[math-fun] Unfolding questions
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?
I think the requisite cuts will always form a spanning tree of the edge graph of the polyhedron in question. If this is indeed the case, then the number of edges that need to be cut will be V-1. This agrees with the data of 3 for the tetrahedron and 7 for the cube; 5 cuts will flatten the octohedron, for an additional example. If this is true, then the minimal number is unique. My intuitions about query 5 are mixed. On Sat, Jun 24, 2017 at 8:57 PM, 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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<< 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 << 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 ... 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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<< 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 ...
If you swap the existential quantifier with a universal one, you get an unsolved problem: "Does every convex polyhedron have at least one net (unfolding without overlaps)?" ...so I imagine that, for this to be unsolved, there must be convex polyhedra for which there is at least one unfolding with overlaps. Best wishes, Adam P. Goucher
See however http://mathworld.wolfram.com/Unfolding.html etc. --- << Shephard's conjecture states that every convex polyhedron admits a self-unoverlapping unfolding (Shephard 1975). This question is still unsettled (Malkevitch), though most mathematicians believe that the answer is yes. It is known that Shephard's conjecture is false for non-convex 3-dimensional polyhedra (Bern et al. 1999, Malkevitch). >> WFL On 6/25/17, Adam P. Goucher <apgoucher@gmx.com> wrote:
<< 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 ...
If you swap the existential quantifier with a universal one, you get an unsolved problem:
"Does every convex polyhedron have at least one net (unfolding without overlaps)?"
...so I imagine that, for this to be unsolved, there must be convex polyhedra for which there is at least one unfolding with overlaps.
Best wishes,
Adam P. Goucher
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Delete "however" in previous post! Incidentally, note that the polytore is a counterexample to a claim on Weisstein's page that << Furthermore, for a polyhedron with no coplanar faces, at least one edge cut must be made from each vertex or else the polyhedron will not flatten. >> Possibly that page was intended to refer to _convex_ polyhedra alone? [ If anyone out there knows his email address, please forward this to him! ] WFL On 6/25/17, Fred Lunnon <fred.lunnon@gmail.com> wrote:
See however http://mathworld.wolfram.com/Unfolding.html etc. ---
<< Shephard's conjecture states that every convex polyhedron admits a self-unoverlapping unfolding (Shephard 1975). This question is still unsettled (Malkevitch), though most mathematicians believe that the answer is yes.
It is known that Shephard's conjecture is false for non-convex 3-dimensional polyhedra (Bern et al. 1999, Malkevitch). >>
WFL
On 6/25/17, Adam P. Goucher <apgoucher@gmx.com> wrote:
<< 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 ...
If you swap the existential quantifier with a universal one, you get an unsolved problem:
"Does every convex polyhedron have at least one net (unfolding without overlaps)?"
...so I imagine that, for this to be unsolved, there must be convex polyhedra for which there is at least one unfolding with overlaps.
Best wishes,
Adam P. Goucher
_______________________________________________
-----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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<< 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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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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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?
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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?
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-- Forewarned is worth an octopus in the bush.
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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I recently saw in a book an example of two disjoint sets of integers whose sums of nth powers are equal for n = 1, 2, 3, 4, 5.* So given any integer M >= 5, I'm wondering whether there necessarily exist two disjoint sets of integers whose sums of nth powers are equal for all n in the range 1 <= n <= M. If this does not hold for all M >= 5, for which M does it hold? —Dan ———————————————————————————————————————————————————————————— * Given as {1, 6, 7, 17, 18, 23} and {2, 3, 11, 13, 21, 22}.
I recently saw in a book an example of two disjoint sets of integers whose sums of nth powers are equal for n = 1, 2, 3, 4, 5.
https://en.wikipedia.org/wiki/Prouhet–Tarry–Escott_problem Click on the French version of the wiki for more examples.
Fascinating! Thanks. —Dan
On Jun 28, 2017, at 1:50 PM, Hans Havermann <gladhobo@bell.net> wrote:
I wrote:
I recently saw in a book an example of two disjoint sets of integers whose sums of nth powers are equal for n = 1, 2, 3, 4, 5.
https://en.wikipedia.org/wiki/Prouhet–Tarry–Escott_problem
Click on the French version of the wiki for more examples.
A slightly clumsy omission from the Wikipedia pages is that these power-sums are equal also for 0-th powers: so `ideal' solutions satisfy just as many equations as there are members of each set. The `odious/evil' solution with Thue-Morse sequence connection seems astonishing, and remains at present inexplicable to me! WFL On 6/28/17, Dan Asimov <asimov@msri.org> wrote:
Fascinating! Thanks.
—Dan
On Jun 28, 2017, at 1:50 PM, Hans Havermann <gladhobo@bell.net> wrote:
I wrote:
I recently saw in a book an example of two disjoint sets of integers whose sums of nth powers are equal for n = 1, 2, 3, 4, 5.
https://en.wikipedia.org/wiki/Prouhet–Tarry–Escott_problem
Click on the French version of the wiki for more examples.
