Agnes Scott College
Larry Riddle, Agnes Scott College

Twindragon Boundary

Recall that the twindragon consists of two copies of the Heighway dragon drawn side by side with the tail of one corresponding to the head of the other. The boundary of the twindragon is the union of 4 self-similar pieces. The first two are part of the boundary from one Heighway dragon and the last two are part of the boundary from the other Heighway dragon.

upperleft upperright lowerright lowerleft

 
 

Put them together by clicking on the buttons to the left in sequence

Image4-0

To construct the boundary, start with the dashed figure (Level 0) in the following image, then replace each pair of segments with the four patterns shown to get Level 1.

TwindragonBoundaryDesign

Notice that all four motifs are the same except for scale and how they are rotated. In addition, each motif consists of 3 self-similar parts. For example, the red motif consists of the following three parts.

redmotif

Now iterate this construction by replacing each of those three parts by copies of that color's motif. Here is Level 2. The dotted line is Level 1.

TwindragonBoundaryLevel2

Repeat ad infinitum to construct the entire boundary around the twindragon.


Iterated
Function
System

To generate the basic motif for the four parts of the boundary as shown above in Level 1 requires three functions for each motif. Each motif will involve the same scaling/rotation matrices. The only difference will be the translations.

Red Boundary:

\({R_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & { - 1/2} \\ {1/2} & {1/2} \\ \end{array}} \right]{\bf{x}} \)
 
   scale by \(\frac{1}{\sqrt 2}\), rotate by 45°
 
\({R_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { - 1/4} & { - 1/4} \\ {1/4} & { -1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/2 \\ 1/2 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{2 \sqrt 2}\), rotate by 135°
 
\({R_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 1/4} & { 1/4} \\ {-1/4} & { 1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/2 \\ 1/2 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{2 \sqrt 2}\), rotate by −45°
 

Green Boundary:

\({G_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & { - 1/2} \\ {1/2} & {1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/2 \\ -1/2 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{\sqrt 2}\), rotate by 45°
 
\({G_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { - 1/4} & { - 1/4} \\ {1/4} & { - 1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1 \\ 0 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{2 \sqrt 2}\), rotate by 135°
 
\({G_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 1/4} & {1/4} \\ {-1/4} & { 1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/2 \\ 1/2 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{2 \sqrt 2}\), rotate by −45°
 

Orange Boundary:

\({O_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & { - 1/2} \\ {1/2} & {1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/2 \\ -1/2 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{\sqrt 2}\), rotate by 45°
 
\({O_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { - 1/4} & { - 1/4} \\ {1/4} & { - 1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 3/4 \\ -3/4 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{2 \sqrt 2}\), rotate by 135°
 
\({O_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 1/4} & {1/4} \\ {-1/4} & { 1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/4 \\ -1/4 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{2 \sqrt 2}\), rotate by −45°
 

Blue Boundary:

\({B_1}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} {1/2} & { - 1/2} \\ {1/2} & {1/2} \\ \end{array}} \right]{\bf{x}} \)
 
   scale by \(\frac{1}{\sqrt 2}\), rotate by 45°
 
\({B_2}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { - 1/4} & { - 1/4} \\ {1/4} & { - 1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/4 \\ -1/4 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{2 \sqrt 2}\), rotate by 135°
 
\({B_3}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 1/4} & {1/4} \\ {-1/4} & { 1/4} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} 1/4 \\ -1/4 \\ \end{array}} \right]\)
 
   scale by \(\frac{1}{2 \sqrt 2}\), rotate by −45°
 

Similarity
Dimension

The boundary of the twindragon consists of 4 parts, each of which is self-similar with 2 non-overlapping pieces scaled by \(\frac{1}{2 \sqrt 2}\) and one piece scaled by \(\frac{1}{ \sqrt 2}\). According to Moran's equation, the similarity dimension d must satisfy \[2{\left( {\frac{1}{{2\sqrt 2 }}} \right)^d} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^d} = 1\] By letting \(x = {\left( {\frac{1}{{\sqrt 2 }}} \right)^d}\) we can rewrite this equation as \[2{x^3} + x - 1 = 0\]

which has the solution

\[x = \frac{1}{6}{\left( {54 + 6\sqrt {87} } \right)^{1/3}} - \frac{1}{{{{\left( {54 + 6\sqrt {87} } \right)}^{1/3}}}}\]

The dimension of the boundary is therefore

\[d = - \frac{{\ln x}}{{\ln \sqrt 2 }} = - \frac{{\ln \left( {\frac{1}{6}{{\left( {54 + 6\sqrt {87} } \right)}^{1/3}} - \frac{1}{{{{\left( {54 + 6\sqrt {87} } \right)}^{1/3}}}}} \right)}}{{\ln \sqrt 2 }} \approx 1.523627\]

Special
Properties

The construction of the boundary as described above gives a way to calculate the area of the twindragon directly. Suppose we start with a figure like Level 0 but for which the length of the bold segment (representing the initial segment for the twindragon) is of length b.

AreaLevel0

Let a be the length of one of the short sides of Level 0 for the construction of the boundary. Then b2 = a2 + (3a)2= 10a2. The area inside the Level 0 boundary is 7a2.

Now consider what happens with the area inside the Level 1 boundary as shown below.

AreaLevel1

The smaller red, green, blue, and orange triangles contain the same area inside the boundary and outside the boundary of Level 0. Thus in the iteration from Level 0 to Level 1, we only need to add the additional area from the light shaded regions but subtract the area from the dark shaded regions. This additional area is

\[2 \cdot \left( {\frac{1}{2} \cdot 2a \cdot a} \right) - 2 \cdot \left( {\frac{1}{2} \cdot a \cdot \frac{a}{2}} \right) = 2{a^2} - \frac{1}{2}{a^2} = \frac{3}{2}{a^2}\]

The same thing will happen in the iteration from Level 1 to Level 2, except the new shaded regions would have a side that has been scaled by \(1/\sqrt 2\). Continuing the iteration and adding the additional area at each new level gives the total area inside the boundary of the twindragon to be

\[\begin{array}{l} 7{a^2} + \frac{3}{2}{a^2} + \frac{3}{2}{\left( {\frac{a}{{\sqrt 2 }}} \right)^2} + \frac{3}{2}{\left( {\frac{a}{{{{\sqrt 2 }^2}}}} \right)^2} + \frac{3}{2}{\left( {\frac{a}{{{{\sqrt 2 }^3}}}} \right)^2} + \ldots \\ \\ = 7{a^2} + \frac{3}{2}{a^2}\left( {1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots } \right) \\ \\ = 7{a^2} + \frac{3}{2}{a^2}\left( {\frac{1}{{1 - \frac{1}{2}}}} \right) = 10{a^2} = b^2 \\ \end{array}\]

 


References

  1. Angle Chang and Tianrong Zhang. "The Fractal Geometry of the Boundary of Dragon Curves," J. Recreational Mathematics, Vol. 30, No. 1 (1999-2000), 9-22. [Available at http://poignance.coiraweb.com/math/Fractals/Dragon/Bound.html]
  2. Edgar, Gerald A. Measure, Topology, and Fractal Geometry, Springer-Verlag, 1990.