# Babylonian spiral

Babylonian spiral is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

The Babylonian spiral is a sequence of points in the plane that are created so as to continuously minimally increase in vector length and minimally bend in vector direction, while always moving from point to point on strictly integral coordinates. Of the two criteria of length and angle, the length has priority.

Examples

P(1) and P(2) are defined to be at (x = 0, y = 0) and (x = 0, y = 1). The first vector is from P(1) to P(2). It is vertical and of length 1. Note that the square of that length is 1.

Next in sequence is the vector from P(2) to P(3). This should be the smallest distance to a point with integral (x, y) which is longer than the last vector (that is, 1). It should also bend clockwise more than zero radians, but otherwise to the least degree.

The point chosen for P(3) that fits criteria is (x = 1, y = 2). Note the length of the vector from P(2) to P(3) is √2, which squared is 2. The lengths of the vectors thus determined can be given by a sorted list of possible sums of two integer squares, including 0 as a square.

Find and show the first 40 (x, y) coordinates of the Babylonian spiral.

Show in your program how to calculate and plot the first 10000 points in the sequence. Your result should look similar to the page at https://oeis.org/plot2a?name1=A297346&name2=A297347&tform1=untransformed&tform2=untransformed&shift=0&radiop1=xy&drawlines=true".

## J

It's convenient, here, to use complex numbers to represent the coordinate pairs.

Implementation:

<lang J> require'stats' bspir=: {{

``` r=. 0
e=. 2>.<.+:%:y
for_qd.
(y-1){.(</.~ *:@|) (/:|) (#~ (=<.)@:*:@:|) j./"1 (2 comb e),,.~1+i.e
do.
d=. ~.(,+)(,-)(,j.);qd
ar=. 12 o. -~/_2{.r
ad=. (- 2p1 * >:&ar) 12 o. d
-r=. r, ({:r)+d{~ (i. >./) ad
end.
```

}} </lang>

<lang J> 4 10\$bspir 40

```   0   0j1   1j2   3j2   5j1   7j_1   7j_4   6j_7  4j_10  0j_10
```

_4j_9 _7j_6 _9j_2 _9j3 _8j8 _6j13 _2j17 3j20 9j20 15j19 21j17 26j13 29j7 29 28j_7 24j_13 17j_15 10j_12 4j_7 4j1

``` 5j9  7j17 13j23 21j26 28j21  32j13   32j4  31j_5 29j_14 24j_22
```

</lang>

Also:

<lang J>

```  require'plot'
plot bspir 40
plot bspir 1e4
```

</lang>

There's an online example of these two cases here (follow the link, select the right pane, then hit control-R or select Edit Run > All Lines from the menu, this might take a couple seconds, depending on your machine, and this url may need to be updated in the future).

This could be made significantly faster with a better estimator (or with a better implementation of J compiled to javascript).

## Julia

Translation of: Python

<lang ruby>""" Rosetta Code task rosettacode.org/wiki/Babylonian_spiral """

using GLMakie

const squarecache = Int[]

""" Get the points for a Babylonian spiral of `nsteps` steps. Origin is at (0, 0) with first step one unit in the positive direction along the vertical (y) axis. See also: oeis.org/A256111, oeis.org/A297346, oeis.org/A297347 """ function babylonianspiral(nsteps)

```   if length(squarecache) <= nsteps
append!(squarecache, map(x -> x * x, length(squarecache):nsteps))
end
# first line segment is 1 unit in vertical direction, with y vertical, x horizontal
xydeltas, δ² = [(0, 0), (0, 1)], 1
for _ in 1:nsteps-2
x, y = xydeltas[end]
θ = atan(y, x)
candidates = Tuple{Int64,Int64}[]
while isempty(candidates)
δ² += 1
for (k, a) in enumerate(squarecache)
a > δ² ÷ 2 && break
for j in isqrt(δ²)+1:-1:1
b = squarecache[j+1]
a + b < δ² && break
if a + b == δ²
i = k - 1
push!(
candidates,
(i, j),
(-i, j),
(i, -j),
(-i, -j),
(j, i),
(-j, i),
(j, -i),
(-j, -i),
)
end
end
end
end
_, idx = findmin(p -> mod1(θ - atan(p[2], p[1]), 2π), candidates)
push!(xydeltas, candidates[idx])
end
return accumulate((a, b) -> (a[1] + b[1], a[2] + b[2]), xydeltas)
```

end

println("The first 40 Babylonian spiral points are:") for (i, p) in enumerate(babylonianspiral(40))

```   print(rpad(p, 10), i % 10 == 0 ? "\n" : "")
```

end

lines(babylonianspiral(10_000), linewidth = 1)

