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Solution: Here is my current 3×3×3 solution technique,

derived mainly from the algorithms I developed and learned from friends in 1981 and 1982.

1. Get the first layer. I used to do this by the intuitive technique, but I now start with the white cross just as speedsolvers do, because I hope to teach myself some of Fridrich-style F2L algorithms soon.

1b. Position edges on the middle layer. I sometimes flip a pair as well.

2. Position and orient the other 4 corners. I hold the now-complete F2L facing up, and use the corner algorithms from my 2×2×2 technique (see above), all of which move edges on the D layer only.

2a. To cycle (DRF → DLF → DLB), use F'RF' L^{2} FR'F'
L^{2} F^{2} (9f)

2b. To exchange the corners at DFL and DFR, use L^{2} B D
B' L^{2} F U' F U F^{2} (10f)

2c. To exchange the corners at DRF and DLB, use L^{2} U' R
F' R' U L^{2} D B D B' (11f)

2c. To rotate the corners at DRF, DRB and DLB, use R'D'RD'
R'D^{2}RD^{2} (8f)

2d. If there are exactly two corners rotated properly, I position them and a third corner at DRF, DRB and DLB, and use the previous algorithm once or twice until there is exactly one corner oriented properly, then turn D until that corner is at DFL and use the algorithm again one or two more times.

3. Position the remaining edges.
3a. To cycle (UF, FR, FL), use R_{2}F R_{2}'F^{2} R_{2}F

R_{2}' (7s)

3b. To swap two pairs (UR, DR) and (UF, DF), use
(R^{2}F^{2})^{3} (6f).

4. There may be one or two pairs of edges to flip.

4a. To flip edges at DF and DR, use
R_{2}'D^{2}R_{2}D'R_{2}'D R_{2}D^{2} R_{2}'D^{2}R_{2}D
R_{2}'D'R_{2}D^{2} (16s).

4b. To flip edges at UF and UB, use (R_{2}U)^{2}
R_{2}U^{2} (R_{2}'U)^{2} R_{2}'U^{2} (12s)

## The Centerless 3×3×3 "Void Cube"

The Void Cube is a 3×3×3 with no center pieces or core — it has actual physical holes going right through the middle in all three dimensions. It works just like a 3×3×3, due to an ingeneous mechanism that is even more surprising to those who understand how a normal 3×3×3 mechanism works.

The void cube can be emulated on a 3×3×3 simply by covering the centers with black:

*Void Cube emulated by a 3×3×3*

As shown here, apparent parity errors can arise on the void cube: the entire cube is solved, except for a single swapped pair of corners. A single swapped pair of edges is also possible.

Number of Combinations: Assuming a single corner as a fixed
reference, the corners alone have the same number of combinations as
the 2×2×2, which is 7!×3^{6} = 3674160. Because any
center slice turn creates an odd permutation of edges, the edges can
be permuted into any of the 12! combinations. As on the normal 3×3×3
cube, they can also be flipped in pairs, giving 12!×2^{11} =
980995276800. Combined, the Void Cube has 7!×3^{6} × 12!×2^{11}
= 3604333606207488000 ≈ 3.6×10^{18} combinations.

Solution: I use corners as the orientation reference when solving a Void cube.

1. Get the first 4 corners in place using normal intuitive "first layer" techniques.

1a. (You might also choose to get the first layer edges too, as it does no harm to do them now.)

2. Solve the other four corners using the algorithms from the 2×2×2 above.

3. Solve the rest of the edges as far as possible.

3a. I use the edge movement algorithms shown in the following discussion of the 3×3×3.

3b. Try to notice a pair-swap parity error as soon as possible. It is best to try to notice this when there are 4 left to put in place, then check if the 4 edges are out of place in a cycle of 4, rather than as two swapped pairs. If this happens, there will definitely be a parity problem. If you reach a point where there are exactly 3 edges out of place, then there will definitely not be a parity problem.

