christoph bandt and nguyen viet hung- fractal n-gons

8/3/2019 Christoph Bandt and Nguyen Viet Hung- Fractal n-Gons

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FRACTAL n-GONS

CHRISTOPH BANDT AND NGUYEN VIET HUNG

Abstract. We consider all self-similar sets in the plane for which the cyclic groupacts transitively on the pieces, as is the case for n-gon Sierpinski gaskets, for squareand terdragon. For each n our family is parametrized by the points in the unitdisk. Due to a connectedness criterion, we can define Mandelbrot sets and useto them to find various new examples with interesting properties. We study aninfinite class of simple n-gons which contains Sierpinski n-gons as a special case,and determine resistance exponents for all members of this class.

MSC classification: Primary 28A80, Secondary 52B15, 34B45

1. Introduction

Most of the classical fractals, like Cantors middle-third set, von Kochs curve,Sierpinskis gasket and carpet, Mengers sponge etc. [18, 8, 12] are self-similar setswith certain symmetries. That is, they are compact sets which fulfil an equation

A = f1(A) ... fn(A) (1)where f1,...,fn are contracting similarity maps on R

d. And there is at least onesymmetry map s with s(A) = A which permutes the pieces Ak = fk(A), so thats(Ak) = Aj where j depends on k and s. The self-similarity and symmetry makethe shapes look attractive, but what is more, they simplify the description andmathematical treatment of such fractals. An analytic theory including Brownianmotion [17], Dirichlet forms [16], spectrum of the Laplacian [1] and geodesics [23]has been developed only on symmetric fractals. Recently, Falconer and OConnor[13] have classified and counted certain symmetric fractals.

This paper studies a large family of fractals with rich symmetry. A compact subset

A of the complex plane is called a fractal n-gon ifA fulfils equation (1) for similaritymappings fk(z) = kz + ck, k = 1,...,n, and there is a rotation in the plane whichacts transitively on the pieces, permuting them in an n-cycle.

Some fractal n-gons which generalize the Sierpinski gasket for arbitrary n havebeen considered by several authors [19, 23]. Some others like twindragon, terdragon,Gospers snowflake and the fractal cross [18, Ch. 6] are known as tiles. Figures 1and 2 show some examples which seem to be new.

For each n, all fractal n-gons are parametrized by one complex parameter run-ning through the unit disk (Section 2). Since connectedness of A means that neigh-boring pieces intersect (Section 3), we can define a Mandelbrot set Mn for fractaln-gons in Section 4. M2 was introduced by Barnsley and Harrington in 1985 [9, 8]1


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2 CHRISTOPH BANDT AND NGUYEN VIET HUNG

Figure 1. 3-gon fractals where pieces intersect in one point. Upperrow: The point has addresses 01 102 , 01 10202 (Example 1).Lower row: Corresponding reverse fractals, see Section 5.

and studied by several authors [10, 11, 2, 20, 21, 22]. We shall see that Mn hassimilar properties. In section 5, we describe a fast algorithm to generate Mn, anddiscuss the overlap set of n-gons which determines their geometry. Finally, we con-sider the simplest n-gons with one-point intersection set, classify them and derivethe basic analysis for this infinite family.

2. Parametrization

We want to derive a very simple form for the mappings fk which generate a fractaln-gon. We start with the simple observation that a similarity map, that is, a changeof the Cartesian coordinate system, will not change the structure of a self-similar

set.

Remark 1. LetA =

fk(A) be a self-similar set inRd andh : Rd Rd a similarity

map. Then B = h(A) is the self-similar set generated by the mappings gk = hfkh1.

Proof. B = h(A) =

hfk(A) =

hfkh1(B)

In the complex plane, when h consists of translation, scaling and rotation, i.e.h(z) = z + c, then gk will have the same factor k as fk. If h includes a reflectionand thus reverses orientation, h(z) = z + c, then the gk will have the conjugatefactors k. In the present paper we consider all similitudes h as isomorphisms anddo not distinguish between a self-similar set and its mirror image.


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FRACTAL n-GONS 3

Proposition 1. Any fractal n-gon is similar to a self-similar set A =n

k=1 fk(A)where

fk(z) = z + bk for k = 1,...,n where b = cos

2

n+ i sin

2

nand || < 1.

Thus for each n, the points in the open unit disk parametrize fractal n-gons.

Proof. By definition, a fractal n-gon A has n pieces Ak, k = 1,..,n, and a rotations transforms the pieces cyclically into each other. We choose the numbering so thats(Ak) = Ak+1 where + is taken modulo n. In the sequel we shall often write A0 andf0 instead of An, fn.

By applying a translation h(z) = z +v we push the origin of our coordinate system

to the fixed point of s so that s(z) = bz where bn

= 1 since sn

preserves all piecesof A and so must be the identity map. Since bk runs through all roots of unity, it isno loss of generality (but requires another renumbering of the Ak) to assume thatb = cos 2

n+ i sin 2

n.

We took the fk to be orientation-preserving, so f0(z) = z + u for some , u Cwith || < 1. Applying the rotation h(z) = z/u we transform A into a positionwhere f0(z) = z + 1.

Now we note that skf0sk(A) = skf0(A) = fk(A) for k = 1,...,n 1. Thus

A =n1

k=0 skf0sk(A) so due the uniqueness of self-similar sets [12, 8] the set A is

the self similar set with respect to the mappings skf0sk(z) = z+bk, k = 0,...,n1.

These mappings will now be called fk.

The points z in a self-similar set are often described by their addresses u1u2...,sequences of symbols um {0,...,n 1} which denote pieces and subpieces of A towhich z belongs [8, 12, 5]. We have

z = p(u1u2...) = limm

fu1 fum(0) (2)

where p : {0,...,n 1} A is the so-called address map [8]. In our case, theaddress map has a particularly simple form.

