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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, and also shares several techniques with the Roux method. In fact, if you use this method on the 3x3 it is just Roux. I call it QTPI (cutie pie) Method, or Quarter Turn Parity Intuitive Method.

Here is a full demonstration solve with some explanation:

Here is a photo gallery of the method, with brief descriptions of the steps:

Here is a YouTube playlist showing short examples from steps 2-6:

https://www.youtube.com/playlist?list=PLxYYfdYl2gTjEI143CEwWfGd2uhOH0A2m

This method is for intermediate intuitive solvers who can already solve a 3x3, and want to start solving the 4x4 without memorizing algorithms.

This is not intended for complete beginners, since it assumes knowledge of concepts such as commutators and blockbuilding. But, you can still use it as a beginner if you want to learn about how the cube works. If you don't know about commutators, I suggest this video, as it goes over 4x4 commutators in particular: 4x4 Commutator Video

This is not really a speedsolving method, but perhaps it could be adapted for speedsolving, as it has a reasonably efficient move count.

My goals in making this method:

The goal is to end up with two 1x3x4 blocks around two opposite centers, similar to the Roux method. I usually build two opposite centers first, then hold the solved centers on L and R, and use the M slices to pair edges. But, you can really do whatever you want - user @dbeyer likes to build the perimeter of the blocks first and then drop in the centers.

In the advanced version explained below, you will instead build a 1x3x4 block and a 1x3x3 block.

If you know a set of corner algorithms such as CMLL, you can use them here and skip to the next step. But this method is for people who haven't memorized any algorithms, so I have developed a heuristic for solving last-layer corners efficiently using commutators.

First, determine corner parity like so:

Odd-parity cases cannot be solved directly with commutators, but every quarter-turn of any face changes corner parity between odd and even (180-degree turns like U2 count as 2 quarter turns and do not change parity). This gives us two basic options for solving odd parity cases:

This heuristic will solve last-layer corners on any cubic puzzle. If you want to save some moves, you can use an advanced version that starts with 5 corners unsolved.

Spoiler: Advanced 5-corner solution, and a long explanation of parity

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 this is the most efficient way I have found:

1. Place a bottom-layer wing edge piece in the top layer on the L or R side, oriented such that the color you want to pair is on the U face.

2. Rotate the slices to join center pieces with the edge, making a bar that goes from the L to the R face - perpendicular to the slice where it will be inserted. Rotate the U layer 180 degrees to shift the edge piece between the L and R side, depending on where the center pieces are located.

3. Once the two center pieces are joined to the edge piece, rotate the U layer 90 degrees to place the three pieces in the correct slice, then rotate that slice to insert the trio.

It's not obvious, but this step sets up the last step to use efficient 4-move commutators.

It took me a while to get the hang of this as well, but it is more efficient than using commutators for all 10 edge pieces. Instead of pairing adjacent edges, you pair opposite edges in the unsolved slice, then insert them into the top face. The principal is the same as that used in step 4b of the Roux method.

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 edge pieces with one commutator.

There are a lot of different ways to do this, but I like to solve all of the edge pieces except for those in the unsolved slice. It's not guaranteed to end up with only 3 pieces unsolved if you do this, but the odds are favorable - 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. 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 wing 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 do a single quarter turn of the unsolved slice 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.

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 and conjugate, similarly to how we construct a pair 3-cycle on a 3x3 cube. This is easiest to see 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.

If you like algorithms, note that all cases in which the last 3 edges are in the same slice can be solved with the same commutator or its mirror. For example, if the unsolved slice is u, the unsolved edges are in the columns FL, FR, and BR, and the FL edge belongs in the FR column, use [L': [F' U' F, u]].

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. Examples of 4-move center commutators include r u2 r' u2 and r2 u2 r'2 u2.

One of the strengths of this method is that step 3 sets up the centers for 4-move commutators in step 6. Because of this, 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.

Once you complete the centers, the cube is solved!

Since we are solving pieces directly and ensuring even edge parity in step 4, we will never get PLL or OLL edge parity problems. We avoid corner permutation parity problems in step 2.

