Formulas for the Number of Spanning Trees in a

Applied Mathematical Sciences, Vol. 5, 2011, no. 64, 3147 - 3159

Formulas for the Number of Spanning Trees in a Maximal Planar Map A. Modabish, D. Lotfi and M. El Marraki Department of Computer Sciences, Faculty of Sciences University of Mohamed V, P.O. Box 1014, Rabat, Morocco hafi[email protected], doun.lotfi@ gmail.com, [email protected] Abstract The number of spanning trees of a map C denoted by τ (C) is the total number of distinct spanning subgraphs of C that are trees. A maximal planar map is a simple graph G formed by n vertices, 3(n − 2) edges and all faces having degree 3 [2]. In this paper, we derive the explicit formula for the number of spanning trees of the maximal planar map and deduce a formula for the number of spanning trees of the crystal planar map.

Mathematics Subject Classification: 05C85, 05C30 Keywords: graphs; maps; maximal planar map; crystal planar map; fan planar map; complexity; spanning trees

1

Introduction

First of all, let’s recall some necessary definitions related to our work, an undirected graph G is a triplet (VG ,EG ,δ) where VG is the set of vertices of the graph G, EG is the set of edges of the graph G and δ is the application δ : EG → P(v), ei → δ(ei ) = {vj , vk } with vj and vk are end vertices of the edge ei . A loop is an edge ei ∈ EG with vj = vk , if δ(ei ) = δ(ej ) with i = j then the edges ei and ej are called multiple. A graph which contains neither multiple edges nor loops is called a simple graph. The degree of a vertex v noted deg(v) is the number of edges incident to it. The sum of the degrees of all vertices of a graph is equal to twice the number of its edges i.e. v∈VG deg(v) = 2|EG |. A graph G is called connected if any two of its vertices may be connected by a path [12]. The graphs that we consider are in most cases connected but may contain multiple edges. A map C is a graph G drawn on a surface X

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or embedded into it (that is, a compact variety 2-dimensional orientable) in such a way that: the vertices of graph are represented as distinct points of the surface, the edges are represented as curves on the surface that intersect only at the vertices, if we cut the surface along the graph thus drawn, what remains (that is, the set X\G) is a disjoint union of connected components, called faces, each homeomorphic to an open disk [6]. A planar map is a map drawn on the plane. Through this paper, all maps are planar and connected. Euler’s formula for maps is: |VC | + |FC | − |EC | = 2 − 2g, where FC is the set of faces of the map C and g is the genus of a map C, in the planar case g = 0. [6]. A tree is a connected graph without cycle. A plan tree is a tree designed in the plane. The distance between two distinct vertices vi and vj of a tree, denoted by d(vi , vj ) is equal to the length of the shortest path (number of edges in) that connects vi and vj , by Conventionally, d(vi , vi ) = 0. We define a complete vertex in a graph G by the vertex v0 such that d(u, v0 ) = 1 for each u ∈ VG \ {v0 }. In a complete graph all the vertices are completes. The Wiener index of a connected graph is the sum of distances between all pairs of vertices [3], [4], the Wiener index of a connected graph G is defined as: W (G) = d(vi , vj ). {vi ,vj }⊆V (G)

The particular case of trees has been echoed by several people. For a tree with n vertices, the Wiener index W (Tn ) is minimized by the star tree with n vertices and maximized by the path with n vertices [12]. In [3] we worked on the Wiener index in the case of planar maps and gave an inequality, which minimizes and maximizes any map by the maximal planar map with n vertices and the path of n vertices and also we gave a formula which calculates the Wiener index of maximal planar map En , our results given in the following theorem: Theorem 1. Let Cn be a map with n vertices, then (n − 2)2 + 2 = W (En ) ≤ W (Cn ) ≤ W (Pn ) =

n(n2 − 1) . 6

The complexity of a map C denoted by τ (C) is the number of spanning trees of the map C. A spanning tree is a subgraph which is a tree that visits all vertices (see Fig. 1). The number of spanning trees in a graph (network) G is an important invariant of the graph (network). It is also an important measure of reliability of a graph (network). A classic result of Kirchhoff [5], [9] can be used to determine the number of spanning trees for C = (VC , EC ). Let VC = {v1 , v2 , ..., vn }, to state the result, we

