# Kernels of critical graph components

This post is motivated by G(N,p), the classical Erdos-Renyi random graph, specifically its critical window, when $p=p(N)=\frac{1}{N}(1+\lambda N^{-1/3})$.

We start with the following observation, which makes no restriction on p. Suppose a component of G(N,p) is a tree. Then, the graph geometry of this component is that of a uniform random tree on the appropriate number of vertices. This is deliberately informal. To be formal, we’d have to say “condition on a particular subset of vertices forming a tree-component” and so on. But the formality is broadly irrelevant, because at the level of metric scaling limits, if we want to describe the structure of a tree component, it doesn’t matter whether it has $\log N$ or $\frac{1}{7}N$ vertices, because in both cases the tree structure is uniform. The only thing that changes is the scaling factor.

In general, when V vertices form a connected component of a graph with E edges, we define the excess to be E-V+1. So the excess is non-negative, and is zero precisely when the component is a tree. I’m reluctant to say that the excess counts the number of cycles in the component, but certainly it quantifies the amount of cyclic structure present. We will sometimes, in a mild abuse of notation, talk about excess edges. But note that for a connected component with positive excess, there is a priori no way to select which edges would be the excess edges. In a graph process, or when there is some underlying exploration of the component, there sometimes might be a canonical way to classify the excess edges, though it’s worth remarking that the risk of size-biasing errors is always extremely high in this sort of situation.

Returning to the random graph process, as so often there are big changes around criticality. In the subcritical regime, the components are small, and most of them, even the largest with high probability, are trees. In the supercritical regime, the giant component has excess $\Theta(N)$, which is qualitatively very different.

It feels like every talk I’ve ever given has begun with an exposition of Aldous’s seminal paper [Al97] giving a distributional scaling limit of the sizes of critical components in the critical window, and a relation between the process on this time-scale and the multiplicative coalescent. And it remains relevant here, because the breadth-first exploration process can also be used to track the number of excess edges.

In a breadth-first exploration, we have a stack of vertices we are waiting to explore. We pick one and look its neighbours restricted to the rest of the graph, that is without the vertices we have already fully explored, and also without the other vertices in the stack. That’s the easiest way to handle the total component size. But we can simultaneously track how many times we would have joined to a neighbour within the stack, which leads to an excess edge, and Aldous derives a joint distributional scaling limit for the sizes of the critical components and their excesses. (Note that in this case, there is a canonical notion of excess edge, but it depends not just on the graph structure, but also on the extra randomness of the ordering within the breadth-first search.)

Roughly speaking, we consider the reflected exploration process, and its scaling limit, which is a reflected parabolically-drifting Brownian motion (though the details of this are not important at this level of exposition, except that it’s a well-behaved non-negative process that hits zero often). The component sizes are given by the widths of the excursions above zero, scaled up in a factor $N^{1/3}$. Then conditional on the shape of the excursion, the excess is Poisson with parameter the area under the excursion, with no rescaling. That is, a critical component has $\Theta(1)$ excess.

So, with Aldous’s result in the background, when we ask about the metric structure of these critical components, we are really asking: “what does a uniformly-chosen connected component with fixed excess look like when the number of vertices grows?”

I’ll try to keep notation light, but let’s say T(n,k) is a uniform choice from connected graphs on n vertices with excess k.

[Note, the separation of N and n is deliberate, because in the critical window, the connected components have size $n = \Theta(N^{2/3})$, so I want to distinguish the two problems.]

In this post, we will mainly address the question: “what does the cycle structure of T(n,k) look like for large n?” When k=0, we have a uniform tree, and the convergence of this to the Brownian CRT is now well-known [CRT2, LeGall]. We hope for results with a similar flavour for positive excess k.

2-cores and kernels

First, we have to give a precise statement of what it means to study just the cycle structure of a connected component. From now on I will assume we are always working with a connected graph.

There are several equivalent definitions of the 2-core C(G) of a graph G:

• When the excess is positive, there are some cycles. The 2-core is the union of all edges which form part of some cycle, and any edges which lie on a path between two edges which both form part of some cycle.
• C(G) is the maximal induced subgraph where all degrees are at least two.
• If you remove all the leaves from the graph, then all the leaves from the remaining graph, and continue, the 2-core is the state you arrive at where there are no leaves.

It’s very helpful to think of the overall structure of the graph as consisting of the 2-core, with pendant trees ‘hanging off’ the 2-core. That is, we can view every vertex of the 2-core as the root of a (possibly size 1) tree. This is particular clear if we remove all the edges of the 2-core from the graph. What remains is a forest, with one tree for each vertex of the 2-core.

