Agnes Scott College
Larry Riddle, Agnes Scott College

The Lakes of the Z2 Heighway Dragon


The Z2 Heighway dragon has symmetric spirals in the second and the fourth quadrants that spiral into the points (−1,1) in the second quadrant and (1,−1) in the fourth quadrant. The figure below (left) shows a close-up zoom of the spiral in the second quadrant (right figure). The figure on the left is on top of a grid of size 1/6 x 1/6.

heighwayZ2spiralzoomQ2  lakes

I call these white areas the "lakes" of the Z2 Heighway dragon. The lakes spiral through 45° angles, with each lake just touching the one before and the one after as indicated at the dots in the figure. The dot in the upper right where the first lake begins is located at the point (−2/3,1). To see why, consider the following figure which shows the fourth iteration of the Z2 Heighway dragon IFS starting with the line segment −1 ≤ x ≤ 1 as the initial set. In this figure the grid is of size 1/4 x 1/4.

iteration: ...

After the fourth iteration we have two scaled copies (shown in blue) in the second quadrant of the initial line segment. As we continue to iterate the IFS, these blue segments can generate scaled versions of the Z2 dragon, with one rotated by 90°. (Strictly speaking, the IFS must be changed to reflect working on the smaller blue versions; the scaling/rotations for each function will remain the same, but the translation vectors would change.) Click on the buttons underneath the figure to see the growth of these mini-dragons through the 12th iteration of the larger dragon, and the final "infinite" iteration (each of the blue segments is also a segment in the red dragon at each iteration.) As you do so, you will begin to see the development of the lakes in the second quadrant. In particular, the edges of the blue copies are getting closer.

Will the two blue mini-dragons touch? As shown in the section on the Z2 Heighway Dragon Size, the iterations starting with the initial symmetric line segment −1 ≤ x ≤ 1 will expand by 4/3 in the horizontal direction and by 5/3 in the vertical direction. For the blue dragons we are starting with a line segment that is 1/4 the length of the original segment. Thus the iterations that produce the horizontal blue mini-dragon will expand by (1/4)(4/3) = 1/3 in the horizontal direction. Since the right most endpoint of the initial horizontal blue segment in iteration 4 is at x = −1, the right most edge of the horizontal blue mini-dragon will be at x = −2/3. The other blue mini-dragon has been rotated by 90°. It will expand in the horizontal direction by (1/4)(5/3) = 5/12. Since the initial blue segment for this iteration is at x = −1/4, the left most edge of the vertical blue mini-dragon will be at x = −1/4 − 5/12 = −2/3. Hence the two blue mini-dragons will come together at x = −2/3.

grid: 1/4 x 1/4

For the horizontal blue mini-dragon, the touch point corresponds to the right most boundary point with the smallest y-coordinate. For the regular Z2 dragon, this is the point with y = −1 [see Z2 Heighway Dragon Size]. That is a distance of 1 below the center horizontal axis. For the blue mini-dragon, it would therefore be a distance 1/4 below the center horizontal axis at y = 1.25, hence at y = 1. For the vertical blue mini-dragon, the touch point corresponds to the top most boundary point of the dragon with the largest x-coordinate, when viewed in a horizontal position. For the regular Z2 dragon, this is the point with x = 1. That is a distance of 1 to the right of center vertical axis. For the vertical blue mini-dragon, it would therefore be a distance of 1/4 above the center horizontal axis at y = 3/4, hence at y = 1. Therefore the two blue mini-dragons just touch at y = 1. Combined with the earlier result about the x-coordinate, this means that they just touch at the point (−2/3, 1).

grid: 1/6 x 1/6

The other points that separate the lakes as they spiral by 45° clockwise towards (−1,1) are part of the orbit of the first point (−2/3,1) under iteration by the function h4 in the IFS for the Z2 Heighway dragon, where \[{h_4}({\bf{x}}) = \left[ {\begin{array}{*{20}{c}} { 1/2} & { 1/2} \\ {-1/2} & { 1/2} \\ \end{array}} \right]{\bf{x}} + \left[ {\begin{array}{*{20}{c}} -1 \\ 0 \\ \end{array}} \right]\] If we group these points in groups of 4, then we have the following forms for n ≥ 0.

4n      \(\left(-\dfrac{3 \cdot 4^n +(-1)^{n+1}}{3 \cdot 4^n},1 \right) \)
4n+1      \(\left(-\dfrac{6 \cdot 4^n +(-1)^{n+1}}{6 \cdot 4^n}, \dfrac{6 \cdot 4^n +(-1)^{n+1}}{6 \cdot 4^n} \right) \)
4n+2      \(\left( -1, \dfrac{6 \cdot 4^n +(-1)^{n+1}}{6 \cdot 4^n} \right) \)
4n+3      \(\left(-\dfrac{12 \cdot 4^n +(-1)^n}{12 \cdot 4^n}, \dfrac{12 \cdot 4^n +(-1)^{n+1}}{12 \cdot 4^n} \right) \)

Every fourth point will have a y-coordinate of 1. Every fourth point starting with the third will have an x-coordinate of −1.