Monthly Archives: November 2013

The Combinatorial Nullstellensatz

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I’ve been taking a TCC course this term on Additive Combinatorics, delivered via video link from Bristol by Julia Wolf. At some point once the dust of this term has settled, I might write some things about the ideas of the course I’ve found most interesting, in particular the tools of discrete Fourier analysis to get a hold on some useful combinatorial properties of subsets of \mathbb{Z}/n\mathbb{Z} for example.

For this post, I want to talk instead about a topic that was merely mentioned in passing, the Combinatorial Nullstellensatz. The majority of this post is based on Alon’s original paper, which can be found here, and Chapter 9 of Tao and Vu’s book Additive Combinatorics. My aim is to motivate the theorem, give a proof, introduce one useful application from additive combinatorics, and solve Q6 from IMO 2007 as a direct corollary.

What does Nullstellensatz mean? Roughly speaking, it seems to mean ‘a theorem specifying the zeros’. We will be specifying the zeros of a polynomial. We are comfortable with how the zeros of a complex-valued polynomial of one variable behave. The number of zeros is given precisely by the degree of the polynomial (allowing appropriately for multiplicity). It is generally less clear how we might treat the zeros of a polynomial of many variables. The zero set is likely to be some surface, perhaps of dimension one less than the number of variables. In particular, it no longer really makes sense to talk about whether this set is finite or not. The Combinatorial Nullstellensatz gives us some control over the structure of this set of zeros.

The idea behind the generalisation is to view the Fundamental Theorem of Algebra as a statement not about existence of roots, but rather about (combinatorial) existence of non-roots. That is, given a polynomial P(x) of degree n, for any choice of (n+1) complex numbers, at least one of them is not a root of P. This may look like a very weak statement in this context, where we only expect finitely many roots anyway, but in a multivariate setting it is much more intuitively powerful.

Recall that the degree of a monomial is given by the sum of the exponents of the variables present. So the degree of 4x^2 y^3 z is 6. The degree of a polynomial is then given by the largest degree of a monomial in that polynomial. A polynomial P(x_1,\ldots,x_n) over a field F with degree d might have lots of monomial terms of degree d. Suppose one of these monomials is x_1^{d_1}\ldots x_n^{d_n}, where \sum d_i=d. Then one version of the Combinatorial Nullstellensatz asserts that whenever you take subsets of the base field S_i\subset F with |S_i|\ge d_i+1, then there is a point with x_i\in S_i such that P(x_1,\ldots,x_n)=0.

In other words, you can’t have a box (ie product of sets) of dimension d_1+1 \times d_2+1 \times\ldots\times d_n+1 on which the polynomial is zero.

Unsurprisingly, the proof proceeds by induction on the number of variables. Alon’s result proceeds via a more general theorem giving information about the possibility of writing multinomial polynomials as linear combinations of polynomials in one variable.

We would like to start this induction by fixing the x_n co-ordinate, then viewing P as a polynomial in x_1,\ldots,x_{n-1} only. One problem with this approach is that the largest degree monomials in P are not necessarily still the largest degree monomials in P with x_n fixed. So we need to apply a division algorithm argument.

I’m going to miss some steps so as to keep this of suitable blog post length. The key idea is to apply the division algorithm to P with respect to the simplest polynomial that is zero on all of S_n, which we define as:

g(x_n)=\prod_{s_n\in S_n}(x_n-s_n).

We can decompose as

P(x_1,\ldots,x_n)=q_n(x_1,\ldots,x_n)g(x_n)+\sum_{j=0}^{|S_n|-1}r_{n,j}(x_1,\ldots,x_{n-1})x_n^j.

So now we ask where the term x_1^{d_1}\ldots x_n^{d_n} is coming from, bearing in mind that d_n<|S_n|. The lower order terms in g cannot contribute to this, as  they cannot be of maximal degree. Also, the first term in q_n(\mathbf{x})g(x_n) cannot contribute as the exponent of x_n is too large. So the term in question must be coming from r_{n,d_n}(x_1,\ldots,x_{n-1})x_n^{d_n}. So now we can apply the induction hypothesis to the polynomial r_{n,d_n} to find $x_1\in S_1,\ldots, x_{n-1}\in S_{n-1}$ such that r_{n,d_n}(x_1,\ldots,x_{n-1} is non-zero. With these values, we can view the remainder as a polynomial in x_n of degree |S_n|>d_n, and so there is an x_n\in S_n such that

\sum_{j=0}^{|S_n|}r_{n,j}(x_1,\ldots,x_{n-1})x_n^j)\neq 0.

