941e719e000814d6f528fddb9727741d.ppt

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Inequalities Involving the Coefficients of Independence Polynomials Vadim E. Levit 1, 2 1 Ariel University Center of Samaria, Israel Eugen Mandrescu 2 2 Holon Institute of Technology, Israel Combinatorial and Probabilistic Inequalities Isaac Newton Institute for Mathematical Sciences Cambridge, UK - June 23 -27, 2008

O u t l i n e I(G; x) = the independence polynomial of graph G Results and conjectures on I(G; x) for some graph classes… … some more open problems

Introduction The Main Inequality Quasi-Regularizable Graphs Well-Covered Graphs Perfect Graphs König-Egerváry Graphs Corona Graphs Palindromic Graphs

Some definitions A set of pairwise non-adjacent vertices is called a stable set or an independent set. (G) = stability number is the maximum size of a stable set of G. Example Stable sets in G : , {a}, {a, b, c}, {a, b, c, d} , … (G) = |{a, b, c, d, e, f}| = 5 a, b, c, d, e, f a G d b c e

If sk denotes the number of stable sets of size k in a graph G with (G) = , then I(G) = I(G; x) = s 0 + s 1 x +…+ s x is called the independence polynomial of G. I. Gutman, F. Harary , Generalizations matching Gutman & of. Hararypolynomial Harary - ‘ 83 Utilitas Mathematica 24 (1983)

Example All the stable sets of G : 1 6 …… 8 …… {a}, {b}, {c}, {d}, {e}, {f} …. . {a, b}, {a, d}, {a, e}, {a, f}, 3 {b, c}, {b, e}, {b, f}, {d, f}, …… { a, b, e }, { a, b, f }, { a, d, f } I(G) = 1 + 6 x + 8 x 2 + 3 x 3 G b a d c e f

There are non-isomorphic graphs with I(G) = I(H) Example I(G) = I(H) = 1+6 x+4 x 2 H G

ALSO non-isomorphic trees can have the same independence polynomial ! T 1 I(T 1) = I(T 2) = 1+10 x+36 x 2+58 x 3+42 x 4+12 x 5+x 6 T 2 K. Dohmen, A. Ponitz, P. Tittmann, Discrete Mathematics and Theoretical Computer Science 6 (2003)

… for historical reasons The line graph of G = (V, E) is LG = (E, U) where ab U whenever the edges a, b E share a common vertex in G. e f Example G a b c d E = { a, b, c, d, e, f } a LG {a, b, d} = matching in G {a, b, d} = stable set in LG e b f c d

… recall for historical reasons If G has n vertices, m edges, and mk I. Gutman, F. Harary, Generalizations of matching polynomial matchings. Mathematica 24, then the of size k (1983) Utilitas matching polynomial of G is If G has n vertices, m edges, and mk matchings of size k, then while is the positive matching polynomial of G.

E x a m p l e 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. b a LG c d e LG = (E, U) f c G a b d e f G = (V, E) M+(G; x) = I(LG; x) = 1+6 x+7 x 2+1 x 3

Some “relatives” of I( ; ) : “Dependence polynomial” : D(G; x) = I(H; -x), where H is the complement of G Fisher & Solow - 1990 Twin: - “Independent set polynomial” Hoede & Li - 1994 “Clique polynomial”: C(G; x) = I(H; x), where H is the complement of G Hajiabolhassan & Mehrabadi - 1998 “Clique polynomial”: C(G; x) = I(H; -x), where H is the complement of G, Goldwurm & Santini - 2000 Go “Vertex cover polynomial of a graph”, where the coefficient ak is the number of vertex covers V’ of G with |V’| = k, Dong, Hendy & Little - 2002

Connections with other polynomials: I. Gutman, F. Harary, Utilitas Mathematica 24 (1983) Chebyshev polynomials of the first and second kind: G. E. Andrews, R. Askey, R. Roy, Special functions (2000) Hermite polynomials:

Connections with other polynomials: The generalized chromatic polynomial : P(G; x, y) is equal to the number of vertex colorings : V {1; 2; …, x} of the graph G = (V, E) such that for all edges uv E the relations (u) y and (v) y imply (u) (v). K. Dohmen, A. Ponitz, P. Tittmann, A new two-variable generalization of the chromatic polynomial, Discrete Mathematics and Theoretical Computer Science 6 (2003) 69 -90. Re ma rk P(G; x, y) is a polynomial in variables x, y, which simultaneously generalizes the chromatic polynomial, the matching polynomial, and the independence polynomial of G, e. g. , I(G; x) = P(G; x + 1, 1).

