Breaking the $2^n$ Barrier for Counting Linear Extensions with a Short Elementary Algorithm
TL;DR
A deterministic exact algorithm is given that counts the linear extensions of an arbitrary $n$-element poset in time $O^*(1.89^n)$ where $O^*(\cdot)$ suppresses polynomial factors.
Abstract
A linear extension of a finite partially ordered set is a total ordering that respects the partial order. We give a deterministic exact algorithm that counts the linear extensions of an arbitrary $n$-element poset in time $O^*(1.89^n)$, where $O^*(\cdot)$ suppresses polynomial factors. This breaks the $2^n$ barrier for the general problem and resolves a question explicitly posed by Koivisto at Dagstuhl 2013. The proof refines an argument of Kozma for two-dimensional posets. A chain partition handles the case in which the poset is sufficiently far from an antichain. Otherwise, fix a maximum antichain (a largest set of pairwise incomparable elements). For each of its elements that has a comparable element above it outside the antichain, we record only which such element appears first. A decoding lemma enumerates the resulting patterns from their multiplicities. Once a pattern is fixed, each antichain element has a release condition and at most one deadline, so the dynamic program stores only the number of released elements in each deadline class. A stars-and-bars count bounds the total number of states.