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

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

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Introduction Quasi-Regularizable Graphs König-Egerv á ry Graphs The Main Inequality Well-Covered Graphs Perfect Graphs Corona Graphs Palindromic Graphs

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

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If s k 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. Gutman & Harary - ‘83 Harary I. Gutman, F. Harary, Generalizations of matching polynomial Utilitas Mathematica 24 (1983)

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All the stable sets of G : {a}, {b}, {c}, {d}, {e}, {f} ….. {a, b}, {a, d}, {a, e}, {a, f}, {b, c}, {b, e}, {b, f}, {d, f}, …… { a, b, e }, { a, b, f }, { a, d, f } All the stable sets of G : …… …… {a}, {b}, {c}, {d}, {e}, {f} ….. {a, b}, {a, d}, {a, e}, {a, f}, {b, c}, {b, e}, {b, f}, {d, f}, …… { a, b, e }, { a, b, f }, { a, d, f } 1683 1683 G c b d a f e I (G) = 1 + 6 x + 8 x 2 + 3 x 3 Example

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

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ALSO non-isomorphic trees can have the same independence polynomial ! T1T1 T2T2 K. Dohmen, A. Ponitz, P. Tittmann, Discrete Mathematics and Theoretical Computer Science 6 (2003) + I (T 1 ) = I (T 2 ) = 1+10 x +36 x 2 +58 x 3 +42 x 4 +12 x 5 + x 6

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, each edge { a }, { b }, … pairs as { a, b }, { a, c }, … and also { a, b, e }. , each edge { a }, { b }, … pairs as { a, b }, { a, c }, … and also { a, b, e }. A matching in a graph G is a set of edges that A matching in a graph G is a set of edges that have no endpoints in common. … for historical reasons, recall G a e b f c d but {a, d}, {b, f}, {a, b, c} … are not matchings Examples

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

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

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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 + 6 x+ 7 x 2 + 1 x 3 G c b d a f e LG c b d a f e G = (V,E)LG = (E,U) ExampleExample

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

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

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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 : 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). P(G;x,y) The generalized chromatic polynomial : P(G;x,y) 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. 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). Connections with other polynomials: Re ma rk

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

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I (K 3 + K 4 ) = I (K 3 ) + I (K 4 ) – 1 = = (1+3x) + (1+4x) – 1 = 1+7x K3K3 K4K4 + ExampleExampleExampleExample

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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. G H G○HG○HG○HG○H GHHH Example

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

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

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N[ v ] = { a,c,d } { v } c G a db v I( G ) = I (G-v) + x I (G-N[v]) = P4P4 a c d b = I (P 4 ) + x I ({b}) = = 1 + 4 x + 3 x 2 + x ( 1 +x) = I (P 4 ) = 1 + 4 x + 3 x 2 = 1 + 5 x + 4 x 2 G - v = P 4 G -N[ v ] = { b } Example

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

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

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A polynomial P ( x ) = a 0 + a 1 x + … + a n x n is unimodal (log-concave) if its sequence of coefficients a 0, a 1, a 2,..., a n is unimodal (log-concave, respectively). P (x) = 1 + 4x + 50x 2 + 2x 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 Example

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Is there a (connected) graph G with (G) = whose sequence s 0, s 1, s 2, …, s is NOT unimodal ? Recall that s k 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 s 0, s 1, s 2, …, s is NOT unimodal ! Y. Alavi, P. Malde, A. Schwenk, P. Erdös Congressus Numerantium 58 (1987)

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I (H) = 1+64 x +634 x 2 +500 x +625 x is not unimodal I (H) = 1+64 x +634 x 2 +500 x 3 +625 x 4 is not unimodal unimodal I ( G ) = 1 + 6x + 8x 2 + 2x 3 is unimodal Examples K5K5 K 22 K5K5 K5K5 K5K5 H G

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{1, 2, …, }, G (G) = s (1) < s (2) < s (3) < … < s ( ) s k stable sets G k. 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 s k is the number of stable sets in G of size k. Moreover, any deviation from unimodality is possible! Theorem Y. Alavi, P. Malde, A. Schwenk, P. Erdös Congressus Numerantium 58 (1987)

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

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If all the roots of a polynomial with positive coefficients are real, then the polynomial is log-concave. Sir I. Newton, Arithmetica Universalis (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)

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

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

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

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There exist There exist unimodal independence polynomials whose product is not unimodal. =++ + + + + I (G G) = I (G) I (G) = = 1+232x + 13750x 2 + 34790x 3 + 101185 x 4 + 100842x 5 + 117649 x 6 I (G) = 1+116 x +147 x 2 +343 x 3 Example G K7K7 K 95 K7K7 K7K7 G = K 95 +3 K 7

