Babylonian spiral: Difference between revisions
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Show in your program how to calculate and plot the first 10000 points in the sequence. Your result |
Show in your program how to calculate and plot the first 10000 points in the sequence. Your result |
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should look similar to [[File:From_oies_A@97346_A297347_plot2a.png] |
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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". |
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; See also |
; See also |
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Revision as of 02:23, 28 August 2022
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.
- Task
Find and show the first 40 (x, y) coordinates of the Babylonian spiral.
- Stretch task
Show in your program how to calculate and plot the first 10000 points in the sequence. Your result should look similar to [[File:From_oies_A@97346_A297347_plot2a.png]
- See also
J
It's convenient, here, to use complex numbers to represent the coordinate pairs.
Implementation:
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.
}}
Task example:
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
Also:
require'plot'
plot bspir 40
plot bspir 1e4
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
""" 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)
- 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
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;
- 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
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
""" 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.
See also: oeis.org/A256111, oeis.org/A297346, oeis.org/A297347
"""
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 '')
# stretch portion of task
plot(*zip(*points10000))
axis('scaled')
show()
- 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.
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()
Raku
Translation
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
}
# The task
say "The first $_ Babylonian spiral points are:\n",
(babylonianSpiral($_).map: { sprintf '(%3d,%4d)', @$_ }).batch(10).join("\n") given 40;
# 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)")
],
],
);
- 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.
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 };
}
}
# The task
say "The first $_ Babylonian spiral points are:\n",
@spiral[^$_].map({ sprintf '(%3d,%4d)', |$_ }).batch(10).join: "\n" given 40;
# 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)")
],
],
);
- Same output:
Wren
Generates an image similar to the OEIS one.
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) }
dxys.add(candidates[0])
}
var accs = []
var sumx = 0
var sumy = 0
for (dxy in dxys) {
sumx = sumx + dxy[0]
sumy = sumy + dxy[1]
accs.add([sumx, sumy])
}
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()- 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]