On the edge cover polynomial of a graph

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1 On the edge cover polynomial of a graph Saieed Akbari a,b, Mohammad Reza Oboudi a,b a Department of Mathematical Sciences, Sharif University of Technology, P.O. Box , Tehran, Iran b School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O. Box , Tehran, Iran Abstract Let G be a simple graph of order n and size m. An edge covering of a graph is a set of edges such that every vertex of the graph is incident to at least one edge of the set. Here we introduce a new graph polynomial. The edge cover polynomial of G is the polynomial E(G, x) = m i=1 e(g, i)xi, where e(g, i) is the number of edge covering sets of G of size i. Let G and H be two graphs of order n such that δ(g) n, where δ(g) is the minimum degree of G. If E(G, x) = E(H, x), then we show that the degree sequence of G and H are the same. We show that cycles and complete bipartite graphs are determined by their edge cover polynomials. Also we determine all graphs G for which E(G, x) = E(P n, x), where P n is the path of order n. We show that if δ(g) 3, then E(G, x) has at least one non-real root. We prove that the real roots of edge cover polynomial of trees are dense in the interval [ 4,0]. Finally, we characterize all graphs whose edge cover polynomials have exactly one or two distinct roots. In fact their roots are contained in { 3,, 1, 0}. 010 AMS Subject Classification: 05C31; 05C70. Keywords: Edge covering set; Edge cover polynomial; Root. 1 Introduction Throughout this paper we will consider only graphs without loops and multiple edges. Let G = (V (G), E(G)) be a simple graph. The order and the size of G denote the number of vertices and the number of edges of G, respectively. The complete graph, the cycle, and the path of order n, are denoted by K n, C n and P n, respectively. We denote the complete bipartite graph with part sizes m and n, by K m,n. Also we call K 1,n a star. For every vertex v V (G), the degree of v is the number of edges adjacent to v and is denoted by d G (v). For simplicity we write d(v) instead of d G (v). By a i (G) we mean the number of vertices of G with degree i. A pendant vertex is a vertex with degree one. We denote the minimum degree of vertices of G by δ(g). For simplicity we use Addresses: s akbari@sharif.edu (S. Akbari), m r oboudi@math.sharif.edu (M.R. Oboudi). 1

2 δ instead of δ(g). The line graph L(G) of G, is the graph whose vertices are the edges of G, and two vertices of L(G) are adjacent if they share an end in G. The decycling number (or the feedback vertex number) of a graph G is the minimum number of vertices that need to be removed in order to eliminate all its cycles. An independent set in a graph is a set of vertices no two of them are adjacent. The cardinality of a maximum independent set in a graph G is called the independence number of G and denoted by α(g). A covering of a graph G is a set of vertices of G such that every edge of G is incident to at least one vertex of the set. The minimum number of vertices in a covering of a graph G is called the covering number of G and denoted by β(g). Indeed, the independent sets and coverings are related in a very simple way. A set S is an independent set of a graph G if and only if V (G) \ S is a covering of G. Therefore we have the following identity which first observed by Gallai: α(g) + β(g) = n, where n is the order of G [7, p. 96]. An r-matching of G is a set of r edges of G which no of them have common vertex. For every natural number r, we denote the number of r-matchings of G by m(g, r) and we let m(g, 0) = 1. The maximum number of edges in a matching of a graph G is called the matching number of G and denoted by α (G). A perfect matching of G is a matching with cardinality n, where n is the order of G. We denote the number of perfect matchings of G by pm(g). For every graph G with no isolated vertex, an edge covering of G is a set of edges of G such that every vertex is incident to at least one edge of the set. In other words, an edge covering of a graph is a set of edges which together meet all vertices of the graph. A minimum edge covering is an edge covering of the smallest possible size. The edge covering number of G is the size of a minimum edge covering of G and denoted by ρ(g) or β (G). For a detailed treatment of these parameters, the reader is referred to [7]. By the celebrated König -Eger vary identity [7, Theorem 8.3], for every bipartite graph G we have α (G) = β(g). Also for every graph G, α(g) ρ(g), and for bipartite graphs the equality holds [7, Theorem 8.30]. There is a useful relation between ρ(g) and α (G). In fact for every graph G of order n, with no isolated vertex [7, p. 43], ρ(g) + α (G) = n. Let u V (G). By N u we mean the set of all edges of G incident with u while by N u we mean the set E(G) \ N u. If A X is a set, then we denote the complement of A by A. There are many polynomials associated to graphs. For example domination polynomial, chromatic polynomial, clique polynomial, characteristic polynomial and Tutte polynomial, see [], [6], [11], [], and [3], respectively. Let G be a graph of order n and size m. One of the important polynomials introduced by Gutman which is called acyclic polynomial [13], or matching polynomial [1] defined as, n µ(g, x) = ( 1) r m(g, r)x n r. r=0 For more details see [14, 17]. Another polynomial is the independence polynomial that introduced by Gutman and Harary [18], as I(G, x) = α(g) i=0 s(g, i)x i,

3 where s(g, i) is the number of independent sets of G with size i. We let s(g, 0) = 1. For more details see [15, 16, 1]. It is easy to see that for every graph G, µ(g, x) = x n I(L(G), 1 x ). The vertex-cover polynomial was defined in [10] as follows: ψ(g, τ) = n cv(g, r)τ r, r=0 where cv(g, r) is the number of covering of G with cardinality r. There is a simple relationship between the vertex cover polynomial and the independence polynomial. Clearly, a set S is an independent set of a graph G if and only if V (G) \ S is a covering of G. Therefore cv(g, r) = s(g, n r), for r = 0,..., n. Thus we have ψ(g, x) = x n I(G, 1 x ). Now, in a similar way we intend to define the edge cover polynomial of a graph G. Let G be a graph without isolated vertex. By E(G, i) we mean the family of edge covering sets of G with cardinality i. Let e(g, i) = E(G, i). Now, we define the edge cover polynomial E(G, x) of G as follows: E(G, x) = m i=ρ(g) e(g, i)x i. Also, for a graph G with an isolated vertex we define E(G, x) = 0. We let E(G, x) = 1, when the order and the size of G are zero. For example one can see that E(K 4, x) = x 6 +6x 5 +15x 4 +16x 3 + 3x. Two graphs G and H are said to be E-equivalent, written G H, if E(G, x) = E(H, x). For every graph G, the E-equivalence class of G is defined as [G] = {H : G H}. A graph G is said to be E-unique, if [G] = {G}. In the second section we obtain some properties of the edge cover polynomial. We show that by having the edge cover polynomial one can obtain some part of the degree sequence of the graph. We state an explicit formula for the edge cover polynomial by expanding the polynomial at 1. We prove that for every forest F, E(F, 1) { 1, 0, 1}. In third section we find some recursive formulas for the edge cover polynomial and we show that among all trees the coefficients of the edge cover polynomial of paths are maximum while the coefficients of the edge cover polynomial of stars are minimum. In fourth section we consider the E-equivalence class of some special graphs such as paths, cycles, and complete bipartite graphs. Finally, in the last section we consider the roots of the edge cover polynomial and we characterize all graphs whose the edge cover polynomials have only two roots. Indeed those roots are in the set { 3,, 1, 0}. Some properties of the edge cover polynomial of a graph In this section we obtain some properties of the edge cover polynomial of a graph. It is clear that the edge cover polynomial of a graph G with no isolated vertex is a monic polynomial with degree m, where m is the size of G. We begin by a simple lemma which omit its proof. Lemma 1. Let G be a graph with connected components G 1,..., G k. Then E(G, x) = k i=1 E(G i, x). 3

