Math::PlanePath::KochSnowflakes -- Koch snowflakes as concentric rings
use Math::PlanePath::KochSnowflakes; my $path = Math::PlanePath::KochSnowflakes->new; my ($x, $y) = $path->n_to_xy (123);
This path traces out concentric integer versions of the Koch snowflake at successively greater iteration levels.
48 6 / \ 50----49 47----46 5 \ / 54 51 45 42 4 / \ / \ / \ 56----55 53----52 44----43 41----40 3 \ / 57 12 39 2 / / \ \ 58----59 14----13 11----10 37----38 1 \ \ 3 / / 60 15 1----2 9 36 <- Y=0 / \ \ 62----61 4---- 5 7---- 8 35----34 -1 \ \ / / 63 6 33 -2 \ 16----17 19----20 28----29 31----32 -3 \ / \ / \ / 18 21 27 30 -4 / \ 22----23 25----26 -5 \ / 24 -6 ^ -9 -8 -7 -6 -5 -4 -3 -2 -1 X=0 1 2 3 4 5 6 7 8 9
The initial figure is the triangle N=1,2,3 then for the next level each straight side expands to 3x longer and a notch like N=4 through N=8,
*---* becomes *---* *---* \ / *
The angle is maintained in each replacement, for example the segment N=5 to N=6 becomes N=20 to N=24 at the next level.
The X,Y coordinates are arranged as integers on a square grid per "Triangular Lattice" in Math::PlanePath, except the Y coordinates of the innermost triangle which is
N=3 X=0, Y=+2/3 * / \ / \ / \ / o \ / \ N=1 *-----------* N=2 X=-1, Y=-1/3 X=1, Y=-1/3
These values are not integers, but they're consistent with the centring and scaling of the higher levels. If all-integer is desired then rounding gives Y=0 or Y=1 and doesn't overlap the subsequent points.
Counting the innermost triangle as level 0, each ring is
Nstart = 4^level length = 3*4^level many points
For example the outer ring shown above is level 2 starting N=4^2=16 and having length=3*4^2=48 points (through to N=63 inclusive).
The X range at a given level is the initial triangle baseline iterated out. Each level expands the sides by a factor of 3 so
Xlo = -(3^level) Xhi = +(3^level)
For example level 2 above runs from X=-9 to X=+9. The Y range is the points N=6 and N=12 iterated out. Ylo in level 0 since there's no downward notch on that innermost triangle.
Ylo = / -(2/3)*3^level if level >= 1 \ -1/3 if level == 0 Yhi = +(2/3)*3^level
Notice that for each level the extents grow by a factor of 3 but the notch introduced in each segment is not big enough to go past the corner positions. They can equal the extents horizontally, for example in level 1 N=14 is at X=-3 the same as the corner N=4, and on the right N=10 at X=+3 the same as N=8, but they don't go past.
The snowflake is an example of a fractal curve with ever finer structure. The code here can be used for that by going from N=Nstart to N=Nstart+length-1 and scaling X/3^level Y/3^level to give a 2-wide 1-high figure of desired fineness. See examples/koch-svg.pl in the Math-PlanePath sources for a complete program doing that as an SVG image file.
The area of the snowflake at a given level can be calculated from the area under the Koch curve per "Area" in Math::PlanePath::KochCurve which is the 3 sides, and the central triangle
* ^ Yhi / \ | height = 3^level / \ | / \ | *-------* v <-------> width = 3^level - (- 3^level) = 2*3^level Xlo Xhi triangle_area = width*height/2 = 9^level snowflake_area[level] = triangle_area[level] + 3*curve_area[level] = 9^level + 3*(9^level - 4^level)/5 = (8*9^level - 3*4^level) / 5
If the snowflake is conceived as a fractal of fixed initial triangle size and ever-smaller notches then the area is divided by that central triangle area 9^level,
unit_snowflake[level] = snowflake_area[level] / 9^level = (8 - 3*(4/9)^level) / 5 -> 8/5 as level -> infinity
Which is the well-known 8/5 * initial triangle area for the fractal snowflake.
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
$path = Math::PlanePath::KochSnowflakes->new ()
Create and return a new path object.
($n_lo, $n_hi) = $path->level_to_n_range($level)
Return per "Level Ranges" above,
(4**$level, 4**($level+1) - 1)
As noted in "Level Ranges" above, for a given level
-(3^level) <= X <= 3^level -(2/3)*(3^level) <= Y <= (2/3)*(3^level)
So the maximum X,Y in a rectangle gives
level = ceil(log3(max(abs(x1), abs(x2), abs(y1)*3/2, abs(y2)*3/2)))
and the last point in that level is
Nlevel = 4^(level+1) - 1
Using this as an N range is an over-estimate, but an easy calculation. It's not too difficult to trace down for an exact range
Entries in Sloane's Online Encyclopedia of Integer Sequences related to the Koch snowflake include the following. See "OEIS" in Math::PlanePath::KochCurve for entries related to a single Koch side.
http://oeis.org/A164346 (etc)
A164346 number of points in ring n, being 3*4^n A178789 number of acute angles in ring n, 4^n + 2 A002446 number of obtuse angles in ring n, 2*4^n - 2
The acute angles are those of +/-120 degrees and the obtuse ones +/-240 degrees. Eg. in the outer ring=2 shown above the acute angles are at N=18, 22, 24, 26, etc. The angles are all either acute or obtuse, so
A178789 + A002446 = A164346
Math::PlanePath, Math::PlanePath::KochCurve, Math::PlanePath::KochPeaks
Math::PlanePath::QuadricIslands
http://user42.tuxfamily.org/math-planepath/index.html
Copyright 2011, 2012, 2013, 2014, 2015 Kevin Ryde
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