xuding zhu zhejiang normal university
DESCRIPTION
Circular c olouring of graphs. Xuding Zhu Zhejiang Normal University. A distributed computation problem. V: a set of computers. D: a set of data files. a. b. e. c. d. 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 - PowerPoint PPT PresentationTRANSCRIPT
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
For which rational , r there is a graph G with ?)( rG
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
For which rational , r there is a graph G with ?)( rG
planar
Four Colour Theorem .4,3,2
For which rational , r there is a graph G with ?)( rGc
Answer [Moser, Zhu, 1997]: [2,4] allfor r
planar
Four Colour Theorem implies ].4,2[ r
For which rational , r there is a graph G with ?)( rG
freeminor Kn
Hadwiger Conjecture 1,,3,2 integers n
For which rational , r there is a graph G with ?)( rGc
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
For which rational , r there is a graph G with ?)( rGc
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
For which rational , r there is a graph G with ?)( rG
line
Trivially: all positive integers
For which rational , r there is a graph G with ?)( rGc
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
What happens in the interval [3,4]?
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
What happens in the interval [3,4]?
Theorem [Lukot’ka-Mazak,2010]
graph a ofindex chromaticcircular theis 3
10,3 rEvery
2 4 5 6 7 3
Theorem [Lukot’ka-Mazak,2010]
graph a ofindex chromaticcircular theis 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
Theorem [Lukot’ka-Mazak,2010]
graph a ofindex chromaticcircular theis 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
Give G an arbitrary orientation.
nvvvGV ,,,)( Assume 21
)(),,(),(
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
Give G an arbitrary orientation.
nvvvGV ,,,)( Assume 21
)(),,(),(
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 perfect if for every induced subgraph H of G, )()( HH
A graph G is circular perfect if for every induced subgraph H of G, )()( HH cc
A graph G is perfect if for every induced subgraph H of G, )()( HH
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.
A graph G is perfect if for every induced subgraph H of G, )()( HH
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
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
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
)()()(, GGGG )()()(, GGGG cc
2
5)()( ,5)( 555 CCC cc
A key step in the proof is calculating the Lovasz theta number of circular cliques and their complements.
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
Theorem [Grotschel-Lovasz-Schrijver, 1981]
For perfect graphs, the chromatic number is computable in polynomial time.
Theorem [Bachoc Pecher Thiery, 2011] circular circular
.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]
Alternative Kneser Colouring Theorem [Chen, 2011]
5,5K
Alternative Kneser Colouring Theorem [Chen, 2011]
5,5K
Alternative Kneser Colouring Theorem [Chen, 2011]
5,5K 5,5K
Alternative Kneser Colouring Theorem [Chen, 2011]
5,5K 5,5K
1
5
44
33
22
1
5
5,5 ofcopy colourfulA K
qqK , ofcopy colourfulA
Alternative Kneser Colouring Theorem [Chen, 2011]
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
)( 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