How to compute the independence polynomial ? If V(G) V(H) = , then 1. I(G H) = I(G) I(H) G H = disjoint union of G and H 2. I(G+H) = I(G) + I(H) – 1 where G+H = (V(G) V(H); E) E = E(G) E(H) {uv: u V(G), v V(H)} G+H = Zykov sum of G and H I. Gutman, F. Harary, Utilitas Mathematica 24 (1983)

I(K 3+K 4) = I(K 3) + I(K 4) – 1 = = (1+3 x) + (1+4 x) – 1 = 1+7 x E x a m p l e K 3 + K 4

How to compute the independence polynomial ? The corona of the graphs G and H is the graph G○H obtained from G and n = |V(G)| copies of H, so that each vertex of G is joined to all vertices of a copy of H. Example H H G ○H G G

Theorem I(G○H; x) = (I(H; x))n I(G; x / I(H; x)) I. Gutman, Publications de l’Institute Mathematique 52 (1992) Example I(H) = 1+2 x H G G ○H I(G) = 1+3 x+x 2 I(G○H; x) = (1+2 x)3 I(G; x / (1+2 x)) = = 1 + 9 x + 25 x 2 + 22 x 3

Proposition If G = (V, E), v V and uv E, then the following assertions are true: (i) I(G) = I(G – v) + x I(G – N[v]) (ii) I(G) = I(G – uv) – x 2 I(G – N(u) N(v)), where N(v) = { u : uv E } is the neighborhood of v V and N[v] = N(v) {v}. I. Gutman, F. Harary, Utilitas Mathematica 24 (1983)

I(G) = I(G-v) + x I(G-N[v]) = Example = I(P 4) + x I({b}) = = 1 + 4 x + 3 x 2 + x (1+x) = = 1 + 5 x + 4 x 2 G-v = P 4 N[v] = {a, c, d} {v} b G-N[v] = {b} a c b P 4 a c d G v I(P 4) = 1 + 4 x + 3 x 2 d

Some properties of the coefficients of independence polynomial, as … - unimodality log-concavity palindromicity … - definitions & examples results & conjectures … -

A sequence of reals a 0, a 1, . . . , an is: (i) unimodal if a 0 a 1 . . . am . . . an for some m {0, 1, . . . , n}, (ii) log-concave if ak-1 ak+1 (ak)2 for every k {1, . . . , n-1}. Examples (1, 4, 5, 2) is both uni & log-con (1, 2, 5, 3) is unimodal, NON-log-concave: 1 5 > 22 (-1, 2, -3, 4) is NON-unimodal, but it is log-concave: (-1) (-3) 22, 2 4 (-3)2 However, every log-concave sequence of positive numbers is unimodal.

A polynomial P (x) = a 0 + a 1 x +…+ anxn is unimodal (log-concave) if its sequence of coefficients a 0 , a 1 , a 2 , . . . , an is unimodal (log-concave, respectively). Example P(x) = 1 + 4 x + 50 x 2 + 2 x 3 is unimodal with mode k = 2 P(x) = (1 + x)n is unimodal with the mode k = n/2 and is also log-concave

Recall that sk denotes the number of stable sets of size k in a graph. Question Answer For there athere is a (connected) Is 3, (connected) graph G = (G) whose with (G) with whose= sequence s 0, s 1, s 2 , … , s is NOT unimodal ! is NOT unimodal ? Y. Alavi, P. Malde, A. Schwenk, P. Erdös Congressus Numerantium 58 (1987) H. Wilf

Examples G I(G) = 1 + 6 x + 8 x 2 + 2 x 3 is unimodal I(H) = 1+64 x +634 x 2 +500 x 3 +625 x 4 is not unimodal H K 5 K 22 K 5

Moreover, any deviation from unimodality is possible! Theorem For any permutation of the set {1, 2, …, }, there is a graph G such that (G) = and s (1)< s (2)< s (3)< … < s ( ) where sk is the number of stable sets in G of size k. Y. Alavi, P. Malde, A. Schwenk, P. Erdös Congressus Numerantium 58 (1987)

A graph is called claw-free if it has no claw, ( i. e. , K 1, 3 ) as an induced subgraph. claw Theorem K 1, 3 I(G) is log-concave for every claw-free graph G. Y. O. Hamidoune Journal of Combinatorial Theory B 50 (1990) Remark There are non-claw-free graphs with log-concave independence polynomial. I(K 1, 3 ) = 1 + 4 x + 3 x 2 + x 3

Theorem Moreover, I(G) has only real roots, for every claw-free graph G. M. Chudnovsky, P. Seymour, J. Combin. Th. B 97 (2007) Theorem If all the roots of a polynomial with positive coefficients are real, then the polynomial is log-concave. Sir I. Newton , Arithmetica Universalis (1707)

What is known about I(T), where T is a tree?

