Draw lines between every pair of distinct points for n points arranged in a circle, producing something like the below result. Shortest code (in bytes) wins! Your lines don't have to be transparent, but it looks better that way. The output must be a vector graphic, or be an image at least 600 pixels by 600 pixels (either saved to a file or displayed on the screen). To complete the challenge you must draw at least 20.
Mathematica, 13 bytes
CompleteGraph
Looks like this only fails to give a circular embedding for n=4, but the question states n>=20
n (I had the answer ready from the fixed 37) :(
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Commented
Dec 1, 2016 at 18:23
As I'm not terribly fluent with MATL I expect that this is somewhat more golfable. (Would be nice to at least beat Mathematica :-) I.e. the the flip w is not optimal, it could probably be avoided...
:G/4*Jw^2Z^!XG
Test it Online! (Thanks @Suever for this service, thanks @DrMcMoylex for -2 bytes.)
Explanation (for N=3):
: Generate Range 1:input: [1,2,3]
G/ Divide By the first input [0.333,0.666,1]
4* Multiply by 4 [1.33,2.66,4.0]
Jw^ i ^ (the result so far) [-0.49+ 0.86i,-.5-0.86i,1.00]
(This results in a list of the n-th roots of unity)
2Z^ Take the cartesian product with itself (i.e. generate all 2-tuples of those points)
!XG Transpose and plot
It is worth noting that for generating the N-th roots of unity you can use the formula exp(2*pi*i*k/N) for k=1,2,3,...,N. But since exp(pi*i/2) = i you could also write i^(4*k/N) for k=1,2,3,...,N which is what I'm doing here.
G thank you very much!
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I wasn't really sure if I'd be breaking any rules, but I did it anyway!
Golfed
p={}for i=0,19 do add(p,{64+64*cos(i/20),64+64*sin(i/20)})end for x in all(p)do for y in all(p)do line(x[1],x[2],y[1],y[2])end end
Ungolfed
points={}
for i=0,19 do
x=64+64*cos(i/20)
y=64+64*sin(i/20)
add(points,{x,y})
end
for x in all(points) do
for y in all(points) do
line(x[1],x[2],y[1],y[2])
end
end
PICO-8 is a Lua based fantasy console with a native resolution of 128x128. I made the circle as big as I could...
Creates a set of 37 points arranged in a circle, and then draws lines between all possible subsets of two points. Someone posted a shorter answer that takes advantage of CompleteGraph, but I believe this is the shortest one aside from those relying on CompleteGraph.
Graphics@Line@Subsets[CirclePoints@37,{2}]
Tuple. You also need to update this to accept arbitrary n, but conveniently that won't cost you any bytes.
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Tuples
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Using the HTML5 Canvas API.
f=n=>{with(Math)with(c.getContext`2d`)for(translate(S=300,S),O=n;O--;)for(rotate(a=PI*2/n),N=n;N--;)beginPath(stroke()),lineTo(0,S),lineTo(sin(a*N)*S,cos(a*N)*S)}
/* Demo */
f(20)<canvas id=c width=600 height=600>-13 bytes: Removed closePath(), moved stroke() inside beginPath()
-2 bytes: Defined variable a inside rotate()
This solution works for all n>1, even though the original post didn't require that, it does.
My favorite is n=5, don't ask why, also, if you want a cooler GUI, use:
int a=Math.min(this.getHeight(),this.getWidth())/2;
In place of the hard-coded 300, it'll use the width or height of the frame as the diameter.
Saved 8 bytes thanks to Shooqie. Saved 21 bytes thanks to Geobits.
import java.awt.*;void m(final int n){new Frame(){public void paint(Graphics g){Point[]p=new Point[n];int a=300;for(int i=1;i<n+1;i++){p[i-1]=new Point(a+(int)(a*Math.cos(i*2*Math.PI/n)),a+(int)(a*Math.sin(i*2*Math.PI/n)));for(int j=0;j<i;j++){g.drawLine(p[i-1].x,p[i-1].y,p[j].x,p[j].y);}}}}.show();}
Output for n=37:
Frame x= and final (I think?)