In general, the k-core is the maximal induced subgraph where all degrees are at least k. The core is generally taken to be something rather different. For this post (and any immediate sequels) I will never refer to the k-core for k>2, and certainly not to the traditional core. So I write ‘core’ for ‘2-core’.

As you can see in the diagram, the core consists of lots of paths, and topologically, the lengths of these paths are redundant. So we will often consider instead the kernel, K(G), which is constructed by taking the core and contracting all the paths between vertices of degree greater than 2. The resulting graph has minimal degree at least three. So far we’ve made no comment about the simplicity of the original graphs, but certainly the kernel need not be simple. It will regularly have loops and multiple edges. The kernel of the graph and core in the previous diagram is therefore this:

Kernels of critical components

To recap, we can deconstruct a connected graph as follows. It has a kernel, and each edge of the kernel is a path length of some length in the core. The rest of the graph consists of trees hanging off from the core vertices.

For now, we ask about the distribution of the kernel of a T(n,K). You might notice that the case k=1 is slightly awkward, as when the core consists of a single cycle, it’s somewhat ambiguous how to define the kernel. Everything we do is easily fixable for k=1, but rather than carry separate cases, we handle the case $k\ge 2$.

We first observe that fixing k doesn’t confirm the number of vertices or edges in the kernel. For example, both of the following pictures could correspond to k=3:

However, with high probability the kernel is 3-regular, which suddenly makes the previous post relevant. As I said earlier, it can introduce size-biasing errors to add the excess edges one-at-a-time, but these should be constant factor errors, not scaling errors. So imagine the core of a large graph with excess k=2. For the sake of argument, assume the kernel has the dumbbell / handcuffs shape. Now add an extra edge somewhere. It’s asymptotically very unlikely that this is incident to one of the two vertices with degree three in the core. Note it would need to be incident to both to generate the right-hand picture above. Instead, the core will gain two new vertices of degree three.

Roughly equivalently, once the size of the core is fixed (and large) we have to make a uniform choice from connected graphs of this size where almost every vertex has degree 2, and $\Theta(1)$ of the rest have degree 3 or higher. But the sum of the degrees is fixed, because the excess is fixed. If there are n vertices in the core, then there are $\Theta(n)$ more graphs where all the vertices have degree 2 or 3, than graphs where a vertex has degree at least 4. Let’s state this formally.

Proposition: The kernel of a uniform graph with n vertices and excess $k\ge 2$ is, with high probability as $n\rightarrow\infty$, 3-regular.

This proved rather more formally as part of Theorem 7 of [JKLP], essentially as a corollary after some very comprehensive generating function setup; and in [LPW] with a more direct computation.

In the previous post, we introduced the configuration model as a method for constructing regular graphs (or any graphs with fixed degree sequence). We observe that, conditional on the event that the resulting graph is simple, it is in fact uniformly-distributed among simple graphs. When the graph is allowed to be a multigraph, this is no longer true. However, in many circumstances, as remarked in (1.1) of [JKLP], for most applications the configuration model measure on multigraphs is the most natural.

Given a 3-regular labelled multigraph H with 2(k-1) vertices and 3(k-1) edges, and K a uniform choice from the configuration model with these parameters, we have

$\mathbb{P}\left( K \equiv H \right) \propto \left(2^{t(H)} \prod_{e\in E(H)} \mathrm{mult}(e)! \right)^{-1},$

where t(H) is the number of loops in H, and mult(e) the multiplicity of an edge e. This might seem initially counter-intuitive, because it looks we are biasing against graphs with multiple edges, when perhaps our intuition is that because there are more ways to form a set of multiple edges we should bias in favour of it.

I think it’s most helpful to look at a diagram of a multigraph as shown, and ask how to assign stubs to edges. At a vertex with degree three, all stub assignments are different, that is 3!=6 possibilities. At the multiple edge, however, we care which stubs match with which stubs, but we don’t care about the order within the multi-edge. Alternatively, there are three choices of how to divide each vertex’s stubs into (2 for the multi-edge, 1 for the rest), and then two choices for how to match up the multi-edge stubs, ie 18 in total = 36/2, and a discount factor of 2.

We mention this because in fact K(T(n,k)) converges in distribution to this uniform configuration model. Once you know that K(T(n,k)) is with high probability 3-regular, then again it’s probably easiest to think about the core, indeed you might as well condition on its total size and number of degree 3 vertices. It’s then not hard to convince yourself that a uniform choice induces a uniform choice of kernel. Again, let’s state that as a proposition.