Conjecture 1 If T is a tree, then I(T) is unimodal. Y. Alavi, P. Malde, A. Schwenk, P. Erdös Congressus Numerantium 58 (1987) I(T) = 1+7 x + 15 x 2 +14 x 3 +6 x 4 +x 5 Example Still open … T

Conjecture 2 If F is a forest, then I(F) is unimodal. Y. Alavi, P. Malde, A. Schwenk, P. Erdös Congressus Numerantium 58 (1987) Example F I(F) = I(K 1, 3 ) I(P 4) = 1+8 x+22 x 2+25 x 3+13 x 4+3 x 5 Still open …

There exist unimodal independence polynomials whose product is not unimodal. Example G = K 95+3 K 7 G K 7 K 95 K 7 I(G) = 1+116 x +147 x 2+343 x 3 I(G G) = I(G) = = 1+232 x + 13750 x 2 + 34790 x 3 + 101185 x 4 + 100842 x 5 + 117649 x 6

Theorem (i) log-concave unimodal = unimodal; (ii) log-concave = log-concave. J. Keilson, H. Gerber Journal of American Statistical Association 334 (1971) Ho we ver G = K 40 + 3 K 7 , H = K 110 + 3 K 7 P 1 = I(G) = 1+61 x +147 x 2+343 x 3 … log-concave P 2 = I(H) = 1+131 x+147 x 2+343 x 3 … not log-con P 1 P 2= 1+192 x +8285 x 2+28910 x 3+ +87465 x 4+100842 x 5+117649 x 6 1008422 – 87465 117649 = – 121 060 821 i. e. , log-concave unimodal is not necessarily -concave log

Consequence The unimodality of independence polynomials of trees does not directly implies the unimodality of independence polynomials of forests !

Conjecture 1* If T is a tree, then I(T) is log-concave. Hence, independence polynomials of forests are log-concave as well !

Introduction The Main Inequality Quasi-Regularizable Graphs Well-Covered Graphs Perfect Graphs König-Egerváry Graphs Corona Graphs Palindromic Graphs

Def 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 0 = 0, = n, 1 0 = 1 < = 2 1 = = 2 2 = 3 < 3 = 6 -k = 2 ( -k ) H G

Theorem If G (G) = is a graph with and (G) = , then (i) (k+1) sk+1 -k sk , 0 k (ii) s 1 s -1. V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006)

P r o o f Let H = (A, B, E) be a bipartite graph with X A X is a k-stable set in G (|X|=k) Y B Y is a (k+1)-stable set in G, and XY E X Y any Y B has k+1 k-subsets |E| = (k+1)sk+1 if X A and v V(G) N[X] X {v} B hence, deg. H(X) -k and (k+1) sk+1= |E| -k sk

Introduction The Main Inequality Quasi-Regularizable Graphs Well-Covered Graphs Perfect Graphs König-Egerváry Graphs Corona Graphs Palindromic Graphs

Definition G is called quasi-regularizable if |S| |N(S)| for each stable set S. C. Berge, Annals of Discrete Mathematics 12 (1982) Examples Quasi-reg Nonquasi-reg

Theorem If G is a quasi-regularizable graph of order n = 2 (G) = 2 , then (i) -k 2 ( -k) (ii) (k+1) sk+1 2 ( -k) sk (iii) sp sp+1 … s -1 s where p = (2 -1)/3. V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006)

Proof (i) If S is stable and |S| = k 2 |S| |S N(S)| 2( -k) = 2( -|S|) n-|N[S]| -k 2 ( -k) (ii) (k+1) sk+1 2 ( -k) sk because (k+1) sk+1 -k sk (iii) sp … s -1 s for p = (2 -1)/3 since by (ii), it follows that sk+1 sk , whenever k +1 2( -k) p = (2 -1)/3

Unimodal ? Log-concave ? sk Unconstrained ? decreasing 1 2 3 p k We found out that (sk) is decreasing in this upper part: if G is quasi-regularizable of order 2 (G), then sp … s -1 s , p = (2 1)/3)

Example (G) = 4 n=8 G G is quasi-regularizable p = (2 -1)/3 = (8 -1)/3 = 3 s 3 = 15 s 4 = 4 I(G) = 1 + 8 x + 19 x 2 + 15 x 3 + 4 x 4 G is quasi-regularizable & I(G) is log-concave

Example G (G) = 5 n=9 p = (2 -1)/3 = (10 -1)/3 = 3 s 3 = 30 s 4 = 17 s 5 = 4 G is not a quasi-reg graph I(G) = 1 + 9 x + 26 x 2 + 30 x 3 + 17 x 4 + 4 x 5 G is not quasi-regularizable & I(G) is log-concave

Example (G) = 6 n = 16 G is a quasi-reg G graph p = (2 -1)/3 = (12 -1)/3 = 4 s 4 = 15 s 5 = 6 s 6 = 1 K 1 K 10 K 1 K 1 G = K 10 + 6 K 1 I(G) = 1 + 16 x + 15 x 2 + 20 x 3 +15 x 4 + 6 x 5 + 1 x 6 G is quasi-reguralizable & I(G) is not unimodal!

