Let $X$ be a finite ultrametric space and $P(X)$ be the space of probability measures on $X$ endowed with the Wasserstein-Kantorovich-Rubinstein metric (briefly WKR-metric) defined by the formula $$\rho(\mu,\eta)=\max\{|\int_X fd\mu-\int_X fd\eta|:f\in Lip_1(X)\}$$ where $Lip_1(X)$ is the set of non-expanding real-valued functions on $X$.

Problem. Is there any fast algorithm for calculating this metric between two measures on a finite ultrametric space? Or at least for calculating some natural distance, which is not "very far" from the WKR-metric?

Added in Edit. There is a simple upper bound $\hat \rho$ for the WKR-metric, defined by recursion on the cardinality of the set $d[X\times X]=\{d(x,y):x,y\in X\}$ of values of the ultrametric on $X$. If $d[X\times X]=\{0\}$, then for any measures $\mu,\eta\in P(X)$ on $X$ put $\hat\rho(\mu,\eta)=0$. Assume that for some natural number $n$ we have defined the metric $\hat\rho(\mu,\eta)$ for any probability measures $\mu,\eta\in P(X)$ on any ultrametric space $(X,d)$ with $|d[X\times X]|<n$.

Take any ultrametric space $X$ with $|d[X\times X]|=n$. Let $b=\max d[X\times X]$ and $a=\max(d[X\times X]\setminus\{b\})$. Let $\mathcal B$ be the family of closed balls of radius $a$ in $X$. Since $X$ is an ultrametric space, the balls in the family $\mathcal B$ either coincide or are disjoint.

Given any probability measures $\mu,\eta$ on $X$, let $$\hat\rho(\mu,\eta)=\tfrac12b\cdot\sum_{B\in\mathcal B}|\mu(B)-\eta(B)|+\sum_{B\in\mathcal B'}\min\{\mu(B),\eta(B)\}\cdot\hat\rho(\mu{\restriction}B,\eta{\restriction}B),$$ where $\mathcal B'=\{B\in\mathcal B:\min\{\mu(B),\eta(B)\}>0\}$ and the probability measurees $\mu{\restriction} B$ and $\eta{\restriction}B$ assign to each subset $S$ of $B$ the numbers $\mu(S)/\mu(B)$ and $\eta(S)/\mu(B)$, respectively.

It can be shown that $\rho\le\hat\rho$.

Question. Is $\rho=\hat\rho$?

  • 1
    $\begingroup$ Idea: associate a tree to $X$ as in section 7 of arxiv.org/pdf/cond-mat/0603685.pdf and apply techniques from arxiv.org/pdf/1907.00257.pdf. I haven't checked to see if the pieces all fit together though, so caveat emptor. $\endgroup$ Sep 23, 2020 at 17:59
  • $\begingroup$ What do you consider a fast algorithm? Because for any finite optimal transport problem there is always the dumb fall-back of just writing down the corresponding linear programming problem and using one of the known polynomial algorithms on it. $\endgroup$
    – mlk
    Sep 25, 2020 at 6:22
  • 1
    $\begingroup$ @mlk By fast I mean something which will work in a reasonable time on practice. The problem has a motivation in genetics. I need to calculate the distance between two measures on a the space 4^n (with a suitable tree ultrametric) for small n (say n=9). Is the linear programming efficient in this case? $\endgroup$ Sep 25, 2020 at 8:03
  • $\begingroup$ @TarasBanakh Sadly most of my knowledge on the topic is from a single class on discrete optimization I had almost a decade ago, so take everything I say with a grain of salt. With your problem size my guess would be no for the naive implementation as you would have to consider a problem on the set of all pairs of points, i.e. almost (4^n)^2, which are far too many. However you have a lot more structure, which maybe allows you to discard almost all of them. so there might be a chance in this or in the related graph matching algorithms. $\endgroup$
    – mlk
    Sep 25, 2020 at 9:58

2 Answers 2


This is a rather more fun problem than I thought. I must apologize though, as your question is a reference request and I have no references apart from pointing at any textbook on discrete optimization. It turns out, the key is that one can rewrite your problem into a flow problem on a tree, which then is almost trivial to solve. Thus, if I am not mistaken, not only is your upper bound $\hat{\rho}$ the correct value for $\rho$, but the same is true for many other heuristic ways to construct an upper bound. The ultrametric seems to try its best to actively prevent you from accidentally choosing bad solutions and you can use this to define some algorithms which should be almost optimal.


I think the problem is easier to understand in the transport formulation (which is the dual of the one used in the question): $$ \rho(\mu,\eta) := \min \left\{ \int_{X \times X} d(x,y) \,dT : T \in P(X\times X), T(.,X) = \mu,T(X,.)=\eta\right\} $$ i.e. $T(A,B)$ tells us how much mass is transported from $A$ to $B$. I will mostly use this and some derived formulation, but it is good to have both around. In particular, if you have an $f$ for the formulation in the question and a $T$ for this formulation which both give you the same value, you know that both have to be optimal.

Futhermore, we can assume that $\operatorname{supp} \mu \cap \operatorname{supp} \eta = \emptyset$, as transporting from a point to itself is free. In fact, I will not assume that $\mu$ and $\eta$ are probability measures but only that $\mu(X) = \eta(X)$, which works equally well with all definitions and allows us to easily substract similar amounts from both without having to renormalize in every step. In fact in this context it can be useful to consider the signed measure $\nu = \mu -\eta$ instead, which sufficiently describes both of them.

