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 $1^+ 2^+ 3^+ 4^+ 5^+ 6^+$:
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.

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    $\begingroup$ If you don't mind some comments on possible reasons why this question might not have received the attention you wanted, the title is simultaneously forbiddingly wordy and does not really convey the spirit of the question. Perhaps a friendlier title like "Can we find polyhedra with faces of area $1, 2, 3, ...$?" might be better? $\endgroup$ Apr 28, 2014 at 0:41
  • $\begingroup$ @QiaochuYuan: I take your fine suggestion! $\endgroup$ Apr 28, 2014 at 0:46
  • $\begingroup$ Is the situation for the $2$ dimensional version of the question clear? Otherwise, this might be a good warm-up. $\endgroup$ Apr 28, 2014 at 1:13
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    $\begingroup$ @ChristianRemling: If I understand your meaning of "the 2 dimensional version," those are golygons: lattice polygons with edge lengths 1,2,3,... $\endgroup$ Apr 28, 2014 at 1:17
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    $\begingroup$ Nigin went on to find an 11-face golyhedron, yadi.sk/d/Zzw_Q6gKTZjwk $\endgroup$ Mar 9, 2016 at 22:58

2 Answers 2


I found a 32-face example with face areas $\{ 1, 2, \dots, 32 \}$:

32-face gollyhedron

It took a reasonable amount of experimentation to stop it from self-intersecting.

  • $\begingroup$ Wow!!! ${}{}{}$ $\endgroup$ Apr 29, 2014 at 18:17
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    $\begingroup$ It's a brain-bending activity to verify. You might add a few labels to indicate where these faces are. (I gave up after finding faces 1 through 11.) Gerhard "Would Make A Good Puzzle" Paseman, 2014.04.29 $\endgroup$ Apr 29, 2014 at 18:21
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    $\begingroup$ Especially clever how you terminated either end with those "club-foot" constructions that capture $\{1,2,3,4,5\}$. $\endgroup$ Apr 29, 2014 at 18:23
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    $\begingroup$ A labelled version is available at cp4space.wordpress.com/2014/04/30/golygons-and-golyhedra $\endgroup$
    – Boris Bukh
    Apr 30, 2014 at 14:46
  • $\begingroup$ @JosephO'Rourke You should have received an e-mail now...? $\endgroup$ Aug 15, 2015 at 17:03

By using the Euler characteristic $\chi = L-E+V$ of the graph $\Gamma$ corresponding to the polyhedron (each cube is a vertex and we link adjacent cubes). One can show that $$ S = 4N - 2L + 2\chi $$ where $S$ is the total area of the uncovered faces, $N$ the number of cubes and $L$ the number of loops in $\Gamma$.

This gives a constraint on the possible $A(P)$. For example $A(P)=1^n 2^n 3^n \dots p^n$ can only be constructed if $4\mid np(p+1)$.

For the interesting case of golyhedra ($n=1$), our constraint reduces to $p \equiv 0,3 \:\mathrm{mod}\: 4$.

$p$-face golyhedra have been constructed by Alexey and Adam for $p=12, 15, 32$. Our constraint suggests that $p=11$ should also be possible. And indeed, we have found an 11-face golyhedron (see image). Because $p=7,8$ are easily ruled out, this has to be the smallest golyhedron.

We are led to conjecture that a $p$-face golyhedron exists if and only if $p \equiv 0,3 \:\mathrm{mod}\: 4$.

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    $\begingroup$ Your first formula looks like the Euler characteristic of a planar graph. Is it clear that the adjacency graph you describe is always planar (what about a cube with sidelength 3 - I counted $V=9$, $E=54$, so $L=46$ - which loops are those)? Or is it clear that the Euler-like formula continues to hold if the graph is not planar? Anyway, the example in the other answer has a planar graph, so there your argument works. $\endgroup$ Mar 9, 2016 at 19:25

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