Skew B

This is a cube I don't own, so I cannot show you an image. However, solving this cube was a great exercise and a first step to solve the Skew B ultimate, which I own.

Here an interactive 3D version of the cube. Drag around the canvas to rotate. Tab on corners to make cube movements.

First, let's introduce a notation for all the pieces and faces this cube has.

Cube notation

Before focus on the pieces themselves, let's set a name for the faces and corners of the cube it self.

We will refer to its six faces as

Up
Down
Front
Back
Left
Right

and we will call them U, D, F, B, L and R respectively.

For the eight corners, we will refer to them as a, b, ..., h. Each one is identified as the unique common corner of a triplet of faces. Concretely:

a with (U, B, L).
b with (U, B, R).
c with (U, F, L).
d with (U, B, R).
e with (D, B, L).
f with (D, B, R).
g with (D, F, L).
h with (D, B, R).

Here is an ASCII art diagram:

                .a -------.b
              .'    U   .' |
             c ------- d   |
             |         | R |
             |    F    |  .f
             |         |.'
             g ------- h

Pieces notation

Now let's set a name for the pieces and pieces' faces that compose the puzzle.

There is one center piece for each cube's face, and one piece's face for each center piece. We will denote both by the cube's face they belong. That is: U, D, ..., R.

With respect corners, there are one corner piece for each cube's corner. For each corner piece we will use the same notation as its associated cube's corner: a, b, ..., h.

Each corner piece has three faces. Each corner piece's face lies in one cube's face. We will denote each corner piece's face as the "corner underscored face". That is a_U, a_B, a_L, b_U, ..., h_R.

Here is a diagram:

           .+------------:------------+
         .'  a_U .  ·  '  '.  b_U   .'|
       .' ·  '              '.    .'  |
     .''.          U          '..' b_R|
   .'    '.           _ . · ' .':'.   |
 .' c_U    '.  . · '  d_U   .' .'   '.|
+------------:------------+'   :     :|
|          .' '.          |d_R:     · |
| c_F   .'       '. d_F   |  :  R  :  |
|    .'             '.    | :     ·   |
| .'                   '. |.'     :f_R+
|:           F           :|:     .' .'
| '.                   .' |  ' . '.'
|    '.             .'    | h_R :'
| g_F   '.       .'  h_F  |   .'
|          '. .'          | .'
+------------:------------+'

Permutations

We are using the usual cycle notation. However, we introduce some additional notation for some permutations:

We denote a counter-clockwise rotation of the corner pieces a, b, ..., h by the same name. For example d := (d_U d_F d_R). We will call those permutation corner orientation permutations, or simply orientation permutations.
We use the greek letters (α,β,...,θ) to denote the counter-clockwise rotation of the corresponding half of the cube. We will call these permutation atomic permutations, since they correspond to a single movement of the puzzle.

In the usual cycle notation, they read:

α := a (c_U b_B e_L) (c_F b_R e_D) (c_L b_U e_B) (L U B)
β := b (a_U d_R f_B) (a_L d_F f_D) (a_B d_U f_R) (B U R)
δ := d (b_U c_F h_R) (b_B c_L h_D) (b_R c_U h_F) (R U F)

Here is a diagram of the atomic permutations correspondence with the corners.

                .α -------.β
              .' |      .' |
             γ --+---- δ   |
             |   |     |   |
             |  .ε-----+--.ζ
             |.'       |.'
             η ------- θ

Orbits

As most of the cubes, no all pieces can go to all places in the cube.

The trivial example is no corner pieces can be placed at cube's center spots, or the other way around, no matter the permutations we apply. We express that by saying that the set of corner pieces (and the set of center pieces) are closed by permutations.

An easy but not trivial observation is that the sets

{a, d, f, g}

{b, c, e, h}

are also closed by permutations.

Because each of them consist in exactly four elements, and we all (including the cube) live in three dimensions, just by physically rotating the cube on our hands we can place all the pieces within that set in they correct position. Of course, their faces would be not oriented. Also, this procedure would determine where the other six faces and four corners should be placed.

Solving strategy (Big picture)

We start from a situation where we have complex combinations of orientation permutations and piece's permutations in terms of single atomic permutations. Coming up with an strategy from here is extremely difficult.

On the other hand, in an ideal situation, we would have orientation permutations and really simple piece's permutations in terms of (as complex as needed) combinations of atomic permutations. In this ideal situation, coming up with an strategy would be trivial.

Our procedure to come up with a solving strategy will be to reach some halfway point between the initial and the ideal situation, where we will have simple combinations of orientation permutations and more or less simple piece's permutations in terms of complex atomic permutations.

In this intermediate situation, we will be able to found not a trivial, but a simple enough solving strategy.

