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Inequalities Involving the Coefficients of Independence Polynomials Vadim E. Levit 1, 2 1 Ariel 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 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 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 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 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}, 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 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) 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 … 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 … 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 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; 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 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 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. 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 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 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 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 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}) 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 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: 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 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 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 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 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. , 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, 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? 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. 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. 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 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 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 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 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 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 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 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 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 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. 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) = 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)| 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 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 = 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 = 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 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!

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

Definitions A graph G is called well-covered if all its maximal stable sets are 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 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. 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 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 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 = 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) 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 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 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 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+ 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. 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. 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 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 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) 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 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 = 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; 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 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 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!

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

C. Berge, 1961 G is called perfect if (H) = (H) for any induced 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) = , 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 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 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 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 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 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 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 … 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 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)

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

G is called a König-Egerváry (K -E) graph if (G) + (G) = |V(G)|. 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), 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 Example Proof G H

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

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

Recall : “Corona” operation P 3 K 1 2 K 1 K 3 P 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 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 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 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. 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 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 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 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 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 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)

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

Definition characteristic independence matching A (graph) polynomial P(x) = a 0 + a 1 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 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 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 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 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) = 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 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| = 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 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 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) = 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 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 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 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 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 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 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 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 = 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 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 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 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 … 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 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 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 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 ! … Thank you ! ! . . . תודה Thank you very much ! ! ת ו ד ה רבה