MPE Mathematical Problems in Engineering 1563-5147 1024-123X Hindawi Publishing Corporation 10.1155/2014/750618 750618 Research Article The Number of Spanning Trees of the Cartesian Product of Regular Graphs Wu Mei-Hui 1 Chung Long-Yeu 2 Yau Her-Terng 1 The Center of General Education, Information Technology Section Chia Nan University of Pharmacy and Science Tainan 71710 Taiwan chna.edu.tw 2 Department of Applied Informatics and Multimedia Chia Nan University of Pharmacy and Science Tainan 71710 Taiwan chna.edu.tw 2014 1472014 2014 18 03 2014 07 05 2014 14 7 2014 2014 Copyright © 2014 Mei-Hui Wu and Long-Yeu Chung. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The number of spanning trees in graphs or in networks is an important issue. The evaluation of this number not only is interesting from a mathematical (computational) perspective but also is an important measure of reliability of a network or designing electrical circuits. In this paper, a simple formula for the number of spanning trees of the Cartesian product of two regular graphs is investigated. Using this formula, the number of spanning trees of the four well-known regular networks can be simply taken into evaluation.

1. Introduction

In this paper, we deal with simple undirected graphs having no self-loop or multiple edges and consider the Cartesian product of two regular graphs only. It is well known that, for designing large-scale interconnection networks, the Cartesian product is an important method to obtain large networks from smaller ones, with a number of parameters that can be easily calculated from the corresponding parameters for those small initial graphs. The Cartesian product preserves many nice properties such as regularity, transitivity, super edge-connectivity, and super point-connectivity of the initial regular graphs . In fact, many well-known networks can be constructed by the Cartesian products of simple regular graphs, for example, Boolean n-cube networks, hypercube networks, and lattice networks.

Alternatively, the study of the number of spanning trees in a graph has a long history and has been very active because the problem has different practical applications in different fields. For example, the number characterizes the reliability of a network and, in physics, designing electrical circuits, analyzing energy of masers, and investigating the possible particle transitions . The larger degree of points a network has, the more I/O ports and edges are needed and the more cost is required.

The number of spanning trees of some special network has been taken into evaluation . Recently, some authors derived results about the counting where the number of spanning trees can be found from . However, the study for spanning trees of the Cartesian product of regular graphs remains an open and important invariant.

The number of spanning trees of Boolean n-cube networks, lattice networks, and generalized Boolean n-cube networks has been taken into account [13, 17, 18]; these networks belong to the class of networks Qn with two regular graphs Q1 and G which is defined recursively by Qn=Qn-1×G for n2. In this paper, we will present the formula of the number of spanning trees of the Cartesian product of regular graphs. Using this present formula, the main results in [13, 17, 18] can be obtained much more simply and will be extended.

2. The Number of Spanning Trees Definition 1.

Let G be a graph with n points labeled 1,2,,n. The adjacency matrix of G,A(G), is an n×n matrix with the ith row and jth column entry given by(1)[A(G)]ij={1ifpointsiandjareadjacent0ortherwise. The Kirchhoff matrix of G,H(G), is equal to D(G)-A(G), where D(G) is an n×n diagonal matrix whose diagonal entries are the degree of point n and A(G) is the adjacency matrix. Thus the ith row and jth column entry is given by (2)[H(G)]ij={deg(i)ifi=j-1ifijandpointsiandjareadjacent0ortherwise.

Lemma 2 (see [<xref ref-type="bibr" rid="B18">30</xref>]).

If G is a graph on n points with Kirchhoff matrix H(G) and Hij(G) is the submatrix of H(G) obtained by removing the ith row and jth column then the number of spanning trees of G,  t(G), is any cofactor of H(G). That is, t(G)=(-1)i+jdet(Hij(G)).

Lemma 3.

If A is an n×n triangulable matrix, which has n eigenvalues, then the sum of product of any n-1 eigenvalues of A is the sum of all principal minors of A.

Proof.