Proposition: For any H a 3-regular labelled multigraph H with 2(k-1) vertices and 3(k-1) edges as before,

$\lim_{n\rightarrow\infty}\mathbb{P}\left( K(T(n,k)) \equiv H \right) \propto \left(2^{t(H)} \prod_{e\in E(H)} \mathrm{mult}(e)! \right)^{-1}.$

As we said before, the kernel describes the topology of the core. To reconstruct the graph, we need to know the lengths in the core, and then how to glue pendant trees onto the core. But this final stage depends on k only through the total length of paths in the core. Given that information, it’s a combinatorial problem, and while I’m not claiming it’s easy, it’s essentially the same as for the case with k=1, and is worth treating separately.

It is worth clarifying a couple of things first though. Even the outline of methods above relies on the fact that the size of the core diverges as n grows. Again, the heuristic is that up to size-biasing errors, T(n,k) looks like a uniform tree with some uniformly-chosen extra edges. But distances in T(n,k) scale like $n^{1/2}$ (and thus in critical components of G(N,p) scale like $N^{1/3}$). And the core will be roughly the set of edges on paths between the uniformly-chosen pairs of vertices, and so will also have length $\Theta(n^{1/2})$.

Once you have conditioned on the kernel structure, and the (large) number of internal vertices on paths in the core (ie the length of the core), it is natural that the assignment of the degree-2 vertices to core paths / kernel edges is uniform. A consequence of this is that if you record $(Y_1,\ldots,Y_m)$ the lengths of paths in the core, where m=3(k-1), then

$\frac{(Y_1,\ldots,Y_m)}{\sum Y_i} \stackrel{d}\rightarrow \mathrm{Dirichlet}(1,1,\ldots,1).$

This is stated formally as Corollary 7 b) of [ABG09]. It’s worth noting that this confirms that the lengths of core paths are bounded in probability away from zero after the appropriate rescaling. In seeking a metric scaling limit, this is convenient as it means there’s so danger that two of the degree-3 vertices end up in ‘the same place’ in the scaling limit object.

To recap, the only missing ingredients now to give a complete limiting metric description of T(n,k) are 1) a distributional limit of the total core length; 2) some appropriate description of set of pendant trees conditional on the size of the pendant forest. [ABG09] show the first of these. As remarked before, all the content of the second of these is encoded in the unicyclic k=1 case, which I have written about before, albeit slightly sketchily, here. (Note that in that post we get around size-biasing by counting a slightly different object, namely unicyclic graphs with an identified cyclic edge.)

However, [ABG09] also propose an alternative construction, which you can think of as glueing CRTs directly onto the stubs of the kernel (with the same distribution as before). The proof that this construction works isn’t as painful as one might fear, and allows a lot of the other metric distributional results to be read off as corollaries.

References

[ABG09] – Addario-Berry, Broutin, Goldschmidt – Critical random graphs: limiting constructions and distributional properties

[CRT2] – Aldous – The continuum random tree: II

[Al97] – Aldous – Brownian excursions, critical random graphs and the multiplicative coalescent

[JKLP] – Janson, Knuth, Luczak, Pittel – The birth of the giant component

[LeGall] – Le Gall – Random trees and applications

[LPW] – Luczak, Pittel, Wierman – The structure of a random graph at the point of the phase transition

# Random transpositions

We study a procedure for generating a random sequence of permutations of [N]. We start with the identity permutation, and then in each step, we choose two elements uniformly at random, and swap them. We obtain a sequence of permutations, where each term is obtained from the previous one by multiplying by a uniformly-chosen transposition.

Some more formality and some technical remarks:

• This is a Markov chain, and as often with Markov chains, it would be better it was aperiodic. As described, the cycle will alternate between odd and even permutations. So we allow the two elements chosen to be the same. This laziness slows down the chain by a factor N-1/N, but removes periodicity. We will work over timescales where this adjustment makes no practical difference.
• Let $\tau_1,\tau_2,\ldots$ be the sequence of transpositions. We could define the sequence of permutations by $\pi_m= \tau_m\cdot\tau_{m-1}\cdot \ldots\cdot \tau_1$. I find it slightly more helpful to think of swapping the elements in places i and j, rather the elements i and j themselves, and so I’ll use this language, for which $\pi_m = \tau_1\cdot \tau_2\cdot\ldots \cdot \tau_m$ is the appropriate description. Of course, transpositions and the identity are self-inverse permutations, so it makes no difference to anything we might discuss.
• You can view this as lazy random walk on the Cayley graph of $S_N$ generated by the set of transpositions. That is, the vertices of the graph are elements of $S_N$, and two are connected by an edge if one can be obtained from the other by multiplying by a transposition. Note this relation is symmetric. Hence random transposition random walk.
• Almost everything under discussion would work in continuous time too.