4 Lemma. Let G be a graph of order n and size m with no isolated vertex. If the edge cover polynomial of G is E(G, x) = m i=ρ(g) e(g, i)xi, then the following hold: i) n ρ(g). ii) If i 0 = min{ i e(g, i) = ( m i ) }, then δ = m i iii) If G has no connected component isomorphic to K, then a δ (G) = ( m m δ) e(g, m δ). Proof. i) The proof is trivial since every edge covers just two vertices. ii) Let i m δ + 1. Thus every subset S E(G) with cardinality i is an edge covering of G. Therefore e(g, i) = ( ) m i. Now, suppose that i m δ. Consider a vertex v of degree δ. Let A N v, such that A = i. Clearly, A is not an edge covering of G. So e(g, i) < ( ) m i. iii) Since G has no connected component isomorphic to K, then for every two distinct vertices u and v, N u = N v implies that u = v. Let A = { N u d(u) = δ } and B = {S E(G), S = m δ, S / E(G, m δ)}. It is not hard to see that A = B and this completes the proof. Remark 1. For every natural number n, E(K 1,n, x) = E(G n, x) = x n, where G n is a disjoint union of n copies of K. This example shows that for δ = 1, the number of pendant vertices and the order of a graph are not uniquely determined by the edge cover polynomial. Theorem 1. Let G be a k-regular graph of order n with k. Suppose that E(H, x) = E(G, x), for some graph H. Then H is a k-regular graph of order n. Proof. First note that E(H) = E(G) = deg(e(g, x)) = nk. By Parts (ii) and (iii) of Lemma we obtain that δ(h) = k and H has n vertices of degree k. Thus we conclude that H is a k-regular graph of order n. The next theorem shows that we can determine some coefficients of E(G, x) by having the degree sequence of G. Also we can obtain some vertex degrees of G using E(G, x). Theorem. Let G be a graph with no isolated vertex. Let E(G, x) = m i=1 e(g, i)xi be the edge cover polynomial of G. Then the following hold: i) For every natural number i, e(g, i) ( ) m i v V (G) ( ) m d(v). i 4

5 ii) For every i, i m δ(g) +, e(g, i) = iii) For every i, i m δ(g) +, e(g, i + 1) = ( m i i + 1 ( ) m i iv) Let δ = δ(g). For every k, 1 k δ, Proof. v V (G) ) e(g, i) + 1 i + 1 ( ) m d(v). i v V (G) ( ) m k + 1 a k (G) = e(g, m k + 1) e(g, m k) 1 k k In particular a δ (G) = ( m m δ) e(g, m δ). ( ) m d(v) d(v). i k 1 j=1 ( ) m j ja j (G). m k i) Let {v 1,..., v n } be the vertex set of G. For every natural number i and j = 1,..., n, let A i,j = { A A = i, A N vj }. Note that N vi = m d(v i ), for every j = 1,..., n. Thus A i,j = ( m d(v j) ) i. Since every subset of N vj is not an edge covering of G, we obtain Hence we conclude the result. E(G, i) = n A i,j. ii) Let A i,j be the set that was defined in the previous part. To obtain the result it is sufficient to prove that if i m δ(g) +, then for every distinct numbers j and k, A i,j Ai,k =. Assume that S A i,j Ai,k. Therefore S = i and S N vi N vj. Since N vi N vj 1, this implies that S δ(g) 1, a contradiction. Therefore we are done. iii) By the part(ii), and using the equality ( ) j k+1 = j k j k+1( k), the result follows. iv) By taking i = m k in part(iii) we obtain the formula for a k (G). Also by Lemma we know that e(g, m δ + 1) = ( m m δ+1) which completes the proof. j=1 Corollary 1. Let G and H be two graphs of order n such that δ(g) n. If E(G, x) = E(H, x), then the degree sequence of G and H are the same. 5

6 Proof. Since the size of G and H are equal, by Lemma, δ(g) = δ(h) = δ. Using Theorem, we obtain a k (G) = a k (H), for δ k n. By considering the equality v V (G) d(v) = v V (H) d(v), we conclude that a n 1(G) = a n 1 (H). Hence the proof is complete. We continue this section by the next theorem and some of its corollaries. Theorem 3. Let G be a graph. Then E(G, x) = S V (G) ( 1) S (x + 1) E(G\S). Proof. Let A V (G). The number of subsets B E(G) with cardinality i which do not cover any vertices of A is ( ) E(G\A) i. Now, by inclusion-exclusion principle for every i we have ( ) E(G) e(g, i) = + ( ) E(G \ S) ( 1) S. i i =S V (G) Therefore So we are done. e(g, i) = S V (G) ( ) E(G \ S) ( 1) S. i By Theorem 3 one can obtain the following corollary. Corollary. Let m, n be natural numbers. Then: i) E(K n, x) = n k=0 ( 1)n k( n k) (x + 1) ( k ). ii) E(K m,n, x) = m k=0 ( 1)k( m k ) ((x + 1) m k 1) n. The value of some polynomials related to graphs at x = 1 has been considered in many papers. In [1] it was proved that for every forest F, D(F, 1) {±1}, in fact D(F, 1) = ( 1) α(f ), where D(G, x) is the domination polynomial of G. Here we intend to consider E(G, x) at x = 1. First we state an upper bound for E(G, 1). To obtain an upper bound for E(G, 1) we need the following theorem. Theorem 4. [0] For every graph G, where ϕ(g) is the decycling number of G. I(G, 1) ϕ(g), The above theorem shows that for every forest F, I(F, 1) { 1, 0, 1}. The next corollary shows that a similar fact holds for the edge cover polynomial. 6