Yes! It needs a few modifications for odd-layer cubes, but it certainly works. Adding the middle slice on the 5x5 actually enables us to solve proportionally more of the center pieces before the last step: 65% on the 5x5 vs about 58% on the 4x4. However, the method gets progressively less efficient as the cubes get larger than 5x5, since you always have to leave 5/8 of the wing slice centers unsolved, and there are proportionally more center pieces versus edge pieces as cubes get lager. But hey, at least you don't have to learn any parity algorithms!

Here is the modified method for 5x5:

1. 1x4x4 block and 1x4x5 block around opposite centers.

2. 3/4 of the middle slice, matching the 1x4x5 block (this makes a cool-looking pattern!)

3. Pair (not solve) the last edge of the second 1x4x4 block

4. Solve all middle edges and corners exactly as in Heise steps 3-4

5. 3/4 of a wing slice, creating a 3x4x5 block

6. 7 wing edges, using the unsolved wing slice

7. Commutator to solve last 3 wing edges

8. 3-5 center commutators

I will probably do a detailed tutorial of the 5x5 version of this method at some point, since it is really fun. You get all of the problems and techniques from the 4x4 method, plus more. You also don't need to keep track of corner parity, since you can simply use Heise steps 3 and 4 to solve the corners and middle edge pieces at the same time. You can also use different variations, such as modifying steps 1-3 to orient edges earlier: make two 1x4x4 blocks, solve 1/2 of the middle slice to match these blocks, do Petrus-style middle-edge orientation, extend one 1x4x4 block into a 1x4x5 block while solving the last middle edge piece on the bottom layer, solve another 1/4 of the middle slice to extend it to a 3/4 slice matching the 1x4x5 block, proceed with step 4.

Maybe, but I can't say for sure because I'm not a speedsolver. You could easily use existing algorithm sets for corners, last edges, and last centers. The method relies heavily on slice moves, though Roux solvers may not mind this. Last center commutators are not known for speed, but the Sandwich method uses them, so there is some precedent for their use in speedsolving. The method that is most similar is Roux, so perhaps a Roux solver could find common ground.

The biggest problem is likely the fact that this method uses several techniques that are not useful elsewhere. It would require lots of specific practice that would not transfer to other methods, and would not leverage other speedcubing skills. I designed the method this way on purpose. I wanted many different challenges that could be solved intuitively in a variety of ways, and were different than the challenges of the 3x3. This is the opposite of what you want in a speedsolving method.

As for my personal times, I can solve the 3x3 in about 3 minutes 1 and a half minutes and I can solve the 4x4 using this method in about 5 3 and a half minutes. My personal best is 3:034:20, which is recorded in my demonstration video. In my demonstration video, which took 4:20, it took me 2:12 to solve first blocks and corners, which is about 4 times longer than a competent solver using the Lewis method, which uses the same steps up to that point. So, if a speedy solver is 4 times faster than I am, they should be able to get to 1 minute pretty easily.

For the 4x4, My personal best move count is 105, and the average of my last 5 solves (at least, those in which I counted my moves) is 110. My personal worst, including mistakes, is 137. I have only counted my moves on 5x5 twice, and I came up with 208 and 212. This is an average of 2.14 moves per piece solved, which is only slightly less efficient than my average of 1.96 moves per piece solved on the 4x4. The difference is probably due to the fact that I am just not as good at solving 5x5.

Those might seem like pretty good numbers, but keep in mind that I am not a speed cuber, so I take a really long time to complete these solves. I'm sure that many other methods would become much more efficient if you spent lots of time deliberately trying to be efficient. In my demonstration video, during which I was not trying to be efficient, my move count was 123. But a more competent cuber could probably be much more efficient than me, particularly with blockbuilding. Anyhow, I don't think that this method is amazingly move-efficient, but at least it is not inefficient.

So, that is my method. If you have any suggestions for improvements I'd love to hear them. Please also let me know if any part of the explanation was unclear or confusing. Thanks for reading!

