Lovsz theta function and its relationships with perfect graph theory - - PowerPoint PPT Presentation

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Lovász theta function and its relationships with perfect graph theory

Arnaud Pêcher joint work with C. Bachoc and A. Thiery

• Univ. Bordeaux (LaBRI / INRIA RealOpt)

Shanghai Jiao Tong University October 24th, 2016

• A. Pêcher (Univ. Bordeaux)

Shanghai, 2016 1 / 27

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Outline

1

Lovasz’ theta function and perfectness

2

A closed formula.

3

Separating the values.

4

Proving the closed formula.

• A. Pêcher (Univ. Bordeaux)

Motivation Shanghai, 2016 2 / 27

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Powers of chordless cycles

Cq

p = qth power of the chordless cycle Cp with p vertices

C9 C2

9

C3

9

• A. Pêcher (Univ. Bordeaux)

Motivation Shanghai, 2016 3 / 27

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Complements = circular-cliques

Circular-Clique Kp/q: vertices {0, 1, · · · , p − 1} and edges ij, s.t. q ≤ |i − j| ≤ p − q. Hence Cq−1

p

= Kp/q. K9/1 K9/2 K9/3 K9/4 K9/2 = C9 K9/3 = C2

9

K9/4 = C3

9

• A. Pêcher (Univ. Bordeaux)

Motivation Shanghai, 2016 4 / 27

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Some properties of Lovász’s theta function ϑ

Lovász’s theta function is a real function such that, for every graph G: ϑ(G) is computable in polynomial time with given (polynomial space encoding) accuracy ω(G) ≤ ϑ(G) ≤ χf(G) ≤ χ(G) (Sandwich Theorem) if G is homomorphic to H then ϑ(G) ≤ ϑ(H) Explicitly known for a few families of graphs: perfect graphs (Sandwich Theorem), cycle graphs (Lovász 1978), Kneser graphs (Lovász 1979), square

• f cycle graphs (Brimkov et al 2000)

Let ϑp/q = ϑ(Kp/q)

• A. Pêcher (Univ. Bordeaux)

Motivation Shanghai, 2016 5 / 27

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From Perfect to Circular-perfect graphs

Circular chromatic number (Vince, 1988)

χc(G) = inf { k/d|G → Kk/d }

Circular clique number (Zhu, 2000)

ωc(G) = sup { k/d|Kk/d → G } χ(G) = ⌈χc(G)⌉ ω(G) = ⌊ωc(G)⌋ ω(G) ≤ ωc(G) ≤ ωf(G) = χf(G) ≤ χc(G) ≤ χ(G)

Perfect Graph (Berge, 1960)

A graph G is perfect if ∀H ⊆ G, χ(H) = ω(H). Examples: bipartite graphs, chordal graphs, comparability graphs … If G is perfect then ϑ(G) = ω(G). Hence ω(G) = χ(G) is polytime. (even weighted: Grötschel, Lovász, Schrijver 1981)

• A. Pêcher (Univ. Bordeaux)

Motivation Shanghai, 2016 6 / 27

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From Perfect to Circular-perfect graphs

Circular chromatic number (Vince, 1988)

χc(G) = inf { k/d|G → Kk/d }

Circular clique number (Zhu, 2000)

ωc(G) = sup { k/d|Kk/d → G } χ(G) = ⌈χc(G)⌉ ω(G) = ⌊ωc(G)⌋ ω(G) ≤ ωc(G) ≤ ωf(G) = χf(G) ≤ χc(G) ≤ χ(G)

Circular-Perfect Graph (Zhu, 2000)

A graph G is circular-perfect if ∀H ⊆ G, χc(H) = ωc(H). Examples: perfect graphs, circular-cliques, outerplanar graphs … If G is circular-perfect then ϑ(G) = ϑ ( Kp/q ) where χc(G) = p/q. Aim: use this equality to prove that χc is polytime.

