maksim zhukovskii – zero-one k-laws for g(n,n−α)

Zero-one k-laws for G(n, n−α)

Maksim Zhukovskii

MSU, MIPT, Yandex

Workshop on Extremal Graph Theory6 June 2014

Maksim Zhukovskii Zero-one k-laws for G(n, n−α)

First-order properties.

First-order formulae:relational symbols ∼, =;logical connectivities ¬,⇒,⇔,∨,∧;variables x, y, x1, ...;quanti�ers ∀,∃.

• property of a graph to be complete

∀x ∀y (¬(x = y) ⇒ (x ∼ y)).

• property of a graph to contain a triangle

∃x ∃y ∃z ((x ∼ y) ∧ (y ∼ z) ∧ (x ∼ z)).

• property of a graph to have chromatic number k

∀x ∀y (¬(x = y) ⇒ (x ∼ y)).

∃x ∃y ∃z ((x ∼ y) ∧ (y ∼ z) ∧ (x ∼ z)).

∀x ∀y (¬(x = y) ⇒ (x ∼ y)).

∃x ∃y ∃z ((x ∼ y) ∧ (y ∼ z) ∧ (x ∼ z)).

∀x ∀y (¬(x = y) ⇒ (x ∼ y)).

∃x ∃y ∃z ((x ∼ y) ∧ (y ∼ z) ∧ (x ∼ z)).

Limit probabilities.

G(n, p): n � number of vertices, p � edge appearanceprobability

G � an arbitrary graph, LG � property of containing G.LG is the �rst order property.

maximal density ρmax(G) = maxH⊆G{ρ(H)}, ρ(H) = e(H)v(H)

p0 = n−1/ρmax(G);

p�p0 ⇒ limn→∞

Pn,p(LG) = 1,

Pn,p(LG) = 0,

p=p0 ⇒ limn→∞

Pn,p(LG)/∈ {0, 1}.

If G � set of graphs, L = {LG, LG, G ∈ G},p = n−α, α is irrational

then ∀L ∈ L limn→∞ Pn,p(L) ∈ {0, 1}.

G � an arbitrary graph, LG � property of containing G.

LG is the �rst order property.

Pn,p(LG) = 1,

Pn,p(LG) = 0,

p=p0 ⇒ limn→∞

Pn,p(LG)/∈ {0, 1}.

then ∀L ∈ L limn→∞ Pn,p(L) ∈ {0, 1}.

Pn,p(LG) = 1,

Pn,p(LG) = 0,

p=p0 ⇒ limn→∞

Pn,p(LG)/∈ {0, 1}.

then ∀L ∈ L limn→∞ Pn,p(L) ∈ {0, 1}.

Pn,p(LG) = 1,

Pn,p(LG) = 0,

p=p0 ⇒ limn→∞

Pn,p(LG)/∈ {0, 1}.

then ∀L ∈ L limn→∞ Pn,p(L) ∈ {0, 1}.

Pn,p(LG) = 1,

Pn,p(LG) = 0,

p=p0 ⇒ limn→∞

Pn,p(LG)/∈ {0, 1}.

then ∀L ∈ L limn→∞ Pn,p(L) ∈ {0, 1}.

Pn,p(LG) = 1,

Pn,p(LG) = 0,

p=p0 ⇒ limn→∞

Pn,p(LG)/∈ {0, 1}.

then ∀L ∈ L limn→∞ Pn,p(L) ∈ {0, 1}.

Pn,p(LG) = 1,

Pn,p(LG) = 0,

p=p0 ⇒ limn→∞

Pn,p(LG)/∈ {0, 1}.

then ∀L ∈ L limn→∞ Pn,p(L) ∈ {0, 1}.

Zero-one law

L � the set of all �rst-order properties.

De�nition

The random graph obeys zero-one law if ∀L ∈ L

limn→∞

Pn,p(n)(L) ∈ {0, 1}.

P is a set of functions p such that G(n, p) obeys zero-one law.

Theorem [J.H. Spencer, S. Shelah, 1988]

Let p = n−α, α ∈ (0, 1].If α ∈ R \Q then p ∈ P.If α ∈ Q then p /∈ P.

