If a Rubik Cube is scrambled using only half-turns (i.e., through 180º), then every face will show at most two colours - the original plus the one belonging to the opposite side. A puzzle in that state is not too difficult to restore. Furthermore, this is still a challenging and enjoyable exercise. At the same time you will be learning and practising techniques which remain useful when it comes to solving a completely mixed up cube. These same techniques may in turn be supplemented by various problem-solving games which should provide extra stimulation in that general context.

Indeed, that was how our children mastered it, working up through the following three levels:

[The notation employed below is widely recognised, and is described in the Appendix].
 
 

LEVEL ONE: SOME PATTERNS EXHIBITING TWO-WAY SYMMETRY

Start by learning this operator:

1a. L²R²F².L²R²B² - exchanging two pairs of edge-pieces, and leaving an H-pattern on the top and bottom of the cube.

You can of course keep changing the colours of the faces which get subjected to that operator. While doing that, every now and again apply one or more of the following three move-combinations:
(i) U² followed immediately by D² ...
and/or sometimes: (ii) L² followed immediately by R² ...
or perhaps: (iii) F² followed immediately by B².

Judicious applications of operator 1a in between one or other of the above three move-combinations, will produce a number of symmetrical patterns. Try varying the faces on which your operator is acting; (i.e., by turning the cube around, for example so as to bring the Right Face to the Front).

Eventually, you should come across (and be able to correct) the following configurations:
Chequers on two, four or six faces,
Crosses on four faces,
Dots on four faces (i.e. the original centre-piece surrounded by a different colour),
Bars on four faces - sometimes parallel (like Pillars), sometimes skewed.
Several mixtures of those patterns are also feasible.

At this early stage, it is best not to try any extra exploratory experiments other than these mentioned above.

However, later on - when you feel confident - take a look at this other operator:

1b. R²U².R²U².R²U²- which also exchanges two pairs of edge-pieces, and in addition it produces what our youngsters called the "Eyes" pattern on opposite faces. This operator can also be written as (R²U²)³. This is perhaps the easiest to learn of all operators, although in Level One it is not as useful as operator 1a.

You can then practise and utilize this second operator 1b for creating or rectifying the chequered pattern on opposite faces. While in Level One, however, it is essential to preserve horizontal and vertical symmetry, so every application of 1b should be followed immediately by another one on the opposite sides. For example, straightaway after a (R²U²)³, turn (L²D²)³... (that yields a chequered or partial chessboard configuration on the front and back faces). If you wish, such a pattern may then be corrected with 1a followed by another 1a after turning the cube around - which is essentially the same as applying L²R²F².L²R²B² and then F²B²R².F²B²L².
 
 

LEVEL TWO: A "PARTIALLY SCRAMBLED" PUZZLE

> Take a cube which has been 'mixed up' by subjecting it only to 180-degree turns; (i.e., apart from that restriction, you may turn all or any of its layers at random)...

Now, try and rectify just the corner-pieces using trial and error.

> Occasionally, the following sequence will be helpful:
R²B²R² or F²R²F² [it does not matter much if you choose the wrong one]: they exchange two pairs of corner-pieces along parallel diagonals. Figure 8 helps to illustrate this (see later). Admittedly, those sequences do disturb a few edge-pieces, but these can all be dealt with later....
 

> ... Once all the corners are correct, tackle the edge-pieces by employing operators 1a and 1b. However, to manage that, it is essential to master the art of the "preliminary manoeuvre":
For example, you will often find it necessary to perform a swap like that shown in Figure 1 below:

FIGURE 1. This configuration is easily adapted for operator 1b.
 

In order to apply operator 1b, a preliminary R-move is necessary to make parallel the two lines of exchange. Note that this "R" is only a quarter-turn, i.e. through 90 degrees. Afterwards, you need to reverse the preliminary move - with an anticlockwise quarter-turn of that right layer, which is written R^-1. So the entire sequence is:
R.R²U²R²U²R²U².R^-1
 

> Sometimes your preliminary manoevre will consist of two or three moves. Look at Figure 2 below:

FIGURE 2: Crosswise exchange of two pairs of edge-pieces.
 

Here, a useful preliminary would be
R²L²D - thereby lining up the pieces for operator 1a. Afterwards, you reverse that manoeuvre - which involves working it backwards whilst performing the inverse of every move (when clockwise quarter-turns are changed to anticlockwise ones). In this instance, that requires D^-1L²R². The complete sequence is therefore:
R²L²D.L²R²F²L²R²B².D^-1L²R². [Clockwise 180-degree turns are of course no different from anticlockwise ones, so their "inverse" is the same as the original].
 