4. Fix edge parity, if any. This is done by making any single
slice turn: L_{2}, R_{2}, F_{2}, B_{2}, U_{2}, or D_{2}.
If possible, choose one that will cause one of the edges to be moved
to where it belongs. The rest can then be solved with one or two
3-cycles or a double-pair swap.

## 3×3×4

As of summer 2009, the 3×3×4 is in commercial production; it can also be emulated easily by a bandaged 4×4×4:

*A real 3×3×4 can be emulated by a 4×4×4*

Watch Tony Fisher's 3×3×4; another video here. There are photos here.

Here is another that also shows part of the internal mechanism. This video shows a 3×3×4 being assembled from a kit.

Since this cube has two odd-numbered axes, a full six-face
checkerboard pattern is possible (so long as you do the two odd-number
axes first: F_{2}^{2}R_{2}^{2}U_{24}^{2} works, but
F_{2}^{2}U_{24}^{2}R_{2}^{2} and U_{24}^{2}F_{2}^{2}R_{2}^{2} put
stripes on two and four faces respectively)

The corners of this puzzle are equivalent to the
permutation-only 2×2×2. Like the 2×2×3, it
cannot be turned further after one of the side faces (L, F, R or
B) is turned 90^{o}:

*The 3×3×4 in a locked position*

This cube has eight single edge pieces, located on the 3×3 faces, and four pairs of double-edges that behave like the edges of the 4×4×4, located on the 4×4 faces. The algorithm for flipping a double-edge on the 3×3×4 is also the best known algorithm for erforming the same task on a 4×4×4.

This puzzle can emulate the 2×3×3 by bandaging the 1^{st}
layer to the 2^{nd} and the 3^{rd} to the 4^{th}.

Notation: This puzzle has the ability to emulate a normal Rubik's
Cube with the restriction that four faces can only be turned by
180^{o} turns. When emulating a 3×3×3 this way, the 3×3×4 is oriented
with the 3×3 faces Up and Down, and the four 3×4 faces as the
Left, Front, Right and Back faces.

In addition to the previously used notation for face turns, I use
two more turns for the middle two layers. If the U layer is the 1^{st}
layer and the D layer is the 4^{th} layer, then U_{2} means turning
the 2^{nd} layer in the same direction as U, and D_{2} means
turning the 3^{rd} layer in the same direction as D. The 2 subscript
refers to the 2^{nd} layer away from the face that is named by the
letter, so U_{2} means Up, layer 2. We can also use a 3 subscript, to
refer to the 3^{rd} layer away from the face, and so U_{3} means the
same as D_{2}'.

Pieces on the U and D layers can be named the same way as on the
3×3×3 cube, but the pieces on the other two layers need more unique
names. There are two face-center pieces on the F face, named F_{U} and
F_{D}. The two edge pieces between the F and R faces are FR_{U} and
FR_{D}. They can also be referred to as RF_{U} and RF_{D}. If an edge
piece moves from one location to the other, the name can be chosen to
make it clear the piece has also flipped: for example, if FR_{U} and
FR_{D} exchange places, we say that FR_{U} moved to RF_{D} and FR_{D}
moved to RF_{U}.

Number of Combinations: If bandaged like a 2×3×3, the puzzle can be
put in the same number of combinations as that puzzle, which is
8!×8!/4 = 406425600. The inner two layers have distinguishable edge
pieces (corresponding to the corners on an inner 2×3×3) and 4 pairs of
indistinguishable centers. In addition, parity of the 2^{nd} and
3^{rd} layer edge pieces is linked to that of the corners, so it is
impossible to get an odd permutation of one without having an odd
permutation of the other. The same relationship holds true for the
centers and the 1^{st} and 4^{th} layer edge pieces, however the
ability to make an invisible swap of center pieces nullifies this
restriction. (These points is is described near the end of
this assembly video.) This makes the number of
combinations (8!×8!)/(4×4×2) × 8!×8!/2^{4} =
5161930260480000 ≈ 5.16×10^{15}.