Remark 2. The point z with address u1u2..., um {0,...,n 1} in the fractal n-gon A() has the representation z = m=1 bumm1. Thus A() is the support of arandom series of powers of with coefficients chosen from the n-th roots of unity.Proof. fu1 fum(0) = bu1 + bu2 + + m1bum .

Example 1. We explain how to determine for the first fractal in Figure 1.The intersection point z should have the addresses 01 and 102 where 1 = 111...This is indicated by the relation 01102 which by the above remark turns into anequality for :

p(01) = 1 + b( + 2 + ...) = 1 + b

1 = p(102) = b + + b2

2

1


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Figure 2. Some 4-gon fractals. In the upper row, pieces intersectin two points, in the lower row in four points. See Example 2.

which leads to the quadratic equation 2(1 + b)

2 + 1 = 0. The solution with

|| 1 is =

1

2+ (1

3

2)i .

For 0110202 we have to consider p(10202) = b + + b22 + 3 + b44/(1) whichleads to the equation z4 z3 + z2 + 2bz = b. Again there is only one solution with|| 1 which is determined numerically: 0.5135 + 0.1004i. In both cases 1/is a Pisot number over Q(b). This also holds for the lower row of Figure 1, where was replaced with (see Section 5).

3. Connectedness

Connectedness of fractals is a very important property, which is well-known fromthe dynamics of complex quadratic maps. For self-similar sets, it implies localconnectedness and arcwise connectedness [14, 6].

A self-similar set with n pieces Ak is connected if and only if the graph withvertices k = 1,...,n and with edges {j,k} for intersecting pieces Aj Ak = isconnected [14, 6]. Thus a self-similar set A with two pieces A0, A1 is connected ifand only ifA0 A1 = [8, Ch. 8].

We shall see that this also holds for fractal n-gons. Due to the rotational symme-try, A0 A1 = implies Ak Ak+1 = for each k which implies connectedness ofAdue to the above criterion. The necessity of A0 A1 = is more difficult to prove.


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FRACTAL n-GONS 5

Theorem 2. A fractal n-gon A is connected if and only if A0 A1 = . If A isdisconnected, then it is totally disconnected, and all pieces Ak are disjoint.

Proof. Let B(x, r) denote the ball around x with radius r. Let be the smallest

positive number such that A B(0, ) = B0. For any word u = u1...um {0,...,n1}m of length |u| = m let fu = fu1 fum. For m = 1, 2,... let

Bm =|u|=m

fu(B(0, )) =|u|=m

B(fu(0), ||m)

be the m-th level ball approximation of A. We shall use the following fact.(*) IfBm is connected and f0(Bm) f1(Bm) = , it follows that fj(Bm) intersects

fj+1(Bm) for j = 1,...,n 1 and Bm+1 =

n1j=0 fj (Bm) is connected.

Now we distinguish two cases.Case one. A0 A1 = . In this case (*) implies by induction that all Bm are

connected. Then A must be connected, too: if A would split into two disjoint non-empty closed sets F0, F1, then the distance of F0 and F1 is a positive and for||m < the set Bm could not be connected.

Case two. A0 A1 = . Then A0 and A1 have some positive distance , and bythe argument just given there is a smallest m with f0(Bm) f1(Bm) = . By (*),Bl is connected for all l = 0, 1,...,m. We show that all Aj are disjoint and hence Ais totally disconnected.

Since Bm is a connected finite union of closed balls of radius ||m, the outerboundary of Bm (those points which can be connected with infinity in the comple-

ment ofBm) forms a closed curve consisting of finitely many arcs of circles of thisradius. may have finitely many double points where balls touch each other, but noother multiple points. Thus we can consider as a curve oriented in clockwise direc-tion, without self crossings, only touching points. Let j = fj () for j = 0,...,n 1.We assumed 0 1 = and want to show that all j are disjoint.

Now let 1 be the smallest positive number with 0 B(0, 1), and C the circlearound 0 with radius 1. Then C intersects 0 in a point c0 and j in the pointcj = b

jc0.The origin 0 is not inside 0, because then it would also be inside 1, which

implies that 0 and 1 intersect. Now there is a largest 2 > 0 such that the open

ball B(0, 2) does not intersect 0. Let D be the circle around 0 with radius 2.Then D intersects 0 in a point d0 and each j in the point dj = bj d0.

In the ring formed by D and C, the j sit like spokes in a wheel, connecting djwith cj. Of course 0 may touch D in other points than d0, but they all must liebetween dn1 and d1 because otherwise the Jordan curves 0 and 1 (or n1) wouldintersect. But this means that 0 is enclosed by 1, n1 and the arcs dn1d1 on Dand cn1c1 on C. Since 0 does not intersect the neighbor curves and does not crossthe circles D and C, it can not intersect the curves 2, 3, ..., n2. Thus all j, andhence all Aj are pairwise disjoint.

When dealing with self-similar sets, it is very natural to require the open set

condition,OSC [12, 8, 7]. We say that OSC holds for the mappings

f1,...,fnif there


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Figure 3. 4-gon fractals where pieces intersect in a Cantor set

is a nonempty open set U with

fk(U) U and fj(U) fk(U) = for j, k {1,...,n}, j = k. (3)If all Ak are disjoint, with pairwise distance > 0, then such U exists: we cantake U =

xA B(x,

2

). For connected sets A in the plane, Bandt and Rao [7] provedthat OSC holds when pieces intersect in a finite set. Thus all examples in Figures 1and 2 fulfil OSC.

Proposition 3. If || < sin n1+sin

nthen A is totally disconnected with disjoint Ak.

If || > 1n

for n 2, then the open set condition fails, and A is connected.

Proof. Taking any z =

m=0 mbkm A, we have

|z|

m=0

|m| = 11 r with r = || .

