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IMO resources for Graph Theory

I will be participating as a trainer for Canada’s 2018 IMO Summer Training camp. I’m giving a session on graph theory. As I prepared my notes I found many resources online that already cover some aspects of graph theory. So here are those resources:

“IMO Training 2008: Graph Theory” by Adrian Tung.

This is an in-depth description of the basic combinatorial and geometric techniques in graph theory. It is a very thorough and helpful document with many Olympiad level problems for each topic. (No solutions are given.)

Topics include:

  1. Trees and Balancing
  2. Friends, Strangers and Cliques
  3. Directed Graphs and Tournaments
  4. Matchings
  5. Hamiltonian/Eulerian Paths/Cycles

“Graph Theory” by Po-Shen Lo. (2008)

A large collection of problems and topics almost all of which have solutions or hints.

Topics include:

  1. Basic facts
  2. Extremal Graph Theory
  3. Matchings
  4. Ramsey Theory
  5. Planarity

“Graph Theory” by Matthew Brennan. (Canada Winter Camp, 2014)

Contains a concise list of important results together with a guided discussion to five example problems that use graph theory.

“Probabilistic Method/Graph Theory” by James Rickards. (Canada Summer Camp, 2015)

An introduction to the probabilistic method in graph theory along with 10 problems.

“SIMO Graph Theory Training”. (SIMO training 2003)

A list of about 30 problems and solutions in graph theory.

Topics:

  1. Graph Theory
  2. Coloring problems

“Ramsey Theory and the IMO” by Ben Green. (2008)

This is a 4 page article that introduces Ramsey Theory for graphs and arithmetic progressions and its historical relation to the IMO.

“Coloring Points” at Cut-the-knot

A collection of 12 topics about coloring graphs and planes. There are many problems with solutions.

“Equivalence of seven major theorems in combinatorics” by Robert Borgersen (2004).

This series of slides states 7 results in extremal combinatorics that are really the same.

Topics:

  1. Dilworth’s Theorem
  2. Konig’s Bipartite Theorem
  3. Hall’s Marriage Theorem
  4. Menger’s Theorem
  5. (Others)

Stevo’s Forcing Class Fall 2012 – Class 9

(This is the ninth lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the eighth lecture here. Quotes by Stevo are in dark blue; some are deep, some are funny, some are paraphrased so use your judgement. As always I appreciate any type of feedback, including reporting typos, in the comments below.)

Summary

  • We define the notion of properness.
  • Show that the poset \in -Collapse is proper.
  • Justin Moore gives some intuition.
  • Explain the Open Colouring Axiom.
  • Give a related proper poset.

Continue reading Stevo’s Forcing Class Fall 2012 – Class 9

The Delta-System Lemma

This is a delta system of rivers. I didn't know this until last year.

Now that you know the basics of countable elementary submodels (CESM), you might think that you are in the clear. “Mike”, you say arrogantly, “I know all the most basic properties of CESMs, without proof I remind you, what else could I possibly want?”. I gently and patiently remind you that CESMs are worthless unless you know how to apply them properly.

So let’s do that.

Here are two theorems whose proofs you might already know, but that can be proved using elementary submodels. I will show you a proof of the \Delta -system lemma (a fundamental lemma in infinitary combinatorics) and a topological theorem of Arhangel’skii. Both of these proofs are taken from Just & Weese’s book “Discovering Modern Set Theory 2”, chapter 24.

Continue reading The Delta-System Lemma

The Secret Santa Problem (Part 2)

Last time, just in time for Christmas, we looked at the Secret Santa Problem. Basically the problem is to set up a secret santa type gift exchange without using any external aids like random number generators. A similar problem given to me by Sam Coskey is the following:

Sam’s Problem. Is it possible for two people to each choose a natural number so that both numbers are exactly 1 apart and neither person knows who has the larger of the two numbers?

When Sam gave the problem to me he intended that each player would choose a natural number and then they would sequentially ask questions to each other, possibly refining their original numbers. In this sense it is more like a game of Guess Who than secret santa.

After much thinking, it turns out that there is a fairly easy solution to this problem.

Part 1. Can either player choose the number 0?

Well no, because that person would know that they have a smaller number than the other player.

Part 2. If both players know that 1, 2, \dots, k cannot be chosen then k+1 cannot be chosen (as k+1 would have to be the smaller of the two numbers). So by induction, no number can be chosen by either player.

The lesson here is that induction is a very useful technique! This sounds naive but, when problem solving for contests, induction is often overlooked. Here is another related problem:

Father/Son problem. Is it possible for two players to each choose a human being so that the two humans are father and son, but neither player knows who has the father and who has the son?

The solution is again fairly simple, and uses induction. This time we need a different type of induction. We observe that no player can pick the first ever human being (as they would have the father of the other player’s choice). Now if there is a set S of humans that cannot be chosen, then the sons of people in S cannot be chosen either.

There you go, induction wins again!

 

Creeping Along

In my ongoing love affair with compactness I am constantly revisiting a particular proof of the Heine-Borel theorem, a characterization of compactness in \mathbb{R} . There are two proofs that I know of: the standard “subdivision” proof and the “creeping along” proof. I am going to focus on the creeping along proof.

Heine-Borel Theorem. A subset A \subseteq \mathbb{R} is compact if and only if it is closed and bounded.

To do some creeping we need to collect some useful facts.

Fact 1. A subset \mathcal{A} \subseteq \mathbb{R} is bounded if and only if is contained in some closed interval [a,b]

Fact 2. The set \mathbb{R} is complete (as a linear order) because every non-empty set A \subseteq \mathbb{R} with an upper bound has a least upper bound, called \sup A .

Fact 3. Closed subsets of compact subsets of \mathbb{R} are in fact themselves compact. With fact 1 this means that it is enough to show that closed and bounded intervals in \mathbb{R} are compact. (In general closed subsets of compact T_2 spaces are compact.)

So now let us creep:

Continue reading Creeping Along