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Frame x was from another solution that involved a thread. You need the final though as it's an internal class reference to an external variable in the owning class.
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Commented
Dec 1, 2016 at 18:27
int declarations outside the for loops
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import itertools as T,math as M
from PIL import Image as I,ImageDraw as D
s=300
n=input()
t=2*M.pi/n
o=I.new('RGB',(s*2,)*2)
for x in T.combinations([(s*M.cos(t*i)+s,s*M.sin(t*i)+s)for i in range(n)],2):D.Draw(o).line(x)
o.show()
Output for n=37
from PIL import* shorter?
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Commented
Dec 2, 2016 at 19:57
import *, depending on how you install, you could skip the PIL and import Image/ImageDraw directly
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N=input('');t=0:2*pi/N:N;k=nchoosek(1:N,2)';line(cos(t)(k),sin(t)(k))
Output for N=37:
Output for N=19:
gplot too, but I didn't manage to make it short enough...
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Commented
Dec 2, 2016 at 9:05
It uses the same formula as most languages that don't have convenient builtin for this challenge (even if I didn't look at them to find it, but that's a fairly easy to find formula). So not very interesting, but there are usually not a lot of Perl answers to this kind of challenges, so I just wanted to propose one.
$i=new Imager xsize=>700,ysize=>700;for$x(1..$_){for$y(1..$_){$i->line(color=>red,x1=>350+300*cos($a=2*pi*$x/$_),x2=>350+300*cos($b=2*pi*$y/$_),y1=>350+300*sin$a,y2=>350+300*sin$b)}}$i->write(file=>"t.png")
And you'll need -MImager (9 bytes), -MMath::Trig (providing pi, 13 bytes), and -n (1 byte) ==> + 23 bytes.
To run it :
perl -MImager -MMath::Trig -ne '$i=new Imager xsize=>700,ysize=>700;for$x(1..$_){for$y(1..$_){$i->line(color=>red,x1=>350+300*cos($a=2*pi*$x/$_),x2=>350+300*cos($b=2*pi*$y/$_),y1=>350+300*sin$a,y2=>350+300*sin$b)}}$i->write(file=>"t.png")' <<< 27
It will create a file named t.png which contains the image.
You'll need to install Imager though, but no worries, it's quite easy :
(echo y;echo) | perl -MCPAN -e 'install Imager'
(The echos will configure you cpan if you've never used it before. (actually that will only work if your perl is recent enough, I think for most of you it will be, and I'm sorry for the others!)).
And the more readable version (yes, it's fairly readable for a Perl script!) :
#!/usr/bin/perl -n
use Imager;
use Math::Trig;
$i=Imager->new(xsize=>700,ysize=>700);
for $x (1..$_){
for $y (1..$_){
$i->line(color=>red,x1=>350+300*cos($a=2*pi*$x/$_), x2=>350+300*cos($b=2*pi*$y/$_),
y1=>350+300*sin($a), y2=>350+300*sin($b));
}
}
$i->write(file=>"t.png");
-1 byte thanks to Titus.
for loops, then yes, they are mandatory.
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y2. I bet you don´t need that. And can you write to STDOUT?
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a=polygon((0,0),(1,0),20)
sequence(sequence(segment(vertex(a,i),vertex(a,j)),j,1,20),i,1,20)
Each line is separately entered into the input bar. Here is a gif showing the execution:
How it works
The polygon command creates a 20-sided polygon, with the vertices of the baseline at (0,0) and (1,0). The next command then iterates over each vertex of the polygon with index i, using the sequence and vertex commands, and for each vertex with index i, draws a line segment to every other vertex with index j using the segment command.