At a very general level, this sort of model is interesting because sometimes the only practical way to introduce ‘global randomness’ is repeatedly to apply ‘local randomness’. This is not the case for permutations – it is not hard to sample uniformly from $S_N$. But it is a tractable model in which to study relevant questions about the generating randomness on a complicated set through iterated local operations.

Since it is a Markov chain with a straightforward invariant distribution, we can ask about the mixing time. That is, the correct scaling for the number of moves before the random permutation is close in distribution (say in the sense of total variation distance) to the equilibrium distribution. See this series of posts for an odd collection of background material on the topic. Diaconis and Shahshahani [DS81] give an analytic argument for mixing around $\frac{N\log N}{2}$ transpositions. Indeed they include a constant because it is a sharp cutoff, where the total variation distance drops from approximately 1 to approximately 0 in O(N) steps.

Comparison with Erdos-Renyi random graph process

In the previous result, one might observe that $m=\frac{N\log N}{2}$ is also the threshold number of edges to guarantee connectivity of the Erdos-Renyi random graph G(N,m) with high probability. [ER59] Indeed, there is also a sharp transition around this threshold in this setting too.

We explore this link further. We can construct a sequence of random graphs simultaneously with the random transposition random walk. When we multiply by transposition (i j), we add edge ij in the graph. Laziness of RTRW and the possibility of multiple edges mean this definition isn’t literally the same as the conventional definition of a discrete-time Erdos-Renyi random graph process, but again this is not a problem for any of the effects we seek to study.

The similarity between the constructions is clear. But what about the differences? For the RTRW, we need to track more information than the random graph. That is, we need to know what order the transpositions were added, rather than merely which edges were added. However, the trade-off is that a permutation is a simpler object than a graph in the following sense. A permutation can be a described as a union of disjoint cycles. In an exchangeable setting, all the information about a random permutation is encoded in the lengths of the these cycles. Whereas in a graph, geometry is important. It’s an elegant property of the Erdos-Renyi process that we can forget about the geometry and treat it as a process on component sizes (indeed, a multiplicative coalescent process), but there are other questions we might need to ask for which we do have to study the graph structure itself.

Within this analogy, unfortunately the word cycle means different things in the two different settings. In a permutation, a cycle is a directed orbit, while in a graph it has the usual definition. I’m going to write graph-cycle whenever relevant to avoid confusion.

A first observation is that, under this equivalence, the cycles of the permutation form a finer partition than the components of the graph. This is obvious. If we split the vertices into sets A and B, and there are no edges between them, then nothing in set A will ever get moved out of set A by a transposition. (Note that the slickness of this analogy is the advantage of viewing a transposition as swapping the elements in places i and j.)

However, we might then ask under what circumstances is a cycle of the permutation the same as a component of the graph (rather than a strict subset of it). A first answer is the following:

Lemma: [Den59] The permutation formed by a product of transpositions corresponding in any order to a tree in the graph has a single cycle.

We can treat this as a standalone problem and argue in the following predictable fashion. (Indeed, I was tempted to set this as a problem during selection for the UK team for IMO 2017 – it’s perfectly suitable in this context I think.) The first transposition corresponds to some edge say ab, and removing this edge divides the vertices into components $A \ni a, B\ni b$. Since no further transposition swaps between places in A and places in B, the final permutation maps a into B and b into A, and otherwise preserves A and B.

This argument extends to later transpositions too. Now, suppose there are multiple cycles. Colour one of them. So during the process, the coloured labels move around. At some point, we must swap a coloured label with an uncoloured label. Consider this edge, between places a and b as before, and indeed the same conclusion holds. WLOG we move the coloured label from a to b. But then at the end of the process (ie in the permutation) there are more coloured labels in B than initially. But the number of coloured labels should be the same, because they just cycle around in the final permutation.

We can learn a bit more by trying thinking about the action on cycles (in the permutation) of adding a transposition. In the following pair of diagrams, the black arrows represent the original permutation (note it’s not helpful to think of the directed edges as having anything to do with transpositions now), the dashed line represents a new transposition, and the new arrows describe the new permutation which results from this product.

It’s clear from this that adding a transposition between places corresponding to different cycles causes the cycles to merge, while adding a transposition between places already in the same cycle causes the cycle to split into two cycles. Furthermore the sizes of the two cycles formed is related to the distance in the cycle between the places defining the transposition.

This allows us to prove the lemma by adding the edges of the tree one-at-a-time and using induction. The inductive claim is that cycles of the permutation exactly correspond to components of the partially-built tree. Assuming this claim guarantees that the next step is definitely a merge, not a split (otherwise the edge corresponding to the next step would have to form a cycle). If all N-1 steps are merges, then the number of cycles is reduced by one on each step, and so the final permutation must be a single cycle.

