Communicated by Alireza Abdollahi. 1. Introduction. For a set S V, the open neighborhood of S is N(S) = v S

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1 Transactions on Combinatorics ISSN print: , ISSN on-line: Vol. 01 No , pp c 2012 University of Isfahan ON THE VALUES OF INDEPENDENCE AND DOMINATION POLYNOMIALS AT SPECIFIC POINTS S. ALIKHANI AND M. H. REYHANI Communicated by Alireza Abdollahi Abstract. Let G be a simple graph of order n. We consider the independence polynomial and the domination polynomial of a graph G. The value of a graph polynomial at a specific point can give sometimes a very surprising information about the structure of the graph. In this paper we investigate independence and domination polynomial at 1 and Introduction Let G = V, E be a simple graph. For any vertex v V, the open neighborhood of v is the set Nv = {u V uv E} and the closed neighborhood is the set N[v] = Nv {v}. For a set S V, the open neighborhood of S is NS = v S Nv and the closed neighborhood of S is N[S] = NS S. The corona of two graphs G 1 and G 2, as defined by Frucht and Harary in [12], is the graph G = G 1 G 2 formed from one copy of G 1 and V G 1 copies of G 2, where the ith vertex of G 1 is adjacent to every vertex in the ith copy of G 2. The corona G K 1, in particular, is the graph constructed from a copy of G, where for each vertex v V G, a new vertex v and a pendant edge vv are added. The join of two graphs G 1 and G 2, denoted by G 1 + G 2 is a graph with vertex set V G 1 V G 2 and edge set EG 1 EG 2 {uv u V G 1 and v V G 2 }. Graph polynomials are a well-developed area useful for analyzing properties of graphs. The value of a graph polynomial at a specific point can give sometimes a very surprising information about the structure of the graph. Balister, et al. in [9] proved that for any graph G, qg, 1 is always a power MSC2010: Primary: 05C60 ; Secondary: 05C31. Keywords: Independence polynomial; Domination polynomial; Value. Received: 11 June 2012, Accepted: 16 July Corresponding author. 49

2 50 S. Alikhani and M. H. Reyhani of 2, where qg, x is interlace polynomial of a graph G. Stanley in [14] proved that 1 n P G, 1 is the number of acyclic orientations of G, where P G, λ is the chromatic polynomial of G and n = V G. In this paper we study independence and domination polynomial of a graph at 1 and 1. For convenience, the definition of the independence and domination polynomial of a graph will be given in the next sections. We denote the path of order n, the cycle of order n, and the wheel of order n, by P n, C n, and W n, respectively. 2. The independence polynomial of a graph at 1 and 1 An independent set of a graph G is a set of vertices where no two vertices are adjacent. The independence number is the size of a maximum independent set in the graph and denoted by αg. For a graph G, let i k denote the number of independent sets of cardinality k in G k = 0, 1,..., α. The independence polynomial of G, α IG, x = i k x k, is the generating polynomial for the independent sequence i 0, i 1, i 2,..., i α. IG; 1 = i 0 + i 1 + i i α equals the number of independent sets of G, and IG; 1 = i 0 i 1 + i α i α = f 0 G f 1 G, where f 0 G = i 0 + i 2 + i , f 1 G = i 1 + i 3 + i are equal to the numbers of independent sets of even size and odd size of G, respectively. IG; 1 is known as the alternating number of independent sets. We need the following theorem: Theorem 2.1. [13] If w V G and uv EG, then the following equalities hold: i IG; x = IG w; x + xig N[w]; x, ii IG; x = IG uv; x x 2 IG Nu Nv; x. A vertex v is pendant if its neighborhood contains only one vertex; an edge e = uv is pendant if one of its endpoints is a pendant vertex. Theorem 2.2. If u V G is a pendant vertex of G and v Nu, then IG; 1 = IG N[v]; 1. Proof. Since u V G is a pendant vertex of G and v Nu, Theorem 2.1 assures that IG; x = IG v; x + xig N[v]; x = 1 + xig {u, v}; x + xig N[v]; x. Therefore IG; 1 = IG N[v]; 1. k=0

