Complexity of Products of Some Complete and Complete Bipartite Graphs

Abstract

The number of spanning trees in graphs (networks) is an important invariant; it is also an important measure of reliability of a network. In this paper, we derive simple formulas of the complexity, number of spanning trees, of products of some complete and complete bipartite graphs such as cartesian product, normal product, composition product, tensor product, and symmetric product, using linear algebra and matrix analysis techniques.

1. Introduction

In this work we deal with simple and finite undirected graphs𝐺=(𝑉,𝐸), where𝑉is the vertex set and𝐸is the edge set. For a graph𝐺, a spanning tree in𝐺is a tree which has the same vertex set as𝐺. The number of spanning trees in𝐺, also called the complexity of the graph, denoted by𝜏(𝐺), is a well-studied quantity (for long time). A classical result of Kirchhoff [1], can be used to determine the number of spanning trees for𝐺=(𝑉,𝐸). Let𝑉={𝑣₁,𝑣₂,…,𝑣_𝑛}; then the Kirchhoff matrix𝐻defined as𝑛×𝑛, characteristic matrix,𝐻=𝐷−𝐴, where𝐷is the diagonal matrix whose elements are the degrees of the vertices of𝐺. While𝐴is the adjacency matrix of𝐺,𝐻=[𝑎_𝑖𝑗]is defined as follows:(i)𝑎_𝑖𝑗=−1𝑣_𝑖and𝑣_𝑗are adjacent and𝑖≠𝑗,(ii)𝑎_𝑖𝑗equals the degree of vertex𝑣_𝑖if𝑖=𝑗,(iii)𝑎_𝑖𝑗=0otherwise. All of the cofactors of𝐻are equal to𝜏(𝐺). There are other methods for calculating𝜏(𝐺). Let𝜇₁≥𝜇₁≥⋯≥𝜇_𝑝denote the eigenvalues of𝐻matrix of a𝑝point graph. Then it is easily shown that𝜇_𝑝=0. Furthermore, Kelmans and Chelnokov [2] have shown that𝜏(𝐺)=(1/𝑝)∏^𝑝−1_𝑘=1𝜇_𝑘. The formula for the number of spanning trees in a d-regular graph𝐺can be expressed as𝜏(𝐺)=(1/𝑝)∏^𝑝−1_𝑘=1(𝑑−𝜆_𝑘),where𝜆₀=𝑑,𝜆₁,𝜆₂,…,𝜆_𝑝−1are the eigenvalues of the corresponding adjacency matrix of the graph. However, for a few special families of graphs there exist simple formulas that make it much easier to calculate and determine the number of corresponding spanning trees especially when these numbers are very large. One of the first results is due to Cayley [3] who showed that the complete graph on𝑛vertices,𝐾_𝑛has𝑛^𝑛−2spanning trees,𝑛≥2. Another result is that𝜏(𝐾_𝑝,𝑞)=𝑝^𝑞−1𝑞^𝑝−1,𝑝,𝑞≥1, where𝐾_𝑝,𝑞is the complete bipartite graph with bipartite sets containing𝑝and𝑞vertices, respectively. It is well known, as in, for example, [4, 5]. Another result is due to Sedlać̌ek [6] who derived a formula for the wheel on𝑛+1vertices,𝑊_𝑛+1; he showed that𝜏(𝑊_𝑛+1)=((3+√5)/2)^𝑛+((3−√5)/2)^𝑛−2, for𝑛≥3. Sedlacek [7] also later derived a formula for the number of spanning trees in a Mobius ladder,𝑀_𝑛,𝜏(𝑀_𝑛)=(𝑛/2)[(2+√3)^𝑛+(2−√3)^𝑛+2]for𝑛≥2. Another class of graphs by Boesch et al., for which an explicit formula has been derived, is based on a prism [8, 9].

Now, we can introduce the following lemmas.

Lemma 1 (see [10]). Consider𝜏(𝐺)=(1/𝑛²)det(𝑛𝐼−𝐷+𝐴)where𝐴and𝐷are the adjacency and degree matrices of𝐺and the complement of𝐺, respectively, and𝐼is the𝑛×𝑛unit matrix.

Lemma 2. Let𝐸_𝑛(𝑥)be𝑛×𝑛matrix,𝑥≥2such that 𝐸_𝑛(𝑥)=𝑥1⋯⋯⋯11⋱⋱⋱⋱⋮⋮⋱⋱⋱⋱⋮⋮⋱⋱⋱⋱⋮⋮⋱⋱⋱⋱11⋯⋯⋯1𝑥. (1) Then, det𝐸_𝑛=(𝑥+𝑛−1)(𝑥−1)^𝑛−1. (2)

Proof. From the definition of the circulant determinants, we have det𝐸_𝑛(𝑥)==det𝑥1⋯⋯⋯11⋱⋱⋱⋱⋮⋮⋱⋱⋱⋱⋮⋮⋱⋱⋱⋱⋮⋮⋱⋱⋱⋱11⋯⋯⋯1𝑥_𝑛𝑗=1𝑥+𝜔_𝑗+𝜔²_𝑗+𝜔³_𝑗+⋯+𝜔_𝑗^𝑛−1=()×𝑥+1+1+⋯+1_{𝑛𝑗=1,𝜔𝑗≠1}𝑥+𝜔_𝑗+𝜔²_𝑗+𝜔³_𝑗+⋯+𝜔_𝑗^𝑛−1₌₋₁=(𝑥+𝑛−1)×(𝑥−1)^𝑛−1. (3)

We can generalize the previous lemma as follows.

Lemma 3. Let𝐴,𝐵∈𝐹^𝑛×𝑛and𝐹∈𝐹^{𝑘𝑛×𝑘𝑛}such that 𝐹=𝐴𝐵⋯⋯⋯𝐵𝐵⋱⋱⋱⋱⋮⋮⋱⋱⋱⋱⋮⋮⋱⋱⋱⋱⋮⋮⋱⋱⋱⋱𝐵𝐵⋯⋯⋯𝐵𝐴. (4) Then, det𝐹=[det(𝐴−𝐵)]^𝑘−1det[𝐴+(𝑘−1)𝐵]. (5)

Lemma 4 (see [11]). Let𝐴∈𝐹^𝑛×𝑛, let𝐵∈𝐹^𝑛×𝑚, let𝐶∈𝐹^𝑚×𝑛,and let𝐷∈𝐹^𝑚×𝑚;assume that𝐴,𝐷are nonsingular matrices. Then det=()𝐴𝐵𝐶𝐷−1^𝑛𝑚det𝐴−𝐵𝐷⁻¹𝐶=()det𝐷−1^𝑛𝑚det𝐴det𝐷−𝐶𝐴⁻¹𝐵. (6)

Formulas in Lemmas 2, 3, and 4 give some sort of symmetry in some matrices which facilitates our calculation of determinants.