Let p(λ) be the character polynomial of A and λ1,λ2,,λn are n eigenvalues of A. Then (3)p(λ)=(λ-λ1)(λ-λ2)(λ-λn). From (3), we obtain the following.

(a) The coefficient of λ=(-1)n-1·(the  sum  of  product  of  anyn-1eigenvaules  ofA).

On the other hand, (4)p(λ)=det(λI-A)=|λ(1)-a11-a12-a1n-a21λ(2)-a22-a2n-a21-a21λ(n)-ann|, where A=[aij] and λ(i)=λ for i=1,2,,n.

So we only need to prove that the coefficient of λ in det(λI-A) is the sum of all principal minors of A. Let Aii denote the principal minor of A obtained by removing the ith row and ith column from A.

By (4), we obtain the following:(5)the  coefficient  ofλ(1)=(-1)n-1A11(takingλ(2)=λ(3)==λ(n)=0)the  coefficient  ofλ(2)=(-1)n-1A22(takingλ(1)=λ(3)=λ(4)=λ(n)=0)the  coefficient  ofλ(n)=(-1)n-1Ann(takingλ(1)=λ(2)==λ(n-1)=0).

(b) So the coefficient of λ=i=1n  the coefficient of  λ(i)=(-1)n-1(A11+A22++Ann).

Hence the theorem is proved due to (a) and (b).

Since a real symmetric matrix is with the property that the sum of its rows (and its columns) is zero, the rank of H(G)n-1. So 0 is the smallest eigenvalue. We write the eigenvalues of H(G) as an ordered list: (6)0=λ0(G)λ1(G)λ2(G)λn-1(G). The main result in Kelmans and Chelnokov  can also be obtained by the following method.

Lemma 4 (see [<xref ref-type="bibr" rid="B17">31</xref>]).

If the eigenvalues of the Kirchhoff matrix H(G) of the n points graph G are 0=λ0(G)λ1(G)λ2(G)λn-1(G) then t(G), the number of spanning trees of G, is given by (7)t(G)=1ni=1n-1λi(G).

Proof.

By Lemmas 2 and 3, (8)i=1n-1λi(G)=the  sum  of  product  of  any  n-1  eigenvaules  of  H(G)=the  sum  of  all  principal  minors  of  H(G)=n·t(G). Hence t(G)=(1/n)i=1n-1λi(G).

Lemma 5.

Let the eigenvalues of the adjacency matrix A(G) of the regular graph G be written by u1u2un-1un=n, where r is the degree of the regular graph G; then, the number of spanning trees of G is given by (9)t(G)=1ni=1n-1(r-ui).

Proof.

We know H(G)=rIn-A(G), where In is the identity n×n matrix. Since ui is the eigenvalue of A(G) for i=1,2,,n-1, there exists eigenvector xi for i=1,2,,n-1, such that A(G)xi=uixi. So (rIn-H(G))xi=uixi,  rInxi-H(G)xi=uixi, rxi-H(G)xi=uixi, and  H(G)xi=rxi-uixi.

We obtain H(G)xi=(r-ui)xi. Thus r-ui is the eigenvalue of H(G) for i=1,2,,n-1.

Hence the lemma is proved by Lemma 4.

3. Cartesian Product and Kronecker Product Definition 6.

Let G=(N,E) denote a connected graph with N set of all points and E set of all edges in G and let {u,v} denote edge joining points u and v. Let Gi=(Ni,Ei) for i=1,2; the Cartesian product of G1 and G2 is defined by G1×G2=(N,E), where N=N1×N2,  E=E1×E2, and {(u1,v1),(u2,v2)}E if and only if u1=u2 and {v1,v2}E2 or v1=v2 and {u1,u2}E1.

Definition 7 (see [<xref ref-type="bibr" rid="B9">32</xref>]).