Formulas for the number of spanning trees

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Figure 1: A map C gives rise to 16 spanning trees

define the n × n characteristic matrix L = [aij ] as follows: (i) aij = −p(vi , vj ) if i = j, where p(vi , vj ) is the number of edges that connects vi with vj , (ii) aij equals the degree of vertex vi if i = j, and (iii) aij = 0 otherwise. The Kirchhoff matrix tree theorem states that all cofactors of L are equal, and their common value is τ (C). The matrix tree theorem can be applied to any map C to determine τ (C), but this requires evaluating a determinant of a corresponding characteristic matrix. However, for a few special families of graphs there exist simple formulas which 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 such results is due to Cayley who showed that complete graph on n vertices, Kn , has nn−2 spanning trees [1], that is he showed τ (Kn ) = nn−2 for n ≥ 2. Theorem 2. [12] If L∗ (C) is a matrix obtained by deleting row i and column j of the Laplacian matrix L(C), then τ (C) = (−1)i+j det L∗ (C). Another result is due to Sedlacek [10] who derived a formula for the wheel on n + 1 vertices, Wn+1 , which is formed from a cycle Cn on n vertices by adding a vertex adjacent to every vertex of Cn (see Fig. 2). In particular, he showed that √ √ 3+ 5 n 3− 5 n τ (Wn+1 ) = ( ) +( ) − 2 for n ≥ 3. 2 2 In [8], we gave a formula to calculate the complexity of the n-Grid chains Gn where n is the number of squares (|VGn | = 2n + 2) (see Fig. 2) is as follows: √ n+1 √ n+1 1 τ (Gn ) = √ (2 + 3) − (2 − 3) , n ≥ 1, 2 3

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also another formula to calculate the complexity of the n-Fan chains Fn where n is the number of triangles (|VFn | = n + 2) (see Fig. 2) is as follows: √ √ 3 + 5 n+1 1 3 − 5 n+1 ) ) −( τ (Fn ) = √ ( , n ≥ 1. 2 2 5

Figure 2: The Wheel, the n-Grid chains and the n-Fan chains

In addition, other formulas are related to our work can also be found in [7], [10], [11]. In this paper, we derive the explicit formula for the number of spanning trees of a maximal planar map and deduce a formula for the number of spanning trees of the crystal planar map.

2

Main Results

Before presenting the main results, we need the following lemmas: Lemma 1. Let C be a map and τ (C) denote the number of spanning trees of C. If e = vi vi+1 ∈ E(C) (e is not a loop), then τ (C) = τ (C − e) + τ (C.e). Proof: See [8], [12]. Let C be a map of type C= C1 | C2 , C is a map such that v1 and v2 two vertices of C connected by an edge e [2](see Fig. 3).

Figure 3: A map C= C1 | C2


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Property 1. Let C be a map of type C = C1 | C2 - C1 and C2 have two common vertices v1 , v2 , a common edge e and a common face (the external face). - VC =VC1 +VC2 -2, EC =EC1 +EC2 -1 and FC =FC1 +FC2 -1. Example 1. Here is an example of maps C, C1 , C2 , C1 − e, C2 − e, C1 .e and C2 .e (see Fig. 4).

Figure 4: An example of maps C, C1 , C2 , C1 − e, C2 − e, C1 .e and C2 .e Lemma 2. Let C be a map of type C= C1 | C2 (see Fig. 3), then τ (C) = τ (C1 ) × τ (C2 ) − τ (C1 − e) × τ (C2 − e). Proof: See [8]. Definition 1. (A maximal planar map [2]) Let En be a family of maps that contains: • n vertices, two complete vertices of degree n − 1, two vertices of degree 3 and n − 4 vertices of degree 4, • 2(n − 2) faces of degree 3 (all faces having degree 3), • 3(n − 2) edges, then the family of this maps is called a maximal planar map. The maps E3 and E4 are presented in the example 3. For E5 , E6 , and En (see Fig. 5) Example 2. In the maps E3 and E4 , all the vertices are complete. In other words, we say that the map is complete (see Fig 6).

