BMO2 2019

Posted on January 27, 2019 by dominicyeo

The second round of the British Mathematical Olympiad was taken on Thursday by the 100 or so top scoring eligible participants from the first round, as well as some open entries. Qualifying for BMO2 is worth celebrating in its own right. The goal of the setters is to find the sweet spot of difficult but stimulating for the eligible participants, which ultimately means it’s likely to be the most challenging exam many of the candidates sit while in high school, at least in mathematics.

I know that lots of students view BMO2 as something actively worth preparing for. As with everything, this is a good attitude in moderation. Part of the goal in writing about the questions at such length is because I think at this level it’s particularly easy to devote more time than needed to preparation, and use it poorly. This year time is tight at the end of semester, and so what follows is closer to a set of complete solutions than usual, for which apologies, although I hope it is still possible to get a sense of how one might have come across the solutions yourself. Of course, this means that what follows will certainly spoil the problems for anyone who hasn’t tried them by themselves already.

The copyright for the problems is held by BMOS, and reproduced here with permission.

Question 1

As if often the case in geometry questions, what you’ve been asked to prove here isn’t the most natural property of the configuration. A good first step would be to see if there are stronger statements which are true.You are asked to show that triangle BPE is isosceles, but you aren’t told which of the three vertices is the apex. In fact, the task is to show that $BP=EP$ or, alternatively, $\angle BEP=\angle PBE$ . It’s not in general true that BE is equal to BP=EP. Unless you’re very unlucky, you can establish this from one diagram.

Now, you don’t immediately know whether it’s going to be easier to show that two lengths are equal, or that two angles are equal. However, you know that P lies on the perpendicular bisector of BC, hence BP=CP, which is a big clue. In particular, this means that P would be the centre of the circle through BCE. This clearly implies the given result, so deciding to prove this instead is a good strategy.

There are now a number of ways to prove this. Note that D lies on the altitude from A, and the feet of the perpendiculars from D to sides AB and AC are both present in the configuration so (just as for the orthocentre diagram) we can calculate most of the angles involving {A,B,C,D,E}.

For example, ABDE is cyclic, so $\angle BED=\angle BAD = 90-\hat{B}$ , hence $\angle AEB=\hat{B},\,\angle EBA=\hat{C}$ . This shows that AB is tangent to the circumcircle of BCE. But then the line L is a radius of this circle, and so its centre must be P, the unique point on L which is equidistant from B and C.

Alternatively, we could directly calculate $\angle BEC=180-\hat{B}$ and $\angle CBP=90-\hat{B}$ . But BPC is isosceles so $\angle BPC=2\hat{B}$ . In general, the converse of ‘angle at centre is twice angle at circumference’ does not hold, but when we know P is equidistant from B and C this does hold, and so the angle relations precisely confirm that P is the centre of the circle through BPE.

My intention had been that the triangle would be acute-angled, to reduce the number of diagram options based on the magnitude of $\hat{B}$ . If pursuing this second approach, one would need to be careful to account for whether P is on the same side or the opposite side of BC to E. That said, unless you do something very exotic, it should be exactly the same argument or calculation, and such a case distinction probably isn’t very important.

Question 2

First, a short remark. As stated, if n=5, a piece could move 3 squares to the left then 4 squares up, by Pythagoras. Handling all such options is likely to be quite annoying, since some values of n can be written in this Pythagorean form, and others cannot. This brings us to some good general principles for olympiad problems which look like this one:

A construction, when one exists, will probably be possible using simple versions of the allowed moves / structures.
An argument why a construction is impossible should probably be based on ideas which treat the simple moves similarly to the more complicated moves.

The setup of the problem encourages you to think about dividing the board into $n^2$ sub-boards, each with dimensions $n\times n$ . Continue reading →

Lecture 10 – the configuration model

Posted on January 15, 2019 by dominicyeo

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

As we enter the final stages of the semester, I want to discuss some extensions to the standard Erdos-Renyi random graph which has been the focus of most of the course so far. Although we will not get far into the details during this course, the overall goal is to develop models which are close to Erdos-Renyi in terms of ease of analysis, while also allowing more of the features characteristic of networks observed in the real world.

