The combinatorics of solving linear equations
Origins: (I. Schur)
n n nx y z (mod p)Does the congruencealways have non-trivial solutions? 1875 -
1941
The combinatorics of solving linear equations
Origins: (I. Schur)
n n nx y z (mod p)Does the congruencealways have non-trivial solutions?
Yes !
Theorem (Schur – 1916)
If the positive integers are partitioned into finitely many classes then at least one of the classes contains solutions to the equation
x y z.
1875 - 1941
Some notation:
N:: {1,2,3,......}, [n] : {1,2,...,n}
χ N [: [r] - an r-coloring of Nχ N [:
χ induces a partition of N 1 2 rC C ... C
into color classes χ 1iC (i)
Objects in a single color class will be called monochromatic.
Schur’s theorem restated:
In any r-coloring of there is always a monochromatic solution
to the equation x y z.
N
Schur’s theorem restated:
Schur’s theorem restated:
In any r-coloring of there is always a monochromatic solution
to the equation x y z.
N
Schur’s theorem restated:
B. L. van der Waerden(1903 – 1996)
Theorem (van der Waerden – 1927)
In any r-coloring of there is always a monochromatic
arithmetic progression of length k (= k-AP).
N
Theorem (van der Waerden – 1927)
In any r-coloring of there is always a monochromatic
arithmetic progression of length k (= k-AP).
N
Schur’s theorem restated:
In any r-coloring of there is always a monochromatic solution
to the equation x y z.
N
Special case k = 3.
Schur’s theorem restated:
In any r-coloring of there is always a monochromatic solution
to the equation x y 2z.
N
B. L. van der Waerden(1903 – 1996)
What about the equation x y 3z?
What about the equation x y 3z?
Fact. There is a 4-coloring of with no monochromatic
solution to
N
x y 3z.
What about the equation x y 3z?
Fact. There is a 4-coloring of with no monochromatic
solution to
N
x y 3z.
Proof. Define χ N [: [4] by setting
χ t(5 (5k i)) i, 1 i 4.
Then 31 2
1 2 3
tt t5 (5k i) 5 (5k i) 3 5 (5k i)
32 dd(1 5 3 5 ) i 0 (mod 5)
which is impossible.
In general, call an equation
r-regular if every r-coloring of contains a monochromatic
solution to E.
1 2 n 1 1 2 2 n nE E(x ,x ,...,x ) ax a x .. a x 0
N
Also, call E regular if it is r-regular for every r.
In general, call an equation
r-regular if every r-coloring of contains a monochromatic
solution to E.
1 2 n 1 1 2 2 n nE E(x ,x ,...,x ) ax a x .. a x 0
N
Also, call E regular if it is r-regular for every r.
Which equations are regular ?
Theorem (R. Rado – 1933)
E is regular iff it can be solved with all ix 0 or 1,
and not all 0.
In general, call an equation
r-regular if every r-coloring of contains a monochromatic
solution to E.
1 2 n 1 1 2 2 n nE E(x ,x ,...,x ) ax a x .. a x 0
N
Also, call E regular if it is r-regular for every r.
Which equations are regular ?
Theorem (R. Rado – 1933)
E is regular iff it can be solved with all ix 0 or 1,
and not all 0.
x y z, x y 2z, x y 3z
1 0 1, 1 1 2 1, ???
Conjecture (Rado – 1933) For every k, there are equations
which are k-regular but are not (k+1)-regular.
Conjecture (Rado – 1933) For every k, there are equations
which are k-regular but are not (k+1)-regular.
Theorem (Alexeev-Tsimerman – 2009)
The equation
i i
i i
k k2 2
i 02 1 2 1i 1 i 1
x ( 1 )x
is k-regular but not (k+1)-regular.
More generally, for an m x n integer matrix A,
let denote a system of homogenous linear equations.
Ax 0
For example, for
1 2 1 0 0
A 0 1 2 1 0 ,
0 0 1 2 1
the corresponding system is:
1 2 3 2 3 4 3 4 5x 2x x 0, x 2x x 0, x 2x x 0,
which corresponds to an arithmetic progression of length 5 (5-AP).
A matrix A is said to satisfy the columns condition (CC)
if it possible to partition the columns of A into blocks1 2 sB ,B ,...,B
so that:
(i) j 1
jc B
c 0;
(ii)j i
jc B
f or i 1, c
can be expressed as a rational linear
combination of the columns of 1 2 i 1B ,B ,...,B .
A matrix A is said to satisfy the columns condition (CC)
if it possible to partition the columns of A into blocks1 2 sB ,B ,...,B
so that:
For example,
2 4 2 1
0 2 2 1
1 3 2 0has CC since
2 4 2 0 2 4 1
0 23
2 0 , 0 2 1
1 3 2 0 1 3 0
12 2
(i) j 1
jc B
c 0;
(ii)j i
jc B
f or i 1, c
can be expressed as a rational linear
combination of the columns of 1 2 i 1B ,B ,...,B .
Theorem (Rado – 1933)The system is regular iff A satifies the columns condition.
Ax 0
Paul Erdös and Richard Rado1913-1996 1906-1989
Rado also had many other conjectures concerning systems
of regular equations, one of which was this.
Call a subset large if for any r, any r-coloring of S always
has monochromatic solutions for every regular system.
NS
Rado also had many other conjectures concerning systems
of regular equations, one of which was this.
Call a subset large if for any r, any r-coloring of S always
has monochromatic solutions for every regular system.
NS
Conjecture. For any partition of a large set 1 2 tS S S ... S
into finitely many parts, at least one of the parts is large.
iS
Rado also had many other conjectures concerning systems
of regular equations, one of which was this.
Call a subset large if for any r, any r-coloring of S always
has monochromatic solutions for every regular system.
NS
Conjecture. For any partition of a large set 1 2 tS S S ... S
into finitely many parts, at least one of the parts is large.
iS
This was finally proved in 1973 by Deuber.
Walter Deuber – (1943 – 1999)
Even though and are both regular, there is a
fundamental difference between them.
x y z x y 2z
Recall van der Waerden’s theorem:In any r-coloring of there is always a monochromatic
arithmetic progression of length k (= k-AP).
N
Erdös and Turán (1936) ask: Which color class has the k-AP’s?
Even though and are both regular, there is a
fundamental difference between them.
x y z x y 2z
Recall van der Waerden’s theorem:In any r-coloring of there is always a monochromatic
arithmetic progression of length k (= k-AP).
N
Erdös and Turán (1936) ask: Which color class has the k-AP’s?They conjectured that the “densest” class should have them.More precisely, they conjectured that if satisfies NS
then S should contain k-AP’s for every k.
| S [n]|limsup 0
n n
Equivalently, define to be the size of the largest subset
of [n] which contains no k-AP.
kr (n)
Conjecture (Erdös -Turán – 1936)
kr (n) o(n).For all k,
Equivalently, define to be the size of the largest subset
of [n] which contains no k-AP.
kr (n)
Conjecture (Erdös -Turán – 1936)
kr (n) o(n).For all k,
It became clear that even was not going to be so simple
to determine because of:
3r (n)
Theorem (Behrend – 1946)
3r (n) n exp( c logn).
For a suitable constant c > 0,
Equivalently, define to be the size of the largest subset
of [n] which contains no k-AP.
kr (n)
Conjecture (Erdös -Turán – 1936)
kr (n) o(n).For all k,
It became clear that even was not going to be so simple
to determine because of:
3r (n)
Theorem (Behrend – 1946)
3r (n) n exp( c logn).
For a suitable constant c > 0,
This shows that for every 1 ε3r (n) n ε 0 as n .
The first non-trivial upper bound was given by Roth:
Theorem (Roth – 1954)
n3 log log nr (n) O( ).
The first non-trivial upper bound was given by Roth:
Theorem (Roth – 1954)
n3 log log nr (n) O( ).