Solution: Here is the solution I came up with right after gettng my 3×3×4:

1. Pair up the center edge pieces into dedges.

1a. Use R^{2}, F^{2} and/or U_{2} to get at least one
pair matched (you can usually get two).

1b. Use L^{2}, F^{2}, and/or R^{2} to get two edges
that need to be matched into the FL and FR columns.

To join the pair at FL_{U} and FR_{D}, and also join
the pair at FL_{D} and FR_{U}, use R^{2} D_{2} R^{2} D_{2}'
(4s)

To join the pair at FL_{U} and FR_{U}, and also join
the pair at FL_{D} and FR_{D}, use U_{2} F^{2} D_{2} B^{2} D_{2}'
F^{2} D_{2} B^{2} D_{2}' U_{2}' (10s)

2. Solve the dedges and corners as on a 2×2×3.

2a. You might be left with one or more dedges in position but flipped:

Double dedge flip: To flip the dedges at FL and FR,
use R^{2} D_{2}^{2} U_{2} F^{2} U_{2}' D_{2}' F^{2} D_{2}' R^{2}
(9s)

To swap the two edge pieces at FR_{U} and FR_{D},
use U_{2} R^{2} U_{2} F^{2} U_{2}^{2} R^{2} U_{2} R^{2} D_{2}^{2}
F^{2} U_{2}' R^{2} D_{2} (13s)

3. If necessary, match the U and D centers with the corners by doing
F_{2}^{2} or R_{2}^{2}.

4. Solve the F,R,B and L centers.

To exchange both of the F centers with both of the B
centers, use (D_{2}^{2}R_{2}^{2})^{2}) (4s)

To 3-cycle (F_{D} → F_{U} → R_{U}), use D_{2}' F_{2}^{2} D_{2}
F^{2} D_{2}' F_{2}^{2} D_{2} F^{2} (8s)

5. Solve the U and D edge pieces.

To exchange (UL, DL) and (UR, DR), use (R^{2}F_{2}^{2})^{2} (4s)

To exchange (UF, UB) and (DF, DB), use (D^{2}F_{2}^{2})^{2} (4s)

To exchange (UF, DF) and (UR, DR), use (R^{2}F^{2})^{3} (6s)

To 3-cycle (UR → UF → UL), use F^{2} U' R^{2}F^{2}R^{2} U^{2}
R^{2}F^{2}R^{2} U' F^{2} (11s)

To 3-cycle (UR → FD → DR), use (R^{2}DR^{2}D')^{5} (20s)

Single pair swap: To exchange UL and UR, use F^{2}
D_{2}^{2} R^{2} F^{2} D_{2}^{2} F^{2} R^{2} U^{2} D_{2}^{2} F^{2}
U^{2} F^{2} U^{2} (13s)

Other Algorithms

Pure dedge flip: to swap the two edge pieces at FR_{U} and FR_{D}
without affecting the rest of the cube, use D_{2}' R^{2} D_{2}
R^{2} U_{2}' R^{2} U_{2}^{2} F^{2} D_{2} F^{2} U_{2}' R^{2} F^{2}
D_{2}^{2} F^{2} (15s). (This algorithm is also applicable to the
4×4×4.)

## 3×3×5

*5×5×5 emulating a 3×3×5*

Here is a video of Tony Fisher's 3×3×5 in action. here are some photos. It is made from an Eastsheen 5×5×5 cube core, bandaged into a 3×3×5 (using the 5→2+1+2→3 bandaging on two axes, like that pictured here). The result has 9+8+8+8+9=42 externally visible pieces. This bandaging and the design of the Eastsheen 5×5×5 allow the edge and corner pieces to be cut fairly deeply, allowing for a modification like that used for the 2×2×3.

The 3×3×5 is the smallest three-axis shapeshifting puzzle with odd dimensions.