Uniform split-merge

This gives another framework for thinking about the RTRW itself, entirely in terms of cycle lengths as a partition of [N]. That is, given a partition, we choose a pair of parts in a size-biased way. If they are different, we merge them; and if it is the same part, with size k, we split them into two parts, with sizes chosen uniformly from { (1,k-1), (2,k-2), …  (k-1,1) }.

What’s nice about this is that it’s easy to generalise to real-valued partitions, eg of [0,1]. Given a partition of [0,1], we sample two IID U[0,1] random variables $U_1,U_2$. If these correspond to different parts, we replace these parts by a single part with size given by the sum. If these correspond to the same part, with size $\alpha$, we split this part into two parts with sizes $|U_1-U_2|$ and $\alpha - |U_1-U_2|$. This is equivalent in a distributional sense to sampling another U[0,1] variable U and replacing $\alpha$ with $(\alpha U, \alpha(1-U))$. We probably want our partition to live in $\ell^1_\searrow$, so we might have to reorder the parts afterwards too.

These uniform split-merge dynamics have a (unique) stationary distribution, the canonical Poisson-Dirichlet random partition, hereafter PD(0,1). This was first shown in [DMZZ04], and then in a framework more relevant to this post by Schramm [Sch08].

Conveniently, PD(0,1) is also the scaling limit of the cycle lengths in a uniform random permutation (scaled by N). The best way to see this is to start with the observation that the length of the cycle containing 1 in a permutation chosen uniformly from $S_N$ has the uniform distribution on {1,…,N}. This matches up well with the uniform stick-breaking construction of PD(0,1), though other arguments are available too. Excellent background on Poisson-Dirichlet distributions and this construction and equivalence can be found in Chapter 3 of Pitman’s comprehensive St. Flour notes [CSP]. Also see this post, and the links within, with the caveat that my understanding of the topic was somewhat shaky then (as presently, for now).

However, Schramm says slightly more than this. As the Erdos-Renyi graph passes criticality, there is a well-defined (and whp unique) giant component including $\Theta(N)$ vertices. It’s not clear that the corresponding permutation should have giant cycles. Indeed, whp the giant component has $\Theta(N)$ surplus edges, so the process of cycle lengths will have undergone $O(N)$ splits. Schramm shows that most of the labels within the giant component are contained in giant cycles in the permutation. Furthermore, the distribution of cycle lengths within the giant component, rescaled by the size of the giant component, converges in distribution to PD(0,1) at any supercritical time $\frac{(1+\epsilon)N}{2}$

This is definitely surprising, since we already know that the whole permutation doesn’t look close to uniform until time $\frac{N\log N}{2}$. Essentially, even though the size of the giant component is non-constant (ie it’s gaining vertices), the uniform split-merge process is happening to the cycles within it at rate N. So heuristically, at the level of the largest cycles, at any supercritical time we have a non-trivial partition, so at any slightly later time (eg $\frac{(1+\epsilon/2)N}{2}$ and $\frac{(1+\epsilon)N}{2}$ ), mixing will have comfortably occurred, and so the distribution is close to PD(0,1).

This is explained very clearly in the introduction of [Ber10], in which the approach is extended to a random walk on $S_N$ driven by a uniform choice from any conjugacy class.

So this really does tell us how the global uniform randomness emerges. As the random graph process passes criticality, we have a positive mass of labels in a collection of giant cycles which are effectively a continuous-space uniform split-merge model near equilibrium (and thus with PD(0,1) marginals). The remaining cycles are small, corresponding to small trees which make up the remaining (subcritical by duality) components of the ER graph. These cycles slowly get absorbed into the giant cycles, but on a sufficiently slow timescale relevant to the split-merge dynamics that we do not need to think of a separate split-merge-with-immigration model. Total variation distance on permutations does feel the final few fixed points (corresponding to isolated vertices in the graph), hence the sharp cutoff corresponding to sharp transition in the number of isolated vertices.

References

[Ber10] – N. Berestycki – Emergence of giant cycles and slowdown transition in random transpositions and k-cycles. [arXiv version]

[CSP] – Pitman – Combinatorial stochastic processes. [pdf available]

[Den59] – Denes – the representation of a permutation as a product of a minimal number of transpositions, and its connection with the theory of graphs

[DS81] – Diaconis, Shahshahani – Generating a random permutation with random transpositions

[DMZZ04] – Diaconis, Mayer-Wolf, Zeitouni, Zerner – The Poisson-Dirichlet distribution is the unique invariant distribution for uniform split-merge transformations [link]

[ER59] – Erdos, Renyi – On random graphs I.

