# MagicTile aspects

MagicTile, PARITY in the SKEW puzzles

theorem | name | odd/even | restore parity with | twist number | puzzle |
---|---|---|---|---|---|

Astrelin | PitDvoRom | odd | turn whole by 60° | 0 | {4,6-3} 30 v020 runcinated |

1st Baumann | PitDeoBom | odd | turn whole by 90° | 0 | {6,4-3} 20 e010 bitruncated |

Nan | PitDeoDom | odd | big X | 14 | {4,4-7} 49 e 1.41 duoprisme |

2nd Baumann | PitDeeDom | even | small square | 4 | {4,4-6} 36 e 1.41 duoprisme |

Astrelin

It looks like 60-deg rotation of the whole puzzle around some face gives even permutation of the edges (10 6-loops) and odd permutation od centers (3 6-loops + 2 fixed points).

1st Baumann

A 90-deg rotation of the whole puzzle "MT skew {4,6|3} 30 v020" (Roice Nelson) around some face gives even permutation of the edges (14 4-loops and 2 2-loops) and odd permutation of centers (7 4-loops + 2 fixed points). This is reducible to a single center swap. The edge 2-loops are about the two opposite colors of the rotation axis (here white and (192,192,0)).

Schumacher

To create this parity, we need to make some turns such that (1) the edges on each horizontal line (in the default view) are turned an even number of times; (2) the edges on each vertical line are turned an even number of times; (3) all the edges in the "\" direction are turned an odd number of times; (4) all the edges in the "/" direction are turned an odd number of times.

(1) and (2) guarantee that the diamond-shaped pieces can be solved by commutators. (3) and (4) create the parity.

The simplest way to create the parity, I think, is to choose one diagonal line in the "\" direction, and turn all the seven pieces on it; then choose one diagonal line "/" and turn all the seven pieces on it. So the total number of moves to create a parity is 14. The total picture would be like a big "X". It's easy to verify that this sequence satisfy (1)~(4), (this argument relies on the fact that 7 is odd).

2nd Baumann

Unlike in odd cases here in the even case the / edges are separated in two orbits. Dito the \ edges! I encountered a bad parity situation where I had exactly one edge swap left in each of these 4 orbits.

If I do 4 twists in most compact constellation (corners of a small square with horizontal and vertical sides), I hit 4 orbits with 12 diamond face elements exactly twice. This can be undone by a 3-cycle. And I get one edge swap in each of the 4 edge orbits (plus edge swap pairs).

This repairs the parity.