# 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 3-regular graphs

A graph is d-regular if every vertex has degree d. Probably the easiest examples of d-regular graphs are the complete graph on (d+1) vertices, and the infinite d-ary tree. A less trivial example is the Petersen graph, which is 3-regular. 3-regular graphs will be the main focus for some of this post, but initially we lose nothing by considering general d.

Throughout, a necessary condition for the existence of a d-regular graph with N vertices is that at least one of d and N is even, as the sum of the degrees of a graph must be even. We will always assume that this holds, so that when d=3, we are always taking N to be even.

A natural pair of questions for a probabilist is ‘can we sample a d-regular graph with N vertices uniformly at random?’ and ‘what does a typical large d-regular graph look like?’

In a rather old post, I addressed some aspects of the first question, but revisit it briefly here. A good idea, due to Bollobas [B80] is to assign to all the vertices d stubs (or half-edges), and choose a matching of the Nd stubs uniformly at random. This works as a method to generate a random graph with any fixed degree sequence.

If you want your graphs to be simple, this can go wrong, because there’s a chance you get loops (that is, an edge from a vertex v to itself) and multiple edges between the same pair of vertices. It would be nice the graph formed in this fashion was simple with high probability when $N\rightarrow\infty$. Unfortunately that’s not the case, however the probability that the graph is simple remains asymptotically bounded away from 0 and 1. Indeed, because the presence of a loop / multiple edge is asymptotically independent of the presence of a loop / multiple edge elsewhere, it’s unsurprising we have a Poisson limit for the number of such occurences. So from a sampling point of view, it’s reasonable to sample a graph in this way until you find a simple one. This takes O(1) steps, and it’s O(N) steps to check whether a given multigraph is simple.

It’s clear that conditional on the graph generated in this fashion being simple, its distribution is uniform on the set of simple graphs with the correct degree distribution. If you are happy for your graphs to have loops, then it’s a little bit more complicated, because if an edge has multiplicity k, these can appear in k! ways in the configuration construction.

Other asymptotic properties

Loops and multiple edges can be thought of as cycles of length 1 and 2 respectively if you want. We might ask about other small cycles. A calculation in expectation is relatively straightforward. Given three vertices, the probability they form a triangle (in at least one way) is $\Theta(N^{-3})$, and there are $\Theta(N^3)$ ways to choose three vertices. Thus the expected number of triangles is $\Theta(1)$. Finally, the edge structure induced on disjoint triples is asymptotically independent, and hence a Poisson limit. (See [J06] for details, including more detail on the general configuration construction.) The same result holds for the same reasons for cycles of any fixed finite length.

We might also ask about connectivity. At a heuristic level, there are two ways for the graph to be disconnected: it could have some small components; or it could have two components of size $\Theta(N)$. The smallest possible component is $K_4$, and an argument like for the cycles above shows that the number of copies of $K_4$ vanishes in expectation. Now, consider having two components of size roughly N/2. There are $\binom{N}{N/2} \sim 2^{2N}$ ways to make this choice. However, given such a choice, we can handle the probability that all the stubs from one class match within that class by going through the class one stub at a time:

$\frac{\frac{3N}{2}-1}{3N-1} \times \frac{\frac{3N}{2}-3}{3N-3} \times \cdots \times \frac{1}{\frac{3N}{2}+1}.$

We approximate this as

$\frac{\sqrt{(3N/2)!}}{\sqrt{ (3N)!}} \sim e^{3N/2} 2^{-3N/2} \left(3N\right)^{-3N/2},$

and this dominates the number of choices powerfully enough that we might believe it remains valid for a broader range of class sizes. In fact we have a much stronger statement, namely that G(N,3) is 3-connected with high probability. This means that the graph cannot be disconnected by removing two vertices, or equivalently that there are three vertex-disjoint paths between any pair of vertices in the graph, essentially one emerging from each stub. See this note by David Ellis for a quick proof. We might return to this later.

You might ask about planarity. It’s clear from degree consideration that there are no induced copies of $K_5$ in any random 3-regular graph, and since $K_{3,3}$ contains a cycle of length 4, and with high probability G(N,3) doesn’t, that takes care of that possibility too. However, there might be minors of this form. This seemed a good example of the Kuratowski criterion not actually being that useful, since I certainly don’t find the minors of the 3-regular graph an obvious structure to handle.

However, we can use Euler’s formula V – E + F = 2 for planar graphs. Here V = N, E = 3N/2. Faces are described by (a subset of the) cycles, and we there are asymptotically O(1) small cycles, so most faces include a large number of edges. But each edge corresponds to at most two faces. So we have $F \ll E$, and so with high probability Euler’s formula can’t hold in G(N,3) for large N.

We can also ask about the local limit of G(N,3). Since the vertices are exchangeable, we don’t need to worry about whether we choose the root uniformly at random (often referred to as the Benjamini-Schramm sense) or by some other method.

The root has up to three neighbours, and with high probability it has exactly three neighbours. These neighbours have at most two other neighbours themselves. However, we’ve already seen that there are asymptotically O(1) cycles, and so with high probability there are no small cycles near a fixed root vertex. So the six neighbours-of-neighbours are with high probability different to the root and the root’s neighbours and to each other. We can make this argument at arbitrary finite radius from the root, to conclude that the local limit of G(N,3) is the infinite 3-ary tree.

Spectral expansion

[Caveat – this is something I read about and wanted to mention, but I really don’t know much at all about any of this theory, and it’s definitely not certain that what follows wouldn’t be better replaced by a set of links.]

This straightforward local limit offers good heuristics on some of the more global properties. Almost by definition, the d-ary tree expands as rapidly as is possible away from the root among infinite d-regular graphs. There are a number of ways to measure the expansion of a graph, and some methods transfer better to the infinite setting than others. The adjacency matrix of an infinite graph can be defined similarly to that of a finite graph, and it remains possible to talk about eigenfunctions and spectrum. As for the finite setting, d is an eigenvalue because the tree is d-regular, and -d is an eigenvalue because it is also bipartite.

The next largest eigenvalue $\lambda_2$ governs the spectral gap $d-\lambda_2$ which is a measure of the expansion of a graph. A graph is a good (spectral) expander if all the non-trivial eigenvalues are close to zero. A priori, all we know is that $|\lambda_2|\le d$. For the infinite d-ary tree, we have $\lambda_2 = 2\sqrt{d-1}$. This blog post by Luca Trevisan gives a very readable proof.

A key result is that finite graphs can have $\lambda_2 \le 2\sqrt{d-1}$, but not asymptotically. That is, taking N to be the number of vertices:

$\lambda_2 \ge 2\sqrt{d-1} - o_N(1).$

This is the content of the Alon-Boppana theorem [Al86]. In fact the error can be quantified as $O(\frac{1}{\log N})$ – the diamater of the graph is relevant here. A finite d-regular graph for which $\lambda_2\le 2\sqrt{d-1}$ is called a Ramanujan graph. The existence of Ramanujan graphs has been much studied, and various constructions often rely on number theoretic properties of N, and lie at the interface of disparate branches of mathematics where my understanding is zero rather than epsilon.

Now return to our view of the d-ary tree as the local limit of a d-regular graph on N vertices for large N. We might expect from everything above that the uniform d-regular graph is a good expander. Bollobas shows that in the sense of edge-expansion, asymptotically almost all d-regular graphs have edge-expansion bounded away from zero. (See Section 2 of [Ell], including history of the d=3 case.) Friedman [Fri08] proves the conjecture of Alon that for every $\epsilon>0$, a.a.s. $\lambda_2$ for G(N,d) is at most $2\sqrt{d-1}+\epsilon$. In this sense, G(N,d) is asymptotically ‘almost Ramanujan’. (See also [Bor17] for another proof and an introduction including history, context and references.)

Some other links: The Wikipedia page on expanders, which includes a discussion of the different descriptions of expansion, and the Cheeger inequalities and other relations between them; slides for a talk by Spielman on spectra and Ramanujan graphs; a survey by Murty on Ramanujan graphs;.

What next?

This post took a slightly different direction from what I had intended, and rather than make a halting U-turn back to my planned finale, I’ll postpone this. However, a short overture is that I’m interested in the structure of critical components of random graphs during the critical window. This is the window during which the largest components first have cycles with probability $\Theta(1)$. Indeed, the critical components have size $\Theta(N^{2/3})$ and $\Theta(1)$ surplus edges. Conditional on their size, and number of surplus edges, the choice of the graph structure on the component is uniform among such (connected) graphs.

Addario-Berry, Broutin and Goldschmidt [ABG09] study scaling limits of such components. Central to this analysis is the 2-core of such components, which can be described in terms of 3-regular (multi)graphs. Various processes we are now interested in running on the critical components of critical RGs can then be studied in terms of related processes on random 3-regular graphs.

References

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

[Al86] – Alon – Eigenvalues and expanders

[B80] – Bollobas – A probabilistic proof of an asymptotic formula for the number of labelled regular graphs

[B88] – Bollobas – The isoperimetric number of random regular graphs

[Bor17] – Bordenave – A new proof of Friedman’s second eigenvalue theorem and its extension to random lifts. Arxiv.

