American Journal of Applied Sciences 9 (2): 202-207, 2012 ISSN 1546-9239 © 2012 Science Publications Complexity of Cocktail Party Graph and Crown Graph Daoud, S.N. 1Department of Applied Mathematics, Faculty of Applied Science, Taibah University, Al-Madinah, K.S.A. 2Department of Mathematics, Faculty of Science, El-Minufiya University, Shebeen El-Kom, Egypt Abstract Problem statement: The number of spanning trees τ(G) in graphs (networks) was an important invariant. Approach: Using the properties of the Chebyshev polynomials of the second kind and the linear algebra techniques to evaluate the associated determinants. Results: The complexity, number of spanning trees, of the cocktail party graph on 2n vertices, given in detail in the text was proved. Also the complexity of the crown graph on 2n vertices was shown to had the value n n-2 (n-1) (n-2) n-1. Conclusion: The number of spanning trees τ(G) in graphs (networks) is an important invariant. The evaluation of this number and analyzing its behavior is not only interesting from a mathematical (computational) perspective, but also, it is an important measure of reliability of a network and designing electrical circuits. Some computationally hard problems such as the travelling salesman problem can be solved approximately by using spanning trees. Due to the high dependence of the network design and reliability on the graph theory we introduced the above important theorems and lemmas and their proofs. Key words: Complexity of graphs, spanning trees, cocktail party graphs, chebyshev polynomials, crown graph INTRODUCTION spanning trees in a d-regular graph G can be expressed p− 1 τ =1 −µ λ λ λ λ as (G)∏ (dk ) where 0 = 1, 2,….., p-1 are the In this introduction we give some basic p k= 1 definitions and lemmas. We deal with simple and eigenvalues of the corresponding adjacency matrix of finite undirected graphs G = (V, E), where V is the the graph. However, for a few special families of vertex set and E is the edge set. For a graph G, a graphs there exist simple formulas that make it much spanning tree in G is a tree which has the same vertex easier to calculate and determine the number of set as G. The number of spanning trees in G, also corresponding spanning trees especially when these called, the complexity of the graph, denoted by τ(G), numbers are very large. One of the first such result is is a well-studied quantity (for long time). A classical due to Cayley (1889) who showed that complete graph n-2 result of Kirchhoff (1847) can be used to determine on n vertices, Kn has n spanning trees that he showed the number of spanning trees for G = (V, E). Let V = τ n-2 ≥ τ =q− 1 p − 1 ≥ (K n)= n , n 2. Another result, (Kp,q ) p q ,p,q 1 , {v , v ,…,v }, then the Kirchhoff matrix H defined as 1 2 n where K is the complete bipartite graph with bipartite n×n characteristic matrix H = D-A, where D is the p,q sets containing P and q vertices, respectively. It is well diagonal matrix of the degrees of G and A is the known, as in e.g., adjacency matrix of G, H = [a ij ] defined as follows: (i) ≠ (Clark, 2003; Qiao and Chen, 2009). Another result is aij = -1 when vi and vj are adjacent and i j, (ii) aij due to Guy (1970) who derived a formula for the wheel equals the degree of vertex vi if i = j and (iii) aij = 0 on n+1 vertices, W , which is formed from a cycle C τ n+1 n otherwise. All of co-factors of H are equal to (G). on n vertices by adding a vertex adjacent to every τ There are other methods for calculating (G). Let vertex of C . In particular, he showed that µ ≥µ ≥ ≥µ n 1 1 … p denote the eignvalues of H matrix of a p + − τ=3 5n + 3 5 n − ≥ point graph. Then it is easily shown that µ = 0. (W)(n+ 1 )( )2 , for n 3. Sedlacek p 2 2 Furthermore, Kelmans and Chelnokov (1974) shown p− 1 (1970) also later derived a formula for the number of τ =1 µ spanning trees in a Mobius ladder. The Mobius ladder that, (G) ∏ k . The formula for the number of p = k 1 Mn is formed from cycle C2n on 