This concludes the proof by induction.

I want to discuss two relatively simple applications. The first is the Cauchy-Davenport Theorem, which one might view as the first non-trivial theorem in additive combinatorics, giving a bound on the size of a sumset.

Theorem (Cauchy-Davenport): Given A, B non-empty subsets of Z_p for p a prime, then

|A+B|\geq \min\{p,|A|+|B|-1\}.

( A+B:=\{c: c=a+b,a\in A,b\in B\} )

Note that the result isn’t especially surprising. Providing some sort of ordering to the elements of A and B might be a sensible way to proceed. Certainly if they were sets in \mathbb{Z}, this would give a proof immediately.

Proof: Only the case |A|+|B| <= p is interesting. Following Alon’s argument, suppose that |A+B| <= |A|+|B|-2, and let C=A+B. Set f(x,y)=\prod_{c\in C}(x+y-c), so f(a,b)=0 for all a\in A,b\in B.

Then the coefficient of x^{|A|-1}y^{|B|-1} in f is \binom{|A|+|B|-2}{|A|-1} as we have to choose which of the terms in the product supply an x and which supply a y. This is non-zero (in Z_p recall) since the upper integer is less than p. The Combinatorial Nullstellensatz then gives a contradiction.

My second example is from the IMO in Vietnam which I attended. I spent a lot of time thinking about this problem, but made no progress.

IMO 2007 Question 6: Let n be a positive integer. Consider

S=\{(x,y,z) | x,y,z\in \{0,1,\ldots,n\}, x+y+z>0\}

as a set of (n+1)^3-1 points in 3D space. Determine the smallest number of planes, the union of which contains S but does not include (0,0,0).

Answer: 3n. Consider the planes x+y+z = k for k varying between 1 and 3n. The aim is to prove that you cannot do it with fewer.

To prove this, suppose we can do with fewer planes, say k. We write the equation of a plane as

ax+by+cz-d=0.

Note that the d’s are non-zero as (0,0,0) must not be a solution. Then take the product of all these degree one polynomials together and subtract a multiple of

\prod_{i=1}^n (x-i)(y-i)(z-i),

with the multiple chosen so the resulting polynomial has a root at (0,0,0). (This constant must be non-zero to cancel the non-zero product of the d’s.) This resulting polynomial is degree 3n by construction, and x^ny^nz^n has a non-zero coefficient, but it is zero on the box [0,n]^3, which contradicts Combinatorial Nullstellensatz.

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Rhombus Tilings and a Nice Bijection

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I want to write a short post giving an example of what seems to me to be a rather nice proof without words. Like all the best proofs without words, they require some words to set everything up, and then even the proof itself is enhanced with a few words.

The goal is a bijection between two combinatorial objects. The first is the family of rhombus tilings. Perhaps the easiest way to define these is to give an example.

DSC_2215 - Copy

As you can see, we have tiled a hexagon with rhombi. The tiles are allowed to be in any of the three possible orientations. It matters that the angles of the hexagon are 120, as we want it to be possible to squeeze rhombi into a corner in two different ways (ie either a single tile or two tiles together), and thus the rhombi should also have angles 120 and 60. The hexagon does not have to be equilateral, as in this example, but obviously all the side lengths should be an integer multiple of the side length of the rhombus, which without loss of generality we may take to be 1.

The other combinatorial object is the class of plane partitions. We again give an example:

4 3 3 2 1
4 2 2 2
3 2 2 1
2 1 1

Notice that all the rows and columns are weakly decreasing. One observation worth making is that the diagonals gives a family of so-called interlaced partitions. In any case, we want to establish this bijection. First I show the idea, that is the proof without words bit. Then I’ll clarify exactly how to make the bijection work.

The first step is to colour a rhombus tiling with a different colour for each orientation, as shown.

