Báo cáo toán hoc:" An Edge-Minimization Problem for Regular Polygons " pps

There will Figure 1: Optimal configuration for 10 heptagons usually be more than one configuration of k polygons which minimizes the number ofsides see Figure 2 so we pose the following

An Edge-Minimization Problem for Regular Polygons

Ralph H Buchholz <teufel pi@yahoo.com>

Warwick de Launey <warwickdelauney@earthlink.net>

Submitted: Aug 19, 2008; Accepted: Jul 11, 2009; Published: Jul 24, 2009

Mathematics Subject Classification: 05B45, 52C20

Abstract

In this paper we will examine the following problem: What is the minimumnumber of unit edges required to construct k identical size regular polygons in theplane if sharing of edges is allowed?

In this paper we will examine the following problem:

Question 1 What is the minimum number of sides required to construct k identical sizeregular polygons in the plane if sharing of sides is allowed?

Below is an optimal configuration of 10 heptagons which reuses 11 sides There will

Figure 1: Optimal configuration for 10 heptagons

usually be more than one configuration of k polygons which minimizes the number ofsides (see Figure 2) so we pose the following harder problem

Question 2 What are all the optimal configurations?

This second question is particularly interesting because these optimal configurations arelikely to arise in nature For example, there are the quasi-periodic Penrose tilings which

have been found to correspond to the arrangement of atoms in certain types of non-sticksurfaces; biological cell growth on a surface around fixed obstacles; growth of soap filmsbetween parallel walls [6]; large-scale convection cells on the surface of the sun and otherstars—even the hexagonal structure at the pole of Saturn may be a portion of an energyminimisation tiling of the surface of that planet.

Figure 2: The two other optimal configurations for 10 heptagons

The earliest reference to Question 1appears to be an article of Harary and Harborth([5]) where they provide a complete answer for the square, the equilateral triangle, and theregular hexagon They introduce a spiral algorithm which (for each of the three shapes)allows them to build up a sequence of optimal configurations C1, C2, , Ck, , such that

Ck+1 is obtained from Ck by adding a single cell

The authors of the present paper have been studying this problem since the early1980s (see [1]) and related papers have recently begun to appear For example, a series

of papers ([12], [13], [14], [16]) describe the impact of minimum perimeter tilings on thedesign of databases and they obliquely make reference to the square spiral algorithm.More explicitly, the spiral algorithm applied to squares can be found in [9] and there isthe suggestion of the same process applied to squares, triangles and hexagons in Sloane’sOnline Integer Sequence Encyclopedia [10], (see A137228, A078633, A135708)—the lastone is derived from the work in [8] While the seminal tome on tilings in the literature([3]) does not contain a direct reference to this problem, it does contain many of the tilings

we consider in this paper Finally, interesting animations of pentagonal near-tilings areprovided on the Wolfram site (see [7] and [11])

It happens that the square, the equilateral triangle and the hexagon are the onlyregular polygons which tessellate the plane The existence of these tessellations seems

to be the key to Harary and Harborth’s success for n = 3, 4 and 6 When the n-gondoes not tessellate the plane, there are some internal unshared sides This leads to com-plications which seem to make the questions for general n much harder In this paper,

we obtain asymptotically optimal configurations for all n Our approach is to constructnear-tessellations and then apply a spiral allgorithm These near tessellations are formed

by putting together minimal cycles of ngons This is possible because the minimal cyclesalways have length 3, 4 or 6 For some n, there are uncountably many near-tessellationseach producing asymptotically optimal configurations To prove the resulting configura-

tions are in fact asymptotically optimal, we use a correspondence between configurationsand planar graphs.

Let n be odd Whenever the bottom side of an n-gon is horizontal, the uppermostextremity of the n-gon consists of a vertex lying directly above the center of the bottomside In this case, we say the n-gon is oriented upwards Similarly, we say the n-gon isoriented downwards if the vertex lies at the bottom of the n-gon and the horizontal sidelies at the top In both cases, we say the n-gons are properly oriented When n is even,

an n-gon is properly oriented if the top and bottom sides are both horizontal Figure 3

illustrates examples of each of these oriented polygons Notice that, after an appropriate

Figure 3: Properly oriented polygons with 3-7 sides

rotation, all configurations contain at least one properly oriented n-gon When n is even,

it is clear by symmetry that any n-gon sharing a side with a properly oriented polygon

is also properly oriented Figure 4 shows two superimposed properly oriented n-gons ofopposite orientations The symmetry of the figure indicates that the side A of the solid

B

A

Figure 4: Superimposed properly oriented polygonsn-gon is parallel to the side B of the dashed n-gon, and that no other side of the dashed

n-gon is parallel to the side A Thus, no matter how one manœuvres the dashed n-gon, if

it shares the side A, then it will be properly oriented with orientation opposite to that ofthe solid n-gon

