Below is a description of how I solve the cube. I hope to write up a bit more about how I came up with this, and figured out all of the algorithms. The intent was to not use a large number of algorithms, but still be moderately efficient at solving (typically around 80 moves).

These are the steps, by name, along with the number of algorithms associated with each. Some of the "algorithms" are a little tough to explain, as they are only one or two turns.

Steps 3, 5, 6, and 8 could clearly be called algorithms, as I have them memorized, even though they are fairly short (the longest algorithm in the entire solution is 7 turns).

Most cube methods involve starting with an aspect of the solution that is fairly visible, typically a cross on one face. As the solutions proceed, the algorithms are more complex so as not to disturb the pieces already in place. There are also quite a number of algorithms to learn (Fridrich involves knowing 57 algorithms, some with as many as 13 steps, followed by 21 algorithms, with as many as 14 steps).

Instead, I have chosen to reduce the cube space instead of visually clear groups, into groups that are easy to understand by restrictions upon the allowed moves. In general, I seek to restrict an opposite pair of faces to half turns. Step 1 of the solution accomplishes this restriction on a single pair of faces.

Throughout this explanation, I will consider the Red and Orange faces to be top and bottom (which in particular is on top isn't particularly relevant, and usually changes throughout a solve). For this initial step, we will be seeking to restrict the further solve to only require 1/2 turns of another pair of faces, in my case, I use the Blue and Green faces.

Until the final two steps of this solution, it will be helpful to think in a slightly colorblind manner. Until step 7 (and as a temporary measure in step 4), we can consider the colors of opposite faces to be equivalent. With a standard cube, this means we will think of red and orange as the same, yellow and white as the same, and blue and green as the same. Because these pairs naturally associate with each other, this is not difficult to do with a standard cube. When solving a cube of non-standard colors, it can be more challenging to remember which faces should be equivalent.

This step is the most difficult to visualize. The end goal is to have flipped all of the edges to a particular parity. That parity has to be defined according to a particular pair of faces, and for this description, I will use the blue and green face.

There are 12 edges on the cube. Start by examining each of the edges adjacent to the 4 non-Blue/Green centers. There will be one of two cases.

Take a few minutes to convince yourself that for the 4 edge pieces where both of these cases apply, the result will always be the same, no matter where the edge is.

After analyzing the 12 edge pieces, you will find that there will be an even number of them that need to be flipped. Start by flipping 4 at a time until either 0 or 2 remain.

To flip 4, manipulate the edges, being careful to not make any 1/4 turns of the blue or green face (but any other turns are fine), positioning these edges along either the blue or the green face. Now, giving that blue or green face a 1/4 turn will change the parity of these four edge pieces. Repeat as long as there are 4 edges that need to be flipped.

(It is usually easiest to just look for 4 edges that need to be flipped, moving them into place, rather than finding all 12 possible ones at a time).

If there are only two edge pieces that need to be flipped, we will need to use our first simple algorithm. Position one of the edges in the upper front position, and the other edge in the upper left edge. "F' L F" will then flip these two faces. Given the simplicity of this algorithm, it is probably faster to just incorporate it, in whatever position is appropriate, when fixing any of the edges.

In summary, we are looking for edges that are flipped, when viewed in respect to the Blue/Green faces. There will be 0-12 that need to be flipped, and they can be flipped 2 or 4 at a time.

The first step is the most difficult to visualize, and also the step that does the least to make the cube look more like it is solved.

For this second step, we will seek to move the various edge pieces into their colorblind appropriate layers. This can be visualized, as the top on bottom face each having a cross made of only red and orange pieces. As long as you are careful to not give the blue or green face a 1/4 turn, these pieces will always remain flipped correctly, and this is just a matter of moving them into position.

Making a 1/4 turn of yellow or white will move 2 edge pieces from the red/orange faces onto the blue/green faces, and move those two pieces from the blue/green faces to the red/orange. There can be 0-4 of these pieces out of position. By holding the cube with green or blue facing you, it should be easy to see how many need to be positioned. Remember that the top faces can be rotated freely. The front, back and sides should be kept to 180 degree turns.

It should be easy to position a group of 4 edges that need to be rotated into one side, and position them with a 1/4 turn of yellow or white.

If there is a single pair that needs to be swapped, position one facing you (by rotating the top in 1/4 turns), and the other in the back/left edge. "L U L'" will swap these pairs. Notice the similarity with the first step, and this algorithm is probably best just understood rather than memorized.

