the tutte polynomial graph polynomials 238900 winter 05/06

The Tutte Polynomial

Graph Polynomials 238900 winter 05/06

The Rank Generation Polynomial

Reminder

GEF

FnEkFk yxyxGS ,;

FkVFr

FkVFFn

Fk

Number of connected components in G(V,F)


Theorem 2

else

loop a is

bridge a is

,;/,;

,;1

,;1

,; e

e

yxeGSyxeGS

yxeGSy

yxeGSx

yxGS

eG

eG

/

Is the graph obtained by omitting edge e

Is the graph obtained by contracting edge e

In addition, S(En;x,y) = 1 for G(V,E) where |V|=n and |E|=0


Theorem 2 – ProofLet us define :

G’ = G – e G’’ = G / e

r’<G> = r<G’>n’<G> = n<G’>r’’<G> = r<G’’>n’’<G> = n<G’’>


Theorem 2 – Proof

eEeE

eFeF

Let us denote :


Observations

FrGr

FreEreFrEr

FnFn

FrFr

eEFEe

''''

''''

'

'

,


Observations

FrGr

FreEreFrEr

FnFn

FrFr

eEFEe

''''

''''

'

'

,

Fe


Observations

FrGr

FreEreFrEr

FnFn

FrFr

eEFEe

''''

''''

'

'

,

FkeFk

eEkEk

else

loop

VV

VV

1


Observations

FrGr

FreEreFrEr

FnFn

FrFr

eEFEe

''''

''''

'

'

,

''''' GG

3


Observations

else

bridge a is

'

1' e

eEr

eErEr

else

loop a is

''

1'' e

Fn

FneFn

kk

VV

EE

1

1

kk

VV

EE

1

1

2


GEF

FnEkFk yxyxGS ,;

Reminder

GEF

FnFrEr yx


yxGSyxGSyxGS ,;,;,; 10

FeGEF

FnFrEr yxyxGS,

0 ,;

FeGEF

FnFrEr yxyxGS,

1 ,;

Let us define


otherwise

bridge a is

,;

'

'''

'

''1'

,0

e

yx

yx

yx

yxyxGS

GEF

FnFreEr

GEF

FnFreEr

eEF

FnFrEr

FeGEF

FnFrEr

1


otherwise

bridge a is

,;

,;

otherwise

bridge a is

,;

'

'''

'

''1'

0

e

yxeGS

yxeGxS

e

yx

yx

yxGS

GEF

FnFreEr

GEF

FnFreEr


otherwise

loop a is

,;

''

''''''

''

1''''''

,1

e

yx

yx

yx

yxyxGS

GEF

FnFrGr

GEF

FnFrGr

eEF

eFneFrEr

FeGEF

FnFrEr

2

3


otherwise

loop a is

,;/

,;/

otherwise

loop a is

,;

''

''''''

''

1''''''

1

e

yxeGS

yxeGyS

e

yx

yx

yxGS

GEF

FnFrGr

GEF

FnFrGr


otherwise

loop a is

,;/

,;/ e

yxeGS

yxeGyS


otherwise

bridge a is

,;

,; e

yxeGS

yxeGxS

+



otherwise

loop a is

bridge a is

,;/,;

,;/,;

,;/,;

e

e

yxeGSyxeGS

yxeGySyxeGS

yxeGSyxeGxS



otherwise

loop a is

bridge a is

,;/,;

,;,;

,;/,;

e

e

yxeGSyxeGS

yxeGySyxeGS

yxeGSyxeGxS

Obvious



otherwise

loop a is

bridge a is

,;/,;

,;1

,;/,;

e

e

yxeGSyxeGS

yxeGSy

yxeGSyxeGxS

Obvious



otherwise

loop a is

bridge a is

,;/,;

,;1

,;,;

e

e

yxeGSyxeGS

yxeGSy

yxeGSyxeGxS

FreErFreEr

FnFneEF

''''''

,'''

kkVVGG 11'''



otherwise

loop a is

bridge a is

,;/,;

,;1

,;1

e

e

yxeGSyxeGS

yxeGSy

yxeGSx

Obvious

Q.E.D


Reminder

4

5

The Universal Tutte Polynomial

yx

GT

yxGU

GrGnGk ,;

,,,,;


otherwise

loop a is

bridge a is

,;/,;