This was subsequently improved by Szemerédi and Heath-Brown to
cn
3 (log n)r (n) O( ) with the current record being held by
Bourgain (2008)
2
23
(loglog n)3 (log n)r (n) O n .
The first non-trivial upper bound was given by Roth:
Theorem (Roth – 1954)
n3 log log nr (n) O( ).
This was subsequently improved by Szemerédi and Heath-Brown to
cn
3 (log n)r (n) O( ) with the current record being held by
Bourgain (2008)
2
23
(loglog n)3 (log n)r (n) O n .
There has also been a very recent breakthrough improvement
of Behrend’s lower bound by Michael Elkin.
Behrend (1946)
M. Elkin (2009)
22 2 log n
3 2r (n) cn 14log n
22 2 log n
3 2r (n) cn 14log n
Behrend (1946)
M. Elkin (2009)
22 2 log n
3 2r (n) cn 14log n
22 2 log n
3 2r (n) cn 14log n
Is this close to the “truth” ?
What about ?
kr (n)
What about ?
kr (n)
4r (n) o(n) Szemerédi - 1969
What about ?
kr (n)
4r (n) o(n) Szemerédi - 1969
kr (n) o(n) Szemerédi - 1974
+ $1000Endre Szemerédi – (1940 - )
What about ?
kr (n)
4r (n) o(n) Szemerédi - 1969
kr (n) o(n) Szemerédi - 1974
+ $1000
kr (n) o(n) Furstenberg – 1979 (ergodic theory)
Endre Szemerédi – (1940 - )
What about ?
kr (n)
4r (n) o(n) Szemerédi - 1969
kr (n) o(n) Szemerédi - 1974
+ $1000
kr (n) o(n) Furstenberg – 1979 (ergodic theory)
c4r (n) O(n/ (loglogn) ) Gowers - 1998
Endre Szemerédi – (1940 - )
What about ?
kr (n)
4r (n) o(n) Szemerédi - 1969
kr (n) o(n) Szemerédi - 1974
+ $1000
kr (n) o(n) Furstenberg – 1979 (ergodic theory)
c4r (n) O(n/ (loglogn) ) Gowers - 1998
kk
cr (n) O(n/ (loglogn) ) Gowers - 2001
Endre Szemerédi – (1940 - )
What about ?
kr (n)
4r (n) o(n) Szemerédi - 1969
kr (n) o(n) Szemerédi - 1974
+ $1000
kr (n) o(n) Furstenberg – 1979 (ergodic theory)
c4r (n) O(n/ (loglogn) ) Gowers - 1998
kk
cr (n) O(n/ (loglogn) ) Gowers - 2001
The last result can be used to obtain the best currently
available bound for the van der Waerden function W(k).
Endre Szemerédi – (1940 - )
A consequence of van der Waerden’s theorem on arithmetic progressions
is that for every n, there is a least number W(n) so that in any
2-coloring of [W(n)], there is always formed a monochromatic n-AP.
A consequence of van der Waerden’s theorem on arithmetic progressions
is that for every n, there is a least number W(n) so that in any
2-coloring of [W(n)], there is always formed a monochromatic n-AP.
n2222For every n, W(n) 2
9
Theorem (Gowers 2001)
$1000
2222
22
2
2
……
…
2
n 2
’s
W(n) <
The best known lower bound is:
nW(n 1) n 2 , (Berlekamp - 1968)
Elwyn Berlekamp - (1940 - )
The best known lower bound is:
nW(n 1) n 2 , (Berlekamp - 1968)
Elwyn Berlekamp - (1940 - )The only known values are:
nW(n)
2 64353
31785
9 1132
The best known lower bound is:
nW(n 1) n 2 , (Berlekamp - 1968)
Elwyn Berlekamp - (1940 - )
(Brave) Conjecture.$1000
For all n,2nW(n) 2 .
The only known values are:
nW(n)
2 64353
31785
9 1132
A new proof of van der Waerden for 3-AP’s
N
N
“corner”
G(N) = N by N triangular grid
TheoremTheorem: (RLG + J. Solymosi - 2006)
There is a universal constant c such that
if the points of G(N) are colored with at
most c log log N colors, then there is always
a monochromatic corner (x,y), (x+d,y), (x,y+d).
A new proof of van der Waerden for 3-AP’s
N
N
Sketch of proof:
Suppose G(N) is r-colored.
L1
Let L denote the set of points on the diagonal.
1
N
N
Sketch of proof:
Suppose G(N) is r-colored.
L1
Let be points with the “most popular” color c .
1 1S L1
Let L denote the set of points on the diagonal.
1
N
N
Sketch of proof:
Suppose G(N) is r-colored.
L1
………
………
………
Let be points with the “most popular” color c .
1 1S L1
Let L denote the set of points on the diagonal.
1
Let be the points in the product
below L .
1 1 1T S xS
1
N
N
Sketch of proof:
Suppose G(N) is r-colored.
L1
………
………
………
Let be points with the “most popular” color c .
1 1S L1
Let L denote the set of points on the diagonal.
1
Let be the points in the product
below L .
1 1 1T S xS
1 Thus,
21 2
11
2r
N| S |r
2 2| T | ~ N
N
N
Sketch of proof:
Suppose G(N) is r-colored.
L1
………
………
………
Let be points with the “most popular” color c .
1 1S L1
Let L denote the set of points on the diagonal.
1
Let be the points in the product
below L .
1 1 1T S xS
1 Thus,
21 2
11
2r
N| S |r
2 2| T | ~ N
Hence, some line L parallel to L
has at least points of T on it.
21
2rN
2 1
1
L2
N
N
Sketch of proof:
Suppose G(N) is r-colored.
L1
………
………
………
Let be points with the “most popular” color c .
1 1S L1
Let L denote the set of points on the diagonal.
1
Let be the points in the product
below L .
1 1 1T S xS
1 Thus,
21 2
11
2r
N| S |r
2 2| T | ~ N
Hence, some line L parallel to L
has at least points of T on it.
21
2rN
2 1
1
L2
Let be points with
the most popular color c
2 2S L
2
and let be the points in the product below L .
2 2 2T S xS2
62 23
2
N2r
2
18r
| S |
2| T | ~ N
Thus, , etc.
Continue for r+1 steps.Then r 1r 1 r2 1 2 1
1| S | N
2 r
62 23
2
N2r
2
18r
| S |
2| T | ~ N
Thus, , etc.
Continue for r+1 steps.
L i
L j
Lk
a
b
c
d
ef
g
Observe that if abc, bde and cfg are corners
then so is adg.
Then r 1r 1 r2 1 2 1
1| S | N
2 r
62 23
2
N2r
2
18r
| S |
2| T | ~ N
Thus, , etc.
Continue for r+1 steps.
L i
L j
Lk
a
b
c
d
ef
g
Observe that if abc, bde and cfg are corners
then so is adg.
Hence, if then we have a
monochromatic corner.
u vc c , u v,
Then r 1r 1 r2 1 2 1
1| S | N
2 r
62 23
2
N2r
2
18r
| S |
2| T | ~ N
Thus, , etc.
Continue for r+1 steps.
L i
L j
Lk
a
b
c
d
ef
g
Observe that if abc, bde and cfg are corners
then so is adg.
Hence, if then we have a
monochromatic corner.
u vc c , u v,
Then r 1r 1 r2 1 2 1
1| S | N
2 r
This works providedcr2N 2
for a suitable absolute constant c.
This proves the Theorem.
N
Notice that we can project a corner onto a 3-AP
on a copy of . N
Notice that we can project a corner onto a 3-AP
on a copy of .
Hence, we have the
Corollary. Corollary. If {1, 2, …, N} is colored with at
most c log log N colors, then a monochromatic
3-AP must be formed.
N
N
Notice that we can project a corner onto a 3-AP
on a copy of .
