*Illustrations will follow some time later...*

There is a basic strategy, how to solve *any* higher order
"puzzle cube" - or whatever you like to call them
without using brand names or trademarks.
If you only turn the outer faces of such a cube, it behaves
*exactly* like a 3×3×3 cube. The individual edge pieces and the
inner face pieces will not get scrambled.
This leads to the strategy of solving the inner face pieces,
and then collecting all matching inner edge pieces together.
After that, it's just a "simple" 3×3×3 solve.
The corner pieces are completely out of scope during
pre-solving. For cubes of two or more layers,
there are always exactly eight corner pieces,
and these are part of the final solve.

The same methods can be used on any higher order cube, it's just a matter of which of the inner layers you turn, the patterns of the moves are always the same. Special situations can occur on cubes with an even number of layers, where the final 3×3×3 solve seems to be unsolvable. Here you need to know two special sequences to resolve these situations.

1. Solve the Inner Face Pieces

2. Collect the Inner Edge Pieces

3. Solve the 3×3×3 (not explained here)

4. Solve Special Situations on Even-Layer Cubes

Summary of Algorithms

The notation might look a bit strange in some cases, but the motivation is to indicate the "freedom" of turning specific sets of inner layers. Because center layer turns are sometimes confusing, I try to use a more intuitive way.

L R F B U D for outer faces: Left, Right, Front, Back, Up, Down.

Regular turns: L 90° clockwise (when looking at the face), L' 90° counter-clockwise, L2 twice (180°).

l r f b u d for any inner Left, Right, Front, Back, Up, Down layer.

r* for all inner layers of one half

r^{+} for any number of inner layers

H V M for Horizontal, Vertical, Middle center layers

HL HR ML MR turned left/right, VU VD turned up/down

( ) brackets to group sequences

{ } curly braces for optional turns

[ ] Coordinates (location) of pieces; uppercase for outer faces,
lowercase for inner layers

The basic strategy for solving the inner face pieces (here simply called "face") is to start with any face, then to build the opposite face, and then any other face and one of its adjacent faces. As long as there are two or more "free" faces, the individual face pieces can easily be assembled there, and then moved to the face(s) in progress. For the last two faces, the inner pieces have to be swapped between these two faces.

The process for building each of the inner faces is to build stripes of matching pieces on the unsolved faces, and to insert the solved stripes to a matching place. Inserting a vertical stripe from the right half of the front face to the respecting left half of the upper face can simply be done with ( r U2 r' ). A special case is the middle stripe on odd-layered cubes, where you collect all pieces except the center. Then, if you hold the destination layer on top, with the unmatched pieces running front to back: Put the incomplete stripe in horizontal orientation to the front face, use ( VD F' VU ) to insert the pieces.

There is a certain freedom, which inner face piece can go where. On a 3×3×3 cube, a center piece has four different orientations. Any of the inner face pieces of a higher order cube can take one of four places on one face.

When there are only two faces left, the method of swapping the pieces is not obvious. But you only have to learn one sequence that will cycle three pieces across two faces. To cycle three inner face pieces [Fur] - [Ubr] - [Ful] use the sequence ( r U' l' U r' U' l {U} ). The last upper layer turn is optional, it just restores the orientation of the upper layer.

Actually, this sequence can cycle any rectangular area,
including the middle layer pieces on one of the sides.
On a 7×7×7 with five inner layers, for example,
you can cycle an area of 2×3 or 3×2 inner face pieces.
The general notation would be
( r^{m} U' l'^{n} U r'^{m} U' l^{n} {U} ).
The number m of right layers is the width, the number n of left layers is
the height of the area on the upper right of the front and upper face.
The area on the upper left of the front layer is rotated 90° left.
The whole truth is even more complicated: The cycled pieces are the
ones on the intersections of the right layers turned,
and the respective horizontal layers for the left layers turned.
But that covers only very rare cases on cubes with seven or more layers.

Swapping larger areas can even be helpful, if the area contains a few non-matching pieces. Swapping these back can be quicker than swapping all matching pieces one at a time. On an even layered cube, you can "swap" the full inner quarters [Fu*r*] and [Ul*f*] simply with ( Rr* U' Rr*' ).