> Complicated preliminary manoevres are best remembered in terms of the colours of the relevant faces. Beginners may find it helpful to temporarily note down those colour sequences: this should help you to reverse the manoeuvre after applying whatever operator is required. With Figure 2, in terms of colours that manoeuvre is:
Red².Black².Green [... the bottom face is Green].

Of course, you could also have restored Figure 2 by first applying operator 1b - thus giving a configuration like Figure 1.

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> Often, you will be confronted with a triangular exchange of edge-pieces, as in Figure 3. You could of course subject it to 1b, leaving you with two pairs to be swapped (for which you will need preliminary manoevre U²R before repeating 1b [on the F/R layers] - look ahead and compare Figure 5).

Indeed, this is one way of tackling a three-way exchange ... namely to bring in a couple of 'alien' pieces (which you can usually choose - i.e., it is best to try and make the task as simple as possible) and then perform a "two pairs swap". That will leave two incorrect pieces remaining in the original triangle. These can be rectified using an additional "two pairs swap", which at the same time puts the 'alien' couple back where they belong.

FIGURE 3: Triangular exchange of three edge-pieces.
Always examine it carefully to see which piece needs to go where.
 

> However, it is better to learn another operator - which restores Figure 3 in just six moves:

2. F²RL^-1.U²LR^-1
You may find it easier to remember RL^-1 as two "wheels" turning together, written Rw ... i.e., right layer together with its opposite 'wheel'. (Note that they are both quarter-turns, through only 90 degrees).

LR^-1 simply moves the two wheels back, and may be shortened to Lw, so the complete sequence becomes
F²Rw.U²Lw

Figure 4 shows this type of move.

FIGURE 4: A "wheels" combination, written Rw.

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> Solving partially scrambled cubes (containing only two colours on each face) will give plenty of practice with operators 1a, 1b and 2. However, sometimes you will need to look very carefully to see which pieces are "partners" for an exchange.

For example, look at the situation depicted (from different angles) by Figure 5. You might be tempted to turn F^-1 as a preliminary manoevre, hoping to align the two pairs of "eyes" for applying 1b, but they will then be paired off incorrectly. Instead, U²R is necessary.

FIGURE 5 (above and below).
Look very carefully before lining up the pieces for operator 1b.

Figure 5 can of course be restored in other ways, e.g. apply operator 1b straightaway without a preliminary manoeuvre - followed by operator 2.
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> Here is another operator which may be learned and practised with a partially scrambled cube (and which should also prove useful later for a complete unscramble):

3. RB.R^-1B^-1.RB.R^-1B^-1.RB.R^-1B^-1... best memorized as (RB.R^-1B^-1)³.
Note that all moves are quarter-turns. Figure 6 below is the result.

FIGURE 6: Exchanging two pairs of corners.
 

> Some people find B-moves difficult. (I manage by looking down from above, so that the right hand can turn the back layer quite naturally. L-moves probably require maximum concentration, however). Thus, instead of the above version of operator 3, you may prefer (RU.R^-1U^-1)³ ... producing Figure 7 below.

FIGURE 7: Swapping two pairs of corners on the R and U-faces.
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> Figure 8 (below) may be restored using operator 3 after preliminary manoevre FD^-1. [Alternatively, you could opt for BD^-1 ... it is then a matter of applying operator 3 to other layers].

This operator 3 is not absolutely essential for restoring a Level Two configuration. Instead (if confronted with Figure 8, say) you could just turn U² - then apply the R²B²R² sequence mentioned earlier, and later attend to all the edge-pieces.

FIGURE 8: This corner-swap can be required in Level 2.
Operator 3 achieves it without twisting the corners.
Otherwise turn U² to make the exchange-lines parallel and apply R²B²R².

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> Before it can accept operator 3, Figure 9 below requires preliminary manoevre RU ... (which in terms of colours could be jotted down here as "Red, Yellow").

FIGURE 9. In Level 2, these corner swaps might
occasionally be encountered (but not by themselves).
 

> At this early stage, do not try and use operator 3 for a crosswise exchange of corner-pieces which occupy the same layer - because you will then end up with a pair of "twisted" corners. We will deal with corner-twisting in Level Three, but in the meantime simply turn the offending layer through 180 degrees - which corrects its corners whilst disrupting two pairs of edge-pieces, leaving a configuration like Figure 2.
 