One of the more obvious deficiencies of the sparse regime of Erdos-Renyi random graphs for modelling ‘real-world phenomena’ concerns the degree sequence. Indeed, the empirical degree distribution of G(n,c/n) converges to Poisson(c). By contrast, in real-world networks, a much wider range of degrees is typically observed, and in many cases it is felt that these should follow a power law, with a small number of a very highly connected agents.

One way around this problem to construct random graphs where we insist that the graph has a given sequence of degrees. The configuration model, which is the subject of this lecture and this post (and about which I’ve written before), offers one way to achieve this.

Definition and notes

Let $n\ge 1$ and let $d=(d_1,d_2,\ldots,d_n)$ be a sequence of non-negative integers such that $\sum_{i=1}^n d_i$ is even. Then the configuration model with degree sequence d is a random multigraph with vertex set [n], constructed as follows:

To each vertex $i\in[n]$ , assign $d_i$ half-edges;
Then, take a uniform matching of these half-edges;
Finally, for each pair of half-edges in the matching, replace the two half-edges with a genuine edge, to obtain the multigraph $CM_n(d)$ , in which, by construction, vertex i has degree $d_i$ .

One should note immediately that although the matching is uniform, the multigraph is not uniform amongst multigraphs with that degree sequence. Note also that the condition on the sums of the degrees is necessary for any graph, and in this context means that the number of half-edges is even, without which it would not be possible to construct a matching.

This effect is manifest in the simplest possible example, when n=2 and d=(3,3). There are two possible graphs, up to isomorphism, which are shown below:

For obvious reasons, we might refer to these as the handcuffs and the theta , respectively. It’s helpful if we, temporarily, assume the half-edges are distinguishable at the moment we join them up in the configuration model construction. Because then there are 3×3=9 ways to join them up to form the handcuffs (think of which half-edge ends up forming the edge between the two vertices) while there are 3!=6 ways to pair up the half-edges in the theta.

In general, for multigraphs H with the correct degree sequence, we have

$\mathbb{P}( CM_n(d)\simeq H) \propto \left( 2^{\# \text{loops}(H)} \prod_{e\in E(H)} \text{mult}(e)! \right),$

where $\text{mult}(e)$ is the multiplicity with which a given edge e appears in H.

Note: it might seem counterintuitive that this procedure is biased against multiple edges and self-loops, but it is really just saying that there are more ways to form two distinct edges than to form two equal edges (ie a multiedge pair) when we view the half-edges as distinguishable. (See this post for further discussion of this aspect in the 3-regular setting.)

However, a consequence of this result is that if we condition on the event that $CM_n(d)$ is simple, then the resulting random graph is uniform on the set of simple graphs satisfying the degree property. Note that the same example as above shows that there’s no guarantee that there exists a simple graph whose degrees are some given sequence.

d-regular configuration model

In general, from a modelling point of view, we are particularly interested in simple, connected graphs, and so it is valuable to study whether the large examples of the configuration model are likely to have these properties. In this lecture, I will mainly focus on the case where the multigraphs are d-regular, meaning that all the vertices have degree equal to d. For the purposes of this lecture, we denote by $G^d(n)$ , the d-regular configuration model $CM_n(d,\ldots,d)$ .

d=1: to satisfy the parity condition on the sums of degrees, we must have n even. But then $G^1(n)$ will consist of n/2 disjoint edges.
d=2: $G^2(n)$ will consist of some number of disjoint cycles, and it is a straightforward calculation to check that when n is large, with high probability the graph will be disconnected.