3 Graph polynomial at 1 and 1 51 Theorem 2.3. For n 1, the following hold: i IP 3n 2, 1 = 0 and IP 3n 1, 1 = IP 3n, 1 = 1 n ; ii IC 3n, 1 = 2 1 n, IC 3n+1, 1 = 1 n and IC 3n+2, 1 = 1 n+1 ; iii IW 3n+1, 1 = 2 1 n 1 and IW 3n, 1 = IW 3n+2, 1 = 1 n 1. Proof. i We prove by induction on n. For n = 1, since IP 1, x = 1 + x, IP 2, x = 1 + 2x and IP 3, x = 1 + 3x + x 2, we have IP 1, 1 = 0 and IP 2, 1 = IP 3, 1 = 1. Suppose that the result is true for any k 3n. By Theorem 2.1i, IP k+1, x = IP k, x + xip k 1, x, therefore IP 3n+1, 1 = IP 3n, 1 IP 3n 1, 1 = 1 n 1 n = 0; IP 3n+2, 1 = IP 3n+1, 1 IP 3n, 1 = 0 1 n = 1 n+1 ; IP 3n+3, 1 = IP 3n+2, 1 IP 3n+1, 1 = 1 n+1 0 = 1 n+1 ii Since IC 3, x = 1 + 3x, IC 4, x = 1 + 4x + 2x 2, IC 5, x = 1 + 5x + 5x 2 and IC 6, x = 1 + 6x + 9x 2 + 2x 3, we have IC 3, 1 = 2 1, IC 4, 1 = 1 and IC 5, 1 = 1 2. By Theorem 2.1 and Part i, IC n, x = IP n 1, x + xip n 3, x. Therefore IC n, 1 = IP n 1, 1 IP n 3, 1. Now it is easy to see that IC 3k, 1 = 2 1 k, IC 3k+1, 1 = 1 k and IC 3k+2, 1 = 1 k+1. iii By Theorem 2.1i, IW n, x = IC n 1, x + x. Now by part ii we have the result. In the following theorem we compute the number of independent sets of paths, cycles and wheels: Theorem 2.4. i For any positive integer n, ii For any integer n 3, iii For any integer n 4, IP n, 1 = n cos 2 sπ. n + 2 n 2 IC n, 1 = cos 2 s 1 2 π. n IW n, 1 = 1 + n cos 2 s 1 2 π n 1 Proof. We know that see [8], IP n, x = n+1 n 2 2x x cos the results. 2s 1π n 2 and IW n, x = IC n 1, x + x.. 2x x cos 2sπ n+2, IC n, x = By substituting x = 1, we have

4 52 S. Alikhani and M. H. Reyhani We recall the following theorem: Theorem 2.5. [3, 11] If G has t connected components G 1,..., G t, then IG, x = t i=1 IG i, x. Theorem 2.6. For any tree T, IT ; 1 { 1, 0, 1}, i.e., the number of independent sets of even size varies by at most one from the number of independent sets of odd size. Proof. By induction on n = V T. It is not hard to see that for any tree with order 1, 2, 3, 4, 5 we have the result. Now suppose that V T = n + 1 5, v is a pendant vertex of T, Nv = {u} and T 1, T 2,..., T k are the trees of the forest T N[u]. According to Theorem 2.1ii, we have: IT ; x = IT uv; x x 2 IT Nu Nv; x = 1 + xit v; x x 2 IT N[u]; x = 1 + xit v; x x 2 IT 1 ; xit 2 ; x... IT k ; x Therefore IT ; 1 = IT 1 ; 1IT 2 ; 1... IT k ; 1 { 1, 0, 1}, since every tree T i has less than n vertices, by the induction hypothesis, IT i ; 1 { 1, 0, 1}. Theorem 2.7. If F is a forest, then IF ; 1 { 1, 0, 1}. Proof. If T 1, T 2,..., T k, k 1, are the connected components of F, then IF ; x = IT 1 ; xit 2 ; x... IT k ; x, because F = T 1 T 2... T k. Now, the conclusions follow from Theorem 2.6. Theorem 2.8. [3, 11] IG 1 + G 2 ; x = IG 1 ; x + IG 2 ; x 1. Theorem 2.9. For any integer k there is some connected graph G such that IG; 1 = k. Proof. First suppose that k is a negative integer. It is easy to check that IK a,a,...,a, x = n1 + x a a 1, and therefore IK a,a,...,a, 1 = 1 a i.e., for any negative integer k there is some connected graph G such that IG; 1 = k. Now suppose that k is a positive integer. Let G i, 1 i k, be k graphs with IG i ; 1 = 2 for every i {1, 2,..., k} for example we can put G i = C 6 for every i {1, 2,..., k} and H = G 1 + G G k. Then IH, x = IG 1, x + IG 2, x IG k, x k 1 and consequently IH, 1 = k + 1. The case of k = 0 follows from Theorem 2.6.