Let B=[bij] be an n×n matrix and C an m×m matrix; then, the Kronecker product B×KC is defined as the mn×mn matrix with block description (10)[b11Cb12Cb1nCb21Cb22Cb2nCbn1Cbn2CbnnC]. The Kronecker sum is defined by B+KC=B×KIm+In×KC, where Ik is the k×k identity matrix for k=m,n. Let M be an mn×mn matrix. M can be partitioned into n2 blocks which are denoted by Bαβ for α=1,2,,n and β=1,2,,n. That is, (11)M=[B11B12B1nB21B22Bn1Bn1Bn2Bnn], where Bαβ is the m×m matrix for α=1,2,,n and  β=1,2,,n.  M is called an n×n    (m×m) block matrix.

Lemma 8.

If A is an n×n matrix and B, C are m×m matrices then

A×K(B+C)=A×KB+A×KC,

(B+C)×KA=B×KA+C×KA.

Proof.

The lemma is easily obtained.

Lemma 9.

If the products AB and CD are defined then (AB)×K(CD)=(A×KC)(B×KD).

Proof.

Let A=[aij] be an m×n matrix and B=[bij] an n×p matrix (12)(A×KC)(B×KD)=[a11Ca1nCam1CamnC][b11Db1pDbn1DbnpD]=[k=1na1kbk1CDk=1na1kbkpCDk=1namkbk1CDk=1namkbkpCD]=(AB)×K(CD).

Lemma 10.

If P and Q are invertible then (P×KQ)-1=P-1×KQ-1.

Proof.

Consider (13)(P×KQ)(P-1×KQ-1)=(PP-1)×K(QQ-1)=Im×KIn=Imn, where P is m×m and Q is n×n.

4. The Number of Spanning Trees of the Cartesian Product of Regular Graphs Lemma 11.

If the points of G1 and G2 are labeled by u1,u2,un and v1,v2,,vm, respectively, and points of G1×G2 are ordered lexicographically, that is, the label of (ui,vj) is smaller than that of (uk,vl) if and only if i<k and j<l, then A(G1×G2)=A(G1)+KA(G2).

Proof.

Since A(G1×G2) is an mn×mn matrix, A(G1×G2) is an n×n  (m×m) block matrix. By the definition of G1×G2, we describe(14)A(G1×G2)=[A(G2)[A(G1)]12Im[A(G1)]13Im[A(G1)]1,n-1Im[A(G1)]1,nIm[A(G1)]21ImA(G2)[A(G1)]23Im[A(G1)]2,n-1Im[A(G1)]2,nIm[A(G1)]n-1,1Im[A(G1)]n-1,2Im[A(G1)]n-1,3ImA(G2)[A(G1)]n-1,nIm[A(G1)]n,1Im[A(G1)]n,2Im[A(G1)]n,3Im[A(G1)]n,n-1ImA(G2)],

where [A(G1)]ij is the (i,j) entry of the adjacent matrix A(G1) of G1 and A(G2) is the m×m adjacent matrix of G2. We know In×KA(G2) is an mn×mn matrix; it can be described as an n×n(m×m) block matrix(15)In×KA(G2)=[A(G2)0m×m0m×m0m×m0m×m0m×mA(G2)0m×m0m×m0m×m0m×m0m×m0m×mA(G2)0m×m0m×m0m×m0m×m0m×mA(G2)],

where 0m×m is the m×m zero matrix. Since A(G1)×KIm is an mn×mn matrix, it can be described as an n×n(m×m) block matrix(16)A(G1)×KIm=[0m×m[A(G1)]12Im[A(G1)]13Im[A(G1)]1,n-1Im[A(G1)]1,nIm[A(G1)]21Im0m×m[A(G1)]23Im[A(G1)]2,n-1Im[A(G1)]2,nIm[A(G1)]n-1,1Im[A(G1)]n-1,2Im[A(G1)]n-1,3Im0m×m[A(G1)]n-1,nIm[A(G1)]n,1Im[A(G1)]n,2Im[A(G1)]n,3Im[A(G1)]n,n-1Im0m×m].

Clearly A(G1×G2)=A(G1)×KIm+In×KA(G2)=A(G1)+KA(G2).

