#### The Finite Number of Interior Component Shapes of the Levy Dragon

Discrete Comput Geom
The Finite Number of Interior Component Shapes of the Levy Dragon
Eytan Alster
The Levy dragon is a connected self-similar tile with disconnected interior. It was previously known that there are at least 16 different shapes of its interior components. Using simple properties of an infinite sequence of curves which converge into the Levy dragon, it is proved that the number of different shapes of the interior components is finite. A detailed description of the buildup of those shapes as unions of various contractions of three convex polygonal shapes is given, and the number of shapes is determined. The Levy dragon K (Fig. 1) is the attractor of an iterated function system F = {f0, f1} where the two R√2→ R2 contracting similarity transformations f0 and f1 include contraction by 1/ 2 and a rotation by π/4 for f0 and by −π/4 for f1. As proved by Huchinson [5], the attractor of such a system is a unique compact set that satisfies For any compact set S, let F (S) = f0(S) ∪ f1(S); then K = F (K), so K is the fixed point of F , and K = lim F k(S) k→∞ If S is a straight line segment C0 whose end points are the fixed points of f0 and f1, the curves Ck = F k(C0) converge (with respect to the Hausdorff metric) as k → ∞
Levy dragon; Interior component shapes
1 Introduction
K = f0(K) ∪ f1(K)
to K . Ck consists of 2k segments of length lk = (1/√2)kl0, where l0 is the distance
between the fixed points of f0 and f1. Figure 2 demonstrates the first iterations in this
sequence with a curve in black and its next iteration in gray. Every Ck segment has a
direction starting with the direction of C0 from the fixed point of f0 to the fixed point
of f1. Ck+1 is obtained from Ck by replacing every segment of Ck by two segments
that form with it an isosceles right triangle with the Ck segment its hypotenuse and
the triangle to the left with respect to its direction. The directions of the two Ck+1
segments replacing a Ck segment are such that they form a path from its start to its
end. For the rest of this paper, a Ck segment means a directed segment as marked
by the arrows in Fig. 2. When there are two opposite arrows on the same segment
as for one of the segments in C4, there are two overlapping segments with opposite
directions. Properties of the curves Ck are discussed in Sect. 2.
Another choice for S used in many treatments of the Levy dragon is the isosceles
right triangle T with end points of the hypotenuse at the fixed points of f0 and f1
and the third vertex to the left with respect to the direction from the fixed point of
f0 to the fixed point of f1. That third vertex is the only common point of the two
contracted triangles in F (T ), and their hypotenuses are on two sides of T (Fig. 3(a)).
The 2k contracted triangles in F k(T ) are connected, but their interiors never overlap
since, as demonstrated in Fig. 3(b), for any square with up to four F k(T ) triangles,
the F k+1(T ) triangles that replace them are within that square, and their interiors do
not overlap.
These two choices for S are thus related: the hypotenuses of the F k(T ) triangles
are the Ck segments, while their other two sides are the Ck+1 segments. In this paper
properties of the curves Ck are used through their relation to the F k(T ) triangles
to determine the shapes of the interior components of the Levy dragon in Sect. 3.
A similar method was used by Ngai and Nguyen [
7
] to prove geometric properties of
the Heighway dragon using properties of the dragon curves.
P. Levy introduced in 1938 [
6
] (annotated English translation in [
4
]) the set now
known as the Levy dragon as the limit of the sequence of curves Ck within a
discussion of the properties of self similar curves like the Koch curve. Many properties
of the Levy dragon are proved in [
6
]. Among them the location and shape of the
largest of the interior components that consists of 22 triangles of F 16(T ) hence
covers 22/216 ≈ 1/3000 of the area of the Levy dragon, which is equal to the area of
triangle T .
The Hausdorff dimension of the boundary of the Levy dragon was first calculated
by Duvall and Keesling [
2
] to be D = 1.934007183 . . . . This dimension was
calculated by Strichartz and Wang [
9
] using a different method.
Deng and Ngai [
3
] calculated the dimensions of subsets of the boundary of the
Levy dragon which are the common intersection of at least a given number of
neighbors in the tiling of the plane by copies of the Levy dragon. Ngai and Tang [
8
] proved
that the closure of each interior component of the Levy dragon is a topological disk.
An F k(T ) triangle with all the 14 triangles intersecting it also in F k(T ), as
demonstrated in Fig. 3(c) for triangle X there, is called in [
2
] covered since it is tiled by two
F k+1(T ) covered triangles and thus is a subset of K .