[Ell] – Ellis – The expansion of random regular graphs

[Fri08] – Friedman – A proof of Alon’s second eigenvalue conjecture and related problems

[J06] – Janson – The probability that a random multigraph is simple by

# 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 $G=K_n$.

# Balkan MO 2017 – Qs 1, 3 and 4

The UK is normally invited to participate as a guest team at the Balkan Mathematical Olympiad, an annual competition between eleven countries from South-Eastern Europe. I got to take part in Rhodes almost exactly ten years ago, and this year the competition was held in Ohrid, in Macedonia. There’s one paper, comprising four questions, normally one from each of the agreed olympiad topic areas, with 4.5 hours for students to address them. The contest was sat this morning, and I’m going to say quite a bit about the geometric Q2, and a little bit about Qs 1 and 3 also. In all cases, this discussion will include most of a solution, with some commentary, so don’t read these if you are planning to try the problems yourself.

I’m not saying anything about Q4, because I haven’t solved it. (Edit: I have solved it now, so will postpone Q2 until later today.)

Question One

Find all ordered pairs of positive integers (x,y) such that

$x^3+y^3=x^2+42xy+y^2.$

The first thought is that if either of x or y is ‘large’, then the LHS is bigger than the RHS, and so equality can’t hold. That is, there are only finitely many solutions. The smallest possible value of y is, naturally, 1, and substituting y=1 is convenient as then $y^2=y^3$, and it’s straightforward to derive $x=7$ as a solution.

Regarding the non-existence of large solutions, you can make this precise by factorising the LHS as

$(x+y)(x^2-xy+y^2) = x^2+42xy+y^2.$

There are 44 terms of degree two on the RHS, and one term of degree in the second bracket on the LHS. With a bit of AM-GM, you can see then that if $x+y>44$, you get a contradiction, as the LHS will be greater than the RHS. But that’s still a lot of possibilities to check.

It struck me that I could find ways to reduce the burden by reducing modulo various primes. 2, 3 and 7 all divide 42, and furthermore cubes are nice modulo 7 and squares are nice modulo 3, so maybe that would bring the number of possibilities down. But my instinct was that this wasn’t the right way to use the fact that we were solving over positive integers.

The second bracket in the factorisation looks enough like the RHS, that it’s worth exploring. If we move $x^2-xy+y^2$ from the right to the left, we get

$(x+y-1)(x^2-xy+y^2) = 43xy.$ (1.1)

Now it suddenly does look useful that we are solving over positive integers, because 43 is a prime, so has to appear as a factor somewhere on the LHS. But it’s generally quite restrictive that $x^2-xy+y^2 | 43xy$. This definitely looks like something that won’t hold often. If x and y are coprime, then certainly $x^2-xy+y^2$ and $y$ are coprime also. But actually if x and y have a non-trivial common factor d, we can divide both sides by $d^2$, and it still holds. Let’s write

$x=dm,\quad y=dn,\quad\text{where }d=\mathrm{gcd}(x,y).$

Then $m^2 -mn+n^2$ really does divide 43, since it is coprime to both m and n. This is now very restrictive indeed, since it requires that $m^2-mn+n^2$ be equal to 1 or 43. A square-sandwiching argument gives $m^2-mn+n^2=1$ iff $m=n=1$. 43 requires a little bit more work, with (at least as I did it) a few cases to check by hand, but again only has one solution, namely $m=7, n=1$ and vice versa.

We now need to add the common divisor d back into the mix. In the first case, (1.1) reduces to $(2d-1)=43$, which gives $(x,y)=(22,22)$. In the second case, after cancelling a couple of factors, (1.1) reduces to $(8d-1)=7$, from which $(x,y)=(7,1),(1,7)$ emerges, and these must be all the solutions.

The moral here seemed to be that divisibility was a stronger tool than case-reduction. But that was just this question. There are other examples where case-reduction is probably more useful than chasing divisibility.

Question Three

Find all functions $f:\mathbb{N}\rightarrow\mathbb{N}$ such that

$n+f(m) \,\big|\, f(n)+nf(m)$

for all $m,n\in\mathbb{N}$.

What would be useful here? There are two variables, and a function. It would be useful if we could reduce the number of variables, or the number of occurences of f. We can reduce the number of variables by taking m=n, to get

$n+f(n) \,\big|\, f(n) [1+n].$ (3.1)

From this, we might observe that $f(n)\equiv 1$ is a solution. Of course we could analyse this much more, but this doesn’t look like a 10/10 insight, so I tried other things first.

In general, the statement that $a\,|\,b$ also tells us that $a\,|\, b-ka$. That is, we can subtract arbitrary multiples of the divisor, and the result is still true. A recurring trope is that the original b is elegant, but an adjusted b-ka is useful. I don’t think we can do the latter, but by subtracting $n^2 +nf(m)$ from the problem statement, we get

$n+f(m) \,\big|\, n^2-f(n).$ (3.2)

There’s now no m on the RHS, but this relation has to hold for all m. One option is that $f(n)=n^2$ everywhere, then what we’ve deduced always holds since the RHS is zero. But if there’s a value of n for which $f(n)\ne n^2$, then (3.2) is a very useful statement. From now on, we assume this. Because then as we fix n and vary m, we need $n+f(m)$ to remain a divisor of the RHS, which is fixed, and so has finitely many divisors. So $f(m)$ takes only finitely many values, and in particular is bounded.

This ties to the observation that $f\equiv 1$ is a solution, which we made around (3.1), so let’s revisit that: (Note, there might be more elegant ways to finish from here, but this is what I did. Also note, n is no longer fixed as in previous paragraph.)

$n+f(n) \,\big|\, f(n) [1+n].$ (3.1)

Just to avoid confusion between the function itself, and one of the finite collection of values it might take, let’s say b is a value taken by f. So there are values of n for which

$n+b \,\big|\, b(1+n).$

By thinking about linear equations, you might be able to convince yourself that there are only finitely many solutions (in n) to this relation. There are certainly only finitely many solutions where LHS=RHS (well, at most one solution), and only finitely many where 2xLHS=RHS etc etc. But why do something complicated, when we can actually repeat the trick from the beginning, and subtract $b(n+b)$, to obtain

$n+b \,\big|\, b^2-b.$

For similar reasons to before, this is a great deduction, because it means if $b\ne 1$, then the RHS is positive, which means only finitely many n can satisfy this relation. Remember we’re trying to show that no n can satisfy this relation if $b\ne 1$, so this is definitely massive progress!

If any of what’s already happened looked like magic, I hope we can buy into the idea that subtracting multiples of the divisor from the RHS is the only tool we used, and that making the RHS fixed gives a lot of information about the LHS as the free variable varies. The final step is not magic either. We know that f is eventually 1. If you prefer “for large enough n, $f(n)=1$,” since all other values appear only finitely often. I could write this with quantifiers, but I don’t want to, because that makes it seem more complicated than it is. We genuinely don’t care when the last non-1 value appears.

Anyway, since we’ve deduced this, we absolutely have to substitute this into something we already have. Why not the original problem statement? Fix m, then for all large enough n

$n+f(m) \,\big|\, 1+nf(m).$ (3.3)

To emphasise, (3.3) has to hold for all large enough n. Is it possible that f(m)=2? Again, it’s easy to convince yourself not. But, yet again, why not use the approach we’ve used so profitably before to clear the RHS? In fact, we already did this, and called it (3.2), and we can make that work [3.4], but in this setting, because f(m) is fixed and we’re working with variable large n, it’s better to eliminate n, to get

$n+f(m)\,\big|\, f(m)^2-1,$

again for all large enough n. By the same size argument as before, this is totally impossible unless f(m)=1. Which means that in fact $f(m)=1$ for all m. Remember ages ago we assumed that f(n) was not $n^2$ everywhere, so this gives our two solutions: $f(n)=1,\, f(n)=n^2$.

Moral: choosing carefully which expression to work with can make life much more interesting later. Eliminating as many variables or difficult things from one side is a good choice. Playing with small values can help you understand the problem, but here you need to think about soft properties of the expression, in particular what happens when you take one variable large while holding another fixed.

[3.4] – if you do use the original approach, you get $n^2-1$ on the RHS. There’s then the temptation to kill the divisibility by taking n to be the integer in the middle of a large twin prime pair. Unfortunately, the existence of such an n is still just a conjecture

Question Four

(Statement copied from Art of Problem Solving. I’m unsure whether this is the exact wording given to the students in the contest.)

On a circular table sit n>2 students. First, each student has just one candy. At each step, each student chooses one of the following actions:

(A) Gives a candy to the student sitting on his left or to the student sitting on his right.

(B) Separates all its candies in two, possibly empty, sets and gives one set to the student sitting on his left and the other to the student sitting on his right.

At each step, students perform the actions they have chosen at the same time. A distribution of candy is called legitimate if it can occur after a finite number of steps.
Find the number of legitimate distributions.

