A new proof of Honeycomb Conjecture by fractal geometry methods

Abstract

Based on fractal geometry, we put forward a concise and straightforward method to prove Honeycomb Conjecture—a classical mathematic problem. Hexagon wins the most efficient covering unit in the two- dimensional space, compared with the other two covering units—triangle and square. From this point of view, honeycomb is treated as a hierarchical fractal structure that fully fills the plane. Therefore, the total side length and area are easily calculated and from the results, the covering efficiency of each possible unit is provided quantitatively.

Key words： Honeycomb Conjecture; fractal geometry; hierarchical fractal structure

Introduction

In 36 B.C., Marcus Terentius Varro observed the hexagonal form of the bee’s honeycomb and wrote in his book, ‘The geometricians prove that this hexagon inscribed in a circular figure encloses the greatest amount of space.” In the 4th century, Pappus of Alexandria formally proposed the Honeycomb Conjecture: a regular hexagonal grid or honeycomb is the best way to divide a surface into regions of equal area with the least total perimeter [1]. Pappus argues that there are only three polygons that could fully tile the plane — the triangle, the square, and the hexagon, and he states that if the same quantity of material is used for the constructions of these figures, it is the hexagon that will be able to hold more honey [1]. Thomas Hales, a mathematician at the University of Michigan, Ann Arbor, held negative viewpoints about Pappus’ work, for it is not based on a rigorous mathematical proof . In fact, Pappus’ intuitive is right and we have proved it [2].

In 1943, L. Fejes Tόth successfully proved Honeycomb Conjecture under the hypothesis that all the covering units should be convex. In 1999, Thomas Hales, finally released an integral proof without the assumption by bringing in the isoperimetric properties. But his proof is based largely on algebra. Now we prove this conjecture more concisely by the method of fractal geometry and algebra.

We regard honeycomb as a hierarchical fractal structure, which is a more briefer and clear way to portray it. Then, by generalizing the conjecture into several cases, we calculate the ratio of total length to area of different cases, and prove that hexagon wins the most efficient covering unit.

A proof of the conjecture

Thomas Hales has proved that when the perimeter of a region is fixed, the region bounded by circular arcs encompasses less area than one bounded by straight lines [1]. Thus, we need only talk about polygons.

L. Fejes Tόth has proved that when the perimeter is fixed, regular polygons occupies the largest area in 2-dimentional plane [3]. Therefore, it left us only problem to solve — what kind of regular polygon could fully tile the plane with the shortest total side length? And we give out the answer.

Requirement for fully covering the plane

Without loss of generality, we consider a vertex around which there are several uniform polygons. Let n be the number of polygons around the vertex, \alpha (N) the degree of internal angle of N-gon,

n={\displaystyle \frac{\mathrm{360}°}{\alpha (N)}}.

To maintain there is neither overlapping nor interstices, n must be a positive integer. Then, we begin with the simplest:

a. Regular triangle is a candidate, for

n={\displaystyle \frac{\mathrm{360}°}{\mathrm{60}°}}=\mathrm{6.}

b. Square can be a candidate, for

n={\displaystyle \frac{\mathrm{360}°}{\mathrm{90}°}}=\mathrm{4.}

c. Regular pentagon can’t be, for

n={\displaystyle \frac{\mathrm{360}°}{\mathrm{108}°}}=\mathrm{3}{\displaystyle \frac{\mathrm{1}}{\mathrm{3}}}\notin N^{+}.

d. Regular hexagon is a candidate, for

n={\displaystyle \frac{\mathrm{360}°}{\mathrm{120}°}}=\mathrm{3.}

e. Regular heptagon,

\alpha (\mathrm{7})\notin N^{+}

, so it can’t be.

f. Regular octagon,

n={\displaystyle \frac{\mathrm{360}°}{\mathrm{135}°}}=\mathrm{2}{\displaystyle \frac{\mathrm{2}}{\mathrm{3}}}\notin N^{+}.

From now on, as \alpha (N) increase, and , n will never be an integer again. Thus, only three candidates survive: the regular triangle, the square and the hexagon. Honeycomb Conjecture reduces to the problem of compare the tiling efficiency between these three, which Pappus claimed many years ago.