The Levy dragon is a connected set with disconnected interior. Bailey, Kim, and
Strichartz [
1
] showed that contrary to the dragon boundary being a fractal, each of
the infinitely many interior components, which are all too small to be seen in Fig. 1,
has a one-dimensional polygonal boundary. By using a computer program that acts
as a “microscope” into F k(T ) and differentiates between an F k(T ) triangle that is
covered or not, the buildup of the interior components is observed. Covered triangles
appear for the first time in F 14(T ), and following the growth of connected sets of
covered triangles, they observed that distinct sets of covered triangles never merge
and that some covered sets of triangles stop growing after a few iterations while
other sets never stop growing. They identified 16 shapes of interior components and
conjectured that each interior component is similar to one of those shapes.
2 Properties of the Curves Ck
The points where Ck segments are joined are called its vertices. By the iteration
process every vertex of Ck is also a vertex of any Ck+m and hence is in K . Every
vertex has a type which describes the direction change of the segment after it with
respect to the segment before it. For two consecutive segments in Ck , if the directions
of the segments before and after the vertex are α and β, respectively, then the type of
the vertex is β − α. In Ck+1 the segment before that vertex is the second segment out
of the two that replace the Ck segment before it, and hence its direction is α − π/4,
while the segment after that vertex is the first segment out of the two that replace the
Ck segment after it, and hence its direction is β + π/4. The type of that Ck+1 vertex
is thus (β + π/4) − (α − π/4) = (β − α) + π/2. The type of a vertex changes by a
right angle after every iteration, and after four iterations it returns to the same type.
There are 2k−1 new vertices in Ck , one for every segment of Ck−1. The type of
all of those new vertices is π/2 to the right since a Ck−1 segment is replaced by two
Ck segments to its left. There are thus only four types denoted by R for a turn to the
right, F for continuing forward, L for a turn to the left, and B for a turn backward.
As demonstrated in Fig. 4, a vertex created with type R in Ck has type F in Ck+1,
type L in Ck+2, type B in Ck+3, and it returns to type R in Ck+4. Every vertex thus
goes through a cyclic change of type with period 4 along the sequence of iterations.
A vertex created in Ck is R type in Ck+4m, F type in Ck+4m+1, L type in Ck+4m+2,
and B type in Ck+4m+3 for any nonnegative integer m. The type of a vertex with type
X that was created in Ck−m is denoted by X[k−m].
At one of the ends of any Ck segment there is a vertex with type R[k]. The vertex at
the other end was created by Ck−m for some m > 0. If two Ck segments overlap, they
must have opposite directions because if two overlapping segments have the same
direction, then at the end where the vertices with type R[k] are, there are two other
overlapping segments with the same direction. This could happen only if there were
two overlapping segments with the same direction in Ck−1. Continuing backward
along the iterations sequence and repeating this argument would lead to the existence
of overlapping segments pair with the same direction in any Ck−m. However, C0 (and
also C1, C2, and C3) do not have overlapping segments at all, and hence there are no
overlapping segments with the same direction in any Ck . Thus, no more than two
segments can overlap because a third would have the same direction as one of them.
For the rest of this paper, such an overlapping segments pair is referred to as an OSP.
Another corollary of this is that no more than four vertices are located at the same
point.
Since all the segments of Ck have the same length and all the turns at the vertices
are multiples of right angles, all the segments of Ck are on the sides of squares of
a lattice of squares tiling R2. These squares are referred to as the Ck squares. All
the new vertices created by the next iteration are at centers of those squares and the
Ck+1 squares are obtained by the diagonals of the Ck squares. Hence, all the vertices
located at one point were created by the same iteration as R type and thus have the
same type after any number of additional iterations.
If there are vertices on the four corners of a Ck square, then the vertices on two
opposite corners were created by Ck at centers of two adjacent Ck−1 squares and thus
have type R. The other two opposite corners are at the two ends of one side of a Ck−1
square, and if there was a Ck−1 segment on it, on one of its ends there would be a
vertex created as R by Ck−1 and thus would have type F in Ck . Hence, the vertices on
two opposite corners of a Ck square are R[k] and on one of the other corners F [k−1].
The vertex at the last corner was created by some iteration up to Ck−2.
Once the types of the vertices in a neighborhood are known, one can determine
which segments are in the same path within that neighborhood. The segments in the
same path either all exist or all are absent.
3 The Shapes of Interior Components
Let P be the reflection through the line perpendicular to C0 at its center. Then P 2 is
the identity, and since the fixed points of f0 and f1 are at the ends of C0 and their
rotation angles are opposite, Pf0P = f1 and Pf1P = f0, and thus
P (K) = Pf0(K) ∪ Pf1(K) = Pf0P 2(K) ∪ Pf1P 2(K) = f1P (K) ∪ f0P (K)
and since the attractor of the iterated function system {f0, f1} is unique [
5
], we have
P (K) = K . Hence, P is a symmetry of the Levy dragon, and thus for every shape of
interior component, the reflected shape also is a shape of an interior component. We
consider such a pair of reflected shapes as one shape.
