Xuding Zhu Zhejiang Normal University

Xuding ZhuXuding ZhuZhejiang NormalZhejiang Normal University University

Circular cCircular colouring olouring of graphsof graphs

A distributed computation problem

V: a set of computers

D: a set of data files

axDaxDVx toaccess has :)(,For

)()( :~ yDxDyx /

a

b

cd

e

If x ~ y, then x and y cannot operate at the same time.

If x ~ y, then x and y must alternate their turns inoperation.

Schedule the operating time of the computersefficiently.

Efficiency: proportion of computers operating on the average.

The computer time is discrete:

time 0, 1, 2, …

1: colouring solution

Colour the vertices of G with k= )(G

colours.

1,...,1,0 : kVf

kxfttx mod )( iff at time operates

a

e b

d c

Colour the graph with 3 colours

0

1

01

2

At time 0

Operatemachines

ac

1

bd

2

e

3

ac

4

bd

5

e

The efficiency is 1/3

In general, at time t, those vertices x with

f(x)=t mod (k) operate.

The efficiency is .)(

1G

Computer scientists solution

The efficiency of this scheduling is 52

Better than the colouring solution

If initially the keys are assigned as above, then no computer can operate.

Once an orientation is given, then the schedulingis determined.

The problem of calculating the efficiency (for a given orientation) is equivalent to a well studied problem in computer science:

the minimum cycle mean problem

Theorem [Barbosa et al. 1989]

There is an initial assignment of keyssuch that the scheduling derived from such anassignment is optimal.

Circular colouring of graphs

1,...,1,0: kVf

)()(~ yfxfyx

G=(V,E): a graph

an integer:1k

3k

An k-colouring of G is

0

1

20

1

A 3-colouring of 5Csuch that

The chromatic number of G is

colouring-k a has G :kmin )( G

Vf :

)()( yfxf

G=(V,E): a graph

an integer:1k

k-colouring of G is

such that yx ~

An

1|)()(|1 kyfxf

a real number

A (circular)

1,...,1,0 k),0[ k

0

1

20.5

1.5

A 2.5-coloring

1r

r-colouring of G is

),0[ r

1|)()(|1 ryfxf

The circular chromatic number of G is

)(Gc { r: G has a circular r-colouring }infmin

f is k-colouring of G

Therefore for any graph G,

)()( GGc

f is a circular k-colouring of G

0=r

3

1

24

x~y |f(x)-f(y)|_r ≥ 1

The distance between p, p’ in the circle is

|'| |,'| min|p'-p| r pprpp

f is a circular r-colouring if

0 r

pp’

Circular coloring method

Let r= )(Gc

Let f be a circular r-coloring of G

x operates at time k iff for some integer m

)1,[)( kkmrxf

0

1

2

3

4

r=4.4

0

1

2

3

4

5

The efficiency of this scheduling is )(

1Gc

Theorem [Yeh-Zhu]

The optimal scheduling has efficiency

)(1Gc

)(Gc )(Gis a refinement of

)(G )(Gcis an approximation of

)(Gc : the real chromatic number

A.Vince, 1988. B.star chromatic number )(* G

More than 300 papers published.

It stimulates challenging problems, leads to better understanding of graph structure in terms of colouring parameters.

“The theory of circular colorings of graphs has become an important branch of chromatic graph theory with many exciting results and new techniques.”

Interesting questions for )(G

are usually also interesting for )(Gc

There are also questions that are not interesting for , but interesting for )(G )(Gc

For the study of , one may needto sharpen the tools used in the study of

)(Gc)(G

Questions interesting for both and )(G )(Gc

For which rational , r there is a graph G with ?)( rG

Answer [Vince 1988]: .2any for r

?)( rGc


gleast at girth of integer fixedany :g

Answer (Erdos classical result): all positive integers.

For which rational , r there is a graph G with ?)( rGc

Answer [Zhu, 1996]: .2 all r

gleast at girth of integer fixedany :g


planar

Four Colour Theorem .4,3,2


Answer [Moser, Zhu, 1997]: [2,4] allfor r

planar

Four Colour Theorem implies ].4,2[ r


freeminor Kn

Hadwiger Conjecture 1,,3,2 integers n


Answer [Liaw-Pan-Zhu, 2003]: .1,2 allfor nr

freeminor Kn

.5 if n

Hadwiger Conjecture implies 1]-n[2, r

?4 if happenswhat n

4K

Answer [Hell- Zhu, 2000]: .3,

3

8

r


Answer [Liaw-Pan-Zhu, 2003]: .1,2 allfor nr

freeminor Kn

.5 if n

Hadwiger Conjecture implies 1]-n[2, r

4K

Answer [Hell- Zhu, 2000]: .3,

3

8

r

[Pan- Zhu, 2004]: .33

8,2 allfor

r


line

Trivially: all positive integers


line

We know very little

2 4 5 6 7 3

2

12

3

12

4

12

5

12

2 4 5 6 7 3

2

12

3

12

4

12

5

12

gap a is 3,2

12

interval in thisindex chromaticcircular hasgraph no

gap a is 2

12,

3

12

2 4 5 6 7 3

What happens in the interval [3,4]?

