coloured faces of an octahedron

wik_chick88 · #1 $Old$ September 9th, 2008, 06:57 AM

how many distinct octahedra (all of the same size) are possible if each face of each octahedron is either red, blue or green? the octahedron is a regular solid with 6 vertices, 12 edges and 8 triangular faces.

im having trouble working out how many symmetries there are of an octahedron. 8 faces and 3 colours so different ways to colour the faces would be 8 choose 3 = 56. but obviously 2 ways might have the same physical result if the octahedron was rotated on a vertice, edge or face.

Please help!!!!

ThePerfectHacker · #2 $Old$ September 9th, 2008, 08:20 AM

Quote:

Originally Posted by wik_chick88 $View Post$

how many distinct octahedra (all of the same size) are possible if each face of each octahedron is either red, blue or green? the octahedron is a regular solid with 6 vertices, 12 edges and 8 triangular faces.

im having trouble working out how many symmetries there are of an octahedron. 8 faces and 3 colours so different ways to colour the faces would be 8 choose 3 = 56. but obviously 2 ways might have the same physical result if the octahedron was rotated on a vertice, edge or face.

This is a way to approach this problem.

Let $G$ be the group of all trasformations of the octahedron. In the other thread it was shown that $|G| = 48$ . And $X$ set of all colorings (even the possibly similar one).
Now put these symettries into conjugacy classes which can be found here. For each conjugacy class, say 90 degree rotation, find which elements in $X$ are fixed and how many of them. This number gets multipled by six since there are 90 degree rotations. And go through each class. After you done the hard part the rest follows by Burnside's lemma - be sure to look at the example with the cube in that link, it is very similar.

wik_chick88 · #3 $Old$ September 16th, 2008, 02:23 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

This is a way to approach this problem.

Let $G$ be the group of all trasformations of the octahedron. In the other thread it was shown that $|G| = 48$ . And $X$ set of all colorings (even the possibly similar one).
Now put these symettries into conjugacy classes which can be found here. For each conjugacy class, say 90 degree rotation, find which elements in $X$ are fixed and how many of them. This number gets multipled by six since there are 90 degree rotations. And go through each class. After you done the hard part the rest follows by Burnside's lemma - be sure to look at the example with the cube in that link, it is very similar.

i am still very confused. any more help you can give me?

wik_chick88 · #4 $Old$ September 17th, 2008, 08:02 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

This is a way to approach this problem.

Let $G$ be the group of all trasformations of the octahedron. In the other thread it was shown that $|G| = 48$ . And $X$ set of all colorings (even the possibly similar one).
Now put these symettries into conjugacy classes which can be found here. For each conjugacy class, say 90 degree rotation, find which elements in $X$ are fixed and how many of them. This number gets multipled by six since there are 90 degree rotations. And go through each class. After you done the hard part the rest follows by Burnside's lemma - be sure to look at the example with the cube in that link, it is very similar.

ok im not SO much confused anymore. ive made my own little octahedron and figured out the conjugacy classes:
- 1 identity
- 9 vertex rotations (90degrees)
- 6 edge rotations (180 degrees)
- 8 face rotations (4 of which are 120 degrees and 4 of which are 240 degrees)
i still dont know how to work out which and how many elements in $X$ are fixed. i understand burnside's lemma buttttt i dont know how many of the $3^8$ elements are unchanged in each conjugacy class...PLEASE HELP?!?!?!

wik_chick88 · #5 $Old$ September 18th, 2008, 08:18 AM

please anyone i really need help understanding this. i looked at the cube example in burnside's lemma in wikipedia but im stuck on how they get $3^3$ and $3^4$ etc for each conjugacy class...please someone help me understand this i even made my own little cube and octahedron but i still dont understand!!

asw-88 · #6 $Old$ September 18th, 2008, 08:30 AM

Does 486 seem like an outrageous answer? because thats the answer that I got...

awkward · #7 $Old$ September 18th, 2008, 05:51 PM

Quote:

Originally Posted by wik_chick88 $View Post$

how many distinct octahedra (all of the same size) are possible if each face of each octahedron is either red, blue or green? the octahedron is a regular solid with 6 vertices, 12 edges and 8 triangular faces.

im having trouble working out how many symmetries there are of an octahedron. 8 faces and 3 colours so different ways to colour the faces would be 8 choose 3 = 56. but obviously 2 ways might have the same physical result if the octahedron was rotated on a vertice, edge or face.

Please help!!!!

wik_chick88,

As pointed out by The Perfect Hacker, the really slick way to solve this problem is to use Burnside's Lemma or, better yet, its cousin the Polya Enumeration Theorem, aka Polya's Theory of Counting. See

Pólya enumeration theorem - Wikipedia, the free encyclopedia.

However, that's not the only way to solve the problem. Consider the count of faces by color, written in the form #red + #blue: 8+0, 7+1, 6+2, 5+3, 4+4, 3+5, etc. There are only 5 cases to consider if you combine cases like 5+3 and 3+5 (meaning 5 red and 3 blue or 3 red and 5 blue faces). Then just work out the distinct possibilities; there aren't that many.