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Thresholds for Ackermannian Ramsey Numbers

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Authors: Menachem Kojman

Gyesik Lee

Eran Omri

Andreas Weiermann

- n = {1..n}
- means:
for every coloring C of the edges of the complete graph Kn, there is a complete Q monochromatic sub-graph of size k.

n–Size of complete graph over which we color edges.

k–Size of homogeneous sub-graph.

c–Number of colors.

2–Size of tuples we color – pairs (edges).

2

Q is homogeneous for C

Example

Let us prove:

- means:
for every coloring C of the edges of the complete graph Kn, there is a complete Q monochromatic sub-graph of size k.

n = 22k-1–Size of complete graph over which we color edges.

k–Size of homogeneous sub-graph.

c = 2–Number of colors.

- Let us, from now on, assume the vertex set of every graph we consider is some initial segment of the natural numbers.
- First step:
Find amin-homogeneouscomplete sub-graph of size2k.

- Definition:A complete graph G = (E,V), with the natural ordering on V, is min-homogeneous for a coloring
if for every v in V, all edges (v,u), for

u > v, are assigned one color.

Let G = (V,E) be the complete graph with V= 22k-1:

x2

x3

x4

x5

x6

x1

…

x22k-1

…

And let C be a coloring of the vertices of G with 2colors (Red, Blue).

C(x1,x2) = redC(x2,x6) = blue

Markxiwith the colorC(x1, xi)

x2

x3

x4

x5

x1

x6

…

…

x22k-1

…

There is a monochromatic complete sub-graph of{x2,x3,…} of size 22k-2. (Sayred)

Mark xi with the colorC(x2, xi)

x2

x3

x6

x1

…

There is a monochromatic complete sub-graph of{x3,x6,…} of size 22k-3. (Sayblue)

Now, we have amin-homogeneoussequence {xi1, xi2, xi3, …, xi2k}

xi2

xi3

xi4

xi5

xi1

xi6

xi2k

…

Mark xiawith C(xia, xib) for allb > a.

There is a monochromatic subset of

{xi1, xi2, xi3, … xi2k} of size k.

- Second (and final) step:
- Find a k-sizedhomogeneouscomplete sub-graph

- We Showed:
- On the other Hand: -Using the Probabilistic Method, we can show:

Denote Rc(k) :=The minimumnto satisfy:

To sum up:

R2(k) – Exponential in k

- More importantly:
- Explanation: - For a k-sized sequence iterate step #1 (repeated division) k times (namely, divide by at most c at each iteration)

A function that can be implemented using only for-loops is calledprimitive recursive.

Ackermann’s function– A simple example of a well defined total function that is computable but not primitive recursive

- g-regressive colorings:
A coloring C is g-regressive if for every (m,n)

C(m,n) ≤ g(min(m,n)) = g(m)

- Can we still demand homogeneity??
- Not necessarily!!! ( e.g C(m,n) = Id(m) )

Observe that

Is true for any g: N N which can be established by means,similar to the regular Ramsey proof and compactness.

The g-regressive Ramsey Number

Denote Rg(k) :=The minimumnto satisfy:

We have seen so far:

For a constant function,g(x) = c

Rg(k) ≤ ck Since

That for g = ID:

Rg(k) is Ackermannian in terms of k.

Namely, DOMINATES every primitive recursive function.

- constant g
Rg(k) < gk (primitive recursive.)

- Threshold g
Rg(k) is ackermannian.

- g = Id

If g(n) is ‘fast’ to go below n1/kthen, Rg(k) is primitive recursive

If g(n)≤n1/kthen,

x1/j - Ackermannian

I will insert a drawing

Suppose B : N N is positive, unbounded and non-decreasing.

Let gB(n) = n1/B−1(n). Where

B-1(n) = min{t : B(t) ≥ n}. Then,

RgB(k) is Ackermannian

iff

B is Ackermannian.

Basic Pointers:

- Assume more colors.
Set c = gB(n)

- Use repeated division to show
min

B(k) ≥ ck (k)gB

Suppose B : N N is positive, unbounded and non-decreasing.

Let gB(n) = n1/B−1(n). Where

B-1(n) = min{t : B(t) ≥ n}. Then,

for every natural number k, it holds that

RgB(k)≤ B(k).

- We show:
- To prove this, we present, given k,
a bad coloring of an Akermannianly large n

C(m,n) = <I,D>

I Largest i s.t m,n are not in same segment.

D Distance between m’s segment and n’s.

(fg)i(µ)

k

(fg)3(2)(µ)

i

(fg)3(µ)

3

2

1

µµ+1 µ+2 …

{(fg)i}

C(m,n) = <I,D> = <2,1>

I Largest i s.t m,n are not in same segment.

D Distance between m’s segment and n’s.

k

i

3

2

1

µµ+1 µ+2 …

m = µ+8

n = µ+29

Given a monotonically increasing function 4g2and a natural number k >2 with

we define a coloring

That is:

- 4g2-regressive on the interval
- Has no min-homogeneous set of size k+1 within that interval.

- So, why is it:
- 4g2-regressive?
- Avoiding a min-homogeneous set?

x1/j - Ackermannian

I will insert a drawing

omrier@cs.bgu.ac.il

Had used means of Model Theory to show that the bound of :

eventually DOMINATES every primitive recursive function.

In 1991 Prömel, Thumser & Voigt and independently in 1999 Kojman & Shelah have presented two simple combinatorial proofs to this fact.

Input:Tuples of natural numbers

Output:A natural number

Basic primitive recursive functions:

- The constant function 0
- The successor function S
- The projection functions Pin(x1, x2,…, xn) = xi

More complex primitive recursive functions are obtained by :

- Composition:h(x0,...,xl-1) = f(g0(x0,...,xl-1),...,gk-1(x0,...,xl-1))
- Primitive recursion:h(0,x0,...,xk-1) = f(x0,...,xk-1) h(S(n),x0,...,xk-1) = g(h(n,x0,...,xk-1),n,x0,...,xk-1)

General Definition – (fg) Hierarchy

Where f0(n) = n and fj+1(n) = f(fj(n))

Ackermann’s Function

Denote: Ack(n) = An(n)

Ackermann’s Function

- Infinite Canonical Ramsey theorem
(Erdös & Rado – 1950)

Definition…

Examples…

- Finite Canonical Ramsey theorem
(Erdös & Rado – 1950)

2. On the other handthere exist t,l such that:

Now, since

We have

And thus:

k

i

3

2

1

µµ+1 µ+2 …

There exists no:

Which is min-homogeneous for Cg... But, for every (m,n) in the interval,

- We show:
- To prove that we used:

- We showed:
To do that we used a general, well known, coloring method…

- The rules:
- Any two players may choose to play Pool or Snooker. (Coloring pairs with two colors)
- A tournament can take place either in Snooker or in Pool. All the couples must choose the same. (Homogeneous set)
- Three players minimum. (Size of subset)

Imagine yourself in a billiard hall…

A tournament is being organized…

6

Ramsey Number for 3 is 6.

Pool=

Snooker =

General Definition – (fg) Hierarchy

Where f0(n) = n and fj+1(n) = f(fj(n))

Suppose g : N N is nondecreasing and unbounded. Then,

Rg(k) is bounded by some primitive recursive function in k

iff

for every t > 0 there is some M(t) s.t for all n ≥ M(t) it holds that g(n) < n1/t and M(t) is primitive recursive in t.