Here, based on constructing method of hierarchical fractal structure (related in the next section), respectively, we let the three covering units grow larger and larger and finally fully fill the plane. Thus their covering efficiency can be expressed quantitatively. We bring in a variable k_i, let

(1)k_i=\underset{n\to \infty }{\lim }{\displaystyle \frac{C_n}{S_n}}.

when C_n is the total side length of n-th order structure and S_n is the area.

Hierarchical fractal structure

Honeycomb can be regarded as a hierarchical fractal structure.

Construction process of this structure could be considered as the uniform hexagons paving step by step, as shown in Fig. 1.

Fig.1 Construction process of the hexagonal hierarchical fractal structure. (a)The 1st order structure is formed by placing six hexagons around the central one. All of them are congruent. (b) The 2nd order; around 1st order structure, six duplications are placed without margin or overlapping. (c) The 3rd order; around the 2nd order, six duplications are placed without margin and overlapping

Full size|PPT slide

Step 1: place a unit hexagon and take it as the center of the whole structure. As shown in Fig. 1 (a) by a green hexagon. We define it as the 0th order structure.

Step 2: place six congruent hexagons tightly around the center hexagon. As shown in Fig. 1 (a). These seven hexagons share the same side length. Then we get the 1st order.

Step 3: treat the group of hexagons in Fig. 1 (a) as a “unit,” and make six duplications of this group around the new “unit,” as shown in Fig. 1(b). The 2nd order comes into being.

Step 4: then repeat the same process. Take the group of hexagons in Fig. 1(b) as a new “unit,” make six duplications and tile them around the new “unit” tightly, as shown in Fig. 1(c). And the 3rd order is formed.

Step 5: take the former group in Fig. 1(c) as a new unit and place six duplications around it.

……

By repeating the same process, we will have any area covered by the same regular hexagons without overlapping or margin.

(I) Hexagon unit

Now, let the side length of a hexagon unit equal to a(a>\mathrm{0}).

From Fig. 1, we generalize the recursive relationships of C_n and S_n as:

(2)S_n=\mathrm{7}{S}_{n-\mathrm{1}}(n\ge \mathrm{1}),S_{\mathrm{0}}={\displaystyle \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}{a}^{\mathrm{2}}.

(3)C_{n+\mathrm{1}}=\mathrm{7}{C}_n-{\mathrm{3}}^n\times \mathrm{2}\times \mathrm{6}a(n\ge \mathrm{0}),C_{\mathrm{0}}=\mathrm{6}a.

To solve the equations, we have

(4)S_n={\mathrm{7}}^n\times {\displaystyle \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}{a}^{\mathrm{2}}.

(5)C_n={\mathrm{3}}^n\times \mathrm{3}a+{\mathrm{7}}^n\times \mathrm{3}a.

Thus,

(6)k_{\mathrm{6}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{C_n}{S_n}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{\mathrm{3}a\times {\mathrm{3}}^n+\mathrm{3}a\times {\mathrm{7}}^n}{{\mathrm{7}}^n\times {\displaystyle \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}{a}^{\mathrm{2}}}}={\displaystyle \frac{\mathrm{2}}{\sqrt{\mathrm{3}}a}}\approx {\displaystyle \frac{\mathrm{1.15}}{a}}.

(II) Regular triangle unit

Construct hierarchical fractal structure with the triangular unit in the same way, as shown in Fig. 2, let the unit’s side length be c. The central triangle with red boundary is a basic unit, defined as the 0th order. The larger triangle with black edge is the 1st order, containing four basic units. The 2nd order, with blue boundary line, is constructed by four 1st-order structures. And following the same rule, the whole plane will be finally covered by such basic triangular covering units.

Fig.2 Construct hierarchical fractal structure with the triangular unit

Full size|PPT slide

We generalize from it the recursive formula of C_n and S_n:

(7)S_n=\mathrm{4}{S}_{n-\mathrm{1}}

(8)C_{n+\mathrm{1}}=\mathrm{4}{C}_n-\mathrm{3}c\times {\mathrm{2}}^n

We solve from them:

(9)S_n={\mathrm{4}}^n{S}_{\mathrm{0}}={\mathrm{4}}^n\times {\displaystyle \frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}{c}^{\mathrm{2}}

(10)C_n={\displaystyle \frac{\mathrm{3}}{\mathrm{2}}}c\times {\mathrm{2}}^n+{\displaystyle \frac{\mathrm{3}}{\mathrm{2}}}c\times {\mathrm{4}}^n

