# 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 $G=K_n$.

# 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.

# Long Paths and Expanders

I’m in Birmingham this week for the LMS-EPSRC summer school on Random Graphs, Geometry and Asymptotic Structure. The event consists of three five-hour mini-courses, a plenary lecture, leaving plenty of time for problem sheet and discussion. I thought it would be worth trying to say a couple of interesting things each day – I do not know whether this will succeed, but I might as well try.

Today, a few thoughts on the first two lectures of Michael Krivelevich’s course on Long Paths and Hamiltonicity in Random Graphs. The aim is to develop tools to investigate the threshold for the presence of a Hamiltonian cycle in G(n,p). In this first part of the course, we were mainly thinking about long paths.

One tool we used a lot was the Depth-First Search algorithm. This is very similar to the exploration process I’ve talked about before. Essentially, here we consider trying to explore the graph in a depth-first way, but instead of viewing all the edges incident to a vertex we have just arrived at, we only look to see whether there is an edge out of the new vertex. If there is, we explore it, then come back eventually to look for more. It really comes down to a difference in the information we are storing. In this DFS, we store the vertices which we haven’t finished exploring, which is the set of vertices on the explored path between the root and the current vertex. So the size of this set evolves like the contour process. In particular, we can read off the sizes of paths from this description. These dynamics are useful in particular because we know there are no edges between the set of vertices we have finished exploring, and the ones we have yet to explore. The stack of ‘processing’ vertices must glue everything else together.

We can translate one of the arguments back into the language for the old exploration process. Recall the increments of the exploration process are $\mathrm{Bin}(\alpha n,\frac{c}{n}) -1$ once we have explored $\alpha n$ vertices. We don’t need to worry about the -1 bit for now. Observe that because we are exploring in a depth-first way, if a subsequence of the Binomial variables of length k are all positive, this corresponds to a path of length (k-1).

So to prove, for example, that the longest path in a subcritical random graph is O(log n), it suffices to prove that there are O(log n) consecutive positive entries in the sequence of n binomial entries. Since the distribution changes continuously, it is convenient to prove that there are O(log n) consecutive positive entries in the first $\epsilon n$ binomial entries. The probability that any of these entries is positive is bounded below by some p, so it suffices to consider instead a sequence of Bernoulli RVs with parameter p. So if we never have clog n consecutive, this gives control of the sequence of geometric random variables corresponding to the gaps between 0s in the sequence. Precisely, these are Geom(q), and we must have $\frac{\epsilon n}{c\log n}$ of them independently being less than clog n. We have to chase a few constants, and use the fact that if $f(n)\rightarrow\infty, \frac{g(n)}{f(n)}\rightarrow\infty$, then

$(1-\frac{1}{f(n)})^{g(n)}\rightarrow 0,$

by comparison with the standard asymptotic result for $e^{-x}$. In any case, we get that this probability tends to 0 if we choose c small enough, and so with high probability there is a path of length clog n.

This is interesting, because we knew already that the largest component in a subcritical random graph had size O(log n). But we also knew that all the components were trees, or ‘almost trees’, and were uniformly chosen from the set of trees (or trees + an edge or two) with appropriate size. And the largest path in a UST on n vertices is $O(n^{1/2})$ with high probability. So we learn that there are enough components of size $\geq c\log n$ that it is actually very probable that one of them will have the unlikely property of being much more path-like than a typical tree.

Krivelevich also showed a pleasant elementary proof of the result that a supercritical random graph has a path of length O(n), using a similar idea.

The other definition of major interest was an expander graph. Often when doing calculations about neighbourhoods of sets of vertices, we run into the problem that the neighbourhoods may overlap, and so we cannot get the total outer neighbourhood (or outer boundary) just by summing over the individual neighbourhood sizes. In an expander graph, we demand that all small sets of vertices have neighbourhood at least as large as some constant multiple of the set size, essentially giving us a bound on the above problem. Concretely, G is a $(k,\alpha)$-expander is for any set of vertices $|U|\leq k, |N(U)|\geq \alpha |U|$.

