wolf wrote:I'm not sure the second point is just a coincidence; it seems to hold for the three VACPs I've seen so far. I can't come up with a counterexample yet either.
Actually, I now agree with you, as the proof is trivial:
From Maekawa's Theorem we know that there must be an even number of vertices extending from a node. Let A be the number of m vertices and B be the number of v vertices. Since A + B is even, A and B must either be both even or both odd, and it follows that A - B must always be an even number.
To also answer Boaz's second point, I agree Eileen's definition that an v vertex is defined as showing one side/color of the paper when folded independently from the model, and similarly, an m vertex would show the other. The definition is with respect to the different sides of the paper and to the orientation of the paper from which we start. I realize the example I posted was convoluted, but it is to demonstrate how one could take advantage of this vertex assignment to convey the wanted model.
bshuval wrote:Third, not all vertices can be designated m or v... Think of a simple twist. The vertices are as much m as they are v. An even simpler example is a kite fold...
I disagree; when one has a vertex on a folded model (a vertex having at least 4 creases emanating from it) it must have an inside and an outside. Think of an ice-cream cone: a folded vertex must have an inside and an outside, just as the cone does.
Even the vertices of a twist fold, when folded independently from the rest of the model have an inside and outside if they indeed be foldable vertices, and thus have the possibility of labeling them as either "m" or "v".
The kite fold raises some questions. When labeling edge vertices, one must question the gender of the edge of the paper itself. Also, one must point out that any set of creases emanating from an edge vertex can be made to be flat foldable, regardless of Maekawa's of Kawasaki's Theorems (I am ignoring the layer overlap problem that exists because the Theorems verify properties of flat-foldable vertices but are not exclusive to all vertices. i.e. flat-foldable vertices follow the theorems, but not all vertices that follow the theorems are flat-foldable.) Since 360 deg of paper does not exist for these points, they prove to be the exception. If the vertices do not have to follow the theorems, can they be labeled?
I propose that they can, and I will define their gender in this way: Were one to reflect an image of the entire crease pattern over the edge in question,
then what
would the node's gender be? Actually, this method of identification should only work for uni-axial bases/vertices, as that will assure that all edge creases lie on the same line when folded.
Thus under this definition, the kite fold has but one definable vertex, the symmetrically middle one, as it's edges lie on the same line when folded. The other two are just single creases through the edge of the paper. I would argue that in single crease cases, one could just label the vertex by the gender of the single crease (which is what I believe I have done in previous circumstances?). Thus, were one to fold a kite fold, I would label all three vertices with the same gender, the choice of which obviously involves how one folded the kite fold.
An interesting point to discuss here, though, is what if one folded one half of the kite fold forward (Valley-fold) and the other back (Mountain-fold). When observed, this resulting point would be not one, but both colors! WTF would this vertex be? Again, this is a special "edge case" that would not be encountered in the center, as one can not create a whole in the center of the paper (many people know that one can not create color changes from the center of the paper). I am not sure how to identify such an edge vertex.
I would like to hear more thoughts on creating a more general definition of vertex assignment. I believe now that all interior vertices are definable and that most normal edge vertices are. What about the weird color changes? Do edge vertices need to be defined at all? Is it helpful?
Thoughts?
) that the vertex is an m-vertex if M-V=2 and a v-vertex if M-V=-2. 