dlitwin@geoworks.com (David Litwin) writes:

While we are on the subject of the Alexander's star, I have never
been entirely satisfied with the arrangment of its solution. I've noticed
that most people who look at it don't even know it is solved and I have to
explain that the solution is that all the stickers of pieces laying in a
common plane around the points are the same color. Hard to say, but I can
point it out to people.
To this end I have spent some time trying to find a more satisfying
solution, one more visually clear and simple. I've only come up with one
alternative that I consider reasonable, and it isn't as pure as I would
like.

This solution involves grouping colors in the depressions of the
star. The main problem lies in the fact that the edges come together in
groups of three, but there are 10 stickers of each color so at some point
having all the depressions of the star a single color breaks down. For
this reason I choose one color (White is my preference) to be an exception
and have it remain in the original configuration, i.e. all in two parallel
planes. With the rest of the colors, I group them in groups of five: three
in one depression, and two in an adjacent depression with the third color
of this second depression being one of a white. The result of this is a
set of interlocking "diamonds" that group visually because the are of the
same color. Unrolled, the star would look like this (this should just fit
on a display of 80 columns):

I haven't found a nice way of having more solid depressions than this.

Has anyone else found any nice solutions?

Dave Litwin

I came up with this solution independently. I also liked on that picked two
opposite depressions on the star and made sure they contained one of each
color. That allowed me to fill each of the other 18 depressions with
homogenous colors.

--
-- Wei-Hwa Huang (whuang@cco.caltech.edu)
Homepage (under construction): http://www.ugcs.caltech.edu/~whuang/
Microsoft: small and limp.