Presented by: Ryan ODonnell Carnegie Melloni by D. H. J. Polymath.
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presented by: Ryan O’Donnell Carnegie Melloni by D. H. J. Polymath
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Transcript of Presented by: Ryan ODonnell Carnegie Melloni by D. H. J. Polymath.
presented by: Ryan O’Donnell
Carnegie Melloni
by D. H. J. Polymath
If A ⊆ {0,1,2}n has density Ω(1),
then A contains a “(combinatorial) line”.
DHJ(3):
point: 1 1 2 1 2 1 0 0 0 1 0 1 1 0 1 2 0 1 2 2
line: 1 0 2 1 1 2 1 1 0 0 0 1 0 0 0 1 1 0 1 1
1 0 2 1 1 2 1 1 0 1 1 1 0 0 1 1 1 1 1 1
1 0 2 1 1 2 1 1 0 2 2 1 0 0 2 1 1 2 1 1
{
}
,
,
1 0 2 1 1 2 1 1 0 ✶ ✶ 1 0 0 ✶ 1 1 ✶ 1 1
[Furstenberg-Katznelson’91]: DHJ(k) is true ∀ k.
(k = 3): ∀ δ > 0, ∃ n0(δ) s.t. ∀ n ≥ n0(δ),
A ⊆ {0,1,2}n has density ≥ δ
⇒ A contains a line
proof: Used ergodic theory.
No effective bound on n0(δ).
Importance in combinatorics:
Implies Szemerédi’s Theorem.
1 0 2 1 1 2 1 1 0 0 0 1 0 0 0 1 1 0 1 1
1 0 2 1 1 2 1 1 0 1 1 1 0 0 1 1 1 1 1 1
1 0 2 1 1 2 1 1 0 2 2 1 0 0 2 1 1 2 1 1
Why I was interested:
0 1 2 0
0 , 1 , 2 , 1
0 1 2 2
A ⊆ {0,1,2}n, no trips w/ all cols
0 0 0 1 1
0 , 0 , 1 , 0 , 1
0 1 0 0 1
A ⊆ {0,1}n, no trips w/ all cols
see [O-Wu’09]
[D.H.J. Polymath]:
∀ δ > 0, ∀ n ≥ 2↑↑ O(1/δ3),
A ⊆ {0,1,2}n has density ≥ δ
⇒ A contains a line
proof: Elementary probability/combinatorics.
also the general k case, with Ackermann-type bounds
Proof sketch
DHJ(2): A ⊆ {0,1}n dens. δ ⇒ A contains line
line: 1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 1 1 0 1 1
1 0 1 1 1 1 1 1 0 1 1 1 0 0 1 1 1 1 1 1
{
}
,
1 0 1 1 1 1 1 1 0 ✶ ✶ 1 0 0 ✶ 1 1 ✶ 1 1
contrapositive: If A has no comparable pairs
— i.e., A is an “antichain” —
then A is very sparse.