Zero-one law

L � the set of all �rst-order properties.

De�nition

The random graph obeys zero-one law if ∀L ∈ L

limn→∞

Pn,p(n)(L) ∈ {0, 1}.

P is a set of functions p such that G(n, p) obeys zero-one law.

Theorem [J.H. Spencer, S. Shelah, 1988]

Let p = n−α, α ∈ (0, 1].If α ∈ R \Q then p ∈ P.If α ∈ Q then p /∈ P.

Quanti�er depth.

Quanti�er depths of x = y, x ∼ y equal 0.

If quanti�er depth of φ equals k then quanti�er depth of ¬φequals k as well.

If quanti�er depth of φ1 equals k1, quanti�er depth of φ2 equalsk2 then quanti�er depths of φ1 ⇒ φ2, φ1 ⇔ φ2, φ1 ∨ φ2,φ1 ∧ φ2 equal max{k1, k2}.

If quanti�er depth of φ(x) equals k, then quanti�er depths of∃xφ(x), ∀xφ(x) equal k + 1.

Quanti�er depth.

Bounded quanti�er depth.

Lk � the class of �rst-order properties de�ned by formulae withquanti�er depth bounded by k.

Random graph G(n, p) obeys zero-one k-law if ∀L ∈ Lk

limn→∞

Pn,p(n)(L) ∈ {0, 1}.

Pk � the class of all functions p = p(n) such that random graphG(n, p) obeys zero-one k-law.

Example: (∀x∃y (x ∼ y)) ∧ (∀x∃y ¬(x ∼ y)) k = 2∀x∃y ∃z ((x ∼ y) ∧ (¬(x ∼ z))) k = 3

Bounded quanti�er depth.

Lk � the class of �rst-order properties de�ned by formulae withquanti�er depth bounded by k.

Random graph G(n, p) obeys zero-one k-law if ∀L ∈ Lk

limn→∞

Pn,p(n)(L) ∈ {0, 1}.

Pk � the class of all functions p = p(n) such that random graphG(n, p) obeys zero-one k-law.

Example: (∀x∃y (x ∼ y)) ∧ (∀x∃y ¬(x ∼ y)) k = 2∀x∃y ∃z ((x ∼ y) ∧ (¬(x ∼ z))) k = 3

Examples

Simple case: α ∈ (0, 1k−1), zero-one k-law holds. The proof is very

simple.Method of proof from Theorem of Glebskii et el.:

Theorem [Y.V. Glebskii, D.I. Kogan, M.I. Liagonkii, V.A. Talanov,1969; R.Fagin, 1976]

Let for any β > 0 min{p, 1− p}nβ →∞, n →∞. Then p ∈ P.

What happens when α ≥ 1k−1? The most dense graph is Kk.

Therefore, the �rst α such that zero-one k-law does not hold forLG is 2

k−1 . Does zero-one k-law hold when α ∈ [ 1k−1 , 2

k−1)?

Examples

Simple case: α ∈ (0, 1k−1), zero-one k-law holds. The proof is very

simple.Method of proof from Theorem of Glebskii et el.:

Theorem [Y.V. Glebskii, D.I. Kogan, M.I. Liagonkii, V.A. Talanov,1969; R.Fagin, 1976]

Let for any β > 0 min{p, 1− p}nβ →∞, n →∞. Then p ∈ P.

What happens when α ≥ 1k−1? The most dense graph is Kk.

Therefore, the �rst α such that zero-one k-law does not hold forLG is 2

k−1 . Does zero-one k-law hold when α ∈ [ 1k−1 , 2

k−1)?

Zero-one k-laws

Zhukovskii; 2010 (zero-one k-law)

Let p = n−α, k ∈ N, k ≥ 3.

If 0 < α < 1k−2 then p ∈ Pk.

Zhukovskii; 2012+ (extension of zero-one k-law)

Let p = n−α, k ∈ N, k ≥ 4, Q = {ab , a, b ∈ N, a ≤ 2k−1}

If α = 1− 12k−1+β

, β ∈ (0,∞) \ Q then p ∈ Pk.