• A. Pêcher (Univ. Bordeaux)

Motivation Shanghai, 2016 6 / 27

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Core of algorithm to compute χc of circular-perfect graphs

For circular perfect graphs, ϑ(G) ̸= χc(G) in general: ϑ(C5) = √ 5 < χc(C5) = 2.5. Strategy to compute χc(G) for a circular-perfect graph G with n vertices in polynomial time: (1) compute ϑ(G) for some precision ϵ > 0 and denote by ϑ this value; (2) for every 1 ≤ p, q ≤ n, if |ϑ − ϑ(Kp/q)| < ϵ, return p/q. Correct provided there is a unique pair (p, q) satisfying (2) and ϵ has polyspace encoding. Hence, roughly speaking, we need to prove that the values ϑp/q are well-separated.

• A. Pêcher (Univ. Bordeaux)

Motivation Shanghai, 2016 7 / 27

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Outline

1

Lovasz’ theta function and perfectness

2

A closed formula.

3

Separating the values.

4

Proving the closed formula.

• A. Pêcher (Univ. Bordeaux)

A closed formula Shanghai, 2016 8 / 27

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Previous solved cases

Theorem - Lovász (1978) - q=2, p odd

ϑ ( Kp/2 ) = ϑ(Cp) = p cos (

π p

) 1 + cos (

π p

) Proofs: Lovász: algebraic arguments Knuth (1994): linear program with two variables

Theorem - Brimkov et al (2000) - q=3, p odd

ϑ ( Kp/3 ) = p  1 −

1 2 − cos

(

2π p ⌊ p 3⌋

) − cos (

2π p

( ⌊ p

3⌋ + 1

)) ( cos (

2π p ⌊ p 3⌋

) − 1 ) ( cos (

2π p

( ⌊ p

3⌋ + 1

)) − 1 )   Proof: linear program with 3 variables + geometrical arguments

• A. Pêcher (Univ. Bordeaux)

A closed formula Shanghai, 2016 9 / 27

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A closed formula

∀0 ≤ k ≤ q − 1, ck = cos (2kπ q ) , ak = cos (⌊kp q ⌋ 2π p ) p = 16, q = 5

Theorem (Bachoc, P., Thiery 2010)

ϑ(Kp/q) = p

q

(∑q−1

i=0 A0(ci) A0(1)

) with A0(x) = 2q−1 ∏q−1

i=1 (x − ai)

• A. Pêcher (Univ. Bordeaux)

A closed formula Shanghai, 2016 10 / 27

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Asymptotic behavior

ϑ(Kp/q) = p

q

( 1 + ∑q−1

i=1 A0(ci) A0(1)

)

Corollary

If q ≥ 3 and p ≥ 4q3/π then p q − 4eπ2 3 q p ≤ ϑ(Kp/q) ≤ p q (Upper bound is trivial as ϑ(Kp/q) ≤ χf(Kp/q) = p

q)

Corollary

For every ϵ > 0, for every positive integer α, there is a positive integer ω such that for every circular-perfect graph G satisfying ω(G) ≥ ω and α(G) ≤ α, we have |ϑ ( G ) − χc(G)| ≤ ϵ.

• A. Pêcher (Univ. Bordeaux)

A closed formula Shanghai, 2016 11 / 27

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Outline

1

Lovasz’ theta function and perfectness

2

A closed formula.

3

Separating the values.

4

Proving the closed formula.

• A. Pêcher (Univ. Bordeaux)

Injectivity Shanghai, 2016 12 / 27

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Separating result

Let p, p′, q, q′ ≤ n such that p

q ̸= p′ q′ .

Let ∆ =

• ϑ(Kp′/q′) − ϑ(Kp/q)
• .

Theorem (Bachoc, P., Thiery (2013)

∆ ≥ c−n5 for some c > 0 Hence, computing χc of circular-perfect graphs is polytime. The proof uses algebraic number theory and is in two steps: (1) ∆ ̸= 0 (2) if ∆ ̸= 0 then ∆ ≥ c−n5 for some c > 0

• A. Pêcher (Univ. Bordeaux)

Injectivity Shanghai, 2016 13 / 27

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∆ = 0: taking advantage of monotonicity

We have p q ≤ p′ q′ if and only if Kp/q → Kp′/q′ (Bondy & Hell ’96) if and only if ϑ(Kp/q) ≤ ϑ(Kp′/q′) Assume ∆ = 0: we have p

q < p′ q′ and ϑ(Kp/q) = ϑ(Kp′/q′) = ϑ.