</lang>
Output:
```The first 40 Babylonian spiral points are:
(0, 0)    (0, 1)    (1, 2)    (3, 2)    (5, 1)    (7, -1)   (7, -4)   (6, -7)   (4, -10)  (0, -10)
(-4, -9)  (-7, -6)  (-9, -2)  (-9, 3)   (-8, 8)   (-6, 13)  (-2, 17)  (3, 20)   (9, 20)   (15, 19)
(21, 17)  (26, 13)  (29, 7)   (29, 0)   (28, -7)  (24, -13) (17, -15) (10, -12) (4, -7)   (4, 1)
(5, 9)    (7, 17)   (13, 23)  (21, 26)  (28, 21)  (32, 13)  (32, 4)   (31, -5)  (29, -14) (24, -22)
```

## Perl

Translation of: Raku

<lang perl>use strict; use warnings; use feature <say state>; use constant TAU => 2 * 2 * atan2(1, 0);

sub B_spiral {

```   my(\$nsteps) = @_;
my @squares = map \$_**2, 0..\$nsteps+1;
my @dxys = ([0, 0], [0, 1]);
my \$dsq  = 1;
```
```   for (1 .. \$nsteps-2) {
my (\$x,\$y) = @{\$dxys[-1]};
our \$theta = atan2 \$y, \$x;
my @candidates;
```
```       until (@candidates) {
\$dsq++;
for my \$i (0..\$#squares) {
my \$a = \$squares[\$i];
next if \$a > \$dsq/2;
for my \$j ( reverse 0 .. 1 + int sqrt \$dsq ) {
my \$b = \$squares[\$j];
next if (\$a + \$b) < \$dsq;
if (\$dsq == \$a + \$b) {
push @candidates, ( [\$i, \$j], [-\$i, \$j], [\$i, -\$j], [-\$i, -\$j],
[\$j, \$i], [-\$j, \$i], [\$j, -\$i], [-\$j, -\$i] );
}
}
}
}
```
```       sub comparer {
my \$i = (\$theta - atan2 \$_[1], \$_[0]);
my \$z = \$i - int(\$i / TAU) * TAU;
\$z < 0 ? TAU + \$z : \$z;
}
```
```       push @dxys, (sort { comparer(@\$b) < comparer(@\$a) } @candidates)[0];
}
```
```   map { state(\$x,\$y); \$x += \$\$_[0]; \$y += \$\$_[1]; [\$x,\$y] } @dxys;
```

}

my @points = map { sprintf "(%3d,%4d)", @\$_ } B_spiral(40); say "The first 40 Babylonian spiral points are:\n" .

```   join(' ', @points) =~ s/.{1,88}\K/\n/gr;</lang>
```
Output:
```The first 40 Babylonian spiral points are:
(  0,   0) (  0,   1) (  1,   2) (  3,   2) (  5,   1) (  7,  -1) (  7,  -4) (  6,  -7)
(  4, -10) (  0, -10) ( -4,  -9) ( -7,  -6) ( -9,  -2) ( -9,   3) ( -8,   8) ( -6,  13)
( -2,  17) (  3,  20) (  9,  20) ( 15,  19) ( 21,  17) ( 26,  13) ( 29,   7) ( 29,   0)
( 28,  -7) ( 24, -13) ( 17, -15) ( 10, -12) (  4,  -7) (  4,   1) (  5,   9) (  7,  17)
( 13,  23) ( 21,  26) ( 28,  21) ( 32,  13) ( 32,   4) ( 31,  -5) ( 29, -14) ( 24, -22)```

## Phix

Library: Phix/pGUI
Library: Phix/online

You can run this online here. Use left/right arrow keys to show less/more edges.