Hence, we have the
Corollary. Corollary. If {1, 2, …, N} is colored with at
most c log log N colors, then a monochromatic
3-AP must be formed.
The best result of this type follows from
a difficult result of Bourgain, which
gives the above conclusion with
23
2
logn
log lognc colors.
N
N
N
N
We can actually continue the preceding argument for (m-1)r+1
steps to get a monochromatic “equilateral binary tree” with m levels.
……
……
……
N
N
We can actually continue the preceding argument for (m-1)r+1
steps to get a monochromatic “equilateral binary tree” with m levels.
……
……
……
We can project these trees onto the real axis to get
monochromatic affine “m-cubes”. These are sets of
the formm
k k kk 1
{a x , 0 or 1}
N
N
We can actually continue the preceding argument for (m-1)r+1
steps to get a monochromatic “equilateral binary tree” with m levels.
……
……
……
We can project these trees onto the real axis to get
monochromatic affine “m-cubes”. These are sets of
the formm
k k kk 1
{a x , 0 or 1}
Similar arguments yield monochromatic
connected crossed diagonals of a
hypercube, etc.
N
N
We can actually continue the preceding argument for (m-1)r+1
steps to get a monochromatic “equilateral binary tree” with m levels.
……
……
……
We can project these trees onto the real axis to get
monochromatic affine “m-cubes”. These are sets of
the formm
k k kk 1
{a x , 0 or 1}
Similar arguments yield monochromatic
connected crossed diagonals of a
hypercube, etc.
The partition regularity of these structures was first established by
Hilbert in 1892 in connection with his work on irreducibility of polynomials.
Combinatorial Combinatorial lineslines
For a finite set denote the set of n-tuples from A.
n1 2 mA {a ,a ,...,a }, let A
A combinatorial line in A is a set of m n-tuples wheren1 2 mX ,X ,...,X
k k k k k(1) (2) (j) (n)X (X ,X ,...,X ,...,X )
either all X (j) are equal, or k k k 1 k m.X (j) a f or
and for each j,
Combinatorial Combinatorial lineslines
For a finite set denote the set of n-tuples from A.
n1 2 mA {a ,a ,...,a }, let A
A combinatorial line in A is a set of m n-tuples wheren1 2 mX ,X ,...,X
k k k k k(1) (2) (j) (n)X (X ,X ,...,X ,...,X )
either all X (j) are equal, or k k k 1 k m.X (j) a f or
and for each j,
Theorem (Hales – Jewett – 1963)For every r, there is an N = N(A, r) such that if is r-colored
then there must always be formed a monochromatic combinatorial line.
NA
Hales – Jewett implies van der Waerden by taking A = {1, 2,…, m} and
identifying the n-tuple with the integer1 2 n(x ,x ,...,x ) 1 2 nx x ... x .
Hales – Jewett implies van der Waerden by taking A = {1, 2,…, m} and
identifying the n-tuple with the integer1 2 n(x ,x ,...,x ) 1 2 nx x ... x .
...a... ...b... ..1 1 .c...
...a... ...b... ..2 2 .c...
...a... ...b... ..m m .c...
… …S 2t
S mt
… …
S 1t
Hales – Jewett implies van der Waerden by taking A = {1, 2,…, m} and
identifying the n-tuple with the integer1 2 n(x ,x ,...,x ) 1 2 nx x ... x .
Of course, Szemerédi’s theorem is a (stronger) density form of
van der Waerden’s theorem since in any r-coloring of , some color
class must have density at least 1/r.
...a... ...b... ..1 1 .c...
...a... ...b... ..2 2 .c...
...a... ...b... ..m m .c...
… …S 2t
S mt
… …
S 1t
van der Waerden
SzemerédiHales - Jewett
We have this picture:
van der Waerden
SzemerédiHales - Jewett
density Hales – Jewett theorem ?
We have this picture:
Is there a
van der Waerden
SzemerédiHales - Jewett
density Hales – Jewett
We have this picture:
Yes !
TheoremTheorem – Density Hales–Jewett (Furstenberg - Katznelson - 1991)For every ε > 0, there is an N = N(ε) such that if
N NR A , with| R| | A| ,
then R must contain a combinatorial line.
TheoremTheorem – Density Hales–Jewett (Furstenberg - Katznelson - 1991)For every ε > 0, there is an N = N(ε) such that if
N NR A , with| R| | A| ,
then R must contain a combinatorial line.
Big Problem:Big Problem: Obtain reasonable (for example, any) bounds on N(ε).
(Furstenberg-Katznelson proof uses ergodic theory tools).
TheoremTheorem – Density Hales–Jewett (Furstenberg - Katznelson - 1991)For every ε > 0, there is an N = N(ε) such that if
N NR A , with| R| | A| ,
then R must contain a combinatorial line.
Big Problem:Big Problem: Obtain reasonable (for example, any) bounds on N(ε).
Warm-up Big Problem: Try this for m = 3.
(Furstenberg-Katznelson proof uses ergodic theory tools).
TheoremTheorem – Density Hales–Jewett (Furstenberg - Katznelson - 1991)For every ε > 0, there is an N = N(ε) such that if
N NR A , with| R| | A| ,
then R must contain a combinatorial line.
Big Problem:Big Problem: Obtain reasonable (for example, any) bounds on N(ε).
Warm-up Big Problem: Try this for m = 3.
Very recent breakthrough !
Tim Gowers and the Polymath Polymath project.
(Furstenberg-Katznelson proof uses ergodic theory tools).
It now appears that the Polymath project has succeeded
in finding 3 combinatorial proofs of the density Hales – Jewett
theorem, and consequently, for the first time we have
explicit (but still quite large!) bounds on N(m,ε).
It now appears that the Polymath project has succeeded
in finding 3 combinatorial proofs of the density Hales – Jewett
theorem, and consequently, for the first time we have
explicit (but still quite large!) bounds on N(m,ε).Is this the wave of the future ??
Back to linear equationsEven though and are both regular, there is a
fundamental difference between them.
x y z x y 2z
x y 2z always has solutions in any set of positive densitywhereas may not. x y z
Back to linear equationsEven though and are both regular, there is a
fundamental difference between them.
x y z x y 2z
x y 2z always has solutions in any set of positive densitywhereas may not. x y z
For example, has no solutions in either x y z
n2
{x [n] : x odd} or {x [n] : x }.
Back to linear equationsEven though and are both regular, there is a
fundamental difference between them.
x y z x y 2z
x y 2z always has solutions in any set of positive densitywhereas may not. x y z
In fact, a system of homogenous linear equations given by
Ax 0 is density regular iff the all 1’s vector is a solution.
For example, has no solutions in either x y z
n2
{x [n] : x odd} or {x [n] : x }.
If [n] is r-colored, how manyhow many monochromatic 3-AP’s must be formed?
It follows from general results of Frankl, RLG, and Rödl that for any regular
system of homogeneous linear equations that a positive fraction of all the
solutions must be monochromatic.
If [n] is r-colored, how manyhow many monochromatic 3-AP’s must be formed?
It follows from general results of Frankl, RLG, and Rödl that for any regular
system of homogeneous linear equations that a positive fraction of all the
solutions must be monochromatic.For the equation let f (n) denote the minimum
possible number of monochromatic solutions formed over
all possible 2-colorings of [n].
1 2 tE(x ,x ,...,x ), E
If [n] is r-colored, how manyhow many monochromatic 3-AP’s must be formed?
It follows from general results of Frankl, RLG, and Rödl that for any regular
system of homogeneous linear equations that a positive fraction of all the
solutions must be monochromatic.For the equation let f (n) denote the minimum
possible number of monochromatic solutions formed over
all possible 2-colorings of [n].
1 2 tE(x ,x ,...,x ), E
For example, for E: x + y = z, it is known that
2
En22(1 o(1))f (n)
Robertson/Zeilberger and Schoen - 1998
If [n] is r-colored, how manyhow many monochromatic 3-AP’s must be formed?
