Ramsey Theory and Classical Graph Ramsey Number

Department of Mathematics, Nanjing University

Email: yaojunc@nju.edu.cn

7 April, 2021

Yaojun CHEN

Ramsey Theory and Classical Graph Ramsey Number

OUTLINE

I. Originations of Ramsey Theory

II. Classical Graph Ramsey Number

What is Ramsey Theory ?

According to a 3500-year-old cuneiform text, an ancient Sumerian scholar once looked to the stars in the heavens and saw

a Lion

a BULL a SCORPION

Today, most stargazers would agree that the night sky appears to be filled with constellations in the shape of straight lines, rectangles and pentagons, and so on.Could it be that such geometric patterns arise from some unknown forces in the cosmos?

Mathematics provides a much more plausible explanation.

In 1928, an English mathematician, philosopher and economist F.P. Ramsey proved that such patterns are actually implicit in any large structure, no matter it is a group of stars or an array of pebbles!

F.P. Ramsey, On a problem of formal logic, Proc. London Math. Soc., 30(1930), 361-376.

1. Ramsey theorem

2. Schur theorem

3. Van der Waerden theorem

4. Erdős-Szekeres theorem

I. Originations of Ramsey Theory

1. Ramsey TheoremPARTY PUZZLE How many people does it take to form a group that always contains either n mutual acquaintances or n mutual strangers?

For it is known that at least 6 people is needed.

n=3,

How to prove it? One method is to list all conceivable combinations, check each one for an acquainted or unacquainted group of three.

Since we would have to check

combinations, this method is neither practical nor insightful.

Now, let points represent people in the party, a red edge joins people who are mutual acquainted, a blue edge joins people who are mutual strangers.

2(62) = 215 = 32768

1 2

3

45

6acquainted

strangers

1

23456

If {2,3,4} are mutual strangers, then the conclusion holds.

If some two people of {2,3,4} who are mutual acquainted, say 2 and 3 are mutual acquainted, then {1,2,3} are acquainted.

On the other hand, 5 people with relations illustrated in the left shows that 5 is not sufficient.

A complete graph : a graph on N vertices, any two vertices are connected by an edge.

KN

u v u vG G

General Form of PARTY PUZZLE: Given positive integers p and q, How large can guarantee that any 2-edge colorings of a complete graph with red and blue, either contains a red , or a blue ? F.P. Ramsey proved that such integer N exists! Let R(p,q) be the least integer N such that any 2-edge colorings of a complete graph with red and blue, either contains a red , or a blue .

p, q≥2.N=N(p, q)

KNKN Kp Kq

KNKN Kp Kq

It is easy to see that R(p,2)=p, R(2,q)=q and R(p,q)=R(q,p).

Theorem 1. For any two integers p,q≥2, R(p,q) exists and if p,q≥3, then the following holds:

R(p, q)≤R(p−1,q)+R(p, q−1) .

Let N=R(p−1,q)+R(p, q−1) .

0

21

R(p−1,q)

R(p−1,q)+1

N−1

KN

0

21

R(p,q−1)

R(p,q−1)+1

N−1

KN

General Form of Ramsey TheoremTheorem 2. For any two given positive integers and there is such that for any and any k-colorings of there exists some i with and a -set such that are in color i.

r, k,q1, q2, . . . , qr ≥r, N=N(q1, q2, . . . , qk),

n≥N, [n](r),1≤ i≤k qi Si ⊆ [n]

S(r)i

denotes all r-subsets of .[n]={1,2,...,n}; S(r)i Si

The smallest integer N satisfies Theorem 2 is called Ramsey number, write as . R(r)(n1, n2, . . . , nk)If then Theorem 2 is drawer principle. If then , that is, multiple colors Ramsey number.

r =1,r =2, R(2)(n1, n2, . . . , nk)=R(n1, n2, . . . , nk)

2. Schur TheoremThe following result, due to Schur (1916), which is viewed as one of the originations of Ramsey theory:

Theorem 3. For any given integer k, there exists an N, such that if then for any k-colorings of there must have of the same color such that

n≥N, [n],x, y, z∈ [n]

x + y = z .Let be the least possible value of N in Theorem 3.

is called Schur number. Some known Schur numbers up to now: The value of is determined by computer in 1965.

SkSk

S1 =2, S2 =5, S3 =14, S4 =45.S4

Schur theorem can be proved by Ramsey theorem.For any given integer k, take

Let be any partition of . For any 2-subset of , we color with if Thus, we get a k-colorings of all edges.

S1, S2, . . . , Sk [N]{i, j} [N] {i, j} ℓ

| i−j |∈Sℓ .By Ramsey theorem, has a 3-subset such that all of its 2-subsets receive the same color.

[N](2) {a, b, c}

Assume and Clearly, for some with , and

a>b>c, x=a−b, y=b−c, z=a−c .x, y, z∈Sℓ ℓ 1≤ℓ≤k

x + y = z .

N = R(2)k

(3,3,...,3) .

a−c∈S t

a−b∈St

1 2

N

j i

If color the edge with , | j−i |∈Sℓ, ij ℓ 1≤ℓ≤k .

a b

c

b−c∈

S t

j−i∈Sℓ

Let and then x=a−b, y=b−c z=a−c, x + y = z .

3. Van der Waerden TheoremVan der Waerden (1928) proved the following.Theorem 4. For any given positive integers there exists a positive integer such that for any k-colorings of , has an arithmetic progressions with terms of the same color.

ℓ, k,W=W(ℓ, k)

[W] [W]ℓ

The least possible value of in Theorem 4, is called Van der Waerden number. Some known Van der Waerden numbers:

W(ℓ, k)

W(3,2)=9, W(3,3)=27, W(3,4)=76,W(4,2)=35, W(5,2)=178.

