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Upper bounds on the size of 4 - and 6 -cycle-free subgraphs of the - - PowerPoint PPT Presentation

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Proof 1 st try Introduction Flag Algebras Flags Upper bounds on the size of 4 - and 6 -cycle-free subgraphs of the hypercube J ozsef Balogh, Ping Hu, Bernard Lidick y and Hong Liu University of Illinois at Urbana-Champaign AMS - March

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SLIDE 1

Introduction Flag Algebras Proof 1st try Flags

Upper bounds on the size of 4- and 6-cycle-free subgraphs of the hypercube

J´

  • zsef Balogh, Ping Hu, Bernard Lidick´

y and Hong Liu

University of Illinois at Urbana-Champaign

AMS - March 18, 2012

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SLIDE 2

Introduction Flag Algebras Proof 1st try Flags

Hypercube

  • Qn is n-dimensional hypercube (n-cube)

Q1 Q2 Q3

  • e(G) := |E(G)|
  • exQ(n, F) := the maximum number of edges of a F-free

subgraph of Qn

  • πQ(F) = lim

n→∞

exQ(n, F) e(Qn)

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SLIDE 3

Introduction Flag Algebras Proof 1st try Flags

Hypercube

  • Qn is n-dimensional hypercube (n-cube)

Q1 Q2 Q3

  • e(G) := |E(G)|
  • exQ(n, F) := the maximum number of edges of a F-free

subgraph of Qn

  • πQ(F) = lim

n→∞

exQ(n, F) e(Qn)

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SLIDE 4

Introduction Flag Algebras Proof 1st try Flags

Hypercube

  • Qn is n-dimensional hypercube (n-cube)

Q1 Q2 Q3

  • e(G) := |E(G)|
  • exQ(n, F) := the maximum number of edges of a F-free

subgraph of Qn

  • πQ(F) = lim

n→∞

exQ(n, F) e(Qn)

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SLIDE 5

Introduction Flag Algebras Proof 1st try Flags

Hypercube

  • Qn is n-dimensional hypercube (n-cube)

Q1 Q2 Q3

  • e(G) := |E(G)|
  • exQ(n, F) := the maximum number of edges of a F-free

subgraph of Qn

  • πQ(F) = lim

n→∞

exQ(n, F) e(Qn)

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SLIDE 6

Introduction Flag Algebras Proof 1st try Flags

πQ(C2t)

Conjecture (Erd˝

  • s [1984])

πQ(C4) = 1/2, πQ(C2t) = 0 for t > 2

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SLIDE 7

Introduction Flag Algebras Proof 1st try Flags

πQ(C2t)

Conjecture (Erd˝

  • s [1984])

πQ(C4) = 1/2, πQ(C2t) = 0 for t > 2

Q7 Q7

πQ(C4) ≥ 1/2

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SLIDE 8

Introduction Flag Algebras Proof 1st try Flags

πQ(C2t)

Conjecture (Erd˝

  • s [1984])

πQ(C4) = 1/2, πQ(C2t) = 0 for t > 2

Q7 Q7

πQ(C4) ≥ 1/2

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SLIDE 9

Introduction Flag Algebras Proof 1st try Flags

πQ(C2t)

Conjecture (Erd˝

  • s [1984])

πQ(C4) = 1/2, πQ(C2t) = 0 for t > 2

Theorem (Chung [1992], Brouwer–Dejter–Thomassen [1993])

πQ(C6) ≥ 1/4

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SLIDE 10

Introduction Flag Algebras Proof 1st try Flags

πQ(C2t)

Conjecture (Erd˝

  • s [1984])

πQ(C4) = 1/2, πQ(C2t) = 0 for t > 2

Theorem (Chung [1992], Brouwer–Dejter–Thomassen [1993])

πQ(C6) ≥ 1/4

Theorem (Conder [1993])

πQ(C6) ≥ 1/3

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SLIDE 11

Introduction Flag Algebras Proof 1st try Flags

πQ(C2t)

Conjecture (Erd˝

  • s [1984])

πQ(C4) = 1/2, πQ(C2t) = 0 for t > 2.

