# 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)).

**Nan**

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.

This repairs the parity.