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 love
Polya's theory, and wouldn't miss a chance to apply it. Now, let's see if I remember how it works. If you're not familiar with the theory, most of what I'm doing will look very arbitrary. So that readers are not completely lost, I will at the very least, explain cycle notation.
Cycle notation is a way to express a permutation. For example, the following permutation of the sequence {1,2,3,4,5,6} is (4)(123)(56).
{2,3,1,4,6,5}
As you might guess, rotations of a solid are permutations, and can therefore be expressed in cycle notation. The above permutation has 3 cycles in it.
1) There are 24 rotations of the cube. Of these rotations, 1 has 6 cycles, 12 have 3 cycles, 8 have 2 cycles, and 3 have 4 cycles. (And I'm sort of skipping my thought process to get these numbers.)
(1*2^6 + 12*2^3 + 8*2^2 + 3*2^4)/24
(64 + 96 + 32 + 48)/24
240/24
10 cubes
2) There are 24 rotations of the octahedron as well. Of these, 1 has 8 cycles, 17 have 4 cycles, and 6 have 2 cycles.
(1*2^8 + 17*2^4 + 6*2^2)/24
(256 + 272 + 24)/24
552/24
23 octahedrons
Already, my solution disagrees with Charlie, which usually isn't a good sign. There's actually a good chance that I made an error somewhere when counting the cycles of all the rotations, so Charlie may still be right. It might be noted that Polya's theory can also tell us how many faces are colored for each possible coloration, but that's more difficult.
3) There are 60 rotations of a dodecahedron. Of these, 1 has 12 cycles, 44 have 4 cycles, and 15 have 6 cycles.
(1*2^12 + 44*2^4 + 15*2^6)/60
(4096 + 704 + 960)/60
5760/60
96 dodecahedrons
Uh oh, more disagreement.
4) 60 roations of the icosahedron. Of these, 1 has 20 cycles, 20 have 8 cycles, 24 have 4 cycles, 15 have 10 cycles.
(1*2^20 + 20*2^8 + 24*2^4 + 15*2^10)/60
(1048576 + 5120 + 384 + 15360)/60
1069440/60
17824 icosahedrons
Umm... I have a feeling that no one's going to prove me wrong here by counting all 17000+ icosahedrons.

Posted by Tristan
on 20060305 10:22:06 