Theorem (Chung [1992])

πQ(n, C2t) = 0 for even t ≥ 4.

Theorem (F¨ uredi–¨ Ozkahya [2009])

πQ(C2t) = 0 for odd t ≥ 7. if πQ(C10) = 0 is still open.

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SLIDE 12

Introduction Flag Algebras Proof 1st try Flags

πQ(C2t)

Conjecture (Erd˝

  • s [1984])

πQ(C4) = 1/2, πQ(C2t) = 0 for t > 2.

Theorem (Chung [1992])

πQ(n, C2t) = 0 for even t ≥ 4.

Theorem (F¨ uredi–¨ Ozkahya [2009])

πQ(C2t) = 0 for odd t ≥ 7. if πQ(C10) = 0 is still open.

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Introduction Flag Algebras Proof 1st try Flags

πQ(C4)

Theorem (Brass–Harborth–Nienborg [1995])

exQ(n, C4) ≥ 1

2(1 + 1 √n)e(Qn) (valid when n is a power of 4)

Theorem (Chung [1992])

πQ(C4) ≤ 0.62284.

Theorem (Thomason–Wagner [2009])

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SLIDE 14

Introduction Flag Algebras Proof 1st try Flags

πQ(C4)

Theorem (Brass–Harborth–Nienborg [1995])

exQ(n, C4) ≥ 1

2(1 + 1 √n)e(Qn) (valid when n is a power of 4)

Theorem (Chung [1992])

πQ(C4) ≤ 0.62284.

Theorem (Thomason–Wagner [2009])

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SLIDE 15

Introduction Flag Algebras Proof 1st try Flags

πQ(C4)

Theorem (Brass–Harborth–Nienborg [1995])

exQ(n, C4) ≥ 1

2(1 + 1 √n)e(Qn) (valid when n is a power of 4)

Theorem (Chung [1992])

πQ(C4) ≤ 0.62284.

Theorem (Thomason–Wagner [2009])

πQ(C4) ≤ 0.62256.

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SLIDE 16

Introduction Flag Algebras Proof 1st try Flags

πQ(C4)

Theorem (Brass–Harborth–Nienborg [1995])

exQ(n, C4) ≥ 1

2(1 + 1 √n)e(Qn) (valid when n is a power of 4)

Theorem (Chung [1992])

πQ(C4) ≤ 0.62284.

Theorem (Thomason–Wagner [2009])

πQ(C4) ≤ 0.62083.

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SLIDE 17

Introduction Flag Algebras Proof 1st try Flags

πQ(C4)

Theorem (Brass–Harborth–Nienborg [1995])

exQ(n, C4) ≥ 1

2(1 + 1 √n)e(Qn) (valid when n is a power of 4)

Theorem (Chung [1992])

πQ(C4) ≤ 0.62284.

Theorem (Thomason–Wagner [2009])

πQ(C4) ≤ 0.62083.

Theorem (Balogh–Hu–L–Liu, ind. Baber [2012+])

πQ(C4) ≤ 0.6068.

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SLIDE 18

Introduction Flag Algebras Proof 1st try Flags

πQ(n, C6)

Theorem (Conder [1993])

πQ(C6) ≥ 1/3.

Theorem (Chung [1992])

πQ(C6) ≤ √ 2 − 1 ≈ 0.41421.

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SLIDE 19

Introduction Flag Algebras Proof 1st try Flags

πQ(n, C6)

Theorem (Conder [1993])

πQ(C6) ≥ 1/3.

Theorem (Chung [1992])

πQ(C6) ≤ √ 2 − 1 ≈ 0.41421.

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SLIDE 20

Introduction Flag Algebras Proof 1st try Flags

πQ(n, C6)

Theorem (Conder [1993])

πQ(C6) ≥ 1/3.

Theorem (Chung [1992])

πQ(C6) ≤ √ 2 − 1 ≈ 0.41421.

Theorem (Balogh–Hu–L–Liu, ind. Baber [2012+])

πQ(C6) ≤ 0.3755.