imagecolorallocate($i=imagecreate(601,601),~0,~0,~0);for(;$a<$p=2*M_PI;)for($b=$a+=$p/=$argv[1];$b>0;)imageline($i,(1+cos($a))*$r=300,$r+$r*sin($a),$r+$r*cos($b-=$p),$r+$r*sin($b),1);imagepng($i);
writes the image to STDOUT
breakdown
// create image with white background
imagecolorallocate($i=imagecreate(601,601),~0,~0,~0);
// loop angle A from 0 to 2*PI
for(;$a<$p=2*M_PI;)
// loop angle B from A down to 0
for($b=$a+=$p/=$argv[1];$b;) // ($a pre-increment)
// draw black line from A to B
imageline($i, // draw line
(1+cos($a))*$r=300,$r+$r*sin($a), // from A
$r+$r*cos($b-=$p),$r+$r*sin($b), // to B ($b pre-decrement)
1 // undefined color=black
);
// output
imagepng($i);
-12 bytes for fixed n=20
Replace $p=2*M_PI with 6 (-8), /=$argv[1] with =M_PI/10 (-2), and $b>0 with $b (-2)
Using exact PI/10 doesn´t hurt. With .3142, the rounding errors from the parametrized version remained, but with M_PI/10 they vanished and I can check $b(<>0) instead of $b>0. I could have saved two bytes with .314, but that would have off-set the points.
The limit $a<6 is sufficiently exact for 20 points.
174 bytes for fixed n=314
imagecolorallocate($i=imagecreate(601,601),~0,~0,~0);for(;$a<314;)for($b=$a++;$b--;)imageline($i,(1+cos($a))*$r=300,$r+$r*sin($a),$r+$r*cos($b),$r+$r*sin($b),1);imagepng($i);
Using 314 points results in a filled circle in that resolution (as do 136,140, every even number above that, and everything above 317).
plot((e=cbind(sin(t<-seq(0,2*pi,l=(n=21)))*2,cos(t)*2)));for(i in 2:n)for(j in 1:i)lines(c(e[i,1],e[j,1]),c(e[i,2],e[j,2]))
Produces :
Nice axis' labels uh ?
-4 bytes thanks to @Titus !
for(i in 2:n){for(j in 1:i)...}. Does Rrequire the braces?
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Tokenised filesize 86 bytes
r=600V.5142;29,r;r;:I.n:t=2*PI/n:F.i=1TOn*n:a=i DIVn*t:b=i MODn*t:L.r*SINa,r*COSa,r*SINb,r*COSb:N.
Dowload interpreter at http://www.bbcbasic.co.uk/bbcwin/bbcwin.html
There's nothing wrong with drawing every line twice, the appearance is identical :-P
Ungolfed
r=600 :REM Radius 600 units. 2 units per pixel, so 300 pixels
VDU5142;29,r;r; :REM Set mode 20 (600 pixels high) and move origin away from screen corner
INPUTn :REM Take input.
t=2*PI/n :REM Step size in radians.
FORi=1TOn*n :REM Iterate through all combinations.
a=i DIVn*t :REM Get two angles a and b
b=i MODn*t :REM by integer division and modlo
LINEr*SINa,r*COSa,r*SINb,r*COSb :REM calculate cartesian coordinates and draw line
NEXT
Output n=21
This looks much better in the original rendering than in the browser.
LINE function. Beats DRAW...
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Commented
Dec 1, 2016 at 22:04
@(N)gplot((k=0:2*pi/N:N)+k',[cos(k);sin(k)]')
This is an anyonmous function that plots the graph we're looking for.
Explanation:
(k=0:2*pi/N:N)+k' Makes a full N+1 x N+1 adjecency matrix and simultaneously defines the vector k of angles, which we we use then for [cos(k);sin(k)]', a matrix of coordinates where each graph node is positioned. gplot just plots the graph that we want.
For N = 29 we get:
document.write('<svg viewBox=-1e3,-1e3,2e3,2e3><path stroke=#000 fill=none d=M1e3,0')
with(Math)for(i=37;--i;)for(j=37;j--;)document.write('L'+1e3*cos(a=i*j*PI*2/37)+','+1e3*sin(a))Only works for prime numbers, such as the original suggestion of 37. You can halve (rounded up) the initial value of i to obtain a fainter image. You can also consistently adjust the 1e3,2e3 to other values to taste (I started with 300,600 but decided that it was too coarse).
@(n)plot(graph(ones(n),'Om'),'La','c')
This is an anoymous function that creates the plot.