Definitions A graph G is called well-covered if all its maximal stable sets are of the same size (namely, (G)). M. L. Plummer, J. of Combin. Theory 8 (1970) If, in addition, G has no isolated vertices and its order equals 2 (G), then G is called very well-covered. O. Favaron, Discrete Mathematics 42 (1982)

Examples H 1 is well-covered H 1 C 4 & H 2 are very well -covered C 4 H 3 & H 4 are not well-covered H 2 H 4

Ex-Conjecture 3 If G is a well-covered graph, then I(G) is unimodal. J. I. Brown, K. Dilcher, R. J. Nowakowski J. of Algebraic Combinatorics 11 (2004) Example G G is a well-covered graph, I(G) = 1+9 x+ 25 x 2 +22 x 3 is unimodal.

Theorem I(G) is unimodal for every i. e. , Conjecture 3 is true well-covered graph G graph for every well-coveredhaving (G) 3. G having (G) 3. T. Michael, W. Traves, Graphs and Combinatorics 20 (2003) They also provided counterexamples for 4 (G) 7.

G = 4 K 10 + K 4, 4, …, 4 1701 times K 10 1701 -partite: each part has 4 vertices G K 10 K 4, 4, …, 4 1701 K 10 Michael & Traves’ counter K 10

G = 4 K 10 + K 4, 4, …, 4 n times 4 G is well-covered, (G) = 4 I(G) = 1+(40+4 n) x+(600+6 n) x 2 + (4000+4 n) x 3 + (10000+n) x 4 I(G) is NOT unimodal iff 4000+4 n min{40+4 n, 1000+n} I(G) is NOT unimodal iff 1701 n 1999 and it is NOT log-concave iff 24 n 2452

G = (4 K 10 + K 4, 4, …, 4) (4 K 1000) 1701 times G K 1000 K 4, 4, …, 4 (G) = 8 K 1000 1701 K 1000 K 10

G is well-covered and (G) = 8, while I(G) = 1 + 14, 844 x + 78, 762, 806 x 2 + 196, 342, 458, 804 x 3 + 235, 267, 430, 443, 701 x 4 + 109, 850, 051, 389, 608, 000 x 5 + 173, 242, 008, 824, 000 x 6 + 173, 238, 432, 000, 000 x 7 + 187, 216, 000, 000 x 8

G = (4 K 10+K 4, 4, …, 4) J. q. K 1000 q 27 (2006) V. E. Levit, E. Mandrescu, European of Combin. 1701 times G is well-covered K 1000 q K 10 0≤q (G ) = q + 4 K 1000 q q times K 4, 4, …, 4 1701 K 10

q=0 Michael & Traves Counter. Example 1≤q≤ 3 New Michael & Traves Counter. Examples 4 ≤q Counter. Examples for 8 ≤ K 1000 q q times K 10 K 4, 4, …, 4 1701 K 10

Gq is well-covered, not connected, (Gq) = q + 4 I(Gq; x) = (1+ 6844 x + 10806 x 2 + 10804 x 3 + 11701 x 4) (1 + 1000 q x)q is not unimodal. Proof: sq+2 > sq+3 < sq+4

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

Conjecture 3* If G is a very well-covered graph, then I(G) is unimodal. V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006) Example G is very well-covered I(G) = 1 + 6 x + 9 x 2 + 4 x 3 G

Theorem If G is a well-covered graph with (G) = , then (i) 0 k ( -k) sk (k+1) sk+1 (ii) s 0 s 1 … sk-1 sk , k = ( +1)/2 . T. Michael, W. Traves, Graphs and Combinatorics 20 (2003) V. E. Levit, E. Mandrescu, Discrete Applied Mathematics 156 (2008)

Proof each (k+1)-stable set includes k+1 stable sets of size k (k+1) sk+1 Every k-stable set Ak is included in some stable set B of size . hence, each B has -k stable subsets of size k+1 that include Ak ( -k) sk thus, ( -k) sk (k+1) sk+1 sk-1 sk , for k ( +1)/2

Theorem If G is a very well-covered graph with (G) = , then (i) ( -k) sk (k+1) sk+1 2 ( -k) sk (ii) s 0 s 1 … s /2 (iii) sp sp+1 … s -1 s , p = (2 -1)/3 (iv) s s -2 (s -1)2 (v) I(G) is unimodal, whenever 9. V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006)

Proof (i) It follows from previous results on quasi-reg graphs, as any well-covered graph is quasi-regularizable (Berge) (i) (ii) s 0 s 1 … s /2 (i) (iii) sp sp+1 … s -1 s where p = (2 -1)/3 (iv) Combining (ii) and (iii), it follows that I(G) is unimodal, whenever 9.