7 Corollary 3. Let G be a graph of order n. Then E(G, 1) = ( 1) n I(G, 1). Proof. Clearly, E(G \ S) = 0 implies that V (G) \ S is an independent set of G and vise versa. So By Theorem 3, E(G, 1) = ( 1) n I(G, 1). Now, by Theorem 4 and Corollary 3 we have the following. Corollary 4. For every graph G, E(G, 1) ϕ(g). 3 Recursive formulas for the edge cover polynomials of graphs In this section we obtain some recursive formulas for the edge cover polynomial of graphs. We recall that E(G, x) = 1, for a graph G with no vertex and no edge. Also E(G, x) = 0, when G has an isolated vertex. We begin this section by the following theorem. Theorem 5. Let G be a graph, u, v V (G) and uv be an edge of G. Then E(G, x) = (x + 1)E(G \ uv, x) + x ( E(G \ u, x) + E(G \ v, x) + E(G \ {u, v}, x) ). Proof. Let S be an edge covering of G with size i. If uv / S, then S is an edge covering of G\uv. Now, suppose that uv S. Thus we have the following cases: i) S N u = S N v = 1. Therefore S \ uv is an edge covering with size i 1 for G \ {u, v}. ii) S N u > 1 and S N v = 1. So S \ uv is an edge covering with size i 1 for G \ v. iii) S N u = 1 and S N v > 1. So S \ uv is an edge covering with size i 1 for G \ u. iv) S N u > 1 and S N v > 1. Therefore S \ uv is an edge covering with size i 1 for G \ uv. Hence we obtain that e(g, i) = e(g \ uv, i) + e(g \ uv, i 1) + e(g \ v, i 1) + e(g \ u, i 1) + e(g \ {u, v}, i 1). This completes the proof. As a corollary we have the following theorem which provides a recursive formula for the determination of edge cover polynomial of trees. Theorem 6. Let G be a graph and u be a pendant vertex of G. Suppose that v is the neighbor of u. Then E(G, x) = xe(g \ u, x) + xe(g \ {u, v}, x). 7

8 Proof. By applying Theorem 5 for two vertices u and v we obtain the result. As an immediate consequence of the above theorem we obtain the following result. Corollary 5. For every natural number n 3, E(P n, x) = xe(p n 1, x) + xe(p n, x). The following theorem shows that among all trees with the same order the coefficients of the edge cover polynomial of stars are minimum and the coefficients of the edge cover polynomial of paths are maximum. Theorem 7. Let T be a tree of order n, then for i = 1,..., n 1, e(k 1,n 1, i) e(t, i) e(p n, i). Moreover, if e(k 1,n 1, i) = e(t, i), for i = 1,..., n 1, then T = K 1,n 1. Also if e(t, i) = e(p n, i) for i = 1,..., n 1, then T = P n. Proof. We prove the theorem by induction on n. For n = 1 there is nothing to prove. Now, let n. Consider the rooted tree of T. Let u be a pendant vertex in the last level of rooted tree and v be its neighbor. Now, by Theorem 6, E(T, x) = xe(t \ u, x) + xe(t \ {u, v}, x). By Corollary 5 and induction hypothesis the proof is complete. The next theorem states a similar recursive formula for the edge cover polynomials of cycles. Corollary 6. For every natural number n 5, E(C n, x) = xe(c n 1, x) + xe(c n, x). Proof. By Theorem 5, for every k 3, we obtain that E(C k, x) = (x + 1)E(P k, x) + xe(p k 1, x) + xe(p k, x). (1) Using the above relation for k = n 1, n and Corollary 5 the result follows. Note that by induction on n or using generating function one can find the coefficients of E(P n, x). Here we use the generating function to obtain an explicit formula for E(P n, x) as follows: Let f(x, y) = i=1 E(P i, x)y i. Using Corollary 5 and some calculations one can obtain that f(x, y) = xy 1 xy xy. 8

9 1 Now, by expanding the fraction 1 xy(1+y), and some computations we get So the coefficient of y n in f(x, y) is f(x, y) = i=1 x i y i+1 (1 + y) i 1. i=1 n 1 ( ) i 1 x i. n i 1 Therefore we obtain that for every natural number n, As an immediate consequence we have the following corollary. E(P n, x) = n 1 ( i=1 n i 1) x i. () Corollary 7. Let n be a natural number. Then ρ(p n ) = n and { 1, e(p n, n ) = n 1, if n is even; otherwise. Using (1) and () one can obtain that for every n, n 5, E(C n, x) = n ( i=3 n n i n i 1) x i. (3) Corollary 8. For every natural number n, n 3, ρ(c n ) = n and {, if n is even; e(c n, n ) = n, otherwise. In the last section of the paper we show that all roots of E(P n, x) and E(C n, x) are real. 4 E-Equivalence class of some graphs In this section we show that complete bipartite graphs, complete graphs and cycles are uniquely determined by their edge cover polynomials. Also, we completely determine the E-equivalence class of paths. We need the following lemma in order to show that the complete bipartite graphs are E-unique. Lemma 3. [4] Let G be a graph. Then pm(g) v V (G) (d(v)!) 1 d(v), where 0! 1 0 =0. If G has no isolated vertices, then equality holds if and only if G is a disjoint union of K n1,n 1,..., K nr,n r, for some natural numbers n 1,..., n r. 9

10 Theorem 8. Let G be a graph and E(G, x) = E(K n,n, x), for some natural number n. Then G = K n,n. Proof. For n=1, there is nothing to prove. Suppose that n. By Theorem 1 we conclude that G is an n-regular graph of order n. Note that e(g, n) = pm(g) = pm(k n,n ) = n!. This shows that the equality holds in Lemma 3, which completes the proof. Theorem 9. Let p and q be two natural numbers such that p q p 1. Let G be a graph and E(G, x) = E(K p,q, x). If p, then G = K p,q. Also if p = 1, then G = k i=1 K 1,p i, for some natural numbers p 1,..., p k and k, where q = k i=1 p i. Proof. First, let p = 1. It is not hard to see that if K is a graph of size m, then ρ(k) = m if and only if every connected component of K is a star. So in this case we are done. Also, by Theorem 8, the case p = q is proved. We prove the case p = at the end of proof. Now, let q > p 3. Clearly, the size of G is pq. Also by Lemma, we have δ(g) = p. We recall that a t (G) is the number of vertices of G with degree t. Let A = {w V (G) d(w) = p} and B = {w V (G) d(w) = q}. To prove the theorem we will show that A and B are independent sets of G, A = q and B = p. Let H K 3 be a graph of size m. One can easily see that for every S V (H) with S 3 the following holds: Now consider three following cases: E(H \ S) m d(u) d(v) for some vertices u and v in S. (4) Case 1. Suppose that q p. So by Theorem we obtain that a p (G) = q and a q (G) = p. Since G has exactly pq edges, we conclude that G is a graph of order p+q. Therefore by Theorem 3 we have E(G, x) = (x + 1) pq q(x + 1) pq p p(x + 1) pq q + ( 1) S (x + 1) E(G\S). S V (G), S Using Theorem 3 for K p,q to obtain the following equality: ( 1) S (x + 1) E(G\S) = ( 1) S (x + 1) E(Kp,q\S). S V (G), S S V (K p,q), S Clearly, Also by (4) we have deg( ( 1) S (x + 1) E(Kp,q\S) ) pq p. S V (K p,q ), S deg( ( 1) S (x + 1) E(G\S) ) pq p. S V (G), S 3 10