To find the shapes of the interior components, we start with the concept of a
covered triangle, as demonstrated in Fig. 3(c), and transfer it to the segment of Ck so that
the hypotenuse of the covered triangle is on it and the triangle is to its left. Thus, the
minimal configuration for a covered segment is as in Fig. 5(a) for segment s there. It
includes seven OSPs and one single segment, which is the hypotenuse of the triangle
that intersects the covered triangle only at its right angle vertex. The opposite segment
to a covered segment is almost covered, missing only this last single segment. Since
there is an R vertex at one of the ends of this single segment because of the existence
of one of the segments in the OSPs perpendicular to the single segment, the segment
opposite to it must also exist. Also, because there is an R vertex at the end of one of
the two OSPs to the right of segment s, there is a segment antiparallel to s there too.
Thus, the minimal configuration for a covered segment s is as in Fig. 5(b).
The significance of covered segments for the purpose of finding the shapes of the
interior components is that if a Ck segment is covered, the two Ck+1 segments that
are opposite to the two segments that replace it are also covered. See Fig. 5(c), where
the Ck+1 segments (in gray) that replace the Ck segments (in black) are shown, and
segments r and t are covered since s is covered. Thus, the F k(T ) triangle to the left
of the covered segment is tiled by two covered F k+1(T ) triangles. Continuing to the
next iterations, for any nonnegative integer m, this triangle is tiled by 2m triangles of
F k+m(T ); hence a covered triangle is a subset of K , and its interior is a subset of the
interior of K .
We say that two F k(T ) triangles are edge connected when they have a common
edge. The interior of the union of a set of edge connected covered triangles is in the
same interior component of K . The buildup of a component of the interior of K
during successive iterations of F k(T ) must start with the appearance after some iteration
of a set of edge connected covered triangles that are not edge connected to any
covered triangle generated in previous iterations. Such a seed of an interior component
grows into its final shape as in the next iterations there are covered triangles edge
connected to it. Finding covered triangles will thus not serve our purpose since they
might be tilling a triangle that was already covered in previous iterations. Therefore
we define the concept of an originally covered segment for a triangle that is not a
subset of a covered triangle.
Definition 3.1 A Ck segment is originally covered if it is covered and the opposite
segment is one of the two segments that replace a Ck−1 segment that is not covered.
Figures 6(a) and (b) show the two configurations for a Ck covered segment s (in
black) with the type R[k] vertex at the beginning and at the end of s, four more Ck
segments in dashed lines and the segments that must be in Ck−1 for those Ck segments
to exist (in gray). The four Ck−1 segments denoted with c, d, e, and f must exist if
the Ck segments in dashed lines exist. Segment b in both cases is the Ck−1 segment
that must be not covered in order that segment s will be originally covered. For b
to be not covered, at least one of the four segments c, d, e, and f must be absent in
Ck−1, and thus at least one of the four segments with dashed lines is absent in Ck .
Lemma 3.2 A Ck originally covered segment has a vertex with type F [k−1] at one of
its end points.
Proof Let the Ck originally covered segment be s in Figs. 6(a) and (b). Consider
the Ck square to the left of s. As any Ck square, it has two R[k] vertices on two
opposite corners and an F [k−1] vertex on one of the other two corners. In both cases,
Figs. 6(a) and (b), the F vertex cannot be at the corner intersected by the four Ck
segments with dashed lines, since if it does, all those four segments are in paths of
the minimal configuration for s to be covered, hence all of them exist, and then s is
not originally covered. Thus, the F [k−1] vertex must be on the last corner, which is
an end point of segment s in both cases.
By the same argument, also the OSP collinear with s at its R end has an F [k−1]
vertex at its other end. The minimal two configurations of an originally covered
segment is demonstrated in Figs. 7(a) and (b). The points at the ends of the OSPs
perpendicular to s at its R end are denoted X and Y (we use those for the vertex and its
type).
Proposition 3.3 The segment opposite to an originally covered segment is covered.
If it is originally covered, then there are only three possible configurations:
• X = B, Y = L
• X = L, Y = B
• X = L, Y = L
• X = B, Y = L
• X = R, Y = L
The segments with end points at X and Y that must be absent are in one path.
If it is only covered, there are only two possible configurations:
Proof The segment opposite to a covered segment is almost covered, missing only
the single segment antiparallel to it on the opposite edge of the square to its left.
Consider the originally covered segment s in the two cases in Figs. 7(a) and (b), and
let s be the opposite segment. Since X, Y , and the two F [k−1] vertices are on the
four corners of a Ck−1 square, either X or Y has type L[k−2].
