Hyperspeedcubing

If anyone is wandering what kind of parities might occur during a 4x4x4x4 solve, here are my findings:

- Pairing 2c elements: the last two 1x2 bars can be tricky (see image below). The trick is to switch around the pairs: instead of solving "ab + cd", try to solve it as "ac + bd".

2cpairs.png


- Pairing 3c elements: trivial, similar to the case of the last edge pairs of a 4x4x4.

- 2c OLL parity (see image below): regular 3d OLL parity.

2collparity.png


- 2c PLL parity: trivial.

- 3c OLL parity: mathematically impossible (*).

- 3c PLL parity (see image below): PLL parity with RKT (preferred center-safe version). r2 debt can be turn into Rw2 debt (same as the R2 debt in a 2x2x2x2), and an R2/U2 debt is fixable with a center rotating RKT algorithm.

3cpll.png


(*) Proof: An r move (aka 2RO) does a 4-cycle on the 3c elements, which is an odd permutation, and leaves an odd RKT debt on the r side. This also means even permutations leave even debt. A solved 3c set has an even permutation on the 3c elements (identity perm is even), so it must have an even debt. PLL parity is two swaps, which is even permutation, so it might be possible (and we saw that it is). However, OLL parity is a single swap: odd parity, meaning odd RKT debt. But a solved cube has to have no debt, which is an even debt. Therefore, pure OLL parity on the 3c pieces (without debt) is impossible.

It is worth mentioning that R moves does two 4-cycles on 3c elements and no changes on r RKT debts: both an even permutation, so it has no effect on permutation parities.

Here is the image of an OLL parity attempt, with a single r debt:

3coll.png
 
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