“Sperner’s Theorem”
DHJ(2): A ⊆ {0,1}n dens. δ ⇒ A contains line
12 345 6 7n
0 0 0 1 1 1 1 1
“equal-slices distribution”
DHJ(2): A ⊆ {0,1}n dens. δ ⇒ A contains line
“equal-slices distribution”
5
0
n
0
2
0
4
1
1
1
3
1
6
1
7
1
1
1
7
1
6
1
3
1
4
1
2
0
n
0
5
0
(each Hamming
wt. equally likely)
DHJ(2): A ⊆ {0,1}n dens. δ ⇒ A contains line
5
0
n
0
2
0
4
1
1
1
3
1
6
1
7
1
1 0 1 1 0 1 1 0
1 2 3 4 5 6 7 n
“equal-slices distribution”
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ ⇒ A contains line
5
0
n
0
2
0
4
1
1
1
3
1
6
1
7
1
1 0 1 1 0 1 1 0
1 2 3 4 5 6 7 n
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ ⇒ A contains line
0 0 0 0 0 0 1 1
5
0
n
0
2
0
4
1
1
1
3
1
6
1
7
1
1 0 1 1 0 1 1 0
1 2 3 4 5 6 7 n
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ ⇒ A contains line
0 0 0 0 0 0 1 10 0 0 0 0 0 1 1
0 0 0 0 0 1 1 0
0 0 0 0 0 0 1 10 0 0 0 0 0 1 1
5
0
n
0
2
0
4
1
1
1
3
1
6
1
7
1
1 0 1 1 0 1 1 0
1 2 3 4 5 6 7 n
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ ⇒ A contains line
0 0 0 0 0 0 1 10 0 0 0 0 0 1 1
0 0 0 0 0 1 1 0
0 0 0 0 0 0 1 1
5
0
n
0
2
0
4
1
1
1
3
1
6
1
7
1
1 0 1 1 0 1 1 0
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ ⇒ A contains line
0 0 0 0 0 0 1 10 0 0 0 0 0 1 1
0 0 0 0 0 1 1 0
0 0 0 0 0 0 1 1
( Pr[degen] = )
“random eq-slices line”
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ
⇒ eq-slice-line in A w.p. ≥ δ2 −
Pr[ x, y ∈ A ]
= Ex y
x
y
∈ A
∈ A
&Pr
= Ex
x ∈ A
2Pr
≥ Ex
x ∈ A
2Pr
= δ2
(independent)
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ
⇒ eq-slice-line in A w.p. ≥ δ2 −
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ
n ≥ 1/δ2 ⇒ A contains a line −
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ
−
Distribution Hackery
1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 1 1 0 1 1unif
unif
rand. coords
unif on
eq-slices on 1 0 0 1
1 0 1 1 1 1 1 0 0 0 1 0 0 1 0 1
[eq-slices(A )]
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ
−
Distribution Hackery
1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 1 1 0 1 1unif
unif 1 0 0 1 1 0 1 1 1 1 1 0 0 0 1 0 0 1 0 1
easy: dTV( unif, unif ) = o(1)unif
unif (A) ≈ δ = E [eq-slices Ay
y
rand. coords
5
0
n
0
2
0
4
1
1
1
3
1
6
1
7
1
1 0 1 1 0 1 1 0
0 0 0 0 0 0 1 10 0 0 0 0 0 1 1
0 0 0 0 0 1 1 0
0 0 0 0 0 0 1 1
1 2 3 4 5 6 7 n
DHJ(2): A ⊆ {0,1}n eq-slices dens. δ
⇒ eq-slice-line in A w.p. ≥ δ2 −
DHJ(3): A ⊆ {0,1,2}n eq-slices dens. δ
⇒ eq-slice-line in A w.p. ≥ δ3 ???
DHJ(3): A ⊆ {0,1,2}n eq-slices dens. δ
⇒ eq-slice-line in A w.p. ≥ δ3 ???
Pr[ x, y ∈ A ]
= Ex y
x
y
∈ A
∈ A
&Pr
= Ex y
x ∈ A
2Pr
≥ Ex y
x ∈ A
2Pr
= δ2
(independent)
5
0
n
0
2
0
4
1
1
1
3
1
6
1
7
1
1 0 1 1 0 1 1 00 0 0 0 0 1 1 0
DHJ(3): A ⊆ {0,1,2}n eq-slices dens. δ
?
0 0 0 0 0 0 1 1
1 2 3 4 5 6 7 n
5
0
n
0
2
0
4
1
1
1
3
1
6
1
7
1
1 0 1 1 0 1 1 0
0 0 0 0 0 0 1 1
0 0 0 0 0 1 1 0
1 2 3 4 5 6 7 n
DHJ(3): A ⊆ {0,1,2}n eq-slices dens. δ
0 0 0 2 2 2 1 1
2 0 2 2 0 1 1 0
eq-slices pt in {0,1,2}n ≡ eq-slices line over {0,1}n
if in A as a line,
& in A as a point,
done!