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Introduction Flag Algebras Proof 1st try Flags

Flag Algebras

Definition

p(H, G): the probability that a random |V (H)|-set U in V (G) induces G[U] isomorphic to H. Razborov [2007] developed flag algebras. Let G be the family of graphs forbidding some structures, then flag algebras can be used to bound lim

G∈G,|V (G)|→∞ p(H, G).

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Introduction Flag Algebras Proof 1st try Flags

Results using Flag Algebras

Theorem (Hladk´ y–Kr´ al’–Norine [2009])

Every n-vertex digraph with minimum outdegree at least 0.3465n contains a triangle.

Theorem (Hatami–Hladk´ y–Kr´ al’–Norine–Razborov [2011], Grzesik [2011])

The number of C5s in a triangle-free graph of order n is at most (n/5)5.

Theorem (Falgas-Ravry–Vaughan [2011])

π(K −

4 , C5, F3,2) = 12/49, π(K − 4 , F3,2) = 5/18.

F3,2 : {123, 145, 245, 345}, C5 : {123, 234, 345, 451, 512}.

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SLIDE 23

Introduction Flag Algebras Proof 1st try Flags

Results using Flag Algebras

Theorem (Hladk´ y–Kr´ al’–Norine [2009])

Every n-vertex digraph with minimum outdegree at least 0.3465n contains a triangle.

Theorem (Hatami–Hladk´ y–Kr´ al’–Norine–Razborov [2011], Grzesik [2011])

The number of C5s in a triangle-free graph of order n is at most (n/5)5.

Theorem (Falgas-Ravry–Vaughan [2011])

π(K −

4 , C5, F3,2) = 12/49, π(K − 4 , F3,2) = 5/18.

F3,2 : {123, 145, 245, 345}, C5 : {123, 234, 345, 451, 512}.

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Introduction Flag Algebras Proof 1st try Flags

Results using Flag Algebras

Theorem (Hladk´ y–Kr´ al’–Norine [2009])

Every n-vertex digraph with minimum outdegree at least 0.3465n contains a triangle.

Theorem (Hatami–Hladk´ y–Kr´ al’–Norine–Razborov [2011], Grzesik [2011])

The number of C5s in a triangle-free graph of order n is at most (n/5)5.

Theorem (Falgas-Ravry–Vaughan [2011])

π(K −

4 , C5, F3,2) = 12/49, π(K − 4 , F3,2) = 5/18.

F3,2 : {123, 145, 245, 345}, C5 : {123, 234, 345, 451, 512}.

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SLIDE 25

Introduction Flag Algebras Proof 1st try Flags

Results using Flag Algebras

Theorem (Hladk´ y–Kr´ al’–Norine [2009])

Every n-vertex digraph with minimum outdegree at least 0.3465n contains a triangle.

Theorem (Hatami–Hladk´ y–Kr´ al’–Norine–Razborov [2011], Grzesik [2011])

The number of C5s in a triangle-free graph of order n is at most (n/5)5.

Theorem (Falgas-Ravry–Vaughan [2011])

π(K −

4 , C5, F3,2) = 12/49, π(K − 4 , F3,2) = 5/18.

F3,2 : {123, 145, 245, 345}, C5 : {123, 234, 345, 451, 512}.

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Introduction Flag Algebras Proof 1st try Flags

Proof by an Example

Example

πQ(C4) ≤ 2/3 Bound infinite problem by a finite piece.

Definition

Hn: the family of spanning subgraphs of Qn not containing C4. Let H ∈ Hs, G ∈ Hn, s < n, p(H, G) is the probability that a random s-hypercube in G induces H. ρ(G) = e(G)/e(Qn). ρ(G) =

  • H∈Hs

ρ(H)p(H, G) ρ(G) ≤ max

H∈Hs ρ(H)

πQ(C4) ≤ max

H∈Hs ρ(H)

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Introduction Flag Algebras Proof 1st try Flags

Proof by an Example

Example

πQ(C4) ≤ 2/3 Bound infinite problem by a finite piece.

