Cale's recognition for square-1 CSP: how it works.

Описание

There wasn't any discussion of why Cale's method of recognising CSP works. I made a video to demonstrate what's happening, and maybe this will help people understand it.

Cale's video: https://www.youtube.com/watch?v=wwtAOoUWui8

How to do this:
There are 6 steps, each of which has you determining the parity of something, and you add them all together (0=even, 1=odd). There are two ways to count:

0 1 2 3 4 5 6
0 1 0 1 0 1 0

I use the second one for BLD tracing, but I instinctively want to use the first one for this. Go figure.

Start at 0. After every step, note what the parity is so far so it's easy to keep track, you don't need to remember them all and add them at the end.

1. Looking at only the relative order of the yellow edges, ignoring the white edges and the exact positions the yellow edges are in, determine the parity of the yellow edges relative to your colour scheme (I use ROGB). I discuss 3 options in the video, but I realised another, possibly quite good way when writing the last part of the description.
2. Repeat step 1, but for white edges instead of yellow.
3. Count the number of [yellow/white] edges in the [odd/even] positions.
4-6. Repeat steps 1-3 for corners instead of edges.

After step 6 you either have a number from 0-6 or 0-1 depending on how you count, then recall that 0=even and 1=odd, and you have the total permutation parity of the scramble.

The process of looking at a scramble might be something like:
1. Ordering of yellow = 3421: odd parity [parity so far = 1]
2. Ordering of white = 1243: odd parity [parity so far = 2]
3. 2 yellows in odd positions: even parity [parity so far = 2]
4-6. ...

Notes on recognising cycle types:
2-cycle: odd (2 pieces wrong, 2 pieces correct)
3-cycle: even (3 pieces wrong, 1 piece correct)
2 2-cycles: even (pairs of locations which had opposite colours still do, either both in same order or both in wrong order. For ROGB, these are: ORBG, GBRO, BGOR. Therefore: if you see pairs of opposites in the correct place in a scramble, just count the parity of each pair separately (RO and GB).
4-cycle: odd (If opposites are in the right places you should follow the info for 2 2-cycles, otherwise it's just 4 wrong.)

The notes for the pairs of opposites suggests an extension to Cale's system for recognising the parity of each set of 4 pieces. Check the relative order of each pair of opposite colours, then count the number of [RO/GB] in [odd/even] positions (taking odd/even positions within the 4 pieces, not over all 8 positions). It works exactly the same as the full system. I may have to make another video now to explain this in more detail ...