The circular chromatic number of circular-perfect graphs is polytime

The circular chromatic number of circular-perfect graphs
is polytime
Arnaud Pêcher
joint work with C. Bachoc and A. Thiery
Univ. Bordeaux (LaBRI)
supported by the ANR/NSC project GraTel ANR-09-blan-0373-01,
NSC98-2115-M-002-013-MY3 and NSC99-2923-M-110-001-MY3
2011 Worshop on Graph Theory
Taipei, March 11th, 2011
A. Pêcher (Univ. Bordeaux)
Bordeaux, 2011
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Outline
1
A closed formula.
2
Separating the values.
3
Proving the closed formula.
4
This is the end
A. Pêcher (Univ. Bordeaux)
A closed formula
Bordeaux, 2011
2 / 28
Some properties of Lovasz's Theta function ϑ
Lovasz'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 )
if
G
is homomorphic to
A. Pêcher (Univ. Bordeaux)
H
then ϑ(G ) ≤ ϑ(H ).
A closed formula
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Powers of chordless cycles
q
p =
C
q
th
power of the chordless cycle Cp with
C9
A. Pêcher (Univ. Bordeaux)
2
C9
A closed formula
p
vertices
3
C9
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Complements = circular-cliques
Circular-Clique Kp/q : vertices {0, 1, · · · , p − 1} and edges ij , s.t.
q −1 = K .
q ≤ |i − j | ≤ p − q . Hence Cp
p/q
K9/1
K9/2
K9/2
A. Pêcher (Univ. Bordeaux)
= C9
K9/3
K9/3
A closed formula
= C92
K9/4
K9/4
= C93
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From Perfect to Circular-perfect graphs
Circular chromatic number
Circular clique number
(Vince, 1988)
χc (G ) = inf
k
(Zhu, 2000)
/d |G → Kk /d
ωc (G ) = sup
χ(G ) = dχc (G )e
k
/d |Kk /d → G
ω(G ) = bωc (G )c
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 :
If
G
bipartite graphs, chordal graphs, comparability graphs ...
is perfect then ϑ(G ) = ϑ
A. Pêcher (Univ. Bordeaux)
p where χ(G ) = p .
K
A closed formula
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From Perfect to Circular-perfect graphs
Circular chromatic number
Circular clique number
(Vince, 1988)
χc (G ) = inf
k
(Zhu, 2000)
/d |G → Kk /d
ωc (G ) = sup
χ(G ) = dχc (G )e
k
/d |Kk /d → G
ω(G ) = bωc (G )c
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 :
If
G
perfect graphs, circular-cliques, outerplanar graphs...
is circular-perfect then ϑ(G ) = ϑ
A. Pêcher (Univ. Bordeaux)
p/q where χc (G ) = p /q .
K
A closed formula
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Basics of polytime algorithm to compute χ of perfect graphs
For a perfect graph
G
, χ(G ) is polytime as ϑ(G ) = χ(G ).
This may be restated as this polytime algorithm to compute χ(G ) for a
perfect graph G with n vertices :
(1) compute ϑ(G ) with precision = 0.5 ;
(2) for every 1 ≤ p ≤ n, if ϑ(G ) = ϑ(Kp ), return p .
Correct as there is a unique
A. Pêcher (Univ. Bordeaux)
p
satisfying (2) and has polyspace encoding.
A closed formula
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Core of algorithm to compute χc of circular-perfect graphs
For circular
√ perfect graphs, ϑ(G ) 6= χc (G ) in general :
5 < χc (C5 ) = 2.5. Hence we can not get χc (G ) directly from
ϑ(C5 ) =
ϑ(G ).
But ... to compute χc (G ) for a circular-perfect graph
polynomial time :
G
with
n
vertices in
(1) compute ϑ(G ) for some precision > 0 ;
(2) for every 1 ≤ p , q ≤ n ((p , q ) = 1, 2q ≤ p ), if |ϑ(G ) − ϑ(Kp/q )| < ,
return p /q .
Correct provided there is a unique pair (p , q ) satisfying (2) and has
polyspace encoding.
Hence, roughly speaking, we need to prove that
0
∃ > 0 such that for every pq 6= qp0 , we have ϑ(Kp/q ) − ϑ(Kp0 /q0 ) ≥ .
A. Pêcher (Univ. Bordeaux)
A closed formula
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q = 2, q = 3 and p odd
Already solved cases :
Theorem - Lovász (1978) - q=2,
ϑ
p
odd
p/2 = ϑ(Cp ) =
p
K
cos πp
1 + cos πp
Proofs :
Lovász : algebraic arguments
Knuth (1994) : linear program with two variables
Theorem - Brimkov et al (2000) - q=3,
p
odd

