Inequalities Involving the Coefficients of Independence Polynomials Combinatorial and Probabilistic Inequalities Isaac Newton Institute for Mathematical

Inequalities Involving the Inequalities Involving the Coefficients of Independence Coefficients of Independence

PolynomialsPolynomials

Combinatorial and Probabilistic Inequalities Isaac Newton Institute for Mathematical

Sciences Cambridge, UK - June 23-27, 2008

Vadim E. LevitVadim E. Levit1,21,2

11Ariel University Center of Samaria, Ariel University Center of Samaria, Israel Israel

Eugen MandrescuEugen Mandrescu22

22Holon Institute of Technology, Holon Institute of Technology,

IsraelIsrael

II(G;x)(G;x) = = the independence the independence

polynomialpolynomial of graph of graph GGResults and conjectures onResults and conjectures on

II(G;x)(G;x) for some graph for some graph classes…classes…

…… some some moremore open open problemsproblems

O u t l i n eO u t l i n e

IntroductionIntroduction

Quasi-Regularizable GraphsQuasi-Regularizable Graphs

König-EgervKönig-Egervááry Graphsry Graphs

The Main InequalityThe Main Inequality

Well-Covered GraphsWell-Covered Graphs

Perfect GraphsPerfect Graphs

Corona GraphsCorona Graphs

Palindromic GraphsPalindromic Graphs

A set ofA set of pairwise non-adjacent pairwise non-adjacent verticesvertices is called ais called a stable setstable set or anor an

independent setindependent set..

(GG) = stability number is the

maximum size of a stable set of GG.

Some Some definitionsdefinitions

Ga

eb

c

d Stable sets in G : , , {a}, {a,b}, {a, b, c}, {a}, {a,b}, {a, b, c},

{a, b, c, d} , … {a, b, c, d} , …

(GG) = |{a,b,c,d,e,fa,b,c,d,e,f}| = 5

ExampleExample

IfIf sskk denotes the number of denotes the number of stable setsstable sets of size of size kk in a graphin a graph

GG withwith (G) = (G) = ,, thenthen

II((GG) = ) = II((GG;;xx) = ) = ss00 ++ ss11xx ++……++ ssxx

is called theis called the

independence polynomialindependence polynomial ofof GG..

Gutman & Harary - ‘83Harary I. Gutman, F. Harary , I. Gutman, F. Harary ,

Generalizations of matching polynomialGeneralizations of matching polynomial Utilitas Mathematica 24 (1983)Utilitas Mathematica 24 (1983)

AllAll the the stable setsstable sets ofof GG : : …… …… {a}, {b}, {c}, {d}, {e}, {f}{a}, {b}, {c}, {d}, {e}, {f} …….. .. {a, b}, {a, d}, {a, e}, {a, f}, {a, b}, {a, d}, {a, e}, {a, f}, {b, c}, {b, e}, {b, f}, {d, f}, {b, c}, {b, e}, {b, f}, {d, f}, ………… { a, b, e{ a, b, e }, { a, b, f}, { a, b, f }, { a, d, f }}, { a, d, f }

1 6 8

3

G c

b

d

a

f

e I(G) = 1 + 6x + 8x2 + 3x3

Example

There are non-isomorphic graphs with I(G) = I(H)

G H

I(G) = I(H) = 1+6x+4x2

Example

ALSO non-isomorphic trees can have the same independence

polynomial !

T1

T2

K. Dohmen, A. Ponitz, P. Tittmann, Discrete Mathematics and Theoretical Computer Science 6 (2003)

I(T1) = I(T2) = 1+10x+36x2+58x3+42x4+12x5++x6

E = E = { a, b, c, d, e, f } { a, b, c, d, e, f }

The The line graphline graph of of G = (V,E)G = (V,E) is is LG = LG = (E,(E,UU)) where where ababUU whenever the whenever the

edges edges a, ba, bEE share ashare a common vertex in G.

…… for historical reasonsfor historical reasons

G a

e

b

f

c d

{a, b, d} = matchingmatching in G {a, b, d} = stable set in LG

LGa

b

e f

cd

Example

If If GG has has nn vertices, vertices, mm edges, and edges, and mmkk

matchings ofmatchings of sizesize kk,, thenthen thethematching polynomialmatching polynomial ofof GG is is

whilewhile

is the is the positive matching polynomialpositive matching polynomial ofof GG..

knk

m

k

k xmxGM 2

0

)1();(

…… recall for historical reasonsrecall for historical reasons

I. Gutman, F. Harary,I. Gutman, F. Harary, Generalizations of matching polynomialGeneralizations of matching polynomial Utilitas Mathematica 24 (1983) Utilitas Mathematica 24 (1983)

If If GG has has nn vertices, vertices, mm edges, and edges, and mmkk

matchings ofmatchings of sizesize kk,, thenthen

km

kk xmxGM

0

);(

Independence polynomial is a generalization of the matching

polynomial, i.e., M+(G;x) = I(LG; x), where LG is the line

graph of G.

M+(G;x) = I(LG;x) = 1+6x+7x2+1x3

G

c

b

d a

f e LGc

b

d

a f

e

G = (V,E) LG = (E,U)

Example

““Clique polynomialClique polynomial””: C(G;: C(G;xx) = ) = II((HH;-;-xx),), where where HH is the is the complement of complement of G,G, GoGoldwurm & Santini - 2000wurm & Santini - 2000

Twin:Twin: - - ““Independent set polynomialIndependent set polynomial”” Hoede & Li - 1994Hoede & Li - 1994

Some “relatives” of I( ; ) :Some “relatives” of I( ; ) :

““Dependence polynomialDependence polynomial”” : D(G; : D(G;xx) = ) = II((HH;-;-xx),), where where HH is the complement of is the complement of GG

Fisher & Solow - 1990Fisher & Solow - 1990

““Clique polynomialClique polynomial””: C(G;: C(G;xx) = ) = II((HH;;xx),), where where HH is the is the complement of complement of GG

Hajiabolhassan & Mehrabadi - 1998Hajiabolhassan & Mehrabadi - 1998

““Vertex cover polynomial of a graphVertex cover polynomial of a graph”, ”, where the coefficient where the coefficient aakk is the number of vertex covers V’ of G with |V’| = k is the number of vertex covers V’ of G with |V’| = k,,

Dong, Hendy & Little - 2002Dong, Hendy & Little - 2002

Chebyshev polynomials of the first and second kind:

Connections with other polynomials:Connections with other polynomials:

Hermite polynomials:

)2

(2; 2 xHxKLI n

n

n

I. Gutman, F. Harary, Utilitas Mathematica 24 (1983)I. Gutman, F. Harary, Utilitas Mathematica 24 (1983)

G. E. Andrews, R. Askey, R. Roy, Special functions (2000)G. E. Andrews, R. Askey, R. Roy, Special functions (2000)

)2(2; 2/11)1( xTxxCI nn

n

)2(14

2; 2/11)2(

2

2

xTx

xxPI n

n

n

P(G; x, y)P(G; x, y) is equal to the number of vertex colorings is equal to the number of vertex colorings : V : V {1; 2; …, x} {1; 2; …, x} of the graph of the graph G = (V,E)G = (V,E) such that such that

for all edges for all edges uv uv E E the relations the relations(u) (u) y y and and (v) (v) y y imply imply (u) (u) (v). (v).