My moral for this question is this: I’m glad I thought about this on the bus first. What I found hardest here was getting the right answer. My initial thoughts:

• Do I know how to calculate the total number of possibilities, irrespective of the algorithm? Fortunately yes I do. Marbles-in-urns = barriers between marbles on a line (maybe add one extra marble per urn first). [4.1]
• What happens if you just use technique a)? Well first you can get into trouble because what happens if you have zero sweets? But fine, let’s temporarily say you can have a negative number of sweets. If n is even, then there’s a clear parity situation developing, as if you colour the children red and blue alternately, at every stage you have n/2 sweets moving from red children to blue and vice versa, so actually the total number of sweets among the red children is constant through the process.
• What happens if you just use technique b)? This felt much more promising.
• Can you get all the sweets to one child? I considered looking at the child directly opposite (or almost-directly opposite) and ‘sweeping’ all the sweets away from them. It felt like this would work, except if for some parity reason we couldn’t prevent the final child having one (or more, but probably exactly one) sweets at the crucial moment when all the other sweets got passed to him.

Then I got home, and with some paper, I felt I could do all possibilities with n=5, and all but a few when n=6. My conjecture was that all are possible with n odd, and all are possible with n even, except those when none of the red kids or none of the kids get a sweet. I tried n=8, and there were a few more that I couldn’t construct, but this felt like my failure to be a computer rather than a big problem. Again there’s a trade-off between confirming your answer, and trying to prove it.

Claim: If n is even, you can’t achieve the configurations where either the red children or the blue children have no sweets.

Proof: Suppose you can. That means there’s a first time that all the sweets were on one colour. Call this time T. Without loss of generality, all the sweets are on red at T. Where could the sweets have been at time T-1? I claim they must all have been on blue, which contradicts minimality. Why? Because if at least one red child had at least one sweet, they must have passed at least one sweet to a blue neighbour.

Now it remains to give a construction for all other cases. In the end, my proof has two stages:

Step One: Given a configuration, in two steps, you can move a candy two places to the right, leaving everything else unchanged.

This is enough to settle the n odd case. For the even case, we need an extra step, which really corresponds to an initial phase of the construction.

Step Two: We can make some version of the ‘sweeping’ move precise, to end up in some configuration where the red number of children have any number of sweets except 0 or n.

Step one is not so hard. Realising that step one would be a useful tool to have was probably the one moment where I shifted from feeling like I hadn’t got into the problem to feeling that I’d mostly finished it. As ever in constructions, working out how to do a small local adjustment, which you plan to do lots of times to get a global effect, is great. (Think of how you solve a Rubik’s cube for example.)

Step two is notationally fiddly, and I would think very carefully before writing it up. In the end I didn’t use the sweeping move. Instead, with the observation that you can take an adjacent pair and continually swap their sweets it’s possible to set up an induction.

Actual morals: Observing the possibility to make a small change in a couple of moves (Step one above) was crucial. My original moral does still hold slightly. Writing lots of things down didn’t make life easier, and in the end the ideas on the bus were pretty much everything I needed.

[4.1] – one session to a group of 15 year olds is enough to teach you that the canon is always ‘marbles in urns’ never ‘balls’ nor ‘bags’, let alone both.

# EGMO 2017 – Paper One – Geometric subconfigurations

I’ve recently been in Cambridge, running the UK’s annual training and selection camp for the International Mathematical Olympiad. My memories of living and studying in Cambridge are very pleasant, and it’s always nice to be back.

Within olympiad mathematics, the UK has traditionally experienced a weakness at geometry. By contrast to comparable nations, for example those from Eastern Europe, our high school curriculum does not feature much Euclidean geometry, except for the most basic of circle theorems and angle equalities, which normally end up as calculation exercises, rather than anything more substantial. So to arrive at the level required to be in with a chance of solving even the easier such questions at international competitions, our students have to do quite a lot of work for themselves.

I’ve spent a bit of time in the past couple of years thinking about this, and how best to help our students achieve this. The advice “go away and do as many problems as you can, building up to IMO G1, then a bit further” is probably good advice, but we have lots of camps and correspondence training, and I want to offer a bit more.

At a personal level, I’m coming from a pragmatic point of view. I don’t think Euclidean geometry is particularly interesting, even though it occasionally has elegant arguments. My main concern is taming it, and finding strategies for British students (or anyone else) to tame it too [1].

Anyway, I’m going to explain my strategy and thesis as outlined at the camp, then talk about Question 1 from EGMO 2017, a competition held in Zurich this year, the first paper of which was sat earlier today (at time of writing). The UK sent a strong team of four girls, and I’m looking forward to hearing all about their solutions and their adventures, but later. I had intended to talk about the other two questions too, but I can’t think of that much to say, so have put this at the end.

My proposed strategy

Before explaining my proposed strategy, let me discuss a couple of standard approaches that sometimes, but rarely, work at this level:

• Angle chase (or length chase) forwards directly from the configuration. Consider lots of intersection points of lines. Consider angles and lengths as variables, and try to find relations.
• Exactly as above, but working back from the conclusion.
• Doing both, and attempting to meet in the middle.

The reason why this doesn’t work is that by definition competitions are competitive, and all participants could probably do this. For similar reasons competition combinatorics problems tend not to reduce instantly to an exhaustive search.

It’s also not very interesting. I’m certainly unlikely to set a problem if it’s known to yield to such an approach. When students do try this approach, common symptoms and side-effects involve a lot of chasing round conditions that are trivially equivalent to conditions given in the statement. For example, if you’re given a cyclic quadrilateral, and you mark on opposing complementary angles, then chase heavily, you’ll probably waste a lot of time deducing other circle theorems which you already knew.

So actually less is more. You should trust that if you end up proving something equivalent to the required conclusion, you’ll notice. And if you are given a cyclic quadrilateral, you should think about what’s the best way to use it, rather than what are all the ways to use it.

On our selection test, we used a problem which essentially had two stages. In the first stage, you proved that a particular quadrilateral within the configuration was cyclic; and in the second stage, you used this to show the result. Each of these stages by themselves would have been an easy problem, suitable for a junior competition. What made this an international-level problem was that you weren’t told that these were the two stages. This is where a good diagram is useful. You might well guess from an accurate figure that TKAD was cyclic, even if you hadn’t constructed it super-accurately with ruler and compasses.

So my actual strategy is to think about the configuration and the conclusion separately, and try and conjecture intermediate results which might be true. Possibly such an intermediate result might involve an extra point or line. This is a standard way to compose problems. Take a detailed configuration, with some interesting properties within it, then delete as much as possible while keeping the properties. Knowing some standard configurations will be useful for this. Indeed, recognising parts of the original diagram which resemble known configurations (possibly plus or minus a point or line) is a very important first step in many settings.

Cyclic quadrilaterals and isosceles triangles are probably the simplest examples of such configurations. Think about how you often use properties of cyclic quadrilaterals without drawing in either the circle or its centre. The moral is that you don’t need every single thing that’s true about the configuration to be present on the diagram to use it usefully. If you know lots of configurations, you can do this sort of thing in other settings too. Some configurations I can think up off the top of my head include: [2]

• Parallelograms. Can be defined by corresponding angles, or by equal opposite lengths, or by midpoint properties of the centre. Generally if you have one of these definitions, you should strongly consider applying one of the other definitions!
• The angle bisector meets the opposite perpendicular bisector on the circumcircle.
• Simson’s line: the feet of the three perpendiculars from a point to the sides (extended if necessary) of a triangle are collinear precisely when the point is on the circumcircle.
• The incircle touch point and the excircle touch point are reflections of each other in the corresponding midpoint. Indeed, all the lengths in this diagram can be described easily.
• The spiral similarity diagram.
• Pairs of isogonal conjugates, especially altitudes and radii; and medians and symmedians.

Note, all of these can be investigated by straightforward angle/length-chasing. We will see how one configuration turned out to be very useful in EGMO. In particular, the configuration is simple, and its use in the problem is simple, but it’s the idea to focus on the configuration as often as possible that is key. It’s possible but unlikely you’d go for the right approach just by angle-analysis alone.

EGMO 2017 Question 1

Let ABCD be a convex quadilateral with <DAB=<BCD=90, and <ABC > <CDA. Let Q and R be points on segments BC and CD, respectively, such that line QR intersects lines AB and AB at points P and S, respectively. It is given that PQ=RS. Let the midpoint of BD be M, and the midpoint of QR be N. Prove that the points M, N, A and C lie on a circle.

First point: as discussed earlier, we understand cyclic quadrilaterals well, so hopefully it will be obvious once we know enough to show these four points are concyclic. There’s no point guessing at this stage whether we’ll do it by eg opposite angles, or by power of a point, or by explicitly finding the centre.

But let’s engage with the configuration. Here are some straightforward deductions.

• ABCD is cyclic.
• M is the centre.

We could at this stage draw in dozens of equal lengths and matching angles, but let’s not do that. We don’t know yet which ones we’ll need, so we again have to trust that we’ll use the right ones when the time comes.

What about N? If we were aiming to prove <AMC = <ANC, this might seem tricky, because we don’t know very much about this second angle. Since R and Q are defined (with one degree of freedom) by the equal length condition, it’s hard to pin down N in terms of C. However, we do know that N is the midpoint opposite C in triangle QCR, which has a right angle at C. Is this useful? Well, maybe it is, but certainly it’s reminiscent of the other side of the diagram. We have four points making up a right-angled triangle, and the midpoint of the hypotenuse here, but also at (A,B,D,M). Indeed, also at (C,B,D,M). And now also at (C,Q,R,N). This must be a useful subconfiguration right?