,;

,;

,,,,;

e

e

yxeGUyxeGU

yxeGUy

yxeGUx

yxGU

Theorem 6

nn yxEU ,,,,;


Theorem 6 - Proof

nn

nVVVVV

n

yxET

yxEU

111

,;

,,,,;

0

5


1

1

bridge a is

GreGr

GneGn

GkeGk

e

If e is a bridge

yx

eGTx

yxGU

eGreGneGk ,;

,,,,;

11

4


If e is a bridge

,,,,;

,;

,;

,,,,;

11

yxeGxU

yxeGxT

yxeGT

x

yxGU

eGreGneGk

eGreGneGk


GreGr

GneGn

GkeGk

e 1loop a is

If e is a loop

yx

eGTy

yxGU

eGreGneGk ,;

,,,,;

1

4


If e is a bridge

,,,,;

,;

,;

,,,,;

1

yxeGyU

yxeGyT

yxeGT

y

yxGU

eGreGneGk

eGreGneGk


1/

/

1

/

loop a is

bridge a is

GreGr

GreGr

GneGn

GneGn

GkeGkeGk

e

e

If e is neither a loop nor a bridge



4

yxeGT

yxeGT

yxGUGrGnGk

,;/,;

,,,,;



yxeGT

yxeGT

yxGU

eGreGneGk

eGreGneGk

,;/

,;

,,,,;

1///

1



,,,,;/

,,,,;

,,,,;

yxeGU

yxeGU

yxGU


otherwise

loop a is

bridge a is

,;/,;

,;

,;

,,,,;

e

e

yxeGUyxeGU

yxeGUy

yxeGUx

yxGU

Theorem 6 - Proof

Q.E.D


otherwise

loop a is

bridge a is

,;/,;

,;

,;

,,,,;..,,,,;

e

e

yxeGVyxeGV

yxeGVy

yxeGVx

yxGVtsyxGV

Theorem 7

nn yxEV ,,,,;

,,,,;,,,,; yxGUyxGV then


otherwise

loop a is

bridge a is

,;/,;

,;

,;

,,,,;

e

e

yxeGVyxeGV

yxeGVy

yxeGVx

yxGV

Theorem 7 – proof

V(G) is dependant entirely on V(G-e) and V(G/e).In addition – ,,,,;,,,,; yxEUyxEV nn

Q.E.D

Evaluations of the Tutte Polynomial

Proposition 8

1,1,GT

Let G be a connected graph.

is the number of spanning trees of G

Shown in the previous lecture


Proposition 8

2,1,GT


is the number of connected

spanning sub-graphs of G

GEF

FnEkFkGSGT 101,0;2,1;

EkFkGEFEkFkGEF

Fn 11


Proposition 8

2,1,GT


is the number of connected

spanning sub-graphs of G

EkFkGEF

GT 12,1;

sub-graphs of G connected


Proposition 8

1,2,GT


is the number of (edge sets forming)

spanning forests of G

GEF


0

1FnGEF Spanning forests


Proposition 8

2,2,GT


is the number of spanning

sub-graphs of G

GEF


GEF

1


Theorem 9

0,1;1; xGTxxG GkGr

The chromatic polynomial and the Tutte polynomial are related by the equation :


Theorem 9 – proof

1,1,,0,

1,; xx

xGUxG

Claim :


Theorem 9 – proof

We must show that :

1,1,,0,

1,; xx

xEUxE nn


Theorem 9 – proofand :

otherwise

loop a is

bridge a is

1;/

1;

0

1;

1

1;

e

e

x

xeG

x

xeG

x

xeG

x

x

x

xG


Theorem 9 – proof

1,1,,0,

1,; xx

xEUxE nn

nn xxE ; Obvious. (empty graphs)

nnn xxx

xEU

1,1,,0,1

,


Theorem 9 – proof

0;loop xGe Obvious

xeGxeGxGe ;/;;loopbridge

Chromatic polynomial property

xeGx

xxGe ;

1;bridge

Chromatic polynomial property


Theorem 9 – proofThus :

otherwise

loop a is

bridge a is

1;/

1;

0

1;

1

1;

e

e

x

xeG

x

xeG

x

xeG

x

x

x

xG


Theorem 9 – proof

And so, due to the uniqueness we shownin Theorem 7 :

1,1,,0,

1,; xx

xGUxG


Theorem 9 – proof

Remembering that :

yx

GT

yxGU

GrGnGk ,;

,,,,;


Theorem 9 – proof

Remembering that :

Q.E.D

0,1,1

0,1,11

1,1,,0,1

,;

xGTx

xGTx

xx

xGUxG

GrGk

GrGnGk


0,2,GT


is the number of acyclic orientations

of G.

