I'm thinking about making an attempt at counting the number of Latin squares of order 12. Currently I'm in (what I call) the "sanity check" stage, where we check that the ranges that need searching through aren't too massive.

The standard method of counting Latin squares involves building up from Latin rectangles. We get an improvement if we "forget" the structure of the Latin rectangle and remember only which symbols occur in which column (in fact, this is what's makes counting Latin squares possible-ish, the idea goes back to Sade). This can be achieved by interpreting the Latin rectangle as a k-regular subgraph of $K_{n,n}$.

So, in order to perform my sanity check, I'm after the number of non-isomorphic $k$-regular subgraphs of $K_{12,12}$ where $k \in \{1,2,\ldots,6\}$ (where subgraphs include all 24 vertices). (Sloane's http://oeis.org/A008327). I'm particularly interested in the case $k=6$.

Question: What is the number of non-isomorphic $k$-regular subgraphs of $K_{12,12}$?

Actually, an estimate would be good enough for what I want. So perhaps it's more realistic to hope for an answer to this question:

Question: What is an easy way to estimate the number of non-isomorphic $k$-regular subgraphs of $K_{12,12}$?

McKay and Wanless (2005) write:

It is unlikely that [the number of Latin squares of order 12] will be computable by the same method for some time, since the number of regular bipartite graphs of order $24$ and degree $6$ is more than $10^{11}$.

Searching through $10^{11}$ graphs is not completely out-of-the-question, but if it were much more than this I would have to reconsider.