In particular, I will focus on the case when d=3, which is the first interesting case. Most of the results we prove here can be generalised (under various conditions) to more general examples of the configuration model. The main goal of the lecture is revision of some techniques of the course, plus one new one, in a fresh setting, and the strongest possible versions of many of these results can be found amongst the references listed at the end.

Connectedness

In the lecture, we showed that $G^3(2n)$ is connected with high probability. This is, in fact, a very weak result, since in fact $G^d(n)$ is d-connected with high probability for $d\ge 3$ [Bol81, Wor81]. Here, d-connected means that one must remove at least d vertices in order to disconnect the graph, or, equivalently, that there are d disjoint paths between any pair of vertices. Furthermore, Bollobas shows that for $d\ge 3$ , $G^d(n)$ is a (random) expander family [Bol88].

Anyway, for the purposes of this course, the main tool is direct enumeration. The matching number $M_{2k}$ satisfies

$M_{2k}=(2k-1)\times (2k-3)\times\ldots\times 3\times 1 = \frac{(2k)!}{2^k \cdot k!},$

and so Stirling’s approximation gives the asymptotics

$M_{2k} = (\sqrt{2}+o(1)) \left(\frac{2}{e}\right)^k k^k,$

although it will be useful to use the true bounds

$c \left(\frac{2}{e}\right)^k k^k \le M_{2k}\le C\left(\frac{2}{e}\right)^k k^k,\quad \forall k,$

instead in some places. Anyway, in $G^3(2n)$ , there are 6n half-edges in total, and so the probability that the graph may be split into two parts consisting of $2\ell,2m$ vertices, with $2\ell+2m=2n$ , and with no edges between the classes is $\frac{\binom{2n}{2\ell} M_{6\ell}M_{6m}}{M_{6n}}.$ Continue reading →

Lecture 9 – Inhomogeneous random graphs

Posted on January 8, 2019 by dominicyeo

I am aiming to write a short post about each lecture in my ongoing course on Random Graphs. Details and logistics for the course can be found here.

As we enter the final stages of the semester, I want to discuss some extensions to the standard Erdos-Renyi random graph which has been the focus of most of the course so far. In doing so, we can revisit material that we have already covered, and discover how easily one can extend this directly to more exotic settings.

The focus of this lecture was the model of inhomogeneous random graphs (IRGs) introduced by Soderberg [Sod02] and first studied rigorously by Bollobas, Janson and Riordan [BJR07]. Soderberg and this blog post address the case where vertices have a type drawn from a finite set. BJR address the setting with more general typespaces, in particular a continuum of types. This generalisation is essential if one wants to use IRGs to model effects more sophisticated than those of the classical Erdos-Renyi model G(n,c/n), but most of the methodology is present in the finite-type setting, and avoids the operator theory language which is perhaps intimidating for a first-time reader.

Inhomogeneous random graphs

Throughout, $k\ge 2$ is fixed. A graph with k types is a graph G=(V,E) together with a type function $V\to \{1,\ldots,k\}$ . We will refer to a $k\times k$ symmetric matrix with non-negative entries as a kernel.

Given $n\in\mathbb{N}$ and a vector $p=(p_1,\ldots,p_k)\in\mathbb{N}_0^k$ satisfying $\sum p_i=n$ , and $\kappa$ a kernel, we define the inhomogeneous random graph $G^n(p,\kappa)$ with k types as:

the vertex set is [n],
types are assigned uniformly at random to the vertices such that exactly $p_i$ vertices have type i.
Conditional on these types, each edge $v\leftrightarrow w$ (for $v\ne w\in [n]$ ) is present, independently, with probability

$1 - \exp\left(-\frac{\kappa_{\mathrm{type}(v),\mathrm{type}(w)} }{n} \right).$

Notes on the definition:

