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8/3/2019 Byungik Kahng and Jeremy Davis- Maximal Dimensions of Uniform Sierpinski Fractals
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MAXIMAL DIMENSIONS OF UNIFORM SIERPINSKI FRACTALS
BYUNGIK KAHNG AND JEREMY DAVIS
Abstract. We study the invariant fractals in convex N-gons given by N identical pure
contractions at its vertices with non-overlapping images, which we will call uniform Sierpin-
ski fractals. We provide the explicit formulae for the maximal contraction ratio RN, and
the maximal Hausdorff dimension hN, for the uniform Sierpinski fractals. We use maximal
N-grams and principal crossing points as the main tool.
1. Introduction
It is well known that the iteration of similarity maps often produces an invariant fractal
with intricate self similarity structure that is aesthetically beautiful. See, for instance, [2, 3, 6]
for more about fractals and self similarities. One of the oldest and also the simplest classes
of such fractals is the class of Sierpinski fractals, which we define as follows.
Definition 1.1 (Sierpinski Fractals). LetPN be a convex N-gon in the Euclidean plane R2,
with the vertices v1, , vN R2. That is,
PN = N
i=1
ivi : 0 i 1,Ni=1
i = 1
.
LetP(PN) be its power set. Furthermore, let
f :P(PN) P(PN)
be a set function in PN given by
(1.1) f(S) = f1(S) fN(S),
where each fi : PN PN is a pure contraction at vi with the contraction ratio ri. Thatis, for each i {1, , N},
(1.2) fi(x) = ri(x vi) + vi, 0 < ri < 1.
Date: December 10, 2009.
2000 Mathematics Subject Classification. Primary: 51N05, 51N20, 28A80; Secondary: 51-01.Key words and phrases. Sierpinski fractal, self similarity, Hausdorff dimension, disturbed control dynam-
ical system, multiple valued iterative dynamics, maximal invariant set, global attractor.1
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2 BYUNGIK KAHNG AND JEREMY DAVIS
Suppose further that the contractions, f1, , fN, have non-overlapping images. That
is, each intersection, fi(PN) fj(PN), i = j, has empty interior. Then, the set
(1.3) S(r1, ,rN)(PN) =
k=0
fk
(PN).
is called the Sierpinski fractal of PN with the contraction ratios r1, , rN. We say a
Sierpinski fractal is uniform if r1 = = rN = r, and we abbreviate it as,
Sr(PN) = S(r, ,r)(PN).
We say a uniform Sierpinski fractal Sr(PN) is maximal if the the common contraction ratio
r is the largest for given PN. Finally, a Sierpinski fractal is calledregular ifPN is a regular
polygon.
Figure 1.1 illustrates the uniform regular Sierpinski fractal Sr(PN) for selected N andr values. Our aim is to determine the condition to achieve the largest r value for given
N, which gives rise to the largest possible regular Sierpinski fractal in terms of Hausdorff
dimension The left hand side column of Figure 1.1 illustrates such maximal cases, where
adjacent images fi(PN) and fi+1(PN) apparently touch each other. In fact, we claim the
following.
Main Theorem (Theorem 3.2 and Corollary 3.3). Let Sr(PN) be a uniform Sierpinski
fractal of a regular N-gon, PN with vertices v1, , vN R2. Then, the largest possible
contraction ratio (r value in Definition 1.1), which we denote as RN, turns out to be,
(1.4) RN =1
2
1
tanN
N14
tan
N
+ N
N14
,
where x stands for the floor function. That is, x is the largest integer that does not
exceed x. As a direct consequence, we conclude that the maximal Hausdorff dimension hN is
attained when r = RN, and
(1.5) hN =ln N
ln(1/RN)=
ln N
ln RN.
The Main Theorem, which is a combination of Theorem 3.2 and Corollary 3.3, will be
proved in Section 3, along with the numerical illustrations. The interactive Mathematicaprograms to plot the regular Sierpinski fractals and to test the results and tools of this paper
will be posted in the first authors website [5].
