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Thresholds for Ackermannian Ramsey Numbers Authors: Menachem Kojman Gyesik Lee Eran Omri Andreas Weiermann Notation… • n = {1..n} • n k c 2 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). Q is homogeneous for C Example Example… Let us prove: •22 k 1 k 2 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… • 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 min-homogeneous complete sub-graph of size 2k. • Definition: A complete graph G = (E,V), with the natural ordering on V, is minhomogeneous for a coloring c:E if for every v in V, all edges (v,u), for u > v, are assigned one color. …Proof… Let G = (V,E) be the complete graph with V= 22k-1: x1 x2 x3 x4 x5 x6 … … x2 And let C be a coloring of the vertices of G with 2 colors (Red, Blue). C(x1,x2) = red C(x2,x6) = blue 2k-1 …Proof… Mark xi with the color C(x1, xi) x1 x2 x3 x4 x5 x6 … There is a monochromatic complete sub-graph of {x2,x3,…} of size 22k-2. (Say red) … x2 2k-1 …Proof… Mark xi with the color C(x2, xi) x1 x2 x3 x6 … There is a monochromatic complete sub-graph of {x3,x6,…} of size 22k-3. (Say blue) …Proof… Now, we have a min-homogeneous sequence {xi , xi , xi , …, xi } 1 xi 1 xi xi 2 2 3 3 xi 2k xi 4 5 xi … 6 xi 2k Mark x with C(x , x ) for all b > a. i i i •Second (and final) step: a a b Find homogeneous complete There isaak-sized monochromatic subsetsub-graph of {xi , xi , xi , … xi } of size k. 1 2 3 2k Ramsey Numbers. Rc(k) := The minimum n to •Denote We Showed: satisfy: ck k n 2 k 2k 2 • On the other Hand: - Using the Probabilistic To sum up: Method, we can show: R2(k) – Exponential in kk k 2 2 k 2 2 • More importantly: k , c c k min k c • 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 A function that can be implemented using only for-loops is called primitive recursive. Ackermann’s function – A simple example of a well defined total function that is computable but not primitive recursive 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 k n n k g established by means, similar to the regular Ramsey proof and compactness. min The g-regressive Ramsey Number Denote Rg(k) := The minimum n to satisfy: n min k g We have seen so far: For a constant function, g(x) = c Rg(k) ≤ ck Since c k min k g 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 • constant g Rg(k) < gk (primitive recursive.) • Threshold g Rg(k) is ackermannian. • g = Id g(x) The Results 1/k IfIfg(n) is ‘fast’ to go below n 1/j - Ackermannian g(n) ≤ n1/kxthen, then, Rg(k) is primitive recursive mininsert a drawing k will I n ( g (n)) k g x The Results Suppose B : N N is positive, unbounded and non-decreasing. 1/B−1(n) Let gB(n) = n . Where B-1(n) = min{t : B(t) ≥ n}. Then, Rg (k) is Ackermannian B iff B is Ackermannian. Min-homogeneity – Lower Threshold Basic Pointers: Suppose B : N N is positive, unbounded •andAssume more colors. non-decreasing. Set c =1/gBB−1(n) (n) Let gB(n) = n . Where • Use repeated division show B-1(n) = min{t : B(t) ≥ n}.toThen, B(k) ≥ ck min (k)g for every natural number k, it holds that B Rg (k) ≤ B(k). B Min-homogeneity – Upper Threshold • We show: 1 j if g(x) x for some j then, Rg (k ) is Ackermanni an. • To prove this, we present, given k, a bad coloring of an Akermannianly large n The=Bad Coloring C(m,n) <I,D> I k Largest i s.t m,n are not in same segment. (fg)i(µ) D Distance between m’s segment and n’s. (2) (fg)3 (µ) i (fg)3(µ) 3 2 1 ( f g )k ( ) µ µ+1 µ+2 … {(fg)i} The=Bad Coloring C(m,n) <I,D> = <2,1> I Largest i s.t m,n are not in same segment. k D Distance between m’s segment and n’s. i 3 2 1 µ µ+1 µ+2 … m = µ+8 n = µ+29 The Coloring – Formal Definition Given a monotonically increasing function 4g2 and a natural number k >2 with min({ t : k g (t )}) we define a coloring c : []2 That is: 1. 