The surprising thing with this puzzle is that, when rotated 90^{o}
into a non-rectangular shape, the central three slices still rotate
and the puzzle behaves like a 3×3×3 with extra cubical protrusions.
This makes the puzzle appear to be just a 3×3×3 with extra cubes glued
on, until you return it to the pure rectangular shape and discover
that now the 1^{st} and 5^{th} layers turn too.

This behavior is duplicated exactly by the bandaged version shown here. Since none of the bandages connect to any central slice, the central three layers can always rotate with respect to each other.

*bandaged cube in a pattern corresponding to a shapeshifted form of the real 3×3×5*

A real cube with proportions similar to the bandaged 5×5×5 is pictured here.

Therefore, by necessity, solving it involves first solving it as if it
were a 3×3×3, then dealing with the 1^{st} and 5^{th} layers. As seen
in his solving video, some of Tony Fisher's
algorithms for restoring these layers involve temporarily putting the
cube into an odd shape.

Here is another 3×3×5 construction; according to
the author this one was made from 3×3×3 cubes, presumably adding the
1^{st} and 5^{th} layers by a technique similar to the
2×2×4. It is also able to shape-shift after 90^{o} turns.

This is the first in a series (not yet complete as of this writing) on how to build one.

Number of Combinations : As with the 2×2×4, we can
consider the outermost layers (the 1^{st} and 5^{th} layers) to be
decorations on a 3×3×3, with 8!×8! combinations (this differs from
the 2×3×3 because different orientations can be
distinguished by using the face centers as a reference). The central
three layers are like a 3×3×3, however 4 pairs of "edge" pieces are
now center pieces that are indistinguishable from each other. These
pieces do permit making what appears to be an odd permutation of the
corners and the other (normal) edges. So the number of combinations
for those 3 layers is 8!×3^{7}×12!×2^{11}/2^{4}. This gives
(8!)^{3}×3^{7}×12!×2^{11}/2^{4} = 8789360691018252209356800000
≈ 8.7894×10^{27}.

## 3×3×6

*6×6×6 emulating a 3×3×6*

Note that with the bandaged version shown here, impossible patterns can be created (similar to shapeshifted patterns, see the 3×3×5, but shapeshifting cannot occur in a real cube because the midplane of the 6-number axis would cut the center slice of cubies on either of the 3-number axes).

As with the 3×3×4, a full six-face checkerboard pattern is
possible, provided you do the two odd-number axes first:
F_{2}^{2}R_{2}^{2}U_{246}^{2}.

http://www.youtube.com/watch?v=iRD2zKh_24s&feature=related

http://www.youtube.com/watch?v=0eYqfi12-ck&feature=related

## 3×3×7

This video shows a fully functional 3×3×7 in action; here is another.

## 3×4×4

"Jin/neochronius" has made a 3×4×4; photos are here. A video of it in action is here.

## 3×4×5

*6×6×6 emulating a 3×4×5*

This puzzle can shape-shift. Hold it with a 3x5 side facing up, then do
(for example) U F_{12}^{2} U_{12}.

This page shows a design for a 3×4×5 that was made with CAD-CAM prototyping tools, the results are seen here and here.

## 3×5×5

This video shows a 3×5×5 in action. It shape-shifts,
because like the bandaged-emulated version, the central 3×3×3 part is
able to function even after one of the 3x5 layers has been turned only
90^{o}.

## 4×4×4

In most 4×4×4 solutions, emulating a 3×3×3 on a 4×4×4 is used as the basis of a large portion of the solving time. The edges are grouped together and the cube (or perhaps a portion of it, such as the first 3 layers, or everything but the centers) is solved as if the 4×4×4 were bandaged to emulate a 3×3×3 or a void cube.

However, on the 4×4×4 it is also possible to flip a single dedge and exchange a single pair of dedges, two things that cannot be done on a 3×3×3.

Notation: I use my extended Singmaster piece labeling and move notation from 1982, described above.