2. Maximal N-grams and Principal Crossing Points
The main tool we use to prove the Main Theorem is the N-gram decomposition ofPN. In
general, an N-gram of a regular polygon PN is a union of selected diagonal line segments of
PN. Figure 2.1 Figure 2.4 exemplify some N-grams of various regular polygons. Clearly,
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MAXIMAL DIMENSIONS OF UNIFORM SIERPINSKI FRACTALS 3
N = 3, r = 1/2.
N = 7, r 0.30798.
N = 6, r = 0.32.
N = 11, r = 0.21.
Figure 1.1. Selected Examples of Uniform Regular Sierpinski Fractals.
Figure 2.1. G5. Figure 2.2. G6.
an N-gram exists only if N 5 and there are more than one N-grams, if N 7. Among
them, we are particularly interested in one specific N-gram for each N 5, which we define
as follows.
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4 BYUNGIK KAHNG AND JEREMY DAVIS
Figure 2.3. G7 (thick). Figure 2.4. G9 (thick).
Definition 2.1 (Maximal N-gram). LetPN be a regular N-gon with vertices
vi = c0 +
cos
2i
N
, sin
2i
N
R2,
where c0 R2, > 0, i = 1, , N, and finally, N 5. Suppose further that
(2.1) m =
N 1
4
+ 1.
Then, we call the union of the diagonal line segments GN given by
(2.2) GN =
Nm
i=1 vi vi+m,
where p q stands for the line segment from p R2 to q R2, the maximal N-gram of PN.
The reason we chose the term maximalin Definition 2.1 is because of the maximal N-grams
are intimately related to the maximal regular Sierpinski fractals, or the regular Sierpinski
fractals given by the maximal contraction ratios. The following lemma justifies our choice.
It will play an important role in the proof of the Main Theorem.
Lemma 2.2 (Principal Crossing Points). Let Sr(PN) be a uniform Sierpinski fractal on a
regular N-gon, PN with vertices v1, , vN, N 5, introduced in Definition 2.1. Suppose further thatf1, , fN and f be as in Definition 1.1. Then, for each i {1, , N},
fi(vim) vi vim GN,(2.3)
fi(vim) vi vim GN,(2.4)
where and are the addition and the subtraction in {1, , N} under the (mod N) iden-
tification, and m is the number defined by the equality (2.1) in Definition 2.1. Furthermore,
the uniform contraction ratio r attains the maximum value, which we denote as RN, if and
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MAXIMAL DIMENSIONS OF UNIFORM SIERPINSKI FRACTALS 5
Figure 2.5. N = 5, 0 < r < RN. Figure 2.6. N = 5, r = R5.
Figure 2.7. N = 7, 0 < r < RN. Figure 2.8. N = 7, r = R7.
only if
(2.5) vi vim vi1 vi(m1) = {fi(vim)},
for some i {1, , N}. Let us call the crossing points between the line segments vi vimand vi1 vi(m1), i {1, , N}, the principal crossing points of GN.
Figure 2.5 Figure 2.8 illustrate the first conclusion of Lemma 2.2, the set inequalities
(2.3) and (2.4). Note that each vertex fi(vim) stays on the diagonal line segment vi vim ofthe maximal N-gram GN, as r value increases from 0 to RN.
The second conclusion (the equivalence part), on the other hand, can be re-expressed
as follows. The uniform contraction ratio r reaches its maximum value if and only if each
fi(PN) touches the adjacent polygons, fi1(PN) andfi1(PN), at the principal crossing points.
Figure 2.6, Figure 2.8 and Figure 2.9 Figure 2.12 depict this claim. Figure 2.13 and Figure
2.14, on the other hand, illustrate what happens if we select wrong N-grams. Note that
the crossing points of the diagonal line segments do not match the touching points of the
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6 BYUNGIK KAHNG AND JEREMY DAVIS
Figure 2.9. N = 8, r = R8. Figure 2.10. N = 9, r = R9.
Figure 2.11. N = 12, r = R12. Figure 2.12. N = 13, r = R13.
Figure 2.13. N = 9, r = R9. Figure 2.14. N = 9, r 0.2835.
adjacent polygons, and compare them with Figure 2.10, which depict the correct maximal
N-gram for the same N value.