So A B(0, 11r ). Therefore,

Ak fk(B(0, 11 r )) = B(b

k,r

1 r )

for k = 0,...,n

1. Note that

|bk

bj

| |b

1

|= 2 sin

nfor k

= j (cf. Proposition

7,(iv)). So ifr < sinn

1+sin n

then |bk bj| 2sin n > 2r1r . This implies that B(bk, r1r )B(bj, r

1r ) = . Hence Ak Aj = for all k = j.Now assume that the open set condition holds (which is true ifA is disconnected).

Then the Hausdorff dimension of A is dimH(A) = lognlog || [12]. On the other hand,

dimH(A) 2 since A is a subset of the plane. It follows that || 1n . Thisbound is sharp for n = 2, 3, 4 where we have two-dimensional fractals (twindragonor rectangle, terdragon, square).

Proposition 4. If n < 25 and a fractal n-gon A fulfils OSC then only consecutive

pieces Ak, Ak+1 can intersect.


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FRACTAL n-GONS 7

Figure 4. Mandelbrot sets for 3- and 4-gons with symmetry lines

Proof. For n 3 all pieces are neighbors, so let n 4. Assume OSC holds andA0 Aj = for some j {2,...,n 2}. Since A0 B(1, r1r ) and Aj B(bj , r1r )this implies |1bj| 2r

1r . Since j {2,...,n2}, it follows that |1b2| 2r1r whichcan be reformulated as r |1b2|

2+|1b2| . However, calculation shows that|1b2|

2+|1b2| >1n

for 5 n 24 which contradicts Proposition 3. For n = 4 we just get equalitywhen r =

|

|= 1

2, and the balls with centers 1 and b2 =

1 meet in zero. Now

1 +m=1 bkmm = 0 is only possible for = 12 which gives the square. We conjecture that the condition n < 25 is not needed here.

4. Mandelbrot set for n-gons

To get an overview over all fractal n-gons, we work in the parameter space.Let us define the Mandelbrot set for fractal n-gons as

Mn = { | || < 1 , A() is connected }.The set

M2 was introduced 1985 by Barnsley and Harrington [9]. It was discussed

in Barnsleys book [8] and investigated by Bousch [10, 11], Bandt [2], Solomyak andXu [22] and Solomyak [20, 21]. So let us consider Mn for n > 2. Figure 4 showsMn for n = 3, 4 as black subset of the region 0 arg() 2 , 0 || < 1n .Proposition 5. Properties of Mn for n 2.

(i) Mn is a closed subset of the open unit disk.(ii) For even n, it has the dihedral group Dn as symmetry group. For odd n,

D2n acts as symmetry group.(iii) Mn includes all with || 1n .(iv) All with

|

|

sin n1+sin

ndo not belong to

Mn.


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Figure 5. Magnification of

M4 in a square of side length 0.003.

For = 0.3021 + 0.3375i taken from the upper right hole, the fractaln-gon must be a Cantor set, for nearby it will be connected.

Proof. (i). A() depends continuously on with respect to the Hausdorff metric[8], and the limit of connected closed sets with respect to the Hausdorff metriccannot be disconnected.

(ii). Since A() is a mirror image of A() (cf. Section 2), all sets Mn aresymmetric under reflection at the real axis, .

Moreover, A() = A(b

) for b = cos 2n

+ i sin 2n

, by an argument in the proofof Proposition 1. Thus Mn is also invariant under multiplication with b, that is,rotation around 2/n, as can be seen in Figure 4.

For odd n, however, Mn is also invariant under the map and hence underrotation around /n. This is proved below in Proposition 6, (i).

(iii) and (iv) follow from Proposition 3.

M2 has an antenna on the real axis since for real [0.5, 0.6] the set A() willbe an interval but for nearby off the real axis it will be a Cantor set. For n > 2we cannot get intervals A() and we have no antenna on Mn. On the whole, Mnseems to become smoother for larger n. Nevertheless, there are apparent holes in

M3 and M4 as were verified for M2 [2]. That means there are Cantor sets A()such that when we decrease continuously to reach = 0 we must get a connectedA on the way. See Figure 5.

Mn seems to be locally connected, a proof for M2 was given by Bousch [10, 11].It is not known whether Mn is the closure of its interior, cf. [22]. It is not knownwhether for each point with || < 1 on the boundary ofMn OSC holds for A(). Onthe other hand, OSC should fail when is in the interior ofMn but there seem tobe a few exceptional parameters, describing twindragons in M2 and the terdragonin M3. Are there other exceptions?

We do not study details of the structure of Mn since we use this set mainly toidentify interesting fractal

n-gons by looking at landmark points in

Mn.


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FRACTAL n-GONS 9

5. The overlap set

Series algorithm for Mn. To draw Mn, we must know when the pieces A0and A1 of A() intersect. By Remark 2, a point z A0 A1 can be written asz = 1 +

m=1 b

kmm = b +

m=1 bjmm where km, jm {0,...,n 1}. Thus

Remark 3. belongs to Mn if and only if is the solution of an equation

q() = 1 b +

m=1

dmm = 0 (4)

where each dm is a difference of powers of b.For fixed , every choice of km, jm, m = 1, 2,... such that the dm = b

km bjm solvethe above equation, corresponds to one point in the intersection set A0 A1.

Of course the choice of dl is restricted since the absolute value of the remainingsum

m=l+1 dm

m is at most 2rl+1

1r where r = || (cf. Proposition 3). If the absolutevalue of the sum up to degree l is larger than this value, then there is no chance toextend the sum to obtain a series with q() = 0.

This argument, with division by at each step, provides an algorithm for generat-ing Mn. IfD denotes the number of possible choices bk bj then we search througha rooted tree with D branches at every vertex. Each vertex is assigned a complexnumber v. If |v| > 2r

1r then the vertex will be considered as dead, and the searchwill not go deeper at this place. We stop the search if either all vertices are dead,in which case Mn, or if we reach a prescribed level lmax in which case belongsto our approximation ofMn. Figures 4 and 5 where made with lmax = 150.Remark 4. (Algorithm to generate Mn)We renumber the set of all bk bj as = {d1,...,dN} with d1 = 1 b. For given and lmax we consider words u = 1k2k3...kl l with l lmax and correspondingnumbers vu C according to the rule

v1 = d1 , vuk =vu

+ dk . (5)

When |vu| > ||1|| then u and its successor words are removed. If at least one wordof length lmax remains, is considered as an element of Mn, otherwise does notbelong to

Mn.