2. Number of Spanning Trees of Cartesian Product of Graphs

The Cartesian product,𝐺₁×𝐺₂, is the simple graph with vertex set𝑉(𝐺₁×𝐺₂)=𝑉₁×𝑉₂and edge set𝐸(𝐺₁×𝐺₂)=[(𝐸₁×𝑉₂)∪(𝑉₁×𝐸₂)]such that two vertices(𝑢₁,𝑢₂)and(𝑣₁,𝑣₂)are adjacent in𝐺₁×𝐺₂if and only if either𝑢₁=𝑣₁and𝑢₂is adjacent to𝑣₂in𝐺₂or𝑢₁is adjacent to𝑣₁in𝐺₁and𝑢₂=𝑣₂[12].

Theorem 5. For𝑛,𝑚≥1,we have 𝜏𝐾₂×𝐾_𝑚,𝑛=𝑚^𝑛−1𝑛^𝑚−1(𝑚+2)^𝑛−1×(𝑛+2)^𝑚−1(𝑛+𝑚+2). (7)

Proof. Applying Lemma 1, we have𝜏𝐾₂×𝐾_𝑚,𝑛=1(2(𝑚+𝑛))²det2(𝑚+𝑛)𝐼−𝐷+𝐴=14(𝑚+𝑛)²=1×det𝑛+21⋯10⋯⋯001⋯⋯⋯⋯⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋱⋱⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱⋱⋱⋱⋱⋮1⋯1𝑛+20⋯⋯0⋮⋱⋱⋱⋱⋱⋱⋮0⋯⋯0𝑚+21⋯1⋮⋱⋱⋱⋱⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋱⋱⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋱⋱⋱⋱10⋯⋯01⋯1𝑚+21⋯⋯⋯⋯⋯1001⋯⋯⋯⋯⋯1𝑛+21⋯10⋯⋯01⋱⋱⋱⋱⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋱⋱⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱⋱⋱⋱⋱⋮1⋯1𝑛+20⋯⋯0⋮⋱⋱⋱⋱⋱⋱⋮0⋱⋱0𝑚+21⋯1⋮⋱⋱⋱⋱⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋱⋱⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯⋯⋯⋯⋯100⋯⋯01⋯1𝑚+24(𝑚+𝑛)²=1det𝑛+22⋯21⋯⋯12⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱2⋮⋱⋱⋮2⋯2𝑛+21⋯⋯11⋯⋯1𝑚+22⋯2⋮⋱⋱⋮2⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱21⋯⋯12⋯2𝑚+2×det𝑛+20⋯0−1⋯⋯−10⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱0⋮⋱⋱⋮0⋯0𝑛+2−1⋯⋯−1−1⋯⋯−1𝑚+20⋯0⋮⋱⋱⋮0⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱0−1⋯⋯−10⋯0𝑚+24(𝑚+𝑛)²𝐵det𝐴𝐵_𝑇𝐶𝐸×det𝐷𝐸_𝑇𝐹=14(𝑚+𝑛)²×det𝐴det𝐶−𝐵^𝑇𝐴⁻¹𝐵×det𝐷det𝐹−𝐸^𝑇𝐷⁻¹𝐸=14(𝑚+𝑛)²det𝑛+22⋯22⋱⋱⋮⋮⋱⋱22⋯2𝑛+2_𝑚×𝑚𝑛()()×det𝑚+2+𝑚2𝑚+3𝑛+2𝑚2𝑛+3𝑚⋯𝑛+2𝑚2𝑛+3𝑚𝑛+2𝑚2𝑛+3𝑚𝑛+2𝑚⋱⋱⋮⋮⋱⋱2𝑛+3𝑚𝑛+2𝑚2𝑛+3𝑚⋯𝑛+2𝑚2𝑛+3𝑚𝑛()()𝑛+2𝑚𝑚+2+𝑚2𝑚+3𝑛+2𝑚_𝑛×𝑛×det𝑛+20⋯00⋱⋱⋮⋮⋱⋱00⋯0𝑛+2_𝑚×𝑚𝑛()+()×det𝑚+2𝑚+4𝑛+2−𝑚⋯𝑛+2−𝑚𝑛+2−𝑚𝑛+2⋱⋱⋮⋮⋱⋱−𝑚𝑛+2−𝑚⋯𝑛+2−𝑚𝑛()+()𝑛+2𝑚+2𝑚+4𝑛+2_𝑛×𝑛=14(𝑚+𝑛)²×2^𝑚det𝑛+221⋯11⋱⋱⋮⋮⋱⋱11⋯1𝑛+22_𝑚×𝑚×2𝑛+3𝑚𝑛+2𝑚^𝑛𝑛()()det𝑚+2+𝑚2𝑚+3𝑛()()2𝑛+3𝑚1⋯11⋱⋱⋮⋮⋱⋱11⋯1𝑚+2+𝑚2𝑚+32𝑛+3𝑚_𝑛×𝑛×det𝑛+20⋯00⋱⋱⋮⋮⋱⋱00⋯0𝑛+2_𝑚×𝑚×−𝑚𝑛+2^𝑛𝑛()+()det𝑚+2𝑚+4𝑛()+()−𝑚1⋯11⋱⋱⋮⋮⋱⋱11⋯1𝑚+2𝑚+4−𝑚_𝑛×𝑛=14(𝑚+𝑛)²×2^𝑚×𝑛+22+𝑚−1𝑛+22−1^𝑚−1×2𝑛+3𝑚𝑛+2𝑚^𝑛×𝑛(𝑚+2)+𝑚(2𝑚+3)×𝑛()()2𝑛+3𝑚+𝑛−1𝑚+2+𝑚2𝑚+32𝑛+3𝑚−1^𝑛−1×(𝑛+2)^𝑚×−𝑚𝑛+2^𝑛×−𝑛(𝑚+2)+(𝑚+4)𝑚×−𝑛()+()+𝑛−1𝑚+2𝑚+4𝑚−1^𝑛−1. (8)
Thus, 𝜏𝐾₂×𝐾_𝑚,𝑛=𝑚^𝑛−1𝑛^𝑚−1(𝑚+2)^𝑚−1(𝑛+2)^𝑚−1×(𝑛+𝑚+2). (9) In particular, 𝜏𝐾₂×𝐾_𝑛,𝑛=2𝑛^2𝑛−2(𝑛+1)(𝑛+2)^2𝑛−2;𝑛≥1. (10)