For both cases, if Y is either R or F , then X = L, and the segments h and g exist
since they are in the paths of segments in the covered configuration, and thus since
s is originally covered, both e and f must be absent. But if Y is R or F , segment
b must exist, and hence the whole path b, c, d , e must exist. Hence, Y is neither R
nor F . Thus, Y is either B or L, and in both cases the one segment needed to make s
covered must exist, hence s is covered.
Thus, if X = L, s is originally covered, and either Y = L or Y = B. If X = L,
then Y = L, and if s is originally covered, then segments i and j exist and segments k
and b are absent. Hence, the path that includes segments b, c, d , and e is absent. But
if X = R or X = F , segment e exists, hence X can only be B.
The path that includes segments b, c, d , and e is absent in both cases, and we refer
to it in these cases as the forbidden path.
If segment s is originally covered while s is only covered, then two of the
segments e, f , g, and h must be absent while all the four segments i, b, j , and k must
exist. Since segment b exists, all the path b, c, d , e exists. If X = L, then h and g
exist, and thus e and f must be absent, hence X = L. Also, X = F since then all four
segments exist, hence X is either B or R. Since X = L, we have Y = L.
We name the three possible configurations of an OSP with the two segments
originally covered by the types appearing in clockwise order around the covered area;
hence we refer to the three possibilities above as the LB, BL, and LL
configurations, respectively.
By Proposition 3.3 an originally covered segment can never be part of the
boundary of an interior component. The hypotenuse of an originally covered F k(T ) triangle
is thus never on the boundary of an interior component, and only the two short edges
of such a triangle can be part of the boundary. The following proposition characterizes
the segments that form the boundary of covered sets.
Proposition 3.4 If a segment is covered and the opposite segment is not covered, then
those segments are on the boundary of the covered set which includes the covered
triangle to the left of the covered segment.
Proof The configuration for such an OSP in Ck with one segment covered and the
other not covered is demonstrated in Fig. 5(a), where segment s is the covered
segment, and the opposite segment is not covered only because one segment is absent. In
Ck+2 every Ck segment is replaced by a C2 to its left. On every half of segment s and
its opposite, there is a Ck+2 OSP where the segment with the same direction as s is
covered while the opposite segment is not covered. Thus, for all p > 0, in Ck+2p on
segment s and its opposite there are 2p OSPs with one segment covered and one not
covered, so all the F k+2p(T ) triangles with hypotenuse on s are not covered; hence
s is on the boundary of the covered area.
Lemma 3.5 From the two Ck+1 segments that replace an originally covered Ck
segment s, one is either originally covered or not covered, while the other is either
covered or not covered depending on whether the segment opposite to s is also originally
covered or only covered and on the type X at the end of the OSP perpendicular to s
at its R end.
Proof Consider Fig. 8(a). Segment s is a Ck originally covered segment that starts
at the R vertex and ends at the F vertex. The gray triangle is the F k(T ) triangle
thus covered. Segments sR and sF are the two Ck+1 segments that replace s. By
Lemma 3.2, since sR starts at an F [k] vertex and ends at an R[k+1] vertex, it can be
originally covered in Ck+1, while sF ends at an L vertex, and hence it cannot be
originally covered.
The segment opposite to sR is covered because s is covered, and therefore sR is
almost covered, and it is covered if the Ck+1 segment h1 exists and not covered if
h1 is absent. For sR to be originally covered, the Ck segment b must be not covered.
Since s is originally covered, two segments at the X end of b are absent, and thus
segment b is not covered; so if sR is covered, it is originally covered. For segment h1
to exist in Ck+1, segment h must exist in Ck . By Proposition 3.3, since s is originally
covered, the opposite segment is covered, and if it is originally covered, then either
X = L, h exists, and sR is originally covered or X = B, h is absent, and sR is not
covered, hence by Proposition 3.4 it is a boundary segment. If the segment opposite to
s is only covered, then by Proposition 3.3 either X = B, h exists, and sR is originally
covered or X = R, h is absent, and sr is a boundary segment.
Since sF cannot be originally covered, either it is not covered and part of the
boundary of the interior component or it is covered. It can only be covered if the
Ck segment c is covered. If c is covered, it is originally covered because of the two
segments that are absent at X since s is originally covered.
In case s starts at the F vertex and ends at the R vertex, the same arguments lead
to the same results, only sR is the second of the two segments replacing s, while sF
is the first.
Theorem 3.6 A Ck OSP with the two segments originally covered develops up to
Ck+2 into one out of three covered shapes in F k+2(T ) depending on whether it is an
LB, BL, or LL.
