# Message #2919

From: andreyastrelin@yahoo.com

Subject: Re: [MC4D] 120Z solved!!!

Date: Thu, 30 Jan 2014 08:12:21 -0800

Melinda, thank you!

Let’s begin with 2C. It’s easy to find that simple commutator gives 3-loop of 2C in one line, so one may try to solve puzzle line by line (like we do with simple edge-rotating MT). But soon he finds that some line have odd permutation of pieces… and there begins the fun!

Flipping of one cell changes parity of permutations of all six lines passing through it. So when you solve one line, you push disorder to five others. And what we can try is to shrink disorder area from whole puzzle to smaller ball… to one cell… to nothing.

Consider one plane (= sphere in 3D, = equator sphere). It contains 6 lines, and every cell in plane belongs to two of them. So sum of parities of these lines is always zero. (First advice: draw a web of lines in one plane and remember it. You will need it very often). Our first task will be to clear one plane.

You don’t want to get lost in 120-cell surface, so I suggest to select "central" cell (for example, the cell that is in the center from the start) and remember it. You may give it some special color if you want. I didn’t. Then you need to find "polar plane" for this cell (that is equator if you consider central cell as a pole). It will be our first plane to work.

Select any line on it. Each of 5 other lines in the plane intersects this one, so you can solve parity disorders by single flips of cells on the selected line. When you solve 5 lines, the sixth will have zero parity.

When the plane is solved, solve one of halves of puzzle (using commutators) by moving disorders to another half.

Select another plane that is close to first one. It intersects with unsolved ball by pentagon, and you can solve its sides one by one. So your unsolved area is shrinking… Soon you will see the single cell where six wrong lines are intersecting. Flip this cell and finish 2C stage.

In 3C situation is similar, but the orbit of 3C piece is the plane. Every cell is intersected by 15 planes, and its flipping changes parity of all these planes. So at first I suggest to select some half-line (5 cells in the line going through the center) and move disorders in all planes to this area or close to it. It is not easy, geometry of the orbit is not intuitively clear, so you should carefully select paths for the pieces when move them to their place. But finally you get most of 3C pieces in place, and only edges of the selected cells may be swapped (plus five other pairs in planes that are not intersected with the selected line). And now there it is the most difficult part.

You need to find some "clusters" flipping of which will not change parities of lines, but changes parities of some planes. Clusters that I found are just "layers" of 120-cell: 12 cells adjacent to one, or 20 cells of the third layer. Flipping of the first cluster changes parity of 32 planes and the second changes 24 planes. In the polar picture they look like 3rd and 4th layers for the first cluster and second and 4th layers for the second one. To apply them to your situation you need to build the polar picture of disordered planes (that is, find polar cells for them) and decompose the set of these cells to spherical clusters. In this stage I got picture with third order symmetry, so centers of clusters were limited to the symmetry axis. After the second attempt I got zero parities of all planes and could finish 2C and 3C stages.

With 4C it’s easy to build 3-cycle jumping diagonally across two cells, but it’s difficult to work with it. So you have to find a path that moves a 4C to adjacent one (5 flips) - it will help you to build 3-cycle for pieces in one face. My macro for it was 232 flips long. Then build a path that returns 4C to its place rotated (8 flips long path exists) - you get rotation of two opposite pieces of one cell. And then you need a path that returns 4C to its place in wrong orientation. I found 17-flip path for it and got macro for flipping of two opposite pieces. Commutator of flipping and rotation gives 3-rotation of the single piece, but it’s long (about 300 flips).

And it’s enough to finish the solve. Some problems will be in the end when you can’t find a pair of opposite cells to flip, but it is solvable by extra moves.

Believe of not, but I consider my solve as short one :) I tried to use short macros when it was possible and leave long ones only for situations when it was difficult to apply short versions.

Andrey