Consider the square $[0,1]^2$. Assume that this region is divided into $N=K^2$ equispaced grid points. How many convex curves can be drawn in terms of $K$?
The points $(0,1)$ and $(1,0)$ are known to be on the convex curve. I am interested in the answer of this questions due to two reasons:
$1.$ I would like to consider $K\rightarrow \infty$ and compare the total number of convex functions to the total number of other types of functions for example (decreasing, etc.)
$2.$ Later I would like to write a program to realize all such discrete convex functions in order to perform optimization over all these functions (for a specific $K$).
I would be also happy to hear any ideas about how to realize this algorithmically.
Here is an example, where one can see examples of three different convex functions:
In this example there are altogether $N=121$ grid points and the distance between each neighboring pair of grid points both in $x$ and $y$ direction is $0.1$.
So, for every value on the $x$ axis, there will be a corresponding value on the $y$ axis.
ADDED (15.07.2018): I programmed and obtained the grid. Using this approach it is impossible to get all convex functions as $N\rightarrow \infty$. I bet we are not even near. Consider the convex function which linearly decreases from $(0,1)$ to $(0.4,0.2)$ and again linearly decreases from $(0.4,0.2)$ to $(1,0)$. It is impossible to get this function with this approach. No mater how fine the grid is, there are infinitely many other functions which are not achieved.
I am not sure that this is what you want, but here is my thinking about it:
If $n$ is the number of divisions of unity ($n=10$ in your drawing), and $x_i, i=1,2,\ldots n$ is the number of divisions traveled downwards by each segment of your curve (for the green curve $x_1,\ldots ,x_{10}$ would be $5,2,1,1,1,0,0,0,0,0$), then your problem could be formulated as the number of solutions of the following integer equation:
$x_1+x_2+\ldots+x_n=n$
subject to:
$x_1\ge x_2\ge\ldots\ge x_n\ge 0$
We can observe that each solution of this integer equation can be put in $1:1$ correspondence with a partition of $n$ (by ignoring the zeros). For the green line, this is:
$10=5+2+1+1+1$
So the number of convex functions equals $p(n)$, which is the number of partitions of $n$.
Unfortunately, $p(n)$ does not have a nice closed-form formula, but you can see the Wikipedia page Partition_(number_theory) for more details, recurrences, asymptotics, etc.
EDIT: To address OP's question about comparing $p(n)$ with the number of non-increasing functions:
If $q(n)$ is the number of non-increasing functions (not necessarily convex), then by reusing the previous notation, we get that $q(n)$ is the number of integer solutions of the following equation:
$x_1+x_2+\ldots+x_n=n$
subject to:
$x_1, x_2,\ldots, x_n\ge 0$
This is a classic Stars and bars problem (theorem two), whose solution is:
$q(n)= {n + n - 1 \choose n - 1}={2n-1\choose n-1}$
A quick check on Wolfram Alpha shows that $\frac{q(n)}{p(n)}\to\infty$, so $p(n)=o(q(n))$