There’s a very nice argument using Posa’s lemma, where we consider all the possible ways to rearrange the vertices in some longest path into a different longest path, and then focus on the endpoints of all these paths. With this so-called rotation-extension technique, we can show that a (k,2)-expander has a path of length at least 3k-1.

There are structural similarities between expander graphs and regular graphs, so it seems natural that there will be some interesting spectral properties. I don’t know much about this, but perhaps it will come up later in the week. But, returning to the random graph long path problem, it now suffices to show subcritical G(n,p) is a (clog n,2)-expander for some c. Expander properties are in some sense the opposite of clustering properties, and independence of a RG inhibit most clustering properties (as discussed in much greater detail in some of the posts about network models). Unfortunately, this doesn’t actually work, as in a subcritical graph, the typical expansion coefficient, even of a small set will be c, for G(n,c/n), which is not large enough. However, if you chose the constants carefully, such an argument should work for c>2, so long as you chose k=an, with a small enough that the probability of a vertex elsewhere in the graph being joined to (at least) two of the k vertices in the set, was small compared with (c-2).

REFERENCES

The course notes are not available, though chapter 3 from these 2010 notes by the same lecturer are related and interesting.

# Diameters of Trees and Cycle Deletion

In the past two posts, we introduced two models of random trees. The Uniform Spanning Tree chooses uniformly at random from the set of spanning trees for a given underlying graph. The Minimum Spanning Tree assigns IID weights to each edge in the underlying graph, then chooses the spanning tree with minimum induced total weight. We are interested to know whether these are in fact the same distribution, and if they are not, what properties can be used to distinguish them asymptotically.

While investigating my current research problem, I was interested in the diameter of large random trees under various models. Specifically, I am considering what happens if you take a standard Erdos-Renyi process on n vertices, where edges appear at constant rate between pairs of vertices chosen uniformly at random, and add an extra mechanism to prevent the components becoming too large. For this particular model, our mechanism consists of removing any cycles as they are formed. Thus all the components remain trees as time advances, so it is not unreasonable to think that there might be some sort of equilibrium distribution.

Now, by definition, any tree formed by the Erdos-Renyi process is a uniform tree. Why? Well, the probability of a configuration is determined entirely by the number of edges present, so once we condition that a particular set of vertices are the support of a tree, all possible tree structures are equally likely. Note that this relies on sampling at a single fixed time. If we know the full history of the process, then it is no longer uniform. For example, define a k-star to be a tree on k vertices where one ‘centre’ vertex has degree k-1. The probability that a uniform tree on k vertices is a k-star is $\frac{k}{k^{k-2}}=k^{-(k-3)}$. But a star can only be formed by successively adding single vertices to an existing star. That is, we cannot join a 3-tree and a 4-tree with a edge to get a 7-star. So it is certainly not immediately clear that once we’ve incorporated the cycle deletion mechanism, the resulting trees will be uniform once we condition on their size.

In fact, the process of component sizes is not itself Markovian. For a concrete example, observe first that there is, up to isomorphism, only one tree on any of {0,1,2,3} vertices, so the first possible counterexample will be splitting up a tree on four vertices. Note that cycle deletion always removes at least three edges (ie a triangle), so the two possibilities for breaking a 4-tree are:

(4) -> (2,1,1) and (4) -> (1,1,1,1)

I claim that the probabilities of each of these are different in the two cases: a) (4) is formed from (2,2) and b) (4) is formed from (3,1). This is precisely a counterexample to the Markov property.

In the case where (4) is formed from (2,2), the 4-tree is certainly a path of length 4. Therefore, with probability 1/3, the next edge added creates a 4-cycle, which is deleted to leave components (1,1,1,1). In the case where (4) is formed from (3,1), then with probability 2/3 it is a path of length 4 and with probability 1/3 it is a 4-star (a ‘T’ shape). In this second case, no edge can be added to make a 4-cycle, so after cycle deletion the only possibility is (2,1,1). Thus the probability of getting (1,1,1,1) is 2/9 in this case, confirming that the process is non-Markovian. However, we might remark that we are unlikely to have O(n) vertices involved in fragmentations until at least the formation of the giant component in the underlying E-R process, so it is possible that the cycle deletion process is ‘almost Markov’ for any property we might actually be interested in.