The generalized chromatic polynomial : P(G;x,y)P(G;x,y)

K. Dohmen, A. Ponitz, P. Tittmann, K. Dohmen, A. Ponitz, P. Tittmann, A new two-variable A new two-variable generalization of the chromatic polynomialgeneralization of the chromatic polynomial, Discrete , Discrete

Mathematics and Theoretical Computer Science 6 (2003) 69-90.Mathematics and Theoretical Computer Science 6 (2003) 69-90.

P(G; x, y)P(G; x, y) is a polynomial in variables is a polynomial in variables x, yx, y, which , which simultaneously generalizes the simultaneously generalizes the chromatic polynomialchromatic polynomial, ,

the the matching polynomialmatching polynomial, and the , and the independenceindependencepolynomialpolynomial of of GG, e.g., , e.g., II(G; x) = P(G; x + 1, 1).(G; x) = P(G; x + 1, 1).

Connections with other polynomials:Connections with other polynomials:

RReemmararkk

How to compute the independence How to compute the independence polynomial ?polynomial ?

where G+H = (V(G)V(H);E) E = E(G)E(H){uv:uV(G),vV(H)}

2.I(G+H) = I(G) + I(H) – 1

If V(G)V(H) = , then

1. I(GH) = I(G) I(H)

GH = disjoint union of G and H

G+H = Zykov sum of G and H


I(K3+K4) = I(K3) + I(K4) – 1 = = (1+3x) + (1+4x) – 1 = 1+7x

K3 K4+EExxaammppllee

How to compute the independence How to compute the independence polynomial ?polynomial ?The The coronacorona of the graphs of the graphs GG andand HH is is

the graph the graph GG○○HH obtained from obtained from GG and and n = n = |V(G)||V(G)| copies of copies of HH, so that each vertex , so that each vertex of of GG is joined to all vertices of a copy is joined to all vertices of a copy

of of HH..

G

HH GG○○HH

GG

HH HH HH Example

I(GG○○H;H;xx) = (I(H;H;xx))n I(G; x x / I(H;xx))

I. Gutman, Publications de lI. Gutman, Publications de l’’Institute Mathematique 52 (1992)Institute Mathematique 52 (1992)

TheoreTheoremm

G

HH

GG○○HH I(G) = 1+3x+x2

I(H) = 1+2x

I(GG○○H;H;x) = (1+2x)3 I(G;x / (1+2x)) = = 1 + 9x + 25x2 + 22x3

Example

IfIf G = (V,E),G = (V,E), vvVV andand uvuv EE, then , then the following assertions are true:the following assertions are true:

PropositionProposition

wherewhere N(N(vv)) = {= { uu : : uuvv EE } } isis thethe neighborhoodneighborhood ofof v v V V andand N[v] = N(v) N[v] = N(v)

{v}.{v}.

(ii)(ii) II(G) = (G) = II(G (G –– uvuv) ) –– xx22 II(G (G –– N( N(uu))N(N(vv)),)),

(i)(i) II(G) = (G) = II(G (G –– vv) + ) + xx II(G (G –– N[ N[vv])])


N[v] = {a,c,d}{v}

c

G a

d b

v

I(G) = I(G-v) + x I(G-N[v]) =

P4 a

c

d

b

= I(P4) + x I({b}) =

= 1 + 4 x + 3 x2 + x (1+x) =

I(P4) = 1 + 4 x + 3 x2

= 1 + 5 x + 4 x2

G-v = P4

G-N[v] = {b}

Example

Some properties of the Some properties of the coefficients of coefficients of

independence polynomial, independence polynomial, as …as …

- - unimodality unimodality - - log-concavitylog-concavity - - palindromicity …palindromicity …

- - definitions & examplesdefinitions & examples - - results & conjectures … results & conjectures …

(-1, 2, -3, 4) is NON-unimodal, but it is

log-concave: (-1)(-3) 22, 24 (-3)2

A sequence of reals aa00, a, a11,..., a,..., ann is: (i) unimodalunimodal ifif aa0 0 aa1 1 ... ... aamm ... ...

a an n

for somefor some mm{0,1,...,n}{0,1,...,n},,

(ii) log-concavelog-concave ifif aak-1k-1 a ak+1k+1 (a (akk))22

for every for every k k {1,...,n-1}. {1,...,n-1}.(1, 4, 5, 2) is both uni & log-con (1, 2, 5, 3) is unimodal, NON-log-concave: 15

> 22

However, every log-concavelog-concave sequence of positive numberspositive numbers is

unimodalunimodal..

Examples

A polynomialA polynomial

P P ((xx)) = = aa0 0 + + aa11xx + +……++ aannxxnn

is is unimodalunimodal (log-concave)(log-concave) if its if its sequencesequence of of coefficientscoefficients aa0 0 , a, a1 1 , ,

aa2 2 , ... , a, ... , ann

is is unimodalunimodal (log-concave, (log-concave, respectively).respectively).

P(x) = 1 + 4x + 50x2 + 2x3 is unimodal with mode k

= 2P(x) = (1 + x)n is unimodal with the

mode k = n/2 and is also log-concave

Example

Is there a (connected) graph G with (G) =

whose sequence

s0, s1, s2 , … , s

is NOT unimodal ?

Recall that sk denotes the number of stable sets of size k

in a graph.Question

H. Wilf

Answer

For 3, there is a (connected) graph G with (G) = whose sequence s0,

s1, s2 , … , s

is NOT unimodal ! Y. Alavi, P. Malde, A. Schwenk, P. ErdY. Alavi, P. Malde, A. Schwenk, P. Erdöös s

Congressus Numerantium 58 (1987)Congressus Numerantium 58 (1987)

II((HH) = ) = 11++6464 xx + +634 634 xx22 ++500500 xx3 ++625625 xx4 is is notnot unimodal unimodal

I(G) = 1 + 6x + 8x2 + 2x3 is unimodalunimodal

Examples

K5K22

K22K5

K5

K5

H

G

For any permutation of the set {1, 2, {1, 2, ……, , }},, there is a graph GG

such that (G) = (G) = and ss(1)(1)< < ss(2)(2)< < ss(3)(3)< < …… < < ss(())

where sskk is the number of stable stable setssets in GG of size kk..