5 Graph polynomial at 1 and The domination polynomial at 1 and 1 A set S V is a dominating set if N[S] = V, or equivalently, every vertex in V \S is adjacent to at least one vertex in S. An i-subset of V G is a subset of V G of cardinality i. Let DG, i be the family of dominating sets of G which are i-subsets and let dg, i = DG, i. The polynomial V G DG, x = dg, ix i is defined as domination polynomial of G [1, 4, 5]. Every root of DG, x i=1 is called a domination root of G. The set of all roots of DG, x is denoted by ZDG, x. Let A B V. Define D A,B G, i as follows D A,B G, i = {S DG, i S B = A}. Let d A,B G, i = D A,B G, i and define D A,B G, x = V i=1 d A,BG, ix i. We shall consider the value of domination polynomial at 1 and 1. Note that DG; 1 = dg, 1 + dg, dg, n equals the number of dominating sets of G, and DG; 1 = dg, 1 + dg, 2 dg, n dg, n = g 0 G g 1 G, where g 0 G = dg, 2 + dg, 4 + dg, , g 1 G = dg, 1 + dg, 3 + dg, are equal to the numbers of dominating sets of even size and odd size of G, respectively. We call DG; 1 as the alternating number of dominating sets. The following theorem gives the value of domination polynomial of complete graphs, complete bipartite graphs and stars at 1 and 1: Theorem 3.1. i DK n, 1 = 1, DK m,n, 1 {±1, 3} and DK 1,n, 1 { 1, 1}. ii DK n, 1 = DK 1,n, 1 = 2 n + 1 and DK m,n, 1 = 2 m 12 n Proof. The results follow easily from DK n, x = 1 + x n + 1 and DK m,n, x = 1 + x m 11 + x n 1 + x m + x n see [4]. Theorem 3.2. Let G be a graph. If G has a vertex v of degree k such that N[v] is a clique, then DG, x = xdg N[v], x x D A,Nv G v, x. =A Nv where N 2,..., N 2 k are all nonempty subsets of Nv. Moreover, DG, 1 = DG N[v], 1. Proof. Suppose that Nv = {v 1,..., v k }. Let N 1 = Ø, N 2,..., N 2 k be all subsets of Nv. Let A be a dominating set of G with cardinality i. We have the following cases: 1 v A and A Nv = N 1, then A {v} is a dominating set of G N[v] with cardinality i 1. 2 v A and there exists an index t, 2 t 2 k such that A Nv = N t. Since N[v] is a clique, A {v} is a dominating set of G v with cardinality i 1 and A {v} Nv = N t. 3 v / A. Therefore A Nv Ø and there exists an index l, 2 l 2 k such that A Nv = N l. Therefore A is a dominating set of G v with cardinality i and A Nv = N l.