Computations

Let us recall some of the atomic permutations:

α := a (c_U b_B e_L) (c_F b_R e_D) (c_L b_U e_B) (L U B)
β := b (a_U d_R f_B) (a_L d_F f_D) (a_B d_U f_R) (B U R)
δ := d (b_U c_F h_R) (b_B c_L h_D) (b_R c_U h_F) (R U F)

The other five could also be described, but they are irrelevant due to the symmetry of the cube. Precisely due to that symmetry there is a multiplicity when studying combinations of atomic movements given by the relative position of the corners. That is: permutations α β and β δ are essentially the same, if you are allowed to physically turn the cube 90 degrees in your hands.

Therefore, when studying combinations of pairs of atomic movements, there are exactly three possibilities: (α, β), (α, δ) and (α, θ). Actually, permutations α and θ act on a disjoint set of pieces and therefore they commute.

So we have only to study combinations from pairs (α, β) and (α, δ).

Alpha - beta pair

We have:

α β = (a_L d_R f_B a_U d_U f_R a_B d_F f_D)
      (b_B e_L c_U b_U e_B c_L b_R e_D c_F) (L R B)

β α = (a_L d_F f_D a_U d_R f_B a_B d_U f_R)
      (b_B e_B c_L b_U e_D c_F b_R e_L c_U) (R L U)

At powering to 3, they result into

(α β)^3 = (β α)^3 = a b c d e f

Because of symmetry

(α β)^3 = (β α)^3 = a b c d e f
(β δ)^3 = (δ β)^3 = a b c d f h
(δ γ)^3 = (γ δ)^3 = a b c d g h
(α γ)^3 = (γ α)^3 = a b c d e g
... 8 more ...

which leads to useful orientations

(α β)^3 (β δ)^6 = (α β)^3 (β^{-1} δ^{-1})^3 = e h^{-1}
(β δ)^3 (δ γ)^6 = f g^{-1}
(δ γ)^3 (γ α)^6 = e^{-1} h
(α γ)^3 (α β)^6 = f^{-1} g
... 8 more ...

Some of the previous permutations.

Another interesting combination is:

α β (β α)^{-1} = a b^{-1} c f^{-1} (L U) (R B)

Alpha - delta pair

If we multiply them we obtain:

α δ = a d c (b_B e_L h_F b_R e_D h_R b_U e_B h_D) (L F R U B)
δ α = a d b (c_U h_F e_D c_F h_R e_B c_L h_D e_L) (R B L U F)

Both elements contains cycles of orders 3, 5 and 9. Let's power them to those powers so we can cancel some elements.

Power three read

(α δ)^3 = b^{-1} e^{-1} h^{-1} (L U F B R)
(δ α)^3 = c^{-1} e^{-1} h^{-1} (R U B F L)

In consequence, power nine read

(α δ)^9 = (L B U R F)
(δ α)^9 = (R F U L B)

So this only permute centers. Actually, we can combine them to obtain a 3-cycle on the centers, instead of a 5-cycle:

(α δ)^9 (δ α)^{-9} = (F U B)

Because of the symmetry of the cube, we deduce the twelve equivalent permutations:

(α δ)^9 (δ α)^{-9} = (F U B)
(α ζ)^9 (ζ α)^{-9} = (B U D)
(α η)^9 (η α)^{-9} = (L U D)
(β δ)^9 (γ β)^{-9} = (L U R)
(β ε)^9 (ε β)^{-9} = (L D R)
(β θ)^9 (θ β)^{-9} = (F R B)
... other six ...

Some of the previous face permutations.

By combining these, we can achieve pairs of face swappings. As an example, we have

(α δ)^9 (δ α)^{-9} (ζ α)^9 (α ζ)^{-9} = (F B) (U D)

In addition to the 3-powered permutations, we can apply 5-powered permutations to remove 5-cycles. If we do so, we obtain:

(α δ)^5 = a^{-1} d^{-1} c^{-1} (b_B h_R e_L b_U h_F e_B b_R h_D e_D)
(δ α)^5 = a^{-1} d^{-1} b^{-1} (c_U e_B h_F c_L e_D h_D c_F e_L h_R)

which may don't look very promising, but with a closer look we realise that can be repeated to obtain this nice expressions for power 15:

(α δ)^15 = b h e
(δ α)^15 = c h e

Observe that by combining these two we obtain

(α δ)^15 (δ α)^{-15} = b c^{-1}

That is: we can orient corners in pairs.

Summarizing, by using this combinations we can:

Orient two or three corners without affecting any other piece.
Cycle among any three faces without affecting any other piece.

Solving strategy

Putting all together, for a solving strategy I propose the following steps:

Choose one closed by permutation corner set as reference. This will determine where all the other pieces should go.
Set the other set's pieces in their corresponding place. I conjecture that the worst case will take 2 movements.

Place center pieces by using combinations of

α β (β α)^{-1}

(α δ)^3

(α δ)^9 (δ α)^{-9}

Orient corners using combinations of
```
(α β)^3 (β^{-1} δ^{-1})^3
```

Summary

Although I don't own this cube, I enjoyed designing a solving strategy. I think the complexity of this cube is in a sweet spot where is straightforward to analyse without being trivial enough to be boring.