Hence for every a

b ∈

[

p q, p′ q′

] , ϑ(Ka/b) = ϑ. Take a

b ∈

[

p q, p′ q′

] such that b is prime; b is coprime with a and a + 1.

• A. Pêcher (Univ. Bordeaux)

Injectivity Shanghai, 2016 14 / 27

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A flavour of algebraic number theory (1/2)

Notations and definitions: for every k, let ζk = exp2iπ/k; let Φ be the Euler phi function: Φ(n) ≤ n ; let Q (ζk) denote the cylotomic field: the smallest complex field containing ζk; for every x ∈ Q (ζk), let polmin(x) ∈ Q[X] be the minimal polynomial of x; x is called an algebraic integer if polmin(x) ∈ Z[X]. Some basic observations: Q (ζk) is a vector space over Q whose dimension is Φ(k) (hence at most k); the set of algebraic integers is a ring.

• A. Pêcher (Univ. Bordeaux)

Injectivity Shanghai, 2016 15 / 27

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A flavour of algebraic number theory (2/2)

Notations: if K and L are two cyclotomic fields s.t. L is an extension of K, let

Gal(L/K)=Aut(L/K) be the Galois group of L over K; TraceL

K(x)=

∑

σ∈Gal(L/K)

σ(x) be the trace of any element x of L; NormL

K(x)=

∏

σ∈Gal(L/K)

σ(x) be the norm of any element x of L.

let σi be the automorphism of Q (ζk) s.t. σi(ζk) = ζi

k (i coprime).

Gal(Q (ζk)) = {σi, (i, k) = 1}; TraceL

K is linear and for every element x of L, TraceL K(x) ∈ K;

for every x ∈ Q (ζk), polmin(x) = ∏

σ∈Gal(Q(ζk)/Q)

(X − σ(x)).

• A. Pêcher (Univ. Bordeaux)

Injectivity Shanghai, 2016 16 / 27

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A useful property of ϑ(Kk/d)

Assume that d is prime: Gal(Q (ζkd) /Q (ζk))={σi, 1 ≤ i ≤ d − 1}; for every i, σi(cos(2π/d)) = cos(2iπ/d) as cos(2iπ/d) = 1

2(ζik kd + ζkd−ik kd

) = σi (

1 2(ζd + ζd−1 d

) ) = σi(cos(2π/d)); hence (setuing L0(X) = A0(X)/A0(1)): ϑ(Kk/d) = k d ( 1 + ∑

i=1...d−1

L0 (σi(cos(2π/d))) ) = k d ( 1 + ∑

i=1...d−1

σi(L0(cos(2π/d))) ) = k d ( 1 + TraceQ(ζkd)

Q(ζk) (L0(cos(2π/d))

) thus ϑ(Kk/d) ∈ Q(ζk)

• A. Pêcher (Univ. Bordeaux)

Injectivity Shanghai, 2016 17 / 27

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∆ = 0: ϑ(Kp/q) is rational

Recall that we have chosen a and b such that ϑ = ϑ(Ka/b) = ϑ(K(a+1)/b), b is prime, (a, b) = 1 and (a + 1, b) = 1. Then ϑ ∈ Q(ζa) and likewise ϑ ∈ Q(ζa+1). Thus ϑ ∈ Q! Let ϑ = k/d. Now take a

b ∈

[

p q, p′ q′

] such that a > d, a/b is not integer, a is prime, (a, b) = 1 , (a, b + 1) = 1 and a/(b + 1) ≥ p/q. ∀i, 2 cos(2iπ/a) and 2 cos(2iπ/b) are algebraic integers of Q(ζab); hence ∀i, A0(cos(2iπ/b)) is an algebraic integer of Q(ζab) as A0(cos(2iπ/b)) = 2b−1 ∏b−1

i=1 (cos(2iπ/b) − ai);

thus x = b ∏b−1

l=1 (2 − 2al)ϑ = a

( 1 + ∑b−1

i=1 A0

( cos ( 2iπ

b

))) is an algebraic integer of Q(ζab).