 

LEVEL THREE: A COMPLETELY SCRAMBLED CUBE

> The RB.R^-1B^-1 sequence learned for operator 3 is also part of operator 6. (Operators 4 and 5 will be given later, in what is perhaps a more logical numbering system).

On a pristine cube, perform RB.R^-1B^-1 twice (not three times). Note that the front layer is in perfect condition except for its upper-front-right corner, which has been twisted clockwise. Apply RB.R^-1B^-1 twice more ... the upper-front-right corner now shows an anticlockwise twist; Figure 10 below gives an example of one. [Another (RB.R^-1B^-1)² then restores the cube - best to do that now, making a total of six RB.R^-1B^-1 sequences, before doing anything else].

FIGURE 10: One corner has been twisted anticlockwise ...
(... so it needs a clockwise twist to correct it).
Also, the upper-right edge-piece needs a "flip".
 

Here then is the complete sequence for the next operator:
6. (RBR^-1B^-1)².F.(RBR^-1B^-1)⁴.F^-1... It twists two of the front corners - one clockwise and another anticlockwise. The F-move in the middle was necessary to bring a different corner into the operating position.

Alternatively, try (RBR^-1B^-1)².F².(RBR^-1B^-1)⁴.F² ... which is essentially the same process, but acting on different corners. It is always a matter of placing both offending corners on the front face - (that may require a preliminary manoevre) - while remembering that the piece being corrected is at upper-front-right. Note too that four applications of (RBR^-1B^-1) are necessary to twist a corner anticlockwise.

Also try (RBR^-1B^-1)².F.(RBR^-1B^-1)².F.(RBR^-1B^-1)².F²  ... which produces three clockwise twists in the front layer.
 

> The Rubik Cube has a "parity law" by which:
Either: Every twisted corner must be matched by an opposite one ...
... unless three corners are incorrectly orientated - in which case they have all been twisted in the same direction.
It is impossible for it to have just one twisted corner; if it does, then the puzzle must have been dismantled and reassembled incorrectly.

Before utilizing this operator 6, it is best to decide in advance whether you are going to tackle three corners or just two. Remember that they must all be in the same layer; (it may be necessary to move them there first). You might have a cube which needs two clockwise and two anticlockwise twists; if so, correct them as two separate pairs (where a "pair" denotes one of each type). If, by mistake, you carry out two clockwise twists - say, with
(RBR^-1B^-1)².F.(RBR^-1B^-1)².F^-1 then (at that stage) you will notice a lot of other pieces out of position - because you have used only four (RBR^-1B^-1) sequences and you always need six (or twelve) to leave the rest of the cube in its original state. If you do find yourself in that sort of trouble, then apply two more (RBR^-1B^-1) sequences straightaway (to bring the total to six), and make a new plan.

> There is also a trick for halving the time spent on producing anticlockwise twists: work with the left layer instead of the front one, using the lower-left-back corner as the operating point. (RBR^-1B^-1)² twists that piece anticlockwise, and L-turns bring other pieces there for treatment.

This strategy is therefore useful when confronted with three anticlockwise twists.

Indeed, you can also use it after (RBR^-1B^-1)² for twisting a second F-corner anticlockwise. Put that other corner piece into the upper-front-right operating point (with F or F²or F^-1). Perhaps write down your front and left colours (White, Red, say), and then topple the whole cube upside down making the White face the left one, with Red at the front - and apply (RBR^-1B^-1)² to the new configuration.
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> One more operator is advisable for tackling a completely scrambled cube. Its function is to "flip" or correct an edge-piece, like the red-yellow one in Figure 10. This next operator involves a new type of turn:

Hold the top and bottom layers still with one hand, and pull the middle or equatorial layer round with the other hand (Figure 11). An anticlockwise move is more natural for a right-hander; E^-1 denotes a quarter-turn.

FIGURE 11. The anticlockwise equator-turn E^-1.
[E denotes the opposite (clockwise) move].

The result is shown below. Here, the F-face was originally entirely white, while the R-face was all red.
The black squares came from the left face.

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7. (RE^-1)⁴ produces Figure 12 below. Four edge-pieces are flipped, three of which are on the equator. The "parity law" for edge-flips stipulates that an even number is always required. If you are tackling two, put them both in the top layer (using a suitable preliminary manoevre) and bring each one in turn to the operating position at upper-right. A second application of this operator 7 will restore the flips on the equator while correcting the remaining flip on top.

FIGURE 12 (above and below):
The four flips produced by operator 7 ...
... viewed from the left and from the right.

