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Can we characterize the following kinds of plane coverings? (Open-ended, but provide some description more "useful" than the constraints given.) For a more answerable question, is there an effective method for continuing any initial finite placement of line segments (or proving it impossible)? Do some initial placements force periodic coverings? I have a more specific question/conjecture at the end, but it requires explanation.

We must cover the plane with line segments such that each is parallel to two other line segments on opposite sides at unit distance. Each line segment must touch one or two others at its endpoints, and the angle between them must be 120 degrees. Moreover, the channels between segments must not have any branching or dead-ends (unit hexagons are allowed).

Several obvious coverings satisfy the above. E.g., a hexagonal grid with unit spacing works, as does a banded pattern of "zig-zag" lines, or a series of concentric hexagons going out to infinity. There are less obvious coverings, such as:
     

In case the above verbal description is inadequate, I think it is equivalent to describe the problem in terms of requirements on local neighborhoods of individual points, illustrated in the following figure:
Local neighborhoods may be rotated as needed.

(1) Every point is on a line segment with parallel segments at unit distance.

(2) Segments can only terminate if they meet one or two other segments at a 120 degree angle.

(3) The channels between segments cannot branch or terminate in a dead end (unit hexagons are allowed). Specifically, the neighborhoods shown with a red dotted boundary cannot appear anywhere in the covering.

Conjecture: Apart from concentrically nested hexagons, only one size of hexagon can be formed in such a covering. That is, if a covering contains a hexagon in which three segments meet at a corner, then there are no larger hexagons in the covering, and any smaller ones are concentrically nested in these. [Update: false as stated, but something similar may hold.] The example above satisfies this. Moreover, if unequal-length hexagon sides are coincident, then these form a 60 degree angle, which is forbidden, so I think the placement of one constrains the entire covering. Can this be proved or is there a counterexample?

An observation that may help is that the plane can be tessellated into regions of parallel line segments.

Assuming these regions are always parallelograms (as above, but not proved), then it may be reducible to a more conventional tiling problem. It looks like three segments join when the 120 degree corners meet and two segments join at adjacent sides. Unit hexagons occur where 60 degree corners meet.

(Note: I don't know the answer. I am actively working on this, but if it's solved, I'll gladly read that.)

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  • $\begingroup$ (I included your images.) $\endgroup$ – Joseph O'Rourke Jun 15 '17 at 14:35
  • $\begingroup$ Calling the stuff in between the line segments a "path" is confusing, since path is often a synonym for line segment. It was only when I saw your illustration of point 3 that I saw what you meant. $\endgroup$ – Jim Conant Jun 15 '17 at 15:18
  • $\begingroup$ I agree. That is why I put "paths" in scare quotes. Can you suggest something more clear? This constraint can even be omitted, though it came up in a different tiling problem that this one generalizes and I wanted to preserve it. $\endgroup$ – PaulC Jun 15 '17 at 15:27
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    $\begingroup$ Instead of "paths," maybe: channels. $\endgroup$ – Joseph O'Rourke Jun 15 '17 at 15:35
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Here is a solution sketch.

As illustrated in the question, we refer to the segments as red, blue, or green, depending on their directions. In what follows, "segment" always refers to a segment appearing in a given covering of the plane, subject to the requirements of the question.

Define a vertex to be a point at which two segments (of different directions) meet. A vertex falls into exactly one of four types, depending on which kinds of segments are adjacent to it: red-blue, red-green, blue-green, red-blue-green. We propose to join the vertices with auxiliary edges, so as to realize the parallelograms illustrated in the question.

To this end, consider (for example) a vertex $v$ incident with blue and red (and possibly also green) edges, as shown. The unit-distance constraint implies that a blue edge and a red edge lie on the grey lines shown. The no-60-degrees constraint implies that these two edges fall into one of the two configurations on the right. In either case, we find a new vertex $v'$ where the two new segments meet. Join $v$ and $v'$ with an auxiliary edge. Repeat this process for all vertices $v$.

Image 1

In the first case, the resulting auxiliary edge $\overline{vv'}$ can't meet any other auxiliary edge in its interior. There will be exactly one more auxiliary edge incident with $v'$, going up and to the left. There may be one or two auxiliary edges incident with $v$, depending on whether $v$ is incident with a green segment. If so, the three auxiliary edges incident with $v$ are at 120-degree angles with each other. If not, then $v$ is incident with exactly one auxiliary edge which is collinear with $\overline{vv'}$.

However, in the second case, the conditions force the appearance of a small hexagon (see below), and three auxiliary edges will meet at their midpoints, i.e. the center of the small hexagon.

Image 2

Consider the union of all the auxiliary edges. From the last two paragraphs, we know that the auxiliary edges are characterized by these local neighborhoods:

  1. A single line (along which segments of exactly two different colors meet).
  2. Three segments meeting at 120-degree angles (where the intersection point is a vertex incident with segments of all three colors).
  3. Six segments meeting at 60-degree angles (where the intersection point is the center of a small hexagon).

The auxiliary edges partition the plane into (possibly infinite) regions, bounded by polygonal curves all of whose sides are at 60 or 120 degree angles to one another. Since no vertex lies in the interior of such a region, all segments inside one region are the same direction (i.e. color).

By the description of 2. and 3. above, each 60-degree angle (resp. 120-degree angle) of the boundary of a region determines the color of the segments inside the region. It is now straightforward to check that the only consistent possibilites for the shape of a region are as follows:

  • parallelogram (angles: two 60's, two 120's)
  • infinite half-strip (angles: one 60, one 120)
  • infinite 60-degree sector (angles: one 60)
  • infinite 120-degree sector (angles: one 120)
  • infinite strip (no angles)
  • half-plane (no angles)

Furthermore, given the shape and orientation of one of the first four types of regions (those with vertices, hence angles), the color of the segments inside it is uniquely determined by the fact that these segments must not be parallel to (any part of) the boundary of the region.

Each side of a region must either be infinite or $(2k+1)/2 \cdot \sqrt{3} / 2$, for a nonnegative integer $k$. Conversely, any decomposition of the plane into regions (of the kinds described in the last paragraph, and subject to constraints 1. 2. 3. above), whose side lengths are either infinite or $(2k+1)/2 \cdot \sqrt{3} / 2$, comes from a desired covering of the plane, which can be easily drawn. Furthermore, the resulting covering of the plane is unique, except in the case where the only regions that appear are infinite strips and half-planes, in which case one needs to specify the color of the segments appearing in one of the regions (two choices).

The previous paragraph implies that no finite configuration can force a periodic covering of the plane.

The most interesting plane coverings arise when every region is finite. In this case, the abstract arrangement of regions is given by the picture in the question statement. The side lengths can be arbitrary chosen, subject to the constraint that opposite sides of paralellograms have the same length. I.e., the yellow segments shown below must have the same length:

Image 3

So it is exactly enough to specify three infinite (two-sided) sequences of nonnegative integers, corresponding to $k$ above. Each sequence corresponds to the lengths of the sides running in one of the three directions. In the case where all these integers are zero, we obtain the tiling of the plane by small hexagons.

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  • $\begingroup$ Thanks for the reply! I have skimmed it and will read it more carefully. I am starting to wonder if it is more natural for the central hexagons to be two units wide instead of one. My reasoning is that you get this by starting with a periodically striped plane tiled by rhombuses all oriented the same, and then rearranging the striped rhombuses so they only meet at 6 60 degree corners or 3 120 degree corners. When I started thinking about this, I was not considering rhombuses at all, only stripes. $\endgroup$ – PaulC Jun 19 '17 at 19:06

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