2n vertices labeled v1, 202 Am. J. Applied Sci., 9 (2): 202-207, 2012 v2,…,v 2n by adding edge vivi+n for every vertex vi where where all other elements are zeros. i≤n. The number of spanning trees in Mn is given by Further we recall that the Chebyshev polynomials n of the first kind are defined by Eq. 1: τ(M) = [(2 + 3)n +− (2 3) n + 2] for n≥2. Another n 2 = class of graphs for which an explicit formula has been Tn (x) cos(narccosx) (1) derived is based on a prism (Boesch and Bogdanowicz, 1987; Boesch and Prodinger, 1986). Let the vertices of The Chebyshev polynomials of the second kind are two disjoint and length cycles be labeled v , v ,…v in 1 2 n defined by Eq. 2: one cycle and w1, w 2,…w n in the other. The prism Rn is defined as the graph obtained by adding to these two 1 d sin(n arccos x) cycles all edges of the form vi,w i. The number of U− (x)= T(x) = (2) n 1n dx n sin(arccos x) spanning trees in Rn is given by the following n formula [(2+ 3)n +− (2 3) n − 2] . 2 It is easily verified that Eq. 3: Lemma 1.1: Temperley (1964): − + = U(x)n 2xU n1− (x) U n2 − (x) 0 (3) 1 τ(G) = det(H + J) p2 It can then be shown from this recursion that by expanding det A n (x) one gets Eq. 4: where, J is the p×p matrix, where all elements are unity. = ≥ We can also deduce the following lemma. Un (x) det(A n (x)),n 1 (4) τ=1 −+ Lemma 1.2: (G)2 det(pI D A) where A , D are Furthermore by using standard methods for solving p the recursion (3), one obtains the explicit formula Eq. 5: the adjacency and degree matrices of G , the complement of G, respectively and I is the p×p unit =1 +−−−−2 n1+ 2 n1 + ≥ (5) Un (x) [(x x 1) (x x 1) ],n 1 matrix. 2 x2 − 1 The advantage of these formulas in lemma1.1, lemma 1.2 is to express τ(G) directly as a determinant where the identity is true for all complex x (except at rather than in terms of cofactors as in Kirchhoff x = ± 1where the function can be taken as the limit). theorem or eigenvalues as in Kelmans and Chelnokov formula. Lemma 1.3: Let B (x) be n×n matrix such that: n 1 τ − Proof lemma 1.2: Since (G)=2 det(H+J),H=D-A x 1 0 p −11x + − 1 0 then D= H + A , since D+D = (p-1)I , then 0− 11x + − 1 ⋱ D= (p − 1)I − D . Thus H+ A = (p − 1)I − D , since B (x) = n ⋱ ⋱ ⋱ ⋱ 0 J− A = I + A . By addition we get H+= J pI − D + A and − + − ⋱ 11 x 1 1 therefore τ=(G ) det(pI −+ D A) . 0− 1 x p2 In our computations we need some lemmas on determinants and some relations concerning Chebyshev Then one can obtain: polynomials. We begin from their definitions, Zhang et + al . (2005). = − 1 x det(B(x))n (x 1)U n− 1 ( ) Let A n(x) be n×n matrix such that: 2 2x− 1 0 Proof: Straightforward induction using properties of −1 2x − 1 0 A (x) = 0 ⋱ ⋱ ⋱ determinants and above mentioned definition. n ⋱ ⋱ ⋱ −1 0− 1 2x Lemma 1.4: Let C (x) be n×n matrix, n ≥3, x>2 then: n 203 Am. J. Applied Sci., 9 (2): 202-207, 2012 x 0 1 Complexity of cocktail party graphs: The cocktail party graph of order n, also called the hyper octahedral 0 x1+ 0 ⋱ graph is the graph consisting of two rows of paired 1 0 x10+ ⋱ = Cn (x) nodes in which all nodes but the paired ones are ⋱ ⋱ ⋱ ⋱ 1 connected with a graph edge. It is the graph ⋱ ⋱ x+ 1 0 complement of the ladder rung graph L and the dual n 1 0 x graph of the hypercube graph Qn. Biggs (1993). See Fig. 1. Then: x det(C (x))= (n + x − 2)U− ( ) n n 1 2 Proof: Straightforward induction using properties of determinants, we have: Fig. 1: Cocktail party graph with 2n-vertices x1− − 1 0 −1 x − 10 Theorem 2.1: Let G be a cocktail party graph with 2n ≥ n+ x − 2 0− 1x − 1 ⋱ vertices, n 3. Its complexity will be: = det(Cn (x)) det x− 2 ⋱ ⋱ ⋱ ⋱ 0 (n− 1) ⋱ ⋱ x− 1 τ(G) = n22n2 (n− 4)((n + 2) 2 − 4) 0− 1x1 − ×+2n − −− 2n − [(n n 4) (n n 4)] Using lemma1.3 yields: ×[((n ++ 2) (n +−−+− 2)2n 4) ((n 2) (n +− 2) 2n 4)] nx2+ − x x Proof: Applying lemmas1.2 ,1.3,1.4 and1.5,we have: det(C(x))= (x − 2)U− ( ) =+− (n x 2)U − ( ) n x2− n1 2 n1 2 1 τ=(G) det(2nI −+ D A) A I (2n) 2 Lemma 1.5: If B = . Then det (B) = det (A-I).