If you want the exactly minimum diameter, then that sounds like a hard combinatorial problem.

But you can get within a constant factor (in degree and diameter) with expander graphs, specifically Ramanujan graphs. A lower bound on the diameter is log(n)/log(d), and Ramanujan graphs have diameter less than 2 log(n)/log(d/4) = O(log(n)/log(d)).

The defining feature of Ramanujan graphs is that they have spectral expansion lambda < 2 srqt(d-1)/d. By considering a random walk starting from one vertex it’s not hard to show this implies a diameter at most log(n)/log(1/lambda) < 2 log(n)/log(d/4).

The Wikipedia article has references.

This is all in the undirected case, so it applies also when the in- and out-degrees are equal. Since the total in-degree equals the total out-degree, I expect it doesn’t help to allow higher in- than out-degrees or vice versa, but I don’t know.

There’s a large body of work on expanders in general, and it’s possible someone has addressed diameter specifically and closed the constant-factor gap left by the spectral approach through Ramanujan graphs.

But you can get within a constant factor (in degree and diameter) with expander graphs, specifically Ramanujan graphs. A lower bound on the diameter is log(n)/log(d), and Ramanujan graphs have diameter less than 2 log(n)/log(d/4) = O(log(n)/log(d)).

The defining feature of Ramanujan graphs is that they have spectral expansion lambda

Seems to me to be that for a diameter of i
the max. number of nodes is
(m+n)(m+n-1)^i

So for a given number of vertices v
the number of nodes remaining at any stage is

v – (m+n)(m+n-1)^i

for a diameter of i
find the biggest i add 1 depending on the result pending.