Definition

Hn: the family of spanning subgraphs of Qn not containing C4. Let H ∈ Hs, G ∈ Hn, s < n, p(H, G) is the probability that a random s-hypercube in G induces H. ρ(G) = e(G)/e(Qn). ρ(G) =

  • H∈Hs

ρ(H)p(H, G) ρ(G) ≤ max

H∈Hs ρ(H)

πQ(C4) ≤ max

H∈Hs ρ(H)

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Introduction Flag Algebras Proof 1st try Flags

Proof by an Example

Example

πQ(C4) ≤ 2/3 Bound infinite problem by a finite piece.

Definition

Hn: the family of spanning subgraphs of Qn not containing C4. Let H ∈ Hs, G ∈ Hn, s < n, p(H, G) is the probability that a random s-hypercube in G induces H. ρ(G) = e(G)/e(Qn). ρ(G) =

  • H∈Hs

ρ(H)p(H, G) ρ(G) ≤ max

H∈Hs ρ(H)

πQ(C4) ≤ max

H∈Hs ρ(H)

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SLIDE 29

Introduction Flag Algebras Proof 1st try Flags

Proof by an Example

Example

πQ(C4) ≤ 2/3 Bound infinite problem by a finite piece.

Definition

Hn: the family of spanning subgraphs of Qn not containing C4. Let H ∈ Hs, G ∈ Hn, s < n, p(H, G) is the probability that a random s-hypercube in G induces H. ρ(G) = e(G)/e(Qn). ρ(G) =

  • H∈Hs

ρ(H)p(H, G) ρ(G) ≤ max

H∈Hs ρ(H)

πQ(C4) ≤ max

H∈Hs ρ(H)

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Introduction Flag Algebras Proof 1st try Flags

Proof by an Example

Example

πQ(C4) ≤ 2/3 Bound infinite problem by a finite piece.

Definition

Hn: the family of spanning subgraphs of Qn not containing C4. Let H ∈ Hs, G ∈ Hn, s < n, p(H, G) is the probability that a random s-hypercube in G induces H. ρ(G) = e(G)/e(Qn). ρ(G) =

  • H∈Hs

ρ(H)p(H, G) ρ(G) ≤ max

H∈Hs ρ(H)

πQ(C4) ≤ max

H∈Hs ρ(H)

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Introduction Flag Algebras Proof 1st try Flags

Is the bound good?

ρ(G) =

  • H∈Hs

ρ(H)p(H, G) ρ(G) ≤ max

H∈Hs ρ(H)

πQ(C4) ≤ max

H∈Hs ρ(H)

H2

H1 H2 H3 H4 H5

πQ(C4) ≤ max ρ(Hi) = ρ(H5) = 3/4 If 0 ≤

i cHip(Hi, G), then

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SLIDE 32

Introduction Flag Algebras Proof 1st try Flags

Is the bound good?

ρ(G) =

  • H∈Hs

ρ(H)p(H, G) ρ(G) ≤ max

H∈Hs ρ(H)

πQ(C4) ≤ max

H∈Hs ρ(H)

H2

H1 H2 H3 H4 H5

πQ(C4) ≤ max ρ(Hi) = ρ(H5) = 3/4 If 0 ≤

i cHip(Hi, G), then

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SLIDE 33

Introduction Flag Algebras Proof 1st try Flags

Is the bound good?

ρ(G) =

  • H∈Hs

ρ(H)p(H, G) ρ(G) ≤ max

H∈Hs ρ(H)

πQ(C4) ≤ max

H∈Hs ρ(H)

H2

H1 H2 H3 H4 H5

πQ(C4) ≤ max ρ(Hi) = ρ(H5) = 3/4 If 0 ≤

i cHip(Hi, G), then

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SLIDE 34

Introduction Flag Algebras Proof 1st try Flags

Is the bound good?

ρ(G) =

  • H∈Hs

ρ(H)p(H, G)

H1 H2 H3 H4 H5

πQ(C4) ≤ max ρ(Hi) = ρ(H5) = 3/4 If 0 ≤

i cHip(Hi, G), then

ρ(G) ≤

  • i

(ρ(Hi) + cHi) p(Hi, G) πQ(C4) ≤ max

i

(ρ(Hi) + cHi) cHi might be negative

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Introduction Flag Algebras Proof 1st try Flags

Example Continued: Flags

Definition

Flags: F=(H, θ), H ∈ Hs, θ : [2k] → V (H) is injective, H[Im(θ)] ∈ Hk.