Proof The covered shape for the OSP of two Ck originally covered segments is a
square tiled by two F k(T ) triangles (lighter gray areas in Fig. 8(b)). By Lemma 3.5,
if X = L, then on an edge of that square there is a Ck+1 OSP with one originally
covered segment and one covered segment, so the covered shape grows by an F k+1(T )
triangle (darker gray areas in Fig. 8(b)). This happens once for an LB or BL and
twice for LL. If X = B, that Ck+1 segment is a boundary segment of the covered
shape, so for an LB or BL, one edge of the square is part of the boundary (solid
black lines in Fig. 8(b)). On the other two edges of the covered square, either the
OSP has one covered segment and one not covered, so it is a boundary segment, or
both are covered when there is another Ck originally covered segment perpendicular
to the originally covered OSP at its F end.
For the Ck+1 OSP with one covered segment and one originally covered, when
X = L, the R vertices are at the beginning of the originally covered one, hence its X
point is at the same point as for the Ck originally covered OSP, only in Ck+1 the type
turns to X = B. By Lemma 3.5 again on one edge of the F k+1(T ) triangle that was
added, there is a Ck+2 OSP with one segment covered and the second originally
covered, hence the covered shape grows by an F k+2(T ) triangle (the darkest gray areas
in Fig. 8(b)). The Ck+2 OSP on the other edge either has one covered segment and
one not covered, so it is a boundary segment, or both are covered when there is
another Ck+1 originally covered segment perpendicular to the Ck+1 originally covered
segment at its F end.
Finally for the Ck+2 OSP with one covered segment and one originally covered,
its X point is again at the same point, only in Ck+2 the type turns to X = R. By
Lemma 3.5, on one edge of the F k+2(T ) triangle that was added, there is a Ck+3
OSP with one segment covered and the second not covered, hence it is a boundary
segment. The Ck+3 OSP on the other edge either has one covered segment and one
not covered, so it is a boundary segment, or both are covered when there is another
Ck+2 originally covered segment perpendicular to the Ck+2 one at its F end. This
originally covered segment exists if there is another Ck originally covered segment
perpendicular to the originally covered OSP at its F end. Thus, the collinear Ck+1
and Ck+3 segments are either both boundary segments or both covered.
To determine the possible interior components shapes, we now first explore the
possible seeds of covered areas and then their possible growth patterns.
Theorem 3.7 The possible Ck configurations for the initial edge connected set of
covered F k(T ) triangles of an interior component include only two configurations
with one originally covered OSP, four configurations with two originally covered
OSPs, and two configurations with four originally covered OSPs.
Proof The first covered triangles of an interior component that appear in Ck are not
edge connected to triangles that were covered before; hence by Proposition 3.3 they
must appear only in OSPs with both segments originally covered.
Consider Fig. 9. It shows the Ck segments configuration for an LB originally
covered OSP with segments denoted by s and b. The square that has the B and L
vertices on two adjacent corners and the F end of the originally covered OSP at its
center is a Ck−2 square; hence the type at the corner opposite the L is another L. The
type at the last corner is denoted by Z. The path of forbidden segments is marked
with circled arrows, and it extends up to the point Z. There is also a path of segments
denoted with a + sign which extends up to Z. All the segments in the minimal LB
configuration are in five paths, and all the segments in one path are denoted by the
same labels s, b, c, d, and e. Also segments in three additional paths are denoted by
α1, α2, and β. The segments in the + path and the α1, α2, and β paths are not in the
minimal configuration of the LB and hence may exist or be absent.
The covered area for an originally covered OSP is a square. For another originally
covered OSP to be with a covered square edge connected to this square, it must be an
OSP perpendicular to the first with a common end. By Lemma 3.2 such an originally
covered OSP must have an F [k−1] at one of its ends. Hence, the only candidates are
the two OSPs at the F end with segments d and e. For one of them to be originally
covered, the β path must exist, while the + path must be absent since it is its
forbidden path. For the other to be originally covered, the α1 and α2 paths must exist and
either Z = B or Z = L (its forbidden path is part of the forbidden path of the LB).
The only additional possible originally covered OSP with covered square edge
connected to the covered squares of those two is the OSP with segments s and b collinear
with the first one with common end at its F end. For this OSP to be originally
covered, the α1, α2, and β paths must exist, the + path must be absent, and Z must be
either L or B. But these are the conditions for the other two OSPs to be originally
covered, hence only 1, 2, or 4 originally covered OSPs with edge connected squares
can appear together.
Figure 10 demonstrates all these possibilities relative to the Ck originally covered
OSPs. The gray shapes are obtained by applying Theorem 3.6 in each case, and there
are solid black lines on the segments already known by Theorem 3.6 to be boundary
segments. Whenever the + path is optional, there is a + near the B.
If the β path exists, the + path must be absent, because otherwise there is a covered
OSP which is not originally covered with its covered square edge connected to the
covered square of the first LB, and this is impossible for the initial appearance of
covered triangles for an interior component.
