Maximum number of intersections between two simple heptagons

Let $f(m,n)$ denote the maximum number of distinct intersection points between the boundaries of two simple polygons with $m$ and $n$ sides in the plane, restricted to pairs whose boundary intersection is finite.

The polygons need not be convex or in general position. Shared vertices, vertex–side contacts, and tangential contacts are allowed, and each distinct intersection point is counted once. Overlapping boundary segments are excluded by the finiteness requirement.

Problem. Determine the exact value of $f(7,7)$. In particular, is

$$f(7,7)=38?$$

The conjectured formula for odd $m,n\geq 3$ is

$$f(m,n)=mn-(m+n)+3,$$

which predicts 38 for two heptagons. A construction with 38 transverse boundary intersections is known, so the question is whether any pair of simple heptagons with finitely many boundary intersections can exceed this number.

For pairs in general position, the upper bound of Černý et al. gives at most 40 intersections. Since transverse crossings of two simple closed curves occur in even numbers, the general-position case reduces to excluding 40 crossings. Extending a bound from general position to the full problem also requires a reduction that preserves simplicity and does not decrease the number of distinct boundary intersection points; degeneracies are addressed in §9 of the journal version of Ackerman–Keszegh–Rote.