Proposition 1 Suppose a configuration contains a single properly oriented n-gon Thenall n-gons in the configuration are properly oriented Moreover, for odd n, adjacent poly-gons have opposite orientation

Given a connected configuration of side-sharing polygons, one can form, by drawing insideeach polygon a maximal inscribed circle, a connected configuration of touching circles.Figure 5 illustrates the construction Notice that reduction to a circle configuration loses

Figure 5: Reduction to a circle configuration

the orientation information Nevertheless, circle configurations capture important features

of the full problem In particular,

Proposition 2 In a circle configuration corresponding to a polygon configuration, twocircles touch iff their corresponding polygons share an edge

In general, a circle configuration is an arrangement of identically-sized circles in the plane

so that circles may touch but not overlap A circle configuration is connected if there is away to move between any pair of circles by passing along a sequence of touching circles.Most circle configurations cannot be obtained by reduction from a polygon configura-tion However, the following question regarding general circle configurations is particularlyrelevant to our investigation of polygonal configurations

Question 3 What is the maximum number of pairs of touching circles possible in a figuration of v circles?

con-2.3 Configurations and Planar Graphs

Each configuration of n-gons in the plane yields a planar graph G where

• each node corresponds to an n-gon in the configuration, and

• two nodes are joined if they correspond to side-adjacent n-gons

Let v denote the number of nodes, f denote the number of faces, and e denote the number

of edges By Euler’s formula,

In general, there are several internal faces and one external face Each internal face isbounded by a closed walk, the boundary of the face Think of the closed walk as orientedclockwise The boundary of the external face is oriented counterclockwise If all of theboundaries of all of the faces are traversed, each edge in the graph is traversed twice, once

in each direction Thus if the i-th face has a boundary walk of length ci, then

13 14 15

7

10 9

Figure 6: A configuration of pentagons and its corresponding planar graph

Example 1 In Figure 6 we show a configuration of 15 pentagons and the ing planar graph G with labeled nodes G contains v = 15 nodes, e = 17 edges and

correspond-f = 4 correspond-faces The boundary walks for the three internal cycles are (1, 2, 3, 8, 14, 15),(3, 4, 5, 6, 7, 8), and (8, 10, 11, 12, 13, 14) The boundary walk for the external cycle is(1, 2, 3, 4, 5, 6, 7, 8, 9, 8, 10, 11, 12, 13, 14, 15) Notice that the edge (8, 9) is traversed twice

- once in each direction

Proposition 3 Let n denote the number of sides of a polygon When n is odd, thecorresponding planar graph is bipartite In particular, all cycles have even length

2.4 Asymptotics

An asymptotic answer to Question 1 is now possible By (1), maximizing the number ofshared sides e in a configuration of v n-gons is equivalent to maximizing the number offaces f Equation (1) also implies

e

f − 1 = 1 +

v − 1

f − 1.Therefore, maximizing f for fixed v is equivalent to minimizing e/(f − 1) In the nextsubsection we will show that for each n there is a minimum length cmin ∈ {3, 4, 6} for theboundary of any face Under the assumption that cmin is well defined for each n, (2) thenimplies

where c1 is the length of the external cycle In order to exhibit configurations which are atleast asymptotically optimal, it is therefore sufficient to find configurations whose internalfaces all have size cmin, and whose external faces have size dominated by f

We use the term necklace to denote a finite collection of non-overlapping congruent circlesfor which every circle touches precisely two others Before determining the value of cmin,

we consider small necklaces of circles It happens that only cycles of length 3, 4 and 6need be considered (see Figure 7 for examples) The 3 cycle is unique On the other

Figure 7: 3, 4, and 6 cycleshand, there are many possible 4 cycles, all of them symmetric in the sense that thequadrilateral subtended by the centers of the circles have alternating internal angles equal.The situation for 6 cycles is markedly different, greatly complicating the situation for n-gons with n odd Note that the internal angles can all be different, and that one of theangles can even be greater than π—thus creating a concave cycle

2.6 Minimal Polygon Necklaces

We extend the notion of a necklace to include regular polygons, but two polygons areallowed to touch only if they share an entire edge The immediate goal is to determinehow the smallest number cmin of polygons in an n-gon necklace depends on the number

n Figure8 shows the three non-isomorphic minimal necklaces for n = 11

Figure 8: The minimal 11-gon necklaces

Consider a face with boundary equal to the closed walk v1, v2, , vc, vc+1 = v1 Let

−π ≤ αi < π denote the angle turned to the right as we pass the node vi Then, because

we wind around the interior of the face exactly once as we traverse the closed walk,