If you draw this subdiagram separately, you have three equal lengths (from the midpoint to every other point), and thus two pairs of equal angles. This is therefore a very rich subconfiguration. Again, let’s not mark on everything just yet – we trust we’ll work out how best to use it later.

Should we start angle-chasing? I think we shouldn’t. Even though we have now identified lots of potential extra pairs of equal angles, we haven’t yet dealt with the condition PQ=RS at all.

Hopefully as part of our trivial equivalences phase, we said that PQ=RS is trivially equivalent to PR=QS. Perhaps we also wrote down RN=NQ, and so it’s also trivially equivalent to PN=NS. Let’s spell this out: N is the midpoint of PS. Note that this isn’t how N was defined. Maybe this is more useful than the actual definition? (Or maybe it isn’t. This is the whole point of doing the trivial equivalences early.)

Well, we’ve already useful the original definition of N in the subconfiguration (C,Q,R,N), but we can actually also use the subconfiguration (A,P,S,N) too. This is very wordy and makes it sound complicated. I’ve coloured my diagram to try and make this less scary. In summary, the hypotenuse midpoint configuration appears four times, and this one is the least obvious. If you found it, great; if not, I hope this gave quite a lot of motivation. Ultimately, even with all the motivation, you still had to spot it.

Why is this useful? Because a few paragraphs earlier, I said “we don’t know very much about this second angle <ANC”. But actually, thanks to this observation about the subconfiguration, we can decompose <ANC into two angle, namely <ANP+<QNC which are the apex angle in two isosceles triangles. Now we can truly abandon ourselves to angle-chasing, and the conclusion follows after a bit of work.

I’m aware I’ve said it twice in the prelude, and once in this solution, but why not labour my point? The key here was that spotting that a subconfiguration appeared twice led you to spot that it appeared a further two times, one of which wasn’t useful, and one of which was very useful. The subconfiguration itself was not complicated. To emphasise its simplicity, I can even draw it in the snow:

Angle-chasing within the configuration is easy, even with hiking poles instead of a pen, but noticing it could be applied to point N was invaluable.

Other questions

Question 2 – My instinct suggested the answer was three. I find it hard to explain why. I was fairly sure they wouldn’t have asked if it was two. Then I couldn’t see any reason why k would be greater than 3, but still finite. I mean, is it likely that $k=14$ is possible, but $k=13$ is not.

In any case, coming up with a construction for $k=3$ is a nice exercise, and presumably carried a couple of marks in the competition. My argument to show $k=2$ was not possible, and most arguments I discussed with others were not overwhelmingly difficult, but didn’t really have any key steps or insight, so aren’t very friendly in a blog context, and I’ll probably say nothing more.

Question 3 – Again, I find it hard to say anything very useful, because the first real thing I tried worked, and it’s hard to motivate why. I was confused how the alternating turn-left / turn-right condition might play a role, so I ignored it initially. I was also initially unconvinced that it was possible to return to any edge in any direction (ie it must escape off to infinity down some ray), but I was aware that both of these were too strong a loosening of the problem to be useful, in all likelihood.

Showing that you can go down an edge in one direction but not another feels like you’re looking for some binary invariant, or perhaps a two-colouring of the directed edges. I couldn’t see any way to colour the directed edges, so I tried two-colouring the faces, and there’s only one way to do this. Indeed, on the rare occasions (ahem) I procrastinate, drawing some lines then filling in the regions they form in this form is my preferred doodle. Here’s what it looks like:

and it’s clear that if the path starts with a shaded region on its right, it must always have a shaded region on its right. As I say, this just works, and I find it hard to motivate further.

A side remark is that it turns out that my first loosening is actually valid. The statement remains true with arbitrary changes of direction, rather than alternating changes. The second loosening is not true. There are examples where the trajectory is periodic. I don’t think they’re hugely interesting though, so won’t digress.

Footnotes

[1] – “To you, I am nothing more than a fox like a hundred thousand other foxes. But if you tame me, then we shall need each other. To me, you will be unique in all the world. To you, I shall be unique in all the world,” said the Fox to the Little Prince. My feelings on taming Euclidean geometry are not this strong yet.

[2] – Caveat. I’m not proposing learning a big list of standard configurations. If you do a handful of questions, you’ll meet all the things mentioned in this list several times, and a few other things too. At this point, your geometric intuition for what resembles what is much more useful than exhaustive lists. And if you’re anxious about this from a pedagogical point of view, it doesn’t seem to me to be a terribly different heuristic from lots of non-geometry problems, including in my own research. “What does this new problem remind me of?” is not unique to this area at all!

# RMM 2017 – Problems 2, 3 and 6

In the previous post, I discussed Problems 1, 4 and 5 from this year’s Romanian Master of Mathematics competition. In this post, I discuss the harder problems (modulo my subjective appreciation of difficulty).

Problem 2

Determine all positive integers n satisfying the following condition: for every monic polynomial P of degree at most n with integer coefficients, there exists a positive integer $k \leq n$, and (k+1) distinct integers $x_1,\ldots,x_{k+1}$ such that

$P(x_1) + P(x_2) + \cdots + P(x_k) = P(x_{k+1}).$

Parsing this question deserve at least a moment. Straight after a first reading, I find it worth writing down any key quantifiers which I might forget later. Here, it’s the words at most. If you want to show the statement holds for n=2, you need to investigate monic polynomials with degree zero, one and two. You should also make sure that any instances of $x_i$ really are always distinct.

This matters in competitions! Two of our contestants failed to get the mark for showing n=2 works, precisely because of not checking the linear case, and a third could have lost it for using examples which are sometimes not distinct. On hard papers, one mark actually is the difference between triumph and frustration. And of course it matters outside competitions too, since small cases are exactly what your reader might examine first, to check they understand the problem posed, so it’s not a good place for awkward errors.

I started by trying to show that it couldn’t possibly happen that every polynomial with degree at most n had this property, for some combinatorial reason. For example, that if every set of distinct integers could only be a solution set for a small number of polynomials, then we would end up with not enough polynomials. But I couldn’t make this work at all; every bound ended up heavily in the wrong direction.

The next natural question is, does a typical polynomial of degree at most n have this property? But choosing a typical polynomial is hard, so in fact I asked, do the simplest polynomials of degree at most n have this property? I think the simplest polynomials of degree at most n are $\{1,x,x^2,\ldots,x^n\}$. Under what circumstances does

$x_1^m + \ldots x_k^m = x_{k+1}^m,$ (1)

have solutions in distinct integers? Famously, when k=2 and $m\ge 3$ this is a very very hard problem indeed. So the first point is that it though it might be useful to use Fermat’s Last Theorem, it would be foolish to pursue a strategy which, if successful, would have a proof of FLT as a sub-problem. At least, it would be foolish if the aim was to finish this strategy within a few hours.

So my main comment on this question is meta-mathematical. If lots of attempts at general arguments don’t work, there must be some special example that does it. And what properties do I want this special example to have? Maybe one might have thought of this from scratch, but my motivation came from (1) in the case m=p-1. Then, by Fermat’s Little Theorem, all the summands are equal to 1 or 0 modulo p. If k>p, then after discounting any uniform factors of p, we obtain a congruence equation which is, in informal terms,

$\left(0\text{ or }1\right)+\ldots+\left(0\text{ or }1\right) \equiv \left(0\text{ or }1\right).$

This looks really promising because it’s quite restrictive, but it’s still just a bit annoying: there are quite a few solutions. But it does give us the right idea, which is to find a polynomial P for which $P(x)\equiv 1$ modulo n. The equation $1+\ldots+1\equiv 1$ modulo n has solutions only if the number of summands on the LHS is 1 modulo n. So in this context, this reduces to showing that P is, additionally, injective on the integers, ie that P(x)=P(y) only when x=y.

It’s a nice exercise to show the existence of polynomials which are constant modulo n, and a good problem to work out how to force injectivity. If a polynomial is increasing everywhere, then it is certainly injective, and so the problem ends up being slightly easier in the case where the degree is odd than when the degree is even, but this is a nice conclusion to a nice problem, so I’ll save it for any interested readers to finish themselves.

Problem 3

Let n be an integer greater than 1 and let X be an n-element set. A non-empty collection of subsets $A_1,\ldots, A_k$ of X is tight if the union $A_1 \cup \dots \cup A_k$ is a proper subset of X and no element of X lies in exactly one of the $A_i$s. Find the largest cardinality of a collection of proper non-empty subsets of X, no non-empty subcollection of which is tight.

Note. A subset A of X is proper if $A\neq X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.

By Neel Nanda:

If |X|=n, there are $2^n$ possible subsets, so at first glance the answer could be a variety of things, from a linear to an exponential function of n, each of which would suggest a different approach. So the first step is to conjecture an answer, and by examining small cases it seems impossible to do better than 2n-2. There are several natural constructions for this bound, such as n subsets of size (n-1) and (n-2) subsets of size 1, so we guess this to be our answer (which later turn out to be right!).

