CSCI 4260MATH 4150

1
CSCI 4260-MATH 4150

• Lecture 18
• Ramsey Numbers

2
An important question

• How many people must be present at a party to
  guarantee that there are three mutual
  acquaintances or three mutual strangers?

3
Ramseys theorem

• Such problems can be addressed by the theory
  developed by Ramsey.
• In general, Ramseys theorem studies properties
  of n objects divided into k classes.
• The special case when the relations are pairwise
  (e.g. acquaintance) modeled by graphs.

4
Red/Blue Coloring

• Any assignment of colors red and blue to edges.
• No constraints (dont confuse with colorings from
  last time).
• Our party can be modeled as
• Party of n people K(n)
• u and v know each other color (u,v) red
• Else color (u,v) blue

5
Ramsey formulation

• Let G be a red/blue colored graph
• Suppose we are given some graph F
• Let H be a subgraph of G
• H is isomorphic to F
• All edges of H are red
• Then we say G contains a red F.
• Similarly for blue F

6
Ramsey formulation (cont)

• Let F1 and F2 be two graphs
• r(F1, F2) Ramsey number of F1 and F2
• Smallest n such that
• Any red/blue coloring of K(n) contains either a
  red F1 or blue F2

7
Lets revisit our party problem

• How many people must be present at a party to
  guarantee that there are three mutual
  acquaintances or three mutual strangers
• Or
• What is r(K(3), K(3))?

8
r(F1, F2)

• To show that r(F1,F2) is n we need to show
• 1. Every red/blue coloring of K(n) contains
  either a red F1 or a blue F2
• This shows that r(F1,F2) ? n
• 2. There exists some red-blue coloring of K(n-1)
  having neither a red F1 or a blue F2
• This shows that r(F1,F2) ? n

9
Lets solve the party problem

• r(K3, K3) 6
• Part I
• Start with any coloring of K6
• Consider v1. It has 5 neighbors. 3 of them must
  be of the same color (say, red)

10
r(K3, K3) 6

• So far
• v1 has 3 red neighbors
• Can you complete the argument?

v1
11
Proof (cont)

• We showed that r(K3,K3) ? 6.
• To show equality, we need to present a coloring
  of K(5) that has no red or blue K3

12
What did Ramsey show

• Given any
• t
• n1, n2, , nk
• There exists a number N such that
• Let N 1, , N
• Assign one of the k colors to each t-subset of
  N
• For some color i,
• there exists a S ? N containing ni elements
  such that every t-subset of S is colored i.

13
Special case t 2

• t-subsets are edges
• Suppose you have k colors
• For any n1, , nk
• There is a number N such that
• For any k-coloring of the edges of K(N)
• There exists a complete subgraph K(ni) whose
  every edge is colored i

14
So far we considered 2 colors

• Ramsey theorem says that Ramsey Numbers, r(Ki,
  Kj), exist.
• The general theorem can be interpreted as
• Every sufficiently large structure, regardless of
  how disorderly it may seem, contains an orderly
  substructure of any prescribed size.

15
Ramsey Numbers

• We only know very few of them
• r(s,t) r(K(s), K(t))
• We just proved that r(3,3) 6
• r(5,5) is unknown
• Its known that 43?r(5,5) ? 49
• Some bounds not as tight
• 798?r(10,10) ? 12,677

16
Some properties

• Let n maxV(F1), V(F2)
• r(F1, F2) ? n
• Can you come up with a graph of size n-1 which
  has no red F1 nor a blue F2?
• Also we have,
• r(F1, F2) r(F2, F1)

17
Another way of looking at ramsey numbers

• Recall the definition of complement of G
• r(F1, F2) is the smallest n such that if G is any
  graph of order n either
• G contains a subgraph isomorphic to F1, or
• G complement contains a subgraph isomorphic to F2

18
For complete graphs

• R(K(s), K(t)) is the smallest n such that every
  graph of order n contains either
• a complete subgraph of order s, or
• an independent set of size t

19
Ramsey numbers of other graphs

• r(F1, F2) where F1 and F2 are not both complete
  graphs have also been studied.
• For example, lets look at the case where
• F1 P(3) path with three vertices
• F2 K(3)

20
r(P3, K3) 3
21
r(P3, K3) 4
22
r(P3, K3) 5

• We already showed that r(P3,K3) 4
• So all we need to show is that it is ? 5
• Given a red/blue coloring of K(5)
• Pick a vertex v1
• If v1 is adjacent to two red edges, done. Why?
• Else v1 is adjacent to at most one red edge
• Suppose (v1, v2), (v1, v3), (v1, v4) are all blue
• If there is a blue edge between any two of v2,
  v3, v4 we have a blue K(3) (together with v1)
• Otherwise we have a red P3 v2 v3 v4

23
Theorem

• For every tree T(m) of order m ? 2, and
• Every integer n ? 2
• r(T(m), K(n)) (m-1)(n-1) 1

24
Lemma

• Let T be a tree of order k.
• If G is a graph with ?(G) ? k 1,
• Then T is isomorphic to some subgraph of G.
• Proof by induction on k.

25
Back to our theorem

• For every tree T(m) of order m ? 2, and
• Every integer n ? 2
• r(T(m), K(n)) (m-1)(n-1) 1

26
PART I r(T(m), K(n)) ? (m-1)(n-1) 1

• Consider the following coloring of K((m-1)(n-1))
• The red subgraph is (n-1)K(m-1)
• There is no red tree of order m
• The blue subgraph is the complete (n-1)-partite
  graph
• There is no blue clique of order n

27
PART II r(T(m), K(n)) (m-1)(n-1) 1

• Proof by induction on n
• Basis n2
• Need to show r(T(m), K(2)) m
• Take any coloring of K(m)
• If there is a blue edge, we have a blue K(2)
• If not, we have a red K(m) which contains a red
  T(m)

28
Inductive step

• Assume, for every tree T(m) and an integer k ?2,
  r(T(m), K(k)) (m-1)(k-1) 1
• Need to show thatr(T(m), K(k1)) (m-1)k 1
• Consider any coloring of K((m-1)k1)
• We will consider two cases

29
Case I

• In K((m-1)k1), there is a vertex u that is
  incident with (m-1)(k-1)1 blue edges.
• Suppose that the edges between u and Wv1,v2,
  , v(m-1)(k-1)1 are blue
• Consider the subgraph induced by W
• By the inductive hypothesis, it contains either
• A red T(m), or
• A blue K(n)

Which is also a T(m) in K((m-1)k1)
Together with u, is a blue K(n1) in K((m-1)k1)
30
Case II

• Every vertex in K((m-1)k1), is adjacent to at
  most (m-1)(k-1) blue edges.
• Which means that every vertex is adjacent to at
  least (m-1)k (m-1)(k-1) m-1 red edges.
• Thus the red subgraph has a minimum degree of at
  least m-1
• By our lemma, the red subgraph contains a red
  T(m).
• So does K((m-1)k1).