We know that (Theorem 9) –

0,1;1; xGTxxG GkGr


0,11,0,2, GTGTSo –

And thus –

1;11;1

1

1;

1

;0,2,

GG

G

x

xGGT

VV

VGkGrGkGr

Acyclic Orientations of Graphs

Let G be a connected graph without loops or multiple edges.

An orientation of a graph is received afterassigning a direction to each edge.

An orientation of a graph is acyclic if it doesnot contain any directed cycles.


xG, is the number of pairs where

is a map

Proposition 1.1

,

xV ,...,3,2,1: and isan orientation of G, subject to the following :

• The orientation is acyclic• vuvu


Proof

The second condition forces the map to be aproper coloring.

The second condition is immediately impliedfrom the first one.


Proof

Conversely, if the map is proper, than thesecond condition defines a unique acyclicorientation of G.

Hence, the number of allowed mappings issimply the number of proper coloring with xcolors, which is by definition xG,


xG,~ be the number of pairs where

is a map

,

xV ,...,3,2,1: and isan orientation of G, subject to the following :

• The orientation is acyclic• vuvu


xGxGNx V ,1,~

Theorem 1.2

0,2,GT


Proof

The chromatic polynomial is uniquelydetermined by the following :

xxG ,0 G0 is the one vertex graph

xHxGxHG ,,, Disjoint union

xeGxeGxG ,/,,


Proof

We now have to show for the new polynomial :

xxG ,~0 G0 is the one vertex graph

xHxGxHG ,~,~,~ Disjoint union

xeGxeGxG ,/~,~,~

Obvious

Obvious


Proof

We need to show that :

xeGxeGxG ,/~,~,~

xeGV ,...,3,2,1: Let :


Proof

xeGV ,...,3,2,1: Let :

vue ,

Let be an acyclic orientation of G-ecompatible with

Let :


Proof

Let be an orientation of G after adding{u v} to

1

Let be an orientation of G after adding{v u} to

2


Proof

We will show that for each pair exactly one of the orientations is acyclic and compatible with ,expect forof them, in which case both are acyclic orientations compatible with

, 21,

21,

xeG ,/~


Proof

Once this is done, we will know that –

due to the definition of

xeGxeGxG ,/~,~,~ xG,~


Proof

For each pair where -

and is an acyclic orientation compatiblewith one of these three scenarios musthold :

,

xeGV ,...,3,2,1:


Proof

vu Case 1 –

Clearly is not compatible with whileis compatible. Moreover, is acyclic :

2 1

1uwwvu 21

uwwvu 21

Impossible cicle


Proof

vu Case 2 –

Clearly is not compatible with whileis compatible. Moreover, is acyclic :

1 2

2vwwuv 21

vwwuv 21

Impossible cicle


Proof

vu Case 3 –

Both are compatible with At least one is also acyclic. Suppose not, then:

vwwuv '2'1

uwwvu 21

21 contains

contains


Proof

vwwuv '2'1

uwwvu 21

21 contains

contains

uwvwu 1'1

contains

Impossible cicle


ProofWe now have to show that both andare acyclic for exactly pairs of with vu

21 xeG ,/~ ,

Let z denote the vertex identifying {u,v} ineG /


ProofZeG /

some acyclic orientation compatible with

uv

impossible to add a circle by the new edge {u,v}

G

two acyclic orientations, compatible with


ProofZeG /

exactly one, necessarily acyclic, compatible with

uv

Some two acyclic orientations, compatible with

G

All other vertices of G remains the same


ProofAnd so both and are acyclic for exactly pairs of with

And so –

vu

21 xeG ,/~ ,

xeGxeGxG ,/~,~,~


ProofIt is obvious that for x = 1 every orientationis compatible with

And so the expression count the number ofacyclic orientations in G

Q.E.D

1: V

the tutte polynomial graph polynomials 238900 winter 05/06

Documents