Different aspects of the accuracy of the algorithm for n = 2 were discussed in[2]. For larger n, the problems are essentially the same, and the calculation timeincreases exponentially with the size of . Figure 4 required two seconds, Figure 5about 10 seconds on a PC.

Overlap set for even n. Now we discuss the size of the overlap set D = A0A1.For n = 2 we have bk {1, 1} so that d = 0 has two representations bk bj whiled = 1 and d = 1 admit a unique representation. If is the root of just one powerseries q() and this q has only coefficients 1 then D is a single point. If N ofthe coefficients are zero, D has cardinality 2N, and if there are infinitely many zerocoefficients,

Dis a Cantor set [21, 5].


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For even n > 2, the situation is very similar. We shall not consider zero coefficientssince we conjecture that cannot be the root of only one q() with coefficients in

when there are zero coefficients in q. (The reason is that for dk = 0 in (5), we finda j = k such that |vuj| < |vuk|.)

However, in a regular n-gon with even n 4, as it is formed by the roots bk, k =0,...,n 1, every side and diagonal, except for the longest ones, has one parallelside or diagonal, and the corresponding d has two representations bk bj. Thus onlythose power series q() where every dm has the form b

km bkm+n/2 can yield a singleintersection point, see Figure 7 below. If we have N coefficients dm which do notcorrespond to the longest diagonals, we shall have 2N points in D, see Figure 2.And when we have infinitely many such d then D will be a Cantor set, as shown inFigure 3.

Example 2. For n = 4 we have b = i and the bk

form a square. The 4-gonin the lower row of Figure 2 was obtained from the relation of addresses u1u2... =01 130303 = v1v2... Here um = vm + 2 mod 4, corresponding to a diagonal of thesquare, for all m > 1 except m = 3 and m = 5 where (uk, vk) = (1, 0) which canbe replaced by (2, 3) since i 1 = i2 i3. Thus there are 4 pairs of addresses whichgive the same equation

p(01) = 1 + i

1 = p(130303) = i i + 2 i3 + 4 i

5

1 .

which results in 5 4 + 3 2 + (1 + 2i) = i with numerical solution 0.4587 + 0.1598i.

Example 3. As a landmark point in M4, we consider the intersection of thesymmetry line in Figure 4 with the boundary ofM4. It turns out that this is

=1 + i

1 +

5.

The corresponding fractal, shown on the left of Figure 3, is mirror-symmetric:A() = A(i) = A(). It seems the only mirror-symmetric 4-gon beside the squarewhich fulfils OSC.

It is possible to determine by identifying addresses u = u1u2... = 010033221 and

v = 133221100. The corresponding equation (4) results in a geometric series with, and the resulting polynomial of degree 4 factors into two quadratic polynomials,of which 2z2+(1+i)zi has as the root with modulus 1. Thus 1/ is a quadraticPisot number over Q(i).

Note that every second combination (uk, vk) can be replaced by one other choice.For instance (0, 3) and (1, 2) give the same d = 1 + i. Thus D = A0 A1 is un-countable, and it is a linear Cantor set. In fact D is a self-similar set with respectto two homotheties with factor r2 = ||2 : For E = f10 (D), inspection of Figure 3shows that E = g1(E) g2(E) with g1(z) = f21 (iz) and g2(z) = f1f0(iz). So theHausdorff dimensions of A and D are log 4 log r and

log2

log r2 , respectively [12], and thedimension of

Dis just one quarter of the dimension of

A.


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FRACTAL n-GONS 11

Figure 6. n-gons for odd n with several series q() : Examples 4,5.

Clearly D fulfils OSC. For A we can explicitly construct an open set - an octagonwith interior angles 3

4 and with alternating side lengths s,rs for some constant s

(for our choice of mappings s = 4r). Drawing this octagon U and the fi(U), we candetermine r in alternative way, by elementary geometry.

Overlap set for odd n. A regular n-gon with odd n has no parallel sides, andno parallel diagonals of the same length. So d = 0 is the only vector in which canbe represented as d = bk bj in different ways - actually in n ways. When we haveOSC, however, coefficients 0 are unlikely to appear in the power series of q(), atleast for n 5. So it seems we have the following alternative: either D is a singleton(see Figure 6), or is the root of several power series q().

Example 4. Considering 2-gons, Solomyak [21] asked whether there is a whichis root of several power series q() and still admits finite intersection set D. Forn = 3, Figure 6 shows an example, with address identifications

0110022221111000 1022111000022221 and 01111100002222111222222211110000 .This is obtained from a landmark point 0.4793 + 0.2767i : the intersection ofthe 30 degree symmetry line with the boundary of M3. Supported by experiment,we conjecture that we have the same situation for the corresponding parameter onthe symmetry line for every odd n, and we do not understand the algebraic reason.For n = 3, the power series derived from the given identifications both lead to

polynomials with the factor 2z4

+ ( 3 + 3i)z3

+ 2 ( 1 + 3i)z2

+ 1 3i, and is aroot of this polynomial.

Also on the symmetry line, but far inside M3, there is the parameter = 12 +36

i

for the terdragon for which we have lots of power series with q() = 0. Instead ofthis well-known figure, we discuss a similar example introduced in [3].

Example 5. For the second picture in Figure 6, we have the identification

012100and many others, since D is a Cantor set [3]. Actually, E = f10 (D) is self-similar:E = f21 (E) f21 (E) where denotes a clockwise 23 rotation with appropriatecenter. In this case,

0

.5222

0

.0893i fulfils the equation 1

= (1

b2)

2

.


