Monochrome Despite Himself

1. Great Theoretical Ideas In Computer Science Monochrome Despite Himself Lecture 26 CS 15-251

2. Pigeonhole Principle: If we put n+1 pigeons into n holes, some hole must receive at least 2 pigeons.

3. Pigeonhole Principle: If we put nk+1 pigeons into n holes, some hole must receive at least k+1 pigeons.

4. Recall: Sprouts Brussels Sprouts Today we will play the Ramsey game.

5. 2 players: green and yellow Start: n points are placed on the page. Players alternate by drawing an edge between some pair of points that has no edge between them. Green draws green edges. Yellow draws yellow edges. If a green triangle is formed, green loses. If a yellowtriangle is formed, yellow loses. If all possible edges get drawn without a monochromatic triangle being formed, green and yellow tie.

6. Try a game on 5 points.

7. Try a game on 6 points.

8. Can you prove this? It is not possible to tie in a game of 6-node Ramsey. equivalently Theorem: Any 2-coloring of the edges of a complete graph on 6 nodes will contain a monochromatic triangle.

9. Let G be a complete graph on 6 nodes with each edge colored green or yellow. Let v be an arbitrary node in G. v If any two of the three nodes are connected by a yellow edge, it forms a yellow triangle with v. Otherwise, the three nodes form a green triangle! v has 5 neighbors. By the pigeonhole principle, at least three of them are connected to v by green edges, or at least three of them are connected to v by yellow edges. Without loss of generality, assume the three edges are yellow.

10. Recall: A k-clique is a set of k nodes with all possible edges present. An independent set of size k is a set of k nodes with no edges between them. independent set of size 3 3-clique 4-clique Theorem: Any graph on 6 nodes contains a 3-clique or an independent set of size 3.

11. Party version: At a party with 6 people, there will be a group of at least 3 mutual acquaintances or at least 3 mutual strangers.

12. How many nodes do we have to have before any 2-coloring of the edges contains a monochromatic 4-clique?

13. Hold on! How do you know that there is such a number?

14. 18 is the magic number!

15. Example: Theorem: Ramsey(g,y) is defined for all F. P. Ramsey [1930] Definition: Let Ramsey(g,y) be the smallest number n such that any green/yellow coloring of the edges of an n-clique will contain a green g-clique or a yellow y-clique.

16. Base case: g = 2, y = 2, g + y = 4. Theorem: Ramsey(g,y) is some specific number. Furthermore, g,y 3 implies Proof: By induction on g+y.

17. If then G v Y By induction hypothesis, Ramsey(g-1,y) and Ramsey(g,y-1) are specific numbers. Let Hbe a graph on Ramsey(g-1,y) + Ramsey(g,y-1) nodes. Let v be an arbitrary node of H. Let G be the set of nodes that have green edges to v. Let Y be the set of nodes that have yellow edges to v.

18. By I.H., this means that we can find large cliques in G or Y. : G + v contains a green g-clique or a yellow y-clique. : Y + v contains a yellow y-clique or a green g-clique. G v Y Thus: or

19. By induction: Binomial Identity: Corollary: Proof: By induction ong + y. Base case: g = y= 2.

20. Binomial Identity: This is an upper bound on Ramsey(k,k). What about a lower bound? Corollary:

21. To argue a lower bound on Ramsey(k,k), we must find a way to color as big a graph as we can while avoiding a monochromatic k-clique.

22. We will use the probabilisitic method!

23. There are possible colorings, each one equally likely. For each possible k-clique Q, define an indicator variable: If all of Q’s edges have the same color. Otherwise. Let G be a complete graph on n nodes. [The value of n to be worked out later.] For each edge of G, randomly color it green or yellow.

24. There are different possible k-cliques. all k-cliques Q  k-cliques Q,

25. Why? It suffices to show , when k > 2 If we choose it follows that

26. Thus if , the expected # of monochromatic k-cliques is < 1. Thus it must be possible to 2-color all the edges so as to avoid a k -clique.

27. The known values of the Ramsey function:

28. Research Problem Compute some new Ramsey number.

29. Philosophical Interlude Ramsey Theory says that any structure of a certain type, no matter how “disordered”, contains a highly ordered substructure of the same type. Complete disorder is impossible.

30. Infinite Pigeonhole Principle: If you put an infinite number of pigeons into a finite number of holes, then some hole will have an infinite number of pigeons.

31. Infinite Ramsey Theorem: Suppose that G is a countably infinite complete graph, each of whose edges is colored green or yellow. It follows that G contains an infinite green clique or an infinite yellow clique.

32. Proof: At stage i (i  1), pick a node vi in G. By the I.P.H.P., there are  many edges of one color incident on vi. Suppose vi has  edges of color c coming from it. • Delete all nodes from G incident on vi with a color other than c that have not yet been examined. • Associate the color c with node vi. We have an  sequence of nodes v1 ,v2 ,v3,… By I.P.H.P. an infinite subset of the nodes have the same color associated with them. These nodes form a monochromatic clique!