Conjecture 4 : “Roller-Coaster” For any permutation of {k, k+1, …, }, k = /2 , there is a well-covered graph G with (G) = , whose sequence (s 0 , s 1 , s 2 , …, s ) satisfies: s (k)< s (k+1)< …< s ( ). T. Michael, W. Traves, Graphs & Combinatorics 20 (2003)

The “Roller-Coaster” Conjecture is valid for (i) every well-covered graph G with (G) 7; T. Michael, W. Traves, Graphs and Combinatorics 20 (2003) (ii) every well-covered graph G with (G) 11. P. Matchett, Electronic Journal of Combinatorics (2004) What about (G) > 11 ? Still open …

P. Matchett (2004) unconstrained sk increasing 1 2 k 3 2 “Roller-Coaster” conjecture: For a well-covered graph, the sequence (sk) is unconstrained with respect to order in its upper part!

V. E. Levit, E. Mandrescu (2006) unconstrained sk increasing 1 2 decreasing 3 2 2 -1 3 k “Roller-Coaster” conjecture*: For a VERY well-covered graph, the sequence (sk) is unconstrained with respect to order in this upper part!

C. Berge, 1961 G is called perfect if (H) = (H) for any induced subgraph H of G, where (H), (H) are the chromatic and the clique numbers of H. E. g. , any chordal graph is perfect.

Theorem If G is a perfect graph with (G) = and (G) = , then sp sp+1 … s -1 s where p = ( 1) / ( 1). Example = 3, p = 2 G I(G) = 2+3 x 3 1+6 x+8 x

Unimodal ? Log-concave ? sk Unconstrained ? decreasing 1 2 3 -1 +1 We found out that the sequence (sk) is decreasing in its upper part: if G is a perfect graph with (G) = , then sp sp+1 … s -1 s for p = ( -1)/( +1). k

Proof If S is stable and |S| = k, then H = G-N[S] has (H) (G)-k. By Lovasz’s theorem |V(H)| (H)( -k) (G)( -k). (k+1) sk+1 (G) ( -k) sk and sk+1 sk is true while k+1 (G)( -k), i. e. , for k ( -1) / ( +1).

Examples G and H are perfect I(G) = 1 + 5 x + 4 x 2 + x 3 is log-concave G H = K 127+3 K 7 K 7 K 127 I(H) = 1+148 x +147 x 2 + 343 x 3 is not unimodal K 7

Remark If G is a minimal imperfect graph, then I(G) is log-concave. Example C 7 I(C 7) = 1 + 7 x + 14 x 2 + 7 x 3

Remark G = K 97+ 4 K 3~ C 5 There is an imperfect graph G whose I(G) is not unimodal. Example I(G) = 1 + 114 x + 603 x 2 + 921 x 3 + 891 x 4 + 945 x 5 + 405 x 6 K 3 G K 3 K 97 K 3 C 5

Corollary If G is a bipartite graph with (G) = , then sp sp+1 … s -1 s where p = (2 -1)/3. Example =5; p=3 I(G) = 1+8 x+19 x 2 +20 x 3+ 10 x 4+2 x 5 G

Corollary If T is a tree with (T) = , then sp sp+1 … s -1 s where p = (2 -1)/3. Example = 6 p = 4 I(T) = 1 + 8 x + 21 x 2 +26 x 3 +17 x 4 + 6 x 5 + x 6 T For P 4 p=1

Unimodal ? Log-concave ? sk 1 2 decreasing 3 2 -1 3 k Conjecture 1: I(T) is unimodal for a tree T. We found out that (sk) is decreasing in this upper part: if T is a tree, then sp sp+1 … s -1 s , p = (2 (T)-1)/3. V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006)

G is called a König-Egerváry (K -E) graph if (G) + (G) = |V(G)|. R. W. Deming, Discrete Mathematics 27 (1979) F. Sterboul, J. of Combinatorial Theory B 27 (1979) G H (G) + (G) = 5 Wellknown ! (H) + (H) < 6 If G is bipartite, then G is a König-Egerváry graph.

