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Consider a cube, with standard topological faces: 6 faces, 12 edges and 8 vertices. Define a cube part as set of faces (possible of different dimension) so that no 2 set elements share same vertex. E.g. there is no cube part consisting of 3 2-dimensional faces, because there have to be 2 faces that share a vertex. Note, cube part doesn't have to contain all cube vertices.

How many different cube parts there are, counting topologically isomorphic cube parts as same?

I think this is application of Redfield–Pólya Theorem, but I'm not familiar with topic and some hints will be usefull.

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A little clarification on "topologically isomorphic": say you take 1 face and two vertices. Case (1) the two vertices are joined by an edge. Case (2) the two vertices are not joinable by an edge. Does your definition of "topologically isomorphic" distinguish between the the two cases? –  Willie Wong Nov 3 '10 at 12:14
    
Yes. These cases are different by definition of cube part. First case is set with one edge, and second case is set with 2 vertices. With isomorphism I want to cover cube rotation and mirror. E.g. there are 3 different cube parts consisting of 2 vertices. On same edge, diagonally on face, diagonally on cube. –  Ante Nov 3 '10 at 12:54

1 Answer 1

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If you want to apply Polya's theorem, you still have to enumerate, say, the number of all "cube parts" that are not identified by a permutation. Note also that I enumerate according to your comment on identifying by rotation and reflection not on your false description in the OP that thease are "topologoically isomorphic".

It is easier to count them directly than to use Polya theory.

Note also that I enumerate according to your comment on identifying by rotation and reflection not on your false description in the OP that these are "topologically isomorphic".

0 things: 1 possibility 1 vertex: 1 possibility 2 vertices: 3 possibilities 3 vertices in a plane: 1 possibility 3 vertices not in a plane (which implies that two of them are diagonally opposite or they are the three neighbours of one vertex): 2+1=3 possibilities 4 vertices in a plane: 1 possibility 4 vertices not in a plane: 5 vertices (=3 vertices): 4 possibilities 6 vertices (=2 vertices): 3 possibilities 7 vertices (=1 vertex): 1 possibility 8 vertices: 1 possibility

1 edge and some vertices (all choices of vertices + symmetric choices of vertices divided by 2): (2^6+2^4)/2=40 possibilities 2 edges in a plane and some vertices (number of vertices 0,1,2,3,4): 1+1+3+1+1=7 2 parallel edges not in a plane and some vertices: 1+1+3+1+1=7 2 non-parallel edges and some vertices: 1+2+4+2+1=10 3 parallel edges and some vertices: 1+1+1=3 possibilities 3 edges (not all parallel) and some vertices: 1+1+1=3 possibilities 4 edges (all parallel or not): 2 possibilities

1 face and some vertices: 1+1+2+1+1=6 possibilites 1 face, 1 edge and some vertices: 1+1+1=3 possibilities 1 face and 2 edges: 1 possibility 2 faces: 1 possibility

Now add everything.

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Thank you. I calculated that way, but I wasn't sure I cover all possibilities and is calculation for each one good. List starts to be quite complicated with more faces (>=2) and some edges and vertices. Because of curiosity, I'm looking for more provable approach. –  Ante Nov 1 '11 at 11:38
    
@Ante: The problem is that you also have to do some case-by-case analysis if you want to apply the Polya theorem. At some point, the action of the group on the set of coloriations has to be analysed. Often, things become easier with Polya theory, but in your problem it is already not totally trivial to list all objects without taking into account symmetries (think of arrangements of three edges). In other applications of Polya theory that part is trivial and then one examines the cycles. Here, I have the impression that both approaches are of roughly the same difficulty. –  Phira Nov 1 '11 at 11:43

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