Here is a full demonstration solve with some explanation:

Here is a photo gallery of the method, with brief descriptions of the steps:

Here is a YouTube playlist showing short examples from steps 2-6:

https://www.youtube.com/playlist?list=PLxYYfdYl2gTjEI143CEwWfGd2uhOH0A2m

**Summary of the Method**- Solve two 1x3x4 blocks around opposite centers (similar to Roux)
- Solve corners with a commutator
- Extend one of the 1x3x4 blocks into a 2x3x4 block by solving one middle slice excluding the top layer
- Use the unsolved slice to solve exactly 7 of the remaining 10 edge pieces
- Solve the remaining 3 edge pieces and some center pieces with a commutator
- Solve the remaining center pieces with 2 or 3 commutators

**Who is this method for?**This method is for intermediate intuitive solvers who can already solve a 3x3, and want to start solving the 4x4 without memorizing algorithms.

This is not intended for complete beginners, since it assumes knowledge of concepts such as commutators and blockbuilding. But, you can still use it as a beginner if you want to learn about how the cube works. If you don't know about commutators, I suggest this video, as it goes over 4x4 commutators in particular: 4x4 Commutator Video

This is not really a speedsolving method, but perhaps it could be adapted for speedsolving, as it has a reasonably efficient move count.

My goals in making this method:

- Completely intuitive - no memorized or written algorithms necessary.
- Ability to solve parity problems at their source with a single quarter-turn.
- Move efficiency.
- Fun, which includes a varied set of problems and solving techniques that are different from a 3x3 solve.

**Detailed Explanation****Step 1: Blockbuilding**(~40-45 moves)The goal is to end up with two 1x3x4 blocks around two opposite centers, similar to the Roux method. I usually build two opposite centers first, then hold the solved centers on L and R, and use the M slices to pair edges. But, you can really do whatever you want - user @dbeyer likes to build the perimeter of the blocks first and then drop in the centers.

In the advanced version explained below, you will instead build a 1x3x4 block and a 1x3x3 block.

**Step 2: Corners**(~9-17 moves)If you know a set of corner algorithms such as CMLL, you can use them here and skip to the next step. But this method is for people who haven't memorized any algorithms, so I have developed a heuristic for solving last-layer corners efficiently using commutators.

First, determine corner parity like so:

- Place one corner in its solved position so that it is correctly permuted.
- Examine the other 3 corners. If exactly 2 are out of place and 1 is twisted in place or solved, the corners have odd parity. All other states have even parity.

Odd-parity cases cannot be solved directly with commutators, but every quarter-turn of any face changes corner parity between odd and even (180-degree turns like U2 count as 2 quarter turns and do not change parity). This gives us two basic options for solving odd parity cases:

- Turn the top face one quarter-turn, then solve using two commutators.
- Solve a different corner using an odd number of quarter-turns, then solve with one commutator.

This heuristic will solve last-layer corners on any cubic puzzle. If you want to save some moves, you can use an advanced version that starts with 5 corners unsolved.

In this solution, you correct corners to even parity while solving the last edge of the second 1x3x4 block and any two corners at the same time. It's like Heise step 3 but without edges to guide you. This requires a more thorough understanding of parity, which will enable you to determine the parity state of any number or type of pieces.

Summary:

Step 1: Instead of solving two 1x3x4 blocks in step 1, leave out one corner/edge pair to make one 1x3x4 block and one 1x3x3 block.

Step 2a: Make the last edge needed to complete the second 1x3x4 block, and store it in the bottom layer.

Step 2b: Check corner parity

Step 2c: Solve two corners using either an odd or even number of quarter-turns, ensuring that corners will have even parity; insert the edge from step 3a into the solution such that it will be solved as well.

Step 2d: Solve the last 2 or 3 corners with a commutator.

If you know the Heise method for the 3x3, you will know that step 3 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 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 (as in Heise), we can end up with odd corner parity, which is also 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 required 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. We can determine corner parity 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 incorrectly permuted 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.

