Thursday, July 11, 2013

Alexander's Star

Alexander's Star is a twisty puzzle in the shape of a great dodecahedron. It is essentially a cornerless megaminx. This fact is the key to understanding how to solve it. This puzzle does not shapeshift. The two main problems with the puzzle are as follows:
  1. The solved state does not look significantly more solved than the scrambled state.
  2. The turning of the puzzle is horrible.
The turning can be made much better with some lubricant.

The Basic Plot
  1. Give It Some Lube!
  2. Solve 5 Lower Equator Pieces
  3. Solve 5 Bottom Face Pieces
  4. Solve 10 Equatorial Pieces
  5. Solve Remaining Pieces
  6. Fix Parity
1. Give It Some Lube!

This step is, of course, not required. But if you want to make your puzzle turn better, it's not hard and well worth doing.

2. Solve Five Lower Equator Pieces

We begin the solve by placing (or finding) a white piece on any face. To line up the remaining white pieces on that face, look at the colour on the underside of the first white piece. Since the Alexander's Star is 10-colour, we know that opposite colours of the puzzle are the same. We turn the pieces in using edge piece series. These pieces are not actually the "first face" pieces, but rather correspond to the lower equatorial edges on a megaminx. We do them first because it is then much simpler to place the bottom face pieces (which are not all the same colour).

3. Solve Five Bottom Face Pieces

The five pieces on the bottom face are solved in much the same way as previously, again, using simple turns and edge piece series.

4. Solve Ten Equatorial Pieces

This stage is the longest of the solve. There are ten equatorial pieces. Each needs to be correctly oriented before being placed. Despite being the longest stage, placing these pieces is no harder than before.

5. Solve Remaining Pieces

To begin this stage, solve four of the five upper equatorial pieces by placing them from the top face into their position. The fifth should be left to help us solve the five top face pieces. Again, all piece placement is done using the edge piece series. When only three pieces are remaining, they will be the 5th upper equator piece as well as two top face pieces. These may be cycled home using the edge piece series. If this is the case, the puzzle is solved!

6. Fix Parity

The parity on this puzzle results from the fact that there are 10 colours rather than 20. It will show up when you finish with the last two pieces needing to be swapped. Obviously there is no way to 3-cycle two pieces. To deal with this, we recognise that we can create a 3-cycle by using a third piece from the lower part of the puzzle, which shares the same colours with one of the two pieces.

We simply move that third piece from the lower part up so that we can essentially swap the position of the two pieces needing to be swapped. It's quite simple, despite being difficult to explain. Once this is done, we return the lower piece to its position. (Note that this lower piece which is returned is actually one of the original two pieces which needed swapping.)

After performing the parity fix, we will end with 2 pieces in their positions but wrongly oriented. To complete the puzzle, we move them out of their position with an edge piece series, and then move them back in with a (different) edge piece series. The video below will make this all extremely clear.

And that's it. Your Alexander's Star is now solved. I trust this site has been helpful. If you have any questions or want some clarifications, please use the comments to do so.

1 comment:

  1. Quote: "The parity on this puzzle results from the fact that there are 10 colours rather than 20."

    Shouldn't this be: "The parity on this puzzle results from the fact that there are 6* colors* rather than 12*." instead? (Three fixes in this sentence, marked by *.)