I think I see what you're saying here. There are nonconvex quadrilaterals that have all sides making an arbitrarily small angle with each other. If we start with an infinitely tall polyhedron, the sum of the angles at the vertex is 0, but as we approach the polyhedron having altitude 0, the sum of the angles must exceed 2pi. So somewhere in between is a polyhedron with a 0-defect vertex there. (Is that what you had in mind?) —Dan
On Jun 27, 2017, at 5:12 PM, David Wilson <davidwwilson@comcast.net> wrote:
I'm thinking that there must be a (non-convex) pyramid with concave quadrilateral base whose apex is a zero-defect vertex. You could cut this figure along three sides of the base (including the concave sides) and unfold into a net. I'm thinking this must be the smallest polyhedron (in terms of faces, vertices or edges) which can be unfolded with fewer than v-1 edge cuts.
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred Lunnon Sent: Sunday, June 25, 2017 9:27 PM To: math-fun Subject: Re: [math-fun] Unfolding questions
<< So if a polyhedron can be unfolded without overlap, it must [have] nonnegative angular defect at every vertex. Wouldn't this be equivalent to the polyhedron being convex? >>
If the polyhedron is convex, then it has positive defect at every vertex. But the reverse implication fails --- take the regular icosahedron and punch one vertex so that it dimples inwards. Or consider the great icosahedron.
Also note that your question 5 concerned whether some unfolding had overlaps, rather than whether some unfolding had no overlaps. Indeed since the Mathworld page exhibits an `unfoldable' tetrahedron --- presumably intending to assert that it has an overlapping unfolding --- it is now plain that my simple-minded intuition about convexity was erroneous.
WFL
On 6/26/17, David Wilson <davidwwilson@comcast.net> wrote:
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Fred Lunnon Sent: Saturday, June 24, 2017 11:18 PM To: math-fun Subject: Re: [math-fun] Unfolding questions
<< Is this minimal number a function of the number of faces, edges and vertices of the polyhedron? Is this minimal number unique? >>
It's unclear over what set these questions are intended to minimise! One possible interpretation is answered negatively by the following.
<< Does any unfolding of the same polyhedron have the same number of edge cuts? >>
No. In an August 2009 math-fun thread instigated by Jim Propp's request for a `polyhedral origami torus' with angular defect zero at every vertex, I proposed a family of `polytores' having 20 faces (4 trapezia + 16 triangles), 12 vertices (8 5-valent + 4 6-valent), 32 edges. Developing the planar net may yield either 14 = 2*7 or 16 = 2*8 free edges, depending on which vertices are interior. See https://www.dropbox.com/s/t8iqaeoe5e86ld1/solitore3.pdf https://www.dropbox.com/s/42grmh6o3re4ulf/flattore3.pdf
I see. We can flatten out the angle at a zero-defect vertex without cutting any of the incident edges. So if a polyhedron has a zero-defect vertex, the free edges need not include that vertex, and we can unfold the polyhedron with fewer edge cuts than would be required if all vertices were no-zero-defect. BTW, your polytores are quite pretty. I love geometric curiosities like Csaszar and Szilassi polyhedral, holyhedra and monstatic polyhedra.
If we required a free edge incident with each vertex, including zero-defect vertices, then I suppose that any simply connected polyhedron would require v-1 free edges forming a spanning tree of the vertices. Non-simply connected polyhedra (genus >= 1) would presumably require more vertices, and the free edge graph would include loops.
I imagine that a simply-connected polyhedron could have a zero-defect vertex, this polyhedron would require fewer than v-1 free edges to
unfold.
I wonder what the smallest such polyhedron might be.
<< is there a polyhedron where some unfolding has more edge cuts, but shorter total edge cut length, than some other unfolding? >>
Yes. A tall, narrow polytore net may exchange 4 short trapezium edges for 3 long triangle edges.
<< Is there a convex polyhedron for which some unfolding exhibits overlapping faces in the plane? >>
Intuitively, `unfolding' can only increase the distance between (given points on) any two faces. However, it's not at the moment obvious to me exactly why this should be a consequence of convexity ...
It's clear that if a polyhedron has positive angular excess at any vertex, it can't be unfolded without overlap local to that vertex. So if a polyhedron can be unfolded without overlap, it must nonnegative angular defect at every vertex. Wouldn't this be equivalent to the polyhedron being convex?
Fred Lunnon
On 6/25/17, David Wilson <davidwwilson@comcast.net> wrote:
1. For a given polyhedron, what is the minimal number edges that need to be cut to unfold it into a connected planar surface? For example, 3 edges are necessary for a tetrahedron, I think 7 for a cube.
2. Is this minimal number a function of the number of faces, edges and vertices of the polyhedron?
3. Is this minimal number unique? Does any unfolding of the same polyhedron have the same number of edge cuts?
4. If (3) is false, is there a polyhedron where some unfolding has more edge cuts, but shorter total edge cut length, than some other unfolding?
5. Is there a convex polyhedron for which some unfolding exhibits overlapping faces in the plane? If so, what is the smallest number of faces on such a polyhedron?
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