TITLE: Complexity of random knot with vertices on sphere
QUESTION [26 upvotes]: Connect $n$ random points on a sphere in a cycle of
segments between succesive points:

     

I would like to know the growth rate, with respect to $n$, of the crossing number
(the minimal number of crossings of any diagram of the knot)
$c(n)$ of such a knot.
I only know that $c(n)$ is $O(n^2)$,
because it is known
that the crossing number is upper-bounded by
the stick number $s(n)$:
$$\frac{1}{2}(7+\sqrt{ 8 c(K) + 1}) \le s(K)$$
for any knot $K$.
And $s(n) \le n$ is immediate.
I feel certain this has been explored but I am not finding it
in the literature.  Thanks for pointers!

REPLY [3 votes]: I did some experiments on the degrees of Alexander polynomials, as suggested by Bill Thurston, and here is a table of values:
{{10, 1}, {20, 3}, {30, 7}, {40, 7}, {50, 19}, {60, 35}, {70, 
  43}, {80, 73}, {90, 73}, {100, 115}, {120, 123}, {140, 175}, {160, 
  233}, {180, 259}}

(the first number is the number of vertices, the second is the degree of the Alexander polynomial).
Or, for those of us more visually-minded, here is a plot:

I would describe the growth rate as at most $N^{3/2},$ consistent with Bill's theory.