Proof of Lemma 2.2. Because fi is a pure contraction at vi, it preserves the line that runs
through vi. Hence, the equalities (2.3) and (2.4) follow. Figure 2.5, Figure 2.6 and Figure
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MAXIMAL DIMENSIONS OF UNIFORM SIERPINSKI FRACTALS 7
2.7, Figure 2.8 illustrate this idea. All we have to prove now is the equivalence between r
reaching its maximum and the equality (2.5) holding for some i {1, , N}. Because of
the symmetry ofPN and the uniformity ofr, the equality (2.5) holds for every i {1, , N}
if it holds for any one i. That is, r attains the maximum if and only if (2.5) holds for everyi {1, , N}.
By symmetry, we conclude that the uniform contraction ratio r reaches its maximum when
and only when the farthest right hand side vertex of fi(PN) and the farthest left hand side
vertex of fi1(PN) coincide, as depicted in Figure 2.15 (the red point in the interior of PN).
Figure 2.16 is the expansion of the lower left hand side copy of the N-gon, fi(PN), of Figure
2.15. Note that the blue, red and green points of Figure 2.15 correspond to those of Figure
2.16. Let us abbreviate Vj = fi(vj) for convenience. Then clearly, Vi = vi, the blue point
in Figure 2.15 and Figure 2.16. Let be the distance between the adjacent vertices Vi and
Vi1, that is, = 1(Vi Vi), where 1() stands for the length.Because PN is a regular N-gon, the size of each outer-angle of PN, that is, the angle
between the pair of vectors vi vi1 = vi1 vi andvi1 vi2 = vi2 vi1, is 2/N. Note
that we used the notion p q to denote q p R2 for better understanding. Because fiis a pure contraction at vi = Vi, the same property holds also for fi(PN). For instance,
(Vi2Vi1M) = 2/N in Figure 2.16, where () means the radian measure of the
angle. Hence, we conclude that,
Vi Vij =
Vi Vi1 +
Vi1 Vi2 + +
Vi(j1) Vij
= (cos 0, sin 0) +
cos2
N, sin2
N
+ +
cos2(j 1)
N , sin2(j 1)
N
,
and that the x-coordinate ofVi Vij keeps on increasing until it reaches its maximum when
j 1 is the largest integer such that cos(2(j 1)/N) > 0. The reason is because the
x-coordinate ofVi Vij is no longer bigger than that of
Vi Vi(j1) from that point on and
keeps decreasing, as depicted as the gray arrow in the upper right hand side of Figure 2.16.
Therefore, we conclude that such j is the largest integer such that
(2.6) 2(j 1)/N < /2, or 4(j 1) < N .
Because j is an integer, we can re-express the condition (2.6) equivalently as 4(j 1) N1,or j 1 + (N 1)/4. Since j is the biggest such integer, we conclude
j =
N 1)
4
+ 1 = m,
as defined by the equality (2.1) of Definition 2.1. Hence, the farthest right hand side vertex
of fi(PN) and the farthest left hand side vertex of fi1(PN) coincide if and only if
vi vim vi1 vi(m1) = Vim = fi(vim),
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8 BYUNGIK KAHNG AND JEREMY DAVIS
vi vi 1
Figure 2.15
Vi
Vim
MVi 1
Vi 2
Figure 2.16
vi vi 1
Figure 3.1
vj
Figure 3.2
as claimed by the equality (2.5).
3. The Maximal Contraction Ratio
Let us make the following elementary observation before proving the Main Theorem.
Proposition 3.1. LetPN be a regular N-gon with vertices v1, , vN, as in Definition 2.1.
Then, (vivjvi1) = /N, for any i, j {1, , N} and j = i, i 1.
Proof. Because PN is a regular polygon, it can be inscribed into a circle as illustrated in
Figure 3.1. Consequently, (vivjvi) is all identical for each j {1, , N} such that
j = i, i 1. Let us denote the common quantity as . Then, from the symmetry, we must
have (vj1vjvj1) = (N 2) , as illustrated in Figure 3.2. Because the external angle of
a regular N-gon is 2/N, we must have (N 2) + 2/N = , as depicted in Figure 3.2.
Consequently, we must have = /N as claimed.
We are now ready to prove the first part of the Main Theorem, which we restate as follows.
