Since tetrahedral chains always have a choice of 3 triangles where the next tetrahedron will be attached, they can be described by words on the 3-letter alphabet, {-. 0, +}. On a 3-letter alphabet, up to reversals and cyclic rotations, there are only twenty-one 4-letter words. To be a helix, a tetrahedral chain must have a dominant twist. Arbitrarily choosing to prefer a '+' twist, the 4-letter words with the number of '+' greater than 0 and greater than the number of '-' are only eight: '++++', '-+++', '000+', '00++', '0+0+', '0+++', '-0++', '-+0+':
A tetrahedral chain is a particular sort of deltahedron that can be formed step-by-step, incrementally adding a single regular tetrahedron face-to-face at each step, and always adding the new tetrahedron at one of the three most recently added triangular faces. Notice that there is only one shape for three tetrahedra attached together face-to-face: the tritetrahedron or boat polyhedron. That shape can indeed be can be made by chaining tetrahedra, but the chaining process will differ according to which of three triangular faces at the stern of the boat is considered to have been the origin. In the space curve diagrams here, each tetrahedron is represented by a 70.53-degree circular arc---the dihedral angle of a tetrahedron---at the beginning of which a discrete twist of {-120, 0, +120} degrees occurs, depending on which of the three candidate faces is chosen for the attachment. The interest here is in periodic patterns in {-, 0, +} so the exact start and finish are of little consequence.
According to Adam Tyc a tetrahedral chain is at most 3-cursal, and a bit more than half are expected to be unicursal.
On Saturday August 8, 2026, I will be leading an activity at the University of Galway, for the Bridges Math Art Conference Family Day. We will be making deltahedra of various shapes from the same strips of shiny cardboard as above by interpreting numerical codes generated by computer. If you succeed in making a deltahedron, you get to take it home.
Saturday, August 8, 1:00–5:00 pm in various locations in Áras na Mac Léinn, on the campus of the University of Galway.
The tetrahedral graph is a Tait graph of a 3-component link.
Many interesting deltahedra, foremost among them the triangle-faced Platonic solids, are not knotted, so it is worthwhile looking at joining up link components to yield knots. Following Q. Xing, E. Akleman, J. Chen, and J. L. Gross in “Single-cycle plain-woven objects,” 2010 Shape Modeling International Conference, IEEE, pp. 90-99, 2010, we find that a link with n components can be converted to a knot by making n-1 small edits. For example, the tetrahedron, octahedron, and icosahedron with 3, 4, and 6 components, become knotted after 2, 3, and 5 edits respectively.
If an edge of a Tait graph carries two different link components (e.g., every edge in the tetrahedral graph illustrated above), then simply subdividing that edge with a 2-valent vertex will join the two link components into one.
A knotted triangle-faced graph based on the tetrahedral graph.
The edited Tait graph will no longer be a triangle-faced, but this is easily fixed. We can add a pair of parallel edges between the two neighbors of the new vertex (because this is a graphical Reidemeister move that does not alter cursality), and then we change the embedding so that the two parallel edges straddle the subdivided edge (Shank's theorem implies that changing embedding in the plane does not alter unicursality.) Now we have recovered a triangle-faced Tait graph with one less link component. Though the graph is no longer 3-connected, simple, or non-degenerate as a deltahedron, we can nonetheless weave it.
In the tetrahedron, performing this edit on any two edges that share a vertex joins the three link components into one. Road code: 3.6 8.1 9.12 2.13 14.7 15.0 17.1011.18 19.16 21.45.22 23.20, where the four edges in boldface get folded to a 180° dihedral angle.
The 2-eared tetrahedron, a knotted degenerate deltahedron.
The minimal set of edges that must be edited always constitutes a spanning tree in the dual of the Petrial, or DP. The Petrial of the tetrahedron is the hemicube, and the dual of the hemicube is the hemi-octahedron. Since the spanning trees of a graph are independent of its embedding, it suffices to draw the DP in the plane in any convenient way, for the hemi-octahedron that is most simply a triangle with doubled edges.
The hemi-octahedron, the DP of the tetrahedron, drawn as an abstract graph.