When we delete a cycle, how many vertices do we lose? Well, for a large tree on n vertices, the edge added which creates the cycle is chosen uniformly at random from the pairs of vertices which are not currently joined by an edge. Assuming that n is w(1), that is we are thinking about a limit of fairly large trees, then the number of edges present is much smaller than the number of possible edges. So we might as well assume we are choosing uniformly from the possible edges, rather than just the possible edges which aren’t already present.

If we choose to add an edge between vertices x and y in the tree, then a cycle is formed and immediately deleted. So the number of edges lost is precisely the length of the path between x and y in the original tree. We are interested to know the asymptotics for this length when x and y are chosen at random. The largest path in a graph is called the diameter, and in practice if we are just interested in orders of magnitude, we might as well assume diameter and expected path length are the same.

So we want to know the asymptotic diameter of a UST on n vertices for large n. This is generally taken to be $n^{1/2}$. Here’s a quick but very informal argument that did genuinely originate on the back of a napkin. I’m using the LERW definition. Let’s start at vertex x and perform LERW, and record how long the resultant path is as time t advances. This is a Markov chain: call the path length at time t $X_t$.

Then if $X_t=k$, with probability $1-\frac k n$ we get $X_{t+1}=k+1$, and for each j in {0,…,k-1}, with probability 1/n we have $X_{t+1}=j$, as this corresponds to hitting a vertex we have already visited. So

$\mathbb{E}\Big[X_{t+1}|X_t=k\Big]=\frac{nk-k^2/2}{n}.$

Note that this drift is positive for $k<< \sqrt n$ and negative for $k>>\sqrt n$, so we would expect $n^{-1/2}$ to be the correct scaling if we wanted to find an equilibrium distribution. And the expected hitting time of vertex y is n, by a geometric distribution argument, so in fact we would expect this Markov chain to be well into the equilibrium window with the $n^{-1/2}$ scaling by the time this occurs. As a result, we expect the length of the x to y path to have magnitude $n^{1/2}$, and assume that the diameter is similar.

So this will be helpful for calculations in the cycle deletion model, provided that the trees look like uniform trees. But does that even matter? Do all sensible models of random trees have diameter going like $n^{1/2}$? Well, a recent paper of Addario-Berry, Broutin and Reed shows that this is not the case for the minimum spanning tree. They demonstrate that the diameter in this case is $n^{1/3}$. I found this initially surprising, so tried a small example to see if that shed any light on the situation.

The underlying claim is that MSTs are more likely to be ‘star-like’ than USTs, a term I am not going to define. Let’s consider n=4. Specifically, consider the 4-star with centre labelled as 1. There are six possible edges in K_4 and we want to see how many of the 6! weight allocations lead to this star. If the three edges into vertex 1 have weights 1, 2 and 3 then we certainly get the star, but we can also get this star if the edges have weights 1, 2 and 4, and the edge with weight 3 lies between the edges with weights 1 and 2. So the total number of possibilities for this is 3! x 3! + 3! x 2! = 48. Whereas to get a 4-path, you can assign weights 1, 2 and 3 to the edges of the path, or weights 1, 2 and 4 provided the 4 is not in the middle, and then you have the 3 joining up the triangle formed by 1 and 2. So the number of possibilities for this is 3! x 3! + 4 x 2! = 44.

To summarise in a highly informal way, in a star-like tree, you can ‘hide’ some fairly low-scoring weights on edges that aren’t in the tree, so long as they join up very low-scoring edges that are in the tree. Obviously, this is a long way from getting any formal results on asymptotics, but it does at least show that we need to be careful about diameters if we don’t know exactly what mechanism is generating the tree!