Theorem 6. For𝑚,𝑛≥1, we have 𝜏𝐾₃×𝐾_𝑚,𝑛=3𝑛^𝑚−1𝑚^𝑛−1(𝑚+3)^2𝑛−2(𝑛+3)^2𝑚−2×(𝑛+𝑚+3)². (11)

Proof. Applying Lemma 1, we have𝜏𝐾₃×𝐾_𝑚,𝑛=19(𝑚+𝑛)²det3(𝑚+𝑛)𝐼−𝐷+𝐴=19(𝑚+𝑛)²=1×det𝑛+31⋯10⋯⋯001⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯1𝑛+30⋯⋯01⋯100⋯⋯0𝑚+31⋯11⋯⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯1𝑚+31⋯⋯101⋯11⋯⋯1𝑛+31⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯101⋯⋯11⋯1𝑛+31⋯⋯101⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯⋯11⋯100⋯⋯001⋯11⋯⋯101⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯101⋯⋯11⋯101⋯⋯101⋯11⋯⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯⋯11⋯101⋯⋯11⋯⋯101⋯11⋯⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯11⋯101⋯⋯101⋯11⋯⋯101⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯101⋯⋯11⋯100⋯⋯001⋯11⋯⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮0⋯⋯01⋯101⋯⋯1𝑚+31⋯11⋯⋯101⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯1𝑚+31⋯⋯11⋯101⋯⋯1𝑛+31⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯⋯11⋯1𝑛+30⋯⋯001⋯10⋯⋯0𝑚+31⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯100⋯⋯01⋯1𝑚+39(𝑚+𝑛)²=1det𝐴𝐵𝐵𝐵𝐴𝐵𝐵𝐵𝐴9(𝑚+𝑛)²[det(𝐴−𝐵)]²[(=1det𝐴+2𝐵)]9(𝑚+𝑛)²det𝑛+30⋯0−1⋯⋯−10⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱0⋮⋱⋱⋮0⋯0𝑛+3−1⋯⋯−1−1⋯⋯−1𝑚+30⋯0⋮⋱⋱⋮0⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱0−1⋯⋯−10⋯0𝑚+3²=1×det𝑛+33⋯32⋯⋯23⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱3⋮⋱⋱⋮3⋯3𝑛+32⋯⋯22⋯⋯2𝑚+33⋯3⋮⋱⋱⋮3⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱32⋯⋯23⋯3𝑚+39(𝑚+𝑛)²𝐵det𝐴𝐵_𝑇𝐶²𝐸×det𝐷𝐸_𝑇𝐹=19(𝑚+𝑛)²×(det𝐴)²det𝐶−𝐵^𝑇𝐴⁻¹𝐵²×det𝐷det𝐹−𝐸^𝑇𝐷⁻¹𝐸. (12)Thus,𝜏𝐾₃×𝐾_𝑚,𝑛=19(𝑚+𝑛)²det𝑛+30⋯00⋱⋱⋮⋮⋱⋱00⋯0𝑛+3_𝑚×𝑚²×det𝑛𝑚+3𝑛+2𝑚+9𝑛+3−𝑚⋯𝑛+3−𝑚𝑛+3−𝑚𝑛+3⋱⋱⋮⋮⋱⋱−𝑚𝑛+3−𝑚⋯𝑛+3−𝑚𝑛+3𝑛𝑚+3𝑛+2𝑚+9𝑛+3_𝑛×𝑛²×det𝑛+33⋯33⋱⋱⋮⋮⋱⋱33⋯3𝑛+3_𝑚×𝑚×det𝑛𝑚+3𝑛+3𝑚²+5𝑚𝑛+3𝑚3𝑛+5𝑚⋯𝑛+3𝑚3𝑛+5𝑚𝑛+3𝑚3𝑛+5𝑚𝑛+3𝑚⋱⋱⋮⋮⋱⋱3𝑛+5𝑚𝑛+3𝑚3𝑛+5𝑚⋯𝑛+3𝑚3𝑛+5𝑚𝑛+3𝑚𝑛𝑚+3𝑛+3𝑚²+5𝑚𝑛+3𝑚_𝑛×𝑛=19(𝑚+𝑛)²(𝑚+3)^2𝑚×−𝑚𝑛+3^2𝑛×det𝑚𝑛+3𝑛+2𝑚+9−𝑚1…11⋱⋱⋮⋮⋱⋱11…1𝑚𝑛+3𝑛+2𝑚+9−𝑚²×3^𝑚×det𝑛+331…11⋱⋱⋮⋮⋱⋱11…1𝑛+33_𝑚×𝑚×3𝑛+5𝑚𝑛+3𝑚^𝑛×det𝑚𝑛+3𝑛+2𝑚²+5𝑚3𝑛+5𝑚1…11⋱⋱⋮⋮⋱⋱11…1𝑚𝑛+3𝑛+2𝑚²+5𝑚.3𝑛+5𝑚 (13)
Using Lemma 2, we have 𝜏𝐾₃×𝐾_𝑚,𝑛=19(𝑚+𝑛)²×(𝑛+3)^2𝑚×−𝑚𝑛+3^2𝑛×−𝑛𝑚+3𝑛+2𝑚+9𝑚+𝑛−1²×−𝑛𝑚+3𝑛+2𝑚+9𝑚−1^2𝑛−2×3^𝑚𝑛+33×+𝑚−1𝑛+33−1^𝑚−1×3𝑛+5𝑚𝑛+3𝑚^𝑛×𝑛𝑚+3𝑛+3𝑚²+5𝑚×3𝑛+5𝑚+𝑛−1𝑛𝑚+3𝑛+3𝑚²+5𝑚3𝑛+5𝑚−1^𝑛−1=19(𝑚+𝑛)²(𝑛+3)^2𝑚×1(𝑛+3)^2𝑛×(3𝑛+3𝑚+9)²×(𝑛𝑚+3𝑛+3𝑚+9)^2𝑛−2×(𝑛+3𝑚)×𝑛^𝑚−1×1(𝑛+3𝑚)^𝑛×6𝑛𝑚+3𝑛²+3𝑚²×𝑛𝑚+3𝑚^2𝑛−1=3𝑛^𝑚−1𝑚^𝑛−1(𝑚+3)^2𝑛−2(𝑛+3)^2𝑚−2(𝑛+𝑚+3)². (14) In particular, 𝜏𝐾₃×𝐾_𝑛,𝑛=3𝑛^2𝑛−2(2𝑛+3)²(𝑛+3)^4𝑛−4;𝑛≥1. (15)

