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**Introduction**

This post gives a simple overview of the different parities that can occur during a reduction solve on 4x4x4 upwards.

Even layered cubes (e.g. 4x4x4 and 6x6x6) have the concept of OLL parity and PLL parity whereas odd layered cubes (e.g. 5x5x5 and 7x7x7) only have the concept of Last Two Edge (L2E) cases.

It may not be immediately apparent how these cases are related so I thought I would try to summarise in simple terms.

**OLL Parity**

Pure algorithm: Rw U2 x Rw U2 Rw R' U2 x' Lw' L U2 Lw L' U2 Rw' R U2 Rw R' U2 Rw' U2 Rw'

The image above shows OLL parity on 4x4x4 and its equivalent on 5x5x5. It is clearly evident during L2E on 5x5x5 but it isn’t easily spotted on 4x4x4 until the OLL stage of a reduction + CFOP solve.

It occurs at the moment you complete your centres during reduction and once the centres have been completed the OLL parity is fixed, until such time as you execute an algorithm which affects OLL parity.

Mathematically speaking it occurs whenever the total number of inner slice turns (QTM) applied whilst scrambling and solving the centres is an odd number. If you were to track all of the t-centre pieces whilst executing a parity algorithm on 5x5x5 you would see that the number of pair swaps is an odd number.

Note: You won’t encounter this parity during a reduction solve on an odd layered super-cube (e.g. 5x5x5 picture cube) because you will solve all of the t-centres correctly, prior to edge pairing.

**Intuitive solution for OLL parity...**

I mentioned above that OLL parity is caused by an odd number of quarter slice turns.

The intuitive solution is simple. Just do a quarter slice turn (e.g. Rw) and re-solve the centres:

e.g. Rw U2 Rw U2 Rw U2 Rw U2 Rw

Notice how the example above uses five Rw turns and therefore switches the OLL parity.

Whilst this will fix the OLL parity on 4x4x4, you'll have to resolve the L4E and F2L.

The intuitive solution is simple. Just do a quarter slice turn (e.g. Rw) and re-solve the centres:

e.g. Rw U2 Rw U2 Rw U2 Rw U2 Rw

Notice how the example above uses five Rw turns and therefore switches the OLL parity.

Whilst this will fix the OLL parity on 4x4x4, you'll have to resolve the L4E and F2L.

**PLL Parity**

Pure algorithm: Rw2 F2 U2 Rw2 R2 U2 F2 Rw2

The image above shows PLL parity on 4x4x4 and its equivalent on 5x5x5. It is clearly evident during L2E on 5x5x5 but it isn’t easily spotted on 4x4x4 until the PLL stage of a reduction + CFOP solve.

It occurs at the moment you complete your edge pairing during reduction on a 4x4x4 and once the edges have been paired the PLL parity is fixed, until such time as you execute an algorithm which affects PLL parity.

The reason that PLL parity doesn’t appear to exist on 5x5x5 is because it presents itself as just another L2E case. It is therefore solved during edge pairing, prior to the 3x3x3 stage.

PLL parity is somewhat simpler than OLL parity and some people do not even consider it to be a parity. It essentially boils down to whether you think of paired wings as a "composite edge" or remaining as individual wings. The arguments are as follows:

- If you restrict yourself to outer layer turns (with solved centres and paired edges) then "PLL parity" is a violation of the regular permutation parity, just like "OLL parity" is a violation of edge orientation parity.
- You can solve "PLL parity" using an even number of inner slice turns, without moving centres. Unlike "OLL parity", "PLL parity" is not a parity when you look at the permutation of the 24 wings.

**Intuitive solution for PLL parity...**

The images above clearly show which pieces need to be exchanged and this can be achieved using the intuitive slice-flip-slice approach.

e.g. Rw' (U' R U R') (F R' F') Rw + Lw (U' R U R') (F R' F') Lw' which affects UF and UB

Whilst this will fix the PLL parity on 4x4x4, you'll have to resolve the F2L and OLL.

e.g. Rw' (U' R U R') (F R' F') Rw + Lw (U' R U R') (F R' F') Lw' which affects UF and UB

Whilst this will fix the PLL parity on 4x4x4, you'll have to resolve the F2L and OLL.

**L2E Cases**

There are 16 distinct cases in total and for illustrative purposes, I have derived them from the following three cases:

I'll refer to the images above as “solved”, “diagonal wing swap” and “adjacent wing swap” but the adjacent swap is essentially the diagonal swap after a tredge flip.

The OLL and PLL parity algorithms were repeatedly applied to these three cases and the resultant cases to produce the full set of L2E cases; 16 in total, including solved.

I have created a web page listing all 16 cases and decent algorithms - http://cubing.mikeg.me.uk/algs/l2e.html

You may find it insightful to execute the algorithms for A2-A6 on a 4x4x4 and compare them to a 5x5x5.

Half of the L2E cases do not have any rotational symmetry and have a probability of 1/12 whereas the other half have rotational symmetry and thus a probability of 1/24.

I hope this summary is useful / interesting to some people.

Last edited: Apr 6, 2017