CS 15-251Lecture 26Lecture 26

Monochrome Despite Monochrome Despite HimselfHimself

Pigeonhole Principle:Pigeonhole Principle:

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

Pigeonhole Principle:Pigeonhole Principle:

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

Recall:SproutsBrussels Sprouts

Today we will play the Ramsey game.

2 players: green and yellowStart: 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 yellow triangle is formed, yellow loses.

If all possible edges get drawn without a monochromatic triangle being formed, green and yellow tie.

Try a game on 5 points.

Try a game on 6 points.

It is not possible to tie in a game of 6-node Ramsey.

equivalentlyTheorem: Any 2-coloring of the edges of a complete graph on 6 nodes will contain a monochromatic triangle.

Can you prove this?

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 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. 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!

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.

3-clique 4-clique independentset of size 3

Theorem: Any graph on 6 nodes contains a 3-clique or an independent set of size 3.

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.

How many nodes do we have to have

before any 2-coloring of the edges contains a monochromatic 4-

clique?

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

18 is the magic

number!

F. P. Ramsey [1930]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.

Ramsey(3,3) 6Ramsey(4,4) 18Ramsey(2,2) 2Ramsey(2,3) 3

Example:

Theorem: Ramsey(g,y) is defined for all , , 2g y

Theorem: Ramsey(g,y) is some specific number. Furthermore,g,y 3 implies

Ramsey( , ) Ramsey( 1, )Ramsey( , 1)

g y g yg y

Proof: By induction on g+y.

Base case: g = 2, y = 2, g + y = 4. Ramsey 2,2 2

By induction hypothesis, Ramsey(g-1,y) and Ramsey(g,y-1) are specific numbers.Let H be a graph on Ramsey(g-1,y) + Ramsey(g,y-1) nodes.

vG Y

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.

Ramsey 1, 1 1 Ramsey , 1 1H g y g y

H H

Ramsey 1, , Ramsey , 1G g y Y g y If then

vG Y

Thus:

Ramsey 1,

Ramsey , 1

G g y

Y g y

or

By I.H., this means that we can find large cliques in G or Y.

Ramsey 1,G g y : G + v contains a green g-clique or a yellow y-clique.

Ramsey , 1Y g y : Y + v contains a yellow y-clique or a green g-clique.

2Ramsey ,

1g y

g yg

Corollary:Proof: By induction on g + y.Base case: g = y= 2. 2 2 2Ramsey 2,2 2 21

Ramsey( , ) Ramsey( 1, ) Ramsey( , 1)3 3

2 1

21

g y g y g yg y g y

g g

g yg

By induction:

11

n n nk k k

Binomial Identity:

2Ramsey ,

1g y

g yg

Corollary:

2 1

1

2 1Ramsey( , )

1

2

4

k

k

kk k

k

This is an upper bound on Ramsey(k,k). What about a lower bound?

? Ramsey ,k k

Binomial Identity: 2nnk

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.

We will use the probabilisitic method!

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.There are possible colorings, each one equally likely.

22n

For each possible k-clique Q, define an indicator variable: 1

0QX

If all of Q’s edges have the same color.Otherwise.

k-cliques Q,

1 2

2

2Pr 1 22

k

Q Q kE X X

There are different possible k-cliques.

nk

1 22

QQ

Q

k

XE E X

nk

all k-cliquesQ

If we chooseit follows that

( 2) / 22 kn

2

2 12

k

nk

Why? It suffices to show ( 2) / 2 22 22

kk

k

( 2) / 2( 2) / 2

2 2

12

2

22

2

2

2 / 2

kk k

kk

k

k

k

, when k > 2

2 / 22 kn Thus if ,the expected # of monochromatick-cliques is < 1. Thus it must be possible to 2-color all the edges so as to avoid a k -clique.

2 / 2 2 12 Ramsey ,

1k k

k kk

The known values of the Ramsey function:Ramsey(3,3) 6Ramsey(3,4) 9Ramsey(4,4) 18Ramsey(3,5) 14Ramsey(3,6) 18Ramsey(3,7) 23Ramsey(3,9) 36Ramsey(2, )Ramsey(3,8) 28Ramsey(4,5) 25Ramsey(5,5) 50

p p

Research ProblemResearch Problem

Compute some new Ramsey number.

Philosophical InterludePhilosophical InterludeRamsey 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.

Infinite Pigeonhole Principle: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.

Infinite Ramsey Theorem: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.

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!