Solution: Here is my current 4×4×4 solution technique, most dating back to May 1982 when I worked it out for myself.

1. Find a center that already has most of the same color on the 4 center pieces; if there is none do a few moves to create one. I use this as the reference "Up" side during the edge-pairing.

2. Pair up edges

2a. I start by positioning four edge-pairs around the
"equator" (U_{2} and U_{3} slices) chosen in such a way that a single
U_{2} or U_{2}' turn will create three matched pairs. Then I "store"
these matched pairs on the U or D face, and continue finding more
pairs to match up. 8 pairs can be created this way, and then a 9^{th} pair
can be created from what remains on the "equator".

2b. To pair up the last few pairs of edges, I use the following algorithms:

Place two pairs to be fixed in the UL and UR
positions. The edge at UL_{B} should match the one at UR_{F}. If the
pair at UR is oriented the wrong way for this to work, flip it over
with something like R'UF'U'.

To exchange UL_{F} with UR_{F}, use R^{2}F_{2}
R^{2}F_{2}' R^{2} (5s). (NOTE: This is just my original 1982
algorithm B^{2}R_{2}B^{2}R_{2}'B^{2} moved to different faces)

The pair at UL is now matched; if there is another broken pair repeat the previous steps again.

3. Solve like a 3×3×3 (ignoring the centers). If you know how to solve the Void Cube, this is essentially the same thing, although you may get a single flipped edge at the end.

3a. My method is here; the important point being that I solve all the corners before the last several edges.

3b. Because there is no reference for the centers, you may get a single swapped edge pair (or corner pair, if you solve the corners last) just like on the void cube. For a corner swap, perform a single quarter-turn of a face that contains the swapped pair of corners, then use 3-cycles to fix the 3 corners that are now out of place and as many edges as possible; a swapped pair of edges will be left.

4. Fix edge "parity". As discussed above, one of the following may be necessary:

4a. To exchange the UR and DR dedges, use F_{2}^{2}R^{2}
F_{2}^{2}R_{12}^{2} F_{2}^{2}R_{2}^{2} (7s)

4b. To flip the dedge at FR, use U_{2} R^{2} U_{2} F^{2}
U_{2}^{2} R^{2} U_{2} R^{2} D_{2}^{2} F^{2} U_{2}' R^{2} D_{2}
(13s)

5. Put the centers in place. I use many variations on the following
"commutator" algorithms, all of which fit a pattern which you will
probably see if you try each of them on a cube (hint: apply tiny
sticky-notes to each of the named facelets, or pairs of facelets in
the 3^{rd} example, do the algorithm, and watch where they move.)
Notice that two center facelets can be solved at once this way.

L_{2}U_{2}L_{2}' U L_{2}U_{2}'L_{2}' U' (8s) : 3-cycle of
(R_{UF} → U_{LB} → U_{LF})

R_{2}'D_{2}'R_{2} U^{2} R_{2}'D_{2}R_{2} U^{2} (8s) :
3-cycle of (R_{DB} → U_{FR} → U_{LB})

L_{2}U_{23}L_{2}' U^{2} L_{2}U_{23}'L_{2}' U^{2} (8s) :
Cycles 3 pairs (R_{F} → U_{L} → U_{R})

More Algorithms: Here are some more algorithms for the 4×4×4, most of them of the "pure" variety that produce no side-effects (except perhaps exchanging identical centers on an otherwise-solved cube).

To cycle the edges (UF_{L}, FU_{R}, UB_{R}), use R_{3}'UF_{2}
D'R_{3}UR_{3}'DR_{3}U' R_{3}'F_{2}'U'R_{3} (14s)

Dedge flip without altering centers: D_{2}'R^{2}D_{2}R^{2}
U_{2}'R^{2}U_{2}^{2} F^{2}D_{2}F^{2}U_{2}' R^{2}F^{2}D_{2}^{2}F^{2}
(15s)

A few more can be found here.

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