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MAXIMAL DIMENSIONS OF UNIFORM SIERPINSKI FRACTALS 9
Theorem 3.2 (Maximal Contraction Ratio). LetSr(PN) be a uniform Sierpinski fractal of
a regular N-gon, PN with vertices v1, , vN. Then, the largest possible contraction ratio,
which we denoted as RN in Lemma 2.2, turns out to be,
(3.1) RN =1
2
1
tanN
N14
tan
N
+ N
N14
.Proof. First, we make some trivial observations to reduce the problem. From symmetry, it is
easy to see that the uniform contraction ratio r must be the biggest when the polygon PN is
regular. For the remainder of the proof, therefore, let us assume PN is a regular N-gon with
the vertices v, , vN introduced in Definition 2.1. Now, observe that the equality (3.1) is
correct when N = 3 or N = 4, in which case, we get R3 = R4 =12 . Hence, we need only to
consider the case, N 5, in which case we can apply Lemma 2.2.
Lemma 2.2 tells us that we must concentrate our attention to the case where the adjacentpolygons fi(PN) and fi1(PN) share a common vertex at a principal crossing point ofGN, as
depicted in the lower region of Figure 2.15. For convenience, let us expand the region as in
Figure 3.3. As in the proof of Lemma 2.2, let Vj = fi(vj). Also, let fi1(vj) = Uj , and let M
be the midpoint of vi and vi1, or equivalently, that of Vi and Ui1. All these are illustrated
in Figure 3.3. Note that Vim is the principal crossing point between fi(PN) and fi1(PN),
as we proved in Lemma 2.2.
And then, let us name the common vertices vi and vi1 of PN as A (blue) and E (green),
respectively. Also, we name the vertices fi(vi1) and fi1(vi1) as B and D, respectively,
and let C be the midpoint of the line segment AE. Finally, let us label the principal crossing
point fi(vim) as F (red) and the midpoint of AE as C.
Now, let s be the side-lengh of PN, that is, s = 1(vi vi1) = 1(Vi Ui1). Then clearly,
(3.2) 1(Vi M) = 1(M Ui1) =s
2.
Also, from Lemma 2.2, we must have
(3.3) 1(Vi Vi1) = 1(Ui Ui1) = s RN, 1(Vi1 M) = 1(Ui D) =s
2 s RN.
On the other hand, using Proposition 3.1, we can express the size of the angle Ui1ViVimas
(Ui1ViVim) = (Vi1ViVim)(3.4)
= (Vi1ViVi2) + + (Vi(m1)ViVim)
=
N+ +
N=
N(m 1) =
N
N 1
4
.
Note that
The equality (3.4) tells us the size of internal angleVi1ViVim of the triangle Vi1ViVimis
N
N14
. Also, Proposition 3.1 tells us that the size of the angle ViVimVi1 is
N
. Hence,
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10 BYUNGIK KAHNG AND JEREMY DAVIS
Ui 1Vi
Vim
MVi 1 Ui
Figure 3.3
we conclude that
(VimVi1M) = (ViVi1Vim)(3.5)
= (ViVimVi1) + (Vi1ViVim)
=
N +
NN 1
4
.
Finally, we put everything together. From (3.2) and (3.4), we get
(3.6) 1(Vim M) = 1(Vi M)tan((Vi1ViVim)) =s
2tan
N
N 1
4
.
Also, from (3.3) and (3.5), we get
1(Vim M) = 1(Vi1 M)tan((VimVi1M))(3.7)
= s2
s RN tan
N+
NN 1
4 .
From (3.6) and (3.7), we get the equation,
(3.8)s
2tan
N
N 1
4
=s
2 s RN
tan
N+
N
N 1
4
,
where the positive constant s cancels itself out. Solving for RN through elementary algebra,
we get the equality (3.1). This completes the proof of Theorem 3.2, which is the first part
of the Main Theorem.
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MAXIMAL DIMENSIONS OF UNIFORM SIERPINSKI FRACTALS 11
The second part of the Main Theorem follows immediately from Hutchinsons Theorem
[4], which can be found in most standard textbooks on fractals, say [2, 3, 6].
Corollary 3.3. Let Sr(PN) be a uniform Sierpinski fractal of a regular N-gon, PN with
vertices v1, , vN. Then, the maximal Hausdorff dimension hN of Sr(PN) is attained when
r = RN, given by the equality (3.1) of Theorem 3.2, and
(3.9) hN =ln N
ln(1/RN)=
ln N
ln RN.
