Heighway Dragon Area
We consider the Heighway dragon as constructed using line segments starting with an initial segment of length 1. An IFS for this construction is given by the following.
where \(r = \frac{1}{{\sqrt 2 }}\). At each iteration, scale the current figure by \(r = \frac{1}{{\sqrt 2 }}\) and rotate by 45 around the origin. This is function \(f_1\). Then make a copy of this figure and rotate it by 90° around the point (1/2,1/2). This will give an equivalent result as the function \(f_2\).
Starting with any compact set, iterations of that set using the two functions in this IFS will always converge to the Heighway dragon. To help find the area of the Heighway dragon, we can start with a square whose diagonal is the initial segment of length 1 as shown below.
Play the video below to see how iterations of this square converge to the dragon.
Let \(S_0\) denote the initial square. The area of \(S_0\) is equal to 1/2. Now let \(S_n\) denote the sequence of iterations of \(S_0\), i.e. \({S_{n+1} = f_1(S_n)\cup f_2(S_n)}\).
The image \(S_1\) consists of 2 squares each with area 1/4. Therefore the area of \(S_1\) is \(2\cdot\frac{1}{4}=\frac{1}{2}\).
The image \(S_2\) consists of 4 squares each with area 1/8. Therefore the area of \(S_2\) is \(4\cdot\frac{1}{8}=\frac{1}{2}\).
The image \(S_3\) consists of 8 squares each with area 1/16. Therefore the area of \(S_3\) is \(8\cdot\frac{1}{16}=\frac{1}{2}\).
The image \(S_4\) consists of 16 squares each with area 1/32. Therefore the area of \(S_4\) is \(16\cdot\frac{1}{32}=\frac{1}{2}\).
The pattern should now be clear. Observe that \(S_{n+1}\) is obtained by first scaling \(S_n\) by \(1/\sqrt{2}\) and rotating by 45°. This means that each of the squares in \(S_n\) is scaled by \(1\sqrt{2}\), and therefore the area of each square in \(S_n\) is scaled by 1/2. This forms half of \(S_{n+1}\). The other half is obtained by rotating the scaled copy of \(S_n\) by 90°. So the iteration for \(S_{n+1}\) doubles the number of squares in \(S_n\) and scales their area by 1/2. Since \(S_0\) has 1 square of area 1/2, this means that \(S_n\) will have \(2^n\) squares, each with area 1/\(2^{n+1}\). Therefore the area of \(S_n\) is \(2^n\cdot\frac{1}{2^{n+1}}=\frac{1}{2}\) for all values of \(n\).
Now the sets \(S_n\) converge in the Hausdorff metric to the Heighway dragon as \(n\) goes to infinity. This observation by itself, however, does not guarantee that the area of the Heighway dragon is equal to 1/2 since Hausdorff convergence alone does not ensure convergence of area [Counterexample]. But in this case, four copies of each set \(S_n\) can tile the plane, and that is sufficient to conclude that the area of the dragon is the same the same as the area of each \(S_n\) [Area Details]. Therefore the area of the Heighway dragon is 1/2.
If the Heighway dragon is constructed using line segments starting with an initial segment of length \(b\), then the area of the dragon is scaled by a factor of \(b^2\). Therefore the area would be \(b^2/2\).
It is also possible to find the area of the Heighway dragon using a method for constructing its boundary.