4g2-regressive on the interval { , ( f g ) k ( )) 2. Has no min-homogeneous set of size k+1 within that interval. Definitions… (d g )i (m, n) l : m ( f g I g (m, n) max i : (d g )i (m, n) 0 Dg (m, n) (d g ) I g ( m,n) (m, n) )i ( ) n (l ) The Coloring… C g (m, n) pr( Dg (m, n), I g (m, n)) a b 1 So, why is it: 2 b 4(max( a, b)) pr( a, b) 2 2 4g -regressive? Avoiding a min-homogeneous set? The Results - Surfing the waves g(x) x1/j - Ackermannian I will insert a drawing x Besides The Asymptotic bounds, We can also establish: RId (82) A53 (2 2 274 ) [email protected] In 1985 Kanamori & McAloon Had used means of Model Theory to show that the bound of : k n n min k 2 ID 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 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 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) General Definition – (fg) Hierarchy Given a function g : N N, denote ( f g )1 (n) n 1 ( f g )i 1 (n) ( f g ) Where 0 f (n) = n and j+1 f (n) ( g ( n )) i = ( n) j f(f (n)) Ackermann’s Function Let g = Id. Now A1 (n) ( f g )1 (n) n 1 Ai1 (n) ( f g )i1 (n) A (n) i ( n) Denote: Ack(n) = An(n) Ackermann’s Function Examples: A1 (n) n 1 A2 (n) A1 (n) 2n ( n) A3 (n) A2 (n) 2 ( n) A4 (n) 2 22 ..2n 22 n 3. Infinite Canonical Ramsey theorem (Erdös & Rado – 1950) Definition… e canonical e Examples… 4. Finite Canonical Ramsey theorem (Erdös & Rado – 1950) k , e n n canonical k e e H X is canonical for f : [ X ] if I e so that e s, t [ H ] f ( s) f (t ) s | I t | I Assume x1 x2 ... xe f ( x1 , x2 ,...xe ) e f ( x1 , x2 ,...xe ) x2 x3 ... xe (mod x1 ) 1. clearly: I g (m, n) k g (m) 2. On the other hand there exist t,l such that: t ( f g )i(l )1 ( ) m n ( f g )i(l 11) ( ) ( f g )i 1 (t ) Now, since We have ( f g )i 1 (t ) ( f g )i( g (t ) ) (t ) g (t ) (d g )i (t , ( f g )i 1 (t )) Dg (m, n) And thus: pr( I g (m, n), Dg (m, n)) 4( g (m)) 2 The Bad Coloring k i 3 2 1 µ µ+1 µ+2 … There exists no: x0 x1 ... xi with (d g ) ( x0 , xi ) 0 i Which is min-homogeneous for Cg... But, for every (m,n) in the interval, (d g ) (m, n) 0 k Homogeneity – Lower Bound • We show: for some f : s.t lim f (n) n lg x if g(x) lg lg x f ( x) then, k n n (k ) 2 g • To prove that we used: c ck k 2 c Homogeneity – Upper Bound • We showed: lg x if g(x) for some s s then, k n n (k ) 2 g To do that we used a general, well known, coloring method… The s-basis coloring Example… The rules: Imagine yourself in a 1. Any two players may choose to play Pool or billiard hall… Snooker. (Coloring pairs with two colors) A tournament being 2. A tournament can takeisplace either in Snooker or in Pool. All the couples must organized… choose the same. (Homogeneous set) 3. Three players minimum. (Size of subset) How many players will ensure a Tournament?? 6 Ramsey Number for 3 is 6. Pool = Snooker = General Definition – (fg) Hierarchy Given a function g : N N, denote ( f g )1 (n) n 1 ( f g )i 1 (n) ( f g ) Where 0 f (n) = n and j+1 f (n) ( g ( n )) i = ( n) j f(f (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.