It follows from general results of Frankl, RLG, and Rödl that for any regular
system of homogeneous linear equations that a positive fraction of all the
solutions must be monochromatic.For the equation let f (n) denote the minimum
possible number of monochromatic solutions formed over
all possible 2-colorings of [n].
1 2 tE(x ,x ,...,x ), E
For example, for E: x + y = z, it is known that
2
En22(1 o(1))f (n)
Robertson/Zeilberger and Schoen - 1998
What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??
What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??
[n] contains 3-AP’s. 2n
4(1 o(1))
What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??
[n] contains 3-AP’s. 2n
4(1 o(1))
Hence, if [n] is randomlyrandomly 2-colored then there will be
essentially monochromatic 3-AP’s. 2n
16(1 o(1))
What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??
[n] contains 3-AP’s. 2n
4(1 o(1))
Hence, if [n] is randomlyrandomly 2-colored then there will be
essentially monochromatic 3-AP’s. 2n
16(1 o(1))
Is this best possible ?
What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??
[n] contains 3-AP’s. 2n
4(1 o(1))
Hence, if [n] is randomlyrandomly 2-colored then there will be
essentially monochromatic 3-AP’s. 2n
16(1 o(1))
Is this best possible ? No!
What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??
[n] contains 3-AP’s. 2n
4(1 o(1))
Hence, if [n] is randomlyrandomly 2-colored then there will be
essentially monochromatic 3-AP’s. 2n
16(1 o(1))
Is this best possible ? No!
It has recently been shown that there is a 2-coloring of [n] which has only 2117
2192(1 o(1)) n monochromatic 3-AP’s.
What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??
[n] contains 3-AP’s. 2n
4(1 o(1))
Hence, if [n] is randomlyrandomly 2-colored then there will be
essentially monochromatic 3-AP’s. 2n
16(1 o(1))
Is this best possible ? No!
It has recently been shown that there is a 2-coloring of [n] which has only 2117
2192(1 o(1)) n monochromatic 3-AP’s.
117 12192 18.735...
What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??
[n] contains 3-AP’s. 2n
4(1 o(1))
Hence, if [n] is randomlyrandomly 2-colored then there will be
essentially monochromatic 3-AP’s. 2n
16(1 o(1))
Is this best possible ? No!
It has recently been shown that there is a 2-coloring of [n] which has only 2117
2192(1 o(1)) n monochromatic 3-AP’s.
117 12192 18.735...
28 6 28 37 59 2828 63759116 116
What is the situation for 3–AP’s, i.e., the equation x + y = 2z ??
[n] contains 3-AP’s. 2n
4(1 o(1))
Hence, if [n] is randomlyrandomly 2-colored then there will be
essentially monochromatic 3-AP’s. 2n
16(1 o(1))
Is this best possible ? No!
It has recently been shown that there is a 2-coloring of [n] which has only 2117
2192(1 o(1)) n monochromatic 3-AP’s.
117 12192 18.735...
In the other direction, it is known that there must
always be at least monochromatic 3-AP’s.2167532768(1 o(1)) n
28 6 28 37 59 2828 63759116 116
1675 11732768 2192
0.05111... 0.05337...
We believe the upper bound is the “truth”, i.e., any
2-coloring of [n] must have at least 21172192
(1 o(1)) n
monochromatic 3-AP’s.
We believe the upper bound is the “truth”, i.e., any
2-coloring of [n] must have at least 21172192
(1 o(1)) n
monochromatic 3-AP’s.
Sketch of our approach (S. Butler, K. Costello, RLG)
Let c :[n] {0,1} be a 2-coloring of [n] and define
jS {x :c(x) j, 1 x n}, j 0, 1,
We believe the upper bound is the “truth”, i.e., any
2-coloring of [n] must have at least 21172192
(1 o(1)) n
monochromatic 3-AP’s.
Sketch of our approach (S. Butler, K. Costello, RLG)
Let c :[n] {0,1} be a 2-coloring of [n] and define
jS {x :c(x) j, 1 x n}, j 0, 1,
0 1V(S ,S ) # of monochromatic 3-AP’s formed
We believe the upper bound is the “truth”, i.e., any
2-coloring of [n] must have at least 21172192
(1 o(1)) n
monochromatic 3-AP’s.
Sketch of our approach (S. Butler, K. Costello, RLG)
Let c :[n] {0,1} be a 2-coloring of [n] and define
jS {x :c(x) j, 1 x n}, j 0, 1,
0 1V(S ,S ) # of monochromatic 3-AP’s formed
j
2πisxj
s S
f (x) e .
We believe the upper bound is the “truth”, i.e., any
2-coloring of [n] must have at least 21172192
(1 o(1)) n
monochromatic 3-AP’s.
Sketch of our approach (S. Butler, K. Costello, RLG)
Let c :[n] {0,1} be a 2-coloring of [n] and define
jS {x :c(x) j, 1 x n}, j 0, 1,
0 1V(S ,S ) # of monochromatic 3-AP’s formed
j
2πisxj
s S
f (x) e .
Thus,
1 2 2
0 1 0 0 1 102V(S ,S ) (f (x)f (2x) f (x)f (2x))dx
2 20 0 1 1f (x)f (2x) f (x)f (2x)
20 1 0 1
0 1 1 0 0 1
0 1 0 1
(f (x) f (x)) (f (2x) f (2x))
(f (x)f (2x) f (x)f (2x))(f (x) f (x))
f (x)f (x)(f (2x) f (2x))
Expanding the integrand, we get
2 20 0 1 1f (x)f (2x) f (x)f (2x)
20 1 0 1
0 1 1 0 0 1
0 1 0 1
(f (x) f (x)) (f (2x) f (2x))
(f (x)f (2x) f (x)f (2x))(f (x) f (x))
f (x)f (x)(f (2x) f (2x))
Expanding the integrand, we get
We can interpret these three terms respectively as:
3
0 1 1 0
0 1
| {(a,b,c) [n] : a b 2c}|
| {(a,b) (S S ) (S S ) :2b a [n]}|
| {(a,b) S S : a b is even}| .
3
0 1 1 0
0 1
| {(a,b,c) [n] : a b 2c}|
| {(a,b) (S S ) (S S ) :2b a [n]}|
| {(a,b) S S : a b is even}| .
The first term is and the third term can be
shown to be bounded above by
2
2n(1 o(1)) ,
2 2nodd even odd even2
(0 0 ) (0 0 )
0 12V(S ,S )
where and denote the number of odd and even numbers,
respectively, in [n] which have color 0.
odd0even0
3
0 1 1 0
0 1
| {(a,b,c) [n] : a b 2c}|
| {(a,b) (S S ) (S S ) :2b a [n]}|
| {(a,b) S S : a b is even}| .
The first term is and the third term can be
shown to be bounded above by
2
2n(1 o(1)) ,
2 2nodd even odd even2
(0 0 ) (0 0 )
If we assume that then this term isodd even0 0 2
8n(1 o(1)) .
0 12V(S ,S )
where and denote the number of odd and even numbers,
respectively, in [n] which have color 0.
odd0even0
Thus, to minimize we should maximize the size of
0 1V(S ,S ),
0 1 1 0T {(a,b) (S S ) (S S ) :2b a [n] }.
Thus, to minimize we should maximize the size of
0 1V(S ,S ),
0 1 1 0T {(a,b) (S S ) (S S ) :2b a [n] }.
Let us color the points (a,b) in [n] by2
red if 0 1 1 0(a,b) {(S S ) (S S )}
and
blue if 0 0 1 1(a,b) {(S S ) (S S ) }
Thus, to minimize we should maximize the size of
0 1V(S ,S ),
0 1 1 0T {(a,b) (S S ) (S S ) :2b a [n] }.
