# Generating uniform trees

A long time ago, I wrote quite a few a things about uniform trees. That is, a uniform choice from the $n^{n-2}$ unrooted trees with vertex set [n]. This enumeration, normally called Cayley’s formula, has several elegant arguments, including the classical Prufer bijection. But making a uniform choice from a large set is awkward, and so we seek more probabilistic methods to sample such a tree, which might also give insight into the structure of a ‘typical’ uniform tree.

In another historic post, I talked about the Aldous-Broder algorithm. Here’s a quick summary. We run a random walk on the complete graph $K_n$ started from a uniformly-chosen vertex. Every time we arrive at a vertex we haven’t visited before, we record the edge just traversed. Eventually we have visited all n vertices, so have recorded n-1 edges. It’s easy enough to convince yourself that these n-1 edges form a tree (how could there be a cycle?) and a bit more complicated to decide that the distribution of this tree is uniform.

It’s worth noting that this algorithm works to construct a uniform spanning tree on any connected base graph.

This post is about a few alternative constructions and interpretations of the uniform random tree. The first construction uses a Galton-Watson process. We take a Galton-Watson process where the offspring distribution is Poisson(1), and condition that the total population size is n. The resulting random tree has a root but no labels, however if we assign labels in [n] uniformly at random, the resulting rooted tree has the uniform distribution among rooted trees on [n].

Proof

This is all about moving from ordered trees to non-ordered trees. That is, when setting up a Galton-Watson tree, we distinguish between the following two trees, drawn extremely roughly in Paint:

That is, it matters which of the first-generation vertices have three children. Anyway, for such a (rooted) ordered tree T with n vertices, the probability that the Galton-Watson process ends up equal to T is

$\mathbb{P}(GW = T) = \prod_{v\in T} \frac{e^{-1}}{C(v)!} = e^{-n} \prod_{v\in T}\frac{1}{C(v)!},$

where $C(v)$ is the number of children of a vertex $v\in T$. Then, since $\mathbb{P}( |GW|=n )$ is a function of n, we find

$\mathbb{P}(GW=T \,\big|\, |GW|=n) = f(n)\prod_{v\in T} \frac{1}{C(v)!},$

where f(n) is a function of n alone (ie depends on T only through its size n).

But given an unordered rooted tree t, labelled by [n], there are $\prod_{v \in t} C(v)!$ ordered trees associated to t in the natural way. Furthermore, if we take the Poisson Galton-Watson tree conditioned to have total population size n, and label uniformly at random with [n], we obtain any one of these ordered trees with probability $\frac{f(n)}{n!} \prod_{v\in t} \frac{1}{C(v)!}$. So the probability that we have t after we forget about the ordering is $\frac{f(n)}{n!}$, which is a function of n alone, and so the distribution is uniform among the set of rooted unordered trees labelled by [n], exactly as required.

Heuristic for Poisson offspring distribution

In this proof, the fact that $\mathbb{P}(C(v)=k)\propto \frac{1}{k!}$ exactly balances the number of orderings of the k children explains why Poisson(1) works out. Indeed, you can see in the proof that Poisson(c) works equally well, though when $c\ne 1$, the event we are conditioning on (namely that the total population size is n) has probability decaying exponentially in n, whereas for c=1, the branching process is critical, and the probability decays polynomially.

We can provide independent motivation though, from the Aldous-Broder construction. Both the conditioned Galton-Watson construction and the A-B algorithm supply the tree with a root, so we’ll keep that, and look at the distribution of the degree of the root as constructed by A-B. Let $\rho=v_1,v_2,v_3,\ldots$ be the vertices [n], ordered by their discovery during the construction. Then $\rho$ is definitely connected by an edge to $v_2$, but thereafter it follows by an elementary check that the probability $\rho$ is connected to $v_m$ is $\frac{1}{n-1}$, independently across all m. In other words, the distribution of the degree of $\rho$ in the tree as constructed by A-B is

$1+ \mathrm{Bin}\left(n-2,\frac{1}{n-1}\right) \approx 1+\mathrm{Poisson}(1).$

Now, in the Galton-Watson process, conditioning the tree to have fixed, large size changes the offspring distribution of the root. Conveniently though, in a limiting sense it’s the same change as conditioning the tree to have size at least n. Since these events are monotone in n, it’s possible to take a limit of the conditioning events, and interpret the result as the Galton-Watson tree conditioned to survive. It’s a beautiful result that this interpretation can be formalised as a local limit. The limiting spine decomposition consists of an infinite spine, where the offspring distribution is a size-biased version of the original offspring distribution (and so in particular, always has at least one child) and where non-spine vertices have the original distribution.

