Cube skeleton bindings Imagine that you have a cube skeleton, like so:

Further imagine that you have three rubber bands that you can loop through any of the faces. However, only one rubber band may go through any particular face at any time. In addition, all 6 faces must be occupied, although any number of rubber bands may be used. I have found 16 solutions, not counting reflections and rotations.








Well...did I get all of the solutions? Also, how would I go about this analytically, preferably with graph theory?
 A: In general we need to combine two perfect matchings on the faces, one for the inside and one for the outside.  For the cube in particular, we have just 6 faces, and each of the three edges in a perfect matching connects either adjacent or opposite faces.  There can be either 0, 1, or 3 pairs of opposite faces matched.
number of matched opposite faces

                   inside
              3      1          0
         3   3a     [1]        [1]        [n] means that n configurations
outside  1   4b   3b,4a,7a    7b,8b           are missing from the pics
         0   6b    1b,6a   2a,5a,1a,5b

(btw, note that pic 2b is not valid, and 7a=8a reflected)

So I agree that there are 16 solutions.  But each of the 9 cases in the table above required some geometrical reasoning, so let's also try the cycle-decomposition approach to see if it is any more tractable.
As described in the comments above, a rubber band arrangement is equivalent to a decomposition of the faces into even-length cycles, where the cycle is the sequence of faces taken by the rubber band oriented so that it enters the polyhedron through the first face (according to some fixed ordering of the faces) in the cycle, then alternates exiting and entering for the rest of the cycle.
There can be 1 cycle of length 6, or 1 of length 4 and one of length 2, or 3 of length 2.
1 cycle of length 6:
    (top, bottom, front, back, left, right)          6b
    (top, bottom, front, left, back, right)          6a
    (top, bottom, front, left, right, back)          7a
    (top, front, bottom, back, left, right)          6a again
    (top, front, bottom, left, back, right)          5b
    (top, front, bottom, left, right, back)          8b
    (top, front, back, bottom, left, right)          7a again
    (top, front, back, left, bottom, right)          8b again
    (top, front, back, left, right, bottom)          2nd missing case
    (top, front, left, bottom, back, right)          5a
    (top, front, left, back, bottom, right)          5b again
    (top, front, left, back, right, bottom)          8b again again
    (top, front, left, bottom, right, back)          5b again again
    (top, front, left, right, bottom, back)          6a again again
    (top, front, left, right, back, bottom)          7a again again
1 cycle of length 2 and 1 cycle of length 4:
    (top, bottom), (back, left, front, right)        4a
    (top, bottom), (back, left, right, front)        1st missing case
    (top, bottom), (back, front, left, right)        4b
    (back, bottom), (left, right, top, front)        1b
    (back, bottom), (left, top, right, front)        1a
    (back, bottom), (left, top, front, right)        7b
3 cycles of length 2:
    (top, bottom), (front, back), (left, right)      3a
    (top, bottom), (front, left), (back, right)      3b
    (top, front), (left, bottom), (back, right)      2a

Well, this one required geometrical reasoning at each step also, so the most we can say is that we are more sure about the cube now, since we got the same 16 solutions again.
For the general case, perhaps something like Pólya's counting method could be adapted to these decompositions into even-length cycles (as opposed to colorings), to handle the symmetries of the polyhedron in question, but it doesn't look straight forward.
