# Calculations on the VIP sphere
# 32 pieces with all distinct colours
# Sphere where each of the 2 hemispheres rotate 180 degrees
#  and the 4 rows (of 8 pieces each) can slide around
#  the circumference
#
#
# Size (vip) = 437,763,136,697,395,052,544,000,000
# No fixed Orientation
# approx.    = 4.4 * 10^26 or 437 septillion!
# trivial centre

vip := Group(
        (1,2,3,4,5,6,7,8),
        (9,10,11,12,13,14,15,16),
        (17,18,19,20,21,22,23,24),
        (25,26,27,28,29,30,31,32) ,
        (1,28)(2,27)(3,26)(4,25)(9,20)(10,19)(11,18)(12,17),
        (5,32)(6,31)(7,30)(8,29)(13,24)(14,23)(15,22)(16,21)
  );;

Within the 2 orbits of 16 pieces any exchange is possible. One
orbit is "polar" and the other is "equatorial".

(28,1) in vip;
true

Thus on the VIP Sphere a single 2-cycle is legal, although I
know of no simple process as yet.

The original calculation by Dr. Singmaster was (16!)^2, and
I have confirmed his result with GAP.

I have also played with the Masterball somewhat. This puzzle is awful!
Just how accurately does this thing have to be lined up to turn it?
It locked up several times on me when I tried to randomize it.

It is the single most difficult puzzle to turn I have ever
encountered, save the Equator puzzle only.

My first thoughts on calculating the number of positions on the
Masterball was it was the same as the VIP sphere divided by
2^16 but I'm not sure. I can't use GAP in the case of the
Masterball (rainbow edition in this case) to verify this because
of the identical pieces. The booklet which came with the
Masterball refers to some number like 350 quadrillion but there
are more zeroes than the american quadrillion, and I get a
totally different number anyways.

Thoughts anyone??

-> Mark <-