Circular colouring of graphs
Xuding Zhu
Zhejiang Normal University
A distributed computation problem
V: a set of computers
D: a set of data files
For x V , D( x)  a  D : x has access to a
x ~ y : D( x)  D( y )  
/
a
b
e
d
c
If x ~ y, then x and y cannot operate at the
same time.
If x ~ y, then x and y must alternate their turns in
operation.
Schedule the operating time of the computers
efficiently.
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.
f : V  0,1,..., k  1
x operates at time t iff t  f ( x) mod k
a
0
The efficiency
is 1/3
e 2
1 b
1
d
0
c
Colour the graph with 3 colours
0
1
2
3
4
5
Operate a
machines c
b
d
e
a
c
b
d
e
At time
In general, at time t, those vertices x with
f(x)=t mod (k) operate.
1
.
The efficiency is
 (G )
Computer scientists solution
The efficiency of this scheduling is
Better than the colouring solution
2
5
If initially the keys are assigned as above,
then no computer can operate.
Once an orientation is given, then the scheduling
is 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 keys
such that the scheduling derived from such an
assignment is optimal.
Circular colouring of graphs
G=(V,E): a graph
k  1 : an integer
An k-colouring of G is
f : V  0,1,..., k  1
such that
x ~ y  f ( x)  f ( y )
0
1
1
2
0
k 3
A 3-colouring of C5
The chromatic number of G is
 (G)  mink : G has a k - colouring
G=(V,E): a graph
kr  1 :
a real
number
an
integer
A (circular)
k-colouring
r-colouring
of G is of G is
An
r ) k  1
f : V  [[000,,1,k,...,
0
1
1.5
0.5
2
A 2.5-coloring
such that x ~ y 
f ( x) 1f(|yf) ( x) 
 f ( y1) || rf (x1)  f ( y ) | k  1
The circular chromatic number of G is
 c (G)  min
inf { r: G has a circular r-colouring }
f is k-colouring of G
f is a circular k-colouring of G
Therefore for any graph G,
c (G)   (G)
0=r
p
p’
0
1
r
4
2
3
The distance between p, p’ in the circle is
| p - p'|r  min | p  p'|, r  | p  p'| 
f is a circular r-colouring if
x~y
|f(x)-f(y)|_r ≥ 1
Circular coloring method
Let r=  c (G )
Let f be a circular r-coloring of G
x operates at time k iff for some integer m
f ( x)  mr  [k , k  1)
r=4.4
4
0
0
4
15
3
3
1
2
2
1
The efficiency of this scheduling is
 c (G )
Theorem [Yeh-Zhu]
The optimal scheduling has efficiency
1
 c (G )
 c (G ) is a refinement of  (G )
 (G ) is an approximation of  c (G )
 c (G ) : the real chromatic number
A.Vince, 1988.
B.star chromatic number
 * (G)
More than 300 papers published.
“The theory of circular colorings of graphs has become
an important branch of chromatic graph theory with
many exciting results and new techniques.”
It stimulates challenging problems, leads to better
understanding of graph structure in terms of colouring
parameters.
Interesting questions for  (G )
are usually also interesting for
c (G)
There are also questions that are not
interesting for  (G ) , but interesting for  c (G)
For the study of  c (G) , one may need
to sharpen the tools used in the study of  (G )
Questions interesting for both  (G ) and
c (G)
For which rational
r,
there is a graph G with
Answer [Vince 1988]:
(cG
(G) )rr??
for any r  2.
For which rational
r,
there is a graph G with
 (G)  r ?
of girth at least g
g : any fixedinteger
Answer (Erdos classical result): all positive integers.
For which rational
r,
there is a graph G with
c (G)  r ?
of girth at least g
Answer [Zhu, 1996]:
g : any fixedinteger
all r  2.
For which rational
r,
there is a graph G with
 (G)  r ?
planar
Four Colour Theorem
2,3,4.
For which rational
r,
there is a graph G with
c (G)  r ?
planar
Four Colour Theorem implies
Answer [Moser, Zhu, 1997]:
r  [2,4].
for all r  [2,4]
For which rational
r,
there is a graph G with
 (G)  r ?
K n  minorfree
Hadwiger Conjecture
integers 2,3,, n  1
For which rational
r,
there is a graph G with
c (G)  r ?
KK4n  minorfree
Hadwiger Conjecture implies
r [2,n - 1]
8 all
for
 r  2, n  1.
r   ,3 .
 3happens
if n if5n.  4 ?
what
Answer [Hell[Liaw-Pan-Zhu,
Zhu, 2000]:
2003]:
For which rational
r,
there is a graph G with
c (G)  r ?
KK4n  minorfree
Hadwiger Conjecture implies
r [2,n - 1]
8 all
for
 r  2, n  1.
r   ,3 .
 3 if n  5.
Answer [Hell[Liaw-Pan-Zhu,
Zhu, 2000]:
2003]:
[Pan- Zhu, 2004]:
 8
for all r  2,   3.
 3
For which rational
r,
there is a graph G with
 (G)  r ?
line
Trivially: all positive integers
For which rational
r,
there is a graph G with
line
We know very little
c (G)  r ?
1
2
3
1
2
2
2
3
1
2
4
1
2
5
4
5