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It seems not possible to use this equation directly to replace finitely many symbolsin the above address identification. But a calculation shows that we can replace 12

and 00 by 1122 and 0001, respectively, which leads to identifications 0(12)k112 1(00)k0001 and corresponding qk() = 0 for k = 1, 2,... Thus we have infinitelymany power series.

The reverse n-gon. For odd n, we call A() the reverse of A(). TheSierpinski gasket is self-reverse, and for Example 4 and the terdragon, the reverse ismirror-symmetric to A(). If we are not on a symmetry line, however, the appearanceof A and its reverse can differ considerably, as Figure 1 shows. Nevertheless, theirstructure is similar. Since the reverse of the reverse is A itself, if in the followingstatements means if and only if.

Proposition 6. Letn be odd and A = A() a fractal n-gon.

(i) A is connected if the reverse is connected.(ii) The cardinalities of the intersection sets D of A and of its reverse coincide.

(iii) A is a p.c.f. fractal if the reverse is.(iv) A satisfies OSC if the reverse does.(v) A is of finite type if the reverse is.

Proof. For (i) and (ii) we use Remark 3. D() is described by the sequencesd1, d2,... from for which 1 b +

m=1 dm

m = 0, and D() is described by thesequences d1, d

2,... from for which 1 b +

m=1 d

m()m = 0. Thus between

both sets there is a one-to-one correspondence, given by dm = dm for even m and

dm = dm for odd m. If one set is empty or finite, then the other is, too.Moreover, when a point of D() has preperiodic addresses, then the correspondingpoint of D() also has. So if A is p.c.f. [16], then the reverse is p.c.f.

For (iv) and (v) we use neighbor maps f1u fv for words u, v {0, 1,...,n1}m, m =1, 2,... [4, 7, 3]. For fractal n-gons we have w = fv(z) =

mz +m

k=1 bvkk1 and

f1u (w) =w

mmk=1 bukkm1. Thus all neighbor maps are translations:

f1u fv(z) = z +m

k=1

(bvk buk)km1 .

Now defining u, v by uk = uk, vk = vk for even k and u

k = vk, v

k = uk we see

that an n-gon and its reverse have just the same neighbor maps. OSC means thatneighbor maps cannot approach the identity map [4], so (iii) is proved. Finite typesays that there are only finitely many neighbor maps f1u fv with Au Av = [7].Au, Av are disjoint iff the dk = b

vk buk can be extended to a sequence d1, d2,... forwhich 1 b + dkk = 0. With the above construction, this implies (iv).

6. Simple n-gons

To conclude this study, we consider an infinite family of fractal n-gons with aparticular simple structure. We say that an n-gon A is simple if A0 A1 is asingleton with two addresses of the form

0

k 1j.


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FRACTAL n-GONS 13

Figure 7. All simple n-gons for n 8 where the fk are not homotheties

For n = 3, we have only the Sierpinski gasket. For every n 3 which is nota multiple of 4, there is a Sierpinski n-gon for which the fk are homotheties [19].

Figure 7 shows all other examples of simple n-gons with n 8. Including Sierpinskin-gons, we have two examples for n = 5, 7, 9, 10 but only one for n = 3, 4, 6, 8, 12,and three for n = 11. Moreover, the number increases with n as we shall see now.Remark 3 gives a necessary condition for a simple n-gon:

1 b + 1 (b

k bj) = 0 , that is, 1

= 1 +bj bk1 b . (6)

Thus all possible parameters of simple n-gons are known. The problem is that formany choices of (k, j) there are further points in the intersection set D.

Proposition 7. .

(i) If A is a simple n-gon with 0k 1 j, fulfils (6) and is an integer linearcombination of 1,b,...bn1.

(ii) If (6) holds for some (k, j), and if

|1 b + bj bk + bk bj| =1 b + bk bj

> ||1 || (7)for allk, j {0,...,n 1} with (k, j) = (k, j), where = max{|d| , d },then A() is a simple n-gon.

(iii) |1 b| = 2 sin n

, and = 2 cos 2n

for odd n while = 2 for even n.

Proof. (i) was proved above. For j < k we have bj

bk

1b = bkj(1 + b + ... + bj1).(ii): If (7) is true, then 1 b + (bk bj) + m=2 dmm = 0 for all dm and(k, j) = (k, j) since the modulus of the last sum is ||2

1|| . Similarly, 1 b +lm=1(b

k bj)m + (bk bj)l+1 +m=l+2 dmm = 0 for l 1 and (k, j) = (k, j)since by (6) 1b+(bkbj) = (1b) and hence 1b+lm=1(bkbj )m = l(1b).Thus D is the singleton with addresses 0k and 1j.

(iii): |1 b| = 2sin n

is obtained by direct calculation, or by looking at theright-angled triangle with vertices 1, 1 and b within the unit circle. Since =|1 bn/2| = |1 + b| for odd n, we similarly get from the triangle with vertices

1

,1 and

b.


14/20


Schlicker and Dennis [19] constructed their Sierpinski n-gons with real for everyn which is not a multiple of 4. They used the interior of the regular n-gon as open

set U and determined as a trigonometric sum, requiring that neighboring fk(U)touch only at a vertex. With (7) we get a simpler formula.

Remark 5. For each n 3 which is not a multiple of 4 there is a simple n-gonwith positive real factor . We have

1

= 1 +

1

2sin 2n

for oddn and1

= 1 +

1

sin n

for n = 4q + 2 .

Proof. For n = 4q, the regular n-gon generated by the bk has a diagonal bjbkof length which is parallel to the side

1b. We obtain from (6) taking bj bk =

|1b|(1

b). For this the right-hand side of (7) equals

|1

b|. So (7) holds because

the circle C() with radius around m() = 1 b + bj bk and the circle C0 withradius |1 b| around 0 just touch in the point 1 b.

Here we give a longer list of simple fractal n-gons.

