[Question]
  [
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.
[![enter image description here](https://i.stack.imgur.com/erAyJ.jpg)](https://i.stack.imgur.com/erAyJ.jpg)
  
[Answer]
**Mathematica, 13 bytes**
```
CompleteGraph
```
[![lined-up-circle-37-points](https://i.stack.imgur.com/BLIGp.png)](https://i.stack.imgur.com/BLIGp.png)
Looks like this only fails to give a circular embedding for `n=4`, but the question states `n>=20`
[Answer]
# [MATL](https://matl.suever.net/?code=%3AG%2F4%2aJw%5E2Z%5E%21XG&inputs=21&version=19.5.1), ~~16~~ 14 bytes
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!](https://matl.suever.net/?code=%3AG%2F4%2aJw%5E2Z%5E%21XG&inputs=21&version=19.5.1) (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.
![](https://i.stack.imgur.com/taK5j.jpg)
[Answer]
## [PICO-8](http://www.lexaloffle.com/pico-8.php), 131 bytes
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
```
[![128x128 madness](https://i.stack.imgur.com/tJqZ8.png)](https://i.stack.imgur.com/tJqZ8.png)
PICO-8 is a Lua based fantasy console with a native resolution of **128x128.** I made the circle as big as I could...
[Answer]
# Mathematica, 42 bytes
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}]
```
[![enter image description here](https://i.stack.imgur.com/7jjWK.png)](https://i.stack.imgur.com/7jjWK.png)
[Answer]
# HTML + JS (ES6), 34 + ~~177~~ ~~164~~ 162 = 196 bytes
Using the [HTML5 Canvas API](https://developer.mozilla.org/en-US/docs/Web/API/Canvas_API).
[See it on CodePen](http://codepen.io/darrylyeo/pen/xRpXYM).
```
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()`
[Answer]
# Java, ~~346~~ ~~338~~ ~~322~~ 301 Bytes
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`:
[![enter image description here](https://i.stack.imgur.com/PCPpz.png)](https://i.stack.imgur.com/PCPpz.png)
[Answer]
## Python 2, ~~258~~ ~~235~~ 229 Bytes
```
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`  
[![n=37](https://i.stack.imgur.com/DVMoT.png)](https://i.stack.imgur.com/DVMoT.png)
[Answer]
# Octave, ~~88~~ 69 bytes
```
N=input('');t=0:2*pi/N:N;k=nchoosek(1:N,2)';line(cos(t)(k),sin(t)(k))
```
Output for `N=37`:
[![enter image description here](https://i.stack.imgur.com/NtKPt.png)](https://i.stack.imgur.com/NtKPt.png)
Output for `N=19`:
[![enter image description here](https://i.stack.imgur.com/eZfQh.jpg)](https://i.stack.imgur.com/eZfQh.jpg)
[Answer]
## Perl, 229 bytes
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 `echo`s 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));
    }
}
$i->write(file=>"t.png");
```
[![enter image description here](https://i.stack.imgur.com/qa4Bf.png)](https://i.stack.imgur.com/qa4Bf.png)
 *-1 byte thanks to Titus.*
[Answer]
# [GeoGebra](https://www.geogebra.org/), 92 bytes
```
a=polygon((0,0),(1,0),20)
sequence(sequence(segment(vertex(a,i),vertex(a,j)),j,1,20),i,1,20)
```
Each line is separately entered into the input bar. Here is a gif showing the execution:
[![Execution](https://i.stack.imgur.com/l1wyn.gif)](https://i.stack.imgur.com/l1wyn.gif)
**How it works**
The `polygon` command creates a 20-sided polygon, with the vertices of the baseline at `(0,0)` and `(1,0)`. The next command then iterates over each vertex of the polygon with index `i`, using the `sequence` and `vertex` commands, and for each vertex with index `i`, draws a line segment to every other vertex with index `j` using the `segment` command.
[Answer]
# PHP, ~~186~~ ~~184~~ 196 bytes
```
imagecolorallocate($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.
[![exact PI graph](https://i.stack.imgur.com/xrTFg.png)](https://i.stack.imgur.com/xrTFg.png)
**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).
[Answer]
## [NetLogo](https://ccl.northwestern.edu/netlogo/) - 44 bytes
```
cro 20[create-links-to other turtles fd 20]
```
![NetLogo output](https://i.stack.imgur.com/OQJGV.png)
[Answer]
## R, ~~127~~ 123 bytes
```
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 :**
[![](https://i.stack.imgur.com/ofkIh.jpg)](https://i.stack.imgur.com/ofkIh.jpg)
Nice axis' labels uh ?