Lemma 12.

Let Gi be the regular graph of degree ri for i=1,2; then, the degree of G1×G2 is r1+r2. If the number of the points of G1 (resp., G2) is n (resp., m) and the points of G1×G2 are ordered lexicographically then H(G1×G2)=H(G1)+KH(G2).

Proof.

By Lemmas 8 and 11, (17)H(G1×G2)  =(r1+r2)Imn-A(G1×G2)=(r1+r2)Imn-(A(G1)+KA(G2))=(r1+r2)Imn-(A(G1)×KIm+In×KA(G2))=(r1Imn-In×KA(G2))+(r2Imn-A(G1)×KIm)=(In×Kr1Im-In×KA(G2))+(r2In×KIm-A(G1)×KIm)=In×K(r1Im-A(G2))+(r2In-A(G1))×KIm=In×KH(G2)+H(G1)×KIm=H(G1)+KH(G2), where Imm is the mn×mn identity matrix.

Lemma 13.

If A and B are triangulable matrices then the eigenvalues of A+kB are given by α+β, respectively, as α and β vary through the eigenvalues of A and B.

Proof.

Since A and B are triangulable, there exist invertible matrices Q and P such that A1=QAQ-1 and B1=PBP-1 are upper triangular. If A and B are n×n and m×m matrices, respectively, by Lemmas 9 and 10, (18)A1+kB1=A1×KIm+In×KB1=(QAQ-1)×K(PImP-1)+(QInQ-1)×K(PBP-1)=(Q×KP)(AQ-1×KImP-1)+(Q×KP)(InQ-1×KBP-1)=(Q×KP)(A×KIm)(Q-1×KP-1)+(Q×KP)(In×KB)(Q-1×KP-1)=(Q×KP)[(A×KIm)+(In×KB)](Q-1×KP-1)=(Q×KP)[(A×KIm)+(In×KB)](Q×KP)-1=(Q×KP)(A+kB)(Q×KP)-1. So A+kB is similar to A1+kB1 and they have the same eigenvalues. Obviously A1+kB1=A1×KIm+In×KB1 is upper triangular with diagonal entries given by α+β, respectively, as α and β vary through the eigenvalues of A1 and B1. Hence the eigenvalues of A+kB are α+β, respectively, as α and β vary through the eigenvalues of A and B.

Theorem 14.

Let G1 and G2 be the regular graphs with degrees m and n, respectively. If the eigenvalues of the adjacency matrix A(G1) are written as α1α2αn-1αm=m and the eigenvalues of the adjacency matrix A(G2) are written as β1β2βn-1βn=n, then the number of spanning trees of the Cartesian product G1×G2 is (19)t(G1×G2)=1mni,j[(m+n)-(αi+βj)], where i and j satisfy (i,j){1,2,,m}×{1,2,,n}-{(m,n)}.

Proof.

We know G1×G2 has mn points and the degree of G1×G2 is m+n. By Lemma 11, A(G1×G2)=A(G1)+KA(G2). By Lemma 13, the eigenvalues of A(G1×G2) are αi+βj for i=1,2,,m and j=1,2,,n. The result follows by Lemma 5.

Theorem 15.

If G1 and G2 are the regular graph of degrees m and n, respectively, the eigenvalues of the Kirchhoff matrix H(G1) are written as 0=λ0λ1λm-1, and the eigenvalues of the Kirchhoff matrix H(G2) are written as 0=γ0γ1γn-1, then the number of spanning trees of the Cartesian product of G1 and G2 is (20)t(G1×G2)=1mni,j(λi+γj), where i and j satisfy (i,j){0,1,,m-1}×{0,1,,n-1}-{(0,0)}.

Proof.

By Lemma 12, H(G1×G2)=H(G1)+KH(G2). By Lemma 13, the eigenvalues of H(G1×G2) are λi+γj for i=0,1,,m-1 and j=0,1,,n-1. Hence by Lemma 4, the result follows.