Theorem 8. The following generate simple fractal n-gons.

For odd n = 2q + 1 :1

= 1 +

bl/2

2sin 2n

where 0 l < n6

+1

2,

for evenn = 4q + 2 :1

= 1 +

bl

sin n

where 0 l < n12

+1

2,

for evenn = 4q :1

= 1 +

bl+12

sin n

where 0 l < n12

.

For even n this list is complete, and for odd n there are no other parameters forsimple n-gons for which (6) is fulfilled with |bk bj | = .Proof. 1. Construction of parameters. All these are solutions of (6) whichwe can rewrite as

m() =1 b

= 1 b + bj bk . (8)

To see this, consider the structure of the set of the differences bkbj . We start withodd n = 2q + 1. The modulus |b

k

bj

| can be any of the values 0, 2sin

n , 2sin2

n , ...,2sin qn

= . Diagonals of all lengths of the regular n-gon spanned by the bk appearin vertical direction, and is invariant under multiplication by b and 1. Thus

= {2ibl/2 sin tn

| t = 0, 1,...,q, l = 0, 1, ..., 2n 1}.For even n the set is defined differently. For n = 4q we get

= {2bl sin tn

| l = 0,...,n 1, t = 0, 2, ..., 2q} {2bl+12 sin t

n| l = 0,...,n 1, t = 1, 3, ..., 2q 1} ,

and for n = 4q + 2 we have t = 1, 3,..., 2q + 1 in the first part and t = 0, 2,..., 2q inthe second.


15/20

FRACTAL n-GONS 15

Geometrically, the necessary condition (8) for simple n-gons says that m() (1

b) + , or, equivalently, (1

b)

m() + . For our list, we took only those

(k, j) with maximal |bk bj | = . Thus the are taken so that m() is on a circle Dof radius around the center1b. By Remark 5, we can start with positive real forn = 4q. For n = 4q we began with a positive multiple of b1/2. Since and yieldmirror-equivalent n-gons, we decided to continue with m() in counter-clockwisedirection on D, and thus with in clockwise direction.

Now let us explain how we got the inequalities in the theorem. We consider thecircles C() around m() with radius , and the circle C0 around 0 with radius|1 b|. In each case 1 b belongs to C() C0. For positive real this is the onlyintersection point, otherwise we have a complete arc C() of C() inside C0.

For odd n, we determine the value for which C() subtends an angle n

at m()and so the other endpoint c ofC() belongs to m() + . In this borderline case weconsider the isosceles triangle cm()(1 b) with angle cm()(1 b) =

n. Since

|(1 b) c|sin

n

=

sin( 2

2n

)=

2cos 2n

cos 2n

= 2 ,

0c(1 b) is an equilateral triangle. From the representation of in the theorem,the angle between vectors 1 b and m() (1 b) is l

n. Adding angles at point

1 b in this borderline case we get 3

+ ( 2

2n

) + ln

= , or l = n6

+ 12

.When m() moves counterclockwise on D then |m()| continuously decreases and

the length of C() increases. Thus for l < n6

+ 12

the set m() + has only the

point 1 b inside C0 while for l n

6 +1

2 there is at least one other point of m() + inside C0.For even n the condition follows the same principle, studying the borderline case

with a second point c (m() + ) C0. But now |c m()| = 2cos n due tothe structure of , so that the angle (1 b)cm() is a right angle, in virtue of|m() (1 b)| = 2 and (1 b)m()c =

n. Again, 0c(1 b) is an equilateral

triangle. Considering angles at point 1 b this leads to 3

+ ( 2

n) + s

n= which

is equivalent to s = n6

+ 1. Here s = sn

is the angle between the lines through 0, 1 band through 1b, m() = 1b

when 1

= 1+r bs/2 with positive real r. The equation

holds for the borderline case. Putting s 2l for n = 4q + 2 and s 2l + 1 forn = 4q, this gives the condition in our theorem.

2. Sufficiency of our condition. We use (7) to show that A() is simple if belongs to the above list. Due to our construction, all points of m() + areoutside the circle C0 around 0 with radius |1 b| which by (6) and |bk bj| = equals |

1 |. Unfortunately, this is not exactly what is required in (7): the radius ofthat circle is larger by the factor |1|

1|| .Let us assume n is odd, and calculate the factor. Our formula for can be written

= 2 sinxbl/2+2sinx

with x = 2n

. Writing N = |bl/2 + 2 sin x| we have |1 | = 1N

and|1|1|| =

1N2 sinx . Clearly,

N

2 = (2 sinx

+ cos 2lx

)2 + sin2 2lx

= (1 + 2 sinx

)2

4(1

cos2

lx)sin

x.


16/20


We have to check whether for each from our list, all elements of m() + ,

except 1

b, have modulus >

|1

b

| |1|1|

|. Due to our construction, it is sufficient

to check this only for the point c() on the outer ring ofm() + which is next to1 b, in other words c() = m() + b1/2(1 b m()). Moreover, it is sufficientto check this point for the largest possible value of l, since |m()| and |c()| willdecrease when m() runs counter-clockwise along D. Note that when m() moveson D, then c() is also rotated around 1 b, about the same angle as m().