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Figure 5: The maps E5 , E6 and En

Figure 6: The maps E3 and E4 Lemma 3. Let En be a maximal planar map with n vertices, then n/τ (En ),

n ≥ 3.

Proof: We first form the Kirchhoff characteristic matrix Ln (n × n) corresponding to the labeling of En shown as follows (in Figure 7):

Figure 7: The principal matrix of En Next, we focus our attention on the principal submatrix Ln−1 which obtained by canceling its last row and column corresponding to vertex vn . So, the number of spanning trees of a maximal planar map En equals τ (En ) = det(Ln−1 ).


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n − 1 −1 −1 −1 −1 −1 . . . −1 −1 −1 n − 1 −1 −1 −1 −1 . . . −1 −1 −1 −1 3 −1 0 0 ... 0 0 −1 −1 −1 4 −1 0 . . . 0 0 −1 0 −1 4 −1 . . . 0 0 τ (En ) = −1 −1 −1 0 0 −1 4 . . . 0 0 .. . .. . . .. .. .. .. ... . . .. . . . . . −1 −1 0 0 0 0 . . . 4 −1 −1 −1 0 0 0 0 . . . −1 4 we denote ri by the i-th row and ci by the i-th column of the determinant. In the previous determinant, we replace c1 by c1 + c2 + ... + cn−1 , i.e., we add to the first column the sum of other (transformation is symbolized as follows: c1 n−1 ← i=1 ci , ; this does not change the determinant, then we obtain: 1 −1 −1 −1 −1 −1 . . . −1 −1 1 n − 1 −1 −1 −1 −1 . . . −1 −1 0 −1 3 −1 0 0 ... 0 0 0 −1 −1 4 −1 0 . . . 0 0 0 −1 4 −1 . . . 0 0 τ (En ) = 0 −1 0 −1 0 0 −1 4 . . . 0 0 .. .. .. .. .. .. . . . ... . . .. . . . . . 0 −1 0 0 0 0 . . . 4 −1 1 −1 0 0 0 0 . . . −1 4 Next, we replace cj by c1 + cj for j = 2, ..., n − 1, i.e., cj ← c1 + cj , we obtain: 1 1 0 0 τ (En ) = 0 0 .. . 0 1

0 0 0 0 0 . . . 0 0 n 0 0 0 0 . . . 0 0 −1 3 −1 0 0 . . . 0 0 −1 −1 4 −1 0 . . . 0 0 −1 0 −1 4 −1 . . . 0 0 −1 0 0 −1 4 . . . 0 0 .. .. .. .. .. . . .. .. . . . . . . . . −1 0 0 0 0 . . . 4 −1 0 1 1 1 1 ... 0 5

Expanding Ln−1 along the first row we obtain the determinant of order (n − 2) × (n − 2) and expanding the determinant obtained along the first row we obtain the determinant of order (n − 3) × (n − 3) as follows:

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3 −1 0 0 . . . 0 0 −1 4 −1 0 . . . 0 0 0 −1 4 −1 . . . 0 0 0 −1 4 . . . 0 0 τ (En ) = n 0 .. .. . . .. .. .. .. . . . . . . . 0 0 0 0 . . . 4 −1 1 1 1 1 ... 0 5 hence the result. Lemma 4. Let En be a maximal planar map with n vertices, then 4τ (En−1 ) τ (En−2 ) τ (En ) = − , n n−1 n−2

n ≥ 5.