In particular, the number of the offspring of the root is size-biased, and it is well-known and not hard to check that size-biasing Poisson(c) gives 1+Poisson(c) ! So in fact we have, in an appropriate limiting sense in both objects, a match between the degree distribution of the root in the uniform tree, and in the conditioned Galton-Watson tree.

This isn’t supposed to justify why a conditioned Galton-Watson tree is relevant a priori (especially the unconditional independence of degrees), but it does explain why Poisson offspring distributions are relevant.

Construction via G(N,p) and the random cluster model

The main reason uniform trees were important to my thesis was their appearance in the Erdos-Renyi random graph G(N,p). The probability that vertices {1, …, n} form a tree component in G(N,p) with some particular structure is

$p^{n-1} (1-p)^{\binom{n}{2}-(n-1)} \times (1-p)^{n(N-m)}.$

Here, the first two terms give the probability that the graph structure on {1, …, n} is correct, and the the final term gives the probability of the (independent) event that these vertices are not connected to anything else in the graph. In particular, this has no dependence on the tree structure chosen on [n] (for example, whether it should be a path or a star – both examples of trees). So the conditional distribution is uniform among all trees.

If we work in some limiting regime, where $pn\rightarrow 0$ (for example if n is fixed and $p=\frac{1}{N}\rightarrow 0$), then we can get away asymptotically with less strong conditioning. Suppose we condition instead just that [n] form a component. Now, there are more ways to form a connected graph with one cycle on [n] than there are trees on [n], but the former all require an extra edge, and so the probability that a given one such tree-with-extra-edge appears as the restriction to [n] in G(N,p) is asymptotically negligible compared to the probability that the restriction to [n] of G(N,p) is a tree. Naturally, the local limit of components in G(N,c/N) is a Poisson(c) Galton-Watson branching process, and so this is all consistent with the original construction.

One slightly unsatisfying aspect to this construction is that we have to embed the tree of size [n] within a much larger graph on [N] to see uniform trees. We can’t choose a scaling p=p(n) such that G(n,p) itself concentrates on trees. To guarantee connectivity with high probability, we need to take $p> \frac{\log n}{n}$, but by this threshold, the graph has (many) cycles with high probability.

At this PIMS summer school in Vancouver, one of the courses is focusing on lattice spin models, including the random cluster model, which we now briefly define. We start with some underlying graph G. From a physical motivation, we might take G to be $\mathbb{Z}^d$ or some finite subset of it, or a d-ary tree, or the complete graph $K_N$. As in classical bond percolation (note G(N,p) is bond percolation on $K_N$), a random subset of the edges of G are included, or declared open. The probability of a given configuration w, with e open edges is proportional to

$p^e (1-p)^{|E(G)| - e} q^{k(w)},$ (*)

where the edge-weight $p\in(0,1)$ as usual, and cluster weight $q\in (0,\infty)$, and $k(w)$ counts the number of connected components in configuration w. When q=1, we recover classical bond percolation (including G(N,p) ), while for q>1, this cluster-reweighting favours having more components, and q<1 favours fewer components. Note that in the case $q\ne 1$, the normalising constant (or partition function) of (*) is generally intractable to calculate explicitly.

As in the Erdos-Renyi graph, consider fixing the underlying graph G, and taking $p\rightarrow 0$, but also taking $\frac{q}{p}\rightarrow 0$. So the resulting graph asymptotically ‘wants to have as few edges as possible, but really wants to have as few components as possible’. In particular, 1) all spanning trees of G are equally likely; 2) any configuration with more than one component has asymptotically negligible probability relative to any tree; 3) any graph with a cycle has #components + #edges greater than that of a tree, and so is asymptotically negligible probability relative to any tree.

