# blueknight cubing

## it makes you faster

### The Square-1... my very favorite puzzle!

Usually when in a scrambled state, this puzzle is not in the shape of a cube. This is a big problem because the corner pieces (or "the big pieces") have a bandaged effect on the puzzle. Basically this means that depending on which state the puzzle is in, your algorithms usually ain't gonna work. For the sake of simplicity, the first step is to bring puzzle back into a cube shape. It will remain in the state before and after every algorithm you use.

Once your square-1 is in a cube shape, you can start solving it with algorithms. Firstly, you will orient the corners and edges. Secondly, you will permute the corners and edges. Lastly, there may be a difficult to recognize parity case. It sucks ass, and so does the algorithm to solve it. Sorry!

Interpreting the images- In keeping with my awesomeness, all the images on this page were created by me. They are easily interpreted. On the square-1, you are always concerned with both layers at once. The square on the left represents the top layer as seen from the top. The square on the right represents the bottom layer as seen from the top. Always make sure to line up both layers before using your algorithms. :)

Always keep the small side of the middle layer on the left. This means that the "flip line" will be pointing toward your left shoulder at all times. This is a much more comfortable way to solve than if you were to hold it the other way.

Regarding the color scheme: for orientation images I used the standard white top green bottom scheme (this is the color scheme that was on all the original square-1s). For the permutation images, however, the original color scheme can be a bit confusing, so I will use the 3x3 colors to make things a bit easier to recognize.

Now that you have an idea of the steps involved in solving a square 1, start by learning notation!

### Square 1 Notation

Unlike the Rubik's cube, notation is complicated on the square-1. It is based on turns and flips. Turns affect the U and D layers, and they look like ordered pairs (3,2). Flips can be compared to R2 on a 3x3, and they are represented by a forward slash /. When you see an algorithm it will be in the format turn, flip, turn, flip, ect. Here's an example: /(3,-3)/(-3,0)/(0,3)/(0,-3)/(0,3)/(-3,0)

To understand the numbers, you have to think in terms of number of edges. The first number tells you which way to spin the top layer, and the second number tells you which way to turn the bottom layer. (3,-3) means to turn the top layer clockwise until 3 edges have crossed over the flip line, and to turn the bottom layer anticlockwise until 3 edges have crossed over the flip line. Corners are worth two, because they are twice as wide as an edge. If the puzzle is in a square shape, you can pretend that 3 is a quarter turn U or D like on a Rubik's cube. Also, 6 can be thought of as a half turn (like U2 or D2).More often than not, though, you'll get an algorithm with 1s, 2s, and 4s in it, and this just takes practice to get used to.

For now, why don't you try to do this algorithm to your solved square 1:

/(3,3)/(1,0)/(-2,-2)/(2,0)/(2,2)/(-1,0)/(-3,-3)/(-2,0)/(3,3)/(3,0)/(-1,-1)/(-3,0)/(1,1)/(-4,-3)

That should swap two edges in the top layer. Do it again to solve.

If you mess up and do the wrong turn during an algorithm, it will probably not be possible to complete the algorithm. This happens lots as a beginner, so don't let it bother you. Try again. As a beginner, your best notation practice is to follow scrambles from CCT instead of scrambling by hand. This takes forever, but it's an excellent way to get used to notation, so do it.

### Making a Square/Cube Shape

The first step for solving a square-1 is to bring it into a square/cube shape. On a normal Rubik's Cube (3x3), the edges and corners have different properties, and behave differently (for example, you can cycle edges with 2 gen, but never corners). On a Square-1, the corners behave exactly the same as edges, but because of the bandaged nature of the corners most algorithms won't work if the puzzle is not first in the shape of a cube. Before and after every algorithm in the solve, the cube must be in this shape.

If your square-1 is scrambled, you are probably having trouble finding a place to turn it. Since the corners are simply 2 edges bandaged together, the puzzle becomes very easy to turn once all the single edges are in pairs. The easiest way to complete this step is to bring all of the edges into pairs, and then bring all the pairs into a massive semi-circular block of edges.

As for pairing single edges: There are two main shapes that unpaired edges like to hang out in. These shapes are a bar and an L. My basic rule is that a pair is better than a bar, and a bar is better than an L.

How to pair single edges:

• All unpaired edges will exist in one of two shapes. Ls and Bars.
• An L is the worst shape to have. You don't want edges to exist in L shapes.
• A bar is better than an L, but a bar is still not as good as a pair.
• 1. You can turn an L into a bar by attaching it onto a group of 4 edges.
• 2. You can turn two Ls into a bar and a pair by simply attaching them.
• 3. You can turn two bars into two pairs by simply attaching them.
• 4. You can attach a single bar to a group of six to make your group of eight.
• It is your goal to make a group of eight.
• Once you make a pair (of 2 edges) you will never have to split it into single edges again.   Once you have all of the edges into a large chunk in the top (We call this the "tulip" shape), place the tulip so that the two corners are facing toward you, and the flip line is between them. Then do this algorithm:

To solve cube shape from tulip-on-top:

/(2,4)/(1,2)/(-3,-3)/

### Orientation of Corners (CO)

The orientation steps can be started only after both layers of the puzzle are square-shaped. Even for beginners, the Orientation of Corners never takes more than five seconds. This is a very simple step. In order to understand this step, however, you must first understand the method for perserving the shape of the square-1. This step can always be done in 2 'flips' or less. Always make sure to have the two layers mis-aligned before doing a flip. This may feel wrong at first, but it's the only way to preserve the square shape of the layers.  Keeping this in mind, you can treat the puzzle as if it were a 2x2 in order to solve this step.