```--
-- demo/rosetta/Babylonian_spiral.exw
-- ==================================
--
with javascript_semantics
function next_step(atom last_distance)
// Find "the next longest vector with integral endpoints on a Cartesian grid"
integer next_distance = 100*last_distance, // Set high so we can minimize
nmax = floor(sqrt(last_distance)) + 2
// ^ The farthest we could possibly go in one direction
sequence next_steps = {}
for n=0 to nmax do
integer n2 = n*n,
mmin = floor(sqrt(max(0,last_distance-n2)))
for m=mmin to nmax do
integer test_distance = n2 + m*m
if test_distance>last_distance then
if test_distance>next_distance then exit end if
if test_distance<next_distance then
next_distance = test_distance
next_steps = {}
end if
next_steps &= {{m,n}}
if m!=n then
next_steps &= {{n,m}}
end if
end if
end for
end for
return {next_steps, next_distance}
end function

sequence x = {0,0},         -- first two points
y = {0,1}          --  taken as given
integer distance = 1,
px = 0, py = 1,     -- position
pdx = 0, pdy = 1    -- previous delta

procedure make_spiral(integer npoints) // Make a Babylonian spiral of npoints.
sequence deltas
atom t4 = time()+0.4
for n=length(x)+1 to npoints do
{deltas,distance} = next_step(distance)
atom max_dot = 0, ldx = pdx, ldy = pdy
for delta in deltas do
integer {tx,ty} = delta
for d in {{tx,ty},{-tx,ty},{tx,-ty},{-tx,-ty}} do
integer {dx,dy} = d
if ldx*dy-ldy*dx<0 then
atom dot = ldx*dx+ldy*dy
if dot>max_dot then
max_dot = dot
{pdx,pdy} = {dx,dy}
end if
end if
end for
end for
px += pdx
py += pdy
x &= px
y &= py
if time()>t4 then exit end if
end for
end procedure

make_spiral(40)
printf(1,"The first 40 Babylonian spiral points are:\n%s\n",
{join_by(columnize({x,y}),1,10," ",fmt:="(%3d,%3d)")})

include pGUI.e
include IupGraph.e
Ihandle dlg, graph, timer
constant mt = {{5,6,1}, -- {number, minmax, tick}
{10,12,2},  -- add more/remove steps as desired
{20,20,4},
{40,32,8},
{60,70,10},
{100,220,40},
{200,350,50},
{400,700,100},
{800,1200,200},
{1000,2000,400},
{10000,12000,3000},
{100000,150000,50000},
{150000,150000,50000},
{200000,400000,100000}}
-- perfectly doable, but test yer patience:
--             {250000,400000,100000},
--             {500000,1200000,400000}}
integer mdx = 4, -- index to mt
max_mdx = 4

function get_data(Ihandle /*graph*/)
integer {n,m,t} = mt[mdx]
string title = sprintf("Babylonian spiral (%,d)", {n})
if length(x)<mt[\$][1] then
title &= sprintf(" (calculating %,d/%,d)",{length(x),mt[\$][1]})
end if
IupSetStrAttribute(graph, "GTITLE", title)
sequence xn = x[1..n],
yn = y[1..n]
IupSetAttributes(graph,"XMIN=%d,XMAX=%d,XTICK=%d",{-m,m,t})
IupSetAttributes(graph,"YMIN=%d,YMAX=%d,YTICK=%d",{-m,m,t})
return {{xn,yn,CD_RED}}
end function

function key_cb(Ihandle /*ih*/, atom c)
if c=K_ESC then return IUP_CLOSE end if -- (standard practice for me)
if c=K_F5 then return IUP_DEFAULT end if -- (let browser reload work)
if c=K_LEFT then mdx = max(mdx-1,1) end if
if c=K_RIGHT then mdx = min(mdx+1,max_mdx) end if
IupUpdate(graph)
return IUP_CONTINUE
end function

function timer_cb(Ihandln /*ih*/)
if max_mdx=length(mt) or not IupGetInt(dlg,"VISIBLE") then
IupSetInt(timer,"RUN",false)
else
integer next_target = mt[max_mdx+1][1]
make_spiral(next_target)
if length(x)=next_target then
max_mdx += 1
if mdx=max_mdx-1 then
mdx = max_mdx
end if
end if
end if
IupRedraw(graph)
return IUP_IGNORE
end function