6
7
1
2
3
1
2
2
1
2
4
1 

 2  ,3  is a gap
2 

2
3
4
1
2
5

1
1

 2  ,2   is a gap
3
2

5
6
7
no graph has circular chromaticindex in thisinterval
What happens in the interval [3,4]?
11
then Petersen graph has  'c 
3
2
3
4
5
6
7
Theorem [Afshani-Ghandehari-Ghandehari-Hatami-Tusserkani-Zhu,2005]
 11 
 ,4  is a gap
3 
What happens in the interval [3,4]?
Maybe it will look like the interval [2,3]:
Gaps everywhere ?
2
3
NO!
4
5
6
7
Theorem [Afshani-Ghandehari-Ghandehari-Hatami-Tusserkani-Zhu,2005]
 11 
 ,4  is a gap
3 
What happens in the interval [3,4]?
2
3
4
5
6
7
Theorem [Lukot’ka-Mazak,2010]
 10 
Everyr   3,  is thecircular chromaticindex of a graph
 3
(a,b)is an Index- interval
Everyr  a, b is thecircular chromaticindex of a graph
2
3
4
5
6
7
Theorem [Lukot’ka-Mazak,2010]
 10 
10 

,  risan
 3Every
,  -isinterval
thecircular chromaticindex of a graph
 3Index
 3
 3
Theorem [Lin-Wong-Zhu,2013]
1

 n, n   is an Index - interval,for everyeven integern  4 .
6

2
3
4
5
6
7
Theorem
Theorem [Lukot’ka-Mazak,2010]
[Lin-Wong-Zhu,2013]
1   10 
 n,10

an- interval
Index
- interval,
everyodd
integer
.
Every
3, n risan
circularfor
chromatic
index
of angraph
is3,S
 is the
 3  4  3 
For the study of  c (G) , one may need
to sharpen the tools used in the study of  (G )
A powerful tool in the study of list colouring graphs is
Combinatorial Nullstellensatz
AssumeV (G)  v1, v2 ,, vn 
Give G an arbitrary orientation.
QG ( x1 , , xn ) 
 (x  x )
i
j
( vi ,v j )E
c is a propercolouringof GFind a proper colouring=
find a nonzero assignment
 QG (c(v1 ),, c(vn ))  0 to a polynomial
What is the polynomial for circular colouring?
AssumeV (G)  v1, v2 ,, vn 
Give G an arbitrary orientation.
QG ( x1 , , xn ) 
 (x  x )
i
( vi ,v j )E
j
0
3
2
1
4
Color set
5
4
3
6
2
1
7
Colors assigned to adjacent
vertices have circular distance
at least q
0
8
9
15
10
14
11
12
13
Color set
5
4
3
6
2
1
7
Colors assigned to adjacent
vertices have circular distance
at least q
1
p
The circle has perimeter
q
0
8
9
1
q
15
10
14
11
12
13
What is the polynomial for circular colouring?
AssumeV (G)  v1, v2 ,, vn 
Give G an arbitrary orientation.
QG ( x1 , , xn ) 
 (x  x )
i
( vi ,v j )E
j
5
4
3
6
j e
2
1
7
0
8
9
15
10
14
11
12
13
( 2j / p ) i
Theorem [Alon-Tarsi]
Suppose D is an orientation of G with
|EE(D)|  |OE(D)|. T henG is

