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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 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).
k cn
Q is homogeneous for C
Example
2
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.
kk
22 12
• 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.
Proof…
• 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.
Ec :
x1 x2 x3 x4 x5 x6 …
Let G = (V,E) be the complete graph with V= 22k-1:
And let C be a coloring of the vertices of G with 2 colors (Red, Blue).
C(x1,x2) = red C(x2,x6) = blue
…Proof…
x22k-1…
Mark xi with the color C(x1, xi)
There is a monochromatic complete sub-graph of {x2,x3,…} of size 22k-2. (Say red)
x1 x2 x3 x4 x5 …x6 …
…Proof…
x22k-1…
Mark xi with the color C(x2, xi)
There is a monochromatic complete sub-graph of {x3,x6,…} of size 22k-3. (Say blue)
x1 x2 x3 x6 …
…Proof…
Now, we have a min-homogeneous sequence {xi
1, xi
2, xi
3, …, xi
2k}
Mark xiawith C(xi
a, xi
b) for all b > a.
There is a monochromatic subset of
{xi1, xi
2, xi
3, … xi
2k} of size k.
xi1
xi2
xi3
xi4
xi5
xi6
…
•Second (and final) step: Find a k-sized homogeneous complete sub-graph
…Proof…
xi2k
Ramsey Numbers.
• We Showed:
• On the other Hand: - Using the Probabilistic Method, we can show:
kkk2
22
kk
k2
222
To sum up:
R2(k) – Exponential in k
Denote Rc(k) := The minimum n to satisfy:
k cn
• More importantly:
• Explanation: - For a k-sized sequence iterate step #1 (repeated division) k times (namely, divide by at most c at each iteration)
kc
kcck min,
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) )
k gnnk min
Observe thatIs true for any g: N N which can be
established by means, similar to the regular Ramsey proof and compactness.
Denote Rg(k) := The minimum n to satisfy:
We have seen so far:
For a constant function, g(x) = c
Rg(k) ≤ ck Since k gck min
k gn min
The g-regressive Ramsey Number
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
The Results
I will insert a drawing
x
g(x)
x1/j - AckermannianIf g(n) is ‘fast’ to go below n1/k then, Rg(k) is primitive recursiveIf g(n) ≤ n1/k then,
k gngn k min))((
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
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 Rg
B(k) ≤ B(k).
Basic Pointers:• Assume more colors.
Set c = gB(n)• Use repeated division to show
min
B(k) ≥ ck (k)gB
Min-homogeneity – Upper Threshold
• We show:
• To prove this, we present, given k, a bad coloring of an Akermannianly large n
an.Ackermanni is )( then,
j somefor g(x) if
g
1
kR
x j
The Bad Coloring
µ µ+1 µ+2 …
1
2
3
i
k
(fg)3(µ)(fg)3
(2)(µ)
(fg)i(µ)
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})()( kgf
The Bad Coloring
µ µ+1 µ+2 …
1
2
3
i
k
m = µ+8
n = µ+29
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.
The Coloring – Formal Definition Given a monotonically increasing function 4g2 and
a natural number k >2 with
we define a coloringThat is:1. 4g2-regressive on the interval2. Has no min-homogeneous set of size k+1
within that interval.
)})(:min({ tgkt
2][:c
))()(,{ kgf
Definitions…
)()(: ),()( )( nifmlnmid lgg
0),()(:max),( nmdinmI igg
),()(),( ),( nmdnmD nmIgg g
The Coloring…
)),(),,(pr(),( nmInmDnmC ggg
So, why is it:
4g2-regressive?
Avoiding a min-homogeneous set?
2)),(max(42
1),pr( bab
baba
I will insert a drawing
x
g(x)
x1/j - Ackermannian
The Results - Surfing the waves
Besides The Asymptotic bounds,We can also establish:
)2()82(2742
53ARId
In 1985 Kanamori & McAloon
eventually DOMINATES every primitive recursive function.
kID
nnk2min
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.
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)
Primitive Recursive Functions
Given a function g : N N, denote
)()()()(
1)()())((
1
1
ni
fnif
nnfng
gg
g
Where f0(n) = n and fj+1(n) = f(fj(n))
General Definition – (fg) Hierarchy
Let g = Id. Now
)()()()( )(
11n
iAnifniA n
g
Denote: Ack(n) = An(n)
Ackermann’s Function
1)()()( 11 nnfnA g
Examples:
Ackermann’s Function
1)(1 nnA
nnAnA n 2)()( )(2 1
nn nAnA 2)()( )(
3 2
2222..2
2)(4
n
nA
3. Infinite Canonical Ramsey theorem (Erdös & Rado – 1950)Definition…
ee canonical
Examples…4. Finite Canonical Ramsey theorem (Erdös & Rado – 1950)
k ennek canonical,
ItIstfsfeHts
eI
eXfXH
||)()( ][,
thatso if
][:for canonical is
) (mod...),...,( 13221 xxxxxxxf ee
exxxf e ),...,( 21
exxx ... Assume 21
1. clearly: )(),( mgknmI g
2. On the other hand there exist t,l such that:)()()()()()( 1
)1(1
)(1 tffnmft ig
lig
lig
2))((4)),(),,(pr( mgnmDnmI gg And thus:
Now, since )()()()( ))((
1 tftf tgigig
),())()(,()()( 1 nmDtftdtg gigig We have
The Bad Coloring
µ µ+1 µ+2 …
1
2
3
i
k
0),()( with ... 010 igi xxi
dxxx
There exists no:
Which is min-homogeneous for Cg... But, for every (m,n) in the interval,
0),()( nmk
dg
Homogeneity – Lower Bound
• We show:
• To prove that we used:
k ccck 2
2)( then,
)( lglg
lgg(x) if
)(lim s.t : somefor n
gknnk
xfx
x
nff
Homogeneity – Upper Bound
• We showed:
To do that we used a general, well known, coloring
method…
2)( then,
somefor lg
g(x) if
gknnk
ss
x
The s-basis coloring
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)
Pool =
Snooker =
How many players will ensure a Tournament?? 6Ramsey Number for 3 is 6.
Given a function g : N N, denote
)()()()(
1)()())((
1
1
ni
fnif
nnfng
gg
g
Where f0(n) = n and fj+1(n) = f(fj(n))
General Definition – (fg) Hierarchy
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.