Let us color the points (a,b) in [n] by2
red if 0 1 1 0(a,b) {(S S ) (S S )}
and
blue if 0 0 1 1(a,b) {(S S ) (S S ) }
If and consist of blocks then this coloring
is a red – blue “checkerboard” coloring of [n] .0S 1S
2
2[n]
A red/blue “checkerboard”
(a,b)
The requirement in T that means that we can
restrict our considerations to the parallelogram P determined
by the lines
2b a [n]
1 12 2
y x and y (1 x).
P
2[n]0S
1S
2[n ] P
The intersection of the checkerboard with the parallelogram P
“Incremental improvement” algorithms produced the following pictures.
x y 2z random start
x y 2z random start
x y 2z another random start
x y 2z all blue start
Converging to a common block structure !
(0,0) (1,0)
(0,1) (1,1)
The “checkerboard” picture for the approximate block sizes obtained by the incremental algorithm (normalized by dividing by n ).
2
(0,0) (1,0)
(0,1) (1,1)
(1, 1/2)
(0,1/2)
Restricting the checkerboard to the relevant parallelogram P.)
kx
kx………
……
Local optimization of the red/blue blocks
kx
kx………
……
Increasing the value of by kx ε
kx
kx
= lost
………
……
kx
kx
= lost
= gained
………
……
*kx
= lost
= gained
Δ = 0 at local minimum………
……
*kx
(0,0) (1,0)
(0,1) (1,1)
We apply the “local minimum” condition to this picture.
This leads to the following set of linear equations:
This leads to the following set of linear equations:
The correspond to the interval widths (normalized so that they sum to 1).
iα
The remaining equations arise from forcing equal amounts of red and blue
along each line.
This leads to the following set of linear equations:
The correspond to the interval widths (normalized so that they sum to 1).
iα
The remaining equations arise from forcing equal amounts of red and blue
along each line. This results in the values etc.,
6 281 2548 548
α ,α ,...,
which gives the current best solution for x + y = 2z, namely there are
2-colorings which have only 21172192(1 o(1)) n monochromatic 3-AP’s.
x 2y 3z all blue start
Another equation
x 2y 3z random start
Two runs of x + y = z
(random starts)
Two runs of x + y = z
(random starts)
411
611
111
Known optimal coloring for x + y = z.
We next show the best currently known patterns for minimizing monochromatic
solutions to certain equations Ax + By = Cz for 2-colorings of [n], where A + B = C.
We next show the best currently known patterns for minimizing monochromatic
solutions to certain equations Ax + By = Cz for 2-colorings of [n], where A + B = C.
These were found by first doing a local perturbation of a coloring until it went to
a locally minimal pattern, in the sense that changing no single block gives any
improvement to the overall structure.
We next show the best currently known patterns for minimizing monochromatic
solutions to certain equations Ax + By = Cz for 2-colorings of [n], where A + B = C.
These were found by first doing a local perturbation of a coloring until it went to
a locally minimal pattern, in the sense that changing no single block gives any
improvement to the overall structure. We then took the resulting (approximate)
structure and optimized so that on each line in the pattern intersected equal
amounts of each color.
We next show the best currently known patterns for minimizing monochromatic
solutions to certain equations Ax + By = Cz for 2-colorings of [n], where A + B = C.
These were found by first doing a local perturbation of a coloring until it went to
a locally minimal pattern, in the sense that changing no single block gives any
improvement to the overall structure. We then took the resulting (approximate)
structure and optimized so that on each line in the pattern intersected equal
amounts of each color.
In the following figures we indicate the coefficient of n corresponding to
random ( ), the coefficient of the current best known configuration ( )
and we show what fraction of random we are currently achieving ( ).
We also show the coloring of the current best known configuration.
αγαγ
2
We collect information on all these patterns in the plot below. On the bottom axis is A/C (which will run between 0 and 1, with 1 corresponding to x + y = 2z) and on the vertical axis is (which again runs from 0 to 1).
αγ
With this “normalization” we would expect that the curve formed from all points would be continuous, i.e., a small change in A/C would result in a small change in since the same pattern should be a good approximation for both.
αγ
A/C
We collect information on all these patterns in the plot below. On the bottom axis is A/C (which will run between 0 and 1, with 1 corresponding to x + y = 2z) and on the vertical axis is (which again runs from 0 to 1).
αγ
With this “normalization” we would expect that the curve formed from all points would be continuous, i.e., a small change in A/C would result in a small change in since the same pattern should be a good approximation for both.
αγ
However, there are some gaps!. The most interesting (as of now) are the ones around 2/5 and 3/5, corresponding to the equation 2x + 3y = 5z (in blue). Something seems to be happening in that region, e.g., we haven’t been able to identify any stable behavior for the equation 7x + 10y = 17z.
A/C
We collect information on all these patterns in the plot below. On the bottom axis is A/C (which will run between 0 and 1, with 1 corresponding to x + y = 2z) and on the vertical axis is (which again runs from 0 to 1).
αγ
With this “normalization” we would expect that the curve formed from all points would be continuous, i.e., a small change in A/C would result in a small change in since the same pattern should be a good approximation for both.
αγ
However, there are some gaps!. The most interesting (as of now) are the ones around 2/5 and 3/5, corresponding to the equation 2x + 3y = 5z (in blue). Something seems to be happening in that region, e.g., we haven’t been able to identify any stable behavior for the equation 7x + 10y = 17z.
What happens as A/C 0 ? Does curve stay below 0.95 or does it tend to 1 ?
A/C
We have primarily just examined equations of the form
Ax + By = Cz with A + B = C. These are all density regular
equations and, in particular, solutions are invariant under
translation.
We have primarily just examined equations of the form
Ax + By = Cz with A + B = C. These are all density regular
equations and, in particular, solutions are invariant under
translation.
In every case except one, the best colorings we found are composed
of a finite number of anti-symmetrically colored blocks, such as
28 6 28 37 59 2828 63759116 116
for x + y = 2z.
We have primarily just examined equations of the form
Ax + By = Cz with A + B = C. These are all density regular
equations and, in particular, solutions are invariant under
translation.
In every case except one, the best colorings we found are composed
of a finite number of anti-symmetrically colored blocks, such as
28 6 28 37 59 2828 63759116 116
for x + y = 2z.
The one exceptional equation we found is 2x + 3y = 5z.
Here is a picture of what the heuristic algorithm always produced.
2x 3y 5z all blue start
This picture also consists of “blocks”, but this time the blocks
are formed by alternating blue/red (and red/blue) strings.Performing a similar local optimizing procedure on these blocks
gives a value of = 0.923755…, which is unexpectedly low.
αγ
This picture also consists of “blocks”, but this time the blocks
are formed by alternating blue/red (and red/blue) strings.Performing a similar local optimizing procedure on these blocks
gives a value of = 0.923755…, which is unexpectedly low.
αγ
Presumably, this is the optimal coloring for 2x + 3y = 5z although
we can’t prove it.
blue odd, red even
red odd, blue even
………
This picture also consists of “blocks”, but this time the blocks
are formed by alternating blue/red (and red/blue) strings.Performing a similar local optimizing procedure on these blocks
gives a value of = 0.923755…, which is unexpectedly low.
αγ
Presumably, this is the optimal coloring for 2x + 3y = 5z although
we can’t prove it. What is different about this equation ?
blue odd, red even
red odd, blue even
………
For the equation E: ax + by = cz with a + b = c, random 2-colorings
are never optimal.
In fact, it can be shown that the optimal 2-coloring of [n]
has at most times as many monochromatic solutions
to E as a random one does.
21
2bc1
For example, for 3-AP’s, (i.e., E: x + y = 2z), this ratio is 78
0.875
while the truth appears to be 0.854… .
Are random colorings ever optimal ?
Yes !Yes !
Are random colorings ever optimal ?
Yes !Yes !
Let E denote the equation x + y = z + w.
Are random colorings ever optimal ?