From here, a solution is deceptively simple, though empirically the five full solutions in the contest show that it was by no means easy to find. We proceed by induction on the size of X, and want to show that any collection of subsets S has size at least (2n-2). By assumption all subcollections are not tight, so if the union of a subcollection is not the whole set X, then there is an element which appears in exactly one subset. This is a useful result, so we’d like to force a subcollection whose union is not the whole set X.

One way to guarantee that the union of a subcollection is not X is by taking the subcollection of all subsets not containing some element b. So there is some element c which appears in only one subset not containing b. If we choose b so that it’s the element contained in the fewest subsets of S, c is in at least as many subsets of S, but in only one subset not containing b. This means that at most one subset containing b doesn’t contain c. This is useful, because after removing at most 2 subsets (the coefficient of n in 2n-2, importantly!), we now have that every subset in S either contains both b and c or neither. This means that we can replace the pair (b,c) with a new element d, to get a new collection of subsets S’ of a set X’, of size n-1, so by induction $|S| \le |S'|+2\le 2n-2$.

There is also the case where all subsets contain b, but we can create an equivalent collection of subsets of X \ {b} by removing b from all subsets. So again by induction we are done.

Problem 6

Let ABCD be any convex quadrilateral and let P, Q, R, S be points on the segments AB, BC, CD, and DA, respectively. It is given that the segments PR and QS dissect ABCD into four quadrilaterals, each of which has perpendicular diagonals. Show that the points P, Q, R, S are concyclic.

I thought this problem was extremely hard. The official solution starts with a ‘magic lemma’, that isn’t quite so magic if you then read how it’s used. The overall claim is that PQ, RS and AC are concurrent (or parallel), and this is proved using the fact that the radical axis of the two circles with diameters PQ and RS also passes through this point of concurrency. Hunting for key properties of subsets of points in the diagram is an important skill in hard olympiad geometry, since it exactly reflects how problem-setters produce the problems. All the more so when there is lots of symmetry in the construction. But this is a hard example – there are a lot of potentially relevant subsets of the configuration.

When you’re really stuck with how to get involved in a synthetic configuration, you might consider using coordinates. Some of the UK students have been reading some chapters of a book (Euclidean Geometry in Mathematical Olympiads by Evan Chen. I’ve only had my own copy for a couple of days, but my initial impression is very positive – it fills a gap in the literature in a style that’s both comprehensive and readable.) focusing on various analytic approaches, so James and I felt it was safer to make sure we knew what the best settings were, and how far we could take them.

You almost certainly want the intersection of PR and QS to be your origin. I wanted to set up the configuration using the language of vectors, referenced by (P,Q,R,S). This was because $PQ\perp BO$ and so on, hence $\mathbf{b}\cdot (\mathbf{q}-\mathbf{p})=0$ and so on. An alternative is to use complex numbers, which makes this condition a bit more awkward, but is more promising for the conclusion. Concyclity is not a natural property in vectors unless you can characterise the centre of the circle, but can be treated via cross-ratios in $\mathbb{C}$. You also have to decide whether to describe the collinearity of A, B and P by expressing $\mathbf{p}=\lambda_{\mathbf{p}} \mathbf{a}+(1-\lambda_{\mathbf{p}})\mathbf{b}$, or via something more implicit. There definitely are not four degrees of freedom here, since specifying A certainly defines at most one valid set of (B,C,D), so one is mindful we’ll have to eliminate many variables later. We also have to account for fact that $\mathbf{r}$ is a negative scalar multiple of $\mathbf{p}$, and it’s not clear whether it’s better to break symmetry immediately, or use this towards the end of a calculation.

The point of writing this was that if your initial thought was ‘this looks promising via coordinate methods’, then I guess I agree. But there’s a difference between looking promising, and actually working, and there are lots of parameterisation options. It’s certainly worth thinking very carefully about which to choose, and in this case, challenging though they were, the synthetic or synthetic-trigonometric methods probably were better.

# Antichains in the grid

In the previous post on this topic, we discussed Dilworth’s theorem on chains and antichains in a general partially ordered set. In particular, whatever the size of the largest antichain in a poset, it is possible to partition the poset into exactly that many chains. So for various specific posets, or the directed acyclic graphs associated to them, we are interested in the size of this largest antichain.

The following example turned out to be more interesting than I’d expected. At a conventional modern maths olympiad, there are typically three questions on each paper, and for reasons lost in the mists of time, each student receives an integer score between 0 and 7 per question. A natural question to ask is “how many students need to sit a paper before it’s guaranteed that one will scores at least as highly as another on every question?” (I’m posing this as a straight combinatorial problem – the correlation between scores on different questions will be non-zero and presumably positive, but that is not relevant here.)

The set of outcomes is clearly $\{0,1,\ldots,7\}^3$, with the usual weak domination partial order inherited from $\mathbb{R}^3$. Then an antichain corresponds to a set of triples of scores such that no triple dominates another triple. So the answer to the question posed is: “the size of the largest antichain in this poset, plus one.”

In general, we might ask about $\{1,2,\ldots,n\}^d$, again with the weak domination ordering. This directed graph, which generalises the hypercube as well as our example, is called the grid.

Heuristics for the largest antichain

Retaining the language of test scores on multiple questions is helpful. In the previous post, we constructed a partition of the poset into antichains, indexed by the elements of some maximal chain, by starting with the sources, then looking at everything descended only from sources, and so on. (Recall that the statement that this is possible was referred to as the dual of Dilworth’s theorem.) In the grid, there’s a lot of symmetry (in particular under the mapping $x\mapsto n+1-x$ in every coordinate), and so you end up with the same family of antichains whether you work upwards from the sources or downwards from the sinks. (Or vice versa depending on how you’ve oriented your diagram…) The layers of antichains also have a natural interpretation – each layer corresponds to a given total score. It’s clear a priori why each of these is an antichain. If A scores the same as B overall, but strictly more on the first question, this must be counterbalanced by a strictly lower score on another question.

So a natural guess for the largest antichain is the largest antichain corresponding to some fixed total score. Which total score should this be? It ought to be the middle layer, that is total score $\frac{(n+1)d}{2}$, or the two values directly on either side if this isn’t an integer. My intuition was probabilistic. The uniform distribution on the grid is achieved by IID uniform distributions in each coordinate, which you can think of as a random walk, especially if you subtract off the mean first. It feels that any symmetric random walk should have mode zero or next-to-zero. Certainly this works asymptotically in a rescaled sense by CLT, and in a slightly stronger sense by local CLT, but we don’t really want asymptotics here.

When I started writing the previous paragraph, I assumed there would be a simple justification for the claim that the middle layer(s) was largest, whether by straight enumeration, or some combinatorial argument, or even generating functions. Perhaps there is, and I didn’t spot it. Induction on d definitely works though, with a slightly stronger hypothesis that the layer sizes are symmetric around the median, and monotone on either side of the median. The details are simple and not especially interesting, so I won’t go into them.

From now on, the hypothesis is that this middle layer of the grid is the largest antichain. Why shouldn’t it, for example, be some mixture of middle-ish layers? (*) Well, heuristically, any score sequence in one layer removes several possibilities from a directly adjacent layer, and it seems unlikely that this effect is going to cancel out if you take some intermediate number of score sequences in the first layer. Also, the layers get smaller as you go away from the middle, so because of the large amount of symmetry (coordinates are exchangeable etc), it feels reasonable that there should be surjections between layers in the outward direction from the middle. The union of all these surjections gives a decomposition into chains.

This result is in fact true, and its proof by Bollobas and Leader, using shadows and compression can be found in the very readable Sections 0 and 1 of [1].

Most of the key ideas to a compression argument are present in the case n=2, for which some notes by Leader can be found here, starting with Proof 1 of Theorem 3, the approach of which is developed over subsequent sections. We treat the case n=2, but focusing on a particularly slick approach that does not generalise as successfully. We also return to the original case d=3 without using anything especially exotic.

Largest antichain in the hypercube – Sperner’s Theorem

The hypercube $\{0,1\}^d$ is the classical example. There is a natural correspondence between the vertices of the hypercube, and subsets of $[d]$. The ordering on the hypercube corresponds to the ordering given by containment on $\mathcal{P}([d])$. Almost by definition, the k-th layer corresponds to subsets of size k, and thus includes $\binom{d}{k}$ subsets. The claim is that the size of the largest antichain is $\binom{d}{\lfloor d/2 \rfloor}$, corresponding to the middle layer if d is even, and one of the two middle layers if d is odd. This result is true, and is called Sperner’s theorem.

I know a few proofs of this from the Combinatorics course I attended in my final year at Cambridge. As explained, I’m mostly going to ignore the arguments using compression and shadows, even though these generalise better.

As in the previous post, one approach is to exhibit a covering family of exactly this number of disjoint chains. Indeed, this can be done layer by layer, working outwards from the middle layer(s). The tool here is Hall’s Marriage Theorem, and we verify the relevant condition by double-counting. Probably the hardest case is demonstrating the existence of a matching between the middle pair of layers when d is odd.

Take d odd, and let $d':= \lfloor d/2\rfloor$. Now consider any subset S of the d’-th layer $\binom{[d]}{d'}$. We now let the upper shadow of S be

$\partial^+(S):= \{A\in \binom{[d]}{d'+1}\,:\, \exists B\in S, B\subset A\},$

the sets in the (d’+1)-th layer which lie above some set in S. To apply Hall’s Marriage theorem, we have to show that $|\partial^+(S)|\ge |S|$ for all choice of S.