Moreover, any deviation from unimodality is possible!

Theorem

Y. Alavi, P. Malde, A. Schwenk, P. ErdY. Alavi, P. Malde, A. Schwenk, P. Erdöös s Congressus Numerantium 58 (1987)Congressus Numerantium 58 (1987)

A graph is called claw-free if it has no clawclaw, ( i.e., K1,3 ) as an induced

subgraph. K1,3Theorem

II((GG)) is log-concave for is log-concave for every claw-free graph every claw-free graph G.G.

RemarkThere are non-claw-free graphs with log-concave independence

polynomial.

I(K1,3 ) = 1 + 4x + 3x2 + x3

Y. O. Hamidoune Y. O. Hamidoune Journal of Combinatorial Theory B 50 (1990)Journal of Combinatorial Theory B 50 (1990)

IfIf all the rootsall the roots of a of a polynomialpolynomial withwith positive positive

coefficientscoefficients areare realreal, , then the then the polynomial ispolynomial is log-concavelog-concave..Sir Sir II. Newton , . Newton , Arithmetica UniversalisArithmetica Universalis (1707) (1707)

Theorem

Moreover

,I(G) has only real roots, for every claw-free graph G.

Theorem

M. Chudnovsky, P. Seymour, J. Combin. Th. B 97 (2007)

What is known aboutWhat is known about II(T)(T),,

wherewhere TT is a treeis a tree??

IfIf TT is a tree, is a tree, thenthen II(T)(T) is unimodalis unimodal..

I(T) = 1+7x + 15x2 +14x3 +6x4 +x5

Still open …

Example

Conjecture 1Conjecture 1


T

IfIf FF is a forest, is a forest, then then II(F)(F) is unimodalis unimodal..

I(F) = I(K1,3 ) I(P4) = 1+8x+22x2+25x3+13x4+3x5

F

Still open …

Example

Conjecture 2Conjecture 2


There existThere exist unimodal independence polynomials whose product is not

unimodal.

I(GG) = I(G) I(G) = == 1++232x + + 13750x2 + + 34790x3

+ + 101185x4 + + 100842x5 + + 117649x6

I(G) = 1+116 x +147 x2+343 x3

Example G

K7 K95K7

K7

G = K95+3K7

(i)(i) log-concave log-concave unimodal =unimodal = unimodal;unimodal;

i.e.,i.e., log-concavelog-concave unimodalunimodal is notis not necessarily necessarily log-concavelog-concave

J. Keilson, H. Gerber J. Keilson, H. Gerber Journal of American Statistical Association 334 (1971)Journal of American Statistical Association 334 (1971)

Theorem

G = K40 + 3K7 , H = K110 + 3K7

P1 = I(G) = 1+61x +147x2+343x3 … log-concave

P2 = I(H) = 1+131x+147x2+343x3 … not log-con

P1 P2= 1+192x +8285x2+28910x3+ +87465x4+100842x5+117649x6

1008422 – 87465 117649 = – 121 060 821

HoHoweweveverr

(ii)(ii) log-concavelog-concave log-concavelog-concave = = log-log-concave.concave.

The unimodality of The unimodality of independence polynomials independence polynomials

ofof trees trees does does notnot directlydirectly implies implies

the unimodality of the unimodality of independence polynomials independence polynomials

ofof forestsforests! !

Consequence

Hence,Hence,

independence polynomials independence polynomials ofof forests are log-forests are log-concaveconcave as well ! as well !

IfIf TT is a tree, is a tree, then then II(T)(T) is log-concaveis log-concave..

Conjecture 1*Conjecture 1*









00 = 0, = 0, = n, = n, 11

G

-k = 2 (-k )

1 = = 2

2 = 3 < 3 = 6

0 = 1 < = 2

Let G be a graph of order n with (G) = and 0 k . Then-k = max{n|N[S]| : S is stable, |S| = k}.

Examples

H

Def

IfIf GG is a graph withis a graph with (G) = andand (G) = , thenthen

(ii) s 1 s-1 s-1.

Theorem

(i) (k+1) sk+1 -k sk , 0 k

V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006)

LetLet HH = = (A(A,,BB,,E)E) be a bipartite graph be a bipartite graph withwith

XXAA X X is ais a kk-stable set in-stable set in GG (|X|=k)(|X|=k)

YYBB Y Y is ais a (k+1)(k+1)--stable set instable set in GG,,

andand XYXYEE X X YY

anyany YYBB hashas k+1k+1 kk-subsets-subsets ||EE|| = (k+1)s = (k+1)sk+1k+1

ifif XXAA andand vvV(G) V(G) N[X] N[X] X X{v}{v}BB

hencehence, , degdegHH(X) (X) -k-k andand

(k+1) s(k+1) sk+1k+1= = ||EE|| -k-k s skk

PPrrooooff









GG is calledis called quasi-regularizablequasi-regularizable ifif |S| |S| |N(S)| |N(S)| for each stable setfor each stable set

SS..

Quasi-regQuasi-reg

Non-Non-quasi-quasi-

regreg

Definition

C. Berge, Annals of Discrete Mathematics 12 (1982)C. Berge, Annals of Discrete Mathematics 12 (1982)

ExamplExampleses

IfIf GG is a quasi-regularizable is a quasi-regularizable graph of ordergraph of order nn = 2 = 2(GG) = 22,

thenthen

(ii)(ii) (k+1)(k+1) ssk+1k+1 2 ( 2 (-k)-k) sskk

(i)(i) -k-k 2 ( 2 (-k)-k)

(iii)(iii) sspp s sp+1p+1 … … s s-1-1 ss

wherewhere p = p = (2(2-1)/3-1)/3..

Theorem


ProofProof ((i)) IfIf SS is stable and is stable and ||SS| = k| = k 2 |S| 2 |S| |S |SN(S)|N(S)| 2(2(-k) = 2(-k) = 2(-|S|) -|S|) n-|N[S]| n-|N[S]| -k -k 2 ( 2 (-k)-k)

(ii)(ii) (k+1) sk+1k+1 22 ((-k)-k) s sk k

becausebecause (k+1) sk+1k+1 -k-k skk

((iii)) ssp p … … s s-1 -1 s s forfor pp = =

(2(2-1)/3-1)/3 sincesince byby (ii),(ii), it follows it follows

thatthat ssk+1k+1 ssk k ,, whenever whenever

kk +1 +1 2( 2(-k) -k) p p = = (2(2-1)/3-1)/3

We found out that (sk) is decreasing in this upper part:

ifif GG is quasi-regularizable of orderis quasi-regularizable of order 22(G), , thenthen

1 2 3 p k

sk

decreasing

Unimodal ? Log-concave ?