*-4 bytes* thanks to @Titus !
[Answer]
# BBC BASIC, 98 ascii characters
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.
[![<code>enter image description here</code>](https://i.stack.imgur.com/bxYnE.png)](https://i.stack.imgur.com/bxYnE.png)
[Answer]
# Octave, ~~50 48 46~~ 45 bytes
```
@(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:
[![enter image description here](https://i.stack.imgur.com/wcxsW.jpg)](https://i.stack.imgur.com/wcxsW.jpg)
[Answer]
## JavaScript (ES5)/SVG (HTML5), 181 bytes
```
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).
[Answer]
# MATLAB, 36 bytes
```
@(n)plot(graph(ones(n),'Om'),'La','c')
```
This is an anoymous function that creates the plot.
```
@(n)                                     Define an anonymous fuction of ùòØ
               ones(n)                   Create an ùòØ√óùòØ 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:
[![enter image description here](https://i.stack.imgur.com/kv3Ob.png)](https://i.stack.imgur.com/kv3Ob.png)
[![enter image description here](https://i.stack.imgur.com/4leFr.png)](https://i.stack.imgur.com/4leFr.png)
[Answer]
## QBasic 4.5, ~~398~~ 271 bytes
```
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`:
[![enter image description here](https://i.stack.imgur.com/JVsDo.png)](https://i.stack.imgur.com/JVsDo.png)
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...
[Answer]
## [QBIC](https://drive.google.com/drive/folders/0B0R1Jgqp8Gg4cVJCZkRkdEthZDQ), ~~98~~ 94 bytes
```
$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:
[![enter image description here](https://i.stack.imgur.com/5wiXo.png)](https://i.stack.imgur.com/5wiXo.png)
[Answer]
# Processing, 274 bytes (239 + `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)
[![50 points](https://i.stack.imgur.com/Gtooa.jpg)](https://i.stack.imgur.com/Gtooa.jpg)
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`.
[![enter image description here](https://i.stack.imgur.com/Sb4Eg.png)](https://i.stack.imgur.com/Sb4Eg.png)
[Answer]
# Bash + Jelly + GraphViz, 52 characters, 52 or 63 bytes
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 ‚Äì none of the characters are outside the range 0 to 255 ‚Äì 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 '∆ìR≈ícj‚Ç¨‚Åæ--Y‚Äú»§?√∏√ü·πáe4·∫°‚Å¥B¬∂¬¶√ór¬∞·∫Ü¬ª;‚Äù}'|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:
```
∆ìR≈ícj‚Ç¨‚Åæ--Y‚Äú»§?√∏√ü·πáe4·∫°‚Å¥B¬∂¬¶√ór¬∞·∫Ü¬ª;‚Äù}
∆ì                               read number from stdin
 R                              produce range from 1 to that number
                                (here used to produce a list with
                                that many distinct elements)
  ≈íc                            select all unordered pairs from that
      ‚Åæ--                       a string consisting of two hyphens
    j‚Ç¨                          join each pair via the string
         Y                      join on newlines
                            ;   prepend (in this context)
          ‚Äú»§?√∏√ü·πáe4·∫°‚Å¥B¬∂¬¶√ór¬∞·∫Ü¬ª    "graph{node[shape=point]"
                             ‚Äù} 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):
[![complete graph on 21 points](https://i.stack.imgur.com/2nVrk.png)](https://i.stack.imgur.com/2nVrk.png)
[Answer]
# PHP + HTML SVG, ~~316~~ 263 bytes
**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:
[![enter image description here](https://i.stack.imgur.com/ZuUVn.png)](https://i.stack.imgur.com/ZuUVn.png)
The attached snippet is limited to 18 points because of the 30k single post limit.
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```
[Answer]
# R, 108 bytes
```
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.
[![N=21](https://i.stack.imgur.com/ztK6z.png)](https://i.stack.imgur.com/ztK6z.png)
# R + igraph, 87 bytes
Another solution using package `igraph`.
```
library(igraph);plot(make_full_graph(21),layout=cbind(cos(t<-seq(0,2*pi,l=21)),sin(t)))
```
[![N=21](https://i.stack.imgur.com/0Nvl0.png)](https://i.stack.imgur.com/0Nvl0.png)
]