sonata_green | Entries tagged with math

A plane has two degrees of freedom. Selecting a 2D plane from a 3D space requires choosing what to hold constant. We can pin a single coordinate, giving an orthogonal plane; we can pin the sum or difference of two coordinates, giving a wide staircase; or we can pin the sum/difference of all three coordinates (±x±y±z), giving a qbert staircase. So the answer is that there is a one-coordinate "diagonal", but not a zero-coordinate plane: the sum of zero variables is already constant, and imposing the requirement that a constant must be constant doesn't give us an additional restriction, meaning it fails to reduce the degrees of freedom.

It seems to me that the key feature of the hexagonal tiling is that there are no cells that share a vertex but not an edge. I think the proper generalization to 3D would then be a honeycomb whose vertex figure is a tetrahedron. There are multiple such, but I'm currently looking at the bitruncated cubic honeycomb, which has the virtue of being cell-transitive.

I'm having trouble wrapping my head around it. Ultimately the question is what "planes" exist[1], but for now I just want to try to understand the nature of the base pattern.

The relationship with the cubic honeycomb seems to mean that a cell at (0,0,0) has cube-face neighbors at the permutations of (±2,0,0) and cube-vertex neighbors at (±1,±1,±1). Which locations are therefore full/empty? For (x,0,0) we have full cube-cells at even x, but it's not simple parity because the vertex operation is odd (and (1,0,0) and (1,1,0) are both empty).

Are (0,1,z) all empty? ...It looks like the cube-edge neighbors of a full cell are necessarily empty; likewise face connections "skip over" an empty cell, so yes.

The question is what invariant (±1,±1,±1) maintains. It's not parity, though it seems like it should have something to do with parity; it's not the mod-3 sum of the coordinates... oh! The rule is that all three coordinates must have the same parity. This explains why (0,1,z) are all empty. This further implies that 2/8=1/4 of the cells are full, which accords with the observation that each edge has exactly one full cube.

[1] I don't quite have a rigorous definition, but it seems to me that a candidate diagonal should separate the space into two congruent half-honeycombs, with the post-smoothing diagonal plane passing through every "surface" cell. However, this is clearly not a sufficient condition, since in 2D this would accept diagonals of every rational slope. Maybe also require the surface to be cell-transitive?

However, the bitruncated cubic honeycomb specifically seems deeply connected to the cubic-octahedral something-or-other, so probably we can just assume that all the diagonals we want have 45-degree slopes. This gives two fundamental diagonal types, the two-coordinate (wide-stair) diagonal and the three-coordinate (qbert) diagonal. ...and the zero-coordinate (orthogonal) plane.

(Why no one-coordinate? Because "diagonal" is a relationship between two coordinates. Except I'm having trouble making that rigorous... something to come back to later, maybe.)

Eyeballing it, it looks like there are two types of planes that we'd want to smooth out: a slope with shallow triangular pyramids (blue/teal), and a wall with deep square pyramids (red).

figure 1: honeycomb of rhombic dodecahedra, with cut-lines showing possible approximate flat planes
Rhombic_dodecahedra.jpg: en:User:AndrewKepert, Rhombic dodecahedra, cut-lines added by Sonata Green, CC BY-SA 3.0

Putting it together, it looks like we have a cuboctahedron (green) augmented with pyramids (red and blue). (These pyramids are relatively shallow, compared to those that would be required in order to form the first stellation of cuboctahedron.)

figure 2: a single rhombic dodecahedron from figure 1, with cut-lines showing the various pieces it might be divided into
Rhombic_dodecahedra.jpg: en:User:AndrewKepert, Rhombic dodecahedra, image cropped and cut-lines added by Sonata Green, CC BY-SA 3.0

For both types of pyramids, we want to override the base's color if the other 3 or 4 faces unanimously disagree. The tricky cases are if anything else happens.

For the triangular pyramid, we can have 0, 1, 2, or 3 of the outer faces agree with the base. 0 and 3 are easy; an obvious idea would be to have 1 act as 0 and 2 act as 3, but then the base has no input at all. What would this look like? Is it what we want?