Collecting the inner edge pieces can easily be done for eight of the twelve edges. The remaining four edges require cycling of edge pieces.

While placing assembled edges in any location of the top and bottom layer, the horizontal layers can freely be moved. All you have to do is to swap edge pieces from and to the upper and lower face by moving the outer faces forth and back, so that the stripes keep their original horizontal position. To move [FR] and [FU] edge pieces, use: ( R U' R' {U} ), to move [FL] and [FU] edge pieces, use: ( L' U L {U'} ). The last upper turn is obsolete, it just restores the orientation of the upper face. Actually, the sequences cycle three edges, but at this stage of solving, the location of the edges in the upper face is not important.

For collecting the individual edge pieces, it's worth knowing that each piece has its dedicated place across the edge. If you look at an incomplete vertical edge, you just have to memorize the color of one side, and then swap matching edge pieces to any other vertical edge in the same orientation. Then turn the horizontal inner layers to combine the matching edge pieces.

If one edge holds edge pieces of different orientation, you have to turn the horizontal layers to separate the pieces of different orientation. Then swap the edge forth and back in proper orientation. If the pieces are in the [FL] and [FR] vertical edge, and an incomplete edge is in [FU] position, use ( R U' R2 U R ) to bring the [FR] edge to [BR] with the pieces in proper orientation. The [FU] edge will go to [RF] position, and the [RB] to [FU], but at this stage of solving, it does not matter.

When all eight edges are collected and placed in the top and bottom layer, you just have to turn the horizontal middle layers back to restore the inner faces. Then you have to learn one sequence that will cycle three inner edge pieces across three edges. Sometimes, only two swapped edge pieces are left. Then you either need one easy to remeber sequence to permute more edge pieces, or one more complicated sequence to swap exactly those two pieces.

To cycle three inner edge pieces between [FUr] - [UBr] - [RUb] use the sequence: ( r B' R B r' {R'} ). Actually, this sequence works for any number of slices on the right half, including the middle layer of odd-layered cubes. Looking at the top layer, the piece from the back face will appear to be flipped on the right face, the piece from the right face will appear unflipped on the front face. This means that you can often fix two locations in one go: Think of moving a piece from [FUr] to [UBr], then find the edge where the [UBr] piece should go, or the edge with the matching [FUr] piece. Bring that edge in proper orientation to [RU].

If there are only two pieces left, use ( ( r2 U2 ) ×4 ) for the layer where the two pieces are located to get more scrambled edge pieces. If you put the two pieces to [FUl] and [BUl] positions, use ( ( r U2 ) ×2 F2 r F2 l' U2 l U2 r2 ) to resolve this situation in one go.

Having only four edges to complete does not sound much, but in worst case, that makes eight wing parts to fix. On a 7×7×7, there are two pieces in each wing part, a total of 16 pieces. One strategy is to start with edges where both wings do not match, because you can only cycle pieces across edges. The next point is to look out for "flipped" pieces, that actually must go to the opposite wing part. You can either find the proper piece to cycle the wrong one to another edge, or intentionally put the proper piece to the opposite wing, and then use the wing flip explained later in step 4.3. But on a 13×13×13, that's a total of 40 pieces to deal with. Here you can try to focus on matching pairs or triplets, and then to cycle these to their final location.

There are different strategies on how to solve a 3×3×3 cube. Although there are "all edges first" and "all corners first" strategies, the common strategy is to solve the cube layer by layer.

Solving the first layer (face and side) can be done straight-forward. For the next layer, only one algorithm (and its mirrored counterpart) for inserting one edge piece from the bottom layer to the middle layer is needed.

A faster strategy is to solve the first layer edges, called "Cross", and then each corner piece together with the matching middle edge piece. This is called F2L "First Two Layers" solving.

For the last layer, there are many different strategies. The main difference between these is the number and complexity of algorithms to learn.

The simple solving schemes use simple operations that have to be done over and over again. There are edges first and corners first schemes, with permutation and orientation for each step. The solution by David Singmaster is edges first, while the solution by Josef Trajber is corners first.