3. Number of Spanning Trees of Normal Product of Graphs

The normal product, or the strong product,𝐺₁∘𝐺₂, is the simple graph with𝑉(𝐺₁∘𝐺₂)=𝑉₁×𝑉₂,where(𝑢₁,𝑢₂)and(𝑣₁,𝑣₂)are adjacent in𝐺₁∘𝐺₂if and only if either𝑢₁=𝑣₁and𝑢₂is adjacent to𝑣₂,𝑢₁is adjacent to𝑣₁and𝑢₂=𝑣₂, or𝑢₁is adjacent to𝑣₁and𝑢₂is adjacent to𝑣₂[13].

Theorem 7. For𝑛,𝑚≥1,we have 𝜏𝐾₂∘𝐾_𝑚,𝑛=2^{2𝑚+2𝑛−2}×𝑛^𝑚−1×𝑚^𝑛−1×(𝑛+1)^𝑚×(𝑚+1)^𝑛. (16)

Proof. Applying Lemma 1, we have𝜏𝐾₂∘𝐾_𝑚,𝑛=14(𝑚+𝑛)²det2(𝑚+𝑛)𝐼−𝐷+𝐴=14(𝑚+𝑛)²×det^{2𝑛+21⋯10⋯⋯001⋯10⋯⋯01⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯12𝑛+20⋯⋯01⋯100⋯⋯00⋯⋯02𝑚+21⋯10⋯⋯001⋯1⋮⋱⋱11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱10⋯⋯01⋯12𝑚+20⋯⋯01⋯10}_{01⋯10⋯⋯02𝑛+21⋯10⋯⋯01⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯100⋯⋯01⋯12𝑛+20⋯⋯00⋯⋯001⋯10⋯⋯02𝑚+21⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱10⋯⋯01⋯100⋯⋯01⋯12𝑚+2}=14(𝑚+𝑛)²=1det2𝑛+22⋯20⋯⋯02⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱2⋮⋱⋱⋮2⋯22𝑛+20⋯⋯00⋯⋯02𝑚+22⋯2⋮⋱⋱⋮2⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱20⋯⋯02⋯22𝑚+2×det2𝑛+20⋯00⋯⋯00⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱0⋮⋱⋱⋮0⋯02𝑛+20⋯⋯00⋯⋯02𝑚+20⋯0⋮⋱⋱⋮0⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱00⋯⋯00⋯02𝑚+24(𝑚+𝑛)²det2𝑛+22⋯22⋱⋱⋮⋮⋱⋱22⋯22𝑛+2_𝑚×𝑚×det2𝑚+22⋯22⋱⋱⋮⋮⋱⋱22⋯22𝑚+2_𝑛×𝑛×det2𝑛+20⋯00⋱⋱⋮⋮⋱⋱00⋯02𝑛+2_𝑚×𝑚×det2𝑚+20⋯00⋱⋱⋮⋮⋱⋱00⋯02𝑚+2_𝑛×𝑛=14(𝑚+𝑛)²×2^𝑚(𝑛+𝑚)𝑛^𝑚−1×2^𝑛(𝑛+𝑚)𝑚^𝑛−1×2^𝑚(𝑛+1)^𝑚×2^𝑛(𝑚+1)^𝑛=2^{2𝑚+2𝑛−2}×𝑛^𝑚−1×𝑚^𝑛−1×(𝑛+1)^𝑚×(𝑚+1)^𝑛. (17)
In particular, 𝜏𝐾₂∘𝐾_𝑛,𝑛=2^4𝑛−2×𝑛^2𝑛−2×(𝑛+1)^2𝑛;𝑛≥1. (18)

Theorem 8. For𝑚,𝑛≥1, we have 𝜏𝐾₃∘𝐾_𝑚,𝑛=3^{3𝑚+3𝑛−2}×𝑛^𝑚−1×𝑚^𝑛−1×(𝑛+1)^2𝑚×(𝑚+1)^2𝑛. (19)