Theorem If G is a König-Egerváry graph, then (i) sk tk, k = (G), where = (G) and (1+2 x) (1+x) = t 0 + t 1 x +…+ t -1 x 1+ t x (ii) the coefficients sk satisfy (iii) sp ≥ sp+1 ≥… ≥ s -1 ≥ s for p = (2 1)/3 . V. E. Levit, E. Mandrescu, Congressus Numerantium 179 (2006)

Example Proof G H

Proof & I(G) = s 0 + s 1 x +…+ s -1 x 1+ s x

Theorem If G has sk stable sets of size k, 1 k (G) = , then D. C. Fisher, J. Ryan, Discrete Mathematics 103 (1992) … and an alternative proof was given by L. Petingi, J. Rodriguez, Congressus Numerantium 146 (2000)

Unimodal ? Log-concave ? sk Unconstrained ? decreasing 1 2 3 p k We found out that the sequence (sk) is decreasing in this upper part: If a König-Egerváry graph G has (G) = , then sp sp+1 … s -1 s for p = (2 1)/3

Example G = K 6 + 7 K 1 (G) = 7 (G) = 6 n = 13 G is a K-E graph G K 1 K 1 K 6 K 1 K 1 p = (2 -1)/3 = (14 -1)/3 = 5 s 5 = 21 s 6 = 7 s 7 = 1 I(G) = 1 + 13 x + 21 x 2 + 35 x 3 +35 x 4+ 21 x 5 + 7 x 6 + 1 x 7 21 21 13 35 < 0 I(G) is not log-concave, but unimodal! K 1

Conjecture 5 I(G) is unimodal for every -Egerváry graph G. Example König unimodal I(G) = 1 + 8 x + 20 x 2 +23 x 3 +20 x 4 +1 x 5 = 5, = 3, p = (2 -1)/3 = 3 G

Recall : “Corona” operation P 3 K 1 2 K 1 K 3 P 4 G = P 4 {P 3 , K 1 , 2 K 1 , K 3}

Particular case of “Corona” K 1 K 1 G = P 4 K 1 P 4 Remark Def. Each stable set of G = H K 1 can be enlarged to a maximum stable set. Equivalently, well-covered if allif each of its stable G is called G is well-covered its maximal stable sets is contained in a maximum stable 1970). sets are of the same size (M. D. Plummer, set.

Theorem Appending a single pendant edge to each vertex of H H*. Let G be a graph of girth > 5, which is isomorphic to neither C 7 nor K 1. Then G is well–covered if and only if G = H* for some graph H. A. Finbow, B. Hartnell, R. Nowakowski, J. Comb. Th B 57 (1993) Remark H* is very well-covered, for any graph H

Theorem If G is a graph of order n, and I(G) = s 0 + s 1 x +…+ s -1 x 1+ s x , then and the formulae connecting the coefficients of I(G) and of I(G*) are: V. E. Levit, E. Mandrescu, Discrete Applied Mathematics (2008)

A spider is a tree having at most one vertex of degree > 2. Well-covered spiders : K 1 K 2 P 4 Sn

Let T* be the tree obtained from the tree T by appending a single pendant edge to each vertex of T. Remark (T*) = the order of T Example (T*) = 4 ( T )*

Theorem Appending a single pendant edge to each vertex of H H*. For a tree T K 1 the following are equivalent: (i) T is well-covered (ii) T is very well-covered (iii) T = L* for some tree L G. Ravindra, Well-covered graphs, J. Combin. Inform. System Sci. 2 (1977) (iv) T is a is well-covered spider or T is obtained from a well-covered tree T 1 and a well-covered spider T 2, by adding an edge joining two nonpendant vertices of T 1, T 2, respectively. V. E. Levit, E. Mandrescu, Congressus Numerantium 139 (1999)

unconstrained sk increasing 1 2 decreasing 3 2 2 -1 3 k For every well-covered tree T, with (T) = , the sequence (sk) is unconstrained with respect to order in this upper part!

Proposition The independence polynomial of any well-covered spider Sn , n 1, is unimodal and mode(Sn) = n- (n-1)/3 V. E. Levit, E. Mandrescu, Congresus Numerantium 159 (2002) I(K 1)=1+x I(K 2) = 1+2 x all are I(P 4) = 1+4 x+3 x 2 unimodal !

