1)I have lots of black and white squares that can be joined together to make cubes. How many distinguishable cubes can I make?

2)Now I try it with triangles and regular octahedrons?

3)Now pentagons and regular dodecahedrons?

4)Triangles again but making regular icosahedrons!?

Note: Distinguishable means rotations are the same, but reflections are not.

I'm going to try to count the rotations more systematically to make less errors.  I thought I might find some errors in my previous answers, but they remain the same.  However, now anyone can check my work

Ok, there are basically 4 different types of permutations that I will count separately: a) the identity, b) rotation about a face, c) rotation about an edge, and d) rotation about a corner. (When I say "about a face", I mean around an axis going through the center of opposite faces.)

1) cube
a) 1 rotation, 6 cycles
b) 6 faces/2 = 3 axes * 2 = 6 rotations, 3 cycles
3 axes * 1 = 3 rotations, 4 cycles
c) 12 edges/2 = 6 axes * 1 = 6 rotations, 3 cycles
d) 8 corners/2 = 4 axes * 2 = 8 rotations, 2 cycles
(1*2^6 + 12*2^3 + 8*2^2 + 3*2^4)/24
= 10, same as before

2) octahedron
a) 1 rotation, 8 cycles
b) 8 faces/2 = 4 axes * 2 = 8 rotations, 4 cycles
c) 12 edges/2 = 6 axes * 1 = 6 rotations, 4 cycles
d) 6 corners/2 = 3 axes * 2 = 6 rotations, 2 cycles
3 axes * 1 = 3 rotations, 4 cycles
(1*2^8 + 17*2^4 + 6*2^2)/24
= 23, same as before

3) dodecahedron
a) 1 rotation, 12 cycles
b) 12 faces/2 = 6 axes * 4 = 24 rotations, 4 cycles
c) 30 edges/2 = 15 axes * 1 = 15 rotations, 6 cycles
d) 20 corners/2 = 10 axes * 2 = 20 rotations, 4 cycles
(1*2^12 + 44*2^4 + 15*2^6)/60
= 96, same as before

4) icosahedron
a) 1 rotation, 20 cycles
b) 20 faces/2 = 10 axes * 2 = 20 rotations, 8 cycles
c) 30 edges/2 = 15 axes * 1 = 15 rotations, 10 cycles
d) 12 corners/2 = 6 axes * 4 = 24 rotations, 4 cycles
(1048576 + 5120 + 384 + 15360)/60
17824, same as before

Posted by Tristan on 2006-03-08 00:16:36