Proof. Applying Lemma 1, we have𝜏𝐾₃∘𝐾_𝑚,𝑛=19(𝑚+𝑛)²det3(𝑚+𝑛)𝐼−𝐷+𝐴=19(𝑚+𝑛)²=1×det3𝑛+31⋯10⋯⋯001⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯13𝑛+30⋯⋯01⋯100⋯⋯03𝑚+31⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯13𝑚+30⋯⋯001⋯10⋯⋯03𝑛+31⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯100⋯⋯01⋯13𝑛+30⋯⋯001⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮0⋯⋯01⋯100⋯⋯001⋯10⋯⋯001⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯100⋯⋯01⋯100⋯⋯001⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮0⋯⋯01⋯100⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯100⋯⋯001⋯10⋯⋯001⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯100⋯⋯01⋯100⋯⋯001⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮0⋯⋯01⋯100⋯⋯03𝑚+31⋯10⋯⋯001⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯13𝑚+30⋯⋯01⋯100⋯⋯03𝑛+31⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮0⋯⋯01⋯13𝑛+30⋯⋯001⋯10⋯⋯03𝑚+31⋯11⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱11⋯100⋯⋯01⋯13𝑚+39(𝑚+𝑛)²=1det𝐴𝐵𝐵𝐵𝐴𝐵𝐵𝐵𝐴9(𝑚+𝑛)²[det(𝐴−𝐵)]²[(=1det𝐴+2𝐵)]9(𝑚+𝑛)²det3𝑛+30⋯00⋯⋯00⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱0⋮⋱⋱⋮0⋯03𝑛+30⋯⋯00⋯⋯03𝑚+30⋯0⋮⋱⋱⋮0⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱00⋯⋯00⋯03𝑚+3²=1×det3𝑛+33⋯30⋯⋯03⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱3⋮⋱⋱⋮3⋯33𝑛+30⋯⋯00⋯⋯03𝑚+33⋯3⋮⋱⋱⋮3⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱30⋯⋯03⋯33𝑚+39(𝑚+𝑛)²det3𝑛+30⋯00⋱⋱⋮⋮⋱⋱00⋯03𝑛+3_𝑚×𝑚²×det3𝑚+30⋯00⋱⋱⋮⋮⋱⋱00⋯03𝑚+3_𝑛×𝑛²×det3𝑛+33⋯33⋱⋱⋮⋮⋱⋱33⋯33𝑛+3_𝑚×𝑚×det3𝑚+33⋯33⋱⋱⋮⋮⋱⋱33⋯33𝑚+3_𝑛×𝑛. (20)
Using Lemma 2, we have 𝜏𝐾₃∘𝐾_𝑚,𝑛=19(𝑚+𝑛)²×(3𝑛+3)^2𝑚×(3𝑚+3)^2𝑛×3^𝑚×(𝑛+𝑚)×𝑛^𝑚−1×3^𝑛×(𝑛+𝑚)×𝑚^𝑛−1=3^{3𝑚+3𝑛−2}×𝑛^𝑚−1×𝑚^𝑛−1×(𝑛+1)^2𝑚×(𝑚+1)^2𝑛. (21) In paricular, 𝜏𝐾₃∘𝐾_𝑛,𝑛=3^6𝑛−2×𝑛^2𝑛−2×(𝑛+1)^4𝑛;𝑛≥1. (22)

4. Number of Spanning Trees of Composition Product of Graphs

The composition, or lexicographic product,𝐺₁[𝐺₂], is the simple graph with𝑉₁×𝑉₂as the vertex set in which the vertices(𝑢₁,𝑢₂)and(𝑣₁,𝑣₂)are adjacent if either𝑢₁is adjacent to𝑣₁or𝑢₁=𝑣₁and𝑢₂is adjacent to𝑣₂in𝐺₂[13].

Theorem 9. For𝑛,𝑚≥1, we have 𝜏𝐾₂𝐾_𝑚,𝑛=4(𝑚+𝑛)²×(𝑚+2𝑛)^2𝑚−2(𝑛+2𝑚)^2𝑛−2. (23)

Proof. Applying Lemma 1, we have𝜏𝐾₂𝐾_𝑚,𝑛=14(𝑚+𝑛)²det2(𝑚+𝑛)𝐼−𝐷+𝐴=14(𝑚+𝑛)²=1×det𝑚+2𝑛+11⋯10⋯⋯01⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯1𝑚+2𝑛+10⋯⋯00⋯⋯0𝑛+2𝑚+11⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯01⋯1𝑛+2𝑚+10⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯00⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯00⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯00⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯0𝑚+2𝑛+11⋯10⋯⋯01⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯1𝑚+2𝑛+10⋯⋯00⋯⋯0𝑛+2𝑚+11⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯01⋯1𝑛+2𝑚+14(𝑚+𝑛)²det𝑚+2𝑛+11⋯11⋱⋱⋮⋮⋱⋱11⋯1𝑚+2𝑛+1_𝑚×𝑚²×det𝑛+2𝑚+11⋯11⋱⋱⋮⋮⋱⋱11⋯1𝑛+2𝑚+1_𝑛×𝑛²=14(𝑚+𝑛)²(2𝑛+2𝑚)²(𝑚+2𝑛)^2𝑚−2×(2𝑛+2𝑚)²(𝑛+2𝑚)^2𝑛−2=4(𝑚+𝑛)²(𝑚+2𝑛)^2𝑚−2(𝑛+2𝑚)^2𝑛−2. (24)
In particular, 𝜏𝐾₂𝐾_𝑛,𝑛=16×3^4𝑛−4×𝑛^4𝑛−4;𝑛≥1. (25)

Theorem 10. For𝑚,𝑛≥1, we have 𝜏𝐾₃𝐾_𝑚,𝑛=3⁴(𝑚+𝑛)⁴(3𝑚+2𝑛)^3𝑛−3(3𝑛+2𝑚)^3𝑚−3. (26)

Proof. Applying Lemma 1, we have𝜏𝐾₃𝐾_𝑚,𝑛=19(𝑚+𝑛)²det3(𝑚+𝑛)𝐼−𝐷+𝐴=19(𝑚+𝑛)²×det^{3𝑛+2𝑚+11⋯10⋯⋯00⋯⋯01⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱⋮1⋯13𝑛+2𝑚+10⋯⋯00⋯⋯00⋯⋯03𝑚+2𝑛+11⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯13𝑚+2𝑛+10⋯⋯00⋯⋯00⋯⋯03𝑛+2𝑚+11⋯1⋮⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯00⋯⋯01⋯13𝑛+2𝑚+1}_{0⋯⋯00⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯00⋯⋯03𝑚+2𝑛+11⋯10⋯⋯00⋯⋯01⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱⋮1⋯13𝑚+2𝑛+10⋯⋯00⋯⋯00⋯⋯03𝑛+2𝑚+11⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮0⋯⋯01⋯13𝑛+2𝑚+10⋯⋯00⋯⋯00⋯⋯03𝑚+2𝑛+11⋯1⋮⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯00⋯⋯01⋯13𝑚+2𝑛+1}=19(𝑚+𝑛)²det3𝑛+2𝑚+11⋯10⋯⋯01⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯13𝑛+2𝑚+10⋯⋯00⋯⋯03𝑚+2𝑛+11⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯01⋯13𝑚+2𝑛+1³=19(𝑚+𝑛)²det3𝑛+2𝑚+11⋯11⋱⋱⋮⋮⋱⋱11⋯13𝑛+2𝑚+1_𝑚×𝑚³×det3𝑚+2𝑛+11⋯11⋱⋱⋮⋮⋱⋱11⋯13𝑚+2𝑛+1_𝑛×𝑛³. (27)Using Lemma 2, we have 𝜏𝐾₃𝐾_𝑚,𝑛=3⁴(𝑚+𝑛)⁴(3𝑚+2𝑛)^3𝑛−3(3𝑛+2𝑚)^3𝑚−3. (28) In particular, 𝜏𝐾₃𝐾_𝑛,𝑛=6⁴×5^6𝑛−6×𝑛^6𝑛−2;𝑛≥1. (29)

5. Complexity of Tensor Product of Graphs

The tensor product, or Kronecker product,𝐺₁⊗𝐺₂, is the simple graph with𝑉(𝐺₁⊗𝐺₂)=𝑉₁×𝑉₂,where(𝑢₁,𝑢₂)and(𝑣₁,𝑣₂)are adjacent in𝐺₁⊗𝐺₂if and only if𝑢₁is adjacent to𝑣₁in𝐺₁and𝑢₂is adjacent to𝑣₂in𝐺₂[13].