11 Now, we show that A is an independent set of G. Let u, v A and uv is an edge of G. So we conclude that deg( ( 1) S (x + 1) E(G\S) ) = pq p + 1, S V (G), S a contradiction. Since G is a graph of order p + q and size pq and a p (G) = q, a q (G) = p, therefore B is an independent set of G so we obtain that G = K p,q. Case. Assume that q = p 1. By Theorem we conclude that a p (G) = q and a p+1 (G) = = a p (G) = 0. By Theorem 3 and similar to the previous case, it is not hard to see that the coefficient of x pq p+1 in the polynomial E(K p,q, x) (x + 1) pq + q(x + 1) pq p and E(G, x) (x + 1) pq + q(x + 1) pq p is p and s a p 1 (G), respectively, where s is the number of edges of G whose end points have degree p. Therefore a p 1 (G) = p + s. By considering v V (G) d(v), we obtain a p 1(G) = p, s = 0 and G is a graph of order p + q. So we are done. Case 3. Now, let q p. It is not hard to see that for p 4, the following holds. deg( ( 1) S (x + 1) E(H\S) ) pq p 1, S V (H), S 3 for H = G or H = K p,q. Note that by (4), for p = 3 and H = G the degree of the above polynomial is at most pq p. The coefficients of x pq p+1 in the polynomials f(x) = E(K p,q, x) (x + 1) pq + q(x + 1) pq p and g(x) = E(G, x) (x + 1) pq + q(x + 1) pq p are 0 and s a p 1 (G), respectively, where s is the number of edges of G with end points of degree p. Hence s = a p 1 (G). The coefficient of x pq p in f(x) is ( q ) p if q = p and ( q ) if q > p. Let t, h be the number of triangles and the number of K 4 in G, respectively, whose vertices have degree p. Therefore the coefficient of x pq p in g(x) is ( ) ( ) q q a p 1 (G)(pq p + 1) a p (G) + s(pq p + 1) + s t 0 = a p (G) s t 0, { t h, if p = 3; where t 0 = 0, if p 4. Now, suppose q = p. Therefore a p (G)+s+t 0 = p. Let p 4, so we obtain a p 1 (G)+a p (G) = p. Thus we can write 4p = d(v) = pa p (G) + (p 1)a p 1 (G) + pa p (G) + d(v). v V (G) So we conclude that a p 1 (G) = v V (G),d(v) p+1 d(v). v V (G),d(v) p+1 If a p 1 (G) 0, then there is a vertex v with d(v) p + 1. Therefore we get a p 1 (G) p + 1. This contradicts the fact that a p 1 (G) p. Thus a p 1 (G) = 0. This shows that A = {w V (G) d(w) = p} is an independent set of G, and a p (G) = p. Hence G = K p,q. 11

12 So it remains to investigate the case (p, q) = (3, 6), (p, q) = (, 3) and (p, q) = (, 4). To prove those cases we use a similar method. First let p = and q = 3. Suppose E(G, x) = E(K,3, x). Therefore G has exactly 6 edges. By Lemma, and Theorem we get δ(g) = and a (G) = 3. By considering v V (G) d(v) = 1 we obtain G has exactly 5 vertices and a 3(G) =. It is not hard to see that G is K,3 or the graph H which has been shown in Figure 4. One can see that E(K,3, x) = x 3 (x + ) 3 x 3 while E(H, x) = x 3 (x + ) 3 x 3. This shows that G = K,3. Now assume that E(G, x) = E(K,4, x). Similarly, we conclude that G has exactly 8 edges, δ(g) = and a (G) = 4. Also by considering v V (G) d(v) = 16, one can see that the order of G is 6 and the degree sequence of G is,,,, 3, 5 or,,,, 4, 4. In first manner G is G 1 while in latter G is G or K,4 (see Figure ). Clearly ρ(g 1 ) = ρ(g ) = 3, but ρ(g) = ρ(k,4 ) = 4. Thus G = K,4. Now, we consider the last case. Let E(G, x) = E(K 3,6 ). Therefore G has exactly 18 edges and δ(g) = 3, a 3 (G) = 6, a 4 (G) = 0. Therefore the degree sequence of G is 3, 3, 3, 3, 3, 3, 5, 5, 8 or 3, 3, 3, 3, 3, 3, 5, 6, 7 or 3, 3, 3, 3, 3, 3, 6, 6, 6. By keeping the notations of the Case 3, in this case we proved that s = a 5 (G) and a 6 (G) + s + t = h + 3. Note that h is the number of connected components of G that are isomorphic to K 4. s 6h and t 4h. Thus we obtain h = 0. This implies that a 5 (G) + a 6 (G) + t = a 6 (G) + s + t = 3. Clearly By the above equality and noting that if t 1, then s 3, we conclude that the degree sequence of G should be 3, 3, 3, 3, 3, 3, 6, 6, 6. Hence s = 0. Therefore A = {v V (G) d(v) = 3} is an independent set. Thus G = K 3,6. Figure 1: E(H, x) = x 3 (x + ) 3 x 3. Remark. By considering the notations of the previous theorem, for the case q > p where p 4, we obtain that a p (G) + s + t 0 = 0. This implies that s = 0. Therefore A is an independent set with cardinality q. Since G has exactly pq edges we conclude that G is bipartite. We think that G = K p,q. Theorem 10. Let G be a graph. Suppose that n and E(G, x) = E(K n, x). Then G = K n. 1