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Pentamagic squares are good examples of sets of integers whose sums of nth powers are equal for n = 1, 2, 3, 4, 5 (and n = 0). Tarry was a specialist on multimagic squares. Christian. www.multimagie.com -----Message d'origine----- De : math-fun [mailto:math-fun-bounces@mailman.xmission.com] De la part de Hans Havermann Envoyé : mercredi 28 juin 2017 22:50 À : math-fun <math-fun@mailman.xmission.com> Objet : Re: [math-fun] Equal sums of powers
I recently saw in a book an example of two disjoint sets of integers whose sums of nth powers are equal for n = 1, 2, 3, 4, 5.
https://en.wikipedia.org/wiki/Prouhet–Tarry–Escott_problem Click on the French version of the wiki for more examples. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Prouhet's Thue-Morse based solution turns out to be easy enough to prove; despite which I managed to explore a number of culs-de-sac before succeeding with the spoiler below, which does not appear available anywhere online that I could discover. Spot the first typo ... WFL _ _ _ _ _ _ _ _ _ _ _ _ _ _ Note the webpages ignore (g = 0)-th powers: with convention 0^0 = 1 , an `ideal' solution satisfies the same number k+1 of equations as the number n of power terms on each side. Prouhet's Thue-Morse based solution (non-ideal with n = 2^k ) equates to the following. Define s(i) == sum of binary digits of i , f(g, h) == \sum_{0 <= i < 2^h} (-1)^{h+s(i)} i^g ; required to prove f(g, h) = 0 for 0 <= g < h . *** Proof via induction on h : trivial for h = 0 ; so suppose true up to h , and let 0 <= g < h+1 . Then f(g, h+1) = (-1)^{h+1} \sum_{0 <= i < 2^{h+1}} (-1)^{h+1+s(i)} i^g via definition, = - \sum_{0 <= i < 2^h} (-1)^{h+s(i)} i^g + \sum_{0 <= i < 2^h} (-1)^{h+s(i)} ( i^g + g 2^h i^{g-1} + ... ) using s(i + 2^h) = 1 + s(i) in range , along with binomial theorem, = ( -f(g, h) + f(g, h) ) + ( g 2^h f(g-1, h) + ... ) = ( 0 ) + ( 0 + ... ) = 0 via inductive hypothesis. QED Incidentally at g = h , the nonzero power sum above evaluates to f(h, h) = h 2^(h-1) f(h-1, h-1) = h(h-1)(h-2) ... 2^( h-1 + h-2 + ... ) = h! 2^(h_C_2) , *** (i.e. factorial times binomial coefficient). Replacing i^g by (i + a)^g , (b i)^g , i_C_g in the definition of f(g, h) yields results analogous to those above. Fred Lunnon On 6/29/17, Christian Boyer <cboyer@club-internet.fr> wrote:
Pentamagic squares are good examples of sets of integers whose sums of nth powers are equal for n = 1, 2, 3, 4, 5 (and n = 0). Tarry was a specialist on multimagic squares.
Christian. www.multimagie.com
-----Message d'origine----- De : math-fun [mailto:math-fun-bounces@mailman.xmission.com] De la part de Hans Havermann Envoyé : mercredi 28 juin 2017 22:50 À : math-fun <math-fun@mailman.xmission.com> Objet : Re: [math-fun] Equal sums of powers
I recently saw in a book an example of two disjoint sets of integers whose sums of nth powers are equal for n = 1, 2, 3, 4, 5.
https://en.wikipedia.org/wiki/Prouhet–Tarry–Escott_problem
Click on the French version of the wiki for more examples.
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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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On Sat, 24 Jun 2017, David Wilson 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.
Allan Wechsler's conjecture that the cuts form a spanning tree is correct for convex polyhedra with no flat edges. (This follows from Lemmas 3 and 4 of http://erikdemaine.org/papers/Ununfoldable/ , the mentioned Bern et al. paper.) So the count is always number of vertices minus 1 for this case. Even for polyhedra homeomorphic to a sphere, the number of cuts can be smaller than the number of vertices minus one, at least in special cases where a closed loop unfolds flat. See Figure 2 of http://erikdemaine.org/papers/Ununfoldable/ I'm pretty sure that this example has other edge unfoldings that use V-1 cuts, so this shows some variation even without tori. I'm pretty sure you'd always need Omega(V) cuts. In general the number of cuts should be at most V-1 + genus, and I expect that "usually" (for small perturbations of a surface) all edge unfoldings have exactly V-1 + genus cuts.
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?
There's a lot of discussion about the "edge unfolding of convex polyhedra" problem in our book Geometric Folding Algorithms [http://gfalop.org/] (Chapter 22). Also e.g. mentions the results above and shows the Mathworld examples, among others. Another known experimental result is that, as n goes to infinity, most unfoldings (approaching 100%) of a random n-vertex convex polyhedron (say, convex hull of random points on a sphere) are *overlapping*. So there are lots of examples. :-) Beyond the old Bern et al. paper, finding a nonoverlapping edge unfolding of a nonconvex polyhedron is in fact NP-complete: http://erikdemaine.org/papers/UnfoldingComplexity_CCCG2011/ Erik -- Erik Demaine | edemaine@mit.edu | http://erikdemaine.org/
participants (9)
-
Adam P. Goucher -
Allan Wechsler -
Christian Boyer -
Dan Asimov -
David Wilson -
Erik Demaine -
Fred Lunnon -
Hans Havermann -
Michael Kleber