2π = α1+ α2+ · · · + αc (3)The angle αi is called the change in bearing as we pass through vertex vi

Consider what happens in the corresponding polygon configuration as we move around

a boundary Each vertex in the planar graph is associated with a polygon in the sponding configuration, therefore each cycle in the planar graph corresponds to a necklace

corre-of polygons If the necklace contains c polygons, then the necklace contains cn sides, corre-ofwhich c are shared If there are I edges on the interior border, the remaining E = cn−I−2care on the outside of the necklace Let Ei and Ii respectively denote the number of ex-ternal and internal polygon sides contributed to the necklace by the the i-th polygon.Then

Figure 9: Change in bearing versus the number of internal and external sides

We can now use this fundamental equation to determine cmin for all possible n-gons.The smallest possible cycle occurs when c = 3 As I1 = I2 = I3 = I, equation (6) becomes

which is impossible to solve over the integers (the left side is odd, and the right side iseven) In particular, no 3-cycles or 5-cycles can occur Next, if 4-cycles are possible then,

as above, symmetry would force I1 = I3, I2 = I4, which means

2m + 1 = 4 + 2I1+ 2I2from equation 7 This is also impossible

Now we simply need to demonstrate the existence of solutions for a 6-cycle Equation(6) becomes

2n = I1+ I2+ I3+ I4+ I5+ I6+ 6

We attempt to solve a two parameter specialization by setting (I1, I2, I3, I4, I5, I6) =(a, a, b, a, a, b) and n = 2m + 1, namely,

All this proves the following:

Proposition 4 The smallest edge-sharing necklace for an n-gon is a

Notice that 4-cycle necklaces occur when n ≡ ±2 (mod 6) The symmetry imposed bythe 4 incircles forces these necklaces to always have at least the cyclic group C2 as asymmetry group However, whenever n is a multiple of four, then C4 is possible as well

If α1, , α4 denote the four bearings of the 4-cycle, then symmetry forces α1 = α3,

α2 = α4 and α1 = π − α2 Translating this into the internal edge counts leads to

(I1, I2, I3, I4) =m,n

2 − m − 2, m,n

2 − m − 2,where m is constrained by the non-overlapping requirement π

3 < (m + 1)2π

n Notice thatthe sum of the internal edges is fixed for each polygon All such m are possible, and toavoid double counting, (m + 1)2πn ≤ π

2 is required, which leads ton

6 − 1 < m ≤ n

4 − 1

Examples of 4-cycles for small n are shown in Table1 The number of 4-cycle necklaces,

N4(n), is now easy to obtain:

N4(n) =

jn

4 − 1k−ln

6 − 1m for n ≡ ±2 (mod 6)

With 6-cycle polygonal necklaces, unlike the 4-cycle case, nothing can be inferred aboutthe symmetry of the discrete case from the continuous case In particular, the possible

6 internal arc-lengths of a 6-circle necklace can all be distinct However, a systematictaxonomy of 6-polygon necklaces for small polygons does not reveal any correspondingasymmetric (or concave) necklaces—all examples to date are convex and have either 2, 3,

or 6-fold symmetry

For a regular n-gon we denote the incircle radius by rn, the circumcircle radius by Rn,and the angle subtended at the centre by a side by θn = 2π/n The search begins byfixing two n-gons with their centres on the x-axis at P1 = (rn, 0) and P2 = (−rn, 0) (seeFigure10) The next three n-gons are centred at P3, P4 and P5, sharing an edge with their

distances between pairs of non-adjacent points in {P1, P2, P3, P4, P5, P6} are computed—aconfiguration fails if any such distance is less than 2rn (due to overlap) and passes if allsuch distances are greater than 2Rn Any remaining “undecided” configurations can besubjected to a check using the roots of unity representation, described in section 2.9.

If the number of interior (non-shared) edges on each n-gon are respectively denoted

by (I1, I2, I3, I4, I5, I6), as one traverses the inside of the 6-cycle starting at P2 say, then

Ii = αi

θn − 1

Table 2shows the results for odd n-gons with up to 13 sides In particular, extendingthis table up to 301 sides reveals that odd n-gons with n ≡ 0 (mod 3) have 6-cycleswith 2, 3 or 6-fold symmetry—i.e., 6-cycles of the form (a, b, c, a, b, c), (a, b, a, b, a, b), or(a, a, a, a, a, a) When n ≡ ±1 (mod 6) the 6-cycles only have 2-fold symmetry—i.e., an

0 50 100 150 200 250

of some 2n-th root of unity for the same purpose, noting in this case the roots of unityalternate their exponent (between odd and even) with the orientation of the underlyingn-gon as the path of edge-sharing n-gons is traversed (see Figure 12)