In other words, the limit of the distribution is the uniform spanning tree of G, and so this (like Aldous-Broder) is a substantial generalisation, which constructs the uniform random tree in the special case where $G=K_n$.

# Gromov-Hausdorff Distance on Trees

This post continues the exposition of Gromov-Hausdorff distance, as introduced in Chapter Four of Steve Evans’ Probability and Real Trees, which we are reading as a group in the Stats Dept in Oxford at the moment. In this post, we consider applications of the Gromov-Hausdorff distance we have just introduced in the context of trees, viewed as metric spaces.

First we consider a direct application of the previous result, which related the Gromov-Hausdorff distance to the infimum of the distortion across the family of correspondences between the two relevant metric spaces. I found this as Corollary 3.7 in notes by Le Gall and Miermont on scaling limits of random trees and maps, which can be found here. I’m not clear whether there is an original source, but the result is simple enough that probably it does not matter hugely.

Proposition – Given f,g excursions above [0,1], and $T_f, T_g$ the real trees associated with these excursions in the standard way, then

$d_{GH}(T_f,T_g)\le 2 ||f-g||.$

Proof: We construct an appropriate correspondence

$\mathcal {R}=\left\{(a,b) \,:\, \exists t\in[0,1]\text{ s.t. } a=p_f(t), b=p_g(t)\right\}.$

In other words, the trees are defined as projections from [0.1] (with some equivalence structure), so taking pairs of projections from [0,1] gives a natural correspondence. Now, suppose $(p_f(s),p_g(s)), (p_f(t),p_g(t))\in \mathcal {R}$. Then

$d_{T_f}(p_f(s),p_f(t)) = f(s) + f(t) - 2\hat f(s,t),$

where $\hat f(s,t):= \min{r\in[s,t]}f(r)$. Obviously, we can replace f by g to obtain

$d_{T_g}(p_g(s),p_g(t)) = g(s) + g(t) - 2\hat g(s,t).$

By thinking slightly carefully about where the functions achieve their minima compared with where the minimum in the sup norm is achieved, we can conclude that

$|d_{T_f}(p_f(s),p_f(t)) - d_{T_g}(p_g(s),p_g(t)) | \le 4||f-g||,$

and so the result follows from the relation between Gromov-Hausdorff distance and the infimum of distortions over the set of correspondences proved at the end of the previous post.

Gromov-Hausdorff Limits of Real Trees

If we are going to consider the Gromov-Hausdorff topology for limits of tree, we want to be sure that the limit of a sequence of real trees is another real tree. In particular, we want to show that this convergence preserves the property of being a geodesic space.

Theorem 4.19 – Given $X_n$ geodesic spaces, and $X$ complete, such that $X_n \stackrel{d_{GH}}\rightarrow X$, then X is geodesic.

Proof: Here, I’ll go in the opposite order to Evans’ book, as I think it’s easier understand why a special case of this implies the whole result once you’ve actually shown that special case.

Anyway, we want to show that given $x,y\in X$, there is a geodesic segment $x\leftrightarrow y$ in X. We will start by showing that there is a well-defined midpoint of the geodesic, that is a point $z \in X$ such that $d(x,z)=d(y,z)=\frac12 d(x,y)$.

Given arbitrary $\epsilon>0$, we can take n such that $d_{GH}(X,X_n) < \frac{\epsilon}{3}$ and then a correspondence $\mathcal{R}$ between $X,X_n$ with $\mathrm{dis}\mathcal {R}<\frac{2\epsilon}{3}$. Now, by definition of correspondence, we have $x',y'\in X_n$ such that $(x,x'),(y,y')\in\mathcal{R}$. But we do have geodesics in $X_n$, so we can take $z'\in X_n$ the midpoint of geodesic $x'\leftrightarrow y'$. Predictably, we now take $z\in X$ such that $(z,z')\in\mathcal {R}$.

