You are correct --- I was trying to start from the base, but that's a mistake. Still, I don't think any old 4 lines will do: eg. they mustn't be at pi/2 to their neighbours! WFL On 6/28/17, Michael Kleber <michael.kleber@gmail.com> wrote:
Of course it's possible with a quadrilateral base -- any origami vertex that has only four edges coming out of it leads to an example of this, right?
--Michael
On Tue, Jun 27, 2017 at 8:34 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I don't think that is possible with a quadrilateral base. Hexagonal is feasible; however the base overlaps in anything I can visualise. Heptagonal might be necessary to avoid overlapping! WFL
On 6/28/17, 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? > > > > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >
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