If α = 1− 12k−1+β

, β ∈ {2k−1 − 1, 2k−1} then p ∈ Pk.

2k − 2, 1

)⋃(1−

2k − 3, 1−

2k − 2

)⋃. . .⋃

2k−1 + 2k−2, 1−

2k−1 + 2k−2 + 1

)⋃(1−

2k−1 + 2k−1−12

, 1−1

2k−1 + 2k−2

⋃. . .⋃1−

2k−1 +2k−1−

[2k−1

, 1−1

2k−1 + 2k−1

⋃ . . .? . . .⋃[

k − 2

Zero-one k-laws

Let p = n−α, k ∈ N, k ≥ 3.

If 0 < α < 1k−2 then p ∈ Pk.

If α = 1− 12k−1+β

, β ∈ (0,∞) \ Q then p ∈ Pk.

If α = 1− 12k−1+β

, β ∈ {2k−1 − 1, 2k−1} then p ∈ Pk.

2k − 2, 1

)⋃(1−

2k − 3, 1−

2k − 2

)⋃. . .⋃

2k−1 + 2k−2, 1−

2k−1 + 2k−2 + 1

)⋃(1−

2k−1 + 2k−1−12

, 1−1

2k−1 + 2k−2

⋃. . .⋃1−

2k−1 +2k−1−

[2k−1

, 1−1

2k−1 + 2k−1

⋃ . . .? . . .⋃[

k − 2

Zero-one k-laws

Let p = n−α, k ∈ N, k ≥ 3.

If 0 < α < 1k−2 then p ∈ Pk.

If α = 1− 12k−1+β

, β ∈ (0,∞) \ Q then p ∈ Pk.

If α = 1− 12k−1+β

, β ∈ {2k−1 − 1, 2k−1} then p ∈ Pk.

2k − 2, 1

)⋃(1−

2k − 3, 1−

2k − 2

)⋃. . .⋃

2k−1 + 2k−2, 1−

2k−1 + 2k−2 + 1

)⋃(1−

2k−1 + 2k−1−12

, 1−1

2k−1 + 2k−2

⋃. . .⋃1−

2k−1 +2k−1−

[2k−1

, 1−1

2k−1 + 2k−1

⋃ . . .? . . .⋃[

k − 2

Zero-one k-laws

Zhukovskii; 2013++ (no k-law)

Let p = n−α, k ∈ N.

If k ≥ 3, α = 1k−2 then p /∈ Pk.

If k ≥ 4,Q̃ =

2k−1 − 2 · 1, . . . , 1;2k−1−2·1−1·2

2, . . . , 1

2; 2k−1−2·1−2·2

3, . . . , 1

3; 2k−1−2·1−3·2

4, . . . , 1

2k−1−2·1−3·2−1·35

, . . . ; . . . ; 2k−1−2·1−3·2−4·38

, . . . ;. . . ; . . . ; . . .

α = 1− 1

2k−1+β, β ∈ Q̃ then p /∈ Pk.

Critical points

α : 0 → 1

Critical points

α : 0 → 1

1k−2

Critical points

α : 0 → 1

1k−2 ?

Large gap: (1/(k − 2), 1− 1/2k−1)

Critical points

α : 0 → 1

1k−2 ? 1− 1

2k−1

Large gap: (1/(k − 2), 1− 1/2k−1)

Critical points

α : 0 → 1

1k−2 ? 1− 1

2k−1 . . .

Large gap: (1/(k − 2), 1− 1/2k−1)

Critical points

α : 0 → 1

1k−2 ? 1− 1

2k−1 . . . 12k−2k−2

Large gap: (1/(k − 2), 1− 1/2k−1)

Critical points

α : 0 → 1

1k−2 ? 1− 1

2k−1 . . . 12k−2k−2 . . . 1

2k−3

Large gap: (1/(k − 2), 1− 1/2k−1)

Critical points

α : 0 → 1

1k−2 ? 1− 1

2k−1 . . . 12k−2k−2 . . . 1

2k−31

2k−2

Large gap: (1/(k − 2), 1− 1/2k−1)

Critical points

α : 0 → 1

1k−2 ? 1− 1

2k−1 . . . 12k−2k−2 . . . 1

2k−31

2k−20

Large gap: (1/(k − 2), 1− 1/2k−1)

Cases k = 3, k = 4

For any α ∈ (0, 1) p ∈ P3.