Proposition Moreover, The independence polynomial of any well–covered spider Sn is log–concave. Pr oo f & “If P, Q are log-concave, then P Q is log-concave. ” V. E. Levit, E. Mandrescu, Carpathian J. of Math. 20 (2004)

Definition characteristic independence matching A (graph) polynomial P(x) = a 0 + a 1 x +…+ anxn is called palindromic if ai = an-i , i = 0, 1, . . . , n/2. J. J. Kennedy – “ vertex palindromic graphs, I. Gutman, Independent. Palindromic graphs” Graph Theory Notes of New York, XXIII (1992) Graph Theory Notes of New York, XXII (1992) Example P(x) = (1 + x)n In fact, (1+x)n = I(n. K 1) n. K 1 v 2 v 3 vn

Theorem Let G = H q. K 1 have (G) = and (sk) be the coefficients of I(G). Then the following are true: (i) |S| q |NG(S)| for every stable set S of G; (ii) q (k+1) sk+1 (q+1) ( -k) sk, 0 k < (iii) sr … s -1 s , r = ((q+1) - q)/(2 q+1) (iv) if q = 2, then I(G) is palindromic and s 0 s 1 … sp , p = (2 +2)/5 sr … s -1 s , r = (3 -2)/5 .

Unimodal ? Log-concave ? sk Unconstrained ? decreasing 1 2 3 r We found out that the sequence (sk) is decreasing in this upper part: if G = H q. K 1 has (G) = , then sr … s -1 s , r = ((q+1) -q)/(2 q+1) k

sk Unimodal ? decreasing increasing 1 2 3 2 +2 5 3 -2 5 k If G = H 2 K 1 , 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 ?

Examples K 1, 3 = the “claw” I(K 1, 3) = 1+4 x+3 x 2+x 3 is not palindromic. I(G) = 1+s 1 x+s 2 x 2 = 1+nx+x 2 G = Kn–e, n 2 Remarks 1. If (G) = 2 and I(G) is palindromic, then n 2, I(G) = 1 + n x + 1 x 2 and I(G) is log-concave, and hence unimodal, as well. 2. If (G) = 3 and I(G) is palindromic, then n 3, I(G) = 1 + n x + nx 2 + 1 x 3 and I(G) is log-concave, and hence unimodal, as well.

I(G) = 1+2406 x+1382 x 2+1382 x 3+2406 x 4+1 x 5 (G) = 5 s 1 = 5+28+1832+539+2 = 2406 K 2 K 539 s 4 = 5+7 7 = 2406 s 2 = 10+2 539+6 7 7 K 1832 = 1382 s 3 = 10+4 7 7 7 = 1382 5 K 1 4 K 7 G = K 1832 + 4 K 7 + (K 2 K 539) + 5 K 1

Theorem If G has a stable set S with: |N(A) S| = 2|A| for every stable set A V(G) – S, then I(G) is palindromic. D. Stevanovic, Graphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV (1998) Example S = { } G I(G) = 1+ 5 x 2 + 1 x 3

Remark The condition that: “G has a stable set S with: |N(A) S| = 2|A| for every stable set A V(G) – S” is NOT necessary! Example G I(G) = 1+6 x+6 x 2+1 x 3 S = { } I. Gutman, Independent vertex palindromic graphs, Graph Theory Notes of New York XXIII (1992)

Corollary If G = (V, E) has s =1, s -1=|V| and the unique maximum stable set S satisfies: |N(u) S| = 2 for every u V-S, then I(G) is palindromic. D. Stevanovic, Graphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV (1998) Example S={ } G I(G) = 1 + 9 x + 27 x 2 + 38 x 3+ + 1 x 6 + 9 x 5 + 27 x 4

How to build graphs with palindromic independence D. Stevanovic, Graphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV (1998) polynomials ? A clique cover of G is a spanning graph of G, each component of which is a clique. RULE 1: If is a clique cover of G, then: for each clique C , add two new non-adjacent vertices and join them to all the vertices of C. The new graph is denoted by {G}. The set S = {all these new vertices} is the unique maximum stable set in the new graph H = {G} and satisfies: |N(u) S| = 2 for any u V(H)-S. Hence, I(H) is palindromic by Stevanovic’s Theorem.

I(G) = 1+6 x+9 x 2+2 x 3 Example ={ } G I(H) = 1+12 x+48 x 2+76 x 3+48 x 4 +12 x 5+1 x 6 H = {G} |N(u) S| = 2, for any u V(H)-S S={ }

In particular: If each clique of the clique cover of G consists of a single vertex, then: the new graph {G} is denoted by G○2 K 1. Example G○m. K 1 is the corona of G and m. K 1. G ={ G○2 K 1 } I(G○2 K 1) = 1 + 12 x + 53 x 2 + 120 x 3+ +156 x 4+1 x 8+ 12 x 7 + 53 x 6 + 120 x 5

How to build graphs with palindromic independence D. Stevanovic, Graphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV (1998) polynomials ? A cycle cover of G is a spanning graph of G, each component of which is a vertex, an edge, or a proper cycle. RULE 2. If is a cycle cover of G, then: (1) add two pendant neighbors to each vertex from ; (2) for each edge ab of , add two new vertices and join them to a & b; (3) for each edge xy of a proper cycle of , add a new vertex and join it to x & y. The new graph is denoted by {G}. The set S = {ALL THESE NEW VERTICES} is stable in the new graph H = {G} and satisfies: |N(v) S| = 2 for any v V(H)-S. Therefore, I(H) is palindromic.