Unconstrained ?

sp … s-1 s , p = p = (2(21)/3)1)/3)

ExampExamplele

G is quasi-regularizable

I(G) = 1 + 8 x + 19 x2 + 15 x3 + 4 x4

p = (2-1)/3 = (8-1)/3 = 3

(G) = 4 n = 8

G is quasi-regularizable & I(G) is log-concave

s3 = 15 s4 = 4

G

ExampExamplele

G is not a quasi-

reg graph

I(G) = 1 + 9 x + 26 x2 + 30 x3 + 17 x4 + 4 x5

p = (2-1)/3 = (10-1)/3 = 3

(G) = 5 n = 9

G is not quasi-regularizable & I(G) is log-concave

s3 = 30 s4 = 17 s5 =

4

G

ExampExamplele G is a

quasi-reg

graph

I(G) = 1 + 16 x + 15 x2 + 20 x3

+15 x4 + 6 x5 + 1 x6

p = (2-1)/3 = (12-1)/3 = 4

(G) = 6 n = 16

G is quasi-reguralizable & I(G) is not unimodal!

G = K10 + 6K1 s4 = 15 s5 = 6 s6 = 1

G K10

K1

K1K1

K1

K1K1









A graph G is called well-covered if all its

maximal stable sets are of the same size (namely, (G)).

If, in addition, G has no isolated vertices and its order equals

2(G), then G is called very well-covered.

M. L. Plummer, J. of Combin. Theory 8 (1970)

O. Favaron, Discrete Mathematics 42 (1982)

Definitions

C4 & H2 are very well-

covered

Examples H1H1 is well-covered

H4

H3 & H4 are

not well-covered

H2C4

H3

G is a well-covered graph, I(G) = 1+9x+ 25x2 +22x3 is

unimodal.

If GG is a well-covered graph, then II(G)(G) is unimodal.

ExEx-Conjecture -Conjecture 33

G

J. I. Brown, K. Dilcher, R. J. Nowakowski J. I. Brown, K. Dilcher, R. J. Nowakowski J. of Algebraic Combinatorics 11 (2004) J. of Algebraic Combinatorics 11 (2004)

Example

i.e., i.e., Conjecture 3Conjecture 3 is is truetrue for every well-covered for every well-covered

graphgraph GG havinghaving (G) (G) 3 3. .

They also provided

counterexamples for 4 (G) 7.

T. Michael, W. Traves, Graphs and Combinatorics 20 (2003)T. Michael, W. Traves, Graphs and Combinatorics 20 (2003)

Theorem I(I(GG) is unimodal) is unimodal for for

everyeverywell-covered graphwell-covered graph GG

havinghaving (G) (G) 3 3. .

K4, 4,…, 4

1701

K10K10

K10K10

GG = 4K10 + K4, 4, …, 4

1701 times

1701-partite: each part has 4 vertices

GG

Michael & Traves’ counter Michael & Traves’ counter exampleexample

G = 4KG = 4K10 10 ++ KK4, 4, 4, 4, ……, 4, 4

n times 4

GG is well-covered, (GG) = 4 I(G) = 1+(40+4n) x+(600+6n) x2 + (4000+4n) x3 +

(10000+n) x4

I(G) is NOT unimodal iff 1701 n 1999

and it is NOT log-concave iff 24 n 2452

I(G) is NOT unimodal iff 4000+4n min{40+4n,1000+n}

KK4, 4,…, 4

1701

KK10KK10

KK10KK10

K1000

K1000

K1000

K1000

GG1701 times

GG = (4K(4K10 + KK4,4,…,4)) (4K1000)

(GG) = 8

G is well-covered and (G) = 8, while

I(G) = 1 + 14,844 x + 78,762,806 x2

+ 196,342,458,804 x3

+ 235,267,430,443,701 x4

+ 109,850,051,389,608,000 x5

+ 173,242,008,824,000,000 x6

+ 173,238,432,000,000,000 x7

+ 187,216,000,000,000,000 x8

1701 times

G = (4K10+K4,4,…,4) qK1000q

GG isis well-well-coveredcovered

K4, 4,…, 4

1701

K1000q

K10

K1000q

K10

K10

K10

q times

((GG)) = q + = q + 44

0 0 ≤≤ q q

V. E. Levit, E. Mandrescu, European J. of Combin. 27 (2006)

qq = = 00

K4, 4,…, 4

1701

K1000q

K10

K1000q

K10

K10

K10

q times

4 4 ≤≤ q q

11 ≤ ≤ qq ≤ ≤ 33

Michael & Traves CounterExample

New Michael & Traves CounterExamples

CounterExamples forCounterExamples for 8 8 ≤≤

Gq is well-covered, not

connected, (Gq) = q + 4

I(Gq;x) = (1+ 6844 x + 10806 x2 +

10804 x3 + 11701 x4) (1 + 1000q x)q

is not unimodal. Proof: sq+2 > sq+3 < sq+4

H = Gq Gq is well-covered, connected I(H) = 2 I(Gq) 1 is not unimodal.

Gq Gq

any

v

any

u

G is very well-covered I(G) = 1 +

6x + 9x2 + 4x3

Example

Conjecture 3*

G

IfIf GG is a very well-covered

graph, then then II(G)(G) is unimodal.


If G is a well-covered graph with (G) = , then

(i) 0 k

(-k) sk (k+1) sk+1

(ii) s0 s1… sk-1 sk , k = (+1)/2 .

Theorem


V. E. Levit, E. Mandrescu, V. E. Levit, E. Mandrescu, Discrete Applied Mathematics 156 (2008) Discrete Applied Mathematics 156 (2008)

eacheach (k+1)--stable set includesstable set includes k+1

stable sets of sizestable sets of size k (k+1) sk+1

EveryEvery kk-stable set-stable set AAkk is included in is included in some stable setsome stable set BB of sizeof size ..

thus, (-k) sk (k+1) sk+1

hence, eachhence, each BB hashas -k-k stable subsetsstable subsets ofof

sizesize k+1k+1 that includethat include AAkk (-k) sk

sk-1 sk , for k (+1)/2

ProofProof

IfIf GG is a is a veryvery well-coveredwell-covered graph with graph with (G) = (G) = , then, then

(v) II(G)(G) is unimodal, whenever 9 9..

(iii) sspp s sp+1p+1 …… s s-1-1 s s,,p = p = (2(2-1)/3-1)/3

(ii) ss0 ss1 … ss /2

(i) ((-k) s-k) skk (k+1) s (k+1) sk+1 k+1 2 (2 (-k) s-k) skk

Theorem


(iv) ss ss-2 (ss-1)2

(iv) CombiningCombining (ii) andand (iii),, it follows it follows that that II(G)(G) is unimodal, whenever is unimodal, whenever 99..

