I know this is an old question, but I have just set a similar problem for myself today and found this question after solving it.
Disclaimer: I was only concerned with the equilibrium shape (the "catenary" for discrete case), so if the OP was after the dynamics of such a system, this won't serve as an answer. However, the last comment from the OP indicates the interest in the following method.
As <PERSON> said in the comments, to compute the equilibrium shape (the "catenary" for discrete case) the most simple way seems to be the method of Lagrange multipliers.
For (my own) convenience, the g vector is up instead of down, so the catenary is standing. 359
Also, I'm using Cartesian coordinates.
The example of the result is above, with the smooth curve representing a catenary in the usual sense, with the arc length equal to the sum of all "rod lengths" between the point masses. I have taken the support points out of consideration and set the supporting rod length to be half of the length between two masses, just for convenience.
The arc length is taken as the fixed value $L=3$ with $l=L/6=1/2$; and the distance between support points is $2$.
Solution for the particular case above, since it's simple enough to infer the general case without additional comments.
Due to the symmetry, it's enough to consider only the part with $x>0$.
The one and only equilibrium condition in this case is the minimum value of the potential energy for the system, and since I have directed the gravitation up, we need to maximize the sum of $y_k$:
$$\text{maximize} \quad f=y_1+y_2+y_3$$
Subject to constraints:
$$\begin{cases} g_1=(y_1-y_2)^2+(l/2-x_2)^2-l^2=0 \ g_2=(y_2-y_3)^2+(x_2-x_3)^2-l^2=0 \ g_3=y_3^2+(x_3-1)^2-l^2/4=0 \end{cases}$$
Lagrangian multiplier method involves introducing additional variables for each constraint, and writing down the Lagrange function:
$$L=f-\sum_k \lambda_k g_k$$
Then the necessary (but not sufficient) condition for the maximum of $f$ will be:
$$\nabla L=0$$
Then the general method requires building a Hessian matrix with the second derivatives, but there's no need in this case! We know that all $x_k,y_k$ should be real and positive, and as far as I can see, there's always a single solution like that (for any arc length I tried I mean).
Considering all the partial derivatives, we recover our original constraint equations, and $5$ additional ones:
$$\begin{cases} (y_1-y_2)^2+(l/2-x_2)^2-l^2=0 \ (y_2-y_3)^2+(x_2-x_3)^2-l^2=0 \ y_3^2+(x_3-1)^2-l^2/4=0 \ 1-2 \lambda_1 (y_1-y_2)=0 \ 1+2 \lambda_1 (y_1-y_2)-2 \lambda_2 (y_2-y_3)=0 \ 1+2 \lambda_2 (y_2-y_3)-2 \lambda_3 y_3=0 \ \lambda_1(l/2-x_2)-\lambda_2(x_2-x_3)=0 \ \lambda_2(x_2-x_3)-\lambda_3(x_3-1)=0 \end{cases}$$
The only solution with real and positive $x_k,y_k$ is:
$$\begin{array}( x_2=<PHONE_NUMBER> \ x_3=0.904152 \ y_1=<PHONE_NUMBER> \ y_2=<PHONE_NUMBER> \ y_3=<PHONE_NUMBER> \ \lambda_1 = 1.59714 \ \lambda_2 = 2.35603 \ \lambda_3 = 6.49642 \end{array}$$
Numerical solutions are obtained with Wolfram Mathematica. The resulting diagram (also created with Mathematica) is presented above. 101