For l = n6

+ 12

the point c will be on C0 and 0(1 b)c is equilateral. Since n isodd, the largest l < l is at least 1

3smaller than l. So we take l = n+1

6and rotate c

clockwise around 1b about the angle 3n

to obtain the point c. Only this point needsto be checked. The triangle 0(1 b)c is still isosceles, and 0(1 b)c =

3+

3n, so

|c

|= 2

|1

b

|sin(

6+

6n) =

|1

b

| (cos x

3+

3sin x

3) with x from above. It remains

to check whether

f(x) = cosx

3+

3sin

x

3>

(1 + 2 sin x)2 4(1 cos2lx)sin x 2sin x1

for 0 < x 10

, that is, n 5. Since 2lx = n+16

n

5

for n 5 and cos 5

0.8 we increase the right-hand side when replacing 4(1 cos2lx) by 0.8. Since

(1 + 2t)2 0.8t > (1 + 2t) 0.4t, the function g(x) = (1 0.4sin x)1 is largerthan the above right-hand side. A plot shows f(x) > g(x), except perhaps for xvery near to 0, since f(0) = g(0) = 1. However, f(0) =

3 > g(0) = 0.4 confirms

f(x) > g(x) also for x near 0.Few modifications are needed for even n. We set y = 2x =

nso that = siny

bl+sin y

and |1|1|| =

1Nsiny where N = |bl + sin y| = (1 + sin y)2 (1 cos2ly)sin y . The

estimate l l 13

, rotation about 3n

and f(x) are the same as above. We have2ly = 2 n+2

12n

4

and cos 2ly 0.7 so that|1 |1 ||

(1 + sin y)2 0.3sin y sin y

1< (1 0.15sin2x)1 = h(x) ,

and it is easy to see that f(x) > h(x) for 0 < x 8

, that is, n 4.3. Necessity of the inequalities. Now assume has the form given in the

theorem, but l is not smaller than the given bound. In this case the arc of C()

inside C0 subtends an angle n at m(), and so we get (k, j) = (k, j) suchthat (7) fails. In terms of the algorithm of Remark 4, we have v1 = 1 b, andv2 =

v1

+ (bk bj) = 1 b but inside C0. It remains to show that we also find

vk+1 =vk

+ dk in C for k = 2, 3,..., with dk . Then there is a second intersection

point in A0 A1, and A() is not simple.If vk is inside or on C0 then |vk | | v1 | = |m()|, and so the circle Ck of radius

around mk =vk

either intersects C0 in an arc which subtends an angle n atmk, or Ck contains the center 0 of C0. (The subtended angle grows if the center mkmoves nearer to C0, as long as |mk| .) In the first case we find mk + dk withdk on the arc. In the second case dk exists by the following remark. Inductioncompletes the proof.


17/20

FRACTAL n-GONS 17

Remark 6. Let C be a circle with center m and radius . If 0 is inside C then onepoint of the set m + is inside C0.

Proof of the remark. The distance from 0 to the next diameter of C withpoints of m + is sin

2nwhich is just half of the radius |1 b| of C0 for odd

n, and less than 0.55 |1 b| for even n 4. Thus C0 intersects this diameter in asegment of length > |1 b| = 2 sin

nwhich always contains a point of m + .

For even n, the discussion of the overlap in section 5 shows that we cannot getsimple n-gons from |bk bj| < . We conjecture that this holds also for odd n.

7. Analysis on simple n-gons

During the last 20 years, several approaches have been used to develop analysison fractals: Brownian motion [17], resistance networks, harmonic functions, Laplaceoperator and associated Dirichlet form [16, 1]. It turned out that all these approachesare more or less equivalent. They require existence of a harmonic structure onthe underlying space, and the corresponding fractal exponents (spectral dimension,random walk dimension, resistance exponent) are tightly related [16].

The theory was developed for p.c.f. fractals, which means that the intersectionsAi Aj are finite and all their points have preperiodic addresses. Simple n-gonsobviously belong to this class. On p.c.f. fractals the finite sets of intersection pointsof m-th level pieces provide tractable approximations of the space. But even in thisclass, not all spaces admit a harmonic structure.

Our aim is to show that all simple n-gons have a harmonic structure, using the

approach of resistance networks. For details, we refer to Kigami [16, Chapter 2].We give a simple scaling scheme which applies to all simple n-gons and allows tocalculate resistance exponents.

Let A denote the simple n-gon given by 0k 1j and hence i(k + i)(i +1)(j + i)for i = 0,...,n1. When considering the pieces Aw ofA as resistances, it is importantto note that each piece has either three or two neighbors, and the boundary pointswhere the neighbors are attached are always in the same position, up to symmetry.When Aw with w = w1...wl is mapped to A by g(z) = b

wl f1w (z), the transformedboundary points have addresses 0, k, and j 1.

While k, j where used to determine by (6), at this point it seems appropriate tochange notation. The standard points with addresses 0, k, j

1 are called X,Y,Z,

respectively, so that Y = bkX, Z = bmY and X = bqZ with m = j 1 k andq = n (j 1). In particular, k + m + q = n.

The resistance structure of an electrical black box which can be measured onlyat three points x,y,z is given by Figure 8a. The resistances rx, ry, rz given hereshould be those of any Ai, with x = fi(X), y = fi(Y), z = fi(Z), and we choosery + rz = 1 as our normalizing condition. Then the effective resistances are (x, y) =rx + ry, (x, z) = rx + rz, and (y, z) = 1.

For the set A, we have exactly the same picture, but with vertices X , Y , Z andresistances RX , RY, RZ, Figure 8d. Our aim is now to find a constant 1 suchthat

RX=

rx , RY=

ry , RZ=

rz .


18/20


X

Y Z

R

RRY Z

Xx

z

r

X

Y Z

y

YZ

X

k

m

qx

x

r

zy

r

xx

xrxxx

xrxxx

xrxxx xrxxx

xrxxx

xrxxx

zr

zryryr

a b c d

Figure 8. Resistance scaling: a scheme for a piece, b connectingpieces, c using ry + rz = 1, d resulting resistances for large set

will be called resistance factor. If it exists then for all points a, b A and all map-pings fw with w = w1...wl, the effective resistance fulfils (fw(a), fw(b)) =

l(a, b).This will provide a harmonic structure on A [16]. Given and the factor || of themappings, we define the resistance exponent as = log log || . Two points in a, b Athen have resistance (a, b) |a b|. This is similar to the interpretation of thefractal dimension as mass-scaling exponent. Note that < 1 since resistance in anetwork grows more slowly than in a straight wire. Table 1 summarizes some resultswhich we get from the theorem below.

