cycle, and three swaps will sort all of the letters into their correct places. This arrangement has odd parity: 4 letters minus 1 cycle equals 3 swaps. Now look at this example:

C D A B

Here, the letters are arranged in two separate cycles, and we can sort them using only two swaps. This arrangement has even parity: 4 letters minus 2 cycles equals 2 swaps. Note that this example is the same as the previous example, except that the letters have all been rotated one spot to the left. This should help to illustrate how a quarter-turn switches pieces between odd and even parity.

We can apply exactly the same analysis to our corners quite easily. Count the number of corners that are out of place (don't count corners twisted in place), subtract the number of cycles it takes to permute them, and that number will tell you whether the corners have odd or even parity.

After we determine parity, our next goal is to solve two corners such that 3 corners remain unsolved with even parity. We can do this by solving 2 of our 5 unsolved corners using either an odd or an even number of quarter turns. 4-move commutators are very useful here. Remember that a 180 degree turn like U2 counts as two quarter turns. Once you have 3 corners unsolved and even parity, solve the last 3 corners with a single commutator, and this step is complete. (If you end up with all unsolved corners twisted in place, this is an even parity situation as well, it is just less efficient.)

Personally I have found this step to be quite challenging but extremely fun and interesting. After practicing with my 2x2 I am starting to do fairly well. This step also works as a method for solving a 2x2 intuitively and reasonably efficiently.

**Step 3b: Solve the remaining edge pair to extend the 1x3x3 block into a 1x3x4 block.**(~6 moves)

This is self-explanatory. You want to end up with two 1x3x4 blocks solved on opposite sides, and all corners solved.

**Step 4: Solve 3/4 of one middle slice, to extend one 1x3x4 block into a 2x3x4 block.**(~14 moves)

This is more block building, but you have to use entirely new tactics that don't apply to 3x3. It took me some time to get the hang of this step, but I can now complete it in 12-15 moves. You can use several techniques to connect center pairs with edge pieces and insert them, but my favorite is to put the edge piece on the L or R side of the top layer, turn the M slices to pair centers with it, then rotate the top layer to insert the trio.

**Step 5: Use the unsolved slice to solve 7 of the remaining 10 edge pieces.**(~14 moves)

It took me a while to get the hang of this as well, but it ends up being more efficient than using commutators. Instead of pairing adjacent edges, you pair opposite edges on the unsolved slice, then insert them into the top face. The goal here is to arrive at a state in which three edge pieces are unsolved, which will mean that we have even edge parity, and we can solve these last three pieces with one commutator.

There are a lot of different ways to do this, but I like to solve all of the top-layer edges except for the two in the unsolved slice. In this case, all of the unsolved edge pieces will be in the same slice. It's not guaranteed to end up with only 3 pieces unsolved, but the odds are favorable here - for the even parity cases, you have a 2/3 chance for 3 unsolved pieces, a 1/6 chance for 4 unsolved pieces (which you can solve using 2 commutators), and a 1/6 chance for a skip in which all 4 pieces are solved. This is my preferred technique because of that sweet and satisfying skip. However, you can frequently save a lot of moves by just looking out for good cases and solving whatever edges are easiest.

It's usually not necessary to check for parity since you will correct it automatically by arriving at a state with 3 unsolved edge pieces, which is an even parity state. 2 unsolved edge pieces means odd parity, and 3 unsolved edge pieces means even parity. If you want to check parity earlier, or if you have more than 3 unsolved pieces, you can trace swaps and cycles just as we did when solving the corners. Note that individual edge pieces cannot be flipped in place, so a flipped edge piece must always be permuted to be solved.

If you have odd edge parity, simply turn the unsolved slice by 90 degrees to get to even parity. That's all it takes to avoid edge parity problems. Once you have even parity, you can check which of the two even-parity positions is most favorable for edges and centers - switch between them with a 180 degree turn of the unsolved slice.

**Step 6: Use a commutator to solve the last 3 edge pieces and some center pieces.**(~10 moves)

Since edges now have even parity, we can solve the last edge pieces with a commutator instead of learning a set of "last edges" algorithms.

It's possible to include either three or six center pieces along with the edge commutator, similarly to how we construct a pair 3-cycle on a 3x3 cube. This is easiest if the edges are not conjugated outside of the M slice layers. You can also include center pieces in your conjugates, of course. Sometimes it even makes sense to conjugate only the centers with a wide S2 or E2 move.

Using tricks like this, it's possible to move a lot of center pieces along with the edge commutators, without increasing the move count much or at all. Solving even one center piece during this step can end up being very helpful, since it could decrease the number of commutators needed to solve the centers in the next step.

**Step 7: Center commutators**(~20-25 moves)

In this last step, we have four types of commutators available:

1. 8-move commutators, moving 3 pieces

2. 8-move pair commutators, moving 6 pieces

3. 4-move commutators, moving 6 pieces

4. 4-move pair commutators, moving 12 pieces

Because center pieces have a 1/4 chance of being solved randomly, you will most often end up with 7 or 8 unsolved pieces, just by random chance, or less if you were able to solve some more center pieces along with your edge commutator. Centers will usually take 2 or 3 commutators to solve, the first of which can usually be a 4-move commutator that moves 6 pieces. Because of the way the previous steps set up the centers, you will often be able to easily conjugate a 4-move commutator with a wide slice move. Note that 4-move commutators can be extremely efficient, even with long conjugates. I frequently solve 5 pieces with a single 4-move commutator, which is very efficient even if it requires a 3-move conjugate for a total of 10 moves. It's worth studying and understanding how these 4-move commutators work.

Center commutators will usually require some kind of conjugate, and you can save moves by trying to conjugate the same face for every commutator. This means that you won't have to undo your conjugate at the end of each commutator - you can just rotate the face to a new conjugate for the next commutator instead.

Once you complete the centers, the cube is solved!

**Wait, what about PLL and OLL parity?**

Since we are solving pieces directly and ensuring even edge parity in step 5, we will never have to solve PLL or OLL edge parity problems as such. We avoid corner permutation parity problems in step 3.

**The End**

So, that is my method. If you have any suggestions for improvements I'd love to hear them! I feel like I should be able to do more work during the corner step, or do it more elegantly, but I haven't come up with any improvements yet.

Please also let me know if the explanation was too confusing, or if you have any interest in a video demonstration. Thanks for reading!

Last edited: Mar 9, 2019