DSC_2216 - Copy

The next step is the proof without words bit. We now look at the diagram as if we were looking into a stack of cubes arranged in the positive orthant of R^3. The colouring makes this much more visually arresting. Black rhombi correspond to the (visible) top sides of cubes, while blue and red faces point out in the x and y directions respectively. The key observation is that a rhombus tiling means we can see at least one face of every cube. Otherwise we would need some smaller rhombi to account for the way that some cubes will be partially hidden between taller but closer piles. So if make a note of the heights of all the piles, we should get a plane partition.

After reordering our definition of plane partition, so it is weakly increasing left-right and down-up, corresponding to the x and y axes drawn on the above diagram, the given rhombus tiling should give the following plane partition:

1 2 3
0 1 3
0 0 2

The only thing we need to sort out is precisely how the dimensions of the hexagon restrict the choice of plane partition. Note that we could keep the heights exactly the same but get a different tiling by adding an extra row of red oriented rhombi above the top-left part, and an extra row of blue oriented rhombi above the top-right part. The point is that this would give us a bigger hexagon.

The first observation is that the dimensions of the plane partition correspond to two of the side lengths of the hexagon, indeed the bottom two sides. The third length of the hexagon corresponds to the maximum possible height (ie z component) of the region we are looking at. This is therefore an upper bound on the heights of the stacks.

So we can conclude our bijective argument. There is a bijection between rhombus tilings of the hexagon with side lengths X, Y, Z and plane partitions with dimension X x Y, (where entries are allowed to be zero) where the largest element (which is by definition also the top-left element, or top-right in our re-definition) is at most Z.

It seems there are plenty of interesting questions to be asked about both deterministic and random tilings and plane partitions, based on talks in Marseille. For now though, I feel ill-qualified even to read about such things, so will leave it at that for today.

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Dispersion in Social Networks

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This post is based on a paper that appeared a couple of weeks ago on the Computer Science section of arXiv. You can find it here. I’m going to write a few things about the ideas behind the paper, and avoid pretty much entirely the convincing data the authors present, as well as questions of implementing the algorithms discussed.

The setting is a social network, which we can describe as a graph. Nodes stand for people, and an edge represents that the two associated people have some social connection. This paper focuses on edges corresponding to friendship in the Facebook graph.

A key empirical feature of the graph topology of such social networks as compared to most mathematical models of random graphs is the prevalence of short cycles, and so-called clustering. Loosely speaking, in an Erdos-Renyi random graph, any potential edges appear in the graph independently of the rest of the configuration. But this does not accord well with our experience of our own Facebook friend circle. In general, knowledge that A is friends with both B and C increases the likelihood that B and C are themselves friends. This effect appears to be more present in other models, such as Preferential Attachment and the Configuration Model, but that is really more a consequence of the degree sequence being less concentrated.

The reason for this phenomenon appearing in social networks is clear. People meet other people generally by sharing common activities, whether that be choice of school, job or hobbies. The question of how readily people choose to add others on Facebook is a worthwhile one, but not something I have the time or the sociological credibility to consider! In any case, it is not a controversial idea that for some typical activity, it is entirely possible that almost all the participants will end up as friends, leading to a large (almost-) ‘clique’ in the graph. Recall a clique is a copy of a complete graph embedded in a larger graph – that is, a set of nodes all of which are pairwise connected.

We could think of much of the structure of this sort of network as being generated in the following way. Suppose we were able to perform the very unlikely-sounding task of listing every conceivable activity or shared attribute that might engender a friendship. Each activity corresponds to a set of people. We then construct a graph on the set of people by declaring that a pair of nodes are connected by an edge precisely if the people corresponding to these nodes can both be found in some activity set.

Another way of thinking about this setup is to consider a bipartite graph, with people as one class of vertices, and activities as the other. Predictably, we join a person to an activity if they engage in that activity. The edges within the class of people are then induced by the bipartite edges. Obviously, under this interpretation, we could equally well construct a graph on the set of activities. Here, two activities would be joined if there is a person who does them both. Graphs formed in this way can be called Intersection Graphs, and there is lots of interest in investigating various models of Random Intersection Graphs.