Z = F because if Z = F , the s path is in the forbidden path.
If Z = R, then α1 is in the forbidden path, and it is absent. Then either the β path
is absent, it is an LB as in Fig. 10(b), and the + path is optional, or the β path exists,
the + path is absent, and it is an LBL as in Fig. 10(d).
If Z = B, then α1 is in the e path and α2 is in the s path, so both of them exist. If
the β path is absent, then it is a BLB as in Fig. 10(f) with the + path optional. If the
β path exists and the + path is absent, then it is an LBLBL as in Fig. 10(g). (The first
L is repeated at the end for a reason that will be clear later.)
If Z = L, then the + path is in the forbidden path, and it is absent. α1 and α2 are
in the same path α. If both β and α are absent, it is an LB as in Fig. 10(a). If β exists
and α is absent, it is an LBL as in Fig. 10(c). If β is absent and α exists, it is an LLB
as in Fig. 10(e). If both β and α exist, it is an LBLLL as in Fig. 10(h).
There is no need to repeat these arguments for the configurations that start with a
BL or LL since these are included within those possibilities rotated or reflected.
The only manner in which these core shapes that are initiated in Ck and reach their
final shape in Ck+2 can grow is if in Ck+1 there is an originally covered OSPs such
that the covered squares they generate have an edge on the Ck+2 segments on the
edge of the shape which might not be a boundary segments. In Fig. 11 these OSPs
for an LB, BL, and LL are denoted with g. If the g segments are covered, they are
originally covered because the forbidden path prevents the relevant Ck segments from
being covered. For g in all three configurations to be covered, three additional Ck+1
segments must exist. Those segments exist in Ck+1 if the Ck segments in path t exist.
Path t exists only if the segments denoted as u1 and v1 exist (u1 starts at the vertex U ,
and v1 ends at the vertex V ).
Since the L and F vertices in Ck turn respectively into B and L in Ck+1, the
originally covered OSP g in an LB is a BL, in a BL it is an LB, and in an LL one
LB (towards U ) and one BL (towards V ). By Theorem 3.6 the covered area initiated
by g grows in Ck+2 and Ck+3 into the same basic shapes, only contracted by 1/√2.
Fig. 13 A tail
We refer to the number of basic shapes in a tail as its tail length and denote the
structures obtained as BnL, LnB, and Ln, mL, where n and m are the tail lengths.
Every pair of consecutive basic shapes in a tail is obtained from the previous pair by
a contraction by 1/2. The points U and V are the fixed points of these contractions
and the tip of the tail in case it is infinite.
For each step of this tail building to go on, the segments equivalent to u1 and v1
must exist. Note in Fig. 11 that for an LB (BL), the first tail shape is a BL (LB),
and its U (V ) point is at the V (U ) point of the LB (BL). For its u1 (v1) segment to
exist so that there will be a second basic shape in the tail, there must be a Ck segment
opposite to v1 (u1). That segment is denoted v2 (u2), so the existence of v1 and v2
(u1 and u2) ensures a tail with length 2 for an LB (BL).
Every odd (even) basic shape in a tail is parallel to the first (second) one. Thus,
for the p basic shape in a tail to exist, there must be a Ck+p−1 segment parallel to
v1 or u1 if p is odd and a Ck+p−2 segment parallel to v2 or u2 if p is even. This
requires the existence of Ck segments denoted by vp or up obtained by rotation by
p right angels counterclockwise from v1 and u2 and clockwise from v2 and u1, as
demonstrated in Fig. 14.
For p > 8, these same Ck segments ensure the existence of the required segments,
so that if the first p − 1 shapes in a tail exist for p = 8i + j if uj or vj exist, then
the p shape in the tail exists. Thus, for any p < 9, if up (or vp) is absent while all
the segments u1 up to up−1 (or v1 up to vp−1) exist, then the tail length is finite and
given by p − 1, while if all eight segments exist, the tail length is infinite. There are
thus only nine possible tail lengths for any tail, and this leads to our main result.
Theorem 3.8 The number of shapes of the interior components of the Levy dragon
is finite.
Proof By Theorem 3.7 there is a finite number of initial sets of edge connected
covered triangles (8 possible configurations). For each of those, there is a finite number
of tails (up to 6). Whatever the tail tip type is, there is a finite number of possible tail
lengths (9), hence the number of possible shapes is finite.
To determine the possible interior components shapes, we first show that the
possible tail lengths for the tails of an LB, BL, or LL are much more limited than those
nine possible lengths. Next, the tail tip types for all tails in all the configurations
obtained for different Z values in Theorem 3.7 are determined. Finally, some shapes
Possible tail lengths
0, 1, 2, 3, ∞
0, 1, 2, 6, ∞
0, 1, 4, 5, ∞
0, 2, 3, 4, ∞
+ path exists
that could exist by these conditions are proved to be impossible because of some
dependencies between paths that provide tail tip segments to different tails within the
same configuration.
