My scenario has various flavors of SAT, constrained quadratic pseudo-Boolean, and integer programming. My attempts to formalize and solve the problem with Z3's QF_LIA optimizer have fallen flat, which leads me to wonder if perhaps there is an alternative formulation of the problem, and a solver that might have better performance characteristics. Any advice is appreciated.

I have:

  • A set $V$ of Boolean variables: $V = \{v_i\}$. For a "large" instance, $|V| \approx 6,000$.
  • A set $P \subseteq V \times V$. For the same "large" instance, $|P| \approx 140,000$.
  • A set of constraints $C \subseteq V \times V$: if $(x,y) \in C$, then $x \wedge y \neq 1$. For the same "large" instance, $|C| \approx 4,000$.
  • A weight function $w_1 : V \mapsto \mathbb{N}$ (i.e., a total function on $V$).
  • A weight function $w_2 : P \mapsto \mathbb{N}$, which is mostly unrelated to $w_1$, except that $\forall v_i,v_j. w_2(v_i,v_j) \leq min(w_1(v_i),w_1(v_j))$. $w_2$ can be thought of as either:
    • A total function on $P$
    • A partial function on $V \times V$
    • A total function on $V \times V$, where $w_2(x,y) = 0$ if $(x,y) \notin P$.

I want: an assignment $Asg : V \mapsto \mathbb{B}$ such that:

  • Satisfies all constraints: $\forall (v_i,v_j) \in C$, $Asg(v_i) \wedge Asg(v_j) \neq 1$.
  • Maximizes $\sum_{v_i \in V} Asg(v_i)*w_1(v_i) + \sum_{(v_i,v_j) \in P} Asg(v_i)*Asg(v_j)*w_2(v_i,v_j)$
    • I.e., maximizes $w_1$ on $V$ and $w_2$ on $P$ simultaneously.

2 Answers 2


Can this be reformulated as follows (careful, I am sleep-deprived with a cranky toddler in the room):

We are looking at a random graph $G$ with 6000 weighted nodes, a subgraph $G'$ induced by about 4000 edges, and 140.000 other weighted edges of positive weight but strictly smaller than the nodes they connect to.

A satisfying assignment for $C$ induces a bipartite subgraph of $G'$ (all nodes with assignment 1 plus their immediate neighbors with assignment 0). So you are looking to optimize over the set of bipartite subgraphs of $G'$?

A few random thoughts:

  • How structured are $G$ and $G'$? I presume a random graph model is not a very good model? What is the distribution of degrees of nodes?
  • Are all nodes in $G'$?

Is there a way we can bound the sum of all $w_2$-induced weights for given node? We know that the node weight $w_1(v)$ is bigger than the edge weight $w_2(v, \centerdot)$, is there something to be said for $\sum_{v_2 \in V} w_2(v, v_2)$? That could give some information about whether "switching a node on" is worth it...

If nodes have a tendency to outweigh the sum of their edges, would that hint at a bigger bipartite subgraph being always better than a smaller one, and at the problem becoming one of maximizing the size of the bipartite subgraph?


It's certainly possible to simplify the presentation:

  • A graph $G = (V, E)$
  • A weight function $w_1 : V \mapsto \mathbb{N}$
  • A weight function $w_2 : E \mapsto (\mathbb{N} \cup \{-\infty\})$. This corresponds to your formulation of $w_2$ as a total function on $V \times V$, where $w_2(x,y) = 0$ if $(x,y) \notin P$, except that $w_2(x,y) = - \infty$ if $(x, y) \in C$

Now, every vertex which doesn't have an edge weighted to negative infinity is automatically included in the subset which maximises the total induced weight, so the graph can be preprocessed to only contain vertices with at least one conflict.

The maximum independent set problem is trivially reduced to this one (set $P$ empty and $w_1$ constant), but there's a chance that your instances, by virtue of their sparsity, will decompose nicely.

  • $\begingroup$ This is a very deep answer, thank you. I'm going to have to contemplate it for a while. $\endgroup$
    – Syzygy
    Aug 21, 2019 at 6:35

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