(i) It follows from previous results on It follows from previous results on quasi-reg graphs, as any well-covered quasi-reg graphs, as any well-covered graph is quasi-regularizable (Berge)graph is quasi-regularizable (Berge)

(i) (ii) s0 s1 … s/2

(i) (iii) ssp p s sp+1 p+1 …… s s-1 -1 s s

wherewhere p = p = (2(2-1)/3-1)/3

ProofProof

For anyany permutation of {k, k+1,…, }, k = /2, there is a well-coveredwell-covered graph G with (G) = , whose sequence

(s0 , s1 , s2 ,…, s) satisfies:

s(k)< s(k+1)< …< s().

Conjecture 4 : Conjecture 4 : “Roller-“Roller-Coaster”Coaster”

T. Michael, W. Traves, Graphs & Combinatorics 20 (2003)T. Michael, W. Traves, Graphs & Combinatorics 20 (2003)

The “The “Roller-CoasterRoller-Coaster” Conjecture” Conjecture isis validvalid forfor

What about (G) > (G) > 1111 ?

Still open …


(i)(i) every well-covered graph GG with (G) (G) 7 7;;

(ii)(ii) every well-covered graph GG with (G) (G) 11. 11.

P. Matchett, Electronic Journal of Combinatorics (2004)P. Matchett, Electronic Journal of Combinatorics (2004)

For a well-covered graph, the sequence (sk) is unconstrained

with respect to order in its upper part!

1 2 3 2

k

sk

increasing

unconstrained

““Roller-CoasterRoller-Coaster” ” conjecture:conjecture:

P. Matchett (2004)P. Matchett (2004)

““Roller-CoasterRoller-Coaster”” conjecture*:conjecture*:For a VERY well-covered graph,

the sequence (sk) is unconstrained with respect to

order in this upper part!

1 2 3 2-1 3

k

sk

increasing

unconstrained

2

decreasing

V. E. Levit, E. Mandrescu (2006)









GG is calledis called perfectperfect ifif (H) = (H) = (H)(H) for any induced for any induced subgraphsubgraph HH ofof GG, , wherewhere (H), (H), (H)(H) are the are the chromatic andand thethe clique numbers of of HH..

C. Berge, 1961C. Berge, 1961

E.g.,E.g., any any chordal graph is chordal graph is

perfect.perfect.

IfIf GG is a perfect graphis a perfect graphwith with (G) = (G) = andand (G) = (G) = , then, then

ssp p s sp+1 p+1 …… s s-1 -1 s s

wherewhere p = p = (( 1) / ( 1) / ( 1) 1)..

= 3,= 3, = 3, p = 2= 3, p = 2

G

II(G)(G) = 1+6x+8x2+3x3

Theorem

Example

We found out that the sequence (sk) is decreasing in its upper part:

ifif GG is ais a perfect graphperfect graph withwith (G) = , (G) = , then then ssp p s sp+1 p+1 …… s s-1 -1 s s for p = =

((-1)/(-1)/(+1)+1)..

1 2 3 -1+1

k

sk

decreasing


Unconstrained ?

IfIf SS is stable andis stable and ||SS| = | = kk, , thenthen H = G-N[S]H = G-N[S] hashas ((HH) ) ((GG)-)-kk..

By Lovasz’s theoremBy Lovasz’s theorem ||VV((HH)| )| (H)(H)(H) (H) ((HH)()(--k) k) (G)((G)(--

k).k).

(k+1) s(k+1) sk+1 k+1 (G) ((G) (-k) s-k) skk

(k+1)(k+1) ssk+1k+1 (G) ((G) (--kk)) sskk andand

ssk+1 k+1 sskk is true whileis true while k+1 k+1 (G)((G)(--

k),k), i.e.,i.e., for for k k ((--1)1) / / ((++1)1)..

ProofProof

II(HH) = = 11++148148 xx + +147 147 xx22 + + 343343 xx3

is not unimodalis not unimodal

I(G) = 1 + 5x + 4x2 + x3 is log-concavelog-concave

G

K127K7

K7

K7

H = K127+3K7

Examples

GG andand HH are are perfectperfect

IfIf GG is ais a minimalminimal imperfect graphimperfect graph, then, then

II(G)(G) is log-concave.is log-concave.

I(C7) = 1 + 7x + 14 x2 + 7x3

C7

Remark

Example

There is anThere is an imperfectimperfect graphgraph GG whosewhose II((GG)) isis notnot

unimodal.unimodal.Example

G = K97+ 4K3~ C5

K97GGK3 K3

K3 K3

C5

I(GG) = 1 + 114x + 603x2 + 921x3 + 891x4 + 945x5 + 405x6

Remark

IfIf GG is a bipartite graph is a bipartite graph with with (G) = (G) = , then , then ssp p ssp+1 p+1 …… ss-1 -1 ss

wherewhere pp = = (2(2-1)/3-1)/3..

I(GG) = 1+8x+19x2

+20x3+10x4+2x5 G

= 5 ; p = 3

Corollary

Example

IfIf TT is a tree withis a tree with (T) = (T) = , thenthen

ssp p s sp+1 p+1 …… s s-1 -1 s s

wherewhere pp = = (2(2-1)/3-1)/3..

= 6 p = 4

I(TT) = 1 + 8x + 21x2

+26x3 +17x4 + 6x5 + x6 T

Corollary

Example

For P4 p=1

We found out that (sk) is decreasing in this upper part:

Conjecture 1: Conjecture 1: I(T)I(T) is unimodal for a is unimodal for a

treetree T. T.

1 2 3 2-1 3

k

sk

decreasing


ifif TT is ais a treetree,then ,then ssp p s sp+1 p+1 …… s s-1 -1 s s , p = = (2(2(T)-(T)-

1)/31)/3..




König-Egerváry GraphsKönig-Egerváry Graphs






G is called a König-Egerváry (K-E) graph if (G) + (G) = |

V(G)|. R. W. Deming, Discrete Mathematics 27 (1979)

If G is bipartite, then G is a König-Egerváry graph.

F. Sterboul, J. of Combinatorial Theory B 27 (1979)

Well-known !

(G) + (G) = 5

G

(H) + (H) < 6

H

If G is a König-Egerváry graph, then

Theorem

(i) sk tk, k = (G), where = (G) and

(ii) the coefficients sk satisfy

;

2

...

32

22

12 3

3

2

21

ssss

(1+2x) (1+x) = t0 + t1x +…+ t…+ t-1-1 x1+ t+ tx

(iii) sp ≥ sp+1 ≥… ≥ ss-1-1 ≥≥ s s for p = (21)/3 .

