The recent discovery of scutoids in biological structures is fascinating. Two scutoids are depicted below (from Scientists Have Discovered an Entirely New Shape, And It Was Hiding in Your Cells), each with a pentagon on one side (top or bottom) and a hexagon on the opposing side. A scutoid could be viewed as a linking of the vertices of the 2-dimensional associahedron in one plane to those of the 2-D permutohedron / permutahedron in a parallel plane through edges extending in the dimension orthogonal to the planes. (The pentagon is also a 2-D stellohedron / stellahedron.)

Is there a natural (hopefully interesting by retaining some optimization properties) generalization of scutoids to higher dimensions?

Two scutoids

  • $\begingroup$ I'm not sure what linking means here. From the context I would guess that linking means take the convex hull of a pentagon and hexagon embedded in parallel planes in $\mathbb{R}^3$. But a scutoid has a vertex between these two planes, so this can't be the right definition of linking. What am I missing? $\endgroup$
    – Aaron Dall
    Aug 2, 2018 at 22:11
  • $\begingroup$ @AaronDall: The links, i.e., edges, for the 3-D scutoid are explicitly drawn and explained in the references to the article. The question is whether a natural extension to a 4-D structure and beyond can be devised, which of course would involve a prescription for making edges to any vertices. $\endgroup$ Aug 2, 2018 at 23:07
  • $\begingroup$ It would be nice to read a precise definition of a scutoid somewhere. For me, the original article is not clear: Figure 1(c) is clearly described, you start with a Voronoi tesselation on a curved surface and translate the seeds in normal direction. But what people seem to call scutoid is the object in figure 1(d) where the base is now flat. In this situation the above construction does not work and it's not clear to me what properties this object should satisfy. For example, whether all the edges should be straight lines or not. $\endgroup$
    – gsa
    Aug 22, 2018 at 15:46

1 Answer 1


It's not clear from the linked article that the hexagon is a permutahedron, so we don't restrict to that case for starters. For our construction of higher dimensional scutoids we need a couple of standard polytope operations:

Prism: For a $d$-polytope $P$ let $\mathrm{prism}(P)$ be the $d+1$-dimensional prism over $P$ given by $P \times [0,1]$.

Vertex Truncation: For any vertex $v$ of $P$ let $E(v)$ be the edges of $P$ incident to $v$ ordered by length. Let $H$ be any affine hyperplane such that $v$ is the only vertex of $P$ lying on one side of $H$. Then the truncation $\mathrm{trunc}_{v,H}(P)$ is the subpolytope of $P$ that lies on the side of $H$ not containing $v$.

A $d$-scutoid is any polytope $Q$ that can be written as $Q = \mathrm{trunc}_{v,H}(\mathrm{prism}(P))$ for some $d-1$-dimensional polytope $P$ (called the base polytope) and some vertex $v$ of $P$. So a $3$-scutoid is combinatorially equivalent to a scutoid (in the sense of the linked article in the OP) when the base polytope is a pentagon.

If the above definition of a $d$-scutoid is not refined enough for some purpose one can always restrict to various families of allowable base polytopes. One chain of families that gets us from arbitrary polytopes to associahedra is $$\text{arbitrary polytopes} \supset \text{fiber polytopes} \supset \text{associahedra}.$$

Edit: As @MarkS points out in a comment below, scutoids are not polytopes: for scutoids to pack correctly some of their faces must be curved. (For visualizations see page 9 of the original article in Nature or this video.) Since the $3$-scutoids with pentagonal base polytopes described above are polytopes, they are technically not scutoids, but rather combinatorially-equivalent polytopal approximations.

Edit: An earlier version of this answer incorrectly identified the base polytope of a (combinatorial) scutoid as a hexagon instead of a pentagon.

  • 2
    $\begingroup$ The scutoid is not a polytope; the side faces are not flat. youtube.com/watch?v=2_NZ1ql8B8Y $\endgroup$
    – Mark S
    Aug 4, 2018 at 12:20
  • 1
    $\begingroup$ @MarkS: Excellent point. I've added an edit explaining that the construction in the answer only gives scutoids up to combinatorial equivalence. $\endgroup$
    – Aaron Dall
    Aug 4, 2018 at 23:19

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