Simple Length-Constrained Expander Decompositions
Greg Bodwin, Bernhard Haeupler, D Ellis Hershkowitz, Zihan Tan
TL;DR
This work addresses the existence of length-constrained expander decompositions by proving a simplified existence result and tightening the sparsity loss. The authors develop a dispersion/counting framework to bound the arboricity of $s$-parallel-greedy graphs, enabling a tighter union-of-cuts sparsity bound and removing reliance on exponential demand structures. They show that for any $h\ge1$, $s\ge2$, $\phi>0$ and graph $G$, there exists an $(h,s)$-length $\phi$-expander decomposition with cut slack $O\left(s\cdot n^{O(1/s)}\right)$ (and length slack $s$), with a generalized form for edge-capacitated graphs and arbitrary node-weightings yielding $O\left(s\cdot|A|^{2/s}\right)$ cut slack. These results enhance the toolkit for fast graph algorithms by enabling near-linear-time constructions of expanders and related decompositions, with direct implications for fast min-cost flow, distance oracles, and dynamic graph problems.
Abstract
Length-constrained expander decompositions are a new graph decomposition that has led to several recent breakthroughs in fast graph algorithms. Roughly, an $(h, s)$-length $φ$-expander decomposition is a small collection of length increases to a graph so that nodes within distance $h$ can route flow over paths of length $hs$ while using each edge to an extent at most $1/φ$. Prior work showed that every $n$-node and $m$-edge graph admits an $(h, s)$-length $φ$-expander decomposition of size $\log n \cdot s n^{O(1/s)} \cdot φm$. In this work, we give a simple proof of the existence of $(h, s)$-length $φ$-expander decompositions with an improved size of $s n^{O(1/s)}\cdot φm$. Our proof is a straightforward application of the fact that the union of sparse length-constrained cuts is itself a sparse length-constrained cut. In deriving our result, we improve the loss in sparsity when taking the union of sparse length-constrained cuts from $\log ^3 n\cdot s^3 n^{O(1/s)}$ to $s\cdot n^{O(1/s)}$.