# Mixing Times 6 – Aldous-Broder Algorithm and Cover Times

In several previous posts, I’ve discussed the Uniform Spanning Tree. The definition is straightforward: we choose uniformly at random from the set of trees which span a fixed underlying graph. But for a dense underlying graph, there are a very large number spanning trees. Cayley’s formula says that the complete graph K_n has $n^{n-2}$ spanning trees, so to select from this list is impractical.

We seek a better algorithm. In a post about a year ago, I presented the result that the path between two fixed points x and y in the UST is distributed as the path generated by Loop-Erased Random Walk, for which we start at x and delete cycles as they appear. An initial problem might be that this only gives us a single path, which might be enough in some contexts, but in general we will want to specify the whole tree. Wilson’s Algorithm is an unsurprising but useful extension to this equivalence which does just that. You start by constructing the LERW between two vertices, then you add the LERW which connects some other vertex to the path you already have. Then you take a further vertex not currently explored and start LERW there, continuing until you hit the tree that you already have. Iterate this process, which must terminate after at most n steps when there are no vertices which to start from. The tree thus obtained is the UST. The tricky part is proving that the method for selecting which unused vertices to start from has no effect on the distribution of paths between two fixed points.

I want to consider a different algorithm, discovered roughly simultaneously by Aldous and Broder. Start a random walk on the underlying graph at some particular vertex. Every time we traverse an edge which takes us to a vertex we haven’t yet explored, add this edge to the tree. For now I don’t want to give a proof that this algorithm works, but rather to talk about how fast it works, because it ties in nicely with something from the Mixing Times book we’ve been reading recently. It is clear that the algorithm terminates at the first time the random walk has visited every vertex. This is a stopping time, called the cover time of the Markov chain. If we are working with an underlying complete, then we notice that this is annoying, because it means that the cover time will increase like n.log n. That is, it will take an increasingly long time to gather the final few vertices into the tree. Perhaps some combination of Aldous-Broder initially then Wilson’s method for the final o(n) vertices might be preferable?

I want to discuss how to treat this cover time. Often we have information about the hitting times of states from other states $\mathbb{E}_x T_y$. A relationship between S, the hitting time, defined to be the maximum of the previous display over x and y, and the expected cover time would be useful, especially for a highly symmetric graph like the complete graph where the expected hitting times are all the same.

Matthews’ Method relates these two for an irreducible finite Markov chain on n states. It says:

$t_{cov}\leq t_{hit}\left(1+\frac12+\ldots+\frac 1 n\right).$

We first remark that this agrees with what we should get for the random walk on the complete graph. There, the hitting time of x from y is a geometric random variable with success probability 1/n, hence expectation is n. The cover time is the standard coupon collector problem, giving expectation n log n, and the sum of reciprocals factor is asymptotically a good approximation.

The intuition is that if we continue until we hit state 1, then reset and continue until we hit state 2, and so on, by the time we hit state n after (n-1) iterations, this is a very poor overestimate of the cover time, because we are actually likely to have hit most states many times. What we want to do really is say that after we’ve hit state 1, we continue until we hit state 2, unless we’ve already done so, in which case we choose a different state to aim for, one which we haven’t already visited. But this becomes complicated because we then need to know the precise conditional probabilities of visiting any site on the way between two other states, which will depend rather strongly on the exact structure of the chain.

Peres et al give a coupling proof in Chapter 11 of their book which I think can be made a bit shorter, at least informally. The key step is that we still consider hitting the sites in order, only now in a random order.

That is, we choose a permutation $\sigma\in S_n$ uniformly at random, and we let $T_k$ be the first time that states $\sigma(1),\ldots,\sigma(k)$ have all been visited. This is a random time that is measurable in the product space, and for each $\sigma$ it is a stopping time.