1 F1 1 F2

Let G ∈ Hn, θ : [1] → V (G).

Definition

p(Fi, θ, G): the probability that a random 1-cube U in G subject to Im(θ) ⊂ U satisfies (U, θ) = Fi. p(Fi, Fj, θ, G): the probability that two random 1-cubes U1, U2 in G subject to U1 ∩ U2 = Im(θ) satisfy (U1, θ) = Fi, (U2, θ) = Fj.

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Introduction Flag Algebras Proof 1st try Flags

Example Continued: Flags

Definition

Flags: F=(H, θ), H ∈ Hs, θ : [2k] → V (H) is injective, H[Im(θ)] ∈ Hk.

1 F1 1 F2

Let G ∈ Hn, θ : [1] → V (G).

Definition

p(Fi, θ, G): the probability that a random 1-cube U in G subject to Im(θ) ⊂ U satisfies (U, θ) = Fi. p(Fi, Fj, θ, G): the probability that two random 1-cubes U1, U2 in G subject to U1 ∩ U2 = Im(θ) satisfy (U1, θ) = Fi, (U2, θ) = Fj.

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Introduction Flag Algebras Proof 1st try Flags

Example Continued: p(Fi, θ, G)

Let G ∈ Hn, θ : [1] → V (G).

Definition

p(Fi, θ, G): the probability that a random 1-cube U in G subject to Im(θ) ⊂ U satisfies (U, θ) = Fi. p(Fi, Fj, θ, G): the probability that two random 1-cubes U1, U2 in G subject to U1 ∩ U2 = Im(θ) satisfy (U1, θ) = Fi, (U2, θ) = Fj.

1 F1 H4 1

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Introduction Flag Algebras Proof 1st try Flags

Example Continued: p(Fi, θ, G)

Let G ∈ Hn, θ : [1] → V (G).

Definition

p(Fi, θ, G): the probability that a random 1-cube U in G subject to Im(θ) ⊂ U satisfies (U, θ) = Fi. p(Fi, Fj, θ, G): the probability that two random 1-cubes U1, U2 in G subject to U1 ∩ U2 = Im(θ) satisfy (U1, θ) = Fi, (U2, θ) = Fj.

1 F1 H4 1 θ

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Introduction Flag Algebras Proof 1st try Flags

Example Continued: p(Fi, θ, G)

Let G ∈ Hn, θ : [1] → V (G).

Definition

p(Fi, θ, G): the probability that a random 1-cube U in G subject to Im(θ) ⊂ U satisfies (U, θ) = Fi. p(Fi, Fj, θ, G): the probability that two random 1-cubes U1, U2 in G subject to U1 ∩ U2 = Im(θ) satisfy (U1, θ) = Fi, (U2, θ) = Fj.

1 F1 H4 1 θ

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Introduction Flag Algebras Proof 1st try Flags

Example Continued: p(Fi, θ, G)

Let G ∈ Hn, θ : [1] → V (G).

Definition

p(Fi, θ, G): the probability that a random 1-cube U in G subject to Im(θ) ⊂ U satisfies (U, θ) = Fi. p(Fi, Fj, θ, G): the probability that two random 1-cubes U1, U2 in G subject to U1 ∩ U2 = Im(θ) satisfy (U1, θ) = Fi, (U2, θ) = Fj.

1 F1 H4 1 θ U

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Introduction Flag Algebras Proof 1st try Flags

Example Continued: p(Fi, θ, G)

Let G ∈ Hn, θ : [1] → V (G).

Definition

p(Fi, θ, G): the probability that a random 1-cube U in G subject to Im(θ) ⊂ U satisfies (U, θ) = Fi. p(Fi, Fj, θ, G): the probability that two random 1-cubes U1, U2 in G subject to U1 ∩ U2 = Im(θ) satisfy (U1, θ) = Fi, (U2, θ) = Fj.

1 F1 H4 1 θ U

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Introduction Flag Algebras Proof 1st try Flags

Example Continued: p(Fi, θ, G)

Let G ∈ Hn, θ : [1] → V (G).

