🤖 AI Summary
This study investigates the expressive power and model-checking complexity of monadic second-order logic (MSO) over permutation structures. While first-order logic (FO) cannot express natural permutation properties—such as the existence of an increasing subsequence or the number of connected components—we show that MSO can capture them; however, key properties like the existence of a fixed point remain undefinable in MSO. Methodologically, we adopt tree-width as a parameterized complexity measure and design a structure-decomposition-based model-checking algorithm, achieving satisfiability testing in time $f(|varphi|, mathrm{tw}(pi)) cdot n$. Our theoretical contributions are threefold: (i) we establish a strict expressive advantage of MSO over FO on permutations; (ii) we characterize the fixed-parameter tractability boundary for MSO model checking on permutations; and (iii) we prove that this problem remains W[1]-hard on several natural permutation classes, even under bounded tree-width.
📝 Abstract
Permutations can be viewed as pairs of linear orders, or more formally as models over a signature consisting of two binary relation symbols. This approach was adopted by Albert, Bouvel and F'eray, who studied the expressibility of first-order logic in this setting. We focus our attention on monadic second-order logic. Our results go in two directions. First, we investigate the expressive power of monadic second-order logic. We exhibit natural properties of permutations that can be expressed in monadic second-order logic but not in first-order logic. Additionally, we show that the property of having a fixed point is inexpressible even in monadic second-order logic. Secondly, we focus on the complexity of monadic second-order model checking. We show that there is an algorithm deciding if a permutation $pi$ satisfies a given monadic second-order sentence $varphi$ in time $f(|varphi|, operatorname{tw}(pi)) cdot n$ for some computable function $f$ where $n = |pi|$ and $operatorname{tw}(pi)$ is the tree-width of $pi$. On the other hand, we prove that the problem remains hard even when we restrict the permutation $pi$ to a fixed hereditary class $mathcal{C}$ with mild assumptions on $mathcal{C}$.