Thresholds for ackermannian ramsey numbers
This presentation is the property of its rightful owner.
Sponsored Links
1 / 47

Thresholds for Ackermannian Ramsey Numbers PowerPoint PPT Presentation


  • 78 Views
  • Uploaded on
  • Presentation posted in: General

Thresholds for Ackermannian Ramsey Numbers. Authors: Menachem Kojman Gyesik Lee Eran Omri Andreas Weiermann. Notation …. n = {1..n} means: for every coloring C of the edges of the complete graph K n , there is a complete Q monochromatic sub-graph of size k .

Download Presentation

Thresholds for Ackermannian Ramsey Numbers

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Thresholds for ackermannian ramsey numbers

Thresholds for Ackermannian Ramsey Numbers

Authors: Menachem Kojman

Gyesik Lee

Eran Omri

Andreas Weiermann


Notation

Notation…

  • 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


Example

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.


Proof

Proof…

  • 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.


Thresholds for ackermannian ramsey numbers

  • 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.


Proof1

…Proof…

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


Proof2

…Proof…

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)


Proof3

…Proof…

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)


Proof4

…Proof…

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


Ramsey numbers

Ramsey Numbers.

  • 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


Thresholds for ackermannian ramsey numbers

  • More importantly:

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


Primitive recursive functions and ackermann s function

Primitive Recursive Functions and Ackermann’s Function

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


Regressive ramsey

Regressive Ramsey….

  • 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.


Thresholds for ackermannian ramsey numbers

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


On the other hand it was known

On the other hand it was known…

That for g = ID:

Rg(k) is Ackermannian in terms of k.

Namely, DOMINATES every primitive recursive function.


The problem

The Problem

  • constant g

    Rg(k) < gk (primitive recursive.)

  • Threshold g

    Rg(k) is ackermannian.

  • g = Id


The results

The Results

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


The results1

The Results

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.


Min homogeneity lower threshold

Min-homogeneity – Lower Threshold

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).


Min homogeneity upper threshold

Min-homogeneity – Upper Threshold

  • We show:

  • To prove this, we present, given k,

    a bad coloring of an Akermannianly large n


The bad coloring

The Bad Coloring

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}


The bad coloring1

The Bad Coloring

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


The coloring formal definition

The Coloring – Formal Definition

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.


Definitions

Definitions…


The coloring

The Coloring…

  • So, why is it:

    • 4g2-regressive?

    • Avoiding a min-homogeneous set?


The results surfing the waves

The Results - Surfing the waves

x1/j - Ackermannian

I will insert a drawing


Besides the asymptotic bounds we can also establish

Besides The Asymptotic bounds,We can also establish:


Thresholds for ackermannian ramsey numbers

[email protected]


In 1985 kanamori mcaloon

In 1985 Kanamori & McAloon

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.


Primitive recursive functions

Primitive Recursive Functions

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


Primitive recursive functions1

Primitive Recursive Functions

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)


Given a function g n n denote

Given a function g : N N, denote

General Definition – (fg) Hierarchy

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


Let g id now

Let g = Id. Now

Ackermann’s Function

Denote: Ack(n) = An(n)


Examples

Examples:

Ackermann’s Function


Thresholds for ackermannian ramsey numbers

  • Infinite Canonical Ramsey theorem

    (Erdös & Rado – 1950)

    Definition…

Examples…

  • Finite Canonical Ramsey theorem

    (Erdös & Rado – 1950)


1 clearly

1. clearly:

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

Now, since

We have

And thus:


The bad coloring2

The Bad Coloring

k

i

3

2

1

µµ+1 µ+2 …


Thresholds for ackermannian ramsey numbers

There exists no:

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


Homogeneity lower bound

Homogeneity – Lower Bound

  • We show:

  • To prove that we used:


Homogeneity upper bound

Homogeneity – Upper Bound

  • We showed:

    To do that we used a general, well known, coloring method…


The s basis coloring

The s-basis coloring


Example1

Example…

  • 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…


How many players will ensure a tournament

How many players will ensure a Tournament??

6

Ramsey Number for 3 is 6.

Pool=

Snooker =


Given a function g n n denote1

Given a function g : N N, denote

General Definition – (fg) Hierarchy

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


Thresholds for ackermannian ramsey numbers

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.


  • Login