V. E. Levit, E. Mandrescu, Congressus Numerantium 179 (2006)

G H

2)()(

4)()(

HG

HG

Example

Proof

kts k

kk

2

k

k

k xk

x

0

2)21(

k

kk xsxGI

0

);(

k

kk xtxxxHI

0

)21()1();(

Proof

&

I(G) = s0 + s1x +…+ s…+ s-1-1 x1+ s+ sx

ks k

k

2

121

kk s

ks

k

1

1

1

1

1

:&

k

k

k

k

k

s

k

sRyanFisher

kk sksk 21 1

12

1

1

k

s

k

s

k

s k

k

k

kk

.3/)12(.,.

),(21)(21

kforei

kkforholdssks kk

If G has sk stable sets of size k, 1 k (G) = ,

then

Theorem

D. C. Fisher, J. Ryan, Discrete Mathematics 103 (1992)

L. Petingi, J. Rodriguez, Congressus Numerantium 146 (2000)

… and an alternative proof was given by

....

321

1

3

1

3

2

1

2

1

1

1

ssss

We found out that the sequence (sk) is

decreasing in this upper part:If a König-Egerváry graph G has (G) = ,

then

1 2 3 p k

sk

decreasing


Unconstrained ?

sp sp+1 … s-1 s for p = (21)/3

ExampExamplele

G is a K-E

graph

I(G) = 1 + 13 x + 21 x2 + 35 x3

+35 x4+ 21 x5 + 7 x6 + 1 x7

p = (2-1)/3 = (14-1)/3 = 5

(G) = 7 (G) = 6 n = 13

21211335 < 0 I(G) is not log-concave, but

unimodal!

G = K6 + 7K1

s5 = 21 s6 = 7 s7 = 1

G K6

K1

K1K1

K1 K1

K1

K1

I(G) = 1 + 8x + 20x2 +23x3 +20x4 +1x5

unimodal Example

= 5, = 3,

p = (2-1)/3 = 3

I(G) is unimodal for every König-Egerváry graph G.

G

Conjecture 5









Recall : “Corona” Recall : “Corona” operation operation

P3

P4

K1 2K1

K3

G = PG = P44 {P {P3 3 , K, K1 1 , 2K, 2K1 1 , K, K33}}

Particular case of Particular case of “Corona”“Corona”K1

P4

K1K1 K1

G = PG = P4 4 K K 11

Each stable set of Each stable set of G = H G = H K K11 can can be enlarged to a maximum be enlarged to a maximum

stable set.stable set.

RemarkRemark

G is called well-covered if all its maximal stable sets are of the same size (M.D.

Plummer, 1970).

Def.Def.Equivalently, G is well-covered if each of its

stable sets is contained in a maximum stable set.

Let G be a graph of girth > 5, which is isomorphic to neither C7 nor K1. Then G is well–covered if and

only if G = H* for some graph H.

Theorem

A. Finbow, B. Hartnell, R. Nowakowski, J. Comb. Th B 57 (1993)

Appending a single pendant edge

to each vertex of H H*.

H* is very well-covered, for any graph H

Remark

If G is a graph of order n, and

I(G) = s0 + s1x +…+ s…+ s-1-1 x1+ s+ sx , then

and the formulae connecting the coefficients of I(G) and of I(G*) are:

)(0,)1(0

Gkkn

jnts

k

jj

jkk

nkkn

jnst

k

jjk

0,0

kGG

k

kk

Gn

knkG

kk

n

j

jj

xxsx

xxsxtxGI

)()(

0

)(

)(

00

*

)1()1(

)1(;

Theorem

V. E. Levit, E. Mandrescu, Discrete Applied Mathematics (2008) V. E. Levit, E. Mandrescu, Discrete Applied Mathematics (2008)

Well-covered spidersWell-covered spiders: :

Sn

A spider is a tree having at most one vertex of degree

> 2.

K2

K1

P4

Let T* be the tree obtained from the tree T by appending a

single pendant edge to each vertex of T.

T

((T**)) = the order ofthe order of T

( )*

ExamplExamplee

RemarkRemark

(T*) = 4

(iv) T is a is well-covered spider or T is

obtained from a well-covered tree T1 and a

well-covered spider T2, by adding an edge

joining two non-pendant vertices of T1,

T2, respectively.

For a tree T K1 the following are equivalent: (i) T is well-covered

(iii) T = L* for some tree L

Theorem

Appending a single pendant edge

to each vertex of H H*.

G. Ravindra, Well-covered graphs, J. Combin. Inform. System Sci. 2 (1977)

V. E. Levit, E. Mandrescu, Congressus Numerantium 139 (1999) V. E. Levit, E. Mandrescu, Congressus Numerantium 139 (1999)

(ii) T is very well-covered

the sequence (sk) is unconstrained with respect to

order in this upper part!

1 2 3 2-1 3

k

sk

increasing

unconstrained

2

decreasing

For every well-covered tree T, with (T) = ,

The independence polynomial The independence polynomial of anyof any wewell-coll-covvered spiderered spider

SSn n , n, n 1, 1, is unimodal andis unimodal and

mode(Sn) = n- (n-1)/3

Proposition

all are unimodal !

I(K1)=1+x

I(P4) = 1+4x+3x2

I(K2) = 1+2x

V. E. Levit, E. Mandrescu, Congresus Numerantium 159 (2002)V. E. Levit, E. Mandrescu, Congresus Numerantium 159 (2002)

The independence The independence polynomial of anypolynomial of any well–well–

covered spidercovered spider S Snn is log–is log–concave.concave.

Proof & “If P, Q are log-

concave, then PQ is log-concave.”

Proposition

V. E. Levit, E. Mandrescu, Carpathian J. of Math. 20 (2004)

Moreover,

n

k

kkn x

k

n

k

nxxSI

1 1

121)1();(









A (A (graphgraph) polynomial) polynomial

PP((xx)) = = aa0 0 + + aa11xx ++……++ aannxxnn is calledis called

palindromicpalindromic ifif aai i = a = an-i n-i , i = 0,1,..., , i = 0,1,..., n/2n/2..

P(x) = (1 + x)n

J. J. Kennedy J. J. Kennedy –– ““Palindromic graphsPalindromic graphs”” Graph Theory Notes of New York, XXII (1992)Graph Theory Notes of New York, XXII (1992)

nK1

v1 v2 v3 vn

In fact, (1+x)n = I(nK1)

DefinitionDefinition characteristicmatchingindependence

I. Gutman, I. Gutman, Independent vertex palindromic graphsIndependent vertex palindromic graphs, , Graph Theory Notes of New York, XXIII (1992)Graph Theory Notes of New York, XXIII (1992)

ExamplExamplee

(i) |S| q|NG(S)| for every stable setfor every stable set S ofof G;

Theorem

(ii) q(k+1)sk+1 (q+1)(-k)sk, 0 0 k k

< <

(iii) sr … s-1 s , r = r = ((q+1)((q+1) - q)/(2q+1) - q)/(2q+1) (iv) ifif q = 2, then then I(G) is palindromic is palindromic andand

LetLet G = HqK1 havehave (G) = andand (sk) be thebe the

coefficients ofcoefficients of I(G). Then the following are true:. Then the following are true:

s0 s1 … sp , p = p = (2(2+2)/5+2)/5

sr … s-1 s , r = r = (3(3-2)/5-2)/5 . .