@(n) Define an anonymous fuction of 𝘯
ones(n) Create an 𝘯×𝘯 matrix of ones
graph( ,'Om') Create a graph object with that adjacency
matrix, omitting self-loops
plot( ,'La','c') Plot the graph with a circular layout
Example:
graph is not part of the bioinformatic toolbox... didn't even know it existed... Nice :)
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Commented
Dec 2, 2016 at 13:08
CLS:SCREEN 11:DEFSTR M-Z:DEFDBL A-L
INPUT"N",A:I=(360/A)*.0175:J=230
Q=",":FOR E=0 TO A
FOR F=E TO A
M=x$(COS(I*E)*J+J):N=x$(SIN(I*E)*J+J):O=x$(COS(I*F)*J+J):P=x$(SIN(I*F)*J+J):DRAW "BM"+M+Q+N+"M"+O+Q+P
NEXT:NEXT
FUNCTION x$(d):x$=LTRIM$(STR$(CINT(d))):END FUNCTION
The screen in QBasic can onl be 640x480, so the circle has a radius of only 230 px, unfortunately. Also, there's some artifacting because of float-to-int precision loss. Looks like this for N=36:
EDIT: I didn't need the storage, the type declaration and all the looping. Calculating all Carthesians from Polars in place is 50% cheaper in byte count...
$SCREEN 11|:i=6.3/a j=230[0,a|[b,a|line(cos(b*i)*j+j,sin(b*i)*j+j)-(cos(c*i)*j+j,sin(c*o)*j+j)
I've converted my original QBasic answer @LevelRiverSt 's answer to QBIC. I thought this would rely too heavily on functions that are not built into QBIC to be feasible, but as it turns out, it saves another 90 bytes. Substituting the
DRAW for LINE saves another 80 bytes. I knew I was forgetting something simple...
When run with a command line parameter of 36, it looks like this:
size call and function call)void d(int s){float a=2*PI/s,x=0,y=-400,m,n;float[][]p=new float[2][s];translate(400,400);for(int i=0;i<s;i++){m=x*cos(a)-y*sin(a);n=x*sin(a)+y*cos(a);x=m;y=n;p[0][i]=x;p[1][i]=y;for(int j=0;j<i;j++)line(p[0][j],p[1][j],p[0][i],p[1][i]);}}
void setup(){size(800,800);d(50);}
I honestly don't know why, but setup had to be on the second line. I used https://en.wikipedia.org/wiki/Rotation_matrix to help me calculate the maths for rotation. This program calculates the points and pushes them to an array, with which we are using to draw lines.
Here is a picture of a polygon with 50 edges (the 100 edges one was almost completely black)
You can add stroke(0,alpha); to have transparent edges, where alpha is the opacity of the line. Here's the same polygon with alpha of 20.
Given that the programs in question disagree on which character encoding to use, the program is full of control characters. Here's what it looks like under xxd, in Latin-1 encoding (which represents each character in one byte):
00000000: 6a65 6c6c 7920 6520 2793 5213 636a 0c8e jelly e '.R.cj..
00000010: 2d2d 59fe 9a3f 1d15 dc65 34d3 8442 7f05 --Y..?...e4..B..
00000020: 1172 80cf fb3b ff7d 277c 6369 7263 6f20 .r...;.}'|circo
00000030: 2d54 7073 -Tps
I couldn't actually get the program to run, though, without converting the input into UTF-8 for some reason (which would make it 63 bytes long). Logically it should work as Latin-1 – none of the characters are outside the range 0 to 255 – but I keep getting "string index out of range" errors no matter how I configure the character encoding environment variables. So this will have to be counted as 63 bytes unless someone can figure out a way to run it without re-encoding it.
The program might be slightly more readable if we interpret it in Jelly's encoding:
jelly e 'ƓRŒcj€⁾--Y“Ȥ?øßṇe4ạ⁴B¶¦×r°Ẇ»;”}'|circo -Tps
The program takes the number of points on standard input and outputs a PostScript image on standard output. (It can trivially be adapted to output in any format GraphViz supports by changing the -Tps at the end; it's just that PostScript has the shortest name. Arguably, you can save five characters by removing the -Tps, but then you get output in GraphViz's internal image format that nothing else supports, which probably doesn't count for the purposes of the question.)