The tree problem

As far as I can gather, any ultrametric can be written in form of a tree (rooted, as used in computer science), where the leaves correspond to the points of $X$ and each subtree to a set of balls containing precisely the points that are its leaves. One can then assign a distance $d_e$ to each edge $e \in E$ of the tree such that the distance between two points in $X$ corresponds to the length of their connecting path through the graph.

One can rewrite finding the WKR-metric into a flow problem on the tree: Extend $\mu$ to the interior nodes by $0$. Now we need to find a flow, i.e. an assignment of a direction and a value $p_e$ to each edge (It is simpler to assume a fixed direction, say upward in the tree and a signed $p_e$ instead) such that in each node $n$ the total of in and outgoing flow corresponds $\nu(n)$. The cost of such a flow then is given by $\sum_e d_e |p_e|$.

The interesting fact about this problem is that on a tree, such a flow is always unique. Also the cost of the unique flow is identical to the WKR-metric. In fact you can recover an $f$ with identical resulting value by assigning a fixed value to a given node $v$ and the recursively setting $f(w) = f(v) \pm d_{(v,w)}$ for all its neighbours, where the sign depends on the direction of the flow. Similarly, you can recover a $T$ by splitting the flow into a sum of weighted paths between leaves and setting $T(\{(x,y)\})$ to the weight of that path. If you take care to never have any cancellation (which is always possible), the corresponding value will again be the same as the cost of the flow.

A fast algorithm given a tree

There are fast algorithms to calculate an optimal flow in graphs, but as we only require the cost of the flow, there is an easy recursive algorithm to calculate it along the tree. For each subtree, we simultaneously construct the internal cost of the flow the flow that leads upwards from it. The total cost then is the internal cost of the whole tree.

  • For each leaf $x$, the internal cost is 0 and the flow upwards is $\nu(x)$.

  • For each subtree, we can recursively calculate internal cost and flow upwards of all of its child trees. The internal cost of the subtree then is the sum of internal costs of its child trees plus the sum of the absolute values of the flows from each of those children multiplied by each respective distance. The flow upwards is simply the sum of all signed flows from the children.

This algorithm only visits each node in the tree once and does a rather simple calculation there, so I'd argue that it is next to optimal. In particular as there are always more children than internal nodes in a tree, it is of order $O(|X|)$. I also believe it is equivalent to the heuristic in the question.

A fast algorithm without a tree

If we do not have the tree structure but are instead only given the distance function, we do not need to calculate the tree. Instead there is a faster way to get to the same value by a simple greedy algorithm:

  • Find the pair of nodes $x,y$ with $\mu(\{x\}) > 0$ and $\eta(\{y\}) > 0$ such that $d(x,y)$ is minimal.
  • Add $d(x,y)\min(\mu(\{x\}),\eta(\{y\}))$ to the total cost and reduce $\mu(\{x\})$ and $\eta(\{y\})$ by $\min(\mu(\{x\}),\eta(\{y\}))$
  • Repeat until $\mu=\eta =0$

If initially one creates a binary heap of all distances this needs a runtime of order $O(|X|^2\log |X|)$. Then in each iteration this algorithm reduces $\operatorname{supp} \mu$ or $\operatorname{supp} \eta$ by a point, so it will run at most for $|X|$ iterations and in doing so remove all elements from the heap again in runtime $O(|X|^2\log |X|)$. As there are a potential $O(|X|^2)$ of distance values to check I'd argue that this again is close to optimal.

The reason why this algorithm returns the right result is evident if one considers the graph in parallel. In each iteration you can add the path between $x$ and $y$ with weight $\min(\mu(\{x\}),\eta(\{y\}))$. When the algorithm finishes, the sum of those paths then gives the flow and one can show that no cancellation occurs. The idea is that the tree is kind of filled from the bottom and a path of minimal distance starting can only ever leave a subtree, if either $\mu$ or $\eta$ is already zero on this subtree, so there will be no future path coming in the opposite direction.

Other distances

A fun observation I had while writing this: At least with Wasserstein-distances, one is generally interested in $d(x,y)^p$ for some $p \in [1,\infty)$ as a cost instead of just $d(x,y)$. But if $d$ is an ultrametric, so is $d^p$, so the whole argument gets adapted easily.

  • $\begingroup$ In the first paragraph of section "The tree problem", you write that the ultrametric can be epressed via the length on the tree. Which metric on the tree do you use? It cannot be the smallest length of the path because it is not an ultrametric. $\endgroup$ Sep 29, 2020 at 18:19
  • 1
    $\begingroup$ @TarasBanakh The path-length metric on the whole is indeed no longer an ultrametric, but it doesn't need to be anymore. It is however identical to the original ultrametric when restricted to the leafs. This is not a contradiction since all counterexamples of path distances not being ultrametrics involve interior nodes. Maybe to clarify, I can choose the edge distances $d_e$ so that in each subtree all leaves are the same path-distance from the root, to be precise half the radius of the smallest ball corresponding to that subtree. $\endgroup$
    – mlk
    Sep 29, 2020 at 19:36

The standard way to quickly approximate Wasserstein distances is to use entropic regularization. Gabriel Peyre and Marco Cuturi wrote a good book on this topic which is available on the Arxiv at https://arxiv.org/abs/1803.00567 (or on Peyre's website). The relevant part is Chapter 4.

However, I'm not sure if there is an extra gain from considering an ultrametric space.

  • 1
    $\begingroup$ Thank you for your answer and the link to the book. Concerning the gain from ultrametric, I think it should be something: if you have two different masses in two disjoint balls, then there is not need to think how to move the difference of masses (it is just multiplied by the distance between any points of these two balls). $\endgroup$ Sep 23, 2020 at 15:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.