For any α ∈ (0, 1/2) p ∈ P4.

For any α ∈ (13/14, 1) p ∈ P4.

Some results for α ∈ (7/8, 13/14) . . .

If α ∈ {1/2, 2/3, 3/4, 5/6, 7/8, 9/10, 10/11, 11/12, 12/13, 13/14},then p /∈ P4.

For any α ∈ (184/277, 2/3) p ∈ P4.

Cases k = 3, k = 4

For any α ∈ (0, 1) p ∈ P3.

For any α ∈ (0, 1/2) p ∈ P4.

For any α ∈ (13/14, 1) p ∈ P4.

Some results for α ∈ (7/8, 13/14) . . .

If α ∈ {1/2, 2/3, 3/4, 5/6, 7/8, 9/10, 10/11, 11/12, 12/13, 13/14},then p /∈ P4.

For any α ∈ (184/277, 2/3) p ∈ P4.

Intervals

α = 2/3,L � the property of containing K4

α = 3/4,

∃x1 ((∃x2∃x3∃x4 ((x3 ∼ x1)∧(x2 ∼ x1)∧(x2 ∼ x3)))∧(∃x2∃x3 ((x1 ∼ x2) ∧ (x1 ∼ x3) ∧ (x2 ∼ x3)))∧

(∀x4 ((x1 ∼ x4) → ((¬(x2 ∼ x4)) ∧ (¬(x3 ∼ x4))))))

α = 4/5,L � the property of containing K4 without one edge

α = 5/6,

∃x1∃x2 ((x1 ∼ x2) ∧ (∃x3 ((x3 ∼ x1) ∧ (x3 ∼ x2)))∧(∃x3∃x4 ((x1 ∼ x3) ∧ (x3 ∼ x4) ∧ (x4 ∼ x2))))

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

k � number of rounds

Spoiler & Duplicator � two players

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht game

EHR(G, H, k)

G, H � two graphs

Ehrenfeucht theorem

Theorem [A. Ehrenfeucht, 1960]

Let G, H be two graphs. For any �rst-order property L expressedby formula with quanti�er depth bounded by a number kG ∈ L ⇔ H ∈ L if and only if Duplicator has a winning strategy inthe game EHR(G, H, k).

Example: ∃x1 ∃x2 ∃x3 ((x1 ∼ x2) ∧ (x2 ∼ x3) ∧ (x1 ∼ x3)).

Ehrenfeucht theorem

Proofs of k-laws

Corollary

Zero-one law holds if and only if for any k ∈ N almost surelyDuplicator has a winning strategy in the Ehrenfeucht game onk rounds.

Random graph G(n, p) obeys zero-one k-law if and only ifalmost surely Duplicator has a winning strategy in theEhrenfeucht game on k rounds.

Open questions

What happens when α ∈(

1k−2 , 1− 1

2k−1

Spencer and Shelah: There exists a �rst-order property and anin�nite set S of rational numbers from (0, 1) such that for anyα ∈ S random graph G(n, n−α) follows the property withprobability which doesn't tend to 0 or to 1.

For any �xed k and any ε > 0 in(1− 1

2k−1 + ε, 1)there is only

�nite number of critical points. Does this property holds for(1− 1

2k−1 , 1)?

What is the maximal k such that there is �nite number ofcritical points?

k = 4, k = 5, . . .

Open questions

1k−2 , 1− 1

2k−1

2k−1 , 1)?

k = 4, k = 5, . . .

Open questions

1k−2 , 1− 1

2k−1

2k−1 , 1)?

k = 4, k = 5, . . .

Open questions

1k−2 , 1− 1

2k−1

2k−1 , 1)?

k = 4, k = 5, . . .

Thank you!

Thank you very much for your

attention!

maksim zhukovskii – zero-one k-laws for g(n,n−α)

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