13 Proof. For n = the proof is clear. Now, suppose that n 3. Using Theorem 1, we obtain that G is an (n 1)-regular graph of order n, therefore G = K n. Now, we wish to show that cycles are E-unique. Theorem 11. Let n 3 and G be a graph. If E(G, x) = E(C n, x), then G = C n. Proof. By applying Theorem 1, we find that G is a -regular graph of order n. Therefore all connected components of G are cycles, say C n1,..., C nk. By Lemma 1, we conclude that E(C n, x) = E(G, x) = k i=1 E(C n i, x). (5) Now, by considering ρ(g) and e(g, ρ(g)) we will show that k = 1. This implies that G = C n. We consider two following cases: Case 1. Suppose that n is an even number. By Corollary 8 we conclude that e(g, n ) =. Also by computation the coefficient of x n in the right side of 5 we obtain k = 1. Case. Now, suppose that n is an odd number. Similarly, by Corollary 8 we obtain that all of the numbers n 1,..., n k are odd number and n = n n k. On the other hand n = n n k, Therefore k = 1 and the proof is complete. Now, we are in a position to determine the E-equivalence class of paths. E(G, x) = E(P n, x), then G = P n or G = P n or G = P n (see Figure 1). We show that if C n 3 C n 3 ' P n '' P n Figure : E(P n, x) = E(P n, x) = E(P n, x). Theorem 1. For every n, n 6, [P n ] contains exactly three following graphs: P n, P n, P n. Proof. Let G be a graph and E(G, x) = E(P n, x). By () we know that n 1 ( ) i 1 E(P n, x) = x i. n i 1 i=1 13

14 We recall that a k (G) is the number of vertices of G with degree k. By Lemma we obtain that δ(g) = 1 and G has n 1 edges. First, suppose that G is connected. In this case we show that G = P n or G = P n. By Lemma we obtain that G has exactly two pendant vertices, say u and v. Consider two following cases: Case 1. Suppose that u and v have no common neighbor. Let u and v be the neighbors of u and v, respectively. Since any edge covering of G contains edges uu and vv, so we have ( ) ( ) n 3 n 4 e(g, n 3) = r 0 = e(p n, n 3) =, n 5 where r 0 = {w V (G) \ {u, v } d(w) = }. Therefore r 0 = n 4. Thus (n 1) = d(z) (n 4) + d(u) + d(u ) + d(v) + d(v ). z V (G) So d(u )+d(v ) 4. Since G is connected, d(u )+d(v ) 4. Thus we obtain that d(u ) = d(v ) = and G is a graph of order n. Since the degree of each vertex of G except u and v is two we obtain that G = P n. Case. Let w be a common neighbor of u and v. Clearly, d(w) 3. Similar to the previous case we obtain ( ) n 3 n 5 a (G) = e(p n, n 3) = ( ) n 4. Thus a (G) = n 4. Since (n 1) = d(z) (n 4) + d(u) + d(v) + d(w), z V (G) we obtain that G is a graph of order n and d(w) = 4. Since all vertices of G \ {u, v} have degree two so this graph is a cycle. Thus G = P n. Now, suppose that G is not connected. Let s be the number of connected components of G that are isomorphic to K. Let H be the subgraph of G obtained by removing all these connected components of G. Clearly, e(g, n ) = e(h, n s ). We have e(h, n s ) = ( n s 1 n s ) a1 (H). Since e(p n, n ) = n 3, thus we obtain that a 1 (H) + s =. Therefore we have the following possibilities: (i). Assume that s = 0. So G has exactly two pendant vertices, say u and v. Therefore in a similar manner as the connected case, first suppose that u and v have no common neighbor. Similarly, we obtain G = P n, which contradicts the non-connectivity of G. Otherwise let w be a common neighbor of u and v. If d(w) 3, we get G = P n, a contradiction. Thus let d(w) =, that is P 3 is a connected component of G. Now, let K be the subgraph of G that obtained by deleting P 3 from G. So e(g, n 3) = e(k, n 5). By Theorem, e(k, n 5) = ( n 3 n 5) a (K) = e(p n, n 3). So we obtain that a (K) = n 4. This contradicts the equality (n 1) = z V (G) d(z). (ii). Suppose that G has exactly one connected component isomorphic to K. Let u be the unique pendant vertex of H and v be the neighbor of that. Thus G has exactly three pendant vertices. Note that ( ) n 3 e(h, n 4) = t = e(p n, n 3), n 5 14

15 where t = {w V (H) \ {v} d(w) = }. So t = n 4. By considering z V (G) d(z), we have (n 1) = d(z) = 3 + (n 4) + d(v) + a, z V (G) where a is the degree summation of other vertices. Therefore a = 0 or a. Thus we get d(v) 3. On the other hand d(v). If d(v) =, then a = 1, a contradiction. Thus d(v) = 3 and a = 0. This also implies that G is a graph of order n. Therefore we obtain that G has three pendant vertices, one vertex of degree three and n 4 vertices of degree two. So G = P n. (iii). Now, let s =, that is G has exactly two connected components isomorphic to K. Then e(h, n 5) = ( n 3 n 5) a (H). Therefore a (H) = n 4. Since a 1 (H) + s =, thus a 1 (H) = 0, that is δ(h) =. Thus (n 1) = d(z) = 4 + (n 4) + d(z). So z V (G) z V (G),d(z) 3 d(z) =, a contradiction. Therefore the case s = does not happen. z V (G),d(z) 3 Notice that by applying Theorem 6, one can easily see that for every n 6, E(P n, x) = E(P n, x) = E(P n, x). This completes the proof. Remark 3. It is not hard to see that P n is E-unique, for n 5. 5 Roots of the edge cover polynomials In this section we will consider the roots of the edge cover polynomials of graphs. For every graph G, zero is a root of E(G, x) with multiplicity ρ(g). We characterize all graphs whose edge cover polynomial have one or two distinct roots. Unlike matching polynomials [19] there exists a tree T (Figure ) such that the non-zero roots of its edge cover polynomial are non-real, indeed we have E(T, x) = x 4 (x + 3x + 3). Moreover, we find an infinite family of trees such that all non-zero roots of their edge cover polynomials are non-real. We show that every graph G with δ(g) 3 has at least two distinct non-real roots and for cases δ(g) = 1, we show that for every n all roots of E(P n, x) and E(C n, x) are real. We show that roots of the edge cover polynomial of paths are dense in interval [ 4, 0]. Let f 1 (x),..., f k (x) be the polynomials in one variable and real coefficients. We say they are compatible if for any non-negative numbers c 1,..., c k, all roots of the polynomial k i=1 c if i (x) are real. For more details on this concept the reader is referred to [9]. We need the following theorem which was proved by Chudnovsky and Seymour in [9]. Theorem 13. Let f 1 (x),..., f k (x) be pairwise compatible polynomials with positive leading coefficients. Then f 1 (x),..., f k (x) are compatible. 15