Name n 1/ dimension Sierpinski gasket 3 2 5/3 1.58 0.74

pentagasket 5 (

5 + 3)/2 9/5 1.67 0.61hexagasket 6 3 7/3 1.63 0.77

4 2 + i 7/4 1.72 0.70Figure 7 5 2.309 + 0.951i 11/5 1.76 0.86

7 3.024 + 0.975i 19/7 1.68 0.86

8 2 +

2 + i 23/8 1.64 0.83

Table 1. Dimensions and resistance exponents of simple n-gons

Theorem 9. All simple fractal n-gons admit a unique harmonic structure. Theresistance factor = RX /rx is

= 1 + n4

1n

for even n,

= 1 + 1n

n22 + n

2 c with c {0, 2} for odd n.

Proof. By inserting the scheme of Figure 8a for each Ai and using ry + rz = 1we obtain Figures 8b and 8c. Now it is easy to determine the effective resistances


19/20

FRACTAL n-GONS 19

between X , Y , Zusing high-school physics:

(X, Y) = 2rx +1k + 1m + q1

= 2rx + k(n k)n(X, Z) = 2rx +

q(n q)n

(Y, Z) = 2rx +m(n m)

n

In Figure 8d we have RX =12

((X, Y)+(X, Z)(Y, Z)). The condition RX = rxgives rx = rx +

12n

(k(n k) + q(n q) m(n m)). This leads to the equations2n( 1)rx = k(n k) m(n m) + q(n q)

2n( 1)ry = k(n k) + m(n m) q(n q)2n( 1)rz = k(n k) + m(n m) + q(n q)Adding the last two equations and using ry + rz = 1 we see that depends only onn and m = j 1 k.

= 1 +m(n m)

nWith this > 1 the values ofrx, ry, rz are uniquely determined. So there is a uniqueharmonic structure [16]. For all even n we have m = n

21 since bjbk was a diameter

of the unit circle. For odd n we either have m = n2 1 and n m = n

2 + 2, or

m = n2, depending on l in Theorem 8. The two cases will alternate, starting at

l = 0 with the first case for n = 4q + 1 and with the second for n = 4q + 3.

Acknowledgement. This work was done when Nguyen Viet Hung visited theUniversity of Greifswald, supported by the Ministry of Education and Training ofVietnam and by the DAAD.

References

[1] B. Adams, S.A. Smith, R.S. Strichartz and A. Teplyaev, The spectrum of the Laplacian onthe pentagasket, Fractals in Graz 2001 (eds. P. Grabner and W. Woess), Birkhauser 2003,1-24.

[2] C. Bandt, On the Mandelbrot set for pairs of linear maps, Nonlinearity 15 (2002), 11271147.

[3] C. Bandt, Self-similar measures, Ergodic Theory, Analysis, and Efficient Simulation of Dy-namical Systems, B. Fiedler (ed.), Springer (2001), 3146.

[4] C. Bandt and S. Graf, Self-similar sets VII. A characterization of self-similar fractals withpositive Hausdorff measure. Proc. Amer. Math. Soc. 114 (1992), 9951001.

[5] C. Bandt and N.V. Hung, Self-similar sets with open set condition and great variety of overlaps,Preprint, 2007.

[6] C. Bandt and K. Keller, Self-similar sets 2. A simple approach to the topological structure ofFractals, Math. Nachr. 154 (1991), 2739.

[7] C. Bandt and H. Rao, Topology and separation of self-similar fractals in the plane, preprint,Greifswald 2006

[8] M.F. Barnsley, Fractals Everywhere, 2nd ed., Academic Press 1993.[9] M.F. Barnsley and A.N. Harrington, A Mandelbrot set for pairs of linear maps, Physica D 15

(1985), 421432.


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[10] T. Bousch Sur quelques problemes de dynamique holomorphe, Ph.D. thesis, Orsay(http://topo.math.u-psud.fr/bousch), 1992.

[11] T. Bousch, Connexite locale et par chemins holderiens pour les systemes iteres de fonctions,Preprint (http://topo.math.u-psud.fr/bousch), 1993.

[12] K.J. Falconer, Fractal Geometry, Wiley 1990.[13] K.J. Falconer and J.J. OConnor, Symmetry and enumeration of self-similar fractals, Bull.

London Math. Soc. electronic version 27 Feb. 2007[14] M. Hata, On the structure of self-similar sets, Japan J. Appl. Math. 2 (1985), 281414.[15] J.E. Hutchinson, Fractals and self-similarity, Indiana Univ. Math. J. 30 (1981), 713747.[16] J. Kigami, Analysis on fractals, Cambridge University Press, 2001.[17] T. Lindstrm, Brownian motion on nested fractals, Mem. Am. Math. Soc. 83 (1990), 1124.[18] B.B. Mandelbrot, The Fractal Geometry of Nature, Freeman, San Francisco 1982.[19] S. Schlicker and K. Dennis, Sierpinski n-gons, Pi Mu Epsilon J. 10 (1995), 8189.[20] B. Solomyak, Mandelbrot set for pairs of linear maps: the local geometry, Analysis in Theory

and Applications 20:2 (2004), 149157.[21] B. Solomyak, On the Mandelbrot set for pairs of linear maps: asymptotic self-similarity,

Nonlinearity 18 (2005), 19271943.[22] B. Solomyak and H. Xu, On the Mandelbrot set for a pair of linear maps and complex

Bernoulli convolutions, Nonlinearity 16 (2003), 17331749.[23] R.S. Strichartz, Isoperimetric estimates on Sierpinski gasket type fractals, Trans. Amer.

Math. Soc. 351 (1999), 1705-1752.

Christoph BandtInstitute for Mathematics and InformaticsArndt University17487 Greifswald, [email protected]

Nguyen Viet HungDepartment of MathematicsHue UniversityHue, [email protected]

christoph bandt and nguyen viet hung- fractal n-gons

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