IupOpen()
graph = IupGraph(get_data,"RASTERSIZE=640x480,XMARGIN=35")
dlg = IupDialog(graph,`TITLE="Babylonian spiral",MINSIZE=270x430`)
IupSetInt(graph,"GRIDCOLOR",CD_LIGHT_GREY)
IupShow(dlg)
IupSetCallback(dlg, "K_ANY", Icallback("key_cb"))
IupSetAttribute(graph,"RASTERSIZE",NULL)
timer = IupTimer(Icallback("timer_cb"), 30)
if platform()!=JS then
IupMainLoop()
IupClose()
end if
```
Output:
```The first 40 Babylonian spiral points are:
(  0,  0) (  0,  1) (  1,  2) (  3,  2) (  5,  1) (  7, -1) (  7, -4) (  6, -7) (  4,-10) (  0,-10)
( -4, -9) ( -7, -6) ( -9, -2) ( -9,  3) ( -8,  8) ( -6, 13) ( -2, 17) (  3, 20) (  9, 20) ( 15, 19)
( 21, 17) ( 26, 13) ( 29,  7) ( 29,  0) ( 28, -7) ( 24,-13) ( 17,-15) ( 10,-12) (  4, -7) (  4,  1)
(  5,  9) (  7, 17) ( 13, 23) ( 21, 26) ( 28, 21) ( 32, 13) ( 32,  4) ( 31, -5) ( 29,-14) ( 24,-22)
```

## Python

<lang python>""" Rosetta Code task rosettacode.org/wiki/Babylonian_spiral """

from itertools import accumulate from math import isqrt, atan2, tau from matplotlib.pyplot import axis, plot, show

square_cache = []

def babylonian_spiral(nsteps):

```   """
Get the points for each step along a Babylonia spiral of `nsteps` steps.
Origin is at (0, 0) with first step one unit in the positive direction along
the vertical (y) axis. The other points are selected to have integer x and y
coordinates, progressively concatenating the next longest vector with integer
x and y coordinates on the grid. The direction change of the  new vector is
chosen to be nonzero and clockwise in a direction that minimizes the change
in direction from the previous vector.

"""
if len(square_cache) <= nsteps:
square_cache.extend([x * x for x in range(len(square_cache), nsteps)])
xydeltas = [(0, 0), (0, 1)]
δsquared = 1
for _ in range(nsteps - 2):
x, y = xydeltas[-1]
θ = atan2(y, x)
candidates = []
while not candidates:
δsquared += 1
for i, a in enumerate(square_cache):
if a > δsquared // 2:
break
for j in range(isqrt(δsquared) + 1, 0, -1):
b = square_cache[j]
if a + b < δsquared:
break
if a + b == δsquared:
candidates.extend([(i, j), (-i, j), (i, -j), (-i, -j), (j, i), (-j, i),
(j, -i), (-j, -i)])
```
```       p = min(candidates, key=lambda d: (θ - atan2(d[1], d[0])) % tau)
xydeltas.append(p)
```
```   return list(accumulate(xydeltas, lambda a, b: (a[0] + b[0], a[1] + b[1])))
```

points10000 = babylonian_spiral(10000) print("The first 40 Babylonian spiral points are:") for i, p in enumerate(points10000[:40]):

```    print(str(p).ljust(10), end = '\n' if (i + 1) % 10 == 0 else )
```

plot(*zip(*points10000)) axis('scaled') show()

</lang>
Output:
```The first 40 Babylonian spiral points are:
(0, 0)    (0, 1)    (1, 2)    (3, 2)    (5, 1)    (7, -1)   (7, -4)   (6, -7)   (4, -10)  (0, -10)
(-4, -9)  (-7, -6)  (-9, -2)  (-9, 3)   (-8, 8)   (-6, 13)  (-2, 17)  (3, 20)   (9, 20)   (15, 19)
(21, 17)  (26, 13)  (29, 7)   (29, 0)   (28, -7)  (24, -13) (17, -15) (10, -12) (4, -7)   (4, 1)
(5, 9)    (7, 17)   (13, 23)  (21, 26)  (28, 21)  (32, 13)  (32, 4)   (31, -5)  (29, -14) (24, -22)
```

### With priority queue

Use a priority queue to generate all x, y combinations. The advantage is that we don't need to do any real math, and it is much faster. <lang python>from itertools import islice, count import matplotlib.pyplot as plt import heapq

def twosquares():

```   q, n = [], 1
```
```   while True:
while not q or n*n <= q[0][0]:
heapq.heappush(q, (n*n, n, 0))
n += 1
```
```       s, xy = q[0][0], []
```
```       while q and q[0][0] == s: # pop all vectors with same length
s, a, b = heapq.heappop(q)
xy.append((a, b))
if a > b:
heapq.heappush(q, (a*a + (b+1)*(b+1), a, b + 1))
```
```       yield tuple(xy)
```

def gen_dirs():

```   d = (0, 1)
for v in twosquares():
# include symmetric vectors
v += tuple((b, a) for a, b in v if a != b)
v += tuple((a, -b) for a, b in v if b)
v += tuple((-a, b) for a, b in v if a)
```
```       # filter using dot and cross product
d = max((a*d[0] + b*d[1], a, b) for a, b in v if a*d[1] - b*d[0] >= 0)[1:]
yield d
```

def positions():

