Latest revision as of 20:51, 11 October 2019
This page is a solution to the task Dragon curve in the Rosetta Code, written in Fōrmulæ.
Description (from Rosetta Code)
Create and display a dragon curve fractal.
(You may either display the curve directly or write it to an image file.)
Algorithms
Here are some brief notes the algorithms used and how they might suit various languages.
 Recursively a right curling dragon is a right dragon followed by a left dragon, at 90degree angle. And a left dragon is a left followed by a right.
*R* expands to * *
\ /
R L
\ /
*
*
/ \
L R
/ \
*L* expands to * *
 The coroutines
dcl and dcr in various examples do this recursively to a desired expansion level.
 The curl direction right or left can be a parameter instead of two separate routines.
 Recursively, a curl direction can be eliminated by noting the dragon consists of two copies of itself drawn towards a central point at 45degrees.
*>* becomes * * Recursive copies drawn
\ / from the ends towards
\ / the centre.
v v
*
 This can be seen in the SVG example. This is best suited to offline drawing since the reversal in the second half means the drawing jumps backward and forward (in binary reflected Gray code order) which is not very good for a plotter or for drawing progressively on screen.
 Successive approximation repeatedly rewrites each straight line as two new segments at a right angle,
*
** becomes / \ bend to left
/ \ if N odd
* *
* *
** becomes \ / bend to right
\ / if N even
*
 Numbering from the start of the curve built so far, if the segment is at an odd position then the bend introduced is on the right side. If the segment is an even position then on the left. The process is then repeated on the new doubled list of segments. This constructs a full set of line segments before any drawing.
 The effect of the splitting is a kind of bottomup version of the recursions. See the Asymptote example for code doing this.
 Iteratively the curve always turns 90degrees left or right at each point. The direction of the turn is given by the bit above the lowest 1bit of n. Some bittwiddling can extract that efficiently.
n = 1010110000
^
bit above lowest 1bit, turn left or right as 0 or 1
LowMask = n BITXOR (n1) # eg. giving 0000011111
AboveMask = LowMask + 1 # eg. giving 0000100000
BitAboveLowestOne = n BITAND AboveMask
 The first turn is at n=1, so reckon the curve starting at the origin as n=0 then a straight line segment to position n=1 and turn there.
 If you prefer to reckon the first turn as n=0 then take the bit above the lowest 0bit instead. This works because "...10000" minus 1 is "...01111" so the lowest 0 in n1 is where the lowest 1 in n is.
 Going by turns suits turtle graphics such as Logo or a plotter drawing with a pen and current direction.
 If a language doesn't maintain a "current direction" for drawing then you can always keep that separately and apply turns by bitabovelowest1.
 Absolute direction to move at point n can be calculated by the number of bittransitions in n.
n = 11 00 1111 0 1
^ ^ ^ ^ 4 places where change bit value
so direction=4*90degrees=East
 This can be calculated by counting the number of 1 bits in "n XOR (n RIGHTSHIFT 1)" since such a shift and xor leaves a single 1 bit at each position where two adjacent bits differ.
 Absolute X,Y coordinates of a point n can be calculated in complex numbers by some powers (i+1)^k and add/subtract/rotate. This is done in the gnuplot code. This might suit things similar to Gnuplot which want to calculate each point independently.
 Predicate test for whether a given X,Y point or segment is on the curve can be done. This might suit linebyline output rather than building an entire image before printing. See M4 for an example of this.
 A predicate works by dividing out complex number i+1 until reaching the origin, so it takes roughly a bit at a time from X and Y is thus quite efficient. Why it works is slightly subtle but the calculation is not difficult. (Check segment by applying an offset to move X,Y to an "even" position before dividing i+1. Check vertex by whether the segment either East or West is on the curve.)
 The number of steps in the predicate corresponds to doublings of the curve, so stopping the check at say 8 steps can limit the curve drawn to 2^8=256 points. The offsets arising in the predicate are bits of n the segment number, so can note those bits to calculate n and limit to an arbitrary desired length or subsection.
 As a Lindenmayer system of expansions. The simplest is two symbols F and S both straight lines, as used by the PGF code.
Axiom F, angle 90 degrees
F > F+S
S > FS
This always has F at even positions and S at odd. Eg. after 3 levels F_S_F_S_F_S_F_S . The +/ turns in between bend to the left or right the same as the "successive approximation" method above. Read more at for instance Joel Castellanos' Lsystem page.
Variations are possible if you have only a single symbol for line draw, for example the Icon and Unicon and Xfractint code. The angles can also be broken into 45degree parts to keep the expansion in a single direction rather than the endpoint rotating around.
The string rewrites can be done recursively without building the whole string, just follow its instructions at the target level. See for example C by IFS Drawing code. The effect is the same as "recursive with parameter" above but can draw other curves defined by Lsystems.

Solution
Case 1. Showing a dragon curve
Case 2. Showing dragon curves, orders 1 to 12