5. The Number of Spanning Trees of the <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M292"><mml:mrow><mml:msup><mml:mrow><mml:mi>r</mml:mi></mml:mrow><mml:mrow><mml:mi>n</mml:mi></mml:mrow></mml:msup></mml:mrow></mml:math></inline-formula>-Lattice Network Definition 16 (see [<xref ref-type="bibr" rid="B27">17</xref>, <xref ref-type="bibr" rid="B28">18</xref>]).

The rn-lattice networks R(r,n) are defined as (21)R(r,n)={Krforn=1R(r,n-1)×Krforn2, where Kr is a complete graph of r points.

When r=2, R(2,n) is well known, the Boolean n-cube network.

We denote R(r,n)=Kr(1)×Kr(2)××Kr(n).

Lemma 17.

The eigenvalues of H(Kr) are r with multiplicity r-1 and 0 with multiplicity 1.

Proof.

Since H(Kr)=rIr-Jr, where Jr is the matrix of all ones, letting PKr(λ) be the character polynomial of H(Kr), we obtain by Gaussian elimination (22)PKr(λ)=λ|11110λ-r-100λ-r0000λ-r|=λ(λ-r)r-1. Hence the result follows.

Lemma 18.

If the distinct eigenvalues of the Kirchhoff matrix H(R(r,n)) are 0=λ0λ1λn-1λn then(23)λ0=withmultiplicity  C(n,n)=1λ1=rwithmultiplicity  (r-1)C(n,n-1)λ2=2rwithmultiplicity  (r-1)2C(n,n)λn-1=(n-1)rwithmultiplicity  (r-1)n-1C(n,1)λn=nrwithmultiplicity  (r-1)nC(n,0),

where C(n,r)=n!/r!(n-r)!.

Proof.

Since R(r,n)=Kr(1)×Kr(2)××Kr(n)and by Lemma 13, we obtain each of eigenvalues of H(R(r,n))=i=1none of eigenvalues of H(Kr(i)).

Hence if we take the eigenvalue 0 of H(Kr(i)) for i=1,2,,n then λ0=0 with multiplicity C(n,n)=1. If we take the eigenvalue r of H(Kr(l)) for some one l{1,2,,n} and the eigenvalue 0 of H(Kr(i)) for each i{1,2,,n}-{l} then λ1=r with multiplicity (r-1)C(n,n-1). If we take the eigenvalue r of H(Kr(l)) and H(Kr(m)), respectively, for l,m{1,2,,n} and the eigenvalue 0 of H(Kr(i)) for each i{1,2,,n}-{l,m}, then λ2=2r with multiplicity (r-1)2C(n,n). We keep performing the same process. Hence the result follows.

The main theorem in [17, 18] can be obtained much more simply by Theorem 19 as follows.

Theorem 19 (see [<xref ref-type="bibr" rid="B27">17</xref>]).

The number of spanning trees of R(r,n) is (24)t(R(r,n))=rrn-n-1i=2niC(n,i)(r-1)i.

Proof.

Since the degree of R(r,n) is n(r-1), the number of points of R(r,n) is rn. By Lemma 18 and Theorem 15, we obtain (25)t(R(r,n))=1rn·rC(n,n-1)(r-1)·(2r)C(n,n-2)(r-1)2·(3r)C(n,n-3)(r-1)3(nr)C(n,0)(r-1)n=1rn·rC(n,n-1)(r-1)+C(n,n-2)(r-1)2+C(n,n-3)(r-1)3++C(n,0)(r-1)n·i=1niC(n,n-i)(r-1)i=1rn·rrn-1·i=1niC(n,n-i)(r-1)i=rrn-n-1i=2niC(n,i)(r-1)i.

Corollary 20 (see [<xref ref-type="bibr" rid="B28">18</xref>]).

The number of spanning trees of the Boolean n-cube network Bn is (26)t(Bn)=22n-n-1i=2niC(n,i).

Proof.

Since Bn=R(2,n), by Theorem 15, the result follows.