Lemma 3.9 The tail lengths of an LB and a BL are limited to the values in Table 2
depending on the tail tip type and existence or absence of the + path.
Proof The eight tail tip segments are always in four pairs of consecutive segments
depending on the tail tip type. The two segments of such a pair either both exist, or
both are absent. Hence, there are only three possible finite nonzero tail lengths for
any tail tip type shown in Table 1.
Actually the possibilities for the tail tips in a given LB or BL configuration are
more limited since some of the tail tip segments are always provided by paths within
the minimal configuration continued towards the tail tips. Figure 15(a) demonstrates
how by using the R and F vertices within the LB configuration and Ck and Ck−1
squares we can determine the R, F , and L vertices on paths towards the V tail tips.
Thus, for the V tail tip of any LB, v7 and v8 always exist. In addition, v6 is in the +
path which for some configurations is forbidden and for some optional. As
demonstrated in Fig. 15(b), this holds also for any BL and its U tail tip segments. Hence, the
possible tail lengths for an LB or BL are those in Table 2.
Lemma 3.10 The only possible tail length combinations for the double tail of an LL
are L0, 0L, L∞, 1L, L1, ∞L, and L∞, ∞L.
Proof As demonstrated in Fig. 15(c), continuations of paths of the minimal LL
configuration provide v7 and v8 for the V tail tip and u7 and u8 for the U tail tip. U , V ,
and the two L vertices are on the four corners of a Ck−2 square; hence either U = L
and V = B , or U = B and V = L. In the first case, u6 and u7 are on the same path,
while in the second case, v7 and v6 are on the same path. But v6 and u6 are on the
same path shown in Fig. 15(c); hence u6 and v6 always exist for any LL. u1 and v1
are on the same path; hence either both are absent and the double tail is L0, 0L, or
both exist, in which case both tails have nonzero length. If U = L, V = B , and u2
is absent, the double tail is L∞, 1L, while if U = B , V = L, and v2 is absent, the
double tail is L1, ∞L. In case U = L if u2 also exists or in case V = L if v2 also
exists, then the double tail is L∞, ∞L.
There are six points that are tail tips in all or some of the eight cases in Fig. 10.
These six tail tips, denoted by V1, U1, U2, V2, V3, and U3, are emphasized in
Fig. 16(a), which is the Ck−2 configuration from which the Ck configuration in
Fig. 9 is obtained by replacing every segment in Fig. 16(a) by a C2. The segments in
Fig. 16(a) are denoted by the labels of the segments that replace them in Fig. 9 and
are extended up to the tail tips. The type in Ck−2 of a vertex with type X in Ck is
denoted by X∗.
Since Z is at a corner opposite a B in a Ck−2 square, we have that if Z = B it was
created before Ck−3, while if Z = B , it might be that Z = B[k−3] and the other B
created before Ck−3. But these two possibilities lead to configurations that are related
by a rotation and thus will lead to the same shapes; hence for the purpose of finding
all shapes, we can assume that Z was created before Ck−3. Hence, Z = R[k−4q], or
Z = B[k−3−4q], or Z = L[k−2−4q] for some integer q > 0.
Proposition 3.11 The tail tip type for all the tails of the configurations in
Theorem 3.7 are determined.
Proof First, consider the case Z = R[k−4]. Figure 16(b) shows the Ck−4
configuration in this case. Either V1 or U1 is an F [k−5]. By Theorem 3.7 the possible shapes
are LB and LBL. For the LB shape, the β path is absent, but if U1 = F , then β is in
the b path; hence U1 = F and V1 = F . For an LBL shape, β exists, and the + path
is absent. U1 = L, so that b is not in the forbidden path, and V1 = L, so that d is not
in the + path. Since at least one out of V1 and U1 is F , the possibilities for (V1, U1)
are (F, F ), (F, B), (F, R), (B, F ), (R, F ). The last two lead to configurations that
are reflected versions of the previous two; hence for the purpose of determining the
possible shapes, we can assume that V1 = F and U1 is F , B, or R.
For the rest of the possibilities, since Z was created before Ck−4, then V1 = U1 =
R[k−4]. U2, V2, and Z are on three corners of a Ck−2 square with an L[k−2] in the
last corner; hence V2 = L[k−2] and U2 = B[k−3]. Also V3, U3, and Z are on three
corners of another Ck−2 square with an L[k−2] in the last corner; hence U3 = L[k−2]
and V3 = B[k−3]. Thus, all six tail tip types are known.
Figure 17 shows the Ck−2 configuration with the paths extended up to the tail tips
and the additional paths γ , δ, μ, ν, η, and χ that might provide the rest of the tail tip
segments.
