How to Complete a Doubling Metric

In recent years, considerable advances have been made in the study of properties of metric spaces in terms of their doubling dimension. This line of research has not only enhanced our understanding of finite metrics, but has also resulted in many algorithmic applications. However, we still do not understand the interaction between various graph-theoretic (topological) properties of graphs, and the doubling (geometric) properties of the shortest-path metrics induced by them. For instance, the following natural question suggests itself: \emph{given a finite doubling metric $(V,d)$, is there always an \underline{unweighted} graph $(V',E')$ with $V\subseteq V'$ such that the shortest path metric $d'$ on $V'$ is still doubling, and which agrees with $d$ on $V$.} This is often useful, given that unweighted graphs are often easier to reason about. We show that for any metric space $(V,d)$, there is an \emph{unweighted} graph $(V',E')$ with shortest-path metric $d':V'\times V' \to \R_{\geq 0}$ such that -- for all $x,y \in V$, the distances $d(x,y) \leq d'(x,y) \leq (1+\eps) \cdot d(x,y)$, and -- the doubling dimension for $d'$ is not much more than that of $d$, where this change depends only on $\e$ and not on the size of the graph. We show a similar result when both $(V,d)$ and $(V',E')$ are restricted to be trees: this gives a simpler proof that doubling trees embed into constant dimensional Euclidean space with constant distortion. We also show that our results are tight in terms of the tradeoff between distortion and dimension blowup.