Fundamentally, this is just a Jelly program that calls into GraphViz to do the drawing; however, Jelly doesn't seem to have any capabilities for running external programs, so I had to use bash to link them together. (This also means that it's cheaper to make Jelly request input from stdin manually; normally it takes input from the command line, but that would mean extra bytes in the bash wrapper.) circo will automatically arrange all the points it's asked to draw in a circle, so the Jelly code just has to ask it to draw a list of points, all of which are connected to each other. Here's how it works:
ƓRŒcj€⁾--Y“Ȥ?øßṇe4ạ⁴B¶¦×r°Ẇ»;”}
Ɠ read number from stdin
R produce range from 1 to that number
(here used to produce a list with
that many distinct elements)
Œc select all unordered pairs from that
⁾-- a string consisting of two hyphens
j€ join each pair via the string
Y join on newlines
; prepend (in this context)
“Ȥ?øßṇe4ạ⁴B¶¦×r°Ẇ» "graph{node[shape=point]"
”} follow output with a "}" character
The use of Jelly lets us slightly compress the string that configures the GraphViz output via its built-in dictionary. The dictionary has graph, node, and point. Annoyingly, it doesn't have shape (it has SHAPE, but GraphViz is case-sensitive), so we have to encode that character-by-character.
Here's the output for input 21 (with a slight modification to the program to make it output in a format that can be uploaded to Stack Exchange):
Golfed version with hardcoded n points and no input n parameter:
<svg height="610" width="610"><?for($i=1;$i<33;$i++){$x[]=300*sin(2*M_PI/32*$i)+305;$y[]=300*cos(2*M_PI/32)+305;}foreach($x as$j=>$w){foreach($y as$k=>$z){echo'<line x1="'.$x[$j].'" y1="'.$y[$j].'" x2="'.$x[$k].'" y2="'.$y[$k].'" style="stroke:red;"/>';}}?></svg>
Previous golfed version with input parameter for n points, 316 bytes:
<svg height="610" width="610"><?$n=$_GET[n];$d=2*M_PI/$n;$r=300;$o=305;for($i=1;$i<=$n;$i++){$x[]=$r*sin($d*$i)+$o;$y[]=$r*cos($d*$i)+$o;}foreach($x as$j=>$w){foreach($y as$k=>$z){echo'<line x1="'.$x[$j].'" y1="'.$y[$j].'" x2="'.$x[$k].'" y2="'.$y[$k].'" style="stroke:rgba(0,0,0,.15);stroke-width:1;" />';}}?></svg>
Usage: save in a file and call from the browser:
http://localhost/codegolf/circle.php?n=32
Ungolfed version with input parameter for n points and CSS:
<style>
line {
stroke: rgba(0,0,0,.15);
stroke-width:1;
}
</style>
<svg height="610" width="610">
<?php
$n=$_GET[n]; // number of points
$d=2*M_PI/$n; // circle parts
$r=300; // circle radius
$o=305; // offset x,y
for ($i=1;$i<=$n;$i++){
$x[]=$r*sin($d*$i)+$o; // store x,y coordinates in array
$y[]=$r*cos($d*$i)+$o;
}
foreach($x as $j => $w){ // iterate all x,y points and connect to each other
foreach($y as $k => $z) {
echo '<line x1="'.$x[$j].'" y1="'.$y[$j].'" x2="'.$x[$k].'" y2="'.$y[$k].'" />'."\n";
}
}
?>
</svg>
Couldn't attach a 32 points fully functional snippet because of the 30k characters limit for a single post. Here is a screenshot:
The attached snippet is limited to 18 points because of the 30k single post limit.