We can show that z is ‘almost’ the midpoint of [x,y] in the sense that

$|d(x,z) - \frac12 d(x,y)| \le |d(x',z')-\frac12 d(x',y')+\frac32 \mathrm{dis}\mathcal{R} \le \epsilon.$

Similarly, we have $|d(y,z)-\frac12 d(x,y)|\le \epsilon$.

Perhaps it’s helpful to think of this point z that we’ve constructed as being like a taut, but not quite rigid string. The midpoint of the string has to be fairly near the midpoint of the endpoints, and in particular, as we let $\epsilon\rightarrow 0$, the z’s we deal with form a Cauchy sequence, and thus converge to some point, which (in a case of poor notation planning) we also call $z \in X$, which is the midpoint of $[x,y]$ as described before.

We can now iterate this iteration, to demonstrate that whenever q is a dyadic rational in [0,1], there exists $z_q \in X$ such that

$d(x,z_q)=q d(x,y),\quad d(y,z_q) = (1-q)d(x,y).$

Again then, if we want the above to hold for some general real r in [0,1], we can approximate r arbitrarily well by dyadic rationals q, and the associated points $z_q$ are Cauchy and thus have a well-defined limit with the required properties. We thus have our geodesic segment $x\leftrightarrow y$.

Rooted Gromov-Hausdorff Distance

In the end, the trees we want to compare might be rooted. For example, we talk about finite trees being invariant under random re-rooting, and we might be interested in similar results for real trees, in particular the BCRT. So we need to compare metric spaces as viewed outwards from particular identified points of each space.

An isometry of rooted spaces must map the root to the root, and so we adjust to obtain rooted Gromov-Hausdorff distance. We might try to consider embeddings into a common space so that the roots are shared, but it will be more convenient to maintain the infimum over metrics on the disjoint union as before. But to ensure the roots are in roughly the ‘same’ place in both set embeddings, we minimise the maximum of the Hausdorff distance between the sets, and the distance between the images of the roots in the common covering space.

Similar results apply as in the unrooted case, and normally the proofs are very similar. As we might expect, when defining a correspondence between rooted spaces, we demand that the pair of roots is one of the roots in the correspondence, and then the same equivalence between the minimal distortion and the rooted G-H distance applies.

Evans shows that $\mathbb{T}^{\mathrm{root}}$, the set of compact rooted trees is complete and separable under the rooted G-H topology. Separability is relatively easy to see. Compact trees have finite $\epsilon$-nets, and there is a canonical way to view the net as the vertices of a finite tree with edge lengths. Approximating these edge lengths by rationals and consider the countable family of isomorphism classes of rooted finite trees gives separability.

If we want, we can also define k-pointed Gromov-Hausdorff distance, where we demand k points in each space are held fixed.

Tree $\eta$-erasure

To show that this is a natural topology to consider for the family of trees, Evans devotes a short section to the operation of $\eta$-erasure, where all points within $\eta$ of a leaf are removed from a given tree.

Formally, $R_\eta$ is a map $\mathbb{T}^{\mathrm{root}}\rightarrow \mathbb{T}^{\mathrm{root}}$ (recalling that these are compact real trees), so that $R_\eta(T)$ consists of the tree

$\{\rho\}\cup\{a \in T \, : \, \exists x\in T,\, a\in[\rho, x],d_T(a,x)\ge \eta\},$

rooted again at $\rho$. We claim that the range of $R_\eta$ is the set of compact rooted trees with a finite number of leaves. In this setting, we want a geodesic definition of leaf, for example a leaf is a point that doesn’t lie in the interior of the (unique) geodesic segment between any other point and the root.

Given a tree T with a finite number of leaves, we can glue disjoint segments of length $\eta$ onto every leaf. Taking $R_\eta$ of this deeper tree will give T. Similarly, suppose $R_\eta(T)$ has infinitely many leaves, which we can label $a_1,a_2,\ldots$. Thus we also have $x_1,x_2,\ldots \in T$ such that $a_i\in[\rho,x_i]$, and the segments $\{[a_i,x_i]\}$ are disjoint, and all have length at least $\eta$, hence T cannot be compact, as it has no finite $\frac{\eta}{2}$ net.

