Logo
All Random Solved Random Open
OPEN
Let $G$ be a chordal graph on $n$ vertices - that is, $G$ has no induced cycles of length greater than $3$. Can the edges of $G$ be partitioned into $n^2/6+O(n)$ many cliques?
Asked by Erdős, Ordman, and Zalcstein [EOZ93], who proved an upper bound of $(1/4-\epsilon)n^2$ many cliques (for some very small $\epsilon>0$). The example of all edges between a complete graph on $n/3$ vertices and an empty graph on $2n/3$ vertices show that $n^2/6+O(n)$ is sometimes necessary.

A split graph is one where the vertices can be split into a clique and an independent set. Every split graph is chordal. Chen, Erdős, and Ordman [CEO94] have shown that any split graph can be partitioned into $\frac{3}{16}n^2+O(n)$ many cliques.