Alternatively, we could assign the types so that vertices $\{1,\ldots,p_1\}$ have type 1, $\{p_1+1,\ldots,p_1+p_2\}$ have type 2, etc etc. This makes no difference except in terms of the notation we have to use if we want to use exchangeability arguments later.
An alternative model considers some distribution $\pi$ on [k], and assigns the types of the vertices of [n] in an IID fashion according to $\pi$ . Essentially all the same results hold for these two models. (For example, this model with ‘random types’ can be studied by quenching the number of each type!) Often one works with whichever model seems easier for a given proof.
Note that the edge probability given is $\approx \frac{\kappa_{\mathrm{type}(v),\mathrm{type}(w)}}{n}$ . The exponential form has a more natural interpretation if we ever need to turn the IRGs into a process. Additionally, it avoids the requirement to treat small values of n (for which, a priori, $k/n$ might be greater than 1) separately.

In the above example, one can see that, roughly speaking, red vertices are more likely to be connected to each other than blue vertices. However, for both colours, they are more likely to be connected to a given vertex of the same colour than a vertex of the opposite colour. This might, for example, correspond to the kernel $\begin{pmatrix}3&1\\1&2\end{pmatrix}$ .

The definition given above corresponds to a sparse setting, where the typical vertex degrees are $\Theta(1)$ . Obviously, one can set up an inhomogeneous random graph in a dense regime by an identical argument.

From an applications point of view, it’s not hard to imagine that an IRG of some flavour might be a good model for many phenomena observed in reality, especially when a mean-field assumption is somewhat appropriate. The friendships of boys and girls in primary school seems a particularly resonant example, though doubtless there are many others.

One particular application is to recover the types of the vertices from the topology of the graph. That is, if you see the above picture without the colours, can you work out which vertices are red, and which are blue? (Assuming you know the kernel.) This is clearly impossible to do with anything like certainty in the sparse setting – how does one decide about isolated vertices, for example? The probabilities that a red vertex is isolated and that a blue vertex is isolated differ by a constant factor in the $n\rightarrow\infty$ limit. But in the dense setting, one can achieve this with high confidence. When studying such statistical questions, these IRGs are often referred to as stochastic block models, and the recent survey of Abbe [Abbe] gives a very rich history of this type of problem in this setting.

Poisson multitype branching processes

As in the case of the classical random graph G(n,c/n), we learn a lot about the IRG by studying its local structure. Let’s assume from now on that we are given a sequence of IRGs $G^n(p^n,\kappa)$ for which $\frac{p^n}{n}\rightarrow \pi$ , where $\pi=(\pi_1,\ldots,\pi_k)\in[0,1]^k$ satisfies $||\pi||_1=1$ .

Now, let $v^n$ be a uniformly-chosen vertex in [n]. Clearly $\mathrm{type}(v^n)\stackrel{d}\rightarrow \pi$ , with the immediate mild notation abuse of viewing $\pi$ as a probability distribution on [k].

Then, conditional on $\mathrm{type}(v^n)=i$ :

when $j\ne i$ , the number of type j neighbours of $v^n$ is distributed as $\mathrm{Bin}\left(p_j,1-\exp\left(-\frac{\kappa_{i,j}}{n}\right)\right)$ .
the number of type i neighbours of $v^n$ is distributed as $\mathrm{Bin}\left( p_i-1,1-\exp\left(-\frac{\kappa_{i,i}}{n}\right)\right)$ .

Note that $p_j\left[1-\exp\left(-\frac{\kappa_{i,j}}{n}\right)\right]\approx \frac{p_j\cdot \kappa_{i,j}}{n} \approx \kappa_{i,j}\pi_j$ , and similarly in the case j=i, so in both cases, the number of neighbours of type j is distributed approximately as $\mathrm{Poisson}(\kappa_{i,j}\pi_j)$ .

This motivates the following definition of a branching process tree, whose vertices have k types. Continue reading →

Eventually Almost Everywhere

A blog about probability and olympiads by Dominic Yeo

Monthly Archives: January 2019

BMO2 2019

Lecture 10 – the configuration model

Lecture 9 – Inhomogeneous random graphs