Alexander's Star

Alexander's Star
Type	Puzzle
Inventor(s)	Adam Alexander
Company	Ideal Toy Company
Country	United States
Availability	1982–present

Alexander's Star is a puzzle similar to the Rubik's Cube, in the shape of a great dodecahedron.

Alexander's Star in a solved state.

History

Alexander's Star was invented by Adam Alexander, an American mathematician, in 1982. It was patented on 26 March 1985, with US patent number 4,506,891, and sold by the Ideal Toy Company. It came in two varieties: painted surfaces or stickers. Since the design of the puzzle practically forces the stickers to peel with continual use, the painted variety is likely a later edition.

Description

The puzzle has 30 moving pieces, which rotate in star-shaped groups of five around its outermost vertices. The purpose of the puzzle is to rearrange the moving pieces so that each star is surrounded by five faces of the same color, and opposite stars are surrounded by the same color. This is equivalent to solving just the edges of a six-color Megaminx. The puzzle is solved when each pair of parallel planes is made up of only one colour. To see a plane, however, one has to look "past" the five pieces on top of it, all of which could/should have different colours than the plane being solved.

If considering the pentagonal regions as faces, like in the great dodecahedron represented by Schläfli symbol {5,5/2}, then the requirement is for all faces to be monochrome (same color) and for opposite faces to share the same color.

The puzzle does not turn smoothly, due to its unique design.[1]

Permutations

There are 30 edges, each of which can be flipped into two positions, giving a theoretical maximum of 30!×2³⁰ permutations. This value is not reached for the following reasons:

Only even permutations of edges are possible, reducing the possible edge arrangements to 30!/2.
The orientation of the last edge is determined by the orientation of the other edges, reducing the number of edge orientations to 2²⁹.
Since opposite sides of the solved puzzle are the same color, each edge piece has a duplicate. It would be impossible to swap all 15 pairs (an odd permutation), so a reducing factor of 2¹⁴ is applied.
The orientation of the puzzle does not matter (since there are no fixed face centers to serve as reference points), dividing the final total by 60. There are 60 possible positions and orientations of the first edge, but all of them are equivalent because of the lack of face centers.

This gives a total of ${\frac {30!\times 2^{15}}{120}}\approx 7.24\times 10^{34}$ possible combinations.

The precise figure is 72 431 714 252 715 638 411 621 302 272 000 000 (roughly 72.4 decillion on the short scale or 72.4 quintilliard on the long scale).

gollark: One of my attempted code guessing submissions accidentally became sentient and erased itself. Very sad.

gollark: Really, this is just prime factorisation with extra obfuscation.

gollark: It's actually already too late.

gollark: The code guessing challenge is easy, it's just graph coloring.

gollark: You're probably doing something like a byte a second now. Perfectly fine.

External links

Description and solution

References

Wray, C. G. (1981). The cube: How to do it. Totternhoe (, Church Green, Totternhoe, Beds. ): C.G. Wray.

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[1] Wray, C. G. (1981). The cube: How to do it. Totternhoe (, Church Green, Totternhoe, Beds. ): C.G. Wray.