It is clear that for fixed T, the family of these maps applied to T is continuous with respect to $\eta$ in the G-H topology. When $\eta$ is changed a small amount, the amount of extra tree removed is locally small, and so approximating correspondences by points in what is left is absolutely fine. Indeed, the operations $T_{\eta_1}, T_{\eta_2}$ commute to give $T_{\eta_1+\eta_2}$.

We want to show that for fixed $\eta$ this map $R_\eta$ is continuous with respect to G-H topology. Suppose two compact rooted trees S,T are covered by a rooted correspondence $\mathcal R$ with distortion $\epsilon \ll \eta$. We can’t immediately restrict $\mathcal R$ to $R_\eta(S)\times R_\eta(T)$, as it won’t necessarily be a surjection under the projection maps any more.

But we can get around this. Note that if $(x,y)\in\mathcal{R}$, then $|d_S(\rho_S,x) - d_T(\rho_T,y)| \le \epsilon$ by assumption. We will construct a correspondence $\mathcal R'$ on the erased trees as follows. Given $(x,y)\in\mathcal{R}$, if $(x,y)\in R_\eta(S)\times R_\eta(T)$, we keep it, and if $x\not\in R_\eta(S), y\not \in R_\eta(y)$, we throw it away. Suppose we have $(s,t) in \mathcal{R}$ with $s\in R_\eta(S), t\not \in R_\eta(T)$. Let $l_s$ be a leaf of S such that $s \in [\rho_S,l_s]$, and let $\bar t$ be the farthest point from the root of the geodesic $[\rho_T,t]$ restricted to $R_\eta(T)$. So the tree above $\bar t$ includes t and has height at most $\eta$.

In words, if a point appears in a pair in the correspondence but is removed by the erasure, we replace it in the pair with the point closest to the original point that was not removed by the erasure.

It remains to prove that this works. My original proof was short but false, and its replacement is long (and I hope true now), but will postpone writing this down either until another post, or indefinitely. The original proof by Evans and co-authors can be found as the main content of Lemma 2.6 in the original paper [1].

REFERENCES

[1] Evans, Pitman, Winter – Rayleigh processes, real trees, and root growth with regrafting.

# Random Mappings for Cycle Deletion

In previous posts here and here, I’ve talked about attempts to describe a cycle deleting process. We amend the dynamics of the standard random graph process by demanding that whenever a cycle is formed in the graph we delete all the edges that lie on the cycle. The aim of this is to prevent the system growing giant components, and perhaps give a system that displays the characteristics of self-organised criticality. In the posts linked to, we discuss the difficulties caused by the fact that the tree structure of components in such a process is not necessarily uniform.

Today we look in the opposite direction. It gives a perfectly reasonable model to take a multiplicative coalescent with quadratic fragmentation (this corresponds to cycle deletion, since there are $O(n^2)$ edges which would give a cycle if added to a tree on n vertices) and a fragmentation kernel corresponding to adding an extra edge to a uniform spanning tree on n vertices then deleting the edges of the unique cycle. The focus of the rest of this post, we consider this fragmentation mechanism, in particular thinking about how we would sample from it most practically. Not least, without going through Prufer codes or some other clever machinery, it is not trivial to sample a uniform spanning tree.

First, we count the number of unicyclic graphs on n labelled vertices. If we know that the vertices on the cycle are $v_1,\ldots,v_k$, then the number of cycles with an identified edge is

$u_1=1,\quad u_k=\frac{k!}{2},\, k\ge 2.$

If we know that the tree coming off the cycle from vertex v_i has size m, say, then each of the possible rooted labelled trees with size m is equally likely. So taking $w_j=j^{j-1}$, the number of rooted trees on j labelled vertices, we get $B_n(u_\bullet,w_\bullet)$ for the number of such unicyclic graphs on [n]. Recall $B_n$ is the nth Bell polynomial, which gives the size of a compound combinatorial structure, where we have some structure on blocks and some other structure within blocks. Then the random partition of [n] given by the tree sizes has the distribution $\text{Gibbs}_n(u_\bullet,w_\bullet)$.

