My 'proof sketch' was all wet. Basically, the lengths of the struts don't matter, so we're really talking about a sequence of 3D rotations (think Quaternions). In 2D, any sequence of rotations preserves the axis, which is always in the same direction. In 3D, we can achieve all manner of twists with a sequence of rotations about different axes. A product of quaternions can be collapsed into a single quaternion uniquely because quaternion multiplication is associative. No Twist theorem applies only to planar cycles where all the twists have parallel axes. So George Hart & Tom Verhoeff are correct; in particular, it is easy to make a 'Mobius pipe' in 3D on which you can draw a line along all the struts & elbows & the line will traverse all of the struts twice -- once on either side. At 10:27 AM 6/4/2013, Henry Baker wrote:
I conjecture that the number of 'degrees of freedom' = the number of connected components of the undirected graph.
Proof sketch. Consider any simple cyclic path in the graph & throw away the rest of the graph (for the moment). Each vertex in the path now has degree exactly 2. But the same 'angle bisector' argument for the triangle (see below) shows that once the angle choice is made for one of the struts incident on the vertex, the same choice is required for the other incident strut. By induction, the choice on one strut constrains the choice on every other strut in the cycle. We thus have at most one consistent labelling for the entire cycle.