How to draw irregular hendecagon in asymptote

Question

I am sorry if this is a duplicate question but how would I draw an irregular hendecagon using asymptote? I am new to it and I know how to draw a regular one but not an irregular one.

[asy]
size(9cm);
pair cis(real magni, real argu) { return (magni*cos(argu*pi/180),magni*sin(argu*pi/180)); }

for(int b=1; b<2; b+=1){
for(int a=6; a<7; a+=1){
int n=a+5*b;
pair ctr=(2.5*a,-2.5*b);
real r=1; pair offs=(0,0);
if(n==3) {r=2/3; offs=(0,-1/3);}
for(int i=1; i<2*n; i+=2){
draw(ctr+cis(r/cos(pi/n),270+(i-2)*180/n)+offs---ctr+cis(r/cos(pi/n),270+i*180/n)+offs,blue);
};
};
};
[/asy]

That is my code for a regular hendecagon, but I don't know how to draw an irregular one. Thanks!

Answer 1

A related question could be How to draw random simple closed smooth curves but with the same perimeter?

With asymptote and the hull_pi.asy package it is easy to have random points and to create the associated convex polygon. You can find hull_pi.asy (not official package) at the following address https://github.com/pivaldi/asymptote-packages/blob/master/hull_pi.asy

Please find a possible solution

    import hull_pi;
    size(10cm);
    // to have a different picture at each launch
    srand(seconds());

    pair[] cloud;
    int nbpt=11;

    path poly_point(pair [] a)
    {
    guide g;
    for (int i; i<size(a);++i)
    g=g--a[i];
    return g--cycle;
    }

    // Generate a first set of random points.
    for (int i=0; i < nbpt; ++i)
      cloud.push((10*unitrand(),10*unitrand()));

    int l=size(hull(cloud));
    while (l<11)
    {
    cloud.push((10*unitrand(),10*unitrand()));
    pair[] phull=hull(cloud);
    l=size(phull);
    }

    pair[] phull=hull(cloud);
    draw(poly_point(phull),1bp+blue);
    dot(phull,red);

and the result