We found out that the sequence (sk) is decreasing in this upper part:

ifif G = G = HqK1 has has (G) = , then, then

1 2 3 r k

sk

decreasing


Unconstrained ?

sr … s-1 s , r = r = ((q+1)((q+1)-q)/(2q+1)-q)/(2q+1)

IfIf G = G = H2K1 , then , then I(GG) is palindromic and its sequence (sk) is increasing in its first part

and decreasing in its upper part !

1 2 3 3-2 5

k

sk

increasing

Unimodal ?

2+2 5

decreasing

Question:

Is I(GG) unimodal ?

K1,3 K1,3 = the “claw”

I(K1,3) = 1+4x+3x2+x3

is not palindromic.I(G) = 1+s1x+s2x2 = 1+nx+x2

1. If (G) = 2 and I(G) is palindromic, then n2, I(G) = 1 + n x + 1x2 and I(G) is log-concave,

and hence unimodal, as well.

Remarks

G = Kn–e, n2

2. If (G) = 3 and I(G) is palindromic, then n3, I(G) = 1 + n x + nx2 + 1x3 and I(G) is log-concave,

and hence unimodal, as well.

ExampleExampless

(G) = 5

G = K1832 + 4K7 + (K2K539) + 5K1

II(G) (G) = 1= 1++24062406xx++13821382xx22++13821382xx33++24062406xx44++11xx55

K1832

5K1

K2K539

4K7

s2 = 10+2539+677 =

1382

s4 = 5+7777 = 2406

s3 = 10+4777

= 1382

s1 = 5+28+1832+539+2 =

2406

IfIf GG has a stable sethas a stable set SS with:with: |N(A)|N(A)S| = 2|A|S| = 2|A| for every stable setfor every stable set

A A V(G) V(G) –– S S, , then then II((GG)) isis palindromicpalindromic..D. Stevanovic, D. Stevanovic, Graphs with palindromic independence polynomialGraphs with palindromic independence polynomial

Graph Theory Notes of New York XXXIV Graph Theory Notes of New York XXXIV (1998)(1998)

Theorem

S = { } II(G) (G) = 1+ = 1+ 55x x + + 55xx2 2 + 1+ 1xx33

ExamplExamplee

G

The condition that: “The condition that: “GG has a stable has a stable setset SS with:with: |N(A)|N(A)S| = 2|A|S| = 2|A| for every for every

stable stable set set A A V(G) V(G) –– S S”” isis NOTNOT necessarynecessary!!

Remark

G S = { }

II(G) (G) = 1+= 1+66xx++66xx22+1+1xx33

I. Gutman, I. Gutman, Independent vertex palindromic graphs,Independent vertex palindromic graphs, Graph Theory Notes of New York XXIII (1992) Graph Theory Notes of New York XXIII (1992)

ExamplExamplee

IfIf G = (V,E)G = (V,E) has has ss=1,s=1,s-1-1=|=|VV|| and the and the unique maximum stable setunique maximum stable set SS satisfies: satisfies: ||N(u)N(u)S| = 2S| = 2 for everyfor every uuV-SV-S,, then I(G) isis

palindromic.

Corollary

GS={ } II(G) (G) = 1 + = 1 + 99x x + + 2727xx2 2 + + 3838xx33+ +

+ 1+ 1xx6 6 + + 99xx5 5 + + 2727xx4 4

D. Stevanovic, D. Stevanovic, Graphs with palindromic independence polynomialGraphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV Graph Theory Notes of New York XXXIV

(1998)(1998)

ExamplExamplee

RULE 1:RULE 1: If If is a is a clique coverclique cover of of GG, then: , then: for each clique for each clique CC,, addadd two new non- two new non-

adjacent vertices adjacent vertices andand join them to all the join them to all the vertices of vertices of CC.. The new graph is The new graph is

denoted bydenoted by {G}.{G}.

A A clique coverclique cover of of GG is a spanning graph of is a spanning graph of GG, , each component of which is a each component of which is a cliqueclique..

The set The set SS = { = {all these new verticesall these new vertices} is the unique } is the unique maximum stable set in the new graph maximum stable set in the new graph H = H = {G}{G}

and satisfies: and satisfies: |N(u)|N(u)S| = 2S| = 2 for any for any uuV(V(HH)-S)-S..Hence, Hence, II((HH)) isis palindromic by Stevanovic’s palindromic by Stevanovic’s

TheoremTheorem..

D. Stevanovic, Graphs with palindromic independence polynomial D. Stevanovic, Graphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV Graph Theory Notes of New York XXXIV

(1998)(1998)

How to build graphs with palindromic independence polynomials ?

S={ }

|N(u)|N(u)S| = 2S| = 2, for any , for any uuV(H)-SV(H)-S

G

H = {G}

II(G) (G) = = 11++66xx++99xx22++22xx33

II(H) (H) == 11++1212xx++4848xx22++7676xx33++4848xx4 4 ++1212xx55++11xx66

= { }

ExamplExamplee

In particular:In particular: If If each cliqueeach clique of the clique of the clique cover cover of of GG consists of a consists of a single vertexsingle vertex, ,

then: the new graph then: the new graph {G}{G} is denoted by is denoted by GG○○2K2K11 . .

GG○○mKmK11 is theis the coronacorona ofof GG andand

mKmK11..

G○2K1

G

= { } II(G○2K(G○2K1) ) = 1 + = 1 + 1212x x + + 5353xx2 2 + + 120120xx33+ +

+ +156156xx44+1+1xx88+ + 1212xx7 7 + + 5353xx6 6 + + 120120xx55

ExamplExamplee

RULE 2.RULE 2. If If is a is a cycle covercycle cover of of GG, then:, then:(1)(1) add two pendant neighbors to add two pendant neighbors to each vertexeach vertex from from ;;(2) for (2) for each edge abeach edge ab of of , add two new vertices and join , add two new vertices and join them them to to aa & & bb;;(3) for (3) for each edge xyeach edge xy of a of a proper cycleproper cycle of of , add a new , add a new vertex vertex and join it to and join it to xx & & yy..

A A cycle covercycle cover of of GG is a spanning graph of is a spanning graph of GG, each , each component of which is a component of which is a vertexvertex, an , an edgeedge, or a , or a proper proper

cyclecycle..