In summary, this step is completed when the red and orange faces each have a cross consisting of only red or orange pieces. In addition, the parity described in the first step should still hold.

In this step, we will twist all of the corner pieces so that the red and orange sides contain only red and orange faces. Aside from the given algorithm, restrict white, yellow, blue and green to only 1/2 turns. This will avoid undoing the previous steps.

There is a single algorithm to this step. Using 1/2 turns of yellow, white, blue, green and 1/4 turns of red and orange, position a pair of cubies such that there is a red/orange face in the LFU cubie, with the red on the L side, and another in the DFR cubie, with the face on the F side. Execute "R' D L2 D' R" to rotate these two cubies. This move also shuffles a few other cubies, but does not rotate any of the corners or edges.

If there is an odd number of corners that need to be twisted, you will get into a situation where the cube cannot be placed in the above configuration. Simply place one of the two cubies correctly, and execute the algorithm. This will fix one of them, and leave a remaining pair. It is also possible that none can be placed in this position. You can either put them in place, but oriented incorrectly, and perform this move, or perform the above setup and moves in a L-R mirror fashion. I tend to just do the extra moves, rather than try to remember the extra algorithm.

When this step is completed, the red and orange faces each should have only red and orange faces. From this point on, we will restrict ourselves to 1/4 turns of orange and red, and 1/2 turns of the other faces (aside from algorithms).

This step is strictly unnecessary, but I've found that without it, determine the next sequence of moves is visually difficult.

Our goal here is to position the corner pieces such that all of the red corners are on the same face, and all of the orange faces on the other. They don't need to match the center piece. Because of our restriction to half turns of the y/w/b/g faces, there should need to be no twisting of the corners.

This step is best solved by reasoning, just being sure to restrict yourself to 1/2 turns on all of the non red/orange faces.

We will look at each pair of corner that aren't the red/orange sides. There are 8 of these pairs. Our goal here is to get all of the pairs to have the same colors, or none of them. If this is the case, you are done.

The algorithm is "R' F R' B2 R F' R".

If you find the cube in a state other than the above, let me know. I don't believe it should be possible.

At this point, we are close to having the cube solvable with only 1/2 turns of any face. The last part of corner shuffling is to give the U or D a quarter turn so that the corner pieces have a colorblind match with the center piece. When that is done, the only pieces that may have a colorblind mismatch with the center pieces would be edges on the r/o/g/b faces.

For this move, only use 1/2 turns of any face to get pieces in position for solving.

This step uses a single algorithm that shuffles 4 edges pieces cyclically around the LFRB faces. There should be an even number of these incorrect edge pieces.

If there are at least 4 edge pieces, position two of them on L and R against the U face, and two of them on F and B against the D face. Then execute "D' M2 D" (M is the middle slice, between L and R).

If you find only two out of place, position on on F on the B edge, and one on R also on the B edge, and execute the algorithm above. You can then do "U2 F2" and rotate the entire cube along U to get into the above position, where the algorithm repeated will get the edges in position.

At the end of this step, the cube should be "solved" according to our colorblind view. Now we have to take our colorblinders off and look at the colors.

For this step, we want to solve the corners fully. This doesn't take very many moves and can always be done with 1/2 turns. It is usually best to pick a corner and put it in the right place, and then solve a few others. You will find putting just a few in the right place will make the rest fall into place, thanks to all of the shuffling we did earlier.

The final stage fixes problems with any remaining edges. Although this can be solved with only 1/2 turns, there is a common situation that is easier to solve with a 1/4 turn move. We will use 3 algorithms at this stage.

It may be easiest to position the pieces in these places by making some "illegal" moves (1/4 turns). As long as you are careful to make sure that the result is indeed only the pieces you want to move, and can remember how to undo the sequence that got you here, this is fine. It isn't necessary to do this, but can take a lot of applications of the above to get the right edges into place.

The one remaining move is when there are three edges that need to be cycled. There will always be one that mismatches on both edges, and two that mismatch just on one edge. Place the dual mismatch piece in FU and the other 2 on the B face, and use "F2 M' F2 M" (M means to rotate the middle slice in the same direction as an "L" turn, so "M'" pushes the top of the slice away from you.

Hopefully, at this point, the cube is solved.