Front. Mech. Eng. 2013, 8(4): 367–370 DOI 10.1007/s11465-013-0273-7
RESEARCH ARTICLE
Tong ZHANG, Kai DING
A new proof of Honeycomb Conjecture by fractal geometry methods
© Higher Education Press and Springer-Verlag Berlin Heidelberg 2013
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 twodimensional 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. Keywords Honeycomb Conjecture, fractal geometry, hierarchical fractal structure
1
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 Received April 19, 2013; accepted June 5, 2013
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Tong ZHANG ( ) Research Center for Solid Mechanics, Beihang University, Beijing 100191, China; School of Engineering, Brown University, Providence, RI 02912, USA E-mail:
[email protected] Kai DING School of Instrument Science and Optical Engineering, Beihang University, Beijing 100191, China
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, Fejes Tóth 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.
2
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. Fejes Tóth 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.
3
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
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number of polygons around the vertex, αðN Þ the degree of internal angle of N-gon, n¼
360° : αðN Þ
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 n¼
360° ¼ 6: 60°
b. Square can be a candidate, for n¼
360° ¼ 4: 90°
c. Regular pentagon can’t be, for n¼
360° 1 ¼3 2 = N þ: 108° 3
d. Regular hexagon is a candidate, for n¼
360° ¼ 3: 120°
e. Regular heptagon, αð7Þ 2 = N þ , so it can’t be. f. Regular octagon, n¼
360° 2 ¼2 2 = N þ: 135° 3
From now on, as αðN Þ increase, and 120° < αðN Þ < 180°, 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 ki , let ki ¼ lim
n↕ ↓ 1
Cn , Sn
Cnþ1 ¼ 7Cn – 3n 2 6a ðn³0Þ, C0 ¼ 6a: To solve the equations, we have pffiffiffi 3 3 2 n a, Sn ¼ 7 2 Cn ¼ 3n 3a þ 7n 3a:
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.
(3)
(4) (5)
Thus, Cn 3a 3n þ 3a 7n 2 pffiffiffi ¼ lim ¼ pffiffiffi ↓ 1 Sn n↕ ↓ 1 3 3 2 3a 7n a 2
k6 ¼ lim n↕
(1)
when Cn is the total side length of n-th order structure and Sn is the area.
4
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 > 0). From Fig. 1, we generalize the recursive relationships of Cn and Sn as: pffiffiffi 3 3 2 a, (2) Sn ¼ 7Sn – 1 ðn³1Þ, S0 ¼ 2
1:15 : a
(6)
(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.
Tong ZHANG et al. A proof of Honeycomb Conjecture by fractal geometry methods
Fig. 2 unit
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Construct hierarchical fractal structure with the triangular
We generalize from it the recursive formula of Cn and Sn : Sn ¼ 4Sn – 1 ,
(7)
Cnþ1 ¼ 4Cn – 3c 2n :
(8)
We solve from them: pffiffiffi 3 2 c, Sn ¼ 4 S0 ¼ 4 2 n
Cn ¼
n
3 3 c 2n þ c 4n : 2 2
(9)
(10)
Then we calculate k3 : 3 3 pffiffiffi c 2n þ c 4n Cn 2 3 2 2 pffiffiffi ¼ lim ¼ k3 ¼ lim : (11) n↕ ↓ 1 Sn n↕ ↓ 1 c 3 2 c 4n 4 Then we need to compare k3 and k6 , We set the area of triangular unit equals to that of hexagonal unit. So we have equation: pffiffiffi c ¼ 6a: (12)
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
Express k3 with a, pffiffiffi pffiffiffi 2 3 1:41 2 k3 ¼ pffiffiffi ¼ > k6 : a a 6a
(13)
(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 > 0), so the area of it is b2 . The configuration of the
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Table 1 Construct hierarchical fractal structure with square unit The 0th order
The 1st order
The 2nd order
first several orders is show in Table1. From Table 1 above, we generalize the recursive relationships of Cn and Sn as: Sn ¼ 9Sn – 1 ðn³1Þ, S0 ¼ b2 , Cnþ1 ¼ 4Cn – 4 2n b ðn³0Þ, C0 ¼ 4b:
(14) (15)
To solve the equations, we get: n
n
Finally, we have k3 > k4 > k6 :
Thus, any partition of the plane into regions of equal area has perimeter at least that of the regular hexagonal honeycomb tiling.
5
Sn ¼ 9Sn – 1 ¼ 9 S0 ¼ 9 b , 2
n
Cnþ1 – 3Cn ¼ 9 12b:
(21)
Prospection
(16) (17)
The method we bring forward might be effective in solving optimization problems of regular 3-dimensional fractal structure, such as Kepler Conjecture.
Thus, Cn 3n 2b þ 9n 2b 2 ¼ lim ¼ : ↓ 1 Sn n↕ ↓ 1 b 9n b2
k4 ¼ lim n↕
(18)
To compare k4 and k6 , we set the area of square unit is equal to that of hexagonal one. And we have such relationship: sffiffiffiffiffiffiffiffiffi pffiffiffiffi 3 3 b¼ a, (19) 2 2 1:24 k4 ¼ rffiffiffiffiffiffiffiffiffi > k6 : pffiffiffi a 3 3 a 2
(20)
Acknowledgements The authors thank the National Natural Science Foundation of China (Grant No. 10602028).
References 1. Hales T C. (8 Jun 1999). The honeycomb conjecture. Discrete and Computational Geometry, 2001, 25: 1–22 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