Since the last post I have been using Gilles Schaeffer's C program planarmap to randomly sample larger polyhedral triangulations (3-vertex-connected triangulations on the sphere.) For example it took less than half an hour for planarmap to output 108,223 random samples of the 256-triangle polyhedral triangulations, and a bit more than 2 hours for SageMath to sort through them and find that 1,416 of these were unicursal: punicursal ≈ 1.3%.
That sounds disappointing until you look at the actual totals. The integer sequence A000109 at the Online Encyclopedia of Integer Sequences counts polyhedral triangulations by the number of vertices up to 23 vertices (which is 2v-4 = 42 triangles). But at that size and larger, Tutte's asymptote, Q(n), (n is half the number of triangles), is a close approximation of the number of polyhedral triangulations.
For example, the last term available in A000109 is 28615703421545 (2.86 E13) at 23 vertices (42 triangles, n=21), Q(n) = 2.77 E13.
Since we can use Q(n) as an estimate of the total number of polyhedral triangulations of any size, we just need a statistically significant random sample to estimate the total number of those polyhedral triangulations that are unicursal. This approach does come to a stop eventually as the generation of random samples (ever more samples being needed for significance as punicursal -> 0) and testing for non-singularity on larger and larger Laplacian minors bogs down.
But look a the largest size I was able to sample from, triangulations with 256 triangles. That is quite a modest-sized triangulation in the shape modeling world, yet there are on the order of 10115 such triangulations (each topologically distinct) of which on the order of 10113 are unicursal!!
The number of polyhedral triangulations of the sphere having n vertices, starting at n=3, is given by OEIS A000109: 1, 1, 1, 2, 5, 14, 50, 233, 1249, 7595, 49566, 339722, 2406841, 17490241,... The corresponding numbers that are z-knotted, are: 0, 0, 1, 0, 3, 4, 22, 70, 482, 2955, 17907, 114642, 825097, 5843386,... ???
Polyhedral triangulations that are z-knotted (a.k.a., unicursal or single left-right cycle) occur more frequently than intuition might expect. Plantri can generate all polyhedral triangulations up to, say, a couple dozen vertices, and we can test them for unicursality using standard software to count spanning trees. Shank's Theorem (1974) says: Given a connected planar graph G having H(G) spanning trees, G is unicursal if and only if H(G) is odd.
For 3-connected planar triangulations the chart above tells the tale up to 16 vertices, both for the general case and with various caps on maximum vertex degree (that often being a practical necessity.) In the 16-vertex case, the z-knotted fraction is 0.334 in the general case, and nearly unchanged until the cap on vertex degree falls to 6, where the knotted fraction falls to 0.221.
Though these sizeable fractions come as a jolt to intuition, one might also reasonably expect that counts of spanning trees would be odd 50% of the time. Well, not quite.
What happens to these fractions as vertex counts go upward of 10,000 or more? It is hard to reach higher vertex counts while examining each and every graph because the numbers of graphs in each vertex cohort are already immense and growing rapidly. Plantri generates graphs in a fixed order, so there is no way we can get a random sample other than by randomly selecting from the entire generated set. Also, knotted-ness shows a tendency to be more common in the graphs that Plantri generates first.
Though Shank's Theorem applies only to connected planar graphs, let's back out and look at a more general class of graphs. Kirchhoff's Matrix Tree Theorem (1847) applies to any sort of graph, connected, disconnected, multigraph or simple. The Matrix Tree Theorem says we can get the count of spanning trees in a graph by calculating the determinant of any minor of a matrix called the Laplacian (which is just the graph's adjacency matrix multiplied by -1, with the diagonal elements filled in with the degree of that respective vertex.) Since we are only interested in the parity of the determinant, all the values in the Laplacian can be mod 2 reduced to 0 or 1, and likewise the calculations of the determinate can be done in mod 2 arithmetic (where, for example, there is no distinction between addition and subtraction!)
The Laplacian of an arbitray graph on n vertices, even after it is reduced to mod 2, not a random integer matrix since the elements on the diagonal have the parity of the sum of the other elements sharing that row or column. For that reason a minor of the Laplacian is not truly random but close.
The probability that the determinant of a random n x n integer matrix is odd rapidly approaches 0.28879 as n goes to infinity.
