I have a general (undirected) graph with a set of nodes, a set of edges, and a weight for each edge. I want to find a minimum clique cover of the vertices of the graph, that is, a partition of the graph into the smallest number of cliques. A clique is a set of nodes such that each pair of nodes is connected. I also want to maximize the sum of edge weights over the cliques. I want to use an integer programming approach for this problem.

Can any one give me some hints or some references that use mixed integer linear programming for the (maximum weight) minimum clique partition?

Thank you very much.

  • $\begingroup$ Yes I mean a partition of the vertices of the graph. I want to find the smallest number of cliques. A clique is a set of vertices such that each pair of vertices is connected. (a clique is a complete subgraph) $\endgroup$
    – nesrin_21
    Oct 20, 2019 at 11:33

1 Answer 1


I think I have some suggestions. You can make a binary variable $e_{ij}$ per edge corresponding to if those two vertices are in a clique together or not. Being in a clique is a transitive property. $e_{ij} \wedge e_{jk} = e_{ik}$. This "and" constraint is expressible as a MIP. This can be achieved with the constraint $e_{ik} <= e_{ij}$, $e_{ik} <= e_{jk}$, $ e_{ij} + e_{jk} - 1 <= e_{ik}$ link. This is the polytope generated by the truth table vertices (0,0,0) (0,1,0) (1,0,0) (1,1,1). This makes only vertex clique covers feasible. For maximizing edge weights you're good to go.

I don't have an elegant way to minimize the total number of cliques. Just maximizing the sum of all edges is a reasonable heuristic if heuristics are acceptable. Another possibility is to take the sum of all the edges coming out of a vertex. This is the size of the clique it belongs to and is bounded by the total number of edges connected to that vertex. $\sum_j e_{ij} = c_i - 1$, where $c_i$ is the size of the clique vertex $i$ belongs to. The sum over all vertices $\sum_j \frac{1}{c_i} = C$ is the total number of cliques $C$. I don't have a good way to perform this inversion, except by MIP brute force.

This can be done by generating as many binary variables $a_{in}$ per vertex as it has neighbors, setting $\sum_n n a_{ni}=c_i$, enforcing that only one $a_{ni}$ is nonzero $\sum_n a_{ni} = 1$, and then making the objective $\sum_{jn}\frac{1}{n}a_{nj}$. This is rather dissatisfying to me and I hope there is a better way.


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