The set The set SS = { = {ALL THESE NEW VERTICESALL THESE NEW VERTICES} is stable in the } is stable in the new graph new graph H = H = {G}{G} and satisfies: and satisfies: |N(v)|N(v)S| = 2 for any S| = 2 for any

vvV(V(HH)-S)-S.. Therefore,Therefore, II((HH)) isis palindromicpalindromic..

The new graph is denoted byThe new graph is denoted by {G}.{G}.

D. Stevanovic, Graphs with palindromic independence polynomial D. Stevanovic, Graphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV Graph Theory Notes of New York XXXIV

(1998)(1998)

How to build graphs with palindromic independence polynomials ?

S={ }

|N(u)|N(u)S| = 2S| = 2, for any , for any uuV(H)-SV(H)-S

II(G) (G) = = 11++77xx++1313xx22++55xx33

II(H) (H) == 11++1515xx++8383xx22++218218xx33++298298xx44+218+218xx55++8383xx66++1515xx77++11xx88

= { } is a cycle cover

G

H = {G}

ExamplExamplee

Proposition

LetLet G = H2K1 havehave (G) = andand

(sk) be the coefficients ofbe the coefficients of I(G).

s0 s1 … sp , p = p =

(2(2+2)/5+2)/5 sr … s-1 s , r = r =

(3(3-2)/5-2)/5 . .

ThenThen I(G) is palindromic is palindromic andand

V. E. Levit, E. Mandrescu, 39th Southeastern Intl. Conf. on Combininatorics, Graph Theory, and Computing,

Florida Atlantic University, March 3-7, 2008

IfIf G = G = H2K1 , then , then I(GG) is palindromic and its sequence (sk) is increasing in its first part and decreasing in its upper part !

1 2 3 3-2 5

k

sk

increasing

Unimodal ?

2+2 5

decreasing

Question:

Is I(GG) unimodal ?

I(G) = 1 + 12x + 55x2 + 128x3 + 168x4 + 128x5 + 55x6 + 12x7 + x8

I(P4) == 1+1+44x++33x22

s0 = 1 s1 = 12 s2 = 55 s3 = 128 (p

= 3)

p = (2(G)+2)/5 = 3, r = (3(G)-3)/5 = 5

(G) = 8

s5 = 128 s6 = 55 s7 = 12 s8 = 1 (r

= 5)

ExamplExamplee

G = P4o2K1

P4

Theorem I(Pn2K1) has only real roots, and consequently is log-concave. Zhu Zhi-Feng, Australasian Journal of Combinatorics 38 (2007)

IfIf GG is is quasi-regularizablequasi-regularizable of order of order 22(G), , then then ssp p s sp+1 p+1 …… s s-1 -1 s s ,, pp = = (2(2--

1)/31)/3..

Theorem

ExampExamplele

G K6

K1K1

K1K1

K1

K1(G) = 6 n = 12

G is quasi-regularizableis quasi-regularizable

pp = = (12-1)/3(12-1)/3 = 4 = 4

ss4 4 = 15= 15 s s5 5 = 6= 6 s s6 6 = 1= 1

I(G) = 1 + 12 x + 15 x2 + 20 x3 + 15 x4 + 6 x5 + 1 x6


Theorem 2

(s1)2 s0 s2 , (s2)2 s1 s3 andand

(s-1)2 s s-2 , (s-2)2 s-1 s-3

I(G) is palindromic, while its coefficients is palindromic, while its coefficients (sk) satisfy: satisfy:

If If is a cycle cover of is a cycle cover of H without without vertex-cyclesvertex-cycles and and G = {H}

has has (G) = , then , then G is quasi-regularizable of order is quasi-regularizable of order 2 and and

s0 s1 … sp , p = (+1)/3 sq … s-1 s , q = (2-1)/3

V. E. Levit, E. Mandrescu, 39th Southeastern Intl. Conf. on Combinatorics, Graph Theory, and Computing, Florida Atlantic University, March 3-7, 2008

(i)

(ii)

1 2 3 2-1 3

k

sk

increasing

Unimodal ?

+1 3

decreasing

If is a cycle cover of H without vertex-vertex-

cyclescycles, , G = {H} has (G) = , then I(G) is palindromic and its sequence (sk) is increasing in its first part and decreasing in its upper part !

Question: Is I(G) unimodal ?

G = {H}

II(H) (H) = = 11++88xx++1919xx22++1313xx33++xx44

II(G) (G) == 11++1616xx++9595xx22++265265xx33++371371xx44+265+265xx55++9595xx66++1616xx77++11xx88

S={ }

= { } H

s0 s1 s2 s3 , p = (+1)/3 = 3

s5 s6 s7 s8 , q = (2-1)/3 = 5

(G) = 8

Example

Trees

Bipartite

König-Egervár

y graphs

Perfect

Very well-

covered

well-covere

d

quasi-regularizab

le

GG○○qKqK11

GG○○2K2K11 & GG perfect

GG○○KK11

GG○○KKpp

{G} & is a cycle cover of

G

Some family relationships

TT○○2K2K11 &

T =T = a tree

==

s o m es o m e

p r o b l e m sp r o b l e m s

o p e no p e n

… f i n a l l y, recall

Problem 1Problem 1

Find an inequality leading Find an inequality leading to partial log-concavity of to partial log-concavity of

the independence the independence polynomial.polynomial.

For very-well covered graphs: (S(S-1-1))22 S S S S-2-2

ExampleExample


P(x) = (1 + x)n

Problem 22Characterize polynomials Characterize polynomials that that areare independence independence

polynomials.polynomials.

P(x) = I(nK1)

but, there is no graph G whose I(G) = 1 + 4x + 17x2

C. Hoede, X. Li C. Hoede, X. Li Discrete Mathematics 125 (1994)Discrete Mathematics 125 (1994)

Example

Problem 3Problem 3

Characterize the graphs whose Characterize the graphs whose independence polynomialsindependence polynomials

are are palindromicpalindromic.. D. Stevanovic D. Stevanovic

Graph Theory Notes of New York XXXIV (1998)Graph Theory Notes of New York XXXIV (1998)A graph A graph GG with with (G) = 2(G) = 2 has has

a palindromic a palindromic independence polynomialindependence polynomial

iff iff G = KG = Knn- e- e.. I(G) == 11 ++ nn x ++ 11 x2 (G) = 2

Example

…… Thank you !Thank you !

! ! תודהתודה ......

Thank you very Thank you very much !much !ת ו ד ה רבהת ו ד ה רבה !!

Documents

Inequalities Involving the Coefficients of Independence Polynomials Combinatorial and Probabilistic Inequalities Isaac Newton Institute for Mathematical