I appreciate that this is not exactly a trivial problem. My approach so far has been to create a graph with $B+W$ vertices and initially no edges, making the first $B$ vertices black and the next $W$ vertices white. Then, connect each of the black vertices to two of the white vertices, such that each white vetex is connected to three black vertices (at least two of which are distinct). Check that the resulting graph is connected. Then use a cycle algorithm (such as the one here) to identify all the cycles of the graph starting at white vertices which do not backtrack, i.e. which do not go from a white vertex to a black vertex and back to a white vertex again, unless the intermediate black vertex is only connected to the initial white vertex. If those cycles have edge numbers given by the $2r_{\infty}\left(k\right)$, then draw the graph and move onto the next permutation of connections between the black vertices and white vertices. Else, just move onto the next permutation of connections between the black vertices and white vertices.

I appreciate that this is not exactly a trivial problem. My approach so far has been to create a graph with $B+W$ vertices and initially no edges, making the first $B$ vertices black and the next $W$ vertices white. Then, connect each of the black vertices to two of the white vertices, such that each white vetex is connected to three black vertices (at least two of which are distinct). Check that the resulting graph is connected. Then use a cycle algorithm (such as the one here) to identify all the cycles of the graph starting at white vertices which do not backtrack, i.e. which do not go from a white vertex to a black vertex and back to a white vertex again, unless the intermediate black vertex is only connected to the initial white vertex. If those cycles have edge numbers given by the $2r_{\infty}\left(k\right)$, then draw the graph and move onto the next permutation of connections between the black vertices and white vertices. Else, just move onto the next permutation of connections between the black vertices and white vertices.

I want to write a function which constructs all possible dessins d'enfant (bipartite graphs fulfilling certain criteria, given below) from a collection of data presented in the following form:

(A useful simplification is that in most cases of interest, for all $i$ and $j$, $r_{0}\left(i\right)=3$ and $r_{1}\left(j\right)=2$ - I've been focussing on this specific case in my own attempted solutions).

I would like to write a Mathematica function which reproduces these graphs, when the corresponding 3-vector of data is input. As a further complication, it is sometimes the case that for one set of data there are multiple dessins. For example, consider the following data:

I want to write a function which constructs all possible dessins d'enfants (bipartite graphs fulfilling certain criteria, given below) from a collection of data presented in the following form:

(A useful simplification is that in most cases of interest, for all $i$ and $j$, $r_{0}\left(i\right)=3$ and $r_{1}\left(j\right)=2$ - I've been focusing on this specific case in my own attempted solutions).

I would like to write a Mathematica function which reproduces these graphs, when the corresponding 3-vector of data is input. As a further complication, it is sometimes the case that for one set of data there are multiple dessins. For example, consider the following data:

To illustrate, the graphs at the top of page 28 in this article correspond to the data given at the top of page 25 of the same article. Taking the specific example of Ia, we see that:

I appreciate that this is not exactly a trivial problem. I would be happy to share my own attempted solutions if needed. I think the two hardest things are imposing condition (4), and finding all the dessins. Any help with this would be very much appreciated (and a solution would, I think, be hugely beneficial to everyone in the field).

(A useful simplification is that in most cases of interest, for all $i$ and $j$, $r_{0}\left(i\right)=3$ and $r_{1}\left(j\right)=2$ - I've been focussing on this specific case in my own attempted solutions).

To illustrate, the graphs at the top of page 28 in this article correspond to the data given at the top of page 25 of the same article. Taking the specific example of Ia, we see that:

I appreciate that this is not exactly a trivial problem. My approach so far has been to create a graph with $B+W$ vertices and initially no edges, making the first $B$ vertices black and the next $W$ vertices white. Then, connect each of the black vertices to two of the white vertices, such that each white vetex is connected to three black vertices (at least two of which are distinct). Check that the resulting graph is connected. Then use a cycle algorithm (such as the one here) to identify all the cycles of the graph starting at white vertices which do not backtrack, i.e. which do not go from a white vertex to a black vertex and back to a white vertex again, unless the intermediate black vertex is only connected to the initial white vertex. If those cycles have edge numbers given by the $2r_{\infty}\left(k\right)$, then draw the graph and move onto the next permutation of connections between the black vertices and white vertices. Else, just move onto the next permutation of connections between the black vertices and white vertices.

I'll post the explicit code for this algorithm ASAP (i.e. after some reworking, which will hopefully get it to do something!), but I don't think it's likely to be the most efficient way of going about the problem, and I am sure a more elegant solution is possible.