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Added the 11-face golyhedron example
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Max Lonysa Muller
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I asked this question two months ago on MSE, where it earned the rare Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days. Perhaps justifiably so! Here I repeat it with slight improvements.


Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$, all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere. Let $A(P)$ be the *area sequence*, the sorted list of areas of $P$'s faces. For example:
           
Using regular expression notation, this sequence can be written as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with $A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e., I conjecture there are no golyhedra. Q1. Can anyone prove or disprove this?

Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more $a$'s. For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
      OrthoPolyhedronTwisted6
Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$? The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$, but end effects destroy the regularity.

The broadest question is: Q3. Which sequences $A(P)$ are achievable? Can they be characterized? Or at least constrained?


***Update*** (*30Apr14*). **Q1** and **Q2** are answered by Adam Goucher's brilliant example that achieves $1^1 2^1 3^1 \cdots 32^1$. In light of this advance, a more specific version of **Q3** may be in order: **Q3a**: Identify some sequence that is *not* realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron, described on Adam Goucher's blog. And later a 12-face golyhedron. Finally, he found a 11-face golyhedron.

I asked this question two months ago on MSE, where it earned the rare Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days. Perhaps justifiably so! Here I repeat it with slight improvements.


Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$, all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere. Let $A(P)$ be the *area sequence*, the sorted list of areas of $P$'s faces. For example:
           
Using regular expression notation, this sequence can be written as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with $A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e., I conjecture there are no golyhedra. Q1. Can anyone prove or disprove this?

Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more $a$'s. For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
      OrthoPolyhedronTwisted6
Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$? The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$, but end effects destroy the regularity.

The broadest question is: Q3. Which sequences $A(P)$ are achievable? Can they be characterized? Or at least constrained?


***Update*** (*30Apr14*). **Q1** and **Q2** are answered by Adam Goucher's brilliant example that achieves $1^1 2^1 3^1 \cdots 32^1$. In light of this advance, a more specific version of **Q3** may be in order: **Q3a**: Identify some sequence that is *not* realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron, described on Adam Goucher's blog. And later a 12-face golyhedron.

I asked this question two months ago on MSE, where it earned the rare Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days. Perhaps justifiably so! Here I repeat it with slight improvements.


Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$, all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere. Let $A(P)$ be the *area sequence*, the sorted list of areas of $P$'s faces. For example:
           
Using regular expression notation, this sequence can be written as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with $A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e., I conjecture there are no golyhedra. Q1. Can anyone prove or disprove this?

Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more $a$'s. For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
      OrthoPolyhedronTwisted6
Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$? The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$, but end effects destroy the regularity.

The broadest question is: Q3. Which sequences $A(P)$ are achievable? Can they be characterized? Or at least constrained?


***Update*** (*30Apr14*). **Q1** and **Q2** are answered by Adam Goucher's brilliant example that achieves $1^1 2^1 3^1 \cdots 32^1$. In light of this advance, a more specific version of **Q3** may be in order: **Q3a**: Identify some sequence that is *not* realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron, described on Adam Goucher's blog. And later a 12-face golyhedron. Finally, he found a 11-face golyhedron.

replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Source Link

I asked this question two months ago on MSEthis question two months ago on MSE, where it earned the rare Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days. Perhaps justifiably so! Here I repeat it with slight improvements.


Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$, all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere. Let $A(P)$ be the *area sequence*, the sorted list of areas of $P$'s faces. For example:
           
Using regular expression notation, this sequence can be written as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with $A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e., I conjecture there are no golyhedra. Q1. Can anyone prove or disprove this?

Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more $a$'s. For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
      OrthoPolyhedronTwisted6
Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$? The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$, but end effects destroy the regularity.

The broadest question is: Q3. Which sequences $A(P)$ are achievable? Can they be characterized? Or at least constrained?


***Update*** (*30Apr14*). **Q1** and **Q2** are answered by Adam Goucher's brilliant example that achieves $1^1 2^1 3^1 \cdots 32^1$. In light of this advance, a more specific version of **Q3** may be in order: **Q3a**: Identify some sequence that is *not* realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron, described on Adam Goucher's blog. And later a 12-face golyhedron.

I asked this question two months ago on MSE, where it earned the rare Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days. Perhaps justifiably so! Here I repeat it with slight improvements.


Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$, all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere. Let $A(P)$ be the *area sequence*, the sorted list of areas of $P$'s faces. For example:
           
Using regular expression notation, this sequence can be written as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with $A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e., I conjecture there are no golyhedra. Q1. Can anyone prove or disprove this?

Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more $a$'s. For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
      OrthoPolyhedronTwisted6
Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$? The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$, but end effects destroy the regularity.

The broadest question is: Q3. Which sequences $A(P)$ are achievable? Can they be characterized? Or at least constrained?


***Update*** (*30Apr14*). **Q1** and **Q2** are answered by Adam Goucher's brilliant example that achieves $1^1 2^1 3^1 \cdots 32^1$. In light of this advance, a more specific version of **Q3** may be in order: **Q3a**: Identify some sequence that is *not* realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron, described on Adam Goucher's blog. And later a 12-face golyhedron.