p c + 1
2π p
2π
−
cos
b
c
−
cos
b
p
3
p
3

ϑ Kp/3 = p 1 − cos 2pπ b p3 c − 1 cos 2pπ b p3 c + 1 − 1

Proof :
1
2
linear program with 3 variables + geometrical arguments
A. Pêcher (Univ. Bordeaux)
A closed formula
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A closed formula
∀0 ≤ k ≤ q − 1, ak = cos
kp
2π
q
p
qY
−1
q
−1
A0 (x )
= 2
(x − ai )
i =1
Theorem (Bachoc, P., Thiery 2010)
ϑ(Kp/q ) =
A. Pêcher (Univ. Bordeaux)
p
q
1+
1
A0
(1)
q−1
X
i =1
A0
A closed formula
cos
2i π
!
q
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Outline
1
A closed formula.
2
Separating the values.
3
Proving the closed formula.
4
This is the end
A. Pêcher (Univ. Bordeaux)
Injectivity
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11 / 28
The main result
0
Let p , p 0 , q , q 0 ≤ n such that qp 6= qp0 .
Let ∆ = ϑ(Kp0 /q0 ) − ϑ(Kp/q ).
Theorem (Bachoc, P., Thiery (2011)
∆ ≥ c −n for some
5
c
>0
Hence, computing χc of circular-perfect graphs is polytime.
The proof is in two steps :
(1) ∆ 6= 0
5
(2) if ∆ 6= 0 then ∆ ≥ c −n for some
A. Pêcher (Univ. Bordeaux)
c
Injectivity
>0
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∆ = 0 : taking advantage of monoticity
We have
p
q
≤
p
q
0
0
if and only if
K
p/q → Kp0 /q0 (Bondy & Hell '96)
if and only if
ϑ(Kp/q ) ≤ ϑ(Kp0 /q0 )
0
Assume ∆ = 0 : ϑ(Kp/q ) = ϑ(Kp0 /q0 ) = ϑ for some pq < qp0 .
i
h
0
Hence for every ba ∈ pq , qp0 , ϑ(Ka/b ) = ϑ.
0
Take ba ∈ qp , pq0 such that
b is prime ;
b is coprime with a and a + 1.
h
i
A. Pêcher (Univ. Bordeaux)
Injectivity
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A avour of algebraic number theory (1/2)
Notations and denitions :
for every k , let ζk = exp2i π/k ;
let Φ be the Euler function ;
let Q (ζk ) denote the cylotomic eld : the smallest complex eld
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
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A avour of algebraic number theory (2/2)
Notations :
if K and
L
are two cyclotomic elds s.t.
L
is an extension of
L/K )=Aut(X
L/K ) be the Galois group of L over K ;
L
TraceK (x )=
σ(x ) be the trace of any element x
Gal(
L
σ∈Gal(L/K )
x )=
NormK (
Y
σ∈Gal(L/K )
σ(x )
be the norm of any element
x
K
, let
of
L;
of
L.
let σi be the automorphism of Q (ζk ) s.t. σi (ζk ) = ζki (i coprime).
Gal(Q (ζk )) = {σi , (i , k ) = 1} ;
TraceLK is linear and for every element
for every
x
∈ Q (ζk ), polmin(x ) =
x
of L, TraceLK (x ) ∈ K ;
(X − σ(x )).
Y
σ∈Gal(Q(ζk )/Q)
A. Pêcher (Univ. Bordeaux)
Injectivity
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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(
2i π/d ) as
d
−i
1
i
cos(2i π/d ) = 2 (ζd + ζd ) = σi 12 (ζd + ζdd −1 ) ;
hence (setting
( ) = A0 (X )/A0 (1)) :
L0 X
ϑ(Kk /d ) =
=
=
!
k
1+
d
1+
d
d
i =1...d −1
L0
(σi (cos(2π/d )))
!
k
k
X
X
i =1...d −1
σi (L0 (cos(2π/d )))
kd )
1 + TraceQ(ζ
Q(ζk ) (L0 (cos(2π/d ))
thus ϑ(Kk /d ) ∈ Q(ζk )
A. Pêcher (Univ. Bordeaux)
Injectivity
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∆ = 0 : ϑ(Kp/q ) is rational
Recall that we have chosen a and b such that ϑ = ϑ(Ka/b ),
(a, b) = 1 and (a + 1, b) = 1.
