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)− 2xU (x) + U (x) = 0 (3) n n1− n2 − 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). I A  = 1 2 det det (A+I). (2n)  n101+ ⋯ 1    Proof: Using the fact that:  0 n2+ 0 ⋱ ⋮   10⋱ ⋱ 1 I    A B   det(A).det(D− CA−1 B)  ⋮ ⋱ ⋱ n+ 2 0  det   =  , −1  +  C D  det(D).det(A− BD C)  1⋯ 10n1   n101+ ⋯ 1    n+ 2 where, A, D are non singular, Marcus and Minc (1964).  0 0 ⋱ ⋮      We have:  I 10⋱ 1   n+ 2   ⋮ ⋱ ⋱ 0  A I  = − −1   det  det(A).det(A A )   I A   1⋯ 10n1+  =2 − det(A I) n01⋯ 1  =det(A − I).det(A + I).   0 n+ 1 0 ⋱ ⋮  1 = det 1 0⋱ ⋱ 1  This formula gives some sort of symmetry in some (2n) 2   + matrices which facilitates our calculation of ⋮ ⋱ ⋱ n 1 0    determinants. 1⋯ 10n  204 Am. J. Applied Sci., 9 (2): 202-207, 2012

n201+ ⋯ 1  x11⋯ ⋯ 1      1 x 1 ⋱ ⋮ 0 n3+ 0 ⋱ ⋮      1⋱⋱ 1 ⋱⋮  × det 1 0⋱ ⋱ 1 D (x) =     n + ⋮ ⋱ ⋱ ⋱ ⋱ 1  ⋮ ⋱ ⋱ n 3 0      ⋮ ⋱ ⋱ x 1  1⋯ 10n2+    1⋯ ⋯ 11x  n110− − ⋯ 0    −1 n − 1 ⋱ ⋮  Then we obtain: 1 2n− 2 = det 0− 1⋱ ⋱ 0  (2n)2 n− 2   det(D )= (x +− n 1)(x − 1) n− 1 ⋮ ⋱ ⋱ n− 1  n   0⋯ 0− 1n1 −   Proof: From the definition of the circulant n110+ − ⋯ 0  determinants, we have:   −1 n2 + − 1 ⋱ ⋮  × −  x11⋯ ⋯ 1  2det 0 1⋱ ⋱ 0     1 x 1 ⋱ ⋮ ⋮ ⋱ ⋱ n+ 2 − 1    1⋱⋱ 1 ⋱⋮  − +  =   0⋯ 0 1n1  det(Dn (x)) det ⋮ ⋱ ⋱ ⋱ ⋱ 1  − + =12n2 − n × n2 ⋮ ⋱ ⋱ x 1  [(n 2)U− ( )] [2nU − ( )]   2 − n1 n1   (2n)n2 2 2 1⋯ ⋯ 11x  1 2(n− 1) 1 n = × ×− 23 n1− 2 (n 2) =∏ (x +ω+ω+ω+ ...... +ω ) (2n) n− 2 n 2 − jjj j 2 n 4 j= 1 [(n+ n2n − 4) −− (n n 2n − 4)] =(x +++ 1 1 ...... +× 1) n 1 (x+ω+ω+ω+23 ...... +ω n1− ) ×2n [((n + 2) + ∏ jjj j n 2 2 j1,= ω ≠ 1 2 (n+ 2) + 4 j =− 1 =(x +−× n 1) (x − 1)n− 1 . (n−− 2)2n 4) −+− ((n 2) (n +− 2) 2n 4)]

(n− 1) = Theorem 2.3: Let G be a crown graph with 2n vertices. n22n2 (n− 4)((n + 2) 2 − 4) Then the number of spanning trees of G is: ×+[(n n2n − 4) −− (n n 2n − 4)] τ=n2− −− n1 − ×++[((n 2) (n −−−+− 2)2n 4) ((n 2) (n +− 2) 2n 4)] (G) n (n 1)(n 2)