[Sch08] – Schramm – Compositions of random transpositions [book link]

# Parking on a ring, linear hashing

I’ve spent most of my doctorate trying to analyse how adding destructive dynamics affects the behaviour of a particular random growth process, the classical random graph. In this post I’m going to talk about another random growth process, which is slightly less natural, but for which one can show some similar qualitative properties.

The model, and the additive coalescent

Consider m places arranged in a circle, and for consistency of analogy we think of these as parking spaces. Some number n of cars will arrive one at a time. Each car will arrive at a space chosen uniformly at random. If it is empty they will park in it, otherwise they will look clockwise until they find an empty space, and park there. For now we are only interested in growth, so we assume cars never leave. We are interested in the sizes of blocks of consecutively parked cars.

The reason to consider this slightly unnatural statement is its equivalence to the problem of hashing with linear probing, apparently a key topic in computer science, which I won’t pretend that I know anything about. In any case, it’s a nice model, and it seems reasonable that it would have a basis in more realistic search algorithms.

So, how does the sequence of sizes of blocks of consecutively parked cars grow? Well, given the sequence of block sizes, it is reasonably easy to convince yourself that the order of the blocks around the circle is uniformly random, and the number of empty spaces between adjacent blocks is also uniformly random.

Assume for now that there are at least three blocks. A block of size x can merge with a block of size y with the arrival of the next car only if the blocks are adjacent, with exactly one empty space between them. The chance of this is uniform among all pairs of blocks. Now suppose this is the case, and that the block of size y lies clockwise from the block of size x. Then they will merge precisely if the next car arrives at any of the x occupied spaces in that block, or at the empty space between the pair of blocks. This has probability $\frac{x+1}{m}$. There’s also the opposite ordering to consider, where the block of size x lies clockwise from the other. The total probability of this merge $\{x,y\}\mapsto \{x+y+1\}$ is therefore proportional to (x+y+2).

So the process of block sizes looks a bit like the additive coalescent, at least for large blocks. This is in contrast to the random graph process, where the sequence of component sizes behaves exactly like a multiplicative coalescent, where blocks merge at a rate proportional to the product of their sizes.

Asymptotics

As in the random graph process, it’s interesting to ask roughly how large the largest block will be in such a configuration. Pittel [3] considers the case where the number of empty places $\ell = m-n \approx \beta m$, for some $\beta\in (0,1)$.

A less interesting model would be to choose the positions of the n cars uniformly at random. But then the size of a block is roughly geometric with parameter $\beta$, and there are $\Theta(m)$ blocks with high probability. Relatively straightforward calculations in extreme value theory suggest that the largest block is likely to have size on the order of $\log m$ in this setting.

Of course, the actual model is slightly more complicated, because the size of a block is self-reinforcing, since larger blocks are more likely to grow than smaller blocks. However, we can still get somewhere with naïve estimates. Let’s label the places clockwise. Then in order for there to be a block starting at 0 and stretching beyond $\alpha \log m$, a necessary condition is that at least $\alpha \log m$ cars arrive at those places. The number of cars which arrive at those places is binomial, since there are n cars, and each arrives at a place chosen uniformly, and independently of the other cars. So this event corresponds to

$\mathrm{Bin}(n,\frac{\alpha \log m}{m}) \ge \alpha \log m.$

Then, since $n\approx (1-\beta)n$, this event corresponds approximately to

$\mathrm{Po}((1-\beta)\alpha \log m) \ge \alpha \log m.$

The probability that a Poisson RV is at least a constant multiple larger than its mean decays exponentially with the mean, hence in this case the probability is asymptotically some negative power of m, depending on the value of $\alpha$. But there are $O(m)$ possible places for such a block to start, so whether we can apply a union bound usefully or not depends on whether the power of m is strictly less than -1.

Since all of this depends on $\alpha$, it is reasonable that everything is fine, and the largest block does have size at least $\alpha \log m$ when $\alpha$ is small, and very unlikely when $\alpha$ is large. This heuristic argument fits with Pittel’s theorem. Indeed, his result shows much stronger concentration: that the fluctuations of the size of the largest block are O(1).

Critical regime and empirical processes

The following is a paraphrase of the introduction and some methods from [2].

Obviously, once m=m cars have arrived, there’s no room for manoeuvre and definitely all the places are taken in one giant block. But it’s not obvious in general what scaling for the number of gaps will give rise to giant blocks of $\Theta(m)$ cars.

As for the random graph, we can find a process similar to the exploration process of a (random) graph which encodes much of the information we care about. Let $Y_k$ be the number of cars which arrive at place k. So the sum of the $Y_k$s will be n, the total number of cars. Now consider the process

$C_0=0, \ldots, C_{k+1}=C_k + Y_{k+1}-1.$

A block has the property that the number of arrivals within that set of places is equal to the number of places. So every time this *empirical process* C drops below its previous running minimum, this indicates the end of a block. To make this equivalence precise, we need to be a bit careful about where we start counting. It works exactly if we start at the beginning of a block. If not, it might introduce some unwanted divisions within the first block.