Yes !Yes !
Let E denote the equation x + y = z + w.
If we count permutations of solutions as different then
it easy to compute that E has solutions in [n]. 3 22n O(n )
3
Are random colorings ever optimal ?
Yes !Yes !
Let E denote the equation x + y = z + w.
If we count permutations of solutions as different then
it easy to compute that E has solutions in [n]. 3 22n O(n )
3
Thus, in a random 2-coloring of [n] we would expect
to find monochromatic solutions.
33 2 22n1 O(n ) n12
O(n )8 3
Are random colorings ever optimal ?
Yes !Yes !
Let E denote the equation x + y = z + w.
If we count permutations of solutions as different then
it easy to compute that E has solutions in [n]. 3 22n O(n )
3
Thus, in a random 2-coloring of [n] we would expect
to find monochromatic solutions.
33 2 22n1 O(n ) n12
O(n )8 3
In fact, anyany 2-coloring of [n] must produce
at least this many monochromatic solutions.
Sketch of proof:
Let {red, blue} be a 2-coloring of [n] = {1,2,…,n}.
c :[n]
Think of where N > 2n.N[n] Z
Define by f(x) = 1 if x is red, and 0 otherwisef :[n] {0,1}
and by g(x) = 1 if x is blue, and 0 otherwise.
g:[n] {0,1}
Sketch of proof:
Let {red, blue} be a 2-coloring of [n] = {1,2,…,n}.
c :[n]
Think of where N > 2n.N[n] Z
Define by f(x) = 1 if x is red, and 0 otherwisef :[n] {0,1}
and by g(x) = 1 if x is blue, and 0 otherwise.
g:[n] {0,1}
Thus, [n]f g 1
The number of monochromatic solutions to x + y = z + w
is given by the expression
N
2 2
x
C: ((ff )(x) (g g)(x) )
Z
where denotes convolution.
With2πijk 2πi(N 1)k
N N
N1 0 k N,ke (1,...,e ,...,e ),
as an orthonormal basis for our Fourier transform , we have
F G F G N ||F|| ||F|| andN
2πixξN1
Nx
F(ξ) F(x)e
Z
With2πijk 2πi(N 1)k
N N
N1 0 k N,ke (1,...,e ,...,e ),
as an orthonormal basis for our Fourier transform , we have
F G F G N ||F|| ||F|| and
Thus,
2 2|| ff || || g g|| N
2 2
x
C: ((ff )(x) (g g)(x) )
Z
N
2πixξN1
Nx
F(ξ) F(x)e
Z
2 2|| ff || || g g|| 2 2 2 2( || f || || f || || g || || g|| ) N
With2πijk 2πi(N 1)k
N N
N1 0 k N,ke (1,...,e ,...,e ),
as an orthonormal basis for our Fourier transform , we have
F G F G N ||F|| ||F|| and
Thus,
2 2|| ff || || g g|| N
2 2
x
C: ((ff )(x) (g g)(x) )
Z
2 2|| ff || || g g|| 2 2 2 2( || f || || f || || g || || g|| ) N
4 4( || f || || g || ) N
N
4 4
ξ
(| f (ξ)| | g(ξ)| ) N
Z
N
2πixξN1
Nx
F(ξ) F(x)e
Z
Continuing,
N
4 4
x
(| f (x)| | g(x)| ) N
Z
C
N
4
x
18
(| f (x)| | g(x)|) N
Z
(by C-S twice)
Continuing,
N
4 4
x
(| f (x)| | g(x)| ) N
Z
C
N
4
x
18
(| f (x)| | g(x)|) N
Z
(by C-S twice)
N
4
x
18
(| f (x) g(x)|) N
Z
N
4
x
18
(| (f g)(x)|) N
Z
Continuing,
N
4 4
x
(| f (x)| | g(x)| ) N
Z
C
N
4
x
18
(| f (x)| | g(x)|) N
Z
(by C-S twice)
N
4
x
18
(| f (x) g(x)|) N
Z
N
4
x
18
(| (f g)(x)|) N
Z
218
|| (f g) (f g)||
2[n] [n]
18
||1 1 ||
3 32 22n n18 3 12
O(n ) O(n )
and the claim is proved.
In fact, none of the colorings we have produced for any ofthe equations ax + by = cz considered have been proved to be optimal (except for x + y = z).
What happens for other equations ?
In fact, none of the colorings we have produced for any ofthe equations ax + by = cz considered have been proved to be optimal (except for x + y = z).
What happens for other equations ?
In fact, none of the colorings we have produced for any ofthe equations ax + by = cz considered have been proved to be optimal (except for x + y = z).
Computational evidence strongly suggests that for the equation
E: x + ay = z, with \{1}, we havea N\
2
2E
1 o(1)2a(a 2a 3)
f (n) n .
What happens for other equations ?
In fact, none of the colorings we have produced for any ofthe equations ax + by = cz considered have been proved to be optimal (except for x + y = z).
Computational evidence strongly suggests that for the equation
E: x + ay = z, with \{1}, we havea N\
2
2E
1 o(1)2a(a 2a 3)
f (n) n .
What are the corresponding results for 4-AP’s4-AP’s (or k-AP’s) ??
A good coloring for avoiding monochromatic 4-AP’s
In this case, the number of monochromatic 4-AP’s
is at most where
24c n O(n)
A good coloring for avoiding monochromatic 4-AP’s
In this case, the number of monochromatic 4-AP’s
is at most where
24c n O(n)
c = 4
A good coloring for avoiding monochromatic 4-AP’s
In this case, the number of monochromatic 4-AP’s
is at most where
24c n O(n)
c =
Is this the “truth”?
4
A good coloring for avoiding monochromatic 4-AP’s
In this case, the number of monochromatic 4-AP’s
is at most where
24c n O(n)
c =
Is this the “truth”?
4
Sensational result (Ben Green and Terry Tao – 2003):
There are arbitrarily long arithmetic progressions of primes.
Finding k-AP’s in sparse sets.
Sensational result (Ben Green and Terry Tao – 2003):
There are arbitrarily long arithmetic progressions of primes.
Ben Green – Cambridge Univ.
Terry Tao - UCLA
Finding k-AP’s in sparse sets.
In fact, they show that there exists a
k-term arithmetic progression of primes
222
222
2100 k
which are all less than
In fact, they show that there exists a
k-term arithmetic progression of primes
222
222
2100 k
(This is probably not best possible !)
which are all less than
An even tougher challenge.
Show that there exist arbitrarily long AP’s of consecutive primes.
The current record is 10, given be the following numbers:
k 0, 1,...,9,P 210 k,
An even tougher challenge.
Show that there exist arbitrarily long AP’s of consecutive primes.
The current record is 10, given be the following numbers:
k 0, 1,...,9,P 210 k,
where
P = 100 99697 24697 14247 63778 66555 87969 84032 95093 24689 19004 18036 03417 75890 43417 03348 88215 90672
29719(found by M. Toplic in 1998)
An even tougher challenge.
Show that there exist arbitrarily long AP’s of consecutive primes.
The current record is 10, given be the following numbers:
k 0, 1,...,9,P 210 k,
where
P = 100 99697 24697 14247 63778 66555 87969 84032 95093 24689 19004 18036 03417 75890 43417 03348 88215 90672
29719(found by M. Toplic in 1998)
Will we ever find 11 ??
Schur’s theorem (finite version):
For all r, there is a least Sc(r) such that any r-coloring
of [Sc(r)] contains a monochromatic solution of x y z.
Some open Some open problemsproblems
Schur’s theorem (finite version):
For all r, there is a least Sc(r) such that any r-coloring
of [Sc(r)] contains a monochromatic solution of x y z.
It is known that 1 2rc π Sc(r) c r !
Some open Some open problemsproblems
Schur’s theorem (finite version):
For all r, there is a least Sc(r) such that any r-coloring
of [Sc(r)] contains a monochromatic solution of x y z.