D
(d  1 )  choosable
An eulerian mapping

is even if
 (e)
eE ( D )
is even
Theorem [Norine-Wong-Zhu, 2008]
Suppose D is an orientation of G with
w


is even
p ,q
( ) 
w


p ,q
( ).
is odd

D
If L is a p-list with |L(v)| d (v)( 2q  1 )  1,
thenG is L-(p,q)- colourable.
Theorem [Norine-Wong-Z, (JGT 2008)]
Suppose G is bipartite,l is a list - size assignment
such thatfor any subgraph H of G,
 (l ( x)  1) | E ( H ) | (2q  1)
xV ( H )
T henG is l-(p,q)- choosable.
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
GH
f : V (G)  V ( H )
edge-preserving
Such a mapping is a homomorphism from G to H
K4
For a G, the chromatic
number
clique number
of G is
K3
((G
G))  min
Gn KGn }.
max {n :K
G
K2
K1
A graph G is perfect if for every induced subgraph
H of G,
 (H )  (H )
p
For
 2,
q
let K p be the graph with vertex set
q
V={0, 1, …, p-1}
i~j
q | i  j | p  q.
The chain
K1 , K 2 , K3 ,..., K n, ...,
is extended to a dense chain.
K
K
K
9
2
11
3
10
3
K
G
K4
K3
5
2
K2
K1
K
The circular chromatic number
K
K
 p

 c (G)  min
inf  : G  K p 
 q
q

9
2
11
3
10
3
K
G
K4
K3
5
2
K2
K1
K
clique number
The circular chromatic
 p
K
:
G
pKG

p

(G
)

min
cc(G max  q
q
q

K
K



9
2
11
3
10
3
K
G
K4
K3
5
2
K2
K1
A graph G is perfect if for every induced subgraph
H of G,
 (H )  (H )
A graph G is circular perfect if for every induced subgraph
H of G,
c ( H )  c ( H )
A graph G is perfect if for every induced subgraph
H of G,
 (H )  (H )
A graph G is circular perfect if for every induced subgraph
H of G,
c ( H )  c ( H )
Theorem [Grotschel-Lovasz-Schrijver, 1981]
For perfect graphs, the chromatic number
is computable in polynomial time.
A graph G is perfect if for every induced subgraph
H of G,
 (H )  (H )
A graph G is circular perfect if for every induced subgraph
H of G,
c ( H )  c ( H )
Theorem
[Bachoc Pecher Thiery, 2011]
Theorem
[Grotschel-Lovasz-Schrijver,
1981]
circular
circular
For perfect graphs, the chromatic number
is computable in polynomial time.
A key step in the proof is calculating the Lovasz theta
number of circular cliques and their complements.



 G   max




BR ,B  0 

B(x,x) 1

 B2 ( x, y) : 
xV
( x , y )V

B( x, y )  0 xy  E 
V V
G, (G)  (G)   (G)
For perfectgraph G, estimating( G ) with
errorless than
1
2
determinesits chromaticnumber.
Theorem
[Bachoc Pecher Thiery, 2011]
Theorem
[Grotschel-Lovasz-Schrijver,
1981]
circular
circular
For perfect graphs, the chromatic number
is computable in polynomial time.
A key step in the proof is calculating the Lovasz theta
number of circular cliques and their complements.



 G   max




BR ,B  0 

B(x,x) 1

 B2 ( x, y) : 
xV
( x , y )V

B( x, y )  0 xy  E 
V V

)G
G
G,,
(c(G
G
((G
G
G
))
(G)))
(G)))((G
(
c
For perfectgraph G, estimating( G ) with
5
1
 (error
C 5 ) less
5 ,than
c (C
)   c (C5 ) its chromaticnumber.
2 5determines
2
Theorem
[Bachoc Pecher Thiery, 2011]
Theorem
[Grotschel-Lovasz-Schrijver,
1981]
circular
circular
For perfect graphs, the chromatic number
is computable in polynomial time.
A key step in the proof is calculating the Lovasz theta
number of circular cliques and their complements.
G circular perfect with  c (G)  k / d