Lemma 11. For𝑚,𝑛≥1, we have 𝜏𝐾₂⊗𝐾_𝑚,𝑛=0. (30)

Theorem 12. For𝑚,𝑛≥1, we have 𝜏𝐾₃⊗𝐾_𝑚,𝑛=3×2^{3𝑚+3𝑛−5}×𝑛^3𝑚−1×𝑚^3𝑛−1. (31)

Proof. Applying Lemma 1, we have𝜏𝐾₃⊗𝐾_𝑚,𝑛=19(𝑚+𝑛)²det3(𝑚+𝑛)𝐼−𝐷+𝐴=19(𝑚+𝑛)²=1×det2𝑛+11⋯11⋯⋯11⋯⋯11⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱⋮1⋯12𝑛+11⋯⋯11⋯⋯11⋯⋯12𝑚+11⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯11⋯12𝑚+10⋯⋯01⋯⋯10⋯⋯02𝑛+11⋯1⋮⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱11⋯⋯10⋯⋯01⋯12𝑛+10⋯⋯01⋯⋯11⋯⋯1⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯⋯11⋯⋯11⋯⋯10⋯⋯01⋯⋯1⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯10⋯⋯01⋯⋯10⋯⋯01⋯⋯10⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯⋯10⋯⋯00⋯⋯01⋯⋯10⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯⋯10⋯⋯01⋯⋯10⋯⋯01⋯⋯1⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯10⋯⋯01⋯⋯11⋯⋯11⋯⋯10⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯11⋯⋯10⋯⋯02𝑚+11⋯10⋯⋯01⋯⋯11⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱⋮1⋯12𝑚+10⋯⋯01⋯⋯10⋯⋯02𝑛+11⋯11⋯⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮0⋯⋯01⋯12𝑛+11⋯⋯11⋯⋯11⋯⋯12𝑚+11⋯1⋮⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱11⋯⋯11⋯⋯11⋯12𝑚+19(𝑚+𝑛)²=1det𝐴𝐵𝐵𝐵𝐴𝐵𝐵𝐵𝐴9(𝑚+𝑛)²[det(𝐴−𝐵)]²[(=1det𝐴+2𝐵)]9(𝑚+𝑛)²det2𝑛0⋯01⋯⋯10⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱0⋮⋱⋱⋮0⋯02𝑛1⋯⋯11⋯⋯12𝑚0⋯0⋮⋱⋱⋮0⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱01⋯⋯10⋯02𝑚²=1×det2𝑛+33⋯31⋯⋯13⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱3⋮⋱⋱⋮3⋯32𝑛+31⋯⋯11⋯⋯12𝑚+33⋯3⋮⋱⋱⋮3⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱31⋯⋯13⋯32𝑚+39(𝑚+𝑛)²𝐵det𝐴𝐵_𝑇𝐶𝐸×det𝐷𝐸_𝑇𝐹=19(𝑚+𝑛)²×(det𝐴)²det𝐶−𝐵^𝑇𝐴⁻¹𝐵²×det𝐷det𝐹−𝐸^𝑇𝐷⁻¹𝐸. (32)Thus,𝜏𝐾₃⊗𝐾_𝑚,𝑛=19(𝑚+𝑛)²det2𝑛0⋯00⋱⋱⋮⋮⋱⋱00⋯02𝑛_𝑚×𝑚²×𝑚()det4𝑛−12𝑛−𝑚⋯2𝑛−𝑚2𝑛−𝑚2𝑛⋱⋱⋮⋮⋱⋱−𝑚2𝑛−𝑚⋯2𝑛−𝑚𝑚()2𝑛4𝑛−12𝑛_𝑛×𝑛²×det2𝑛+33⋯33⋱⋱⋮⋮⋱⋱33⋯32𝑛+3_𝑚×𝑚𝑛()×det4𝑚+6+6𝑚²+8𝑚2𝑛+3𝑚6𝑛+8𝑚⋯2𝑛+3𝑚6𝑛+8𝑚2𝑛+3𝑚6𝑛+8𝑚2𝑛+3𝑚⋱⋱⋮⋮⋱⋱6𝑛+8𝑚2𝑛+3𝑚6𝑛+8𝑚⋯2𝑛+3𝑚6𝑛+8𝑚𝑛()2𝑛+3𝑚4𝑚+6+6𝑚²+8𝑚2𝑛+3𝑚_𝑛×𝑛=19(𝑚+𝑛)²(2𝑛)^2𝑚×−𝑚2𝑛^2𝑛×𝑚()det4𝑛−1𝑚()−𝑚1⋯11⋱⋱⋮⋮⋱⋱11⋯14𝑛−1−𝑚_𝑛×𝑛²×3^𝑚×det2𝑛+331⋯11⋱⋱⋮⋮⋱⋱11⋯12𝑛+33×6𝑛+8𝑚2𝑛+3𝑚^𝑛𝑛()×det4𝑚+6+6𝑚²+8𝑚𝑛()6𝑛+8𝑚1⋯11⋱⋱⋮⋮⋱⋱11⋯14𝑚+6+6𝑚²+8𝑚6𝑛+8𝑚_𝑛×𝑛. (33)
Using Lemma 2, we have 𝜏𝐾₃⊗𝐾_𝑚,𝑛=19(𝑚+𝑛)²(2𝑛)^2𝑚×𝑚2𝑛^2𝑛×(−3𝑛)²×(−4𝑛)^2𝑛−2×3^𝑚2𝑛+3𝑚3×2𝑛3^𝑚−1×6𝑛+8𝑚2𝑛+3𝑚^𝑛×4𝑛𝑚+6𝑛+6𝑚²+8𝑚×6𝑛+8𝑚+𝑛−14𝑛𝑚+6𝑛+6𝑚²+8𝑚6𝑛+8𝑚−1^𝑛−1=3×(2𝑛)^{2𝑚−2𝑛}×𝑚^3𝑛−1×𝑛^{2𝑛+𝑚−1}×2^{5𝑛+𝑚−5}=3×2^{3𝑚+3𝑛−5}×𝑛^3𝑚−1×𝑚^3𝑛−1. (34) In particular, 𝜏𝐾₃⊗𝐾_𝑛,𝑛=3×2^6𝑛−5×𝑛^6𝑛−2;𝑛≥1. (35)