Lemma 3.12 If Z = R[k−4], then the shape L2B is impossible.
Fig. 17 The Ck−2
configuration in case Z was
created before Ck−4
Proof By Proposition 3.11 in this case V1 = F . Two paths, γ , which provides v2 and
v5, and δ, which provides v3 and v4, are added in Fig. 16(b) and extended up to U1.
If U1 = L, then b is in the forbidden path, and hence U1 = L. If γ exists, then also
U1 = B; otherwise γ is in the forbidden path. Thus, if γ exists, then U1 = R, but then
δ is in the same path as γ , so both of them exist. For an L2B shape, γ exists and δ is
absent; hence it is impossible in this case.
Note that L2B is possible in other configurations Z = R[k−4q] with q > 1 and
Z = L[k−2−4q].
Lemma 3.13 If Z = B, the shape B0L∞B is impossible.
Proof For B0L∞B to exist, in case Z = B, the + path must exist, but the δ path
must be absent. Consider the vicinity of Z when it was created in Ck−3−4q obtained
from its vicinity in Ck and demonstrated in Fig. 18(a) for q even and Fig. 18(b) for q
odd. In both cases the square with corners at P , Q, S, and T is a Ck−4−4q square and
hence in Ck−3−4q either P = Q = F and one out of S and T is L, or S = T = F and
one out of P and Q is L. If Q = F , then in both cases δ is in the c path, so it always
exists. If S = T = F , then in the even q case, if P = L, the β path is in the e path,
but β is absent, hence P = L and Q = L, and thus δ is in the γ path which is in the
+ path, while in the odd q case, if P = L, then β is in the + and γ path, and if the
+ path exists, P = L and Q = L, and δ is in the e path. Thus, in both cases, if the +
path exists, then δ exists, so B0L∞B is impossible.
Lemma 3.14 If Z = L the shape L2B2L∞, 1L1, ∞l is impossible.
Fig. 18 The Ck−3−4q
configurations in case Z = B
(a) q even (b) q odd
Fig. 19 The Ck−2−4q
configurations in case Z = L
(a) q even (b) q odd
Proof Consider the vicinity of Z when it was created in Ck−2−4q , as demonstrated in
Fig. 19(a) for q even and Fig. 19(b) for q odd. In both cases the square with corners
at P , Q, S, and T is a Ck−3−4q square, and hence in Ck−2−4q either P = Q = F or
S = T = F . In both cases, if T = F , then δ and μ are in the same path, and if Q = F ,
then ν and χ are in the same path, thus it never happens that χ and δ exist while ν
and μ are absent, and thus L2B2L∞, 1L1, ∞L is impossible.
Theorem 3.15 Each interior component of the Levy dragon is similar to one out of
the following 23 shapes:
5 LB shapes: L0B, L1B, L2B, L3B, and L∞B;
4 LBL shapes: L0B0L, L2B0L, L4B0L, and L2B2L;
3 LLB shapes: L∞, ∞L2B, L1, ∞L2B, and L0, 0L2B;
3 BLB shapes: B∞L∞B, B2L4B, and B0L2B;
3 LBLBL shapes: L2B2L0B4L, L2B2L4B4L, and L2B2L0B0L;
5 LBLLL shapes: L2B2L0, 0L0, 0L, L2B2L∞, 1L0, 0L, L2B2L∞, ∞L0,
0L, L2B2L∞, 1L∞, ∞L, and L2B2L∞, ∞L∞, ∞L.
Proof Using all the initial configurations in Theorem 3.7 with the tail tip types
determined in Proposition 3.11 and the corresponding tail lengths in Lemma 3.9 and
Lemma 3.10 and removing the impossible shapes in Lemmas 3.12, 3.13, and 3.14,
these 23 shapes are the only possible interior components shapes of the Levy
dragon.
These 23 possible shapes are demonstrated in Fig. 20.
By using a program that draws the F k(T ) triangles differentiating between
covered and not covered triangles one can see that all these 23 possible shapes appear.
One representative of each shape is initiated up to C35 within distances less than 3l14
from an L0B0L initiated in C14.
The method of this paper can be used to determine many other geometric
properties of the Levy dragon. For example, replacing existing segments by absent segments
and vice versa, one can determine that the possible shapes of the holes of the Levy
dragon are the same 23 shapes. By using a similar program one can see that there
are holes with all those shapes. Finding those hole shapes is a much easier task than
finding the interior components shapes since their configurations appear much earlier
in the Ck sequence. Some of the holes are thus big enough to be seen in Fig. 1. The
first one to appear in C7 is the configuration of an L2B2L∞, ∞L∞, ∞L hole seen
in Fig. 1 at the center of the Levy dragon.
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