line {
stroke: rgba(0,0,0,.15);
stroke-width:1;
}<svg height="610" width="610">
<line x1="407.6060429977" y1="586.90778623577" x2="407.6060429977" y2="586.90778623577" />
<line x1="407.6060429977" y1="586.90778623577" x2="497.83628290596" y2="534.81333293569" />
<line x1="407.6060429977" y1="586.90778623577" x2="564.80762113533" y2="455" />
<line x1="407.6060429977" y1="586.90778623577" x2="600.44232590366" y2="357.09445330008" />
<line x1="407.6060429977" y1="586.90778623577" x2="600.44232590366" y2="252.90554669992" />
<line x1="407.6060429977" y1="586.90778623577" x2="564.80762113533" y2="155" />
<line x1="407.6060429977" y1="586.90778623577" x2="497.83628290596" y2="75.186667064307" />
<line x1="407.6060429977" y1="586.90778623577" x2="407.6060429977" y2="23.092213764227" />
<line x1="407.6060429977" y1="586.90778623577" x2="305" y2="5" />
<line x1="407.6060429977" y1="586.90778623577" x2="202.3939570023" y2="23.092213764227" />
<line x1="407.6060429977" y1="586.90778623577" x2="112.16371709404" y2="75.186667064307" />
<line x1="407.6060429977" y1="586.90778623577" x2="45.192378864669" y2="155" />
<line x1="407.6060429977" y1="586.90778623577" x2="9.5576740963376" y2="252.90554669992" />
<line x1="407.6060429977" y1="586.90778623577" x2="9.5576740963376" y2="357.09445330008" />
<line x1="407.6060429977" y1="586.90778623577" x2="45.192378864668" y2="455" />
<line x1="407.6060429977" y1="586.90778623577" x2="112.16371709404" y2="534.81333293569" />
<line x1="407.6060429977" y1="586.90778623577" x2="202.3939570023" y2="586.90778623577" />
<line x1="407.6060429977" y1="586.90778623577" x2="305" y2="605" />
<line x1="497.83628290596" y1="534.81333293569" x2="407.6060429977" y2="586.90778623577" />
<line x1="497.83628290596" y1="534.81333293569" x2="497.83628290596" y2="534.81333293569" />
<line x1="497.83628290596" y1="534.81333293569" x2="564.80762113533" y2="455" />
<line x1="497.83628290596" y1="534.81333293569" x2="600.44232590366" y2="357.09445330008" />
<line x1="497.83628290596" y1="534.81333293569" x2="600.44232590366" y2="252.90554669992" />
<line x1="497.83628290596" y1="534.81333293569" x2="564.80762113533" y2="155" />
<line x1="497.83628290596" y1="534.81333293569" x2="497.83628290596" y2="75.186667064307" />
<line x1="497.83628290596" y1="534.81333293569" x2="407.6060429977" y2="23.092213764227" />
<line x1="497.83628290596" y1="534.81333293569" x2="305" y2="5" />
<line x1="497.83628290596" y1="534.81333293569" x2="202.3939570023" y2="23.092213764227" />
<line x1="497.83628290596" y1="534.81333293569" x2="112.16371709404" y2="75.186667064307" />
<line x1="497.83628290596" y1="534.81333293569" x2="45.192378864669" y2="155" />
<line x1="497.83628290596" y1="534.81333293569" x2="9.5576740963376" y2="252.90554669992" />
<line x1="497.83628290596" y1="534.81333293569" x2="9.5576740963376" y2="357.09445330008" />
<line x1="497.83628290596" y1="534.81333293569" x2="45.192378864668" y2="455" />
<line x1="497.83628290596" y1="534.81333293569" x2="112.16371709404" y2="534.81333293569" />
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plot(x<-cos(t<-seq(0,2*pi,l=21)),y<-sin(t),as=1);apply(expand.grid(1:21,1:21),1,function(e)lines(x[e],y[e]))
Could shave 5 bytes off if I got rid of argument ,as=1 which forces an aspect ratio of 1. Uses expand.grid to create a matrix with all possible pair of points, and uses apply to loop through it.
Another solution using package igraph.
library(igraph);plot(make_full_graph(21),layout=cbind(cos(t<-seq(0,2*pi,l=21)),sin(t)))
nand draw lines fornpoints. \$\endgroup\$nwould add much to the challenge as I expect most solutions to work with any number anyway, especially because 37 is odd and hence there are no mirror symmetries. \$\endgroup\$nas input or just pick an arbitrarynover 20? \$\endgroup\$