Consider now a related object, the so-called random mapping digraph. What follows is taken from Chapter 9 of Combinatorial Stochastic Processes. We can view any mapping $M_n:[n]\rightarrow[n]$ as a digraph where every vertex has out-degree 1. Each such digraph contains a collection of directed cycles, supported on those elements x for which $M_n^k(x)=x$ for some k. Such an element x is called a cyclic point. Each cyclic point can be viewed as the root of a labelled tree.

In an identical manner to the unicyclic graph, the sizes of these directed trees in the digraph decomposition of a uniform random mapping is distributed as $\text{Gibbs}_n(\bullet !,w_\bullet)$. So this is exactly the same as the cycle deletion kernel, apart from in the probability that the partition has precisely one block. In practice, for large n, the probability of this event is very small in both cases. And if we wanted to sample the cycle deletion kernel exactly, we could choose the trivial partition with some probability p, and otherwise sample from the random mapping kernel, where p is chosen such that

$p+\frac{1-p}{B_n(\bullet !, w_\bullet)}=\frac{1}{B_n(u_\bullet,w_\bullet)}.$

At least we know from the initial definition of a random mapping, that $B_n(\bullet !,w_\bullet)=n^n$. The number of unicyclic graphs with an identified edge is less clear. It turns out that the partition induced by the random mapping has a nice limit, after rescaling, as the lengths of excursions away from 0 in the standard Brownian bridge on [0,1].

The time for a fuller discussion of this sort of phenomenon is in the context of Poisson-Dirichlet distributions, as the above exchangeable partition turns out to be PD(1/2,1/2). However, for now we remark that the jumps of a subordinator give a partition after rescaling. The case of a stable subordinator is particularly convenient, as calculations are made easier by the Levy-Khintchine formula.

A notable example is the stable-1/2 subordinator, which can be realised as the inverse of the local time process at zero of a Brownian motion. The jumps of this process are then the excursion lengths of the original Brownian motion. A calculation involving the tail of the w_j’s indicates that 1/2 is the correct parameter for a subordinator to describe the random mappings. Note that the number of blocks in the partition corresponds to the local time at zero of the Brownian motion. (This is certainly not obvious, but it should at least be intuitively clear why a larger local time roughly indicates more excursions which indicates more blocks.)

So it turns out, after checking some of the technicalities, that it will suffice to show that the rescaled number of blocks in the random mapping partition $\frac{|\Pi_n|}{\sqrt{n}}$ converges to the Raleigh density, which is a size-biased Normal random variable (hence effectively first conditioned to be positive), and which also is the distribution of the local time of the standard Brownian bridge.

After that very approximate description, we conclude by showing that the distribution of the number of blocks does indeed converge as we require. Recall Cayley’s formula $kn^{n-k-1}$ for the number of labelled forests on [n] with a specified set of k roots. We also need to know how many labelled forests there are with any set of roots. Suppose we introduce an extra vertex, labelled 0, and connect it only to the roots of a rooted labelled forest on [n]. This gives a bijection between unlabelled trees on {0,1,…,n} and labelled forests with a specified set of roots on [n]. So we can use Cayley’s original formula to conclude there are $(n+1)^{n-1}$ such forests. We can do a quick sanity check that these are the same, which is equivalent to showing

$\sum_{k=1}^n k n^{-k-1}\binom{n}{k}=\frac{1}{n}(1+\frac{1}{n})^{n-1}.$

This odd way of writing it is well-motivated. The form of the LHS is reminiscent of a generating function, and the additional k suggests taking a derivative. Indeed, the LHS is the derivative

$\frac{d}{dx}(1+x)^n,$

evaluated at $\frac{1}{n}$. This is clearly the same as the RHS.