It is known that 1 2rc π Sc(r) c r !
(Old) Problem. Is
1/ rSc(r)lim ?r r
Some open Some open problemsproblems
Schur’s theorem (finite version):
For all r, there is a least Sc(r) such that any r-coloring
of [Sc(r)] contains a monochromatic solution of x y z.
It is known that 1 2rc π Sc(r) c r !
(Old) Problem. Is
1/ rSc(r)lim ?r r
$100
Some open Some open problemsproblems
Suppose we consider the “off-diagonal” van der Waerden number W(k, l)
defined as the least integer such that if [W(k, l)] is 2-colored red and blue,
then there either is a red k-AP or a blue l-AP.
k 3 4 5 6 7 8 9 10 11 12 13
W(k,3) 9 18 22 32 46 58 77 97 114 135 160
Suppose we consider the “off-diagonal” van der Waerden number W(k, l)
defined as the least integer such that if [W(k, l)] is 2-colored red and blue,
then there either is a red k-AP or a blue l-AP.
k 3 4 5 6 7 8 9 10 11 12 13
W(k,3) 9 18 22 32 46 58 77 97 114 135 160
k
W(k,3)
Suppose we consider the “off-diagonal” van der Waerden number W(k, l)
defined as the least integer such that if [W(k, l)] is 2-colored red and blue,
then there either is a red k-AP or a blue l-AP.
k 3 4 5 6 7 8 9 10 11 12 13
W(k,3) 9 18 22 32 46 58 77 97 114 135 160
k
W(k,3)
Suppose we consider the “off-diagonal” van der Waerden number W(k, l)
defined as the least integer such that if [W(k, l)] is 2-colored red and blue,
then there either is a red k-AP or a blue l-AP.
Any conjectures ?Any conjectures ?
k 3 4 5 6 7 8 9 10 11 12 13
W(k,3) 9 18 22 32 46 58 77 97 114 135 160
k
W(k,3) Any conjectures ?Any conjectures ?
Only known that c' logkck W(k,3) k
Suppose we consider the “off-diagonal” van der Waerden number W(k, l)
defined as the least integer such that if [W(k, l)] is 2-colored red and blue,
then there either is a red k-AP or a blue l-AP.
We can prove the following related result.Consider the modified off-diagonal van der Waerden number W(k, l)
defined as the least integer such that if [W(k, l)] is 2-colored red and blue,
then there is either a red consecutive block of length k or a blue l-AP.
We can prove the following related result.
Theorem (RLG – 2006)
For a suitable constant c > 0,
Consider the modified off-diagonal van der Waerden number W(k, l)
defined as the least integer such that if [W(k, l)] is 2-colored red and blue,
then there is either a red consecutive block of length k or a blue l-AP.
clogkW(k,3) k .
We can prove the following related result.
Theorem (RLG – 2006)
For a suitable constant c > 0,
Consider the modified off-diagonal van der Waerden number W(k, l)
defined as the least integer such that if [W(k, l)] is 2-colored red and blue,
then there is either a red consecutive block of length k or a blue l-AP.
clogkW(k,3) k .
Unfortunately, the best upper bound we have is32c'kW(k,3) k .
We can prove the following related result.
Theorem (RLG – 2006)
For a suitable constant c > 0,
Consider the modified off-diagonal van der Waerden number W(k, l)
defined as the least integer such that if [W(k, l)] is 2-colored red and blue,
then there is either a red consecutive block of length k or a blue l-AP.
clogkW(k,3) k .
Which bound is closer to the truth?
Unfortunately, the best upper bound we have is32c'kW(k,3) k .
Restricted van der Waerden
It was shown by Spencer (1975) that the following restricted
version of van der Waerden’s theorem holds.
For every k, there exists a finite set S(k) with the properties:
(i) any 2-coloring always produces a monochromatic k-AP;
(ii) S(k) does not contain a (k + 1)-AP.
Restricted van der Waerden
It was shown by Spencer (1975) that the following restricted
version of van der Waerden’s theorem holds.
For every k, there exists a finite set S(k) with the properties:
(i) any 2-coloring always produces a monochromatic k-AP;
(ii) S(k) does not contain a (k + 1)-AP.
We have no idea of the sizes s(k) = |S(k)| of the smallest
sets S(k) which satisfy the above conditions.
Restricted van der Waerden
It was shown by Spencer (1975) that the following restricted
version of van der Waerden’s theorem holds.
For every k, there exists a finite set S(k) with the properties:
(i) any 2-coloring always produces a monochromatic k-AP;
(ii) S(k) does not contain a (k + 1)-AP.
We have no idea of the sizes s(k) = |S(k)| of the smallest
sets S(k) which satisfy the above conditions.
For example, is it true that s(k)W(k)
as k ?
Recall that an equation is r-regular if for any partition
of into r color classes, there is a monochromatic solution to this equation.
1 2 n) 0E(x ,x ,...,x
Conjecture: (Rado – 1933)
For each n, there is an M = M(n) so that if the linear
homogeneous equation is M-regular,
then in fact the equation is regular.
1 2 n) 0E(x ,x ,...,x
Recall that an equation is r-regular if for any partition
of into r color classes, there is a monochromatic solution to this equation.
1 2 n) 0E(x ,x ,...,x
Conjecture: (Rado – 1933)
For each n, there is an M = M(n) so that if the linear
homogeneous equation is M-regular,
then in fact the equation is regular.
1 2 n) 0E(x ,x ,...,x
The conjecture holds for n = 1 and n = 2. (Rado – 1933)
Recall that an equation is r-regular if for any partition
of into r color classes, there is a monochromatic solution to this equation.
1 2 n) 0E(x ,x ,...,x
Conjecture: (Rado – 1933)
For each n, there is an M = M(n) so that if the linear
homogeneous equation is M-regular,
then in fact the equation is regular.
1 2 n) 0E(x ,x ,...,x
The conjecture holds for n = 1 and n = 2. (Rado – 1933)
The conjecture also holds for n = 3 with M(3) = 24. (Fox and Kleitman – 2005)
Recall that an equation is r-regular if for any partition
of into r color classes, there is a monochromatic solution to this equation.
1 2 n) 0E(x ,x ,...,x
Conjecture: (Rado – 1933)
For each n, there is an M = M(n) so that if the linear
homogeneous equation is M-regular,
then in fact the equation is regular.
1 2 n) 0E(x ,x ,...,x
The conjecture holds for n = 1 and n = 2. (Rado – 1933)
The conjecture also holds for n = 3 with M(3) = 24. (Fox and Kleitman – 2005)
ChallengeChallenge: Prove that M(n) always exists, and determine (or estimate) its value.
Let W*(k) denote the least integer such that there is a set S(k) of integers
of size W*(k) such that any 2-coloring always produces a monochromatic k-AP.
Thus, W * (k) W(k).
Let W*(k) denote the least integer such that there is a set S(k) of integers
of size W*(k) such that any 2-coloring always produces a monochromatic k-AP.
Thus, W * (k) W(k).
It is known that W*(3) = 9 = W(3) but that 27W * (4) 35 W(4).
The characteristic function of one such S(4) is:
1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 0 1
Let W*(k) denote the least integer such that there is a set S(k) of integers
of size W*(k) such that any 2-coloring always produces a monochromatic k-AP.
Thus, W * (k) W(k).
It is known that W*(3) = 9 = W(3) but that 27W * (4) 35 W(4).
Does W(k) W * (k) as k ?
The characteristic function of one such S(4) is:
1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 0 1
Is this the correct value of W*(4)?
Let W*(k) denote the least integer such that there is a set S(k) of integers
of size W*(k) such that any 2-coloring always produces a monochromatic k-AP.
Thus, W * (k) W(k).
It is known that W*(3) = 9 = W(3) but that 27W * (4) 35 W(4).