3

11' hasgraph Petersen then c

Theorem [Afshani-Ghandehari-Ghandehari-Hatami-Tusserkani-Zhu,2005]

gap a is 4,3

11

2 4 5 6 7 3


Theorem [Afshani-Ghandehari-Ghandehari-Hatami-Tusserkani-Zhu,2005]

gap a is 4,3

11

Maybe it will look like the interval [2,3]:

Gaps everywhere ? NO!

2 4 5 6 7 3


Theorem [Lukot’ka-Mazak,2010]

graph a ofindex chromaticcircular theis 3

10,3 rEvery

2 4 5 6 7 3



10,3 rEvery

interval-Indexan is(a,b)

graph a ofindex chromaticcircular theis , rEvery ba

interval-Indexan is3

10,3

2 4 5 6 7 3



10,3 rEvery

interval-San is

3

10,3

Theorem [Lin-Wong-Zhu,2013]

.n integer oddevery for interval,-Indexan is4

1, nn

Theorem [Lin-Wong-Zhu,2013]

. 4ninteger even every for interval,-Indexan is6

1,

nn

For the study of , one may needto sharpen the tools used in the study of

)(Gc)(G

A powerful tool in the study of list colouring graphs is

Combinatorial Nullstellensatz

Give G an arbitrary orientation.

nvvvGV ,,,)( Assume 21

)(),,(),(

1 jEvv

inG xxxxQji

0))(,),((

G of colouringproper a is

1 nG vcvcQ

c

Find a proper colouring= find a nonzero assignment to a polynomial



)(),,(),(

1 jEvv

inG xxxxQji

What is the polynomial for circular colouring?

0

2

41

3

0

1

2

34

5

6

7

8

9

10

11 12 13

14

15

Color set

Colors assigned to adjacent vertices have circular distanceat least q

0

1

2

34

5

6

7

8

9

10

11 12 13

14

15

Color set

Colors assigned to adjacent vertices have circular distanceat least q

q1

The circle has perimeter qp

1



)(),,(),(

1 jEvv

inG xxxxQji

What is the polynomial for circular colouring?

0

1

2

34

5

6

7

8

9

10

11 12 13

14

15

ipjej )/2(

Theorem [Alon-Tarsi]

choosable1

is Then .

with ofn orientatioan is Suppose

)(d

G|OE(D)||EE(D)|

GD

D

An eulerian mapping is even if is even )(

)(DEe

e

Theorem [Norine-Wong-Zhu, 2008]

.colourable- is then

,112 with a is If

).()(

with ofn orientatioan is Suppose

odd is ,

even is ,

L-(p,q)G

)q(v)(d |L(v)|p-listL

w w

GD

D

qpqp

Theorem [Norine-Wong-Z, (JGT 2008)]

choosable.- is Then

)12(|)(|)1)((

, of subgraph any for such that

assignment size-list a is bipartite, is Suppose

)(

l-(p,q)G

qHExl

GH

l G

HVx

q=1 case was proved by Alon-Tarsi in 1992.

Corollary[Norine]: Even cycle are circular 2-choosable.

The only known proof uses combinatorial nullstellensatz

Circular perfect graphs

HG

)()(: HVGVf edge-preserving

Such a mapping is a homomorphism from G to H

G

1K

2K

3K

4KFor a G, the chromatic number of G is

}. :{min )( nKGnG

clique number

)(G GKn max

A graph G is perfect if for every induced subgraph H of G, )()( HH

For ,2qp

let q

pK be the graph with vertex set

V={0, 1, …, p-1}

i~j .|| qpjiq

The chain

...,,...,,, ,321 nKKKK

is extended to a dense chain.

G

1K

2K

3K

4K

25K

311K

310K

29K

G

1K

2K

3K

4K

25K

311K

310K

29K

The circular chromatic number

)(Gc inf

q

pKGqp

:min

G

1K

2K

3K

4K

25K

311K

310K

29K

The circular chromatic number

)(Gc

:q

pKGqp

min

clique

)(Gc maxGK

q

p


A graph G is circular perfect if for every induced subgraph H of G, )()( HH cc



Theorem [Grotschel-Lovasz-Schrijver, 1981]

For perfect graphs, the chromatic number is computable in polynomial time.





Theorem [Bachoc Pecher Thiery, 2011] circular circular




A key step in the proof is calculating the Lovasz theta number of circular cliques and their complements.