You can solve the last edge of the second 1x3x4 block during step 2c in three different ways, listed here from easiest to most difficult:

Personally I have found this 5-corner solution to be quite challenging, but extremely fun and interesting. Both the 5-corner and the 4-corner solutions work as a method for solving corners on any cubic puzzle reasonably efficiently. Practicing with a 2x2 can be very helpful.

Summary:

Step 1: Instead of solving two 1x3x4 blocks in step 1, leave out one corner/edge pair to make one 1x3x4 block and one 1x3x3 block.

Step 2a: Make the last edge needed to complete the second 1x3x4 block, and store it in the bottom layer.

Step 2b: Check corner parity

Step 2c: Solve two corners using either an odd or even number of quarter-turns, ensuring that corners will have even parity; insert the edge from step 3a into the solution such that it will be solved as well.

Step 2d: Solve the last 2 or 3 corners with a commutator.

If you know the Heise method for the 3x3, you will know that step 3 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 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 (as in Heise), we can end up with odd corner parity, which is also 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 required 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. We can determine corner parity 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 incorrectly permuted 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.

You can solve the last edge of the second 1x3x4 block during step 2c in three different ways, listed here from easiest to most difficult:

- Keep the edge in the bottom layer, solve all corners, then insert the edge afterwards. This is the easiest and least move-efficient solution - it is often less efficient than the 4 corner solution.
- Pair the edge with the corner that will complete the 1x3x4 block, and then solve that pair while also solving one top-layer corner. This is relatively easy to visualize, but restricts your solving options.
- Keep the edge in the bottom layer until the last few moves of your solution, then pair it with whichever corner will end up in the bottom layer, whether or not that corner is solved. This gives you the freedom to solve any two corners while also solving the last edge efficiently, but it is difficult to visualize because the edge and corner may not match.

Personally I have found this 5-corner solution to be quite challenging, but extremely fun and interesting. Both the 5-corner and the 4-corner solutions work as a method for solving corners on any cubic puzzle reasonably efficiently. Practicing with a 2x2 can be very helpful.

**Step 3: 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 this is the most efficient way I have found:

1. Place a bottom-layer wing edge piece in the top layer on the L or R side, oriented such that the color you want to pair is on the U face.

2. Rotate the slices to join center pieces with the edge, making a bar that goes from the L to the R face - perpendicular to the slice where it will be inserted. Rotate the U layer 180 degrees to shift the edge piece between the L and R side, depending on where the center pieces are located.

3. Once the two center pieces are joined to the edge piece, rotate the U layer 90 degrees to place the three pieces in the correct slice, then rotate that slice to insert the trio.

It's not obvious, but this step sets up the last step to use efficient 4-move commutators.

**Step 4: 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 is more efficient than using commutators for all 10 edge pieces. Instead of pairing adjacent edges, you pair opposite edges in the unsolved slice, then insert them into the top face. The principal is the same as that used in step 4b of the Roux method.

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 edge pieces with one commutator.

There are a lot of different ways to do this, but I like to solve all of the edge pieces except for those in the unsolved slice. It's not guaranteed to end up with only 3 pieces unsolved if you do this, but the odds are favorable - 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. 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 wing 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 do a single quarter turn of the unsolved slice 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 5: 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 and conjugate, similarly to how we construct a pair 3-cycle on a 3x3 cube. This is easiest to see 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.

If you like algorithms, note that all cases in which the last 3 edges are in the same slice can be solved with the same commutator or its mirror. For example, if the unsolved slice is u, the unsolved edges are in the columns FL, FR, and BR, and the FL edge belongs in the FR column, use [L': [F' U' F, u]].

**Step 6: Center commutators**(~16-26 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. Examples of 4-move center commutators include r u2 r' u2 and r2 u2 r'2 u2.

One of the strengths of this method is that step 3 sets up the centers for 4-move commutators in step 6. Because of this, 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.

Once you complete the centers, the cube is solved!

**Some additional questions****Wait, what about PLL and OLL parity?**Since we are solving pieces directly and ensuring even edge parity in step 4, we will never get PLL or OLL edge parity problems. We avoid corner permutation parity problems in step 2.