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

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

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

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Introduction Quasi-Regularizable Graphs König-Egerv á ry Graphs The Main Inequality Well-Covered Graphs Perfect Graphs Corona Graphs Palindromic Graphs

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0 = 0, = n, 1 0 = 0, = n, 1 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

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

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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)s k+1 if X A and v V(G) N[X] X {v} B hence, deg H (X) -k and (k+1) s k+1 = |E| -k s k ProofProofProofProof

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Introduction Quasi-Regularizable Graphs König-Egerv á ry Graphs The Main Inequality Well-Covered Graphs Perfect Graphs Corona Graphs Palindromic Graphs

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

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If G is a quasi-regularizable graph of order n = 2 G2 then If G is a quasi-regularizable graph of order n = 2 (G) = 2 , then (ii) (k+1) s k+1 2 ( -k) s k (ii) (k+1) s k+1 2 ( -k) s k (i) -k 2 ( -k) (i) -k 2 ( -k) (iii) s p s p+1 … s -1 s where p = (2 -1)/3 . Theorem V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006)

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

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We found out that (s k ) is decreasing in this upper part: if G is quasi-regularizable of order 2, then if G is quasi-regularizable of order 2 (G), then 123 p k sksk decreasing Unimodal ? Log-concave ? Unconstrained ? p = (2 1)/3) s p … s -1 s , p = (2 1)/3)

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

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

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Example G is a quasi-reg graph I (G) = 1 + 16 x + 15 x 2 + 20 x 3 +15 x 4 + 6 x 5 + 1 x 6 p = (2 -1)/3 = (12-1)/3 = 4 (G) = 6 n = 16 G is quasi-reguralizable & I (G) is not unimodal! G = K 10 + 6 K 1 s 4 = 15 s 5 = 6 s 6 = 1 G K 10 K1K1 K1K1 K1K1 K1K1 K1K1 K1K1

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Introduction Quasi-Regularizable Graphs König-Egerv á ry Graphs The Main Inequality Well-Covered Graphs Perfect Graphs Corona Graphs Palindromic Graphs

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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

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C 4 & H 2 are very well-covered Examples H1H1 H 1 is well-covered H4H4 H 3 & H 4 are not well-covered H2H2 C4C4 H3H3

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

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

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

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G = 4K 10 + K 4, 4, …, 4 G = 4K 10 + K 4, 4, …, 4 n times 4 GG 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 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}

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

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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 + 173,242,008,824,000,000 x 6 + 173,238,432,000,000,000 x 7 + 187,216,000,000,000,000 x 8

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

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q = 0 K 4, 4,…, 4 1701 K 1000q K 10 K 1000q K 10 q times 4 ≤ q 1 ≤ q ≤ 3 Michael & Traves CounterExample New Michael & Traves CounterExamples CounterExamples for 8 ≤

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

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

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G is very well-covered I (G) = 1 + 6x + 9x 2 + 4x 3 Example Conjecture 3* G If G then I (G) 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)

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

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

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

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

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any For any permutation of well-covered {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 ( ). Conjecture 4 : “Roller-Coaster” Conjecture 4 : “Roller-Coaster” T. Michael, W. Traves, Graphs & Combinatorics 20 (2003)

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

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For a well-covered graph, the sequence (s k ) is unconstrained with respect to order in its upper part! 123 22 k sksk increasing unconstrained “Roller-Coaster” conjecture: P. Matchett (2004)

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“Roller-Coaster” conjecture*: For a VERY well-covered graph, the sequence (s k ) is unconstrained with respect to order in this upper part! 123 2 -1 3 k sksk increasing unconstrained 22 decreasing V. E. Levit, E. Mandrescu (2006)

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Introduction Quasi-Regularizable Graphs König-Egerv á ry Graphs The Main Inequality Well-Covered Graphs Perfect Graphs Corona Graphs Palindromic Graphs

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

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

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We found out that the sequence (s k ) is decreasing in its upper part: if G is a perfect graph with then s p s p+1 … s -1 s = ( -1)/( +1) . if G is a perfect graph with (G) = , (G) = , then s p s p+1 … s -1 s for p = ( -1)/( +1) . 123 -1 +1 k sksk decreasing Unimodal ? Log-concave ? Unconstrained ?