Proof of W(3,2)=9.

1 2 3 4 5 6 7 8 9

Case 1. 4 and 6 has the same color, say in red.

4 652 8Case 2. 4 and 6 are in different colors.

1 2 3 4 5 6 7 8 94 652 83 71 9

1 2 3 4 5 6 7 8 94 652 83 71 9

4. Erdős-Szekeres TheoremErdős and Szekeres (1935) published the following:

Theorem 5. Let m be a positive integer. Then there exists a positive integer N, such that for any N points lie in a plane so that no three points form a straight line, there have m points form a convex m-polygon.

For m=4, N=5.Choose three points, say A,B,C such that ∠ABC<1800 and the other two points D, E lie in ∠ABC.

A

B

C

D

A

B

C

DE

Let N(m) be the smallest integer in Theorem 5. We have known that

No other values of N(m) are known up to now.

N(3) = 3,N(4) = 5,N(5) = 9.

Erdős conjectured that

The conjecture is true for m = 3,4,5.

N(m) = 1 + 2m−2 .

II. Classical Graph Ramsey Number

A challenging problem is to calculate R(5,5).

Up to know, all known values of classical Ramsey numbers R(p,q) are as follows:

It is very difficult to evaluate classical Ramsey numbers both exactly and asymptotically. Most of the values in the table are obtained with help of computers.

p 3 3 3 3 3 3 3 4 4q 3 4 5 6 7 8 9 4 5

R(p,q) 6 9 14 18 23 28 36 18 25

It is known that 43≤R(5,5)≤48 .

0 12

3

4

5

6

789

10

11

12

13

14

1516

Bounds for Classical Ramsey numbers

Erdős’ lower bound (1947): for R(p, p)>2

p2 p≥3.

The number of all red-blue edge colorings of isKn

The number of colorings with monochromatic is at most

Kp

2(n2) .

(np) ⋅ 2 ⋅ 2(n

2)−(p2) .

If

then there exists a red-blue colorings such that has no monochromatic .Kn Kp

(np) ⋅ 2(n

2)−(p2)+1 < 2(n

2),

By the definition of , we haveR(p, p)

R(p, p)>n .

It is not difficult to show that if thenn ≤ 2p2,

Thus,

By the argument above, we have

(np) <

np

2p−1≤ 2

p22 −p+1

= 2 12 p(p − 1)−1 ⋅ 2− p

2 +2 ≤ 2(p2)−1 .

(np) ⋅ 2(n

2)−(p2)+1 < 2(n

2) .

R(p, p) > 2p2 .

Probabilistic Method:

Color the edges of a complete graph randomly. That is, color each edge red with probability 1/2, and blue with probability 1/2. Since the probability that a given copy of has all edges red is

KN

Kp

2−(p2),

the expected number of red copies of isKp

2−(p2)(N

p ) .

Similarly, the expected number of blue copies of isKp

2−(p2)(N

p ) .

Therefore, the expected number of monochromatic copies of isKp

21−(p2)(N

p ) .

Since

21−(p2)(N

p ) ≤ 21−(p2) ( eN

p )p

,

take

21−(p2)(N

p ) ≤ 21−(p2) ( eN

p )p

< 1.

N=(1 − o(1)) p

2e ( 2)p,

we have

This implies

R(p, p)≥(1 − o(1)) p

2e ( 2)p

.

Spencer’s lower bound (1975):

R(p, p)≥(1 − o(1)) 2pe ( 2)

p.

Problem 1. Does there exist a positive constant such that

ε

R(p, p)≥ (1+ε)2pe ( 2)

p

for all sufficiently large ?p

Erdős and Szekeres’ upper bound (1935):

R(p + 1,q + 1) ≤ (p + qp ) .

If then it is easy to see the result holds. Since we have

p + q ≤ 3,R(p + 1,q + 1)≤R(p, q + 1)+R(p + 1,q),

R(p + 1,q + 1) ≤ (p + q − 1p − 1 )+(p + q − 1

p )= (p + q

p ) .

Graham and Rödl’s upper bound (1987):

R(p + 1,q + 1) ≤(p + q

p )log log (p + q)

.

It is believed that this upper bound is far more being satisfied!

is called diagonal Ramsey number if and off-diagonal Ramsey number otherwise. For , more advances are obtained recently.

R(p, q)p = q,

R(p, p)

Conlon’s upper bound (2009):

where c is a constant.

R(p + 1,p + 1) ≤ p−clog p/ log log p(2pp ) .

Sah’s upper bound (2020, arXiv:2005.09251):

where c is a constant.

R(p + 1,p + 1)≤e−c(log p)2(2pp ) .

It is known that:

R(p, p) ≤ (2p − 2p − 1 ) = O ( 4p

p ) .

Problem 2. Does

exist?

limp→∞

p R (p, p)

2 ≤ limp→∞

p R (p, p) ≤ 4.

If true, then

For off-diagonal Ramsey number , if is fixed and , then Erdős and Szekeres’ upper bound implies

R(p, q)p q → ∞

R(p, q)≤qp−1 .

Ajtai, Komlós and Szemerédi improved the bound

R(p, q)≤cpqp−1

(log q)p−2 .

where is a constant dependent on cp p .

When it says there is a constant such thatp = 3, c

R(3,q)≤cq2

log q.

Kim’s lower bound for (1995)::R(3,q)

R(3,q)= Θ(q2/ log q) .

R(3,q)≥cq2

log q,

where is a constant. Thus, the asymptotical order of is:

cR(3,q)

Thank you for your attention!