What we have is a process that looks roughly like a random walk that is constrained to pass through the point (m,n-m), which is equal to (m,-l). Even if we aren’t totally precise about how this is like a random walk, we would expect to see Brownian fluctuations after rescaling. Indeed, we might expect to see a Brownian bridge added to a deterministic linear function with negative gradient. But this is only meaningful if the random part is at least as large as the deterministic part, and since the fluctuations have order $\sqrt{m}$, if l is much larger than this, the rescaled empirical process is essentially deterministic, so we won’t see any macroscopic excursions above the minimum.

If l is substantially smaller than $\sqrt{m}$, then there is no real difference between (m,-l) and (m,0), and what we see is just a Brownian bridge. At this point, where we choose to start the process is actually important. If we were to start it at the minimum of the Brownian bridge instead, we would have seen a Brownian excursion, which corresponds to one block occupying (almost) all of the places.

Unsurprisingly, the story is completed by considering $\ell=\Theta(\sqrt{m})$, where the rescaled empirical process looks like a slanted Brownian bridge, that is Brownian motion conditioned to pass through $(1,-\frac{\ell}{\sqrt{m})$. There isn’t an obvious fix to the question of where to start the process, but it turns out that the correct way is now adding a Brownian excursion onto the deterministic linear function with gradient $- \frac{\ell}{\sqrt{m}}$. It’s now reasonable that the excursions above the minimum should macroscopic.

This scaling limit works dynamically as well, where the same Brownian excursion is used for different gradients of the deterministic line, corresponding to $\ell$ moving through the critical window $m-\Theta(\sqrt{m})$. Finally, a direction to Bertoin’s recent paper [1] for the model with an additional destructive property. Analogous to the forest fire, blocks of cars are removed at a rate proportional to their size (as a result, naturally, of ‘Molotov cocktails’…). Similar effects of self-organised criticality are seen when the rate of bombs is scaled appropriately.

References

[1] – Bertoin – Burning cars in a parking lot (paper / slides)

[2] – Chassaing + Louchard – Phase transition for parking blocks, Brownian excursion and coalescence (arXiv)

[3] – Pittel – Linear probing: the probable largest search time grows logarithmically with the number of records

# Critical Components in Erdos-Renyi

In various previous posts, I’ve talked about the phase transition in the Erdos-Renyi random graph process. Recall the definition of the process. Here we will use the Gilbert model G(n,p), where we have n vertices, and between any pair of vertices we add an edge, independently of other pairs with probability p. We are interested in the sparse scaling, where the typical vertex has degree O(1) in n, and so p=c/n for constant c>0, and we assume throughout that n is large. We could alternatively have considered the alternative Erdos-Renyi model where we choose uniformly at random from the set of graphs with n vertices and some fixed number of edges. Almost all the results present work equally well in this setting.

As proved by Erdos and Renyi, the typical component structure of such a graph changes noticeably around the threshold c=1. Below this, in the subcritical regime, all the components are small, meaning of size at most order O(log n). Above this, in the supercritical regime, there is a single giant component on some non-zero proportion of the vertices. The rest of the graph looks subcritical. The case c=1 exhibits a phase transition between these qualitatively different behaviours. They proved that here, the largest component is with high probability O(n^2/3). It seems that they thought this result held whenever c=1-o(1), but it turns out that this is not the case. In this post, I will discuss some aspects of behaviour around criticality, and the tools needed to treat them.

The first question to address is this: how many components of size n^{2/3} are there? It might be plausible that there is a single such component, like for the subsequent giant component. It might also be plausible that there are n^1/3 such components, so O(n) vertices are on such critical components. As then it is clear how we transition out of criticality into supercriticality – all the vertices on critical components coalesce to form the new giant component.

In fact neither of these are correct. The answer is that for all integers k>0, with high probability the k-th largest component is on a size scale of n^2/3. This is potentially a confusing statement. It looks like there are infinitely many such components, but of course for any particular value of n, this cannot be the case. We should think of there being w(1) components, but o(n^b) for any b>0.

The easiest way to see this is by a duality argument, as we have discussed previously for the supercritical phase. If we remove a component of size O(n^2/3), then what remains is a random graph with n-O(n^2/3) vertices, and edge probability the same as originally. It might make sense to rewrite this probability 1/n as

$\frac{1}{n-O(n^{2/3})}\cdot \frac{n-O(n^{2/3})}{n}=\frac{1-O(n^{-1/3})}{n-O(n^{2/3})}.$

The approximation in the final numerator is basically the same as

$1-o\left(n-O(n^{2/3})\right).$

Although we have no concrete reasoning, it seems at least plausible that this should look similar in structure to G(n,1/n). In particular, there should be another component of size

$O\left([n-O(n^{2/3})]^{2/3}\right)=O(n^{2/3})$.

