Dragon curve

Recursive construction of the curve

Recursive construction of the curve

It can be written as a Lindenmayer system with

• angle 90°
• initial string FX
• string rewriting rules
  • X ↦ X+YF+
  • Y ↦ −FX−Y.

That can be described this way: starting from a base segment, replace each segment by 2 segments with a right angle and with a rotation of 45° alternatively to the right and to the left:

The first 5 iterations and the 9th

The Heighway dragon is also the limit set of the following iterated function system in the complex plane:

${\displaystyle f_{1}(z)={\frac {(1+i)z}{2}}}$

${\displaystyle f_{2}(z)=1-{\frac {(1-i)z}{2}}}$

with the initial set of points ${\displaystyle S_{0}=\{0,1\}}$.

Using pairs of real numbers instead, this is the same as the two functions consisting of

${\displaystyle f_{1}(x,y)={\frac {1}{\sqrt {2}}}{\begin{pmatrix}\cos 45^{\circ }&-\sin 45^{\circ }\\\sin 45^{\circ }&\cos 45^{\circ }\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}}$

${\displaystyle f_{2}(x,y)={\frac {1}{\sqrt {2}}}{\begin{pmatrix}\cos 135^{\circ }&-\sin 135^{\circ }\\\sin 135^{\circ }&\cos 135^{\circ }\end{pmatrix}}{\begin{pmatrix}x\\y\end{pmatrix}}+{\begin{pmatrix}1\\0\end{pmatrix}}}$

This representation is more commonly used in software such as Apophysis.

The path of the curve can be more easily seen with each turn curved

Tracing an iteration of the Heighway dragon curve from one end to the other, one encounters a series of 90-degree turns, some to the right and some to the left. For the first few iterations the sequence of right (R) and left (L) turns is as follows:

1st iteration: R

2nd iteration: R R L

3rd iteration: R R L R R L L

4th iteration: R R L R R L L R R R L L R L L.

This suggests a few different patterns. First, each iteration is formed by taking the previous and adding alternating rights and lefts in between each letter. Second, it also suggests the following pattern: each iteration is formed by taking the previous iteration, adding an R at the end, and then taking the original iteration again, flipping it retrograde, swapping each letter and adding the result after the R. Due to the self-similarity exhibited by the Heighway dragon, this effectively means that each successive iteration adds a copy of the last iteration rotated counter-clockwise to the fractal.

This pattern in turn suggests the following method of creating models of iterations of the Heighway dragon curve by folding a strip of paper. Take a strip of paper and fold it in half to the right. Fold it in half again to the right. If the strip was opened out now, unbending each fold to become a 90-degree turn, the turn sequence would be RRL, i.e. the second iteration of the Heighway dragon. Fold the strip in half again to the right, and the turn sequence of the unfolded strip is now RRLRRLL – the third iteration of the Heighway dragon. Continuing folding the strip in half to the right to create further iterations of the Heighway dragon (in practice, the strip becomes too thick to fold sharply after four or five iterations).

This unfolding method can be seen by calculating a number of iterations (for the animation at right, 13 iterations were used) of the curve using the "swapping" method described above, but controlling the angles for the right turns and negating prior angles.

This pattern also gives a method for determining the direction of the nth turn in the turn sequence of a Heighway dragon iteration. First, express n in the form k2^m, where k is an odd number. The direction of the nth turn is determined by k mod 4, i.e. the remainder left when k is divided by 4. If k mod 4 is 1, then the nth turn is R; if k mod 4 is 3, then the nth turn is L.

For example, to determine the direction of turn 76376:

76376 = 9547 × 8,

9547 = 2386×4 + 3,

so 9547 mod 4 = 3,

so turn 76376 is L.

There is a simple one-line non-recursive method of implementing the above k mod 4 method of finding the turn direction in code. Treating turn n as a binary number, calculate the following boolean value:

bool turn = (((n & −n) << 1) & n) != 0;

• "n & −n" leaves only one bit as a "1", the rightmost "1" in the binary expansion of n;
• "<< 1" shifts that bit to the left one position;
• "& n" leaves either that single bit (if k mod 4 = 3), or a zero (if k mod 4 = 1);
• so "bool turn = (((n & −n) << 1) & n) != 0" is TRUE if the nth turn is L, and is FALSE if the nth turn is R.

Another way of handling this is a reduction for the above algorithm. Using Gray code, starting from zero, determine the change to the next value. If the change is a 1, then turn left, and if it is 0, then turn right. Given a binary input B, the corresponding Gray code G is given by "G = B XOR (B >> 1)". Using G_i and G_i−1, turn equals "(not G_i) AND G_i−1".

• G = B ^ (B >> 1) gets gray code from binary.
• T = (~G0) & G1: if T is equal, then to 0 turn clockwise, else turn counterclockwise.

Another approach to generating this fractal can be created using a recursive function and these Turtle graphics functions drawLine(distance) and turn(angleInDegrees). The Python code to draw an (approximate) Heighway dragon curve would look like this.

def dragon_curve(order: int, length) -> None:
    """Draw a dragon curve."""
    turn(order * 45)
    dragon_curve_recursive(order, length, 1)

def dragon_curve_recursive(order: int, length, sign) -> None:
    if order == 0:
        drawLine(length)
    else:
        rootHalf = (1 / 2) ** (1 / 2)
        dragon_curve_recursive(order - 1, length * rootHalf, 1)
        turn(sign * -90)
        dragon_curve_recursive(order - 1, length * rootHalf, -1)

• In spite of its strange aspect, the Heighway dragon curve has simple dimensions. Note that the dimensions 1, and 1.5 are limits and not actual values.

• Its surface is also quite simple : If the initial segment equals 1, then its surface equals ${\displaystyle \textstyle {\frac {1}{2}}}$. This result comes from its paving properties.
• The curve never crosses itself.
• Many self-similarities can be seen in the Heighway dragon curve. The most obvious is the repetition of the same pattern tilted by 45° and with a reduction ratio of ${\displaystyle \textstyle {\sqrt {2}}}$.

• Its fractal dimension can be calculated : ${\displaystyle \textstyle {{\frac {\ln 2}{\ln {\sqrt {2}}}}=2}}$. That makes it a space-filling curve.
• Its boundary has an infinite length, since it increases by a similar factor every iteration.
• The fractal dimension of its boundary has been approximated numerically by Chang & Zhang .[1]).

In fact it can be found analytically:[2] ${\displaystyle \log _{2}\left({\frac {1+{\sqrt[{3}]{73-6{\sqrt {87}}}}+{\sqrt[{3}]{73+6{\sqrt {87}}}}}{3}}\right)\approx 1.523627086202492.}$ This is the root of the equation ${\displaystyle \textstyle {4^{x}(2^{x}-1)=4(2^{x}+1).}}$

The dragon curve can tile the plane in many ways.