However, because some people prefer an algorithmic approach, I include it here: (1,0)/ (1,0)/(0,3)/ (0,-1)/(-3,-3)/ (1,0)/(-3,6)/ (1,0)/(6,6)/

### Orientation of Edges (EO)

For this step, there are only several cases. It would be wise to try and learn all of these cases. The second case in the first column seems to be the most common, so make sure you know it. (1,0)/(0,-3)/(0,-3)/(-1,-1)/(1,4)/(0,3)/ (1,0)/(0,-3)/(0,-3)/(-1,-1)/(0,3)/(0,3)/ (0,2)/(0,3)/(1,1)/(-1,-4)/ (1,0)/(-1,-1)/(3,3)/(1,1)/ (1,0)/(-1,-1)/ There are two more cases,  but I don't know them yet.

### Permutation of Corners (CP)

The permutation of corners is also a quickly performed step in which you will be solving (ideally) all corners in both layers with just one algorithm. This can be thought of as similar to XLL in the 2x2 Ortega method.

The most common case for this is the "double J-perm" (the first alg in the second column), which is very nice because it's so quick. The ones that look like N-perms seem relatively rare. /(3,-3)/(-3,0)/(0,3)/(0,-3)/(0,3)/(-3,0) /(-3,0)/(3,3)/(0,-3)/ /(3,-3)/(0,3)/(-3,0)/(3,0)/(-3,0)/(0,3) /(3,-3)/(-3,3)/ *3 *2 / *3 *2 /

Early in my practicing, I had myself learn all of the CP cases. This seems like the most improvement for the time you put into it. I would recommend doing the same.

### Edge Permutation (EP)

Edge permutation is by far the longest step. Since there are an almost impossible number of cases for this step, most people will solve it using only a few algorithms. This is very difficult to do quickly.

The method I use for this step is to try and change both layers into a recognizable case, particularly one that corresponds to the EP cases on the 3x3. You can do this by applying a adjacent-adjacent if both layers have parity, or using a parity alg if only one layer has parity. This will fix both layers so that they look like 3x3 cases.

Note for beginners: you can solve the whole EP step using just the adjacent-adjacent swap and any parity algorithm. Adjacent-Adjacent Edge Swap(-2,0) / (3,0) / (-1,-1) / (-2,1) / (2,0) Opposite-Opposite Edge Swap(1,0) / (-1,-1) / (6,0) / (1,1) / (5,0)

3x3 lookin' algs: Ua Perm for Top Layer:/ (3,0) / (1,0) / (0,-3) / (-1,0) / (-3,0) / (1,0) / (0,3) / (-1,0) Ub Perm for Top Layer (1,0) / (0,-3) / (-1,0) / (3,0) / (1,0) / (0,3) / (-1,0) / (-3,0) / H Perm for Top Layer: / (3,-3) / (3,-3) / (0,1) / (-3,3) / (-3,3) / (-1,0) Z Perm for Top Layer: (1,0) / (-1,-1) / (3,0) / (1,1) / (-3,0) / (-1,-1) / (0,1) Ua Perm for Bottom Layer: / (0,-3) / (0,-1) / (3,0) / (0,1) / (0,3) / (0,-1) / (-3,0) / (0,1) Ub Perm for Bottom Layer (0,-1) / (3,0) / (0,1) / (0,-3) / (0,-1) / (-3,0) / (0,1) / (0,3) / H Perm for Bottom Layer:/ (3,-3) / (3,-3) / (-1,0) / (-3,3) / (-3,3) / (0,1) Z Perm for Bottom Layer: (1,0) / (-1,-1) / (0,-3) / (1,1) / (0,3) / (-1,-1) / (0,1)

And of course are an enormous amount of combinations between top and bottom layer.

### Permutation Parity, and Middle Layer Fix

On most puzzles, parity refers to a difficult 2-cycle. The Square-1 is no exception to this. On the square-1, the odds of encountering a parity case are fifty-fifty. Parity involves a very long and difficult to learn algorithm. It is difficult to perform this algorithm in less than 10 seconds, so parity usually destroys any hope of a 'good' solve unless you are very skilled at performing the algorithm quickly.

Also, once parity is fixed, there is another 50% chance that the middle layer will be irregular -- in contrast to everything else on the square 1, this is a very easy fix. You should always fix the middle after everything else. Since some algos will change the middle layer, it is not wise to fix this early in the solve. /(3,3)/(1,0)/(-2,-2)/(2,0)/(2,2)/(-1,0)/(-3,-3)/(-2,0)/(3,3)/(3,0)/(-1,-1)/(-3,0)/(1,1)/(-4,-3)

**Also if your bottom layer is solved, and your top layer is a mess but you know it's parity, use this shorter algo:**

/(3,3)/(1,0)/(-2,-2)/(2,0)/(2,2)/(-1,0)/(-3,-3)/(-2,0)/(2,2)/(0,-2) AUF /(-3,0)/(0,3)/(0,-3)/(0,3)/(2,0)/(0,2)/(-2,0)/(4,0)/(0,-2)/(0,2)/(-1,4)/(0,-3)/(0,3) /(6,0)/(6,0)/(6,0)