The key observation is that $\mathbb{P}(T_{k+1}=T_k)=1-\frac{1}{k+1}$. This holds conditional on any path of the Markov chain because the requirement for the event is that $\sigma(k+1)$ is visited after $\{\sigma(1),\ldots,\sigma(k)\}$. The statement therefore holds as stated as well as just pathwise. Then, by the SMP, conditional on $\{T_{k+1}>T_k\}$, we have

$T_{k+1}-T_k \leq_{st} t_{hit}.$

Note that by the definition of $t_{hit}$, this bound on the hitting time $T_{k+1}$ is unaffected by concerns about where the chain actually is at $T_k$ (since it is not necessarily at $\sigma(k)$).

So, removing the conditioning, we have:

$\mathbb{E}\Big[T_{k+1}-T_k\Big]\leq\frac{1}{k+1}t_{hit},$

and so the telescoping sum gives us Matthews’ result.

One example is the cover time of random walk on the n x n torus, which turns out to be

$O(n^2(\log n)^2).$

If anyone remembers that Microsoft screensaver from many years ago which started with a black screen and a snake leaving a trail of white pixels as it negotiated the screen, this will be familiar. The last few black bits take a frustratingly long while to disappear. Obviously that isn’t quite a random walk, but it perhaps diminishes the surprise that it should take this long to find the cover time.

There are a couple of interesting things I wanted to say about electrical networks for Markov chains and analytic methods for mixing times, but the moment may have passed, so this is probably the last post about Mixing Times. Plans are in motion for a similar reading group next term, possible on Random Matrices.

# Minimum Spanning Trees

In my last post, I discussed the Uniform Spanning Tree. To summarise very briefly, given a connected graph on n vertices, a tree is a subgraph, that is a subset of the edges, which is connected, but which contains no cycles. It turns out this requires the tree to have n-1 edges.

We are interested in natural mechanisms for generating randomly chosen spanning trees of a given graph. One way we can always do this is to choose uniformly at random from the set of possible trees. This UST is in some sense canonical, but it is worth knowing about some other measures on trees that might be of interest.

A family of natural problems in operations research concerns an arbitrary complex network, with some weight or cost associated to each connection. The question is how to perform some operation on the network so as to minimise the resulting cost. Perhaps the most famous such problem is that of the Travelling Salesman. The story is that a salesman needs to visit n locations and wants to do the trip as efficiently as possible. This might be thought of as some sort of financial or time cost, but proably the easiest way to set it up is to imagine he is trying to minimise the distance he has to travel. It is not hard to see why this problem might genuinely arise in plenty of real-world situations, where a organisation or agent is trying to be as efficient as possible.

It might be the case that it is not possible to travel between every pair of locations, but we needn’t assume that for now. So if he knows the distance between any pair of cities, he wants to know which of the possible routes gives the shortest overall distance. The problem is that there are n! routes, and this grows roughly like n^n, which is faster than exponential, so for as few as 20 cities it has turned into a comparison which is too large to compute.

There are various algorithms which reduce the number of routes that must be checked, and some approximation methods. But if you want the exact answer, it is not currently possible to calculate this in polynomial time.

Minimal Spanning Trees and Uniqueness

For the travelling salesman, we were looking for the minimal cost spanning path. In the case of the complete graph, this is the same as the minimal cost non-repeating path of length n-1. Such paths are a subset of the set of spanning trees on the underlying graph. So what if we look instead for the minimal cost spanning tree? This exists as after all, there are only finitely many spanning trees.

So far, this has been deterministic, but we were looking for a random spanning tree. We can achieve this by choosing the weights at random. Anything other than assigning the weights as an IID sequence seems likely to be complicated, but there isn’t a canonical choice of the distribution of the weights. Our first question will be whether the distribution of the weights affects the distribution of the induced MST. In fact it will turn out that so long as the distribution is continuous, it has no effect on the distribution of the MST. The continuous condition might seem odd, but it is present only to ensure that the weights almost certainly end up generating a unique MST.

