Recall that a graph is called $k$-chordal if any cycle $C$ of length $> k$ contains a chord, i.e. an edge joining to non-consecutive vertices in $C$. Let $f(n, k)$ be the minimal number of edges s.t. any graph on $n$ vertices can be made $k$-chordal by deleting at most $f(n, k)$ edges.
There is a paper of Erd\H{o}s and Laskar "On maxinal chordal subgraph" where they show that $$ f(n,3) = \frac{n^2}{2} - (1+o(1))\sqrt{2}n^{3/2}. $$
I am wondering if other bounds for $f(n, k)$ are known for bigger $k$. A bound in terms of the number of edges is also of interest. In particular, is is true that there is some bounded $k$ s.t. $$ f(n, k) = o(|E|), $$ where $E$ is the number of edges?
UPD
The answer for the last question is NO, as seen from the following example. Take $k+1$ copies of $K_{n/(k+1)}$, call it $G_0, ..., G_k$. Join all vertices in $G_i$ and $G_{i+1}$ (addition mod $(k+1)$). One can show that for this graph $f(n, k) \geq n^2/4(k+1)^2$. It is interesting to see what is the right constant, though.