That said, having established that the random mapping partition is essentially the same, it is computationally more convenient to consider that instead. By the digraph analogy, we again need to count forests with k roots on n vertices, and multiply by the number of permutations of the roots. This gives:

$\mathbb{P}(|\Pi_n|=k)=\frac{kn^{n-k-1}\cdot k! \binom{n}{k}}{n^n}=\frac{k}{n}\prod_{i=1}^{k-1}\left(1-\frac{i}{n}\right).$

Now we can consider the limit. Being a bit casual with notation, we get:

$\lim \mathbb{P}(\frac{|\Pi_n|}{\sqrt{n}}\in dl)\approx \sqrt{n}dl \mathbb{P}(|\Pi_n|=l\sqrt{n}).$

Since the Raleigh distribution has density $l\exp(-\frac12 l^2)dl$, it suffices for this informal verification to check that

$\prod_{i=1}^{l\sqrt{n}}(1-\frac{i}{n})\approx \exp(-\frac12 l^2).$ (*)

We take logs, so the LHS becomes:

$\log(1-\frac{1}{n})+\log(1-\frac{2}{n})+\ldots+\log(1-\frac{l\sqrt{n}}{n}).$

If we view this as a function of l and differentiate, we get

$d(LHS)=\sqrt{n}dl \log (1-\frac{l}{\sqrt{n}})\approx \sqrt{n}dl \left[-\frac{l}{\sqrt{n}}-\frac{l^2}{2n}\right]\approx -ldl.$

When l is zero, the LHS should be zero, so we can obtain the desired result (*) by integrating then taking an exponential.

# Bijections, Prufer Codes and Cayley’s Formula

I’m currently at the training camp in Cambridge for this year’s UK IMO squad. This afternoon I gave a talk to some of the less experienced students about combinatorics. My aim was to cover as many useful tricks for calculating the sizes of combinatorial sets as I could in an hour and a half. We started by discussing binomial coefficients, which pleasingly turned out to be revision for the majority. But my next goal was to demonstrate that we are much more interested in the fact that we can calculate these if we want than in the actual expression for their values.

Put another way, my argument was that the interpretation of $\binom{n}{m}$ as the number of ways to choose m objects from a collection of n, or the number of up-and-right paths from (0,0) to (m,n) is more useful than the fact that $\binom{n}{m}=\frac{n!}{m!(n-m)!}$. The opening gambit was to prove the fundamental result underlying the famous construction of Pascal’s triangle that

$\binom{n+1}{m+1}=\binom{n}{m}+\binom{n}{m+1}.$

This is not a hard result to prove by manipulating factorials, but it is a very easy result to prove in the path-counting setting, for example.

So it turned out that the goal of my session, as further supported by some unsubtly motivated problems from the collection, was to convince the students to use bijections as much as possible. That is, if you have to count something awkward, show that counting the awkward thing is equivalent to counting something more manageable, then count that instead. For many simpler questions, this equivalence is often drawn implicitly using words (“each of the n objects can be in any subset of the collection of bags so we multiply…” etc), but it is always worth having in mind the formal bijective approach. Apart from anything else, asking the question “is this bijection so obvious I don’t need to prove it” is often a good starting-point for assessing whether the argument is in fact correct!

Anyway, I really wanted to show my favouriite bijection argument, but there wasn’t time, and I didn’t want to spoil other lecturers’ thunder by defining a graph and a tree and so forth. The exploration process encoding of trees is a strong contender, but today I want to define quickly the Prufer coding for trees, and use it to prove a famous result I’ve been using a lot recently, Cayley’s formula for the number of spanning trees on the complete graph with n vertices, $n^{n-2}$.

We are going to count rooted trees instead. Since we can choose any vertex to be the root, there are $n^{n-1}$ rooted trees on n vertices. The description of the Prufer code is relatively simple. Take a rooted tree with vertices labelled by [n]. A leaf is a vertex with degree 1, other than the root. Find the leaf with the largest label. Write down the label of the single vertex to which this leaf is connected, then delete the leaf. Now repeat the procedure, writing down the label of the vertex connected to the leaf now with the largest label, until there are only two vertices remaining, when you delete the non-root vertex, and write down the label of the root. We get a string of (n-1) labels. We want to show that this mapping is a bijection from the set of rooted trees with vertices labelled by [n] to $[n]^{n-1}$.