We double-count the number of edges in the hypercube from $S$ to $\partial^+(S)$. Firstly, for every element $B\in S$, there are exactly d’ relevant edges. Secondly, for every element $A\in\partial^+(S)$, there are exactly d’ edges to some element of $\binom{[d]}{d'}$, and so in particular there are at most d’ edges to elements of S. Thus

$d' |S|=|\text{edges }S\leftrightarrow\partial^+(S)| \le d' |\partial^+(S)|,$

which is exactly what we require for Hall’s MT. The argument for the matching between other layers is the same, with a bit more notation, but also more flexibility, since it isn’t a perfect matching.

The second proof looks at maximal chains. Recall, in this context, a maximal chain is a sequence $\mathcal{C}=B_0\subset B_1\subset\ldots\subset B_d$ where each $B_k:= \binom{[d]}{k}$. We now consider some largest-possible antichain $\mathcal{A}$, and count how many maximal chains include an element $A\in\mathcal{A}$. If $|A|=k$, it’s easy to convince yourself that there are $\binom{d}{r}$ such maximal chains. However, given $A\ne A'\in\mathcal{A}$, the set of maximal chains containing A and the set of maximal chains containing A’ are disjoint, since $\mathcal{A}$ is an antichain. From this, we obtain

$\sum_{A\in\mathcal{A}} \binom{d}{|A|} \le d!.$ (**)

Normally after a change of notation, so that we are counting the size of the intersection of the antichain with each layer, this is called the LYM inequality after Lubell, Yamamoto and Meshalkin. The heuristic is that the sum of the proportions of layers taken up by the antichain is at most one. This is essentially the same as earlier at (*). This argument can also be phrased probabilistically, by choosing a *random* maximal chain, and considering the probability that it intersects the proposed largest antichain, which is, naturally, at most one. Of course, the content is the same as this deterministic combinatorial argument.

Either way, from (**), the statement of Sperner’s theorem follows rapidly, since we know that $\binom{d}{|A|}\le \binom{d}{\lfloor d/2\rfloor}$ for all A.

Largest antichain in the general grid

Instead of attempting a proof or even a digest of the argument in the general case, I’ll give a brief outline of why the previous arguments don’t transfer immediately. It’s pretty much the same reason for both approaches. In the hypercube, there is a lot of symmetry within each layer. Indeed, almost by definition, any vertex in the k-th layer can be obtained from any other vertex in the k-th layer just by permuting the labels (or permuting the coordinates if thinking as a vector).

The hypercube ‘looks the same’ from every vertex, but that is not true of the grid. Consider for clarity the n=8, d=3 case we discussed right at the beginning, and compare the scores (7,0,0) and (2,2,3). The number of maximal chains through (7,0,0) is $\binom{14}{7}$, while the number of maximal chains through (2,2,3) is $\binom{7}{2, 2,3}\binom{14}{4,5,5}$, and the latter is a lot larger, which means any attempt to use the second argument is going to be tricky, or at least require an extra layer of detail. Indeed, exactly the same problem arises when we try and use Hall’s condition to construct the optimal chain covering directly. In the double-counting section, it’s a lot more complicated than just multiplying by d’, as was the case in the middle of the hypercube.

Largest antichain in the d=3 grid

We can, however, do the d=3 case. As we will see, the main reason we can do the d=3 case is that the d=2 case is very tractable, and we have lots of choices for the chain coverings, and can choose one which is well-suited to the move to d=3. Indeed, when I set this problem to some students, an explicit listing of a maximal chain covering was the approach some of them went for, and the construction wasn’t too horrible to state.

[Another factor is that it computationally feasible to calculate the size of the middle layer, which is much more annoying in d>3.]

[I’m redefining the grid here as $\{0,1,\ldots,n-1\}^d$ rather than $\{1,2,\ldots,n\}^d$.]

The case distinction between n even and n odd is going to make both the calculation and the argument annoying, so I’m only going to treat the even case, since n=8 was the original problem posed. I should be honest and confess that I haven’t checked the n odd case, but I assume it’s similar.

So when n is even, there are two middle layers namely $\frac{3n}{2}-2, \frac{3n}{2}-1$ (corresponding to total score 10 and total score eleven in the original problem). I calculated the number of element in the $\frac{3n}{2}-1$ layer by splitting based on the value of the first coordinate. I found it helpful to decompose the resulting sum as

$\sum_{k=0}^{n-1} = \sum_{k=0}^{\frac{n}{2}-1} + \sum_{k=\frac{n}{2}}^{n-1},$

based on whether there is an upper bound, or a lower bound on the value taken by the second coordinate. This is not very interesting, and I obtained the answer $\frac{3n^2}{4}$, and of course this is an integer, since n is even.

Now to show that any antichain has size at most $\frac{3n^2}{4}$. Here we use our good control on the chain coverings in the case d=2. We note that there is a chain covering of the (n,d=2) grid where the chains have 2n-1, 2n-3,…, 3, 1 elements (%). We get this by starting with a maximal chain, then taking a maximal chain on what remains etc. It’s pretty much the first thing you’re likely to try.

Consider an antichain with size A in the (n,d=3) grid, and project into the second and third coordinates. The image sets are distinct, because otherwise a non-trivial pre-image would be a chain. So we have A sets in the (n,d=2) grid. How many can be in each chain in the decomposition (%). Well, if there are more than n in any chain in (%), then two must have been mapped from elements of the (n,d=3) grid with the same first coordinate, and so satisfy a containment relation. So in fact there are at most n image points in any of the chains of (%). So we now have a bound of $n^2$. But of course, some of the chains in (%) have length less than n, so we are throwing away information. Indeed, the number of images points in a given chain is at most

$\max(n,\text{length of chain}),$

and so the number of image points in total is bounded by

$n+\ldots+n+ (n-1)+(n-3)+\ldots+1,$

where there are n/2 copies of n in the first half of the sum. Evaluating this sum gives $\frac{3n^2}{4}$, exactly as we wanted.

References

[1] – Bollobas, Leader (1991) – Compressions and Isoperimetric Inequalities. Available open-access here.

# Chains and antichains

I’ve recently been at the UK-Hungary winter olympiad camp in Tata, for what is now my sixth time. As well as doing some of my own work, have enjoyed the rare diversion of some deterministic combinatorics. It seems to be a local variant of the pigeonhole principle that given six days at a mathematical event in Hungary, at least one element from {Ramsay theory, Erdos-Szekeres, antichains in the hypercube} will be discussed, with probability one. On this occasion, all were discussed, so I thought I’d write something about at least one of them.

Posets and directed acyclic graphs

This came up on the problem set constructed by the Hungarian leaders. The original formulation asked students to show that among any 17 positive integers, there are either five such that no one divides any other, or five such that among any pair, one divides the other.

It is fairly clear why number theory plays little role. We assign the given integers to the vertices of a graph, and whenever a divides b, we add a directed edge from the vertex corresponding to a to the vertex corresponding to b. Having translated the given situation into a purely combinatorial statement, fortunately we can translate the goal into the same language. If we can find a chain of four directed edges (hence five vertices – beware confusing use of the word ‘length’ here) then we have found the second possible option. Similarly, if we can find an antichain, a set of five vertices with no directed edges between them, then we have found the first possible option.

It’s worth noting that the directed graph we are working with with is transitive. That is, whenever there is an edge a->b and b->c, there will also be an edge a->c. This follows immediately from the divisibility condition. There are also no directed cycles in the graph, since otherwise there would be a cycle of integers where each divided its successor. But of course, when a divides b and these are distinct positive integers, this means that b is strictly larger than a, and so this relation cannot cycle.

In fact, among a set of positive integers, divisibility defines a partial order, which we might choose to define as any ordering whether the associated directed graph is transitive and acyclic, although obviously we could use language more naturally associated with orderings. Either way, from now on we consider posets and the associated DAGs (directed acyclic graphs) interchangeably.

Dilworth’s theorem

In the original problem, we are looking for either a large chain, or a large antichain. We are trying to prove that it’s not possible to have largest chain size at most four, and largest antichain size at most four when there are 17 vertices, so we suspect there may some underlying structure: in some sense perhaps the vertex set is the ‘product’ of a chain and an antichain, or at least a method of producing antichains from a single vertex.

Anyway, one statement of Dilworth’s theorem is as follows:

Statement 1: in a poset with nm+1 elements, there is either a chain of size n+1, or an antichain of size m+1.

Taking n=m=4 immediately finishes the original problem about families of divisors. While this is the most useful statement here, it’s probably not the original, which says the following:

Statement 2: in a poset, there exists $\mathcal{C}$ a decomposition into chains, and an antichain $A$ such that $|\mathcal{C}|=|A|$.

Remark 1: Note that for any decomposition into chains and any antichain, we have $|\mathcal{C}|\ge |A|$, since you can’t have more than one representative from any chain in the antichain. So Statement 2 is saying that equality does actually hold.

Remark 2: Statement 1 follows immediately from Statement 2. If all antichains had size at most m, then there’s a decomposition into at most m chains. But each chain has size n, so the total size of the graph is at most mn. Contradiction.