DHJ(3): A ⊆ {0,1,2}n′ eq-slices dens. δ
by distrib. hackery: A ⊆ {0,1}n′ eq-slices dens. δ also
{0,1,2}n
{0,1}n
A
A
DHJ(3): A ⊆ {0,1,2}n eq-slices dens. δ
by distrib. hackery: A ⊆ {0,1}n eq-slices dens. δ also
{0,1,2}n
{0,1}n
A
A L
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
≥ δ2
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}n
{0,1}n
A
A L ≥ δ2
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}n
{0,1}n
A
A L ≥ δ2
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}n
{0,1}n
A
A L ≥ δ2
if L ⋂ A ≠ ∅, done!
≥ δ
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}n
{0,1}n
A
A L ≥ δ2
≥ δ
idea: A has dens. ≥ in
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}n
idea: A has dens. ≥ in
A
≥ δ + δ3
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}n
A
≥ δ + δ3
Suppose, somehow, that
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}log log log n
A
≥ δ + δ3
Suppose, somehow, that
Suppose, somehow, that
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}log log log n
A
≥ δ + δ3
Suppose, somehow, that
Repeat.
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}log log log n
A
≥ δ + δ3
Suppose, somehow, that
Repeat.
≥ δ2
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}log log log n
A
≥ δ + δ3
Suppose, somehow, that
Repeat.
≥ δ2
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}log log log log log log n
A
≥ δ + 2δ3
Suppose, somehow, that
Repeat.
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}log log log log log log log log log n
A
≥ δ + 3δ3
Suppose, somehow, that
Repeat.
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}log (3/δ3)n
A
> 2/3
Suppose, somehow, that
Repeat.
A contains line trivially.
“Density
Increment
Argument”
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
{0,1,2}n
A
≥ δ + δ3
Suppose, somehow, that
L = { x∈{0,1,2}n : x2→1, x2→0 ∈ A }
Suppose, somehow, that
L = { x : x2→1 ∉ A } ⋃ { x : x2→0 ∉ A }
= { x : x2→1 ∉ A } (harmless cheat)pretend
idea: is a 12-insensitive set 12-insensitive set
= { x : x2→1 ∉ A }
= { x : x2→1 ∉ A }
12-insensitive set
12-insensitive set
= { x : x2→1 ∉ A }
12-insensitive set
monkeying with 1’s and 2’s
does not affect
presence in the set
Closer to an isomorphic copy of
key thm: A dense 12-insensitive set can be
almost completely partitioned
into copies of
key thm: A dense 12-insensitive set can be
almost completely partitioned
into copies of
A has dens. ≥ δ + δ3 inside
⇒ A has dens. ≳ δ + δ3 inside some
key thm: A dense 12-insensitive set can be
almost completely partitioned
into copies of
2 1 1 0 ✶1 ✶1 1 0 1 0 ✶2 1 1 ✶2 1 2 ✶2 ✶2 ✶1 1 1 2
lemma 1: Dense 12-insensitive set contains
a copy of
key thm: A dense 12-insensitive set can be
almost completely partitioned
into copies of
lemma 1: Dense 12-insensitive set contains
a copy of
lemma 2: Dense 12-insensitive set contains
a copy of {0, 1, 2} — i.e., a line
lemma 2: Dense 12-insensitive set contains
a copy of {0, 1, 2} — i.e., a line
lemma 2: Dense 12-insensitive set contains
a copy of {0, 1, 2} — i.e., a line
lemma 2: Dense 12-insensitive set contains
a copy of {0, 1, 2} — i.e., a line
(i) Distrib. hack: make it dense in {0,1}n also.
(ii) Apply DHJ(2).
1 0 1 1 1 1 1 1 0 0 0 1 0 0 0 1 1 0 1 1 ∈ A
1 0 1 1 1 1 1 1 0 1 1 1 0 0 1 1 1 1 1 1 ∈ A
1 0 1 1 1 1 1 1 0 2 2 1 0 0 2 1 1 2 1 1 ∈ A
(iii) … by 12-insensitivity.
Questions?
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