Tighter Bounds for Wheeler Determinization

Given a Wheeler NFA $\mathcal{A}$, the Wheeler determinization problem is to construct a Wheeler DFA $\mathcal{D}$ that accepts the same language as $\mathcal{A}$. We use the notation $n_{\mathcal{A}},m_{\mathcal{A}}$ for the number of vertices and edges of $\mathcal{A}$, and equivalently $n_{\mathcal{D}},m_{\mathcal{D}}$ for $\mathcal{D}$. Alanko et al. [SODA 2020, Inf. Comp. 2021] show that we can solve this problem in $O(n_{\mathcal{A}}^3)$ time. In this paper, we show how to improve the running time to $O(n_{\mathcal{A}} + m_{\mathcal{A}} + n_{\mathcal{D}} + m_{\mathcal{D}})$ when given the Wheeler order of $\mathcal{A}$ (which can be computed in $O(m_{\mathcal{A}}\log n_{\mathcal{A}})$ with an algorithm by Becker et al. [ESA 2023]). Our running time is a factor $n_{\mathcal{A}}^2/\sigma$ faster than the state of the art, where $\sigma$ is the size of the alphabet. Furthermore, for $\sigma=O(1)$ we have the first linear time algorithm for this problem. We show that our bound is tight for sorted inputs with any combination of $n$ and $\sigma$, by giving a family of inputs for which our output $\mathcal{D}$ is minimum, and of maximum size $\Theta(n\sigma)$.