Odd n-gons, with minimal 6-cycles and a properly oriented1 n-gon at the origin, mustsatisfy

for integers αi, βi, where ζ2n = e2πi/2n The lack of asymmetry in the computationalresults, and the facts that ζn

2n= −1 and ζ2a

2n = ζa

n lead to the following conjecture

Conjecture 1 For n odd and ζn= e2πi/n,

−ζ b1n

−ζ n

−ζ b3n

+ζ a 1 n

+ζ a3n

+ζ a 2 n

Figure 12: Roots of unity joining centres of n-gons in a 6-cycle

all prime n greater than 6, relies on the properties of cyclotomic polynomials Since this

is completely subsumed by the work in the next section, we omit it

So far, simple arguments have allowed us to firstly, show that the minimal cycles for n-gonsare of length 3, 4, or 6, depending on the residue class of n modulo 6, and secondly, classifythe minimal cycles of length 3 or 4 However, there seems to be no simple argument thatallows us to classify the minimal 6-cycles

In this section, as a first step towards a classification of minimal 6-cycles, we usearguments in cyclotomic rings to characterize all of the solutions of the equation

where ζn denotes a primitive n-th root of unity for any natural number n

Once the choices for the 6-tuple (e1, e2, e3, e4, e5, e6) are classified, the question of which6-tuples map to 6-cycles remains We will address these geometric issues in a separatesection The solutions to the above equation are determined by the primes which divide

n We begin with two simple lemmas, the first of which we will need to generalize in order

to describe the solutions to equation (8)

Lemma 1 Let n be an odd natural number, and suppose that

ζe1

n = (−1)wζe2

n ,where w = 0, 1 and 0 ≤ e1, e2 < n Then w = 0 and e2 = e1

Proof If w = 1, then the multiplicative group of odd order n generated by ζn containsthe element −1, which has order 2 Therefore w = 0 Since ζn has order n, we have

e2 ≡ e1 (mod n) Since 0 ≤ e1, e2 < n, we must have e2 = e1 Lemma 2 Let p be an odd prime, and let s be odd and coprime to p The minimalpolynomial of a complex primitive pk+1-th root of unity over the ring Z[ζs] is the polynomial

f (x) = 1 + xpk+ x2pk+ · · · + x(p−1)pk

Proof Let ω := ζpk+1 and let m(x) denote the minimum polynomial of ω over Z[ζs].Notice that f (ω) is the sum of the complex p-th roots of unity, so f (ω) = 0 Thereforem(x) divides f (x) Moreover, since f (x) is monic, it is sufficient to show that m(x) and

f (x) have the same degree

Now, for any ring R, the minimum polynomial m(x) of ω over R is

Y

α

(x − α(ω)) ,

where α ranges over the automorphisms of the ring R[ω] which fix ω Since s is coprime

to p, the ring R[ω] is Z[ω1/s] when R = Z[ζs] In this case, each automorphism whichfixes ω corresponds to an element ωi = α(ω), where i ∈ {1, 2, , pk+1− 1} is coprime to

p Consequently, m(x) has degree (p − 1)pk= deg(f ) 

We now prove a substantial generalization of Lemma 1

Lemma 3 Let u and v be nonnegative integers, and let n be an odd natural number.Suppose that, for w = 0 or 1,

If n is 1 or a product of primes exceeding u + v2 and v +u2, then either

1 we have w = 0 and {{e1, e2, , eu}} equals {{eu+1, eu+2, , eu+v}} as multisets,or

2 u + v is an odd prime dividing n, and w = 1

Proof We prove this by induction on the number of primes dividing n, proving theinductive step and the initial cases at the same time Let p be the smallest prime dividing

n = spk+1, where s is coprime to p and k ≥ 0 The initial case is when s = 1 and n is aprime power, while the inductive step is when s > 1 and n has at least two distinct primedivisors

We show that if a = 1 then p = u + v is an odd prime dividing n and w = 1 If a = 1,then we have

(u + v)/2 < p ≤ (u + v) (12)Since (u + v)/2 < p, there must be one exponent gi0 which is distinct from the otherexponents gj Therefore, we must have A(x) = ±ζsfxt for some f and t Consider twocases: w = 0 and w = 1 If w = 0, then (11) becomes

Indeed, since u + v2 < p and v + u2 < p, there are i0 ∈ {1, 2, , u} and j0 ∈ {u + 1, u +

2, , u + v} such that gi0 6= gj for all j 6= i0, and gj0 6= gi for all i 6= j0 But then

ζsfi0 = ±ζsf = −ζsfj0,which is impossible, by Lemma 1 This argument applies equally well when s = 1 (aninitial case) and when s > 1 (the inductive step)

Now consider the case w = 1: (11) becomes