Blindfold cubing requires that the problem of solving the cube be split into a number of independent problems using moves that only affect a sub-portion of the cubes. A normal approach would be along these lines:

orient corners place corners (perhaps swapping 2 edges in the process) orient edges place edges

Making the solving of the corners and the solving of the edges 'almost' independent puts the problem of solving the cube blindfold within reach of the average human being.

Normally you solve the cube in the LRFBUD group. This means that you can make turns of any faces. This may result in flipping of edges. Solving the cube in the L2R2FBUD group means that turns of the L and R face must be double turns, whilst all other faces may be single or double moves. This group is of great use in solving blindfold since edge pieces cannot be flipped by moves within this group.

Orientating the corners so they lie in the L2R2F2B2UD group makes it irrelevant in which order the corners are placed in the correct location.

Which edges will require flipping can be determined by seeing whether the pieces can be placed by moving them through the middle layer using F and B turns. Note that the group that the edges are to be solved in means that the cube can be orientated in 3 different ways. Choosing your orientation of the cube affects which pieces are considered flipped, and which aren't.

The PnXsnX'sn'Pn' technique is a simple way of finding moves that swap/rotates edges/corners. What you do is put all the pieces that you are interested in finding an operator on in the same layer, and call this sequence Pn. The inverse of this sequence is called Pn'. Now, by trial and error find a sequence of moves that affects some of the pieces in the layer, but leaves all other pieces in that layer unmoved. This sequence is called X, and the inverse is called X'. Do a turn of the layer, called sn, which has inverse sn'. Doing X' will now unscramble the other 2 layers, and doing sn' and then Pn' will return the cube to its previous state with just a few cubes altered.

It isn't necessary for there to be moves in all of the stages, and if necessary the moves used could be from a subgroup of all moves by, say, restricting moves of the R face to R2.

The simplest method to find algorithms is to use the PnXsnX'sn'Pn' technique for finding moves.

X=LD2L'F'D2F with S1=U, S2=U2, S2=U' and Pn=nothing where n=1,2 leaves U unchanged except FLU rotated clockwise

These algorithms are well known already, and rotate clockwise the corner in the LFU position before X is applied, and rotate anticlockwise the corner in the LFU position before X' is applied. By choosing Pn from (U, U2, U', none) and sn from (U, U2, U', none) appropriately, which corner from the top layer is rotated clockwise and which is rotated anticlockwise can be chosen. Further, by choosing Pn from (R2U, F2U, R2U2, F2U2, R2U', F2U', R2, F2) and sn from (U, U2, U', none) [note that there are no other allowable moves for sn since it is a rotation of the layer chosen earlier] we can choose any pair of corners on the cube (one from the D layer and one from the U layer) to be rotated with one going clockwise and the other anticlockwise.

Again, the simplest method to find algorithms is to use the PnXsnX'sn'Pn' technique for finding moves.

X=R2D' R2D2 R2D R2D' R2D2 R2D R2 with S1=U, S2=U2, S2=U' and Pn=nothing where n=1,2 leaves U unchanged except FRU and BRU swapped

Note that this example has X=X'. X was found by trying to split the corners from the edge piece and then recombining the corners the other way round.

Using this sequence you can determine the corner moving sequences possible. It may be possible to utilise a sequence that uses moves other than L2R2F2B2UD that takes less moves than using moves L2R2F2B2UD; if you have shown that the result of the sequence is achievable using those moves then that is allowable. The whole point is to find allowable swaps; to speed the solving up and shorten the solution length use shorter algorithms that briefly take the cube out of the group before returning the cube back to a state within the group.

If an even number of swaps will place the corners in the correct place then the corners can be placed without swapping the edges. If an odd number of swaps is required then an odd number of pairs of edges will also need swapping. A rotation of a slice will achieve a change of parity. Since L2R2F2B2UD is being used to solve the corners, in this method the parity can be changed by rotating U or D a total of an odd number of turns. The other way to swap the parity of the corners is to do a swap of 2 corners and 2 edges.

How to determine the parity:Number the corners positions 1,2,...,8 in any fashion that you find easy to remember and work out which position the corner that should be in that position actually is.

As an example: 12345678 83265417Convert this to express in terms of cycles. In this example (187)(23)(46)(5). For each n cycle the parity of the cycle is n-1, so the parity of the position described is 2+1+1+0=4. So this example has even parity, therefore no U or D turns are required. If the parity was odd then a single turn of U or D either way would change the parity to even.

Here is step-by-step how I would determine the cycles in disjoint notation for the example quoted above (write the numbers down until you get the hang of it, then move on to doing it in your head):

We have some corners we haven't dealt with yet so write down (

You have (

Find the lowest numbered corner not written down. Since we haven't written down any it will always be 1. Write that down

You have (1

Look at the piece in position 1 and see where it belongs. It belongs in position 8. Write that down

You have (18

Look in position 8 and see where the piece there belongs. It belongs in position 7. Write that down

You have (187

Look in position 7 and see where the piece there belongs. It belongs in position 1, but we have already written down that number so we write down ) instead

You now have (187)

There are some corners we haven't dealt with yet so we write down (

You have (187)(

Now find the lowest numbered corner we haven't written down, we have written down 1, so the next lowest is 2. Write that down

You have (187)(2

Look in position 2 and see where that piece belongs. It belongs in position 3. Write that down

You have (187)(23

Look in position 3 and see where that piece belongs. It belongs in position 2, but we have already written down that number so we write down ) instead

You have (187)(23)

There are some cornerw we haven't dealt with yet so we write down (

You have (187)(23)(

The lowest numbered corner we haven't written down yet is 4, so write that down

You have (187)(23)(4

Look in position 4 and see where that piece belongs. It belongs in position 6 so write that down

You have (187)(23)(46

Look in position 6 and see where that piece belongs. It belongs in position 4, but we have written down that number so we write down ) instead

You have (187)(23)(46)

We have corners that we haven't dealt with yet so write down (

You have (187)(23)(46)(

The lowest numbered corner we haven't written down is 5, so write that down

You have (187)(23)(46)(5

Look in position 5 and see where that piece there belongs. It belongs in position 5, but we have written down that number so we write down 5 instead

You have (187)(23)(46)(5)

You have written down 8 numbers so you are finished, and the cycles in disjoint notation are (187)(23)(46)(5)

To determine the twist of a layer, for each of the four corners in that layer determine how many clockwise rotations of that corner from untwisted would put it in that rotational orientation (which will be 0, 1 or 2), and add up each twist. Discard any multiples of 3, and you are left with the twist of the layer.

As an example, on a solved cube, do RU2R2U'R2U'R2U2R. Considering the twist of the upper layer: the twist of the FLU corner is 2, for the BLU it is 1, for BRU it is 1, for FRU it is 2, giving a total of 6=(3+3+0), so U twist is 0. The twist of the L layer in this example is FLD+BLD+BLU+FLU=0+0+1+2=3=(3+0)=0.