= Solution
Every facial boundary <walk> has length at least five. Bridge excursions can repeat an <edge>, but deleting those excursions leaves a boundary containing a <cycle in a graph>; the finite girth assumption ensures the <graph> is not a <tree>. Thus the shortest-cycle condition still gives the lower bound five even when bridges occur. Summing boundary lengths counts each <edge> twice, so $5f\le2m$. Using <Euler formula for a connected planar graph>,
$$
2=n-m+f\le n-m+\frac{2m}{5}=n-\frac{3m}{5}.
$$
Hence
$$
\boxed{m\le\frac53(n-2).}
$$
Back to article page