Unsuccessful proof strategies for Dilworth

Since various smart young people who didn’t know the statement or proof of Dilworth’s theorem attempted to find it (in the form of Statement 1, and in a special case) in finite time conditions, it’s easy to talk about what doesn’t work, and try to gain intellectual value by qualifying why.

• Forgetting directions: in general one might well attack a problem by asking whether we have more information than we need. But ignoring the directions of the edges is throwing away too much information. After doing this, antichains are fine, but maybe you need to exhibit some undirected ‘chains’. Unless these undirected chains are much longer than you are aiming for, you will struggle to reconstruct directed chains out of them.
• Where can the final vertex go?: in a classic trope, one might exhibit a directed graph on nm vertices with neither a chain of size n+1 nor an antichain of size m+1. We attempt to argue that this construction is essentially unique, and that it goes wrong when we add an extra vertex. As a general point, it seems unlikely to be easier to prove that exactly one class of configurations has a given property in the nm case, than to prove no configurations has the same property in the nm+1 case. A standalone proof of uniqueness is likely to be hard, or a disguised rehash of an actual proof of the original statement.
• Removing a chain: If you remove a chain of maximal length, then, for contradiction, what you have left is m(n-1)+1 vertices. If you have a long chain left, then you’re done, although maximality has gone wrong somewhere. So you have an antichain size n in what remains. But it’s totally unclear why it should be possible to extend the antichain with one of the vertices you’ve just removed.

An actual proof of Dilworth (Statement 1), and two consequences

This isn’t really a proof, instead a way of classifying the vertices in the directed graph so that this version of Dilworth. As we said earlier, we imagine there may be some product structure. In particular, we expect to be able to find a maximal chain, and a nice antichain associated to each element of the maximal chain.

We start by letting $V_0$ consist of all the vertices which are sources, that is, have zero indegree. These are minima in the partial ordering setting. Now let $V_1$ consist of all vertices whose in-neighbourhood is entirely contained in $V_0$, that is they are descendents only of $V_0$. Then let $V_2$ consist of all remaining vertices whose in-neighourhood is entirely contained in $V_0\cup V_1$ (but not entirely in $V_0$, otherwise it would have already been treated), and so on. We end up with what one might call an onion decomposition of the vertices based on how far they are from the sources. We end up with $V_0,V_1,\ldots,V_k$, and then we can find a chain of size k+1 by starting with any vertex in $V_k$ and constructing backwards towards the source. However, this is also the largest possible size of a chain, because every time we move up a level in the chain, we must move from $V_i$ to $V_j$ where j>i.

It’s easy to check that each $V_i$ is an antichain, and thus we can read off Statement 1. A little more care, and probably an inductive argument is required to settle Statement 2.

We have however proved what is often called the dual of Dilworth’s theorem, namely that in a poset there exists a chain C, and a decomposition into a collection $\mathcal{A}$ of antichains, for which $|C|=|\mathcal{A}|$.

Finally, as promised returning to Erdos-Szekeres, if not to positive integers. We apply Dilworth Statement 1 to a sequence of $m^2+1$ real numbers $a_0,a_1,\ldots,a_{m^2}$, with the ordering $a_i\rightarrow a_j$ if $i\le j$ and $a_i\le a_j$. Chains correspond to increasing subsequences, and antichains to decreasing subsequences, so we have shown that there is either a monotone subsequence of length m+1.

# Turan’s Theorem

Turan’s theorem gives bounds on the number of edges required in a graph on a fixed number of vertices n to guarantee it contains a complete graph of size r+1. Equivalently, an upper bound on the number of edges in a $K_{r+1}$-free graph. For some of the applications and proofs, it may be more natural to look instead at the complement graph, for which the theorem becomes a statement about the existence or otherwise of an independent set of size r+1.

Rather than give an expression for the bound immediately, it is more natural to consider the Turan graph T(n,r), the maximal graph on n vertices without a copy of $K_{r+1}$. This is constructed by dividing the vertices into r classes with ‘as equal size as possible’. That is, some classes have size $\lfloor \frac{n}{r}\rfloor$ and others have size $\lfloor \frac{n}{r}\rfloor +1$. Then connect any pair of vertices which are not in the same class by an edge. This gives a complete r-partite graph on these classes. Since any collection of r+1 vertices contains at least two in the same class, it can’t contain a $K_{r+1}$. Note that the complement of the complete r-partite graph is the union of r disjoint complete graphs on the classes.

There are a number of ways to enumerate the edges in T(n,r), and some can get quite complicated quite quickly. After a moderate amount of thought, this is my favourite. Let $n=\ell r+k$, so T(n,r) has k classes of size (l+1) and (r-k) classes of size l. Pick an ordered pair of vertices uniformly at random. (So picking the same vertices is indeed an option, and is counted twice.) Then the probability they are the same class is

$\frac{k}{r}\cdot\frac{\ell+1}{n}+\frac{r-k}{r}\cdot \frac{\ell}{n} = \frac{1}{r}.$

So the probability they are in different classes is $\frac{r-1}{r}$, and we can treat all of the $2n^2$ ordered pairs in this way, noting a) that we count everything twice; and b) we know a priori that we don’t have loops, so the fact that we’ve included these in the count doesn’t matter. We end up with the enumeration $(1-\frac{1}{r})\frac{n^2}{2}$ for the edges in T(n,r).

A standard proof

For both proofs, I find it slightly easier to work in the complement graph, where we are aiming for the largest number of edges with an independent set of size (r+1). Suppose we have a graph with the minimal number of vertices such that there’s no independent set of given size. Suppose also that there is an edge joining vertices v and w, such that $d(v)> d(w)$. Then if we change v’s neighbourhood $\Gamma(v)$ so that it becomes the same as $\Gamma(w)$, (that is, we replace v with a copy of w, and maintain the original edge vw), then it is easily checked that we still do not have an independent set of that size, but fewer edges.

Note that by attempting to make the neighbourhoods of connected vertices equal, we are making the graph look more like a union of complete components. We can do a similar trick if we have three vertices u,v,w such that there are edges between u and v and v and w, but not u and w. Then we know the degrees of u,v,w are the same by the previous argument, and so it can again be checked that making $\Gamma(u),\Gamma(w)$ the same as $\Gamma(v)$, and adding the edge uw reduces the number of edges, and maintains the non-existence of the independent set.

The consequence of this is that we’ve shown that the minimum can only be attained when presence of edges is an equivalence relation (ignoring reflexivity). Thus the minimum is only attained for a union of at most r complete graphs. Jensen (or any root-mean-square type inequality) will then confirm that the true minimum is attained when the sizes of the r components are as equal as possible.

A probabilistic proof

The following probabilistic proof is courtesy of Alon and Spencer. The motivation is that in the (equality) case of a union of complete graphs, however we try to build up a maximal independent set, we always succeed. That is, it doesn’t matter how we choose which vertex (unconnected to those we already have) to add next – we will always get a set of size r. This motivates a probabilistic proof, as an argument in expectation will have equality in the equality case, which is always good.

Anyway, we build up an independent set in a graph by choosing uniformly at random a vertex which is not connected to any we have so far, until this set of vertices is empty. It makes sense to settle the randomness at the start, so give the vertices a uniform random labelling on [n], and at each stage, choose the independent vertex with minimal label.

Thus, a vertex v will be chosen for the independent set if, and only if, it has a smaller label than all of its neighbours, that is, with probability $\frac{1}{1+d(v)}$. So the expected size of the independent set constructed in this fashion is

$\sum_{v\in V(G)} \frac{1}{1+d(v)}\ge \frac{V}{1+\bar d} = \frac{V}{1+\frac{2E}{V}}.$

One can chase through the expressions to get the bound we want back.

The reason I was thinking about Turan’s theorem was a problem which the UK IMO squad was discussing. It comes from an American selection test (slightly rephrased): given 100 points in the plane, what is the largest number of pairs of points with $\ell_1$ distance in (1,2]?

The key step is to think about how large a collection of points can have this property pairwise. It is easy to come up with an example of four points which work, and seemingly impossible to come up with an example with five points. To prove this, I found it easiest to place a point at the origin, then explicitly work with coordinates relative the basis $(1,1),(1,-1)$ for fairly obvious reasons in this metric.

Anyway, once you are convinced that you can’t have five points with this property pairwise, you are ready to convert into a graph-theoretic statement. Vertices correspond to points, and edges link pairs of points whose distance is in (1,2] as required. We know from the previous paragraph that there is no copy of $K_5$ here, so Turan’s theorem bounds the number of edges, ie the number of suitable pairs.

It also tells us under what sort of circumstances the bound is attained, and from this, it’s natural to split the 100 points into four groups of 25, for example by taking four points which satisfy the condition pairwise (eg a diamond around the origin), and placing each group very near one of the points.

Extensions and other directions

The existence of a complete subgraph is reminiscent of Ramsey theory, which in one case is a symmetric version of Turan’s theorem. In Turan, we are adding enough edges to force a complete subgraph, while in the original form of Ramsey theory, we are asking how large the graph needs to be to ensure that for any edge configuration, either the original graph or the complement graph includes a complete subgraph. It makes a lot more sense to phrase this in terms of colours for the purpose of generalisation.