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

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I H = 1+ 148 x +147 x 2 + 343 x is not unimodal I (H) = 1+ 148 x +147 x 2 + 343 x 3 is not unimodal log-concave I (G) = 1 + 5x + 4x 2 + x 3 is log-concave G K 127 K7K7 K7K7 K7K7 H = K 127 +3K 7 Examples G and H are perfect

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

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There is an imperfect graph G whose I ( G ) is not unimodal. Example G = K 97 + 4K 3 ~ C 5 K 97G K3K3 K3K3 K3K3 K3K3 C5C5 G I ( G ) = 1 + 114x + 603x 2 + 921 x 3 + 891 x 4 + 945 x 5 + 405x 6 Remark

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

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

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We found out that (s k ) is decreasing in this upper part: Conjecture 1: I(T) is unimodal for a tree T. 123 2 -1 3 k sksk decreasing Unimodal ? Log-concave ? if T is a tree,then s p s p+1 … s -1 s = (2 (T)-1)/3 . if T is a tree,then s p s p+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)

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

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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

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

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G H Example Proof

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

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If G has s k 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

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We found out that the sequence ( s k ) is decreasing in this upper part: If a König-Egerváry graph G has (G) = , then 123 p k sksk decreasing Unimodal ? Log-concave ? Unconstrained ? s p s p+1 … s -1 s for p = (2 1)/3

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Example G is a K-E graph I ( G ) = 1 + 13 x + 21 x 2 + 35 x 3 +35 x 4 + 21 x 5 + 7 x 6 + 1 x 7 p = (2 -1)/3 = (14-1)/3 = 5 (G) = 7 (G) = 6 n = 13 21 21 13 35 < 0 I (G) is not log-concave, but unimodal! G = K 6 + 7 K 1 s 5 = 21 s 6 = 7 s 7 = 1 G K6K6 K1K1 K1K1 K1K1 K1K1 K1K1 K1K1 K1K1

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

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Introduction Quasi-Regularizable Graphs König-Egerv á ry Graphs The Main Inequality Well-Covered Graphs Perfect Graphs Corona Graphs Palindromic Graphs

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Recall : “Corona” operation P3P3 P4P4 K1K1 2K 1 K3K3 G = P 4 {P 3, K 1, 2K 1, K 3 }

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Particular case of “Corona” K1K1 P4P4 K1K1 K1K1 K1K1 G = P 4 K 1 Each stable set of G = H K 1 can be enlarged to a maximum stable set. Remark G is called well-covered if all its maximal stable sets are of the same size (M.D. Plummer, 1970). Def. Equivalently, G is well-covered if each of its stable sets is contained in a maximum stable set.

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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. 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

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…+ s -1 + s 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: Theorem V. E. Levit, E. Mandrescu, Discrete Applied Mathematics (2008)

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

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

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(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 non-pendant vertices of T 1, T 2, respectively. For a tree T K 1 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) (ii) T is very well-covered

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the sequence (s k ) is unconstrained with respect to order in this upper part! 123 2 -1 3 k sksk increasing unconstrained 22 decreasing For every well-covered tree T, with (T) = ,

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

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

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Introduction Quasi-Regularizable Graphs König-Egerv á ry Graphs The Main Inequality Well-Covered Graphs Perfect Graphs Corona Graphs Palindromic Graphs

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A (graph) polynomial P( x ) = a 0 + a 1 x + … + a n x n is called palindromic if a i = a n-i, i = 0,1,..., n/2 . P (x) = (1 + x) n J. J. Kennedy – “ Palindromic graphs ” Graph Theory Notes of New York, XXII (1992) nK 1 v1v1 v2v2 v3v3 vnvn In fact, (1+ x ) n = I( nK 1 ) Definition characteristicmatchingindependence I. Gutman, Independent vertex palindromic graphs, Graph Theory Notes of New York, XXIII (1992) Example

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

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We found out that the sequence ( s k ) is decreasing in this upper part: if G = has, then if G = H qK 1 has (G) = , then 123 r k sksk decreasing Unimodal ? Log-concave ? Unconstrained ? r = ((q+1) -q)/(2q+1) s r … s -1 s , r = ((q+1) -q)/(2q+1)

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If G =, then G If G = H 2K 1, then I( G ) is palindromic and its sequence (s k ) is increasing in its first part and decreasing in its upper part ! 123 3 -2 5 k sksk increasing Unimodal ? 2 +2 5 decreasing Question: G Is I( G ) unimodal ?

