I am currently working on the paper A branch-and-cut algrithm for the Edge Inderdiction Clique Problem.
Basically, the problem asks to find a subset of at most $k$ edges to remove from a graph $G$, so that the size of maximum clique, clique number $\omega(G)$, in the remaining, called interdiction graph, is minimized.
In this paper, a single-level formulation of the problem is presented and various inequalities are derived. To solve the problem they develop a customized branch and cut algorithm.
Page 12 deals with the separation of the inequalities
$\sum_{e \in E(K^*)} w_e \ge \sum_{l'=l_{\min}+1}^{l^*}f(l',K^*) \cdot z_{l'}$
where $w_e \in \{0,1\}^E$ is the edge interdiction policy with $\sum_{uv \in E} w_{uv} \le k$, $K^*$ a clique, $z_{l'} = 1$ if every clique of size $l'$ or larger in $G$ is beeing interdicted by the choosen edges of $w_e$ and $f(l', K^*)$ is the minimal number of edges that has to be removed from $K^*$ so that the clique number of the remaining graph is at most $l'-1$.
In the corresponding ILP there is another inequality that ensures that only one $z_l$ is set to $1$ and the other to $0$. The second sum only tightens the inequality but could also be omitted for the sake of simplicity.
Unfortunately, I do not understand in the paper how a part of the separation works. On page 12 you will find the exact separation of these inequalities. Thats ok. Then, they describe a heuristic that uses maximal cliques. I just do not understand this part.
During the separation of the inequalities, a non-maximal clique of G can be obtained since the current interdiction policy may prevent some vertices to be included in the clique $K \in \mathcal{K}$.
I understand how to generate and add cuts based on these maximal cliques, but not where to get these maximal cliques from. Sure, I could pick any node and then calculate greedy a maximal clique, but as I understand it, it says that you get an initial set from somewhere.