• A. Pêcher (Univ. Bordeaux)

Injectivity Shanghai, 2016 18 / 27

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∆ = 0: ϑ(Kp/q) is integer, a contradiction

Hence NormQ(ζab)

Q

(x) = NormQ(ζab)

Q

( b ∏b−1

l=1 (2 − 2al)ϑ

) ∈ Z

NormQ(ζab)

Q

( b

b−1

∏

l=1

(2 − 2al)ϑ ) = NormQ(ζab)

Q

( bϑ ) NormQ(ζab)

Q

(b−1 ∏

l=1

(2 − 2al) ) = ( bk d )Φ(ab) ( NormQ(ζab)

Q

(2 − 2a1) )b−1 = ( bk d )Φ(ab) a2(b−1)φ(b)

Recall that ϑ = k/d. Since a is prime, if l is a prime factor of d distinct of a then l divides b. Likewise l divides b + 1. Thus l = 1 and so d is a power of a. Since a > d, we have d = a0 = 1. Therefore ϑ = ϑ(Ka/b) is an integer, and this implies that a/b is an integer, contradiction: ∆ ̸= 0.

• A. Pêcher (Univ. Bordeaux)

Injectivity Shanghai, 2016 19 / 27

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Step 2: if ∆ ̸= 0 then ∆ ≥ c−n5

Recall that ∆ =

• ϑ(Kp′/q′) − ϑ(Kp/q)
• with p, p′, q, q′ ≤ n.

Let ai = cos (⌊

ip q

⌋

2π p

) , a′

i = cos

(⌊

ip′ q′

⌋

2π p′

) , and let α = qq′

q−1

∏

i=1

(2 − 2ai)

q′−1

∏

i=1

(2 − 2a′

i)∆

1

α is an algebraic integer of Q(ζpp′qq′): |Aut(Q(α))| ≤ qq′pp′ ≤ n4

2

for every σ ∈ Aut(Q(α)), |σ(α)| ≤ 4nn3 Since α ̸= 0 and α is an algebraic integer, we have

• ∏

σ∈Aut(Q(α))

σ(α)

• ≥ 1.

|α| ≥ ∏

σ∈Aut(Q(α)),σ̸=Id

1 |σ(α)| ≥ ( 1 4nn3 )n4

• A. Pêcher (Univ. Bordeaux)

Injectivity Shanghai, 2016 20 / 27

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Outline

1

Lovasz’ theta function and perfectness

2

A closed formula.

3

Separating the values.

4

Proving the closed formula.

• A. Pêcher (Univ. Bordeaux)

Proof of the formula Shanghai, 2016 21 / 27

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From a Semi-Definite Program to a Linear Program

B(G) = {B|B is SDP, tr(B) = 1 and ∀ij ∈ E(G) bij = 0} ϑ(G) = max {∑ bij|B ∈ B(G) } Kp/q circulant graph: optimal symmetric circulant matrix B Bij = b(i+j) mod p and ∀i, bi = bp−i, eigenvalues fi = ∑

j bj cos(2ijπ/p)

bi = 1

p

∑

j fj cos(2ijπ/p) (inverse discrete Fourier transform)

tr(B) = 1 ↔ ∑ fi = 1 B SDP ↔ ∀i, fi ≥ 0 and fi = fp−i ∀ij ∈ E(G) bij = 0 ↔ ∀1 ≤ i ≤ q − 1, bi = 0 ↔ ∀1 ≤ i ≤ q − 1, ∑ fj cos(2ijπ/p) = 0 ϑ(G) = max ∑ bij ↔ ϑ(G) = max pf0

• A. Pêcher (Univ. Bordeaux)

Proof of the formula Shanghai, 2016 22 / 27

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Linear programs

Main linear program ϑp/q = sup { pf0 : fi ≥ 0 ∑

i=0..p/2

fi = 1 1 ≤ j ≤ q − 1 ∑

i=0..p/2

fi cos (2ijπ/p) = 0 } Dual ϑ∗(Kp/q) = inf { pg0 : ∑

j=0..q−1

gj ≥ 1 1 ≤ i ≤ p/2 ∑

j=0..q−1

gj cos (2jiπ/p) ≥ 0 }

• A. Pêcher (Univ. Bordeaux)