Then ϑ ∈ Q(ζa ) and likewise ϑ ∈ Q(ζa+1 ). Thus ϑ ∈ Q !
b
is prime,
0
Now take ba ∈ qp , qp0 such that a/b is not integer, a is prime, (a, b) = 1 ,
(a, b + 1) = 1 and a/(b + 1) ≥ p /q .
∀i , 2 cos(2i π/a) and ai = 2 cos(2i π/b) are algebraic integers of
Q(ζab ) ;
hence ∀i , A0 (cos(2i π/b))
Qbis−1an algebraic integer of Q(ζab ) as
b
−
1
A0 (cos(2i π/b )) = 2
(2i π/b) − ai ) ;
i =1 (cos
Qb −1
P
thus x = b l =1 (2 − 2al )ϑ = a 1 + bi =−11 A0 cos 2bi π
is an
algebraic integer of Q(ζab ).
h
A. Pêcher (Univ. Bordeaux)
i
Injectivity
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∆ = 0 : ϑ(Kp/q ) is integer, a contradiction
Let ϑ = k /d .
ab) (x ) = NormQ(ζ ab)
Hence NormQ(ζ
Q
Q
Q(ζ ab )
NormQ
b
bY
−1
l =1
(2 − 2al )ϑ
!
=
Q(ζ ab )
NormQ
=
Q
b−1 (2 − 2a )ϑ ∈ Z
b
l
l =1
b dk
Φ(ab)
bϑ
Q(ζ ab )
NormQ
bY
−1
l =1
(2 − 2al )
!
a2(b−1)φ(b)
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. Taking a big
enough, we must have d = a0 = 1.
Therefore ϑ = ϑ(Ka/b ) is an integer, and a result of Grötschel et. al implies
that a/b is an integer, a nal contradiction : ∆ 6= 0.
A. Pêcher (Univ. Bordeaux)
Injectivity
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∆ ≥ c −n
5
0
0
Recall that ∆j= kϑ(K
p0 /q0 ) − ϑ(Kj
p/q )kwith p , p , q , q ≤ n.
0
Let ai = cos ipq 2pπ , ai0 = cos ipq0 2pπ0 , and let
α = qq
1
2
3
0
qY
−1
0 −1
qY
i =1
i =1
(2 − 2ai )
(2 − 2ai0 )∆
α is an algebraic integer of Q(ζpp0 qq0 )
deg(polmin(α)) ≤ qq 0 pp 0 ≤ n4
for every σ ∈ Gal(Q(α)), |σ(α)| ≤ 4n n3
Y
Since α 6= 0 and α is an algebraic integer, we have σ(α) ≥ 1.
σ∈Aut(Q(α))
And the result follows from
|α| ≥
1
Y
σ∈Aut(Q(α)),σ6=Id
A. Pêcher (Univ. Bordeaux)
|σ(α)|
Injectivity
≥
4
1 n
4n n 3
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Outline
1
A closed formula.
2
Separating the values.
3
Proving the closed formula.
4
This is the end
A. Pêcher (Univ. Bordeaux)
Proof of the formula
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20 / 28
From a Semi-Denite Program to a Linear Program
B(G ) = {B |B is SDP, tr(B ) = 1 and ∀ij ∈ E (G ) bij = 0}
nX
o
ϑ(G ) = max
bij |B ∈ B(G )
ϑp/q circulant graph : optimal symmetric circulant matrix
B
ij = b(i +j ) mod p and ∀i , bi = bp−i ,
P
eigenvalues fi = j bj cos(2ij π/p )
P
1
bi =
p j fj cos(2ij π/p ) (inverse discrete Fourier transform)
B
tr(B ) = 1 ↔
i =1
B SDP
↔ ∀i , fi ≥ 0 and fi = fp−i
∀ij ∈ E (G ) bij = 0 ↔ ∀1 ≤ i ≤ q − 1, bi = 0
X
f
↔ ∀1 ≤ i ≤ q − 1,
ϑ(G ) = max
X
A. Pêcher (Univ. Bordeaux)
X
j cos(2ij π/p ) = 0
f
ij ↔ ϑ(G ) = max pf0
b
Proof of the formula
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Linear programs
Main linear program
ϑp/q = sup
i ≥0
:
pf0
f
X
i =0..p/2
1≤j ≤q−1
X
i =0..p/2
i =1
f
i cos (2ij π/p ) = 0
f
Dual
ϑ∗ (Kp/q ) = inf
:
pg0
1 ≤ i ≤ p /2
A. Pêcher (Univ. Bordeaux)
X
j =0..q−1
X
j =0..q−1
Proof of the formula
g
j ≥1
g
j cos (2ji π/p ) ≥ 0
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Linear system of optimal solution
suggest that there is an optimal solution of
main linear program whose non-zero coecients (f0 , f1∗ , . . . , fq∗−1 ) satisfy :
Computer experimentations