It turns out that there is a straightforward greedy algorithm to find the MST once the weights are known. We will examine some consequences of this algorithm in the random setting. First we check uniqueness. The condition required for uniqueness is that the weights be distinct. Note that this is slightly weaker than the statement that all of sums of (n-1)-tuples be distinct, which immediately implies a unique MST.

We now prove this condition. Suppose we have distinct weights, and an associated MST. If the underlying graph is a tree, then the result is clear. Otherwise, add some extra edge e, with weight w(e). By the definition of a tree, this generates exctly one cycle. Consider the other edges, say $e_1,\ldots,e_k$ in this cycle. If any of $w(e_i)>w(e)$ then we can replace e_i with e to get a spanning tree with smaller weight, a contradiction of the claim that we started with an MST. So by distinctness of weights, we conclude that $w(e)>w(e_i)$ for all i.

Conversely, suppose we remove some edge e which IS in the MST. We end up with exactly two connected components. Consider all the edges in the underlying graph between the two components, and suppose that one of these f satisfies w(f)<w(e). Then if we add in edge f, which is by construction not in the original MST, we end up with a smaller total weight than we started with, a further contradiction.

We can summarise this in a neat form. Given an edge e between x and y, consider the set of all edges in the underlying graph with weight LESS THAN w(e). Then if x and y are in different components, the edge e must be in the MST. Since we have an explicit description of which edges are present, it follows that the MST is unique. The problem is that working out the component structure of the graph with higher weights removed is computationally rather intensive. We want a slightly faster algorithm.

Kruskal’s Algorithm

Several rather similar algorithms were developed roughly simultaneously. Prim’s algorithm is a slight generalisation of what we will discuss. Anyway, for now we consider Kruskal’s algorithm which has the advantage that it can be described without really needing to draw a diagram.

We start by ordering the weights. Without loss of generality, we might as well relabel the edges so that

$w(e_1)< w(e_2)<\ldots< w(e_{|E|}).$

Now, by the condition derived in the argument for uniqueness, we must have e_1 and e_2 in any MST. Now consider e_3. Unless doing so would create a cycle, add e_3. Then, unless doing so would create a cycle, add e_4. Continue. It is clear that the result of this procedure is acyclic. To check it is actually a spanning tree, we show that it is also connected. Suppose not, and two of the components are A and B. Let e be the edge between A and B with minimal weight. According to the algorithm, we should have included e in our MST because at no point would adding it possibly have created a cycle. So we have proved that this greedy algorithm does indeed give the (unique) MST.

A useful consequence of this is that we know the two edges with overall minimum weight are definitely in the MST. In the search for a random measure on spanning trees, what is most important is that we didn’t use the actual values of the weights in this construction, only the order. In other words, we might as well have assumed the weights were a random permutation from $S_{|E|}$. This now answers our original question about how the random weight MST depends in distribution on the underlying edge weight distribution. So long as with probability one the weights are distinct (which holds if the distribution is continuous), then the distribution of the resulting spanning tree is constant.

It’s not too hard to show this isn’t the same as UST: n=4 suffices as a counterexample. But the difference in asymptotic behaviour of properties such as the diameter is of interest, and will be explored in the next post.

# Uniform Spanning Trees

For applications to random graphs, the local binomial structure and independence means that the Galton-Watson branching process is a useful structure to consider embedding in the graph. In several previous posts, I have shown how we can set up the so-called exploration process which visits the sites in a component as if the component were actually a tree. The typical degree is O(1), and so in particular small components will be trees with high probability in the limit. In the giant component for a supercritical graph, this is not the case, but it doesn’t matter, as we ignore vertices we have already explored in our exploration process. We can consider the excess edges separately by ‘sprinkling’ them back in once we have the tree-like backbone of all the components. Again, independence is crucial here.

