For an undirected unweighted graph $G=(V,E)$ with $n$ vertices and $m$ edges, let $d(u,v)$ denote the distance from $u\in V$ to $v\in V$ in $G$. An $(\alpha,\beta)$-stretch approximate distance oracle (ADO) for $G$ is a data structure that given $u,v\in V$ returns in constant (or near constant) time a value $\hat d (u,v)$ such that $d(u,v) \le \hat d (u,v) \le \alpha\cdot d(u,v) + \beta$, for some reals $\alpha >1, \beta$. If $\beta = 0$, we say that the ADO has stretch $\alpha$. Thorup and Zwick~\cite{thorup2005approximate} showed that one cannot beat stretch 3 with subquadratic space (in terms of $n$) for general graphs. P\v{a}tra\c{s}cu and Roditty~\cite{patrascu2010distance} showed that one can obtain stretch 2 using $O(m^{1/3}n^{4/3})$ space, and so if $m$ is subquadratic in $n$ then the space usage is also subquadratic. Moreover, P\v{a}tra\c{s}cu and Roditty~\cite{patrascu2010distance} showed that one cannot beat stretch 2 with subquadratic space even for graphs where $m=\tilde{O}(n)$, based on the set-intersection hypothesis. In this paper we explore the conditions for which an ADO can be stored using subquadratic space while supporting a sub-2 stretch. In particular, we show that if the maximum degree in $G$ is $\Delta_G \leq O(n^{1/2-\varepsilon})$ for some $0<\varepsilon \leq 1/2$, then there exists an ADO for $G$ that uses $\tilde{O}(n^{2-\frac {2\varepsilon}{3}})$ space and has a sub-2 stretch. Moreover, we prove a conditional lower bound, based on the set intersection hypothesis, which states that for any positive integer $k \leq \log n$, obtaining a sub-$\frac{k+2}{k}$ stretch for graphs with maximum degree $\Theta(n^{1/k})$ requires quadratic space. Thus, for graphs with maximum degree $\Theta(n^{1/2})$, obtaining a sub-2 stretch requires quadratic space.

翻译：对于具有 $n$ 个顶点和 $m$ 条边的无向无权图 $G=(V,E)$，令 $d(u,v)$ 表示 $G$ 中从 $u\in V$ 到 $v\in V$ 的距离。$G$ 的一个 $(\alpha,\beta)$-伸缩比近似距离查询器（ADO）是一种数据结构，对于给定的 $u,v\in V$，能在常数（或近似常数）时间内返回一个值 $\hat d (u,v)$，使得 $d(u,v) \le \hat d (u,v) \le \alpha\cdot d(u,v) + \beta$，其中 $\alpha >1, \beta$ 为实数。若 $\beta = 0$，则称该 ADO 具有伸缩比 $\alpha$。Thorup 和 Zwick~\cite{thorup2005approximate} 表明，对于一般图，无法用次二次空间（以 $n$ 计）突破伸缩比 3。P\v{a}tra\c{s}cu 和 Roditty~\cite{patrascu2010distance} 表明，使用 $O(m^{1/3}n^{4/3})$ 空间可以获得伸缩比 2，因此若 $m$ 相对于 $n$ 是次二次的，则空间使用量也是次二次的。此外，P\v{a}tra\c{s}cu 和 Roditty~\cite{patrascu2010distance} 基于集合交假设指出，即使对于 $m=\tilde{O}(n)$ 的图，也无法用次二次空间突破伸缩比 2。本文探索了 ADO 在支持亚2伸缩比的同时使用次二次空间存储的条件。特别地，我们证明若 $G$ 的最大度满足 $\Delta_G \leq O(n^{1/2-\varepsilon})$，其中 $0<\varepsilon \leq 1/2$，则存在一个使用 $\tilde{O}(n^{2-\frac {2\varepsilon}{3}})$ 空间且具有亚2伸缩比的 ADO。此外，我们基于集合交假设证明了一个条件性下界：对于任意正整数 $k \leq \log n$，为最大度为 $\Theta(n^{1/k})$ 的图获得亚$\frac{k+2}{k}$伸缩比需要二次空间。因此，对于最大度为 $\Theta(n^{1/2})$ 的图，获得亚2伸缩比需要二次空间。