6. Number of Spanning Trees of Symmetric Product of Graphs

The symmetric product,𝐺₁⊕𝐺₂, is the simple graph with𝑉(𝐺₁∘𝐺₂)=𝑉₁×𝑉₂,where(𝑢₁,𝑢₂)and(𝑣₁,𝑣₂)are adjacent in𝐺₁⊕𝐺₂if and only if either𝑢₁is adjacent to𝑣₁in𝐺₁and𝑢₂is not adjacent to𝑣₂in𝐺₂, or𝑢₁is not adjacent to𝑣₁in𝐺₁and𝑢₂is adjacent to𝑣₂in𝐺₂[13].

Theorem 13. For𝑛,𝑚≥1, we have 𝜏𝐾₂⊕𝐾_𝑚,𝑛=(𝑚+𝑛)^{2(𝑚+𝑛−1)}. (36)

Proof. Applying Lemma 1, we have 𝜏𝐾₂⊕𝐾_𝑚,𝑛=14(𝑚+𝑛)²det2(𝑚+𝑛)𝐼−𝐷+𝐴=14(𝑚+𝑛)²=1×det𝑚+𝑛+11⋯10⋯⋯01⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯1𝑚+𝑛+10⋯⋯00⋯⋯0𝑚+𝑛+11⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯01⋯1𝑚+𝑛+10⋯⋯01⋯⋯1⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯⋯11⋯⋯10⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯10⋯⋯00⋯⋯01⋯⋯1⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯⋯11⋯⋯10⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯10⋯⋯0𝑚+𝑛+11⋯10⋯⋯01⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯1𝑚+𝑛+10⋯⋯00⋱⋱0𝑚+𝑛+11⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯01⋯1𝑚+𝑛+14(𝑚+𝑛)²=1det𝑚+𝑛+11⋯11⋯⋯11⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯1𝑚+𝑛+11⋯⋯11⋯⋯1𝑚+𝑛+11⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱11⋯⋯11⋯1𝑚+𝑛+1×det𝑚+𝑛+11⋯1−1⋯⋯−11⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯1𝑚+𝑛+1−1⋯⋯−1−1⋯⋯−1𝑚+𝑛+11⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1−1⋯⋯−11⋯1𝑚+𝑛+14(𝑚+𝑛)²(𝑚+𝑛+1+𝑚+𝑛−1)(𝑚+𝑛+1−1)^{𝑚+𝑛−1}𝐵×det𝐴𝐵_𝑇𝐶=12(𝑚+𝑛)^{𝑚+𝑛−2}×det𝐴det𝐶−𝐵^𝑇𝐴⁻¹𝐵=12(𝑚+𝑛)^{𝑚+𝑛−2}×(2𝑚+𝑛)(𝑚+𝑛)^𝑚−1𝑛×det²+(3𝑚+1)𝑛+2𝑚²+𝑚(𝑛+2𝑚)𝑛+𝑚⋯𝑛+2𝑚𝑛+𝑚𝑛+2𝑚𝑛+𝑚𝑛+2𝑚⋱⋱⋮⋮⋱⋱𝑛+𝑚𝑛+2𝑚𝑛+𝑚⋯𝑛+2𝑚𝑛+𝑚𝑛𝑛+2𝑚²+(3𝑚+1)𝑛+2𝑚²+𝑚(𝑛+2𝑚)_𝑛×𝑛=12(𝑚+𝑛)^{2𝑚+𝑛−3}×(2𝑚+𝑛)×𝑛+𝑚𝑛+2𝑚^𝑛𝑛×det²+(3𝑚+1)𝑛+2𝑚²+𝑚𝑛𝑛+𝑚1⋯11⋱⋱⋮⋮⋱⋱11⋯1²+(3𝑚+1)𝑛+2𝑚²+𝑚𝑛+𝑚_𝑛×𝑛. (37)Thus, 𝜏𝐾₂⊕𝐾_𝑚,𝑛=12(𝑚+𝑛)^{2𝑚+𝑛−3}×()×2𝑚+𝑛𝑛+𝑚𝑛+2𝑚^𝑛×(2𝑛+2𝑚)(𝑛+2𝑚)^𝑛−1=(𝑚+𝑛)^{2(𝑚+𝑛−1)}. (38)In particular, 𝜏𝐾₂⊕𝐾_𝑛,𝑛=(2𝑛)^2(2𝑛−1);𝑛≥1. (39)

Theorem 14. For𝑚,𝑛≥1, we have 𝜏𝐾₃⊕𝐾_𝑚,𝑛=3(2𝑚+𝑛)^3𝑚−3×(2𝑛+𝑚)^3𝑛−3𝑚²+𝑛²+3𝑚𝑛². (40)