Does W(k) W * (k) as k ?
The characteristic function of one such S(4) is:
1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 0 1
Is boundedW(k)W*(k)
as k ?
Is this the correct value of W*(4)?
Let W*(k) denote the least integer such that there is a set S(k) of integers
of size W*(k) such that any 2-coloring always produces a monochromatic k-AP.
Thus, W * (k) W(k).
It is known that W*(3) = 9 = W(3) but that 27W * (4) 35 W(4).
Does W(k) W * (k) as k ?
The characteristic function of one such S(4) is:
1 0 0 1 0 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 0 1
Is boundedW(k)W*(k)
as k ?
What happens with r colors?
Is this the correct value of W*(4)?
What about the regularity of nonlinear equations ?
For example, it is known that the equation
k1 2
1 2 k
aa a
x x x0...
is regular if and only if the corresponding equation
1 1 2 2 k kax a x ... a x 0
is regular.
What about the regularity of nonlinear equations ?
For example, it is known that the equation
k1 2
1 2 k
aa a
x x x0...
is regular if and only if the corresponding equation
1 1 2 2 k kax a x ... a x 0
is regular.
Beautiful old Beautiful old questionquestion
Is regular ? 2 2 2x y z
What about the regularity of nonlinear equations ?
For example, it is known that the equation
k1 2
1 2 k
aa a
x x x0...
is regular if and only if the corresponding equation
1 1 2 2 k kax a x ... a x 0
is regular.
Beautiful old Beautiful old questionquestion
Is regular ? 2 2 2x y z$100
Consider the linear equation L:
x + y + z = 4w
It is easy to 4-color so that L has no monochromatic solution
(for example, define ). a(5 (5b c)) c
An unexpected twistAn unexpected twist
Consider the linear equation L:
x + y + z = 4w
It is easy to 4-color so that L has no monochromatic solution
(for example, define ). a(5 (5b c)) c
Must every 4-coloring of produce a monochromatic solution of L ?
R
An unexpected twistAn unexpected twist
Consider the linear equation L:
x + y + z = 4w
It is easy to 4-color so that L has no monochromatic solution
(for example, define ). a(5 (5b c)) c
Must every 4-coloring of produce a monochromatic solution of L ?
R
Well, it depends !
An unexpected twistAn unexpected twist
Consider the linear equation L:
x + y + z = 4w
It is easy to 4-color so that L has no monochromatic solution
(for example, define ). a(5 (5b c)) c
Must every 4-coloring of produce a monochromatic solution of L ?
R
Well, it depends ! The answer is no in ZFC.
An unexpected twistAn unexpected twist
Consider the linear equation L:
x + y + z = 4w
It is easy to 4-color so that L has no monochromatic solution
(for example, define ). a(5 (5b c)) c
Must every 4-coloring of produce a monochromatic solution of L ?
R
Well, it depends ! The answer is no in ZFC.
(LM is the axiom that every set of reals is Lebesgue measurable)
However, the answer is yes in ZF + LM.
Radoičič, Fox, Alexeev, RLG - 2007
An unexpected twistAn unexpected twist
Consider the linear equation L:
x + y + z = 4w
It is easy to 4-color so that L has no monochromatic solution
(for example, define ). a(5 (5b c)) c
Must every 4-coloring of produce a monochromatic solution of L ?
R
Well, it depends ! The answer is no in ZFC.
(LM is the axiom that every set of reals is Lebesgue measurable)
A theorem of Solovay asserts that ZFC and ZF + LM are equally consistent.
However, the answer is yes in ZF + LM.
Radoičič, Fox, Alexeev, RLG - 2007
An unexpected twistAn unexpected twist
Consider the linear equation L:
x + y + z = 4w
It is easy to 4-color so that L has no monochromatic solution
(for example, define ). a(5 (5b c)) c
Must every 4-coloring of produce a monochromatic solution of L ?
R
Well, it depends ! The answer is no in ZFC.
(LM is the axiom that every set of reals is Lebesgue measurable)
A theorem of Solovay asserts that ZFC and ZF + LM are equally consistent.
However, the answer is yes in ZF + LM.
Radoičič, Fox, Alexeev, RLG - 2007
For which (sets of) equations do we have this distinction ?
An unexpected twistAn unexpected twist
A classic conjecture of Erdös
If then X contains k-AP’s for every k.N 1x
x X
X with
$3000
A classic conjecture of Erdös
If then X contains k-AP’s for every k.N 1x
x X
X with
$3000
Warm-up problem. Prove this holds for k = 3.
If then X contains 4 vertices of a square.
A classic conjecture of Erdös
If then X contains k-AP’s for every k.N 1x
x X
X with
$3000
Warm-up problem. Prove this holds for k = 3.
(Two-dimensional analogue - RLG).
N N 2 21
x y(x,y) X
X with
If then X contains 4 vertices of a square.
A classic conjecture of Erdös
If then X contains k-AP’s for every k.N 1x
x X
X with
$3000
Warm-up problem. Prove this holds for k = 3.
(Two-dimensional analogue - RLG).
N N 2 21
x y(x,y) X
X with
$1000
Paul is surprised when finally gets to read The Book
“Very creative. Very imaginative. Logic……that’s what’s missing.”
Front Row: Mrs. van Lint, Robert Rankin, Mrs. Rankin, Gina Birch, J. H. van Lint, L. J. Mordell, Mrs. Mordell, Paul Erdös (with Cheryl (" Che' ") Graham - Ron Graham's daughter), Mrs. Andrews.
Second Row: Mrs. J.B. Muskat (hidden), Mrs. T.R. Parkin (hidden), George Andrews, Stefan Burr, Mrs. S. Burr, Mrs. Pat Nickson, Francis Coghlan, Gérard Ligozat, John Tate, Bryan Birch, A.O.L. Atkin, Helmut Hasse, Marshall Hall Jnr.
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Fifth Row: Ian Anderson, Marie-Nicole Montouchet, M. G. Gras, A. S. Fraenkel, Olga Taussky-Todd, Synolda Butler, ø. Rødseth, Susan Cohn, Debra Baumert, Anthony Cohn, Jean Pierre Serre, John Leech.
Back Row: George Greaves, Lyliane Bouvier, Jean Fresnel, Marvin Wunderlich, Horst Zimmer, Christiane Martinet, Jacques Martinet, John Todd, Albrecht Pfister, S. Mossige, J. Hjelle, John Brillhart.
Front Row: Mrs. Andrews, Mrs. H. Fredricksen, Mrs. K. Kloos, Mrs. Cassels, Jane Pitman, Mrs. J. Merriman, Felice Bateman, Sally Bateman, Harriet Cantor.
Second Row: E. S. Selmer, J. C. P. Miller, Ron Graham , John Selfridge, John Merriman, H. Tverberg, David Cantor, Andrzej Schinzel, Helen Alderson, J. W. S. Cassels.
Third Row: Michael Keates, Peter Pleasants, Elwyn Berlekamp, Emma Lehmer, D. H. Lehmer, Joseph Muskat, C.E. Fröberg, Robert Spira.
Fourth Row: H.G. Apsimon, Peter Swinnerton-Dyer, Albrecht Fröhlich, A. M. Macbeath, O. Hermann, J.C. Herz, Georges Poitou, John Conway, John Chalk.
Fifth Row: V. Felsch, J. Larmouth, O. Kolberg, Bob Laxton, Peter Weinberger, Roger Cook, Jonathan Hall, T.R. Parkin, J.H.E. Cohn, John Cosgrave.
Back Row: H. J. Godwin, Fred Lunnon, T. Kløve, H. Fredricksen, Jack Howlett, A. L. Dulmage, Nathan Mendelsohn, Leonard Baumert, Hans Riesel, Harold Stark
Frank Plumpton Ramsey(1903-1930)
Arthur Michael Ramsey
100th Archbishop of Canterbury
(1904-1988)
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