6 54 S. Alikhani and M. H. Reyhani By considering the above cases we obtain that dg, i = dg N[v], i 1 + Therefore we have 2 k r=2 d Nr,NvG v, i 1 + DG, x = xdg N[v], x x Now, by putting x = 1 in the above equality the proof is complete. 2 k r=2 2 k r=2 D Nr,NvG v, x. d Nr,NvG v, i. As a consequence of the previous theorem we determine the exact value of the domination polynomial of a forest at 1. We recall the following theorem: Theorem 3.3. [4] If G has t connected components G 1,..., G t, then DG, x = t i=1 DG i, x. Theorem 3.4. Let F be a forest. Then DF, 1 = 1 αf. Proof. By Theorem 3.3 it is sufficient to prove the theorem for trees. We prove this theorem by induction on n, where n is the order of tree T. For n = 1, 2 there is nothing to prove. Now, Let n 3. Let u 0 be an arbitrary vertex of T. Assume that w is a vertex with maximum distance of u 0. Clearly dw=1. Let vw ET. Obviously, at most one of the neighbors of v has degree more than one. Since n 3, then dv 2. Assume that dv = t and {u 1,..., u t 1 } Nv and du i =1, for i = 1,..., t 1. Let H = T {v, u 1,..., u t 1 }. Clearly, H is a tree and αt = αh + t 1. Now, By Theorems 3.3 and 3.2 and induction hypothesis we have DT, 1 = DT N[u 1 ], 1 = DH, 1 1 t 2 = 1 αh 1 t 1. Therefore DT, 1 = 1 αt and the proof is complete. By Theorem 3.4 we have the following corollary: Corollary 3.5. For any forest F, DF ; 1 { 1, 1}, i.e., the number of dominating sets of even size varies by at most one from the number of dominating sets of odd size. Lemma 3.6. [2] If n is a positive integer, then 3 if n 0 mod 4; DC n, 1 = 1 otherwise. Corollary 3.7. For any n 4, DW n, 1 { 1, 3}.

7 Graph polynomial at 1 and 1 55 Proof. Since for every n 4, DW n, x = x1 + x n 1 + DC n 1, x [4], the result follows from Lemma 3.6. Since DP n, x = x[dp n 1, x+dp n 2, x+dp n 3, x] and DC n, x = x[dc n 1, x+dc n 2, x+ DC n 3, x] see [6, 7] we have the following theorem: Theorem 3.8. i For every n 4, DP n, 1 = DP n 1, 1+DP n 2, 1+DP n 3, 1 with initial values DP 1, 1 = 1, DP 2, 1 = 3 and DP 3, 1 = 5. ii For every n 4, DC n, 1 = DC n 1, 1 + DC n 2, 1 + DC n 3, 1 with initial values DC 1, 1 = 1, DC 2, 1 = 3 and DC 3, 1 = 7. iii For any n 4, DW n, 1 = DC n 1, n 1. Theorem 3.9. Let G be a unicycle graph. Then DG, 1 {±1, ±3}. Proof. We prove the theorem by induction on n = V G. If n = 3, then G = K 3 and there is nothing to prove. Now, let n 4. If G = C n, then by Lemma 3.6, DG, 1 { 1, 3}. If G C n, then there exists a vertex u such that du = 1. Suppose that Nu = {v}. Clearly, at most one of the components of G N[u] is a unicycle graph. Now, by Theorems 3.3, 3.2 and 3.4 and induction hypothesis the proof is complete. Theorem [10] For every graph, the number of dominating sets is odd. Theorem Let G be a graph. Then DG, r is odd for every odd integer r. In particular DG, 1 is odd. Proof. Suppose that r, s are two odd integers. Clearly, for every graph G, DG, r DG, s mod 2. Therefore DG, r DG, 1 mod 2. So by Theorem 3.10, we conclude DG, r is odd. Therefore DG, 1 is odd. The following corollary is an immediate consequence of Theorem Corollary Every integer domination root of a graph G is even. We need the following theorem to prove another results: Theorem [4] Let G 1 and G 2 be graphs of orders n 1 and n 2, respectively. Then DG 1 + G 2, x = 1 + x n x n DG 1, x + DG 2, x. The following corollary is an immediate consequence of above theorem:

8 56 S. Alikhani and M. H. Reyhani Corollary For any two graphs G 1 and G 2 DG 1 + G 2, 1 = DG 1, 1 + DG 2, Theorem For any odd integer n = 2k 1 k Z, there is some connected graph G such that DG; 1 = n. Proof. Let G i 1 i k be k graphs with DG i, 1 = 1 for every i {1, 2,..., k} and H = G 1 + G G k. Then DH, x = DG 1, x DG k, x + k 1 and consequently, DH, 1 = k + k 1 = 2k 1 = n. So the result is true for positive odd integer. Now we consider negative odd integer. Let G i 1 i k be k graphs with DG i, 1 = 3 for every i {1, 2,..., k} and H = G 1 + G G k. Then DH, x = DG 1, x DG k, x + k 1 and consequently, DH, 1 = 3k + k 1 = 2k 1. As an example for positive odd integer in above theorem we consider complete n-partite graphs. It is easy to check that for the complete n-partite graph K m1,m 2,...,m n we have m n n DK m1,m 2,...,m n, x = x i x m i 11 + x m m i 1 1. i=m 1 i=2 Suppose that for every 1 i n the number m i is even. By above formula we have m n n DK m1,m 2,...,m n, 1 = 1 i + 1 i=m 1 2 Since all of m i are even, we have DK m1,...,m n, 1 = 2n 1. Now we give an example for negative odd integer in above theorem. Suppose that G is a unicycle graph G with DG, 1 = 3 see Theorem 3.9. Consider H = G } + G + {{... + G }. It is easy to see n times that DH, 1 = 2k + 1. In this paper we studied independence and domination polynomial at 1 and 1. There are many graphs which have 2 as domination roots as an example all graphs of the form H K 1 have this property, see [1]. We think that the study of the value of domination polynomial of a graph at 2 is important, because we have the following conjecture: Conjecture If r is an integer domination root of a graph, then r = 0 or r = 2. Acknowledgements. The authors would like to express their gratitude to the referee for helpful comments. The research of the first author was in part supported by a grant from IPM No

9 Graph polynomial at 1 and 1 57 References [1] S. Akbari, S. Alikhani, and Y. H. Peng, Characterization of graphs using domination polynomial, European J. Combin., [2] S. Akbari and M. R. Oboudi, Cycles are determined by their domination polynomials, Ars Combin., to appear. [3] Y. Alavi, P. J. Malde, A. J. Schwenk and P. Erdös, The vertex independence sequence of a graph is not constrained, Congr. Numer., [4] S. Alikhani and Y. H. Peng, Introduction to domination polynomial of a graph, Ars Combin., in press. Available at [5] S. Alikhani, Y. H. Peng, Domination polynomials of cubic graphs of order 10, Turkish J. Math., [6] S. Alikhani and Y. H. Peng, Dominating sets and domination polynomial of paths, Int. J. Math. Math. Sci., 2009, Article ID , 10 pages doi: /2009/ [7] S. Alikhani and Y. H. Peng, Dominating sets and domination polynomials of certain graphs, II, Opuscula Math., 30 No [8] S. Alikhani and Y. H. Peng, Independence roots and independence fractals of certain graphs, J. Appl. Math. Comput., [9] P. N. Balister, B. Bollobas, J. Cutler, and L. Pebody, The interlace polynomial of graphs at 1, European J. Combin., [10] A. E. Brouwer, The number of dominating sets of a finite graph is odd, preprint. [11] J. I. Brown, K. Dilcher and R. J. Nowakowski, Roots of independence polynomials of well covered graphs, J. Algebraic Combin., [12] R. Frucht and F. Harary, On the corona of two graphs, Aequationes Math., [13] I. Gutman and F. Harary, Generalizations of the matching polynomial, Utilitas Math., [14] R. P. Stanley, Acyclic orientations of graphs, Discrete Math., Saeid Alikhani Department of Mathematics, Yazd University, Yazd, Iran School of Mathematics, Institute for Research in Fundamental Sciences IPM, P.O. Box: , Tehran,Iran. alikhani@yazduni.ac.ir Mohammad H. Reyhani Department of Mathematics, Islamic Azad University, Yazd branch, Yazd, Iran mh reihani@iauyazd.ac.ir