Proof of the formula Shanghai, 2016 23 / 27

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Linear system of optimal solution

Computer experimentations suggest that there is an optimal solution of main linear program whose non-zero coefgicients (f0, f∗

1, . . . , f∗ q−1) satisfy:

   

1 1 1 1 1 cos (⌊

p q

⌋

2π p

) . . . cos (⌊

p q

⌋ 2(q−1)π

p

) . . . . . . . . . . . . 1 cos (⌊ (q−1)p

q

⌋

2π p

) . . . cos (⌊ (q−1)p

q

⌋ 2(q−1)π

p

)

         f0 f∗

1

. . . f∗

q−1

     =      1 . . .     

• A. Pêcher (Univ. Bordeaux)

Proof of the formula Shanghai, 2016 24 / 27

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Tchebychev polynomials

Tchebychev polynomials Tq

Let T0 = 1, T1 = x and ∀q ≥ 2, Tq = 2xTq−1 − Tq−2. ∀x, Tq(cos(x)) = cos(qx)      1 1 1 1 1 T1(a1) . . . Tq−1(a1) . . . . . . . . . . . . 1 T1(aq−1) . . . Tq−1(aq−1)     

• 

    f0 f∗

1

. . . f∗

q−1

     =      1 . . .      T We are looking for the first column of the inverse of matrix T.

• A. Pêcher (Univ. Bordeaux)

Proof of the formula Shanghai, 2016 25 / 27

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Lagrange interpolation

Lagrange interpolation polynomials Lk w.r.t. points ai

Lk(y) = ∏q−1

i=0,i̸=k(y − ai)

∏q−1

i=0,i̸=k(ak − ai)

Let λk,j be the coefgicients on the Tchebychev polynomials basis: Lk(y) = λk,0T0(y) + . . . + λk,q−1Tq−1(y) From Lk(ai) = δik, ( λi,j )

0≤i,j,≤q−1 is the inverse of the matrix T.

Candidate is indeed feasible (main difgiculty)

For every i, λi,0 ≥ 0 (difgicult) and L0 (ci) ≥ 0 (easy). Thus (λ0,0, λ1,0, . . . , λq−1,0) is a feasible solution (restricted to non-zero entries) of the main linear program ; (λ0,0, λ0,1, . . . , λ0,q−1) is a feasible solution of the dual linear program.

• A. Pêcher (Univ. Bordeaux)

Proof of the formula Shanghai, 2016 26 / 27

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The closed formula

Hence, ϑ(Kp/q) = pλ0,0 with λ0,0 given by the polynomial identity: L0(y) = λ0,0T0(y) + . . . + λ0,q−1Tq−1(y) We have

q−1

∑

i=0

Tj ( cos (2iπ q )) = 0 if j ̸= 0 = q if j = 0 Hence

q−1

∑

i=0

L0 ( cos (2iπ q )) = qλ0,0 . Together with L0(y) = A0(y)/A0(1), we have finally, ϑ(Kp/q) = p

q

( 1 +

1 A0(1)

∑q−1

i=1 A0

( cos (

2iπ q

)))

• A. Pêcher (Univ. Bordeaux)

Proof of the formula Shanghai, 2016 27 / 27

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The closed formula

Hence, ϑ(Kp/q) = pλ0,0 with λ0,0 given by the polynomial identity: L0(y) = λ0,0T0(y) + . . . + λ0,q−1Tq−1(y) We have

q−1

∑

i=0

Tj ( cos (2iπ q )) = 0 if j ̸= 0 = q if j = 0 Hence

q−1

∑

i=0

L0 ( cos (2iπ q )) = qλ0,0 . Together with L0(y) = A0(y)/A0(1), we have finally, ϑ(Kp/q) = p

q

( 1 +

1 A0(1)

∑q−1

i=1 A0

( cos (

2iπ q

))) Thanks!

• A. Pêcher (Univ. Bordeaux)

Proof of the formula Shanghai, 2016 27 / 27