**Does this work with larger cubes?**Yes! It needs a few modifications for odd-layer cubes, but it certainly works. Adding the middle slice on the 5x5 actually enables us to solve proportionally more of the center pieces before the last step: 65% on the 5x5 vs about 58% on the 4x4. However, the method gets progressively less efficient as the cubes get larger than 5x5, since you always have to leave 5/8 of the wing slice centers unsolved, and there are proportionally more center pieces versus edge pieces as cubes get lager. But hey, at least you don't have to learn any parity algorithms!

Here is the modified method for 5x5:

1. 1x4x4 block and 1x4x5 block around opposite centers.

2. 3/4 of the middle slice, matching the 1x4x5 block (this makes a cool-looking pattern!)

3. Pair (not solve) the last edge of the second 1x4x4 block

4. Solve all middle edges and corners exactly as in Heise steps 3-4

5. 3/4 of a wing slice, creating a 3x4x5 block

6. 7 wing edges, using the unsolved wing slice

7. Commutator to solve last 3 wing edges

8. 3-5 center commutators

I will probably do a detailed tutorial of the 5x5 version of this method at some point, since it is really fun. You get all of the problems and techniques from the 4x4 method, plus more. You also don't need to keep track of corner parity, since you can simply use Heise steps 3 and 4 to solve the corners and middle edge pieces at the same time. You can also use different variations, such as modifying steps 1-3 to orient edges earlier: make two 1x4x4 blocks, solve 1/2 of the middle slice to match these blocks, do Petrus-style middle-edge orientation, extend one 1x4x4 block into a 1x4x5 block while solving the last middle edge piece on the bottom layer, solve another 1/4 of the middle slice to extend it to a 3/4 slice matching the 1x4x5 block, proceed with step 4.

**Can this be adapted for speedsolving?**Maybe, but I can't say for sure because I'm not a speedsolver. You could easily use existing algorithm sets for corners, last edges, and last centers. The method relies heavily on slice moves, though Roux solvers may not mind this. Last center commutators are not known for speed, but the Sandwich method uses them, so there is some precedent for their use in speedsolving. The method that is most similar is Roux, so perhaps a Roux solver could find common ground.

The biggest problem is likely the fact that this method uses several techniques that are not useful elsewhere. It would require lots of specific practice that would not transfer to other methods, and would not leverage other speedcubing skills. I designed the method this way on purpose. I wanted many different challenges that could be solved intuitively in a variety of ways, and were different than the challenges of the 3x3. This is the opposite of what you want in a speedsolving method.

As for my personal times, I can solve the 3x3 in about 3 minutes 1 and a half minutes and I can solve the 4x4 using this method in about 5 3 and a half minutes. My personal best is 3:034:20, which is recorded in my demonstration video. In my demonstration video, which took 4:20, it took me 2:12 to solve first blocks and corners, which is about 4 times longer than a competent solver using the Lewis method, which uses the same steps up to that point. So, if a speedy solver is 4 times faster than I am, they should be able to get to 1 minute pretty easily.

**Just how move-efficient is this method?**For the 4x4, My personal best move count is 105, and the average of my last 5 solves (at least, those in which I counted my moves) is 110. My personal worst, including mistakes, is 137. I have only counted my moves on 5x5 twice, and I came up with 208 and 212. This is an average of 2.14 moves per piece solved, which is only slightly less efficient than my average of 1.96 moves per piece solved on the 4x4. The difference is probably due to the fact that I am just not as good at solving 5x5.

Those might seem like pretty good numbers, but keep in mind that I am not a speed cuber, so I take a really long time to complete these solves. I'm sure that many other methods would become much more efficient if you spent lots of time deliberately trying to be efficient. In my demonstration video, during which I was not trying to be efficient, my move count was 123. But a more competent cuber could probably be much more efficient than me, particularly with blockbuilding. Anyhow, I don't think that this method is amazingly move-efficient, but at least it is not inefficient.

**The End**So, that is my method. If you have any suggestions for improvements I'd love to hear them. Please also let me know if any part of the explanation was unclear or confusing. Thanks for reading!

Last edited: Apr 7, 2019

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