Let’s record informally how we would recover a tree from the Prufer code. First, observe that the label of any vertex which is not a leaf must appear in the code. Why? Well, the root label appears right at the end, if not earlier, and every vertex must be deleted. But a vertex cannot be deleted until it has degree one, so the neighbours further from the root (or ancestors) of the vertex must be removed first, and so by construction the label appears. So know what the root is, and what the leaves are straight away.

In fact we can say slightly more than this. The number of times the root label appears is the degree of the root, while the number of times any other label appears is the degree of the corresponding vertex minus one. Call this sequence the Prufer degrees.

So we construct the tree backwards from the leaves towards the root. We add edges one at a time, with the k-th edge joining the vertex with the k-th label to some other vertex. For k=1, this other vertex is the leaf with maximum label. In general, let $G_k$ be the graph formed after the addition of k-1 edges, so $G_1$ is empty, and $G_n$ is the full tree. Define $T_k$ to be the set of vertices such that their degree in $G_k$ is exactly one less than their Prufer degree. Note that $T_1$ is therefore the set of leaves suggested by the Prufer code. So we form $G_{k+1}$ by adding an edge between the vertex with label appearing at position k+1 in the Prufer sequence and the vertex of $T_k$ with maximum label.

Proving that this is indeed the inverse is a bit fiddly, more because of notation than any actual mathematics. You probably want to show injectivity by an extremal argument, taking the closest vertex to the root that is different in two trees with the same Prufer code. I hope it isn’t a complete cop out to swerve around presenting this in full technical detail, as I feel I’ve achieved by main goal of explaining why bijection arguments can reduce a counting problem that was genuinely challenging to an exercise in choosing sensible notation for proving a fairly natural bijection.

# Random Walks and Spanning Trees

Introduction

In this post, I’m going to talk about probability distributions on graphs. In particular, consider paths on a graph, and how to assign a distribution to these in a natural way. One way is to consider a standard random walk, treating the vertices as states of a discrete Markov chain. But, in many situations, this isn’t really enough. We might want paths, that is, walks which are self-avoiding. But even in regular lattices like $\mathbb{Z}^d$, it is hard to say a great deal about the set of self-avoiding walks (SAWs) of length n. We would prefer a distribution which has a natural product form, or which at least we can sample from without large combinatorial calculations.

A spanning tree is a connected graph without cycles. The set of edges is maximal, in the sense that adding any further edge creates a cycle, and also minimal, as removing one will disconnect the graph. Counting the number of spanning trees on labelled vertices is harder than one might suspect, and is possibly worth a post by itself. However, in general the uniform distribution on spanning trees is a useful object to consider. Any spanning tree contains a unique path between two vertices of the graph, and so a Uniform Spanning Tree (UST) induces a distribution on paths.

An alternative construction is to take a random walk and remove the cycles. This is not well-defined unless you specify a canonical order in which to remove them, and the obvious option is to remove the cycles in the order in which they appear. That is, every time you end up at a vertex which you have already visited, you remove all the edges traversed since you were last at v. This gives a Loop-Erased Random Walk (LERW), and from this another measure on paths between two vertices (subject to connectivity conditions which guarantee that the LERW almost surely hits the target vertex eventually).

At an informal level, the difference between these measures is significant. Inducing a measure down from a natural but potentially unwieldy uniform distribution is theoretically natural, but hard to work with. On the other hand a computer could sample from LERW, just by performing a random walk and storing the history suitably, which immediately makes it useful. So, the following theorem is elegant in its own right, but in particular as a bridge between these two frameworks.

Theorem: The measures on paths from to in a finite graph G induced by UST and generated by LERW are the same.

Proof: Some books give proofs which involve coupling a LERW construction with a sequence of STs directly, driven by an underlying random walk on the graph. The difficulty in this approach is that to prove that the uniform distribution on spanning trees is stationary often requires the random walk to be in equilibrium, a criterion which causes difficulties when it come to starting at x later in the proof. Essentially, the difficulty in many LERW constructions is that the edges being removed are incident to the wrong vertex in the random walk, and so RW equilibrium is required to show that the spanning tree transitions are reversible. Instead, we proceed by an argument motivated by the fact that LERWs appear ‘backwards’ in UST constructions. Continue reading