Since the odds that the determinant a (large) random integer matrix is odd is 0.28879 (a.k.a., φ(1/2), the Euler function evaluated at 1/2,) Kirchhoff's Theorem is a solid reason to expect that unicursal graphs of any variety will not be scarce, nonetheless, not as common as the 50% that Shank's Theorem might lead us naively to expect.
Taking the dual does not change the cursality of a plane graph; it will be more convenient to study the duals of the 3-connected planar triangulations. Quoting from the OEIS page for A000109, "Every planar triangulation on n >= 4 vertices is 3-connected, and its dual graph is a 3-connected cubic planar graph on 2n-4 vertices." So we will be looking at 7-3 through a Laplacian matrix with 2*7-4 = 10 columns and rows, and ultimately through a minor with 9 columns and rows.
For Plantri 7-3: the Laplacian matrix and a submatrix of the Laplacian along with the value of its determinant.
For Plantri 7-3: the Laplacian matrix, minor, and determinant after modding by 2.
The Laplacian of a 3-regular graph is not random; nor is it after modding by 2; nor after forming the submatrix. The submatrix is what concerns us. Like the Laplacian, the minor has 1's along its diagonal and plus 3 additional 1's in each row (and column), but only 2 additional 1's in the 3 rows (and 3 columns) that got robbed when a row and column were taken away. The 1's on the diagonal just make the problem smaller at a given n, so they probably do not affect the limit. Likewise the effect of trimming off a row and column is unlikely to materially affect the properties of large matrices. The other big thing that separates 3-regular graphs' Laplacian minors from random binary matrices of the same size is sparseness. The Plantri 7-3 minor is already down to (4*9 - 3)/9^2 or 41% non-zero elements, so a sparsity of 59%. For Plantri 16-3 (below), there are 28 rows in the Laplacian and the density of the minor is down to 27*4-3/27^2 = 0.144, so a sparsity of 86%.
In a triangulation of the sphere, a triangle strip can be seen as sort of topological geodesic, an attempt to travel as straight as possible within the given constraints. Notice that both the sequence of shared triangle edges (black) and the edges of the dual graph that connect the triangle centers (green) are zigzag paths, alternating hard-right with hard-left turns.
In a triangulation of the sphere, a triangle strip (above) is a sort of topological geodesic, an attempt to traverse the triangulation along a path as straight as possible without regard to lengths and angles.
Suppose we start at an arbitrary triangle in the triangulation and head off in one of the three possible directions, and from then on we stay in our triangle lane. What happens?
Clearly the strip must eventually loop back on itself since the sphere has no boundary that the strip could end at, and the number of triangles is not infinite. What is the chance our randomly chosen lane will visit every triangle exactly three times and then repeat? Seems a long shot...but apparently not. The chance of this happening is about one in three no matter how complicated the triangulation. A triangulation with such a triangle strip is termed z-knotted (z for zigzag.)
About 1/3 of spherical triangulations are z-knotted (a.k.a., unicursal). The integer sequence of spherical triangulations on n vertices is OEIS A000109.
Most of the numbers in the chart above were first published in 2004 in "Zigzags, Railroads, and Knots in Fullerenes," by M. Deza, M. Dutour, and P. W. Fowler. Below is their chart for triangulations of the sphere.
Z-knottedness in trivalent polyhedra and their dual triangulations. Chart from Deza, Dutour and Fowler.
Our 'n' counts vertices in the triangulation, while the 'n' in Deza, Dutour, and Fowler counts vertices in the dual 3-valent graph: a triangulation with n vertices has a dual with 2n-4 vertices (see the OEIS link above,) so the two charts are identical as far as they go.
Deza, Dutour and Fowler also looked at the same question restricted to fullerenes which is equivalent to restricting spherical triangulations to vertex degrees 5 and 6. The triangulations dual to the largest fullerenes they investigated have (74+4)/2 = 39 vertices, and all vertex degrees in {5, 6}.
Z-knottedness in fullerenes and their dual triangulations. Chart from Deza, Dutour and Fowler.
In this restricted class of spherical triangulations they found a z-knotted fraction of 1970/14246 = 0.138, about 14%. So z-knotted triangulation on the sphere are quite common even if technological constraints on vertex degrees make them somewhat less common.