1
1
cos
j 1 k
.
.
.
1
p
q
.
.
.
cos
j
1
2π
(q −1)p
q
...
p
k
cos
j k1
.
.
.
2π
p
A. Pêcher (Univ. Bordeaux)
...
cos
j
p
q
2(q −1)π
q
p
.
.
.
(q −1)p
k
2(q −1)π
p
Proof of the formula






f0
∗
f1




 = 


..
.
∗

q−1
f
Bordeaux, 2011
1
0
..
.
0





23 / 28
Tchebychev polynomials
Tchebychev polynomials
Let
T0
= 1,
T1
Tq
= x and ∀q ≥ 2,
T
q = 2xTq−1 − Tq−2 .
∀x , Tq (cos(x )) = cos(qx )

1
1
..
.
1
1
1
 




q−1 (a1 )
..
..
..
.
.
.
1 T1 (aq−1 ) . . . Tq−1 (aq−1 )




|
{z
}
( )
T1 a1
...
T
f0
∗
f1




..
.
∗
q −1
f




=





1
0
..
.
0





T
A. Pêcher (Univ. Bordeaux)
Proof of the formula
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Lagrange interpolation
Lagrange interpolation polynomials
Lk
w.r.t. points
Qq−1
ai
i =0,i 6=k (y − ai )
k (y ) = Qq−1
i =0,i 6=k (ak − ai )
L
Let λk ,j be the coecients on the Tchebychev polynomials basis :
k (y ) = λk ,0 T0 (y ) + . . . + λk ,q−1 Tq−1 (y )
L
Hence (λi ,j )0≤i ,j ,≤q−1 is the inverse of the matrix
T
.
Lemma (key point of the proof)
For every i , λi ,0 ≥ 0 and
L0
(cos(2i π/p )) ≥ 0. 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
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The closed formula
Hence, ϑ(Kp/q ) = p λ0,0 with λ0,0 given by the polynomial identity :
( ) = λ0,0 T0 (y ) + . . . + λ0,q−1 Tq−1 (y )
L0 y
Since
q−1
X
i =0
j cos
T
2i π
q
= 0 if j =
6 0
=
Together with
q
if j = 0
( ) = A0 (y )/A0 (1), we have nally,
L0 y
ϑ(Kp/q ) =
A. Pêcher (Univ. Bordeaux)
p
q
1+
1
A0
(1)
q −1
X
i =1
A0
Proof of the formula
cos
2i π
!
q
Bordeaux, 2011
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Outline
1
A closed formula.
2
Separating the values.
3
Proving the closed formula.
4
This is the end
A. Pêcher (Univ. Bordeaux)
This is the end
Bordeaux, 2011
27 / 28
A. Pêcher (Univ. Bordeaux)
This is the end
Bordeaux, 2011
28 / 28
A. Pêcher (Univ. Bordeaux)
This is the end
Bordeaux, 2011
28 / 28