Components

Components are encoded in the following order:

Component ID	Meaning	Valid Range	# of bits
EP	edge permutation	0 to 12! - 1	29 bits
EO	edge orientation	0 to 2¹²-1	12 bits
CP	corner permutation	0 to 8! - 1	16 bits
CO	corner orientation	0 to 3⁸-1	13 bits
PO_U	puzzle orientation (see below)		3 bits
PO_L	puzzle orientation (see below)		2 bits
MO_Q	whether center ("middle") orientation is supported	0 to 1	1 bit
MO	center ("middle") orientation	0 to 4⁶-1	12 bit

Validation

Note that this encoding includes the full pattern of each of these components, without relying on parity to save bits.

For some applications, it is useful to be be able to encode "invalid" patterns.
This allows encoders and decoders to be written more easily.
This provides a parity check.

Guidelines:

Applications SHOULD validate that each of the components are in the valid range.
Applications MAY choose to validate parities (and reject encoded values with invalid parity) if this is important to their correctness.

Byte Layout

Each component is stored with binary bits contiguously from high to low. Example: the high bit of EP is at bit index 0 (first bit of the first byte), and the low bit is at index 28 (5th bit of the third byte).

If puzzle orientation is not encoded:

PO_U is set to 0b111.
PO_L is ignored.

If center orientation is not encoded:

MO_Q is set to 0b0.
The MO bits are ignored.

Puzzle Orientation

PO_U: The center on the U face is specified as a value from 0 to 5 (according to its original zero-indexed position in ULFRBD).
PO_L: The center on the L face is specified as a value from 0 to 3 (according to its original zero-indexed position in ULFRBD, skipping the centers currently on U and D).

The other components are encoded relative to centers. That is, reorienting the puzzle changes PO_U and PO_L, but does not affect any other components.

Puzzle Orientation Example

Apply z y' to a solved cube. Now:

The L center has moved to U.
- L is at (zero-indexed) position 1 in ULFRBD, so PO_U = 1
The B center has moved to L.
- The L and R centers are on U and D (respectively), so remove L and R from ULFRBD to get UFBD.
- B is at (zero-indexed) position 2 in UFBD, so PO_L = 2