Proof: Applying lemmas1.2 and 1.5 we have: Complexity of crown graph: A crown graph on 2n vertices is an undirected graph with two sets of vertices 1 ≠ τ=(G) det(2nI −+ D A) ui and vi and with an edge from ui to vj whenever i j. (2n) 2 The crown graph can be viewed as a complete bipartite n 1⋯ 1  graph from which the edges of a perfect matching have   1⋱ ⋮ I been removed, Biggs (1993) Fig. 2.   ⋱ 1    1⋯ 1 n τ = 1   (G ) 2   (2n) n 1⋯ 1  1 ⋱ ⋮    I⋮ ⋱ 1    1⋯ 1 n  n111+ ⋯ 1    1 n1+ 1 ⋱ ⋮  1 = det 1 1⋱ ⋱ 1  Fig. 2: Corwn graph with 2n-vertices (2n) 2   ⋮ ⋱ ⋱ n+ 1 1    1⋯ 11n1+  Lemma 2.2:Let Dn (x) be n×n matrix such that: 205 Am. J. Applied Sci., 9 (2): 202-207, 2012

n111− ⋯ 1   n11⋯ 1      1 n 1 ⋱ ⋮ 1 n1− 1 ⋱ ⋮     11⋱ ⋱ 1 I  × det 1 1⋱ ⋱ 1      −  ⋮ ⋱ ⋱ n 1  ⋮ ⋱ ⋱ n 1 1      1 1⋯ 11n 1⋯ 11n1− = det     (2n) 2  n101+ ⋯ 1    0 n2+ 0 ⋱ ⋮ Applying lemma 2.2 with x = n+1 and n-1    I 10⋱ 1  respectively, we have:    ⋮ ⋱ ⋱ n+ 2 0    1⋯ 10n1+ 1 −   τ=(G) ××−− 2nn 2(n 1)(n 2) n 1 (2n) 2 (n 2 +   2n− 1 3n − 2⋯ ⋯ 3n − 2  − − − =nn2 (n − 1)(n − 2) n1  2n 3)    (n 2 +  2n1− 2n13n2 − − ⋯ ⋮  Complexity of open cocktail party graphs: The  3n− 4)    open cocktail party graph of order n, is the graph 1  3n2−− 2n1⋱ 2n13n2 −− ⋮  = det consisting of cocktail party graph by deleting n-1 (2n) 2  ⋮ ⋱ ⋱ ⋱ ⋱ 3n− 2    edges from one of their rows Fig. 3.  (n 2 +   ⋮ ⋯ ⋱ ⋱ 2n− 1   3n− 4)   2 +  − − − (n  3n 2⋯ ⋯ 3n 2 2n 1   2n− 3) 

 2n− 1 3n − 2 3n − 2  n+ 3 ⋯ ⋯  − − −   n1n1 n1   2n −1 2n13n2 − −  n+ 4 ⋯ ⋮  n1− n1n1 − −    Fig. 3: Open cocktail party graph with 2n-vertices 3n2−− 2n1 2n13n2 −−  ⋱ ⋮  − n   = (n 1) n1n1−− n1n1 −− Lemma 2.4: If A and B are symmetric matrices and 2 det   (2n) 3n− 2  ⋮ ⋱ ⋱ ⋱ ⋱  A I   −  C =   . n 1 I B  2n− 1     ⋮ ⋯ ⋱ ⋱ n+ 4   n− 1   − − −  Then: 3n 2 3n 2 2n 1 +  ⋯ ⋯ n 3   n1− n1n1 − −  det(C)= det(AB −= I) det(BA − I) Straight forward induction by using properties of Proof: Applying the fact that used in lemma 1.5. determinants.