16 T Figure 3: E(T, x) = x 4 (x + 3x + 3). Theorem 14. For every natural number n, all roots of E(P n, x) are real. Proof. The proof is trivial for n = 1,. Now, we claim the following for every n, n : i) E(P n, x) and E(P n 1, x) are compatible. ii) E(P n, x) and xe(p n 1, x) are compatible. We prove two parts of this claim by induction on n simultaneously. The case n = is trivial. Now, suppose that n 3. First we prove the Part (i). Let a, b be two non-negative numbers. We show that all roots of ae(p n, x) + be(p n 1, x) are real. By Corollary 5, ae(p n, x) + be(p n 1, x) = axe(p n 1, x) + axe(p n, x) + be(p n 1, x). Therefore by Theorem 13 it is sufficient to show that the polynomials xe(p n 1, x), xe(p n, x) and E(P n 1, x) are pairwise compatible. By considering the induction hypothesis we get the result. The proof of the second part is similar. By Corollary 5 and the first part of the claim proof is complete. As a consequence of Theorem 14 we prove that all roots of E(C n, x) are real. Corollary 9. For every natural number n 3, all roots of E(C n, x) are real. Proof. The case n = 3 is trivial. Now, assume that n 4. By Theorem 5 and Corollary 5 we obtain that E(C n, x) = (x + )E(P n, x) + xe(p n 1, x). To prove the theorem it is enough to show that polynomials xe(p n, x), xe(p n 1, x) and E(P n, x) are compatible. So by Theorem 13, it suffices to show that xe(p n, x), xe(p n 1, x) and E(P n, x) are pairwise compatible. Noting the proof of Theorem 14, completes the proof. Now, we show that the roots of the sequence {E(P n, x)} n=1 are dense in interval [ 4, 0]. For investigating about the roots of the edge cover polynomial of some families of graphs, we need to some analytic theorems. For a family {f n (x)} n=1 of (complex) polynomials, we say that a complex number z is a limit point roots of {f n (x)}, if there is sequence {n k } k=1 such that for every k, f nk (z) = 0 or z is a limit point of R{f n (x)}, where R{f n (x)} is the union of roots of f n (x). 16

17 Consider a recursive family of polynomials f n (x) = k i=1 a i(x)f n i (x), for every n k + 1, where a i (x) are fixed non-zero polynomials. Now, form the characteristic equation λ k a 1 (x)λ k 1 a k (x) = 0. If λ 1 (x),..., λ k (x) are the distinct roots of the above equation then [5] for every n, f n (x) = α 1 (x)λ n 1 (x) + + α k (x)λ n k (x). (6) We can find α 1 (x),..., α k (x) by f 1 (x),..., f k (x). Beraha et al. in [5] proved the following result on recursive families of polynomials and their roots. Theorem 15. Let {f n (x)} n=1 be a recursive family of polynomials. Then a complex number z is a limit point roots of {f n (x)} n=1 if and only if there is a complex sequence {z n } n=1 such that f n (z n ) = 0 for all n and z n z as n. The main result of their paper characterizes precisely the limit point roots of a recursive family of polynomials. Theorem 16. [5] Under the non-degeneracy requirements that in relation (6) no α i (x) is identically zero and that for no pair i j is λ i (x) ωλ j (x), for some complex number ω with modulus one, we have a complex number z is a limit point roots of {f n (x)} n=1 if and only if either i) two or more of the λ i (z) are of equal modulus, and strictly greater (in modulus) than the others; or ii) for some j, λ j (z) has modulus strictly greater than all the other λ i (z) have, and α j (z) = 0. By using theses theorems in [8] Brown et al. proved that: Theorem 17. The complex roots of independence polynomials are dense in all complex plane, while real independence roots are dense in interval (, 0]. Now, we are ready to prove the following result. Theorem 18. The real roots of the edge cover polynomials are dense in interval [ 4, 0]. Proof. We show that the roots of paths are dense in interval [ 4, 0]. Let f n (x) = E(P n, x). Consider the sequence {f n (x)} n=1. By Corollary 5, f n (x) = xf n 1 (x) + xf n (x), for n 3, and f 1 (x) = 0, f (x) = x. By solving the characteristic equation λ = xλ + x, we obtain λ 1 (x) = x + x + 4x, λ (x) = x x + 4x. 17

18 Since λ 1 (x), λ (x) are distinct, we conclude that f n (x) = α 1 (x)λ n 1 (x) + α (x)λ n (x), where α 1 (x) = 4x (x + x + 4x) + 4x, α 4x (x) = (x x + 4x) + 4x. Therefore {f n (x)} n=1 satisfies in the conditions of Theorem 16. So by the first part of this theorem among the limit point roots are those z for which λ 1 (z) = λ (z), which implies that z + z + 4z = z z + 4z. Since for any real number z, 4 z 0, z + 4z 0, this shows that any z [ 4, 0] satisfies in this equation. So we are done. The following theorem shows that if all roots of E(G, x) are real then δ(g) = 1 or δ(g) =. Theorem 19. Let G be a graph. Then E(G, x) has at least δ(g) non-real roots (not necessary distinct). In particular when δ(g) = 3, E(G, x) has at least two distinct non-real roots. Proof. When δ(g) = 1 or δ(g) =, there is nothing to prove. Now, suppose δ(g) 3. Let m be the size of G and δ = δ(g). By f (k) (x) we mean the k-th derivative of f(x) with respect to x. We claim that E(G, x) (m δ) = m! δ! (x + 1)δ a δ (G)(m δ)!. Since m δ m δ +, by Theorem, we have ( ) m e(g, m δ) = a δ (G), m δ while for i = m δ + 1,..., m, e(g, i) = ( ) m. i Therefore there is a polynomial g(x) with degree m δ, and leading coefficient a δ (G), such that E(G, x) = (x + 1) m g(x). This proves the claim. Let r and s be the number of real roots and non-real roots of E(G, x), respectively. Thus r + s = m. Using Rolle s Theorem to obtain that E(G, x) (m δ) has at least r (m δ) real roots. Clearly E(G, x) (m δ) has at most real roots, hence r m + δ. Therefore s δ. If δ = 3, then s 1. Thus E(G, x) has a non-real root, say z. Since the conjugate of z is also a non-real root of E(G, x), the proof is complete. Remark 4. For case δ = 1 or δ = we show that for every n, the roots of E(P n, x) and E(C n, x) are real. But it is not true that if δ = 1 or δ =, then all roots of E(G, x) are real. Here we construct an infinite family of trees T n such that every non-zero root of E(T n, x) is non-real (see Figure 4). 18