```   p = (0, 0)
for d in gen_dirs():
yield p
p = (p[0] + d[0], p[1] + d[1])
```

print(list(islice(positions(), 40)))

plt.plot(*zip(*list(islice(positions(), 100000))), lw=0.4) plt.gca().set_aspect(1) plt.show()</lang>

## Raku

Translation of: Wren

### Translation

<lang perl6>sub babylonianSpiral (\nsteps) {

```   my @squareCache = (0..nsteps).hyper.map: *²;
my @dxys = [[0, 0], [0, 1]];
my \$dsq  = 1;
```
```   for ^(nsteps-2) {
my \Θ = atan2 |@dxys[*-1][1,0];
my @candidates;
```
```       until @candidates.elems {
\$dsq++;
```

for @squareCache.kv -> \i, \a { last if a > \$dsq / 2; for reverse (0 .. \$dsq.sqrt.ceiling) -> \j {

```                   last if ( a + my \b = @squareCache[j] ) < \$dsq;
if ((a + b) == \$dsq) {
@candidates.append: [ [i, j], [-i, j], [i, -j], [-i, -j],
[j, i], [-j, i], [j, -i], [-j, -i] ]
}
}
}
}
@dxys.push: @candidates.min: { ( Θ - atan2 |.[1,0] ) % τ };
}
```
```   [\»+«] @dxys
```

}

say "The first \$_ Babylonian spiral points are:\n", (babylonianSpiral(\$_).map: { sprintf '(%3d,%4d)', @\$_ }).batch(10).join("\n") given 40;

1. Stretch

use SVG;

'babylonean-spiral-raku.svg'.IO.spurt: SVG.serialize(

```   svg => [
:width<100%>, :height<100%>,
:rect[:width<100%>, :height<100%>, :style<fill:white;>],
:polyline[ :points(flat babylonianSpiral(10000)),
:style("stroke:red; stroke-width:6; fill:white;"),
:transform("scale (.05, -.05) translate (1000,-10000)")
],
],
```

);</lang>

Output:
```The first 40 Babylonian spiral points are:
(  0,   0) (  0,   1) (  1,   2) (  3,   2) (  5,   1) (  7,  -1) (  7,  -4) (  6,  -7) (  4, -10) (  0, -10)
( -4,  -9) ( -7,  -6) ( -9,  -2) ( -9,   3) ( -8,   8) ( -6,  13) ( -2,  17) (  3,  20) (  9,  20) ( 15,  19)
( 21,  17) ( 26,  13) ( 29,   7) ( 29,   0) ( 28,  -7) ( 24, -13) ( 17, -15) ( 10, -12) (  4,  -7) (  4,   1)
(  5,   9) (  7,  17) ( 13,  23) ( 21,  26) ( 28,  21) ( 32,  13) ( 32,   4) ( 31,  -5) ( 29, -14) ( 24, -22)```
Stretch goal:
(offsite SVG image - 96 Kb) - babylonean-spiral-raku.svg

### Independent implementation

Exact same output; about one tenth the execution time.

<lang perl6>my @next = { :x(1), :y(1), :2hyp },;

sub next-interval (Int \$int) {

```    @next.append: (0..\$int).map: { %( :x(\$int), :y(\$_), :hyp(\$int² + .²) ) };
@next = |@next.sort: *.<hyp>;
```

}

my @spiral = [\»+«] lazy gather {

```   my \$interval = 1;
take [0,0];
take my @tail = 0,1;
loop {
my \Θ = atan2 |@tail[1,0];
my @this = @next.shift;
@this.push: @next.shift while @next and @next[0]<hyp> == @this[0]<hyp>;
my @candidates = @this.map: {
my (\i, \j) = .<x y>;
next-interval(++\$interval) if \$interval == i;
|((i,j),(-i,j),(i,-j),(-i,-j),(j,i),(-j,i),(j,-i),(-j,-i))
}
take @tail = |@candidates.min: { ( Θ - atan2 |.[1,0] ) % τ };
}
```

}

say "The first \$_ Babylonian spiral points are:\n", @spiral[^\$_].map({ sprintf '(%3d,%4d)', |\$_ }).batch(10).join: "\n" given 40;

1. Stretch

use SVG;