I asked this question two months ago on MSE, where it earned the rare Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days. Perhaps justifiably so! Here I repeat it with slight improvements.


Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$, all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere. Let $A(P)$ be the *area sequence*, the sorted list of areas of $P$'s faces. For example:
           
Using regular expression notation, this sequence can be written as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with $A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e., I conjecture there are no golyhedra. Q1. Can anyone prove or disprove this?

Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more $a$'s. For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
      OrthoPolyhedronTwisted6
Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$? The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$, but end effects destroy the regularity.

The broadest question is: Q3. Which sequences $A(P)$ are achievable? Can they be characterized? Or at least constrained?


***Update*** (*30Apr14*). **Q1** and **Q2** are answered by Adam Goucher's brilliant example that achieves $1^1 2^1 3^1 \cdots 32^1$. In light of this advance, a more specific version of **Q3** may be in order: **Q3a**: Identify some sequence that is *not* realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron, described on Adam Goucher's blog. And later a 12-face golyhedron.

replaced http://meta.mathoverflow.net/ with https://meta.mathoverflow.net/
Source Link

I asked this question two months ago on MSE, where it earned the rare TumbleweedTumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days. Perhaps justifiably so! Here I repeat it with slight improvements.


Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$, all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere. Let $A(P)$ be the *area sequence*, the sorted list of areas of $P$'s faces. For example:
           
Using regular expression notation, this sequence can be written as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with $A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e., I conjecture there are no golyhedra. Q1. Can anyone prove or disprove this?

Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more $a$'s. For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
      OrthoPolyhedronTwisted6
Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$? The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$, but end effects destroy the regularity.

The broadest question is: Q3. Which sequences $A(P)$ are achievable? Can they be characterized? Or at least constrained?


***Update*** (*30Apr14*). **Q1** and **Q2** are answered by Adam Goucher's brilliant example that achieves $1^1 2^1 3^1 \cdots 32^1$. In light of this advance, a more specific version of **Q3** may be in order: **Q3a**: Identify some sequence that is *not* realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron, described on Adam Goucher's blog. And later a 12-face golyhedron.

I asked this question two months ago on MSE, where it earned the rare Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days. Perhaps justifiably so! Here I repeat it with slight improvements.


Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$, all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere. Let $A(P)$ be the *area sequence*, the sorted list of areas of $P$'s faces. For example:
           
Using regular expression notation, this sequence can be written as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with $A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e., I conjecture there are no golyhedra. Q1. Can anyone prove or disprove this?

Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more $a$'s. For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
      OrthoPolyhedronTwisted6
Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$? The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$, but end effects destroy the regularity.

The broadest question is: Q3. Which sequences $A(P)$ are achievable? Can they be characterized? Or at least constrained?


***Update*** (*30Apr14*). **Q1** and **Q2** are answered by Adam Goucher's brilliant example that achieves $1^1 2^1 3^1 \cdots 32^1$. In light of this advance, a more specific version of **Q3** may be in order: **Q3a**: Identify some sequence that is *not* realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron, described on Adam Goucher's blog. And later a 12-face golyhedron.

I asked this question two months ago on MSE, where it earned the rare Tumbleweed badge for garnering zero votes, zero answers, and 25 views over 61 days. Perhaps justifiably so! Here I repeat it with slight improvements.


Let $P$ be a polyhedron, all of whose vertices are at points of $\mathbb{Z}^3$, all of whose edges are parallel to an axis, with every face simply connected, and the surface topologically a sphere. Let $A(P)$ be the *area sequence*, the sorted list of areas of $P$'s faces. For example:
           
Using regular expression notation, this sequence can be written as $1^4 2^2 3^2$.

In analogy with golygons, I wondered if there is a $P$ with $A(P)= 1^1 2^1 3^1 4^1 5^1 \cdots$. I don't think so, i.e., I conjecture there are no golyhedra. Q1. Can anyone prove or disprove this?

Easier is to achieve $A(P)= 1^+ 2^+ 3^+ \cdots$, where $a^+$ means one or more $a$'s. For example, this polyhedron achieves $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
      OrthoPolyhedronTwisted6
Q2. But can $A(P)= 1^n 2^n 3^n \cdots$ be achieved, for some $n$? The above example is in some sense close, with $A(P) = \cdots 4^4 5^4 6^4 \cdots$, but end effects destroy the regularity.

The broadest question is: Q3. Which sequences $A(P)$ are achievable? Can they be characterized? Or at least constrained?


***Update*** (*30Apr14*). **Q1** and **Q2** are answered by Adam Goucher's brilliant example that achieves $1^1 2^1 3^1 \cdots 32^1$. In light of this advance, a more specific version of **Q3** may be in order: **Q3a**: Identify some sequence that is *not* realized by any $A(P)$.

Update (9Jun14): Alexey Nigin has constructed a 15-face golyhedron, described on Adam Goucher's blog. And later a 12-face golyhedron.

Linked to Nigin's 15-face example. And now 12-faces.
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Joseph O'Rourke
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Addendum re Adam's solution; and Q3a.
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Joseph O'Rourke
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edited title
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Joseph O'Rourke
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Source Link
Joseph O'Rourke
  • 150.8k
  • 36
  • 358
  • 958
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