More sophisticated solving schemes have a larger number of more complicated algorithms. The most famous solving scheme is the "Fridrich" method by Jessica Fridrich. The strategy is to solve the orientation of the last layer pieces first (OLL), and then to solve the permutation of the last layer pieces (PLL). Each of these two tasks can be divided in substeps, but with a large number of algorithms, OLL and PLL can be done with only one appropriate alogorithm for each step. This is called "Full-Fridrich".

Solving the last layer in one go would require to recognize 1212 different states, and to learn the respective algorithms for solving these states. If you want to know more about this, search for "uppertable.zip".

It's up to you to find out which solution suits you best. Good Luck!

A situation that can be avoided in step 1. are swapped inner faces. This situation will be obvious when trying to solve the first layer, because the corner pieces won't match. The other two situations only become obvious during the last layer solve, depending on which solving strategy is used. If you solve the cube as far as possible, with priority for corner pieces, swapped and/or flipped edges will remain. If your solving scheme solves the last layer edges before the corners, you have to try edge piece operations to get solvable corners. If you can't find the appropriate step in your solving scheme, it could be any combination of swapped edges and/or flipped edge piece. The "unsolvable" cube must be solvable in one of these three states: Swap any two edges, flip one edge, swap back the edges.

Because cubes with an even number of layers have no fixed center pieces,
you have to take care that you arrange the inner faces properly.
Without knowledge of the color scheme,
each corner piece indicates the relative location of
three of the inner faces. For two corner pieces showing two common
colors, the other colors indicate opposing faces.

To swap front and upper inner faces, use ( r* U2 r*' l'* U2 l* ). To swap opposite faces, e.g. up and down, use ( Rr*2 U2 D2 Rr*2 ).

The first "unsolvable" situation are swapped edges, where there is no matching algorithm in your solving scheme. There is a simple "universal" algorithm that will swap two opposite edges. You can simply re-arrange the cube to get any two "misplaced" edges at opposite positions, apply the algorithm, and then reverse the arrangement. To swap [UF] and [UB] edges, use the sequence ( u*2 r*2 Uu*2 r*2 U2 r*2 ). This sequence is of no real use on odd-layered cubes, because it will swap inner face front and back pieces because of asymetry in u and r slice turns.

The second "unsolvable" situation is an inner edge
that looks like a flipped edge piece. Actually, these are
swapped left/right inner edge pieces. On a cube with an odd
number of layers, this does not happen, because the center
edge pieces shows the correct orientation of the other edge pieces.
There is a "universal" algorithm, which can also
solve swapped left/right edge parts on odd layer cubes, that
are very uncomfortable to resolve with simple edge piece cycling.
To "flip" the complete [FU] edge, use the sequence
( r*2 B2 U2 l* U2 r*' U2 r* U2 F2 r* F2 l*' B2 r*2 ).
If you turn matching left and right slices, this will swap
the respective left and right inner edge pieces.
( r^{+}2 B2 U2 l^{+} U2 r^{+}' U2 r^{+} U2 F2 r^{+} F2 l^{+}' B2 r^{+}2 ).
This works on odd-layered cubes as well.

This is a summary of all non-trivial algorithms needed on top of a 3×3×3 solve to solve any higher order cube. The algorithms are in simplified notation. For more detail, see the respective sections above.

- 1. Cycle three inner face pieces: [Fur] [Urb] [Ful]
- r U' l' U r' U' l {U}
- 2. a) Cycle three inner edge pieces: [FUr] [BUr] [RUb]
- r B' R B r' {R'}
- 2. b) Swap [FUl] [BUl] inner edge piece
- ( r U2 ) ×2 F2 r F2 l' U2 l U2 r2
- 4. a) Swap [FU] [BU] edge pieces
- u*2 r*2 Uu*2 r*2 U2 r*2
- 4. b) "Flip" [FU] edge pieces (swap [FUl] [FUr] pieces)
- r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2

Yes, that's it! Only five algorithms, where 2b and 4b need a lot of practice...