'babylonean-spiral-raku.svg'.IO.spurt: SVG.serialize(

```   svg => [
:width<100%>, :height<100%>,
:rect[:width<100%>, :height<100%>, :style<fill:white;>],
:polyline[ :points(flat @spiral[^10000]),
:style("stroke:red; stroke-width:6; fill:white;"),
:transform("scale (.05, -.05) translate (1000,-10000)")
],
],
```

);</lang>

Same output:

## Wren

Translation of: Python
Library: DOME
Library: Wren-trait
Library: Wren-seq
Library: Wren-fmt
Library: Wren-plot

Generates an image similar to the OEIS one. <lang ecmascript>import "dome" for Window import "graphics" for Canvas, Color import "./trait" for Indexed, Stepped import "./seq" for Lst import "./fmt" for Fmt import "./plot" for Axes

// Python modulo operator (not same as Wren's) var pmod = Fn.new { |x, y| ((x % y) + y) % y }

var squareCache = []

"""

```   Get the points for each step along a Babylonian spiral of `nsteps` steps.
Origin is at (0, 0) with first step one unit in the positive direction along
the vertical (y) axis. The other points are selected to have integer x and y
coordinates, progressively concatenating the next longest vector with integer
x and y coordinates on the grid. The direction change of the  new vector is
chosen to be nonzero and clockwise in a direction that minimizes the change
in direction from the previous vector.
```
```   See also: oeis.org/A256111, oeis.org/A297346, oeis.org/A297347
```

""" var babylonianSpiral = Fn.new { |nsteps|

```   for (x in 0...nsteps) squareCache.add(x*x)
var dxys = [[0, 0], [0, 1]]
var dsq = 1
for (i in 0...nsteps) {
var x = dxys[-1][0]
var y = dxys[-1][1]
var theta = y.atan(x)
var candidates = []
while (candidates.isEmpty) {
dsq = dsq + 1
for (se in Indexed.new(squareCache)) {
var i = se.index
var a = se.value
if (a > (dsq/2).floor) break
for (j in dsq.sqrt.floor + 1...0) {
var b = squareCache[j]
if ((a + b) < dsq) break
if ((a + b) == dsq) {
candidates.addAll([ [i, j], [-i, j], [i, -j], [-i, -j],
[j, i], [-j, i], [j, -i], [-j, -i] ])
}
}
}
}
var comparer = Fn.new { |d| pmod.call(theta - d[1].atan(d[0]), Num.tau) }
candidates.sort { |a, b| comparer.call(a) < comparer.call(b) }
}
```
```   var accs = []
var sumx = 0
var sumy = 0
for (dxy in dxys) {
sumx = sumx + dxy[0]
sumy = sumy + dxy[1]
}
return accs
```

}

// find first 10,000 points var Points10000 = babylonianSpiral.call(9998) // first two added automatically

// print first 40 to terminal System.print("The first 40 Babylonian spiral points are:") for (chunk in Lst.chunks(Points10000[0..39], 10)) Fmt.print("\$-9s", chunk)

class Main {

```   construct new() {
Window.title = "Babylonian spiral"
Canvas.resize(1000, 1000)
Window.resize(1000, 1000)
Canvas.cls(Color.white)
var axes = Axes.new(100, 900, 800, 800, -1000..11000, -5000..10000)
axes.draw(Color.black, 2)
var xMarks = Stepped.new(0..10000, 2000)
var yMarks = Stepped.new(-5000..10000, 5000)
axes.label(xMarks, yMarks, Color.black, 2, Color.black)
axes.line(-1000, 10000, 11000, 10000, Color.black, 2)
axes.line(11000, -5000, 11000, 10000, Color.black, 2)
axes.lineGraph(Points10000, Color.black, 2)
}
```
```   init() {}
```
```   update() {}
```
```   draw(alpha) {}
```

}

var Game = Main.new()</lang>

Output:
```The first 40 Babylonian spiral points are:
[0, 0]    [0, 1]    [1, 2]    [3, 2]    [5, 1]    [7, -1]   [7, -4]   [6, -7]   [4, -10]  [0, -10]
[-4, -9]  [-7, -6]  [-9, -2]  [-9, 3]   [-8, 8]   [-6, 13]  [-2, 17]  [3, 20]   [9, 20]   [15, 19]
[21, 17]  [26, 13]  [29, 7]   [29, 0]   [28, -7]  [24, -13] [17, -15] [10, -12] [4, -7]   [4, 1]
[5, 9]    [7, 17]   [13, 23]  [21, 26]  [28, 21]  [32, 13]  [32, 4]   [31, -5]  [29, -14] [24, -22]
```