In fact, the formal proof of this proceeds by an identical argument, only using the exploration process. Because I’ve described this several times before, I’ll be brief. We track how far we have gone through each component in a depth-first walk. In both the supercritical and subcritical cases, when we scale correctly we get a random path which is basically deterministic in the limit (in n). For exactly the same reasons as visible CLT fluctuations for partial sums of RVs with expectation zero, we start seeing interesting effects at criticality.

The important question is the order of rescaling to choose. At each stage of the exploration process, the number of vertices added to the stack is binomial. We want to distinguish between components of size $O(n^{2/3})$ so we should look at the exploration process at time $sn^{2/3}$. The drift of the exploration process is given by the expectation of a binomial random variable minus one (since we remove the current vertex from the stack as we finish exploring it). This is given by

$\mathbb{E}=\left[n-sn^{2/3}\right]\cdot \frac{1}{n}-1=-sn^{-1/3}.$

Note that this is the drift in one time-step. The drift in $n^{2/3}$ time-steps will accordingly by $sn^{1/3}$. So, if we rescale time by $n^{2/3}$ and space by $n^{1/3}$, we should get a nice stochastic process. Specifically, if Z is the exploration process, then we obtain:

$\frac{1}{n^{1/3}}Z^{(n)}_{sn^{2/3}} \rightarrow_d W_s,$

where W is a Brownian motion with inhomogeneous drift -s at time s. The net effect of such a drift at a fixed positive time is given by integrating up to that time, and hence we might say the process has quadratic drift, or is parabolic.

We should remark that our binomial expectation is not entirely correct. We have discounted those $sn^{2/3}$ vertices that have already been explored, but we have not accounted for the vertices currently in the stack. We should also be avoiding considering these. However, we now have a heuristic for the approximate number of these. The number of vertices in the stack should be $O(n^{1/3})$ at all times, and so in particular will always be an order of magnitude smaller than the number of vertices already considered. Therefore, they won’t affect this drift term, though this must be accounted for in any formal proof of convergence. On the subject of which, the mode of convergence is, unsurprisingly, weak convergence uniformly on compact sets. That is, for any fixed S, the convergence holds weakly on the random functions up to time $sn^{2/3}$.

Note that this process will tend to minus infinity almost surely. Component sizes are given by excursions above the running minimum. The process given by the height of the original process above the running minimum is called reflected. Essentially, we construct the reflected process by having the same generator when the current value is positive, and forcing the process up when it is at zero. There are various ways to construct this more formally, including as the scaling limit of some simple random walks conditioned never to stay non-negative.

The cute part of the result is that it holds equally well in a so-called critical window either side of the critical probability 1/n. When the probability is $\frac{1+tn^{-1/3}}{n}$, for any $t\in \mathbb{R}$, the same argument holds. Now the drift at time s is t-s, though everything else still holds.

This result was established by Aldous in [1], and gives a mechanism for calculating distributions of component sizes and so on through this critical window.

In particular, we are now in a position to answer the original question regarding how many such components there were. The key idea is that because whenever we exhaust a component in the exploration process, we choose a new vertex uniformly at random, we are effectively choosing a component according to the size-biased distribution. Roughly speaking, the largest components will show up near the beginning. Note that a critical $O(n^{2/3})$ component will not necessarily be exactly the first component in the exploration process, but the components that are explored before this will take up sufficiently few vertices that they won’t show up in the scaling of the limit.

In any case, the reflected Brownian motion ‘goes on forever’, and the drift is eventually very negative, so there cannot be infinitely wide excursions, hence there are infinitely many such critical components.

If we care about the number of cycles, we can treat this also via the exploration process. Note that in any depth-first search we are necessarily only interested in a spanning tree of the host graph. Anyway, when we are exploring a vertex, there could be extra edges to other vertices in the stack, but not to vertices we’ve already finished exploring (otherwise the edge would have been exposed then). So the expected number of excess edges into a vertex is proportional to the height of the exploration process at that vertex. So the overall expected number of excess edges, conditional on the exploration process is the area under the curve. This carries over perfectly well into the stochastic process limit. It is then a calculation to verify that the area under the curve is almost surely infinite, and thus that we expect there to be infinitely many cycles in a critical random graph.

REFERENCES

[1] Aldous D. – Brownian excursions, critical random graphs and the multiplicative coalescent