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

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Is there a connected graph G with (G) = whose I( G ) is palindromic, but NOT unimodal ? Question G = K 52 + 4K 1 + 3K 4 Answer I (G) = 1 + 68 x + 54 x 2 + 68 x 3 + 1 x 4 4K 1 3K 4 K 52 S 2 = 6+3 4 4 = 54 S 3 = 4+4 4 4 = 68 = s 1 ( G ) = 4

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

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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) Theorem S = { } I (G) = 1+ 5 x + 5 x 2 + 1 x 3 Example G

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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 ! Remark G S = { } I (G) = 1+6 x +6 x 2 +1 x 3 I. Gutman, Independent vertex palindromic graphs, Graph Theory Notes of New York XXIII (1992) Example

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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, is 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. Corollary G S ={ } I (G) = 1 + 9 x + 27 x 2 + 38 x 3 + + 1 x 6 + 9 x 5 + 27 x 4 I (G) = 1 + 9 x + 27 x 2 + 38 x 3 + + 1 x 6 + 9 x 5 + 27 x 4 D. Stevanovic, Graphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV (1998) Example

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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}. A clique cover of G is a spanning graph of G, each component of which is a clique. 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. D. Stevanovic, Graphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV (1998) How to build graphs with palindromic independence polynomials ?

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

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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○2K 1. G○mK 1 is the corona of G and mK 1. G○2K 1 G = { } I (G○2K) = 1 + 12 x + 53 x 2 + 120 x 3 + +156 x 4 +1 x 8 + 12 x 7 + 53 x 6 + 120 x 5 I (G○2K 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 Example

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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. A cycle cover of G is a spanning graph of G, each component of which is a vertex, an edge, or a proper cycle. 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. The new graph is denoted by {G}. D. Stevanovic, Graphs with palindromic independence polynomial Graph Theory Notes of New York XXXIV (1998) How to build graphs with palindromic independence polynomials ?

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

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

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If G =, then G If G = H 2K 1, then I( G ) is palindromic and its sequence (s k ) is increasing in its first part and decreasing in its upper part ! 123 3 -2 5 k sksk increasing Unimodal ? 2 +2 5 decreasing Question: G Is I( G ) unimodal ?

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I (G) = 1 + 12 x + 55 x 2 + 128 x 3 + 168 x 4 + 128 x 5 + 55 x 6 + 12 x 7 + x 8 = 1+4+3 2 I( P 4 ) = 1+4 x +3 x 2 s 0 = 1 s 1 = 12 s 2 = 55 s 3 = 128 (p = 3) p = (2 ( G )+2)/5 = 3, r = (3 ( G )-3)/5 = 5 ( G ) = 8 s 5 = 128 s 6 = 55 s 7 = 12 s 8 = 1 (r = 5) Example G = P 4 o2K 1 P4P4 Theorem I (P n 2K 1 ) has only real roots, and consequently is log-concave. Zhu Zhi-Feng, Australasian Journal of Combinatorics 38 (2007)

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If G is quasi-regularizable of order 2, then s p s p+1 … s -1 s , p = (2 -1)/3 . If G is quasi-regularizable of order 2 (G), then s p s p+1 … s -1 s , p = (2 -1)/3 . Theorem Example G K6K6 K1K1 K1K1 K1K1 K1K1 K1K1 K1K1 (G) = 6 n = 12 is quasi-regularizable G is quasi-regularizable p = (12-1)/3 = 4 s 4 = 15 s 5 = 6 s 6 = 1 I ( G ) = 1 + 12 x + 15 x 2 + 20 x 3 + 15 x 4 + 6 x 5 + 1 x 6 V. E. Levit, E. Mandrescu, Graph Theory in Paris: Proceedings of a Conference in Memory of C. Berge (2006)

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

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123 2 -1 3 k sksk increasing Unimodal ? +1 3 decreasing vertex-cycles, If is a cycle cover of H without vertex-cycles, G = {H} has (G) = , then I( G ) is palindromic and its sequence (s k ) is increasing in its first part and decreasing in its upper part ! Question: Is I( G ) unimodal ?

124
G = {H} I (H) = 1+8 x +19 x 2 +13 x 3 + x 4 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 S ={ } = { } H s 0 s 1 s 2 s 3, p = ( +1)/3 = 3 s 5 s 6 s 7 s 8, q = (2 -1)/3 = 5 (G) = 8 Example

125
Trees Bipartite König- Egerváry graphs Perfect Very well- covered well- covered quasi- regularizable G○qK 1 G○2K 1 G G○2K 1 & G perfect G○K1G○K1G○K1G○K1 G○KpG○KpG○KpG○Kp {G} & is a cycle cover of G Some family relationships T○2K 1 T = T○2K 1 & T = a tree =

126
s o m e s o m e p r o b l e m s p r o b l e m s o p e n … f i n a l l y, recall

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

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

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

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