19 Consider paths P n1,..., P nk, with vertex sets {a 1 1,..., a 1 n 1 },..., {a k 1,..., a k n k } such that vertices a 1 1,..., a k 1 and a 1 n 1,..., a k n k have degree one and the others have degree two. By T (n 1,..., n k ) we mean the tree obtaining by identifying vertices a 1 1,..., a k 1. Therefore T (n 1,..., n k ) is a tree of order n n k k, which is called starlike. By Theorem 6 one can see that if n 1 = = n k = 3, then (see Figure 3) E(T (n 1,..., n k ), x) = x k ((x + 1) k 1). This shows that if k is an odd number, then all non-zero roots of E(T (n 1,..., n k ), x) are non-real.... Figure 4: T (3, 3,..., 3). Now, we are in a position to characterize all graphs G, such that E(G, x) has exactly one distinct root or two distinct roots. We consider all graphs with no isolated vertex. If G has a isolated vertex, then E(G, x) = 0. Note that zero is one of the roots of E(G, x) with multiplicity ρ(g). The next theorem characterize all graph G whose edge cover polynomials have exactly one distinct root. Note that E(K 1,n, x) = x n. Theorem 0. Let G be a graph. Then E(G, x) has exactly one distinct root if and only if every connected component of G is star. Proof. Let m be the size of G. First, suppose that E(G, x) has exactly one distinct root. Therefore E(G, x) = x m. Let s be the number of all connected components of G which are isomorphic to K. If s = m, then we are done. Otherwise let H be the graph obtaining by removing all these connected components. By Lemma 1, E(H, x) = x m s. Also, by Lemma, we obtain δ(h) = 1 and H has exactly m s pendant vertices. It is not hard to see that H is a disjoint union of finitely many stars. Conversely, Suppose that each connected components of G is star. Then by Lemma 1, E(G, x) = x m. This completes the proof. Here we define a graph H(r) which is useful for characterizing graphs whose the edge cover polynomials have exactly two distinct roots. Let H be a graph of order n and size m. Suppose {v 1,..., v n } is the vertex set of H. By H(r) we mean the graph obtained by joining r i 1 pendant vertices to vertex v i, for i = 1,, n such that n i=1 r i = r. If m is the size of H, then H(r) is a graph of order n + r and size m + r. 19

20 Theorem 1. Let G be a graph. Then E(G, x) = x r (x + 1) m, for some natural numbers r and m, if and only if there exists a graph H with size m such that G = H(r). Proof. First let H be a graph with vertex set {v 1,..., v n } and size m. Let r be a natural number. We want to show that E(H(r), x) = x r (x + 1) m. Let A be the set of all new pendant edges that is added for constructing H(r). In fact A = r n. Let S be an edge covering of H(r). Since every pendant edge should be existed in S so A S. On the other hand every vertex of H appeared in at least one pendant edge of A, so A is an edge covering of H(r). Thus S E(H(r)) is an edge covering of H(r) if and only if A S. Therefore we obtain e(h(r), i) = ( m i r), for every i. Thus we have the result. Now suppose that E(G, x) = x r (x + 1) m. Let s be the number of connected component of G those are isomorphic to K. Let K be the graph obtained by deleting all components K from G. Therefore E(K, x) = x r s (x + 1) m. By Lemma, δ(k) = 1 and K has exactly r s pendant vertices. Thus K has exactly r s pendant edges. Since ρ(k) = r s, this shows that every vertex of K appears in at least one of those pendant edges. This completes the proof. We need the following theorem to obtain an upper bound for ρ(g). Theorem. [3] Let G be a graph with no odd cycle component. Then all edges of G can be colored by two colors such that for each vertex v with d(v), s(v) =, where s(v) denotes the number of different colors appearing on the edges incident with v. Corollary 10. Let G be a graph with no odd cycle component. Let δ(g). Then there exists S E(G), such that S and S are two edge covering sets of G. In particular, ρ(g) m. Proof. By Theorem we obtain an edge coloring with colors a and b for G, such that s(v) =, for every vertex v V (G). Now, let S be the set of all edges with color a. This completes the proof. Theorem 3. Let G be a graph with no odd cycle component. Let m and t be the size and the number of pendant vertices of G, respectively. Then ρ(g) m + t. Proof. We prove the theorem by induction on n, where n is the order of G. For n = 1, the proof is clear. Now, suppose that n 3. If δ(g), then by Corollary 10, the result follows. Thus let δ(g) = 1. Assume that u is a pendant vertex of G and v is the vertex adjacent to u. If d(v) = 1, then G has a connected component K. Using induction hypothesis completes the proof. 0

21 Now, let d(v). Using Theorem 6 to obtain Now, consider two following cases: ρ(g) = min{ρ(g \ u) + 1, ρ(g \ {u, v}) + 1}. Case 1. Let d(v) =. If G \ {u, v} has an odd cycle as a connected component, then G is the graph K (see Figure?). It is not hard to see that ρ(g) = m. So we are done. Otherwise, by induction hypothesis we obtain ρ(g) ρ(g \ {u, v}) + 1 m + t + 1. Case. Let d(v) 3. If G \ u has an odd cycle connected component, then G is the graph L (see Figure?). Therefore ρ(g) = m+1. Thus we are done. Otherwise, similar to Case 1, we get ρ(g) ρ(g \ u) + 1 m 1 + t The next theorem characterizes all graphs G for which E(G, x) has exactly two distinct roots. Theorem 4. Let G be a connected graph whose edge cover polynomial has exactly two distinct roots. Then one of the following holds: i) G = H(r), for some connected graph H and natural number r. ii) G = K 3. iii) δ(g) = 1, E(G, x) = x m+s (x + ) m s, where s is the number of pendant vertices of G. iv) δ(g) =, E(G, x) = x m (x + ) m, a (G) = m, and G has a triangle. Proof. Let E(G, x) = x ρ(g) (x + a) k and m be the size of G. Therefore k = m ρ(g). First we claim that δ(g). Let δ(g) 3. Using Theorem for determination e(g, m 1) and e(g, m ), we conclude that ak = m and ( ) k a = ( ) m. By calculation one can see that a = 1 which implies that ρ(g) = 0, a contradiction. So δ(g). Now, consider the following cases: 1

