Ultrametric notes


Draft
Posted on October 13, 2025 by Yu Cong
Tags: ,

https://drops.dagstuhl.de/entities/document/10.4230/LIPIcs.APPROX/RANDOM.2023.2

1 ultrametrics

Definition1ultrametrics

A metric space \((X,d)\) is a ultrametric if it satisfies the strong triangle inequality: \[ d(x,y)\leq \max\{d(x,z),d(z,y)\} \quad \forall x,y,z\in X \]

Ultrametric is a special case of tree metric. To see this, consider the equivalence relation \(x \sim_{\leq r} y\) if \(d(x,y)\leq r\). For each fixed \(r\) there will be many equivalent classes. We collect all equivalent class for all possible \(r\). Let this collection of subsets be \(L\). Note that we dont really have to consider all values of \(r\) but only smallest \(r\) such that the set of equivalent classes changes.

Observation2

\(L\) is laminar.

Now we have a tree-like structure. The root of the tree is the unique equivalent class of \(\sim_{\leq \max d(x,y)}\). The leaves will be equivalent classes of \(\sim_{\leq 0}\), i.e., singleton of elements in \(X\). Then the laminar strucuture of our collection \(L\) naturally forms a tree.

Note that

  1. every leaf in this tree has the same height.
  2. the LCA of \(x\) and \(y\) represents a equivalent class of \(\sim_{\leq d(x,y)}\).

We have to decide the edge weights. Let \(e\) be an edge between nodes \(u\) and \(v\) which are some equivalent class of \(\sim_{\leq a}\) and \(\sim_{\leq b}\). Then the weight of this edge is \(\frac{1}{2}(a-b)\) (assuming \(a\geq b\)).

Now as (2) shows, the distance of leaves in the tree is \[\begin{equation*} \begin{aligned} d_T(x,y)&=d_T(\mathop{\mathrm{LCA}}(x,y),x) + d_T(\mathop{\mathrm{LCA}}(x,y),y)\\ &=\frac{1}{2}(d(x,y) -0)+\frac{1}{2}(d(x,y) -0)\\ &=d(x,y) \end{aligned} \end{equation*}\]

Now i guess we have invented the so-called Hierarchically Separated Trees (HST). Any ultrametric can be converted into a HST in this way.

1.1 exact \(c\)-HST

In the paper the input ultrametric is limited to a special kind called exact \(c\)-HST. Let \(c>1\) be a constant. The tree structure is the same as HST but the edge weight is \(w((u,v))=\frac{1}{2}(c^{depth(u)}-c^{depth(v)})\). Let \(x,y\) be two leaves in a exact \(c\)-HST, then \(d(x,y)=c^{depth(\mathop{\mathrm{LCA}}(x,y))}\).

The key to restricting input to an exact \(c\)-HST is the following lemma.

Lemma3

Given a metric space and its embedding into a distribution over ultrametrics with distortion \(\alpha\), there is an embedding into a distribution over exact \(c\)-HSTs with distortion \(\alpha\cdot c\).

An intuitive but failed proof method We try to show that any ultrametric can be embedded into an exact \(c\)-HST with distortion \(c\). Proving the distortion can be reduced to the following combinatorial problem. Given a sequence of \(n\) strictly increasing numbers \(\{a_i\}_{i\in [n]}\), find another sequence \(c^{x_1},\dots, c^{x_n}\) where \(x_i\leq x_j\) for all \(i<j\), such that \(\max_i \frac{a_i}{c^{x_i}}\leq c \min_i \frac{a_i}{c^{x_i}}\). Now one can see that this is not always true for any fixed \(c\)… so embedding into a distribution may be important.