Theorem 2.5: Let G be an open cocktail party graph We obtain: with 2n vertices, n≥3. Its complexity will be: + (n− 1) n 1 τ(G) = × det (2n− 3)n 2 (n− 1)n n 2 − 2 τ = 2 (G) U(n− 1 )  n− 2  n 2(n− 1)  0 1⋯ ⋯ 1  n− 1 n− 1   = ×−+−+−[(n2 2 n 42 8n 8n) n  n2 − 2  n 4 2  0 (+ 1) 0 1 ⋯ ⋮  n× 2 n − 8n + 8n −  n 1  (n2−− 2 n 42 − 8n + 8n)] n  1 0⋱ 01 ⋮     ⋮ ⋱ ⋱ ⋱ ⋱ 1  Proof: Applying lemmas (1.2), (1.4) and (2.4) we have:  n2 − 2   ⋮ ⋯ ⋱ ⋱ (+ 1) 0   n− 1   2  τ=1 −+ n− 2 (G)2 det(2nI D A)  1⋯ ⋯ 1 0  (2n)  n− 1  206 Am. J. Applied Sci., 9 (2): 202-207, 2012

(n− 1)n+ 1 2n − 3 Boesch, F.T. and H. Prodinger, 1986. Spanning tree = × × det (2n− 3)n2 n − 2 formulas and chebyshev polynomials. Graphs Combinatorics, 2: 191-200. DOI:  n2 − 2  (− 1) − 1 0⋯ ⋯ 0  10.1007/BF01788093 −  n 1  Boesch, F.T. and Z.R. Bogdanowicz, 1987. The number  n2 − 2   −1 − 1 0 ⋯ ⋮  of spanning trees in a Prism. Int. J. Comput. Math.,  n− 1  21: 229-243. DOI: 10.1080/00207168708803568  0− 1⋱ − 10 ⋮  Cayley, G.A., 1889. A theorm on trees. Quart. J. Math.,    ⋮ ⋱ ⋱ ⋱ ⋱ 0  23: 276-378.  n2 − 2  Clark, L., 2003. On the enumeration of multipartite  ⋮ ⋯ ⋱ ⋱ −1  spanning trees of the complete graph. Bull. ICA,  n− 1  38: 50-60.  n2 − 2   0⋯ ⋯ 0− 1 ( − 1)  Guy, R.K., 1970. Combinatorial Structures and their −  n 1  Applications. 1st Edn., Gordon and Breach, New n12+ 2 (n1)− n2 − n2 − York, pp: 508. = ×( −× 2) U− ( ) (n2)n−2 n1 −n 1 2(n1) − Kelmans, A.K. and V.M. Chelnokov, 1974 A certain (n− 1)n n 2 − 2 polynomial of a graph and graphs with an extremal = × U− ( ) nn 1 2(n −1) number of trees. J. Comb. Theory, 16: 197-214. DOI: 10.1016/0095-8956(74)90065-3 n− 1 =[(n2 −+−+ 2 n 42 8n 8n) n Kirchhoff, 1847. Uber die Auflosung der Gleichungen, n 4 2 n× 2 n − 8n + 8n auf welche man beider Untersuchung der Linearen −(n2 −− 2 n 42 − 8n + 8n)] n Verteilung galvanischer Strme gefhrt wird. Ann. Phys. Chem., 72: 497-508.

Marcus, M. and H. Minc, 1964. A Survey of Matrix ACKNOWLEDGMENT Theory and Matrix Inequalities. 1st Edn., Prindle, Weber and Schmidt,, Boston, pp: 180. The author is deeply indebted to the team of work Qiao, N.S. and B. Chen, 2009. The number of spanning at deanship of scientific research at Taibah university, trees and Chains of graphs. J. Applied Math. E- for their continuous helps and the encouragement me to Notes, 9: 10-16. finalize this work. This research work was supported by Sedlacek, J., 1970. Mathematics (Geometry and Graph a grant No. (625/1431) from the deanship of the theory) (Chech), University of Karlova. scientific research at Taibah university, Al-Madinah Temperley, H.N.V., 1964. On the mutual cancellation Al-Munawwarah, K.S.A. of cluster integrals in Mayer's fugacity series. Proc. Phys. Soc., 83: 3-16. DOI: 10.1088/0370- REFERENCES 1328/83/1/302 Zhang, Y., X Yong and M.J. Golin, 2005. Chebyshev polynomials and spanning tree formulas for Biggs, N., 1993. Algebraic Graph Theory. 2nd Edn., circulant and related graphs. Discrete Math., 298: Cambridge University Press, Cambridge, ISBN: 334-364. DOI: 10.1016/j.disc.2004.10.025 0521458978, pp: 205.

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