Proof. From Hutchinsons Theorem on self similar fractals [4],
Ni=1
rhi = 1,
where ri is the contraction ratio of fi. Because r1 = = rN = r, we must have
Nrh = 1, rh =1
N, h = logr(1/N) =
ln N
ln r.
The maximum of h is attained, therefore, when r reaches the maximum, RN. Therefore,
hN = ln N
ln RN.
4. Discussion
In this paper, we set up our problem and the main result using convex polygons, primarily
for convenience. However, the equality (1.4) of the Main Theorem applies in more generalcircumstances as well. As we see from Figure 4.1 Figure 4.4, the Main Theorem applies
to a class of the self similar fractals given by N-grams, which we will call the regular N-
gram fractals. It is but a trivial exercise to set up the precise definition of the regular
N-gram fractals and the appropriate generalization of the Main Theorem, so we leave it to
the interested readers. It is not surprising at all that such generalization is possible. The
uniform contraction ratio r reaches the maximum when the outer vertices of the N-grams
touch each other (Figure 4.2), but the outer vertices of the N gram coincide with the vertices
ofPN (Figure 4.1), and thus the equality (1.4) applies. Indeed, as more iterations are taken,
the regular N-gram fractals gets more similarto the regular Sierpinski fractals, as illustratedin Figure 4.3 and Figure 4.4. Again, we leave the detail to the readers.
If we consider the inner vertices as well, however, the regular N-gram fractal problem
becomes even more interesting. As depicted in Figure 4.5 Figure 4.7, we need to consider
the contractions at the inner vertices as well, with different uniform contraction ratios. For
Figure 4.5 Figure 4.7, the contraction ratios ro = R7 0.308 and ri = 0.136 were selected,
for the contractions at the outer and the inner vertices, respectively. At this moment, we are
not sure what to expect in this case. In any case, it appears that the maximal contraction
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12 BYUNGIK KAHNG AND JEREMY DAVIS
Figure 4.1. N = 7, r = RN 0.30798. Figure 4.2. N = 7, first iteration.
Figure 4.3. N = 7, third iteration. Figure 4.4. N = 7, fourth iteration.
Figure 4.5. ro 0.308, ri = 0.136. Figure 4.6. N = 7, first iteration.
ratio formula (1.4) plays some role, but perhaps not in general. For the time being, therefore,
we leave this problem open for future research.
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MAXIMAL DIMENSIONS OF UNIFORM SIERPINSKI FRACTALS 13
Figure 4.7. N = 7, second iteration. Figure 4.8. N = 7, third iteration.
Acknowledgments. This paper is partly based upon the undergraduate research project [1]
of the second author, supervised by the first author. It was supported in part by Morris Aca-
demic Partnership of University of Minnesota at Morris, and also in part my Mathematical
Association of America. The authors thank Mark Logan, who helped the authors as the sec-
ond reader of this undergraduate research project. They also thank the anonymous referees
of the journal, Fractals, for valuable criticisms and suggestions for the revision of this paper.
References
[1] J. Davis, Sierpinskii Fractals and their dimensionality, B.A. Thesis, University of Minnesota at Morris,
2009.
[2] G. Edgar, Measure, Topology, and Fractal Geometry, Undergraduate Texts in Mathematics, Springer,2007.
[3] K. Falconer, Fractal Geometry: Mathematical Foundations and Application, Wiley, 2nd ed., 2003.
[4] J. Hutchinson, Fractals and self-similarity, Indiana J. Math., 30 (1981), pp. 713747.
[5] B. Kahng, Sierpinski fractal plotter, http://cda.morris.umn.edu/kahng/Sierpinski.html, (2009).
[6] C. A. Rogers, Hausdorff Measures, Cambridge Mathematical Library, Cambridge Univ. Press, 2nd ed.,
1998.
Byungik Kahng, 600 East 4th Street, Division of Science and Mathematics, University
of Minnesota, Morris, MN 56267, USA
E-mail address: [email protected]
Jeremy Davis, 600 East 4th Street, UMM Mail 220, University of Minnesota, Morris, MN
56267, USA
E-mail address: [email protected]