The question addressed by the authors of the paper can be summarised as follows. A social network graph tells us whether two people are ‘friends’, but it does not directly tell us how close their relationship is. It is certainly an interesting question to ask whether the (local) network topology can give us a more quantitative measure of the strength of a friendship.

As the authors explain, a first approach might be to consider how many mutual friends two people have. (We consider only pairs of people who are actually friends. It seems reasonable to take this as a pre-requisite for a strong relationship among people who do actually use Facebook.) However, this can fail because of the way these social networks organise themselves around shared attributes and activities. The size of one of these cliques (which are termed social foci in parts of the literature) is not especially likely to be well correlated to the strengths of the friendships within the clique. In particular, the clique corresponding to someone’s workplace is likely to grow in size over time, especially when people grow towards an age where, on average, they move job much less. So it seems likely that, according to a naive examination of the number of mutual friends, we would predict that a person’s strongest friend is likely to be someone they work with, who perhaps by chance also does some other activity with that person.

The authors phrase this problem slightly differently. They examine algorithms for establishing a person’s spouse or long-term partner with good accuracy from only the local network structure.

Heuristically we might expect that a husband knows many of his wife’s work colleagues, and vice versa. Not all of these ties might be so strong that they actually lead to friendship, in the Boolean sense of Facebook, but we might expect that some noticeable proportion have this property. Naturally, there will be cliques to which both belong. One or more of these might be the reason they met in the first place, and others (eg parents at children’s schools) might have developed over the course of their relationship. However, as we’ve explained, this doesn’t narrow things down much.

(We need not be constrained by this heteronormative scenario. However, as the authors point out in a footnote, there are challenges in collecting data because of the large number of ironic relationship listings on Facebook, mainly among the undergraduate and younger community. This problem is particularly obstructive in the case of same-sex marriage, owing to the smaller numbers of genuine pairings, and larger numbers of false listings for this setting.)

DSC_2212 - Copy

The crucial observation is that if we look at the couple’s mutual friends, we expect to see large parts of the most important cliques from both husband and wife’s lives. Among these mutual friends, there will be some overlap, that is cliques of which both are an integral member. But among the rest, there will be a natural partition into friends who really originate from the husband, and friends who were introduced via the wife. So the induced graph on these mutual friends is likely to split into three classes of vertices, with very poor connectivity between two of them.

This is, up to sorting out scaling and so on, precisely the definition of dispersion, introduced by the authors. The dispersion between two vertices is high if the induced graph on their mutual neighbourhood has poor connectivity. Modulo exact choice of definition, they then exhibit data showing that this is indeed a good metric for determining marriages from the network topology, with success rate of around 50% over a wide range of users.

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Persistent Hubs

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This post is based on the paper “Existence of a persistent hub in the convex preferential attachment model” which appeared on arXiv last week. It can be found here. My aim is to explain (again) the model; the application-based motivation for the result; and a heuristic for the result in a special case. In particular, I want to stress the relationship between PA models and urns.

The preferential attachment model attempts to describe the formation of large complex networks. It is constructed dynamically: vertices are introduced one at a time. For each new vertex, we have to choose which existing vertices to join to. This choice is random and reinforced. That is, the new vertex is more likely to join to an existing vertex with high degree than to an existing vertex with degree 1. It’s clear why this might correspond well to the evolution of, say, the world wide web. New webpages are much more likely to link to an established site, eg Wikipedia or Google, than to a uniformly randomly chosen page.

The model is motivated also by a desire to fit a common property of real-world networks that is not exhibited by, among others, the Erdos-Renyi random graph model. In such a network, we expect a few nodes to have much greater connectivity than average. In a sense these so-called hubs drive connectivity of the system. This makes sense in practice. If you are travelling by train around the South-East of England, it is very likely you will pass through at least one of Reading, East Croydon, or about five major terminus in London. It would be absurd for every station to be of equal significance to the network. By contrast, the typical vertex degree in the sparse Erdos-Renyi model is O(1), and has a limiting Poisson distribution, with a super-exponential tail.

So, this paper addresses the following question. We know that the PA model, when set up right, has power-law tails for the degree distribution, and so has a largest degree that is an order of magnitude larger than the average degree. Let’s call this the ‘hub’ for now. But the model is dynamic, so we should ask how this hub changes in time as we add extra vertices. In particular, is it the case that one vertex should grow so large that it remains as the dominant hub forever? This paper answers this question in the affirmative, for a certain class of preferential attachment schemes.