Then we calculate k_{\mathrm{3}}:

(11)k_{\mathrm{3}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{C_n}{S_n}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{{\displaystyle \frac{\mathrm{3}}{\mathrm{2}}}c\times {\mathrm{2}}^n+{\displaystyle \frac{\mathrm{3}}{\mathrm{2}}}c\times {\mathrm{4}}^n}{{\displaystyle \frac{\sqrt{\mathrm{3}}}{\mathrm{4}}}{c}^{\mathrm{2}}\times {\mathrm{4}}^n}}={\displaystyle \frac{\mathrm{2}\sqrt{\mathrm{3}}}{c}}

Then we need to compare k_{\mathrm{3}} and k_{\mathrm{6}},

We set the area of triangular unit equals to that of hexagonal unit. So we have equation:

(12)c=\sqrt{\mathrm{6}}a

Express k_{\mathrm{3}} with a,

(13)k_{\mathrm{3}}={\displaystyle \frac{\mathrm{2}\sqrt{\mathrm{3}}}{\sqrt{\mathrm{6}}a}}={\displaystyle \frac{\sqrt{\mathrm{2}}}{a}}\approx {\displaystyle \frac{\mathrm{1.41}}{a}}>k_{\mathrm{6}}

(III) Square unit

Construct hierarchical fractal structure with square unit by the same method. Let the side length of a square unit equal to b(b>\mathrm{0}), so the area of it is b^{\mathrm{2}}. The configuration of the first several orders is show in Table1.

Tab.1 Construct hierarchical fractal structure with square unit

The 0th order	The 1st order	The 2nd order
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From Table 1 above, we generalize the recursive relationships of C_n and S_n as:

(14)S_n=\mathrm{9}{S}_{n-\mathrm{1}}(n\ge \mathrm{1}),S_{\mathrm{0}}=b^{\mathrm{2}}

(15)C_{n+\mathrm{1}}=\mathrm{4}{C}_n-\mathrm{4}\times {\mathrm{2}}^n\times b(n\ge \mathrm{0}),C_{\mathrm{0}}=\mathrm{4}b

To solve the equations, we get:

(16)S_n=\mathrm{9}{S}_{n-\mathrm{1}}={\mathrm{9}}^n{S}_{\mathrm{0}}={\mathrm{9}}^n\times b^{\mathrm{2}}

(17)C_{n+\mathrm{1}}-\mathrm{3}{C}_n={\mathrm{9}}^n\times \mathrm{12}b

Thus,

(18)k_{\mathrm{4}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{C_n}{S_n}}=\underset{n\to \infty }{\lim }{\displaystyle \frac{{\mathrm{3}}^n\times \mathrm{2}b+{\mathrm{9}}^n\times \mathrm{2}b}{{\mathrm{9}}^n\times b^{\mathrm{2}}}}={\displaystyle \frac{\mathrm{2}}{b}}

To compare k_{\mathrm{4}} and k_{\mathrm{6}}, we set the area of square unit is equal to that of hexagonal one. And we have such relationship:

(19)b=\sqrt{{\displaystyle \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}}a

(20)k_{\mathrm{4}}={\displaystyle \frac{\mathrm{2}}{\sqrt{{\displaystyle \frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}}}a}}\approx {\displaystyle \frac{\mathrm{1.24}}{a}}>k_{\mathrm{6}}

Finally, we have

(21)k_{\mathrm{3}}>k_{\mathrm{4}}>k_{\mathrm{6}}

Thus, any partition of the plane into regions of equal area has perimeter at least that of the regular hexagonal honeycomb tiling.

Prospection

The method we bring forward might be effective in solving optimization problems of regular 3-dimensional fractal structure, such as Kepler Conjecture.

Acknowledgements

The authors thank National Natural Science Foundation of China (No. 10602028).

References

1 Hales T C. (<day>8</day><month>Jun</month>1999). “The Honeycomb Conjecture”. Discrete and Computational Geometry 25: 1-22 (2001). arXiv:math/9906042

2 Hales T C. Cannonballs and Honeycombs. Notices of the American Mathematical Society, 2000, 47: 440-449

3 Fejes Tóth L. What the bees know and what they do not know. Bulletin of the American Mathematical Society, 1964, 70(4): 468-481