Proof. Applying Lemma 1, we have𝜏𝐾₃⊕𝐾_𝑚,𝑛=19(𝑚+𝑛)²det3(𝑚+𝑛)𝐼−𝐷+𝐴=19(𝑚+𝑛)²×det^{2𝑚+𝑛+11⋯10⋯⋯00⋯⋯01⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱⋮1⋯12𝑚+𝑛+10⋯⋯00⋯⋯00⋯⋯02𝑛+𝑚+11⋯11⋯⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯12𝑛+𝑚+11⋯⋯10⋯⋯01⋯⋯12𝑚+𝑛+11⋯1⋮⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯01⋯⋯11⋯12𝑚+𝑛+1}_{1⋯⋯10⋯⋯00⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯10⋯⋯00⋯⋯00⋯⋯01⋯⋯10⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯01⋯⋯10⋯⋯01⋯⋯10⋯⋯01⋯⋯1⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯00⋯⋯01⋯⋯11⋯⋯10⋯⋯01⋯⋯1⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮1⋯⋯10⋯⋯01⋯⋯10⋯⋯01⋯⋯10⋯⋯0⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯01⋯⋯10⋯⋯00⋯⋯00⋯⋯01⋯⋯1⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮0⋯⋯00⋯⋯01⋯⋯12𝑛+𝑚+11⋯11⋯⋯10⋯⋯01⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮⋮⋱⋱⋮1⋯12𝑛+𝑚+11⋯⋯10⋯⋯01⋯⋯12𝑚+𝑛+11⋯10⋯⋯0⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯⋯11⋯12𝑚+𝑛+10⋯⋯00⋯⋯00⋯⋯02𝑛+𝑚+11⋯1⋮⋱⋱⋮⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱10⋯⋯00⋯⋯01⋯12𝑛+𝑚+1}=19(𝑚+𝑛)²=1det𝐴𝐵𝐵𝐵𝐴𝐵𝐵𝐵𝐴9(𝑚+𝑛)²[det(𝐴−𝐵)]²[(=1det𝐴+2𝐵)]9(𝑚+𝑛)²det2𝑚+𝑛+11⋯1−1⋯⋯−11⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯12𝑚+𝑛+1−1⋯⋯−1−1⋯⋯−12𝑛+𝑚+11⋯1⋮⋱⋱⋮1⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1−1⋯⋯−11⋯12𝑛+𝑚+1²=1×det2𝑚+𝑛+11⋯12⋯⋯21⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱1⋮⋱⋱⋮1⋯12𝑚+𝑛+12⋯⋯22⋯⋯22𝑛+𝑚+11⋯1⋮⋱⋱⋮3⋱⋱⋮⋮⋱⋱⋮⋮⋱⋱12⋯⋯21⋯12𝑛+𝑚+19(𝑚+𝑛)²𝐵det𝐴𝐵_𝑇𝐶²𝐸×det𝐷𝐸_𝑇𝐹=19(𝑚+𝑛)²×(det𝐴)²det𝐶−𝐵^𝑇𝐴⁻¹𝐵²×det𝐷det𝐹−𝐸^𝑇𝐷⁻¹𝐸=(3𝑚+𝑛)²(2𝑚+𝑛)^2𝑚−29(𝑚+𝑛)²×det^{2𝑛2()+𝑛7𝑚+1+3𝑚2+2𝑚𝑛+3𝑚𝑛+2𝑚⋯𝑛+3𝑚𝑛+2𝑚𝑛+3𝑚𝑛+2𝑚𝑛+3𝑚⋱⋱⋮}_{⋮⋱⋱𝑛+2𝑚𝑛+3𝑚𝑛+2𝑚⋯𝑛+3𝑚𝑛+2𝑚𝑛+3𝑚2𝑛2()+𝑛7𝑚+1+3𝑚2+2𝑚𝑛+3𝑚𝑛×𝑛}²×(3𝑚+𝑛)(2𝑚+𝑛)^𝑚−1×det2𝑛²+𝑛(7𝑚+1)+3𝑚²−𝑚𝑛+3𝑚𝑛−𝑚⋯𝑛+3𝑚𝑛−𝑚𝑛+3𝑚𝑛−𝑚𝑛+3𝑚⋱⋱⋮⋮⋱⋱𝑛−𝑚𝑛+3𝑚𝑛−𝑚⋯𝑛+3𝑚𝑛−𝑚𝑛+3𝑚2𝑛²+𝑛(7𝑚+1)+3𝑚²−𝑚𝑛+3𝑚_𝑛×𝑛=(3𝑚+𝑛)²(2𝑚+𝑛)^2𝑚−29(𝑚+𝑛)²×𝑛+2𝑚𝑛+3𝑚^2𝑛det^{2𝑛2()+𝑛7𝑚+1+3𝑚2+2𝑚𝑛+2𝑚1⋯11⋱⋱⋮}_{⋮⋱⋱11⋯12𝑛2()+𝑛7𝑚+1+3𝑚2+2𝑚𝑛+2𝑚𝑛×𝑛}²×(3𝑚+𝑛)(2𝑚+𝑛)^𝑚−1×𝑛−𝑚𝑛+3𝑚^𝑛det2𝑛²+𝑛(7𝑚+1)+3𝑚²−𝑚𝑛−𝑚1⋯11⋱⋱⋮⋮⋱⋱11⋯12𝑛²+𝑛(7𝑚+1)+3𝑚²−𝑚𝑛−𝑚_𝑛×𝑛. (41)
Using Lemma 2, we have 𝜏𝐾₃⊕𝐾_𝑚,𝑛=19(𝑚+𝑛)²(3𝑚+𝑛)²(2𝑚+𝑛)^2𝑚−2×𝑛+2𝑚𝑛+3𝑚^2𝑛1(𝑛+3𝑚)^2𝑛×3𝑛²+3𝑚²+9𝑛𝑚²×2𝑛²+3𝑚²+7𝑛𝑚^2𝑛−2×(3𝑚+𝑛)(2𝑚+𝑛)^𝑚−1×𝑛−𝑚𝑛+3𝑚^𝑛×1(𝑛−𝑚)^𝑛×3𝑛²+3𝑚²×+6𝑛𝑚2𝑛²+3𝑚²+7𝑛𝑚^𝑛−1=3(2𝑚+𝑛)^3𝑚−3(2𝑛+𝑚)^3𝑛−3×𝑚²+𝑛²+3𝑚𝑛². (42) In particular, 𝜏𝐾₃⊕𝐾_𝑛,𝑛=25×3^6𝑛−5×𝑛^6𝑛−2;𝑛≥1. (43)

7. Conclusion

The number of spanning trees𝜏(𝐺)in graphs (networks) is an important invariant. The evaluation of this number is not only interesting from a mathematical (computational) perspective but is also 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.

Acknowledgments

The author is deeply indebted and thankful to the deanship of the scientific research for its help and to the distinct team of employees at Taibah University, Al Madinah Saudi Arabia. This research work was supported by Grant No. 3070/1434.