ExyyxB

B(x,x)yxB

BRB

GVxVyx

VV

0),(

1 :),(

0,

max2),(

)()()(, GGGG

number. chromatic its determines than lesserror

with estimating ,graph perfect For

21

)G(G





ExyyxB

B(x,x)yxB

BRB

GVxVyx

VV

0),(

1 :),(

0,

max2),(

)()()(, GGGG

number. chromatic its determines than lesserror

with estimating ,graph perfect For

21

)G(G

)()()(, GGGG )()()(, GGGG cc

2

5)()( ,5)( 555 CCC cc


d

k

c

KG

dkGG

)(

/)(th perfect wicircular

|)(|2 GVkd

,2 ,1 with , allFor nkd(k,d)(k,d)

encoding. space polynomial with somefor

least at by separated anddistinct all are

dkK




.1),(,2,2 Assume dkdkd

,10For dn

d

nk

ka

d

nc nn

2cos: ,

2cos:

1

0

1

1 1

d

n

d

s s

snd

k

a

ac

d

kK

There are very few families of graphs for which the theta number is known.

Kneser graph KG(n,k)

Vertex set = all k-subsets of {1,2,…,n}

Two vertices X,Y are adjacent iff YX

Petersen graph

= KG(5,2)

12

34

1523

45 35

25

2414

13

There is an easy (n-2k+2)-colouring of KG(n,k):

For i=1,2,…, n-2k+1,

k-subsets with minimum element i is coloured by colour i.

Other k-subsets are contained in {n-2k+2,…,n}and are coloured by colour n-2k+2.

.22)),(( knknKG

Kneser Conjecture [1955]

.22)),(( knknKG

Lovasz Theorem [1978]

There is an easy (n-2k+2)-colouring of KG(n,k):

For i=1,2,…, n-2k+1,

k-subsets with minimum element i is coloured by colour i.

Other k-subsets are contained in {n-2k+2,…,n}and are coloured by colour n-2k+2.

.22)),(( knknKG

Kneser Conjecture [1955]

.22)),(( knknKG

Lovasz Theorem [1978] Johnson-Holroyd-Stahl Conjecture [1997]

.22)),(( knknKGc

Chen Theorem [2011]

Lovasz Theorem

For any (n-2k+2)-colouring c of KG(n,k), each colour class is non-empty

Alternative Kneser Colouring Theorem [Chen, 2011]


5,5K


5,5K


5,5K 5,5K


5,5K 5,5K

1

5

44

33

22

1

5

5,5 ofcopy colourfulA K

qqK , ofcopy colourfulA


For any (n-2k+2)-colouring c of KG(n,k), there exists a colourful copy of ,

qqK , 22 knq

Chang-Liu-Zhu, A simple proof (2012)

qGK

qG

qq

)( then , ofcopy colourful a contains

G of colouring-qevery and ,)( If Theorem

c,

. of colouring-circular a is Assume Gr c

Gqxcxf of colouring- a is )()(

qqqq Kbbaa ,11 ofcopy colourful a is ,,,,,

qr prove toNeed

0)( 1 af

1)( qbf q

0)( 1 bf

1)( qaf q

iaci i )(1 ibci i )(1

2)(1)()(0 321 acbcac

2)(1)()(0 321 bcacbc

even q

1)( qbc q

1)()( 1 racbc q

qr

odd q

1)( qac q

1)( qbc q)( 1ac

1)()( ibfaf ii

0)( 1 af

1)( qbf q

0)( 1 bf

1)( qaf q

2)(1)()(0 321 acbcac

2)(1)()(0 321 bcacbc

Thank you!

Thank you!

Proof (Chang-Liu-Z, 2012)

Consider the boundary of the n-dimensional cube

)]1,1([ nn]1,1[

The proof uses Ky Fan’s Lemma from algebraic topology.

which is an (n-1)-dimensional sphere.

We construct a triangulation of as follows:

)]1,1([ n

The vertices of the triangulation are

nn 0\1,0,1

The vertices of the triangulation are

nn 0\1,0,1

A set of distinct vertices form a simplex if the vertices can be ordered

qxxx ,,, 21

so that

11 1 ij

ij xx

11 1 ij

ij xx

Examples:

n=2

Examples:

n=2

The vertices are

Examples:

n=2

The vertices are

Each vertex is a 0-dimensional simplex (0-simplex)

There are 8 1-simplices

)1,0( )1,1(

Examples:

n=3

A 2-simplex

There are 48 2-simplices

nf nn ,,2,10\1,0,1: Let be such that

).()(, )1( xfxfx

Ky Fan’ Lemma (special case of the lemma needed)

Then there is an odd number of (n-1)-simplices whose vertices are labeled with

).()(,simplex 2 )2( yfxfxy

nn 1)1(,,4,3,2,1

Assume c is a (n-2k+2)-colouring of KG(n,k).

Associate to c a labeling of asfollows:

nn 0\1,0,1

For nnx 0\1,0,1

let }0:{ ix xiS

}1:{ ix xiS

}1:{ ix xiS

Documents

Xuding Zhu Zhejiang Normal University