 
 (G)   K k d
2d  k | V (G) |
For all (k,d), with (k,d) 1, 2d  k  n,
 
 K k d are all distinct and separatedby at least 
for some with polynomialspace encoding.
Assume d  2, k  2d , (k , d )  1.
For 0are
 n very
 d  1few
, families of graphs for which
There
the theta number
 2nisknown.  2  nk  
cn : cos
, an : cos
 d
k
 d 

k d 1 d 1  cn  as 

 K k d   
d n 0 s 1  1  as 
 

 
Theorem
[Bachoc Pecher Thiery, 2011]
Theorem
[Grotschel-Lovasz-Schrijver,
1981]
circular
circular
For perfect graphs, the chromatic number
is computable in polynomial time.
Kneser graph KG(n,k)
Vertex set = all k-subsets of {1,2,…,n}
Two vertices X,Y are adjacent iff
X Y  
12
Petersen graph
= KG(5,2)
45
35
13
25
14
23
24
15
34
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.
  ( KG(n, k ))  n  2k  2.
Kneser
Lovasz Conjecture
Theorem [1978]
[1955]
 ( KG(n, k ))  n  2k  2.
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.
  ( KG(n, k ))  n  2k  2.
Johnson-Holroyd-Stahl
Conjecture [1997]
Kneser
Lovasz
Chen Theorem
Conjecture
Theorem [2011]
[1978]
[1955]
( KG
(n(,nk, ))
nn2k2k2.2.
(
KG
k
))
c
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]
Alternative Kneser Colouring Theorem [Chen, 2011]
K5,5
Alternative Kneser Colouring Theorem [Chen, 2011]
K5,5
Alternative Kneser Colouring Theorem [Chen, 2011]
K5,5
Alternative Kneser Colouring Theorem [Chen, 2011]
1
1
2
2
3
3
4
4
5
5
K5,5

5,5
A colourfulcopyof K
A colourfulcopyof K

q ,q
Chang-Liu-Zhu,
Alternative Kneser
A simple
Colouring
proofTheorem
(2012) [Chen, 2011]
For any (n-2k+2)-colouring c of KG(n,k),
there exists a colourful copy of K  , q  n  2k  2
q,q
T heorem If  (G)  q, and everyq - colouringof G
containsa colourfulcopyof K q,q , then c (G)  q
Assume c is a circular r - colouringof G.
Need to prover  q
f ( x)  c( x) is a q - colouringof G
a ,, a , b ,, b is a colourfulcopyof K
1
q
1
q

q,q
f (ai )  f (bi )  i 1 i 1  c(ai )  i
f (a1 )  0
i  1  c(bi )  i
f (b1 )  0
qq even
odd
c(bq )  c(a1 )  r 1
rq
f (aq )  q 1
f (bq )  q 1
0  c(a1 )  c(b2 ) 1  c(a3 )  2   cc((baqq))qq11
c(a10)  c(b1 )  c(a2 ) 1  c(b3 )  2    c(bq )  q  1
f (a1 )  0
f (aq )  q 1
f (b1 )  0
f (bq )  q 1
0  c(a1 )  c(b2 ) 1  c(a3 )  2  
0  c(b1 )  c(a2 ) 1  c(b3 )  2  
Thank you!
Thank you!
Proof (Chang-Liu-Z, 2012)
The proof uses Ky Fan’s Lemma from algebraic topology.
Consider the boundary ([1,1]n ) of the
n-dimensional cube [1,1]n
which is an (n-1)-dimensional sphere.
We construct a triangulation of
as follows:
([1,1] )
The vertices of the triangulation are
1,0,1 \ 0
n
n
n
The vertices of the triangulation are
1,0,1 \ 0
n
n
A set of distinct vertices form a simplex
if the vertices can be ordered
x1 , x 2 ,, x q
so that
xij  1  xij1  1
xij  1  xij1  1
Examples:
n=2
Examples:
n=2
The vertices are
Examples:
n=2
The vertices are
(0,1)
(1,1)
Each vertex is a 0-dimensional simplex (0-simplex)
There are 8 1-simplices
Examples:
n=3
A 2-simplex
There are 48 2-simplices
Ky Fan’ Lemma (special case of the lemma needed)
Let
f : 1,0,1 \ 0   1,2,,n
n
n
be such that
(1) x, f ( x)   f ( x).
(2) 2  simplex xy, f ( x)   f ( y).
Then there is an odd number of (n-1)-simplices
whose vertices are labeled with
1,2,3,4,, (1) n
n1
Assume c is a (n-2k+2)-colouring of KG(n,k).
Associate to c a labeling of
follows:
For
x 1,0,1 \ 0
let
S x  {i : xi  0}
n
n
Sx  {i : xi  1}

x
S  {i : xi  1}
1,0,1 \ 0
n
n
as