Definition

p(Fi, θ, G): the probability that a random 1-cube U in G subject to Im(θ) ⊂ U satisfies (U, θ) = Fi.

1 F1 H4 1 θ U U

p(F1, θ, H4) = 1/2

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Introduction Flag Algebras Proof 1st try Flags

Proof Continued: Flags

Let M = (mij) be a positive semidefinite 2-by-2 matrix, define pθ = {p(F1, θ; G), p(F2, θ; G)}, then 0 ≤ Eθ[pθMpT

θ ]

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Introduction Flag Algebras Proof 1st try Flags

Proof Continued: Flags

Let M = (mij) be a positive semidefinite 2-by-2 matrix, define pθ = {p(F1, θ; G), p(F2, θ; G)}, then 0 ≤ Eθ[pθMpT

θ ] =

  • 1≤i,j≤2

mijEθ[p(Fi, θ, G)p(Fj, θ, G)]

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SLIDE 45

Introduction Flag Algebras Proof 1st try Flags

Proof Continued: Flags

Let M = (mij) be a positive semidefinite 2-by-2 matrix, define pθ = {p(F1, θ; G), p(F2, θ; G)}, then 0 ≤ Eθ[pθMpT

θ ] =

  • 1≤i,j≤2

mijEθ[p(Fi, θ, G)p(Fj, θ, G)]

Lemma

p(Fi, θ, G)p(Fj, θ, G) = p(Fi, Fj, θ, G) + o(1),

  • (1) → 0 as n → ∞.
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Introduction Flag Algebras Proof 1st try Flags

Proof Continued: Flags

Let M = (mij) be a positive semidefinite 2-by-2 matrix, define pθ = {p(F1, θ; G), p(F2, θ; G)}, then 0 ≤ Eθ[pθMpT

θ ] =

  • 1≤i,j≤2

mijEθ[p(Fi, θ, G)p(Fj, θ, G)] =

  • 1≤i,j≤2

mijEθ[p(Fi, Fj, θ, G)] + o(1)

Lemma

p(Fi, θ, G)p(Fj, θ, G) = p(Fi, Fj, θ, G) + o(1),

  • (1) → 0 as n → ∞.
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Introduction Flag Algebras Proof 1st try Flags

Proof Continued: Flags

Let M = (mij) be a positive semidefinite 2-by-2 matrix, define pθ = {p(F1, θ; G), p(F2, θ; G)}, then 0 ≤ Eθ[pθMpT

θ ] =

  • 1≤i,j≤2

mijEθ[p(Fi, θ, G)p(Fj, θ, G)] =

  • 1≤i,j≤2

mijEθ[p(Fi, Fj, θ, G)] + o(1)

Lemma

Eθ[p(Fi, Fj, θ; G)] =

  • H∈H2

Eθ[p(Fi, Fj, θ; H)]p(H, G)

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SLIDE 48

Introduction Flag Algebras Proof 1st try Flags

Proof Continued: Flags

Let M = (mij) be a positive semidefinite 2-by-2 matrix, define pθ = {p(F1, θ; G), p(F2, θ; G)}, then 0 ≤ Eθ[pθMpT

θ ] =

  • 1≤i,j≤2

mijEθ[p(Fi, θ, G)p(Fj, θ, G)] =

  • 1≤i,j≤2

mijEθ[p(Fi, Fj, θ, G)] + o(1) =

  • 1≤i,j≤2
  • H∈H2

mijEθ[p(Fi, Fj, θ; H)]p(H, G) + o(1)

Lemma

Eθ[p(Fi, Fj, θ; G)] =

  • H∈H2

Eθ[p(Fi, Fj, θ; H)]p(H, G)

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SLIDE 49

Introduction Flag Algebras Proof 1st try Flags

Proof Continued: Flags

0 ≤ Eθ[pθMpT

θ ] =

  • H∈H2
  • 1≤i,j≤2

mijEθ[p(Fi, Fj, θ; H)]p(H, G)+o(1). Let cH(M) =

  • 1≤i,j≤2

mijEθ[p(Fi, Fj, θ; H)], then 0 ≤

  • H∈H2

cH(M)p(H, G) + o(1). So ρ(G) ≤

  • H∈H2

(ρ(H) + cH(M)) p(H, G) πQ(C4) ≤ max

H∈H2 (ρ(H) + cH(M))

