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Subject: Alexander's Star
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1229.5834.0C1F4769@canrem.com>
Date: Wed, 13 Sep 1995 02:21:00 -0400
Organization: CRS Online (Toronto, Ontario)
> From: pbeck@pica.army.mil (Peter Beck)
> Subject: ranking
>
> noticed you did not include
> square-1 or alexanders star
>
> thanks for the compilation
Very little literature exists on Alexander's Star. Ideal Toy published
a solution booklet, and Adam Alexander (the puzzle's inventor) wrote
a book, and David Singmaster wrote his analysis in one of his
Cubic Circulars. Dr. Singmaster also mentions Alexander's Star in
Rubik's Cubik Compendium in a chapter about variations on Rubik's
cube.
The Ideal booklet mentions that Alexander's Star has 24 start
positions, and Dr. Singmaster states there are 12 start positions,
each one with it's own mirror reflection, again giving 24 start
positions in total.
Alexander's Star is akin to an "edge-only" Dodecahedron (Megaminx)
without centres. It has 30 edges, that is 15 pairs of distinct
2 coloured edges.
To calculate the number of permutations of N objects with N1
objects which are like, N2 objects which are like ...
Nrth objects which are like we use the formula:
N!
--------------------
N1! * N2! * ... Nr!
In the case of Alexander's Star it is a bit more complicated because
we must contend with edge flips and no centres, however I believe the
correct calculation is....
30! / 2^15 * 2^29 / 60 = 72,431,714,252,715,638,411,621,302,272,000,000
approx = 7.2 * 10^34
I think the term for this is approximately 72 decillion.
(decillion = 10^33)
Here is a bit of an explanation....
There are 30 total objects, with 15 different pairs of 2 like edges.
29 edges may be flipped in any fashion but the 30th edge is forced.
The Great Dodecahedron has 60 orientations in space.
Here is another way to calculate the same number...
Pick one edge and lock it in position and orientation. We still have
29 edges which can move in position and orientation. There are still
28 edges which can be flipped freely, the 29th edge being forced.
We still have 15 different pairs of 2 like edges.
29! / 2^15 * 2^28 = 7.2 * 10^34 (approx)
To be completely clear the calculation is really
29!/2 / 2^15/2 * 2^28
....because only half of the 29! arrangements are possible and only
half of the 2^15 arrangements are possible but both the /2's cancel.
On the Great Cosmic Ranking List this puzzle has less combinations
than Rubik's Revenge, but more than the VIP sphere.
I know of no Square 1 calculations whatsoever, but would be
interested in seeing anyone's calculations.
-> Mark <-