Understanding the Honeybee Conjecture vs. Sphere Packing Conjecture

Question

I am trying to understand the differences between the Honeybee Conjecture and the Sphere Packing Conjecture (also called the Kepler Conjecture).

As a quick overview:

• "The honeycomb conjecture states that a regular hexagonal grid or honeycomb is the best way to divide a surface into regions of equal area with the least total perimeter." (https://en.wikipedia.org/wiki/Honeycomb_conjecture)

• "The Kepler conjecture, named after the 17th-century mathematician and astronomer Johannes Kepler, is a mathematical theorem about sphere packing in three-dimensional Euclidean space. It states that no arrangement of equally sized spheres filling space has a greater average density than that of the cubic close packing (face-centered cubic) and hexagonal close packing arrangements. The density of these arrangements is around 74.05%." (https://en.wikipedia.org/wiki/Kepler_conjecture)

Reading this, I had the following questions:

1) The Honeycomb Conjecture says that a hexagonal grid is the best way to divide regions into equal area and the least perimeter, whereas the Sphere Packing Conjecture says that closest packing arrangements can be achieved EITHER using Cubic Arrangements OR Hexagonal Arrangements. Do the results from both these conjectures contradict each other - or are both these conjectures addressing fundamentally different problems?

2) From the Wikipedia Page on the Honeycomb Conjecture, (apparently) the official mathematical formulation of the Honeycomb Conjecture can be written as follows:

Why do we want the supremum of the limit of "r" to approach infinity? Why are we interested in the perimeter and area of the "bounded components and the disk of radius r" - does this represent a choice of tiling arrangement? Would the disk B(0,r) represent a choice of tiling like a hexagon? The fourth root of 12 is 1.86 : why is 1.86 relevant here?

3) Finally, for both the Honeybee Conjecture and the Sphere Packing Conjecture - are we able to "bound" the sample space of the solution (i.e. how many possible arrangements can exist)?

For instance, suppose we want to predict the the possible outcomes that can arise from flipping a two-sided coin twice - the sample space corresponding to this problem would include 4 outcomes : (Head, Head), (Head, Tail), (Tail, Head), (Tail, Tail)

• If we were to consider the Sphere Packing Conjecture, we know that the sample space to this problem contains at least 3 possible arrangements: Cubic, Hexagonal and Random. But do there exist more possible arrangements? Are the number of total possible arrangements finite?

• If we were to consider the Honeybee Conjecture, the following video (https://www.youtube.com/watch?v=7edkFs8Vu1E @ 4:03 ) shows us that there the sample space of this problem contains at least 5 possible arrangements:

But do there exist more possible arrangements? Are the number of total possible arrangements finite?

I found this Wikipedia Page (https://en.wikipedia.org/wiki/Uniform_tiling) that contains (beautiful) pictures about different ways to "partition and pack" space:

Reading the Wikipedia page, the Euclidean Plane seems to have 3 types of uniform tiling arrangements : Square, Hexagonal and Triangular. But reading the "Uniform Tilings Using Star Polygons" section - it seems like you can have plenty of other types of uniform tiling arrangements (e.g. Topological 3.12.12 , Topological 4.4.4.4, , Topological 6.6.6, etc.).

"The complete lists of k-uniform tilings have been enumerated up to k=6. There are 20 2-uniform tilings, 61 3-uniform tilings, 151 4-uniform tilings, 332 5-uniform tilings, and 673 6-uniform tilings." (https://en.wikipedia.org/wiki/List_of_k-uniform_tilings) - Is it possible there could existing more tiling arrangements for k>6?

Do there exist finite or infinite Non-Uniform Tiling Arrangements?(https://commons.wikimedia.org/wiki/Category:Non-uniform_tilings_by_regular_polygons)

But to understand these (beautiful) pictures goes far beyond my knowledge, and I am still not sure if the sample space of the Honeycomb Conjecture contains finite or infinite packing arrangements.

Can someone please help me understand this?

Thank you!

Answer 1

Too many questions to answer fully.

Question #1:

They do not contradict each other. One deals with 2-dimensional configurations, the other with 3-dimensional. Similar principles can be applied in approaching them (they are packing problems after all), but a priori there is no reason to assume that the answer to the 3-dimensional question should be related to that of the 2-dimensional question.

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Question #2:

In the writing of the Honeycomb conjecture the parameter $r$ and the related limes superior are used to capture the following.

• We want to look at the big picture - what happens when we cover a large area of the plane.
• We cannot just compute the total length of the edges over the infinite plane because the answer is obviously infinite. Instead, we want to look at the density - what is the length of the edges per unit area (in a very large area).
• We want to ignore border effects (like what happens around the perimeter if we only cover a large area as opposed to the entire plane). More about this with the calculations below.
• Looking at limes superior deals with a few technical issues: we don't know at the beginning whether the limit exists at all. For example, it is (in theory at least) possible that there are better ways that work only for selected intervals of $r$, but those ways are worse for some other intervals of $r$. I failed to make this crystal clear, but on some other measurement problems this kind of difficulties come to the fore, and we have learned to be careful.

So the question is about the asymptotic edge length per unit area. Let's look at squares first:

• If we tile the plane with squares, the length of a side of a single square needs to be $1$ as we want the tiles to have area $1$.
• When we look at a large area, we see that there will be two edges per unit area. After all, each square has four edges, but each edge is shared by two squares, so it comes to two (new) edges per (new) square in average.
• As we only look at the limit when $r\to\infty$ we can ignore the border effect of tiling, say, an $n\times n$ region (or a circular region as prescribed in the formulation of the conjecture), when there are edges used by only a single square in this finite region (or partial edges). The error term disappears when we take limes superior.

Anyway, we conclude that with squares we need edges of length $2$ per unit area when covering the plane.

Hexagons next:

• The area of a hexagon of side $s$ is $A=3\sqrt3 s^2/2$. To get $A=1$ we need to scale $s=(4/27)^{1/4}$.
• When we cover a larger and larger area of the plane with hexagons, there will be three (new) edges per (new) hexagon. As in the case of squares, each of the six edges of hexagon is shared by two tiles, so in the average we need three edges per unit area (again the limes superior taking care of border effects).

The conclusion is that using a hexagonal tiling we need edges of total length $3s=\root4\of{12}$ per unit area. This is where the fourth root of twelve comes from.

The essence of the honeycomb conjecture is that this is best possible. Observe that the conjecture does not require a regular tiling. As long as the plane is partitioned into unit area regions, the shapes can be very odd, and don't all need to be congruent. There are definitely infinitely many ways of doing this. You could start with a single central $13$-gon. Surround it with $13$ identical trapezoids et cetera. Observe that a central $13$-gon needs less edges per unit area for small values of the parameter $r$. Simply because it is closer to the shape of a circle than a hexagon. The catch is that we cannot continue as efficiently from it. I think it is likely that, asymptotically, we can gradually turn it into honeycomb simply by thinking of it as a local disturbance in the honeycomb structure. Limes superior again makes such a ripple effect vanish. But we cannot "globally" do better than the honeycomb.

Question #3:

In Kepler's conjecture the two solutions are "locally identical" in the sense that they both consist of hexagonal layers. The difference is in how the layers are placed on top of each other.