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Thresholds for Ackermannian Ramsey Numbers Authors: Menachem Kojman Gyesik Lee Eran Omri Andreas Weiermann

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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. 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). Q is homogeneous for C Example 2

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Example … Let us prove: 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. n = 2 2k-1 – Size of complete graph over which we color edges. k – Size of homogeneous sub-graph. c = 2 – Number of colors.

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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 a m in-homogeneous complete sub-graph of size 2k. Proof …

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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.

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x1x1 x2x2 x3x3 x4x4 x5x5 x6x6 … Let G = (V,E) be the complete graph with V= 2 2k-1 : And let C be a coloring of the vertices of G with 2 colors (Red, Blue). C(x 1,x 2 ) = red C(x 2,x 6 ) = blue … Proof … x 2 2k-1 …

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Mark x i with the color C(x 1, x i ) There is a monochromatic complete sub-graph of {x 2,x 3,…} of size 2 2k-2. (Say red) x1x1 x2x2 x3x3 x4x4 x5x5 … x6x6 … … Proof … x 2 2k-1 …

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Mark x i with the color C(x 2, x i ) There is a monochromatic complete sub-graph of {x 3,x 6,…} of size 2 2k-3. (Say blue) x1x1 x2x2 x3x3 x6x6 … … Proof …

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Now, we have a min-homogeneous sequence {x i 1, x i 2, x i 3, …, x i 2k } Mark x i a with C(x i a, x i b ) for all b > a. There is a monochromatic subset of {x i 1, x i 2, x i 3, … x i 2k } of size k. xi1xi1 xi2xi2 xi3xi3 xi4xi4 xi5xi5 xi6xi6 … Second (and final) step: Find a k-sized h omogeneous complete sub-graph … Proof … x i 2k

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Ramsey Numbers. We Showed: On the other Hand: - Using the Probabilistic Method, we can show: To sum up: R 2 (k) – Exponential in k Denote R c (k) := The minimum n to satisfy:

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

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Primitive Recursive Functions and Ackermann’s Function A function that can be implemented using only for-loops is called primitive recursive. primitive recursive. Ackermann’s function Ackermann’s function – A simple example of a well defined total function that is computable but not primitive recursive

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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.

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Denote R g (k) := The minimum n to satisfy: We have seen so far: For a constant function, g(x) = c R g (k) ≤ c k Since The g-regressive Ramsey Number

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On the other hand it was known … That for g = ID: R g (k) is Ackermannian in terms of k.Ackermannian Namely, DOMINATES every primitive recursive function.

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The Problem constant g R g (k) < g k (primitive recursive.) Threshold g R g (k) is ackermannian. g = Id

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The Results I will insert a drawing x 1/j - Ackermannian If g(n) is ‘fast’ to go below n 1/k then, R g (k) is primitive recursive If g(n) ≤ n 1/k then,

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The Results Suppose B : N N is positive, unbounded and non-decreasing. Let g B (n) = n 1 / B −1 (n). Where B-1(n) = min{t : B(t) ≥ n}. Then, R g B (k) is Ackermannian iff B is Ackermannian.

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Min-homogeneity – Lower Threshold Suppose B : N N is positive, unbounded and non-decreasing. Let g B (n) = n 1 / B −1 (n). Where B-1(n) = min{t : B(t) ≥ n}. Then, for every natural number k, it holds that R g B (k) ≤ B(k). Basic Pointers: Assume more colors. Set c = g B (n) Use repeated division to show min B(k) ≥ c k (k) g B

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

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The Bad Coloring µ µ +1 µ +2 … 1 2 3 i k (f g ) 3 (µ) (f g ) 3 (2) (µ) (f g ) i (µ) C(m,n) = I Largest i s.t m,n are not in same segment. D Distance between m’s segment and n’s. {(f g ) i }

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The Bad Coloring µ µ +1 µ +2 … 1 2 3 i k m = µ +8 n = µ +29 C(m,n) = = I Largest i s.t m,n are not in same segment. D Distance between m’s segment and n’s.

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The Coloring – Formal Definition Given a monotonically increasing function 4g 2 and a natural number k >2 with we define a coloring That is: 1. 4g 2 -regressive on the interval 2.Has no min-homogeneous set of size k+1 within that interval.

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Definitions …

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The Coloring … So, why is it: 4g 2 -regressive4g 2 -regressive? Avoiding a min-homogeneous setAvoiding a min-homogeneous set?

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I will insert a drawing x 1/j - Ackermannian The Results - Surfing the waves

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Besides The Asymptotic bounds, We can also establish:

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omrier@cs.bgu.ac.il

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In 1985 Kanamori & McAloon eventually DOMINATES every primitive recursive function. Had used means of Model Theory to show that the bound of : In 1991 Prömel, Thumser & Voigt and independently in 1999 Kojman & Shelah have presented two simple combinatorial proofs to this fact.

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Input: Tuples of natural numbers Output: A natural number Basic primitive recursive functions: The constant function 0 The successor function S The projection functions P i n (x 1, x 2,…, x n ) = x i Primitive Recursive Functions

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More complex primitive recursive functions are obtained by : Composition: h(x 0,...,x l-1 ) = f(g 0 (x 0,..., xl-1 ),...,g k-1 (x0,...,x l-1 )) Primitive recursion: h(0,x 0,...,x k-1 ) = f(x 0,...,x k-1 ) h(S(n),x 0,...,x k-1 ) = g(h(n,x 0,...,x k-1 ),n,x 0,...,x k-1 ) Primitive Recursive Functions

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Given a function g : N N, denote Where f 0 (n) = n and f j+1 (n) = f(f j (n)) General Definition – (f g ) Hierarchy

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Let g = Id. Now Denote: Ack(n) = A n (n) Ackermann’s Function

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Examples: Ackermann’s Function

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3.Infinite Canonical Ramsey theorem (Erd ö s & Rado – 1950) Definition … Examples … 4.Finite Canonical Ramsey theorem (Erd ö s & Rado – 1950)

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1. clearly: 2. On the other hand there exist t,l such that: And thus: Now, since We have

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The Bad Coloring µ µ +1 µ +2 … 1 2 3 i k

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There exists no: Which is min-homogeneous for C g... But, for every (m,n) in the interval,

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A natural G ö del ’ s sentence What is a G ö del ’ s sentence?? Paris Harrington – Notion of Relatively large SETS.

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Previous Work 1.Infinite Ramsey ’ s theorem 2.Finite Ramsey ’ s theorem

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Regressive Ramsey Theorem Infinite Finite (*) Examples…

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Homogeneity – Lower Bound We show: To prove that we used:

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Homogeneity – Upper Bound We showed: To do that we used a general, well known, coloring method …

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The s -basis coloring

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Example… Imagine yourself in a billiard hall… A tournament is being organized… The rules: 1.Any two players may choose to play Pool or Snooker. (Coloring pairs with two colors) 2.A tournament can take place either in Snooker or in Pool. All the couples must choose the same. (Homogeneous set) 3.Three players minimum. (Size of subset)

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Pool= Snooker= How many players will ensure a Tournament?? 6 Ramsey Number for 3 is 6.

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Given a function g : N N, denote Where f 0 (n) = n and f j+1 (n) = f(f j (n)) General Definition – (f g ) Hierarchy

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Suppose g : N N is nondecreasing and unbounded. Then, R g (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) < n 1/t and M(t) is primitive recursive in t.

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