22 Case 1. Assume that δ(g) =. We recall that a i (G) denotes the number of vertices of G with degree i. Therefore by Theorem, we obtain that and ( ) k a = ak = m ( ) m a (G). These relations imply that a (G) = m(a 1). So a 1. Since a is a rational number and a is a root of monic polynomial E(G, x), then a is a natural number. Thus a. Since m = v V (G) d(v) a (G), thus m a (G). If m = a (G), then G is a cycle. So G = C m. Also a = 3 and k = m 3. Hence ρ(g) = m 3. On the other hand by Corollary 8, ρ(g) = ρ(c m ) = m. Therefore m 3 = m, which implies that m = 3. Hence G = K 3. On the other hand since E(K 3, x) = x (x + 3), this completes Part (ii). Now, suppose that m > t. By m(a 1) = a (G), we obtain a =. Therefore we find that a (G) = k = ρ(g) = m which implies that E(G, x) = x m (x + ) m. Let H = (V (H), E(H)) be a graph of size m. Suppose ρ = ρ(h) and h i = e(h, i). We claim that (m ρ)h ρ 3h ρ+1, (7) and if H is triangle-free graph, then (m ρ)h ρ h ρ+1. (8) To prove the claim let X and Y be the set of all edge covering sets of H with cardinality ρ and ρ + 1, respectively. Consider the bipartite graph with parts X and Y, such that we join A X to B Y if and only if A B. We can consider every elements of X and Y, as a graph with vertex set V (H). Clearly for every A X, d(a) = (m ρ). Now we show that for every B Y, d(b) 3. Let B Y. First note that every A X is a disjoint union of stars. We have ρ(b) = ρ. Since B has ρ + 1 edges, thus d(b) 1, that is there is A 0 X such that A 0 B. Let B \ A 0 = {e}. Since B \ {e} is a disjoint union of stars it is not hard to see that the coefficient of x ρ in E(B, x) is at most 3, and for triangle-free graph H is at most. This implies that d(b) 3, and d(b), when H is triangle-free. This proves the claim. Suppose that G is triangle-free. It is easy to see that in (8) the equality holds for G. Thus every B has degree. Let E = {v V (G) d(v) = } and F = {v V (G) d(v) 3}, it is not hard to see that E and F are independents sets of G. Therefore G is a bipartite graph with parts E and F. Now, we show that in this case e(g, ρ(g)) < m. Since ρ(g) = m, G has exactly m vertices of degree, and E is an independent set of G, thus for every edge covering set S of G of size m, N u S = 1, for every vertex u E. Hence the number of edge covering set of size m is at most

23 m. Now, consider a vertex v F, and neighbors a 1,..., a k in E. Now, consider a set S E(G) of size m such that for every vertex u E, S N u 1 and S {va 1,..., va k } =. Clearly, S is not an edge covering set of G. This shows that e(g, ρ(g)) < m. But it is a contradiction, since e(g, ρ(g)) = m. Case. If δ(g) = 1. Let s be the number of pendant vertices of G. Since G is connected, then G has not connected component isomorphic to K. Thus by Lemma, s = m ka. Using Theorem 3 to get ρ(g) m+s. Since a is natural, we find that a = 1 or a =. Let a = 1. Then by Theorem 1, there exists a connected graph H such that G = H(s). If a =, so ρ(g) = m+s This implies that E(G, x) = x m+s (x + ) m s. So we are done.. Remark 5. We proved that if δ(g) =, and G has no triangle, then E(G, x) has at lest three distinct roots. We think that if δ(g) = and G K 3, then E(G, x) has at least three distinct roots that is there is no graph G with size m such that E(G, x) = x m (x + ) m. But for every natural number s we can construct a graph G with size m and with s pendant vertices such that E(G, x) = x m+s (x + ) m s (see Figure?). Acknowledgements. The authors indebted to the School of Mathematics, Institute for Research in Fundamental Sciences (IPM) for support. The research of the first author was in part supported by a grant (No ) from School of Mathematics, Institute for Research in Fundamental Sciences (IPM). Some part of this work was done while the second author was visiting the department of mathematics of Eötvös Loránd University. He would like to thank the department for its hospitality and support and Professor László Lovász for his useful comments. References [1] S. Akbari, S. Alikhani, M.R. Oboudi, Y.H. Peng, On the zeros of domination polynomial of a graph, Combinatorics and Graphs. AMS. Contemporary Mathematics 531 (010) [] S. Akbari, S. Alikhani, Y.H. Peng, Characterization of graphs using domination polynomial, European Journal of Combinatorics 31 (010) [3] S. Akbari, M.N. Iradmusa, M. Jamaali, A note on edge coloring of graphs, Ars Combinatoria, to appear. [4] N. Alon, S. Friedland, The maximum number of perfect matchings in graphs with a given degree sequence, Electron. J. Combin. 15 (008) Note 13. [5] S. Beraha, J. Kahane, N. Weiss, Limits of zeros of recursively defined families of polynomials, in Studies in Foundations and Combinatorics, Advances in Math., Supplementary Studies, vol. 1, G. Rota (Ed.), Academic Press, New York, [6] G.D. Birkhoff, D.C. Lewis, Chromatic polynomials. Trans. Amer. Math. Soc. 60 (1946)

24 [7] J.A. Bondy, U.S.R. Murty, Graph Theory, Graduate Texts in Mathematics, 44. Springer, New York, 008. [8] J.I. Brown, C.A. Hickman, R.J. Nowakowski, On the location of roots of independence polynomials, Journal of Algebraic Combinatorics 19 (004) [9] M. Chudnovsky, P. Seymour, The roots of the independence polynomial of a clawfree graph, J. Combin. Theory Ser. B 97 (007) [10] F.M. Dong, M.D. Hendy, K.L. Teo, C.H. Little, The vertex-cover polynomial of a graph, Discrete Math. 50 (00) [11] E.J. Farrell, A note on the clique polynomial and its relation to other graph polynomials. J. Math. Sci. (Calcutta) 8 (1997) [1] E.J. Farrell, An introduction to matching polynomials, J. Combin. Theory Ser. B 7 (1979) [13] I. Gutman, The acyclic polynomial of a graph, Publ. Inst. Math. (Beograd) (N.S.) (36) (1977) [14] I. Gutman, Uniqueness of the matching polynomial, MATCH Commun. Math. Comput. Chem. 55 (006) [15] I. Gutman, Some analytical properties of the independence and matching polynomials, MATCH Commun. Math. Comput. Chem. 8 (199) [16] I. Gutman, An identity for the independence polynomials of trees. Publ. Inst. Math. (Beograd) (N.S.) 50 (64) (1991) [17] I. Gutman, Characteristic and matching polynomials of some bipartite graphs. Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat. 1 (1990) 5 30 (1991). [18] I. Gutman, F. Harary, Generalizations of the matching polynomial, Utilitas Math. 4 (1983) [19] O.J. Heilmann, E.H. Lieb, Theory of monomer-dimer systems, Comm. Math. Phys. 5 (197) [0] V.E. Levit, E. Mandrescu, A simple proof of an inequality connecting the alternating number of independent sets and the decycling number, arxiv: v1 [math.co] [1] V.E. Levit, E. Mandrescu, The independence polynomial of a graph A survey, Proceedings of the 1st International Conference on Algebraic Informatics, 33 54, Aristotle Univ. Thessaloniki, Thessaloniki, 005. [] A. Mowshowitz, The characteristic polynomial of a graph, J. Combin. Theory Ser. B 1 (197) [3] W. T. Tutte, A contribution to the theory of chromatic polynomials, Canadian J. Math. 6 (1954)