6. The Number of Spanning Trees of the <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M347"><mml:mn>2</mml:mn><mml:mo mathvariant="bold"> </mml:mo><mml:mo mathvariant="bold">×</mml:mo><mml:mo mathvariant="bold"> </mml:mo><mml:mn>3</mml:mn><mml:mo mathvariant="bold">⋯</mml:mo><mml:mo mathvariant="bold"> </mml:mo><mml:mo mathvariant="bold">×</mml:mo><mml:mo mathvariant="bold"> </mml:mo><mml:mi>n</mml:mi></mml:math></inline-formula>-Lattice Network Definition 21.

The 2×3×n lattice network Qn can be defined recursively by Q2=K2 and Qn=Kn×Qn-1.

Thus Qn has n! points. We denote Qn=K2×K3××Kn.

Theorem 22.

The number of spanning trees of Qn is (27)t(Qn)=1n!i=1n-12r1<r2<<rin(j=1irj)j=1i(rj-1).

Proof.

Since the eigenvalues of H(Kr) are r with multiplicity r-1 and 0 with multiplicity 1, the distinct i=0n-1C(n-1,i)=2n-1 eigenvalues of Qn are 0 and λi,r1,r2,,ri=j=1irj with multiplicity (r1-1)(r2-1)(ri-1), where r1,r2,,ri, satisfying 2r1<r2<<rin,  i=1,2,,n-1, are nonzero eigenvalues of H(Kr1),H(Kr2),,H(Kri), respectively, and take zero eigenvalues for the remaining H(Kr), where rr1,r2,,ri,  r=2,3,,n-1. By Lemma 13 and Theorem 15, the result follows.

Example 23.

The number of spanning trees of Q3 and Q4 is as shown in Figure 1, where t(Q3)=(1/3!)2(2-1)3(3-1)(2+3)(2-1)(3-1)=75 and t(Q4)=(1/4!)2(2-1)3(3-1)4(4-1)  (2+3)(2-1)(3-1)  (2+4)(2-1)(4-1)(3+4)(3-1)(4-1)(2+3+4)(2-1)(3-1)(4-1) = 1620609272381440.

7. The Number of Spanning Trees of the Generalized Boolean <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M375"><mml:mrow><mml:mi>n</mml:mi></mml:mrow></mml:math></inline-formula>-Cube Network Definition 24.

The generalized Boolean n-cube network BR(r,n) can be defined by (28)BR(r,n)={Crforn=1BR(r,n-1)×K2forn2, where Cr is a cycle with r points. One denotes BR(r,n)=Cr×K2××K2.

Setting En is the n×n matrix by (29)En=,Eni=1i+1i.

Lemma 25.

The eigenvalues of the adjacent matrix A(Kr) are -1 with multiplicity r-1 and r-1 with multiplicity 1.

Lemma 26 (see [<xref ref-type="bibr" rid="B13">33</xref>]).

If B is a sequence matrix, λ is an eigenvalue of B, and f is a polynomial then f(λ) is the eigenvalue of f(B).

Lemma 27.

The eigenvalues of the adjacent matrix A(Cn) are 2cos(2πk/n) for k=0,1,2,,n-1.

Proof.

Since (30)det(λIn-En)=|λ-1000λ-1000000λ-1-100λ|=λ·λn-1+(-1)(-1)n+1(-1)n-1=λn-1, the eigenvalues of En are e(2πk/n)i for k=0,1,2,,n-1. It follows that A(Cn)=En+Enn-1. By Lemma 26, the eigenvalues of A(Cn) are e(2πk/n)i+(e(2πk/n)i)n-1=e(2πk/n)i+(e2πki-(2πk/n)i)=e(2πk/n)i+e-(2πk/n)i=2cos(2πk/n) for k=0,1,2,,n-1.

The main theorem in  can be obtained much more simply as follows.

Theorem 28 (see [<xref ref-type="bibr" rid="B14">13</xref>]).