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Additional Operators

The six listed above are enough to restore a cube. (In other words, numbers 4 and 5 below are not absolutely essential).

However, you will sometimes find yourself having to correct just one edge-pair plus one corner-pair. If you have most or all of those incorrect pieces contained in the same layer, then you could simply give that layer a quarter-turn, enabling you to tackle the edges and corners separately. (In order to get all the pieces into the same layer, you may need to do some preliminary work using operators 1a, 1b and 3). But eventually you will probably decide that it is indeed worth learning operator 5, because it will make the task somewhat easier.

Many hundreds of operators have been discovered, so these extra ones given below represent just a very small selection. Some may suit you, others may not:

2b. Triangular exchange of edge-pieces:
U²F.RwU²Lw.FU² is easily learned once the original operator 2 has been mastered. Pieces are sent clockwise round the front face. Replace F by F^-1 to perform an anticlockwise swap.

4. Triangular exchange of corner-pieces:
UBU^-1F^-1.UB^-1U^-1F (anticlockwise),
and F^-1UBU^-1.FUB^-1U^-1 (clockwise).

That was deduced using the versatile sequence RBR^-1B^-1
- here transferred to other faces as UBU^-1B^-1.

It is really a four-stage process: UBU^-1B^-1 followed by F^-1;
then reverse the sequence giving BUB^-1U^-1 - and restore the front layer with F. The fourth and sixth moves simply cancel each other out. You may or may not prefer to remember it this way.

As explained when operator 2 was introduced, any triangular exchange can also be contrived using a double application of a "two pairs" swap.
_ _ _ _ _ _ _ _

> Here is a third parity law: (two have already been mentioned - involving edge-flips and corner-twists)...
It is impossible to have only two pieces remaining which require a swap.

We can only exchange: [The numbers here below, correspond to those of the operators]
(1) two pairs of edge-pieces,
(2) or three round a triangle,
(3) or two pairs of corner-pieces,
(4) or three round a triangle,
(5) or one edge-pair together with one corner-pair.

To tackle this last situation we have this other operator -

5. U²BU²B^-1U².L²B^-1L².FU²F^-1 - which swaps two corners along with two edge-pieces. Others are available for carrying out an exchange like that, but this one is my favourite because the result is symmetrical.

Furthermore, the above sequence can also help unscramble a "superior" Rubik Cube which contains symbols or emblems on all its faces, (i.e., requiring its centre-pieces need to be orientated correctly !) Operator 5 turns the back centre-piece 90 degrees anticlockwise. An E^-1 move (Figure 11) will then bring the right centre-piece between the swapped edges; now you can repeat or reverse that operator 5 sequence.
 

6b. (F²UR^-1.B²RU^-1)² twists two corners.
 

7b. RF^-1UR^-1F.E.F^-1RU^-1FR^-1.E^-1flips the two edge-pieces on the right equator. You could also use E² instead of E and E^-1 ... that modified sequence works on opposite edges. I also find this operator useful for exchanging edge-pieces on a 4×4×4 cube. E-moves are illustrated in Figure 11.
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My personal strategy

Whereas some people like to restore the top layer completely before moving on to the middle one, others (including myself) prefer to leave one column free, as "parking space".

Thus, I correct the four edge-pieces and three of the top corners, then three or four edges at the base - sometimes together with some on the equator, if convenient. All this can be accomplished just by "common sense". Preliminary manoeuvres for lining up edge-pieces tend to be harder than for corners, so edges are given priority at this early stage. Try and leave at least some of the pieces correctly twisted and flipped, if you can.

After that, operators are indispensable.

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APPENDIX

R = Right; L = Left;
F = Front; B = Back;
U = Up (i.e. Top); D = Down (i.e. Bottom, but the initial "B" has already been taken).

R then denotes a clockwise quarter turn (i.e. through 90 degrees) of the Right Layer, as in Figure 13. R^-1 is an anticlockwise turn. (Some texts write R' instead of R^-1).

"Squared" moves represent 180-degree turns - e.g. F² for the front layer.

FIGURE 13:
An R-move turns the right-hand layer 90º clockwise.
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The "two-wheels" move RL^-1 may be abbreviated as Rw (see Figure 4). Some texts write it as Rs.
LR^-1 or Lw is therefore the same as Rw^-1.

E and E^-1 (Figure 11) both turn the middle layer or "equator" whilst the top and bottom layers remain fixed.
E^-1 may be compared with an Uw-move (where the equator remains fixed).

E-mail: DLMcN@yahoo.com

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