For the square pyramid, we can have 0-4 agreements, with the nontrivial cases being 1, 2, and 3. The obvious choice would be to tiebreak in favor of the base, so that we group {01|234}. Again, I don't have a good sense of what this would look like. Note that for the square grid, the analogous approach would (1) ignore the diagonally opposite cell entirely, and (2) have the diagonal-corners case look like this:

Quarterly sable and argent, a lozenge counterchanged.

[240 words, running total 998/750.]

I've installed Emacs, git, and wc-mode. My $100 Windows laptop is gradually learning to pretend to be a real computer.

A cubic honeycomb meets four cells around an edge, and eight at a vertex. I think (I might be wrong) that a rhombic dodecahedral honeycomb meets three at an edge, six around one type of vertex, and four around the other type of vertex.

If I want an odd number of cells around a vertex, then it seems to me that the arrangement of cells must be described by a polyhedron with an odd number of faces (each face corresponding to the vertex figure of the relevant vertex of one of the cells). (This is not the same operation as forming the dual honeycomb; the cubic honeycomb has a 🐛vertex polyhedron🐛 of an octahedron, while its dual honeycomb is another cubic honeycomb that's syncopated relative to the original.)

What about hexagonal prisms, arranged in a face-centered cubic lattice? No, those meet four (or more) at a vertex. In particular, three hexagons are met by a fourth on the next layer. Syncopated cubes give five to a vertex by the same logic, but where edges cross we necessarily get four cells meeting, so the syncopated-prism strategy in general is out.

Can we make a honeycomb with a vertex polyhedron of a triangular prism? This requires three cells with square vertices and two with triangular vertices.

We've come a long way from the original problem, though. We're only interested in vertices in the context of a half-honeycomb that meshes with its rotation or mirror image. Furthermore, the problematic case with the square tessellation was opposite corners matching across a diagonal (▚); if there's a two-and-two with matching corners being adjacent, the result is fine.

In the rhombic dodecahedral honeycomb, the four-cell vertex has a tetrahedral arrangement, where any two cells are necessarily adjacent. (You can cut a tetrahedron in half by bisecting four of its six edges, making a square cut-face. Each half has two of the original four vertices.) The six-cell vertex's vertex polyhedron is a cube; for three-and-three, there are two possible arrangements: each three can meet each other around a single vertex, or they can make a C shape (the Cs are making out sloppy style). Each of these lends itself to a fairly straightforward choice of plane of bisection, though in the latter case we bisect four of the cube's faces, so it's not as simple as "this corner of the rhombic dodecahedron is a certain color".

[440/250 words, running total 758/500.]

Current Mood: still can't spell tessellation

It's been a decade since the last time I tried something like this. Let's see if this time goes any better.

My goal is 250 words per day. Extra words from previous days can roll over to subsequent days (so I can build up a buffer), but I can't slosh in the other direction (so each day has a hard deadline). In other words, the goal is that by the end of the Nth of November, I'll have published a total of 250N words. For sleep-schedule reasons, the deadline is 23:00 UTC.

...and I'm at like 100 words so far. Not ideal.

The thing I want to talk about right now unfortunately requires pictures to really convey properly, but let's have a go at it.

Suppose you have a square grid where each cell can be in one of two states, and you want to smooth out stairstep diagonals into proper diagonals. You can subdivide each cell into a central diamond and four triangular corners. Let the diamond be colored according to the initial/naive state of the cell; let each corner follow the rule that it's colored according to the majority of the four cells that meet at its right-angled vertex. (Break ties by matching the center.)

The need to break ties makes this slightly ugly. On a grid of hexagons, we can similarly inscribe a smaller hexagon whose vertices are the midpoints of the edges of the cell. This lets us smooth out lines with no special cases.

How do we generalize this to three dimensions? On a cube, we have two different types of diagonals: the "wide staircase" where we cut off an edge, and the "Q*bert staircase" where we cut off a corner. If we pick one or the other, it's not too horrible, but ideally we want to handle both cases.

The 3D analogue of the hexagon might be the rhombic dodecahedron?

[318 words.]

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Sonata

Entries tagged with math

why no one-coordinate diagonal plane

Diagonal-smoothing: the bitruncated cubic honeycomb

Possible progress on rhodod smoothing, with images

More smoothed-diagonal honeycomb thoughts

Let's try this again

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January 2026

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