The number of spanning trees of BR(r,n) is (31)t(BR(r,n))=r2r(2n-1-1)-n+1×i=1n-1(ik=1r-1(i+1-cos2πkr))C(n-1,i).

Proof.

It follows that the points of BR(r,n) are r·2n-1 and the degree of any edge of BR(r,n) is n+1. By Lemma 25, the eigenvalues of the adjacent matrix A(K2) are -1 and 1. By Lemmas 13 and 27, the distinct eigenvalues of the adjacent matrix A(BR(r,n)) are (32)(n-2i-1)+2cos2πkrk=0,1,,r-1with  multiplicityC(n-1,i), where i=0,1,,n-1.

When k=0 and i=0, the eigenvalue is n+1. When k=0 and i=0,1,,n-1, the eigenvalues are (n-2i-1)+2. By Theorem 14, (33)t(BR(r,n))=1r·2n-1i=1n-1((n+1)-((n-2i-1)+2))C(n-1,i)·k=1r-1i=0n-1((n+1)-((n-2i-1)+2cos2πkr))C(n-1,i)=1r·2n-1i=1n-1(2i)C(n-1,i)·k=1r-1i=0n-1(2(i+1-cos2πkr))C(n-1,i). Since i=0n-12C(n-1,i)=2i=0n-1  C(n-1,i) = 22n-1 and i=1n-12C(n-1,i)=22n-1-1, hence (34)t(BR(r,n))=1r·2n-122n-1-12(r-1)2n-1i=1n-1iC(n-1,i)·k=1r-1i=0n-1(i+1-cos2πkr)C(n-1,i). Since k=1r-1sin2(πk/r)=r2 and k=1r-1(1-cos(2πk/r))=k=1r-12sin2(πk/r)=r2/2r-1 as i=1, (35)t(BR(r,n))=r2r(2n-1-1)-n+1i=1n-1iC(n-1,i)·k=1r-1i=1n-1(i+1-cos2πkr)C(n-1,i)=r2r(2n-1-1)-n+1×i=1n-1(ik=1r-1(i+1-cos2πkr))C(n-1,i).

8. The Number of Spanning Trees of the Hypercube Network Definition 29.

The hypercube network H(r,m) can be defined by (36)H(r,m)={Crform=1H(r,m-1)×Crform2, where Cr is a cycle with r points.

Theorem 30.

The number of spanning trees of H(r,m) is (37)t(H(r,m))=2rm-1rml1,l2,,lm(m-i=1mcos2πlir), where l1,l2,,lm satisfy (l1,l2,,lm){0,1,,r-1}×{0,1,,r-1}××{0,1,,r-1}-{(0,0,,0)}.

Proof.

It follows that the points of H(r,m) are rm and the degree of any edge of H(r,m) is 2m. By Lemma 26 and Theorem 14, (38)t(H(r,m))=1rml1,l2,,lm(2m-2i=1mcos2πlir)=2rm-1rml1,l2,,lm(m-i=1mcos2πlir), where l1,l2,,lm satisfy (l1,l2,,lm){0,1,,r-1}×{0,1,,r-1}××{0,1,,r-1}-{(0,0,,0)}.

9. Conclusion

Due to the high dependence of the network design and reliability problem, electrical circuits designing issue are on the graph theory. For example, the larger degree of points a network has, the more I/O ports and edges are needed and the more cost is required. The evaluation of this number not only is interesting from a mathematical (computational) perspective but also is an important issue on practical applications. However, the study for spanning trees of the Cartesian product of regular graphs remains an open and important invariant. In this paper, the eigenvalues of the Kirchhoff matrix of Cartesian product of two regular graphs, G1 and G2, are given by λ+γ as λ and γ vary through the eigenvalues of the Kirchhoff matrices H(G1) and H(G2), respectively. By this result, the formula for the number of spanning trees of the four regular networks can be simply obtained. Using this formula, the main results in [13, 17, 18] can be obtained much more simply and will be extended.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

The authors would like to thank the anonymous reviewers for their valuable suggestions.

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