I am now thinking about a new model. We take an Erdos-Renyi process as before, with edges arriving at some fixed rate, but whenever a cycle appears, we immediately delete all the edges that make up the cycle. Thus at all times the system consists of a collection (or forest) of trees on the n vertices. So initially this process will look exactly like the normal E-R process, but as soon as the components start getting large, we start getting excess edges which destroy the cycles and make everything small again. The question to ask is: if we run the process for long enough, roughly how large are all the components? It seems unlikely that the splitting mechanism is so weak that we will get true giant components forming, ie O(n) sizes, so we might guess that, in common with some other split-merge models of this type, we end up with components of size $n^{2/3}$, as in the critical window for the E-R process.

In any case, the scaling limit process is likely to have components whose sizes grow with n, so we will have a class of trees larger than those we have considered previously, which have typically been O(1). So it’s worth thinking about some ways to generate random trees on a fixed number of vertices.

Conditioned Galton-Watson

Our favourite method of creating trees is inductive. We take a root and connect the root to a number of offspring given by a fixed distribution, and each of these some offspring given by an independent sample from the same distribution and so on. The natural formulation gives no control over the size of the tree. This is a random variable whose distribution depends on the offspring distribution, and which in some circumstances be computed explicitly, for example when the offspring distribution is geometric. In other cases, it is easier to make recourse to generating functions or to a random walk analogue as described in the exploration process discussion.

Of course, there is nothing to stop us conditioning on the total size of the population. This is equivalent to conditioning on the hitting time of -1 for the corresponding random walk, and Donsker’s theorem gives several consequences of a convergence relation towards a rescaled Brownian excursion. Note that there is no a priori labelling for the resulting tree. This will have to be supplied later, with breadth-first and depth-first the most natural choices, which might cause annoyance if you actually want to use it. In particular, it is not obvious, and probably not true unless you are careful, that the distribution is invariant under permuting the labels (having initially assumed 1 is the root etc) which is not ideal if you are embedding into the complete graph.

However, we would like to have some more direct constructions of random trees on n vertices. We now consider perhaps the two best known such methods. These are of particular interest as they are applicable to finding random spanning trees embedded in any graph, rather than just the complete graph.

Uniform Spanning Tree

Given a connected graph, consider the set of all subgraphs which are trees and span the vertex set of the original graph. An element of this set is called a spanning tree. A uniform spanning tree is chosen uniformly at random from the set of spanning trees on the complex graph on n vertices. A famous result of Arthur Cayley says that the number of such spanning trees is $n^{n-2}$. There are various neat proofs, many of which consider a mild generalisation which gives us a more natural framework for using induction. This might be a suitable subject for a subsequent post.

While there is no objective answer to the question of what is the right model for random trees on n vertices, this is what you get from the Erdos-Renyi process. Formally, conditional on the sizes of the (tree) components, the structures of the tree components are given by UST.

To see why this is the case, observe that when we condition that a component has m vertices and is a tree, we are demanding that it be connected and have m-1 edges. Since the probability of a particular configuration appearing in G(n,p) is a function only of the number of edges in the configuration, it follows that the probability of each spanning tree on the m vertices in question is equal.

Interesting things happen when you do this dynamically. That is, if we have two USTs of sizes m and n at some time t, and condition that the next edge to be added in the process joins them, then the resulting component is not a UST on m+n vertices. To see why, consider the probability of a ‘star’, that is a tree with a single distinguished vertex to which every other vertex is joined. Then the probability that the UST on m vertices is a star is $\frac{m}{m^{m-2}}=m^{-(m-3)}$. By contrast, it is not possible to obtain a star on m+n vertices by joining a tree on m vertices and a tree on n vertices with an additional edge.

However, I think the UST property is preserved by the cycle deletion mechanism mentioned at the very start of this post. My working has been very much of the back of the envelope variety, but I am fairly convinced that once you have taken a UST and conditioned on the sizes of the smaller trees which result from cycle deletion. My argument is that you might as well fix the cycle to be deleted, then condition on how many vertices are in each of the trees coming off this cycle. Now the choice of each of these trees is clearly uniform among spanning trees on the correct number of vertices.

However, it is my current belief that the combination of these two mechanisms does not give UST-like trees even after conditioning on the sizes at fixed time.

# 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