Does every drawing of K(7,11) have at least 225 crossings?

How few crossings can a drawing of the complete bipartite graph K(m,n) have? Zarankiewicz's formula gives the count every known drawing reaches, but it is proved only for small cases, the last of them by computer in 1993. The smallest open case: does every drawing of K(7,11) have at least 225 crossings?

In 1944, in a forced labour camp at a brick factory near Budapest, Pál Turán pushed carts of bricks from the kilns to the storage yards on rails that crossed. At every crossing the carts jolted and lost bricks, and he asked how the rails could be laid with as few crossings as possible. In the language of graphs: draw the complete bipartite graph K(m,n), every one of m points joined to every one of n points, with the fewest crossings.

In 1954 Zarankiewicz drew K(m,n) with floor(m/2) floor((m-1)/2) floor(n/2) floor((n-1)/2) crossings and gave a proof that no drawing does better. The proof had a gap, found in the 1960s, and the formula has been a conjecture ever since. Kleitman proved it in 1970 when the smaller side has at most 6 points. In 1993 Woodall settled K(7,7) and K(7,9) by computer, which also gives K(7,8), K(7,10), K(8,8), K(8,9) and K(8,10). No case has been added since.

The smallest open cases are K(7,11) and K(9,9). If every drawing of K(7,11) has at least 225 crossings, the known averaging argument settles K(7,12), K(8,11) and K(8,12) as well: four new exact crossing numbers, the first new cases in 33 years.

Where it stands Just started: the lab first reproduces the 1993 computer proofs, then scales the same method up to K(7,11).