Example ={ I(G) = 1+7 x+13 x 2+5 x 3 } G is a cycle cover I(H) = 1+15 x+83 x 2+218 x 3+298 x 4+218 x 5+83 x 6+15 x 7+1 x 8 H = {G} |N(u) S| = 2, for any u V(H)-S S={ }

Proposition Let G = H 2 K 1 have (G) = and (sk) be the coefficients of I(G). Then I(G) is palindromic and s 0 s 1 … sp , p = (2 +2)/5 sr … s -1 s , r = (3 -2)/5 . V. E. Levit, E. Mandrescu, 39 th Southeastern Intl. Conf. on Combininatorics, Graph Theory, and Computing, Florida Atlantic University, March 3 -7, 2008

sk Unimodal ? decreasing increasing 1 2 3 2 +2 5 3 -2 5 k If G = H 2 K 1 , 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 ?

Example P 4 I(P 4) = 1+4 x+3 x 2 (G) = 8 G = P 4 o 2 K 1 p = (2 (G)+2)/5 = 3, r = (3 (G)-3)/5 = 5 s 0 = 1 s 1 = 12 s 2 = 55 s 3 = 128 (p = 3) Theorem I(Pn 2 K 1) has only real roots, and consequently is log-concave. 5 Zhu Zhi-Feng, Australasian Journal of Combinatorics 38 (2007) 6 7 8 s = 128 s = 55 s = 12 s = 1 I(G) = (r = 5) 1 + 12 x + 55 x 2 + 128 x 3 + 168 x 4 + 128 x 5 + 55 x 6 + 12 x 7 + x 8

Theorem If G is quasi-regularizable of order 2 (G), then sp sp+1 … s -1 s , p = (2 -1)/3. V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006) Example (G) = 6 G is quasi-regularizable p = (12 -1)/3 = 4 s 4 = 15 s 5 = 6 s 6 = 1 n = 12 G K 1 K 1 K 6 K 1 K 1 I(G) = 1 + 12 x + 15 x 2 + 20 x 3 + 15 x 4 + 6 x 5 + 1 x 6

Theorem 2 If is a cycle cover of H without vertex-cycles and G = {H} has (G) = , then G is quasi-regularizable of order 2 and I(G) is palindromic, while its coefficients (sk) (ii) s 0 s 1 … sp , satisfy: p = ( +1)/3 sq … s -1 s , q = (2 -1)/3 (s 1)2 s 0 s 2 , (s 2)2 s 1 s 3 and (s -1)2 s s -2 , (s -2)2 s -1 s -3 V. E. Levit, E. Mandrescu, 39 th Southeastern Intl. Conf. on Combinatorics, Graph Theory, and Computing, Florida Atlantic University, March 3 -7, 2008

sk Unimodal ? decreasing increasing 1 2 3 +1 3 2 -1 3 If is a cycle cover of H without vertex-cycles, 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 ? k

I(H) = 1+8 x+19 x 2+13 x 3+x 4 Example ={ } H G = {H} s 0 s 1 s 2 s 3 , p = ( +1)/3 = 3 S={ } s 5 s 6 s 7 s 8 , q = (2 -1)/3 = 5 (G) = 8 I(G) = 1+16 x+95 x 2+265 x 3+371 x 4+265 x 5+95 x 6+16 x 7+1 x 8

Some family relationships G○q. K 1 König. Egerváry graphs = Very wellcovered G ○K 1 wellcovered G ○K p G○2 K 1 & G perfect Perfect T○2 K 1 & T = a tree Bipartite Trees quasiregularizable {G} & is a cycle cover of G

… f i n a l l y, recall s o m e o p e n p r o b l e m s

Problem 1 Find an inequality leading to partial log-concavity of the independence polynomial. Example For very-well covered graphs: (S -1)2 S S -2 V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006)

Problem 2 Characterize polynomials that are independence polynomials. C. Hoede, X. Li Discrete Mathematics 125 (1994) Example P(x) = (1 + x)n P(x) = I(n. K 1) but, there is no graph G whose I(G) = 1 + 4 x + 17 x 2

Problem 3 Characterize the graphs whose independence polynomials are palindromic. D. Stevanovic Graph Theory Notes of New York XXXIV (1998) Example (G) = 2 A graph G with (G) = 2 has a palindromic independence polynomial iff G = Kn- e. I(G) = 1 + n x + 1 x 2

… Thank you ! ! . . . תודה Thank you very much ! ! ת ו ד ה רבה