We assign a weighting system to possible degrees, that is a function from N to R+. In the case of proportional PA, this function could be f(n)=n. In general, we assume it is convex. Note that the more convex this weight function is, the stronger the preference a new vertex feels towards existing dominant vertices. Part of the author’s proof is a formalisation of this heuristic, which provides some machinery allowing us to treat only really the case f(n)=n. I will discuss only this case from now on.

I want to focus on the fact that we have another model which describes aspects of the degree evolution rather well. We consider some finite fixed collection of vertices at some time, and consider the evolution of their degrees. We will be interested in limiting properties, so the exact time doesn’t matter too much. We look instead at the jump chain, ie those times when one of the degrees changes. This happens when a new vertex joins to one of the chosen vertices. Given that the new vertex has joined one of the chosen vertices, the choice of which of the chosen vertices is still size-biased proportional to the current degrees. In other words, the jump chain of this degree sequence is precisely Polya’s Urn.

This is a powerful observation, as it allows us to make comments about the limiting behaviour of finite quantities almost instantly. In particular, we know that from any starting arrangement, Polya’s Urn converges almost surely. This is useful to the question of persistence for the following reason.

Recall that in the case of two colours, starting with one of each, we converge to the uniform distribution. We should view this as a special case of the Dirichlet distribution, which is supported on partitions into k intervals of [0,1]. In particular, for any fixed k, the probability that two of the intervals have the same size is 0, as the distribution is continuous. So, since the convergence of the proportions in Polya’s Urn is almost sure, with probability one all of the proportions are with \epsilon>0 of their limit, and so taking epsilon small enough, given the limit, which we are allowed to do, we can show that the colour which is largest in the limit is eventually the largest at finite times.

Unfortunately, we can’t mesh these together these finite-dimensional observations particularly nicely. What we require instead is a result showing that if a vertex has large enough degree, then it can never be overtaken by any new vertex. This proved via a direct calculation of the probability that a new vertex ‘catches up’ with a pre-existing vertex of some specified size.

That calculation is nice and not too complicated, but has slightly too many stages and factorial approximations to consider reproducing or summarising here. Instead, I offer the following heuristic for a bound on the probability that a new vertex will catch up with a pre-existing vertex of degree k. Let’s root ourselves in the urn interpretation for convenience.

If the initial configuration is (k,1), corresponding to k red balls and 1 blue, we should consider instead the proportion of red balls, which is k/k+1 obviously. Crucially (for proving convergence results if nothing else), this is a martingale, which is clearly bounded within [0,1]. So the expectation of the limiting proportion is also k/k+1. Let us consider the stopping time T at which the number of red balls is equal to the number of blue balls. We decompose the expectation by conditioning on whether T is finite.

\mathbb{E}X_\infty=\mathbb{E}[X_\infty|T<\infty]\mathbb{P}(T<\infty)+\mathbb{E}[X_\infty|T=\infty]\mathbb{P}(T=\infty)

\leq \mathbb{E}[X_\infty | X_T,T<\infty]\mathbb{P}(T<\infty)+(1-\mathbb{P}(T=\infty))

using that X_\infty\leq 1, regardless of the conditioning,

= \frac12 \mathbb{P}(T<\infty) + (1-\mathbb{P}(T<\infty))

\mathbb{P}(T<\infty) \leq \frac{2}{k+1}.

We really want this to be finite when we sum over k so we can use some kind of Borel-Cantelli argument. Indeed, Galashin gets a bound of O(k^{-3/2}) for this quantity. We should stress where we have lost information. We have made the estimate \mathbb{E}[X_\infty|T=\infty]=1 which is very weak. This is unsurprising. After all, the probability of this event is large, and shouldn’t really affect the limit that much when it does not happen. The conditioned process is repelled from 1/2, but that is of little relevance when starting from k/k+1. It seems likely this expectation is in fact \frac{k}{k+1}+O(k^{-3/2}), from which the result will follow.

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Critical Components in Erdos-Renyi

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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

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