A natural extension is to ask about finding copies of fixed graphs H other than the complete graph. This is the content of the Erdos-Stone theorem. I’d prefer to say almost nothing rather than be vague, but the key difference is that the bound is asymptotic in the number of vertices rather than exact. Furthermore, the asymptotic proportion of vertices depends on the chromatic number of H, which tells you how many classes r are required to embed H in a (large if necessary) r-partite graph. So it is perhaps unsurprising that the limiting proportions end up matching the proportion of edges in the Turan graphs, namely r-1/r as r varies, which leaves the exact scaling open to further investigation in the case where H is bipartite (hence has chromatic number 2).

# EGMO 2016 Paper II

Continuing from yesterday’s account of Paper I, this is a discussion of my thoughts about Paper II of EGMO 2016, happening at the moment in Busteni, Romania. This is not an attempt to describe official solutions, but rather to describe the thought process (well, a thought process) of someone tackling each question. I hope it might be interesting or useful, but for students, it will probably be more useful after at least some engagement with the problems. These are excellent problems, and reading any summary of solutions means you miss the chance to hunt for them yourself.

In actual news, you can follow the scoreboard as it is updated from Romania here. Well done to the UK team on an excellent performance, and hope everyone has enjoyed all aspects of the competition!

Question 4

Circles $\omega_1,\omega_2$ with the same radius meet at two points $X_1,X_2$. Circle $\omega$ is externally tangent to $\omega_1$ at $T_1$, and internally tangent to $\omega_2$ at $T_2$. Prove that lines $X_1T_1,X_2T_2$ meet on $\omega$.

Thought 1: I’m not the biggest fan of geometry ever, but I thought this looked like a nice problem, because it’s only really about circles, so I figured it probably wouldn’t require anything too exotic.

Thought 2: I bet lots of people try inversion. But the equal radius condition means I’m probably happy with the circles as they are. I hope lots of people don’t try to place the diagram in some co-ordinate system, even if it possible to do it sensibly (eg by making $\omega$ the reference circle).

Thought 3: The labelling of $X_1,X_2$ is unrelated to the rest of the indexing. So the intersection of $X_1T_2,X_2T_1$ should also lie on $\omega$, and possibly has some relationship (antipodal?) to the point I actually need to find out. But I can’t think of any reason why it’s easier to prove two points lie on a circle than just one, so let’s leave this as a thought rather than an idea.

Idea 1: I drew a terrible diagram on the back of a draft of my abstract, and for once, this was actually kind of helpful. Forget about radii being equal, one of them wasn’t even a circle. Anyway, while drawing in the later points, I was struggling to make it look convincingly like all the lengths which were supposed to be equal were in fact equal. So the idea was: almost all the segments in the diagram (once I’ve defined the circle centres $O_{\omega_1}$ etc) have one of two lengths (the radii of $\omega_1,\omega$ – red and green-ish in the diagram below), and with this in mind I can forget about the circles. We’ve got a rhombus, which is even better than a parallelogram, which is itself a really useful thing to have in a configuration. Another consequence of the proliferation of equal lengths is that almost all triangles are isosceles, and we know that similarity of isosceles triangles is particularly easy, because you only have to match up one angle.

Idea 2: How to prove it? We have to prove that two lines and a circle concur. This is where I actually need to stop and think for a moment: I could define the point where the lines meet and try to show it’s on the circle, or intersect one line with the circle, and show it’s on the other line. Idea 1 basically says I’m doing the problem using lengths, so I should choose the way that fits best with lengths.

If I define the point P where $X_2T_2$ meets the circle (this was easier to draw in my diagram), then I know $PO_\omega=T_2 O_\omega$ and so on. Then there were loads of isosceles triangles, and some of them were similar, which led to more parallel lines, and from this you could reverse the construction in the other direction to show that P also lay on the other line.

Question 5

Let k, n be integers such that $k\ge 2,\, k\le n\le 2k-1$. Place rectangular k x 1 or 1 x k tiles on an n x n chessboard in the natural way with no overlap until no further tile can be placed. Determine the minimum number of tiles that such an arrangement may contain.

Idea 1: It took me a while to parse the question. Minimum over what? I rephrased it in my head as: “to show the answer is eg n+5, I need to show that whenever you place n+4 tiles legally, you can’t add another. I also need to show that you can place n+5 such that you can’t add another.” This made life a lot easier.

Thought 1: What goes wrong if you take n=2k and beyond? Well, you can have two horizontal tiles on a given row. I’m not really sure how this affects the answer, but the fact that there is still space constraint for n<2k is something I should definitely use.

Diversion: I spent a while thinking that the answer was 4 and it was a very easy question. I spent a bit more time thinking that the answer was n, and it was a quite easy question, then realised that neither my construction nor my argument worked.

Thought 2: can I do the cases n=k,or 2k-1 or k+1? The answers were yes, unsure, and yes. The answer to k+1, which I now felt confident was actually four, was helpful, as it gave me a construction for k+2, …, 2k-1 that seemed good, even though it was clearly not optimal for 2k-1. Therefore, currently my potential answer has three regimes, which seemed unlikely, but this seemed a good moment to start trying to prove it was optimal. From now on, I’m assuming I have a configuration from which you can’t add another block.

Idea 2: About this diagram, note that once I’ve filled out the top-left (k+1)x(k+1) sub-board in this way, there are still lots of ways to complete it, but I do have to have (n-k-1) horizontal and (n-k-1) vertical tiles roughly where I’ve put them. Why? Because I can’t ‘squeeze in’ a vertical tile underneath the blue bit, and I can’t squeeze in a horizontal tile to the right of the blue bit. Indeed, whenever I have a vertical block, there must be vertical blocks either to the left or to the right (*) (or possibly both if we’re near the middle). We need to make this precise, but before doing that, I looked back at where the vertical blocks were in the proposed optimum, and it turns out that all but (k-1) columns include a vertical block, and these (k-1) columns are next to each other.

This feels like a great idea for two reasons: 1) we’ve used the fact that n<2k at (*). 2) this feels very pigeonhole principle-ish. If we had fewer tiles, we’d probably have either at least k columns or least k rows without a vertical (or, respectively, horizontal) tile. Say k columns don’t include a vertical tile – so long as they are next to each other (which I think I know) we can probably include a horizontal tile somewhere in there.

So what’s left to do? Check the previous sentence actually works (maybe it’s full of horizontal tiles already?), and check the numerics of the pigeonhole bound. Also work out how the case n=2k-1 fits, but it seems like I’ve had some (/most) of the good ideas, so I stopped here.

Question 6

I don’t actually want to say very much about this, because I didn’t finish off all the details. I want to talk briefly in quite vague terms about what to do if you think this problem looks scary. I thought it looked a bit scary because it looked similar to two number theoretic things I remember: 1) primes in arithmetic progressions. This is very technical in general, but I can remember how to do 3 mod 4 fairly easily, and 1 mod 4 with one extra idea; 2) if a square-free number can be written as a sum of two squares, this controls its factors modulo 4.

Vague Ideas: It seemed unlikely that this would involve copying a technical argument, so I thought about why I shouldn’t be scared. I think I shouldn’t be scared of the non-existence part. Often when I want to show there are no integer solutions to an equation, I consider showing there are no solutions modulo some base, and maybe this will be exactly what I do here. I’ll need to convert this statement about divisibility into an equation (hopefully) and check that $n\equiv 3,4$ modulo 7 doesn’t work.

For the existence of infinitely many solutions, maybe I’d use Chinese Remainder Theorem [1], or I’ll reduce it to something that I know has lots of solutions (eg Pythagoras), or maybe I can describe some explicit solutions?

Actual Idea 1: We’ve got $n^2+m | n^4$, but this is a very inefficient statement, since the RHS is a lot larger than the LHS, so to be useful I should subtract off a large multiple of the LHS. Difference of two squares is a good thing to try always, or I could do it manually. Either way, I get $n^2+m | m^2$ which is genuinely useful, given I know m=1,2, …, 2n, because the RHS is now comparable in size to the LHS, so I’ve narrowed it down from roughly n^2 possibilities to just three:

$n^2+m=m^2,\quad 2(n^2+m)=m^2,\quad 3(n^2+m)=m^2.$ (*)

I’m going to stop now, because we’ve turned it into a completely different problem, which might be hard, but at least in principle this is solvable. I hope we aren’t actually scared of (*), since it looks like some problems we have solved before. I could handle one of these in a couple of lines, then struggled a bit more with the other pair. I dealt with one by recourse to some theory, and the final one by recourse to some theory after a lot of rearranging which I almost certainly got wrong, but I think I made an even number of mistakes rather than an odd number because I got the correct solution set modulo 7. Anyway, getting to (*) felt like the majority of the ideas, and certainly removed the fear factor of the Q6 label, so to fit the purpose of this discussion I’ll stop here.

[1] During one lunch in Lancaster, we were discussing why Chinese Remainder Theorem is called this. The claim was that an ancient Chinese general wanted to know the size of their army but it was too big to count, so had them arrange themselves into columns of various sizes, and counted the remainders. The general’s views on the efficiency of this algorithm remain lost in the mists of time.