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SLIDE 50

Introduction Flag Algebras Proof 1st try Flags

Proof Continued: Flags

0 ≤ Eθ[pθMpT

θ ] =

  • H∈H2
  • 1≤i,j≤2

mijEθ[p(Fi, Fj, θ; H)]p(H, G)+o(1). Let cH(M) =

  • 1≤i,j≤2

mijEθ[p(Fi, Fj, θ; H)], then 0 ≤

  • H∈H2

cH(M)p(H, G) + o(1). So ρ(G) ≤

  • H∈H2

(ρ(H) + cH(M)) p(H, G) πQ(C4) ≤ max

H∈H2 (ρ(H) + cH(M))

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SLIDE 51

Introduction Flag Algebras Proof 1st try Flags

Proof Continued: Flags

0 ≤ Eθ[pθMpT

θ ] =

  • H∈H2
  • 1≤i,j≤2

mijEθ[p(Fi, Fj, θ; H)]p(H, G)+o(1). Let cH(M) =

  • 1≤i,j≤2

mijEθ[p(Fi, Fj, θ; H)], then 0 ≤

  • H∈H2

cH(M)p(H, G) + o(1). So ρ(G) ≤

  • H∈H2

(ρ(H) + cH(M)) p(H, G) πQ(C4) ≤ max

H∈H2 (ρ(H) + cH(M))

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SLIDE 52

Introduction Flag Algebras Proof 1st try Flags

Computing Eθ[p(Fi, Fj, θ; H)]

H1 H2 H3 H4 H5 1 F1 1 F2

H1 H2 H3 H4 H5 F1, F1 1 1/2 1/4 F1, F2 1/4 1/2 1/4 1/4 F2, F2 1/4 1/2

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SLIDE 53

Introduction Flag Algebras Proof 1st try Flags

Optimizing M

M = m11 m12 m21 m22

  • ρ(H1) + cH1

= 0 + m11 ρ(H2) + cH2 = 1/4 + m11/2 + m12/2 ρ(H3) + cH3 = 1/2 + m12 ρ(H4) + cH4 = 1/2 + m11/4 + m12/2 + m22/4 ρ(H5) + cH5 = 3/4 + m12/2 + m22/2 πQ(C4) ≤ max

i

(ρ(Hi) + cHi)

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SLIDE 54

Introduction Flag Algebras Proof 1st try Flags

Optimizing M

M = m11 m12 m21 m22

  • ρ(H1) + cH1

= 0 + m11 ρ(H2) + cH2 = 1/4 + m11/2 + m12/2 ρ(H3) + cH3 = 1/2 + m12 ρ(H4) + cH4 = 1/2 + m11/4 + m12/2 + m22/4 ρ(H5) + cH5 = 3/4 + m12/2 + m22/2 πQ(C4) ≤ max

i

(ρ(Hi) + cHi)

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SLIDE 55

Introduction Flag Algebras Proof 1st try Flags

Solution

Take M =

  • 2/3

−1/3 −1/3 1/6

  • ,

then max

i

(ρ(Hi) + cHi) = 2/3

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SLIDE 56

Introduction Flag Algebras Proof 1st try Flags

Results

Theorem (Balogh–Hu–L–Liu, ind. Baber [2012+])

πQ(C4) ≤ 0.6068.

Theorem (Balogh–Hu–L–Liu, ind. Baber [2012+])

πQ(C6) ≤ 0.3755. By using H3 and bigger flags. Almost surely can be improved by waiting.

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SLIDE 57

Introduction Flag Algebras Proof 1st try Flags

Results

Theorem (Balogh–Hu–L–Liu, ind. Baber [2012+])

πQ(C4) ≤ 0.6068.

Theorem (Balogh–Hu–L–Liu, ind. Baber [2012+])

πQ(C6) ≤ 0.3755. By using H3 and bigger flags. Almost surely can be improved by waiting.

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SLIDE 58

Introduction Flag Algebras Proof 1st try Flags

Thank you for your attention!