Proof: By lemma 3, we have the determinant of order (n − 3) × (n − 3): 3 −1 0 0 . . . 0 0 −1 4 −1 0 . . . 0 0 0 −1 4 −1 . . . 0 0 τ (En ) 0 0 −1 4 . . . 0 0 = .. n .. . . .. .. .. .. . . . . . . . 0 0 0 0 . . . 4 −1 1 1 1 1 ... 0 5 In the previous determinant, we denote by ci by the i-th column and ri by the i-th row of the determinant, ci ← ci − ci+1 for i = 1, ..., n − 4, we obtain the determinant as follows: 4 −1 0 0 . . . 0 0 −5 5 −1 0 . . . 0 0 1 −5 5 −1 . . . 0 0 τ (En ) 0 1 −5 5 . . . 0 0 = .. n .. . .. . . .. .. . . .. . . . . 0 0 0 0 . . . 5 −1 0 0 0 0 . . . −5 5 In the end, ri ← follows:

i

j=1 rj

for i = 2, ..., n − 3, we obtain the determinant as

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4 −1 0 0 ... 0 0 −1 4 −1 0 . . . 0 0 0 −1 4 −1 . . . 0 0 τ (En ) 0 0 −1 4 . . . 0 0 = .. n . .. .. .. .. . . . . .. . . . . 0 0 0 0 . . . 4 −1 0 0 0 0 . . . −1 4 then τ (Enn ) = 4Δn−1 − Δn−2 , where Δi = det(Di ) (Δi = 4Δi−1 − Δi−2 because Δi is tri-diagonal matrix), and Di is defined as following: ⎤ ⎡ 4 −1 ⎥ ⎢−1 4 −1 ⎥ ⎢ ⎥ ⎢ −1 4 −1 ⎥ ⎢ ⎥ ⎢ −1 4 −1 Di = ⎢ ⎥ ⎥ ⎢ . . . .. .. .. ⎥ ⎢ ⎥ ⎢ ⎣ −1 4 −1⎦ −1 4 i ×i With the same technique, we obtain: hence the result.

3

τ (En−1 ) n−1

= Δn−1 and

τ (En−2 ) n−2

= Δn−2 ,

Applications

Theorem 3. (A maximal planar map) The complexity of the maximal planar map En (see Fig. 5) is given by the following formula: √ n−2 √ n−2 n − (2 − 3) , n ≥ 3. τ (En ) = √ (2 + 3) 2 3 Proof: τ (E3 ) = 3, τ (E4 ) = 16, τ (E5 ) = 75, τ (En ) n−1 ) n−2 ) = 4τ (E − τ (En−2 (by lemma 4), hence we obtain the system: n n−1 ⎧ ⎨

τ (En ) n

n−1 ) = 4τ (E − n−1 τ (E3 ) = 3 ⎩ τ (E4 ) = 16

τ (En−2 ) , n−2

with

Then we get a sequence such that un = 4un−1 − un−2 , n ≥ 3, therefore the characteristic equation is r 2 − 4r + 1 = 0, so the solutions of this equation are:

3156 r1 = 2 −


√

3 and r2 = 2 +

√

3, hence: √ n √ τ (En ) = α(2 − 3) + β(2 + 3)n ,

α, β ∈ R, n ≥ 1.

Using the initial conditions τ (E33 ) = 1 and τ (E44 ) = 4, we obtain: √ √ α = −2 − 76 3, β = −2 + 76 3, hence the result. Let Fn and Gn be the maps as follows (see Figure 8):

Figure 8: The maps Fn and Gn

Theorem 4. (A Fan planar map) The complexity of the Fan planar maps Fn and Gn (see Fig. 8) is given by the following formulas: √ n−1 √ n−1 1 − (2 − 3) τ (Fn ) = √ (2 + 3) , n ≥ 2, 3 √ n−1 √ n−2 1 √ − (2 − 3) , n ≥ 2. τ (Gn ) = ( 3 − 1) (2 + 3) 2 Proof: We put fn = τ (Fn ) and gn = τ (Gn ). f2 = 2, g2 = 1, in the map Fn , we cut the first cycle (see Fig. 8), and we use Lemma 2 (the same goes for the map Gn ), then we obtain: τ (Gn ) = 3τ (Fn−1 ) − τ (Gn−1 ), τ (Fn ) = 2τ (Gn ) − τ (Fn−1 ) therefore, we have the following system: fn = 2gn − fn−1 gn = 3fn−1 − gn−1 with f2 = 2 and g2 = 1,

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we replace by the value of gn in the first equation, we get: fn = 5fn−1 − 2gn−1 gn = 3fn−1 − gn−1 with f2 = 2 and g2 = 1

fn gn

fn−1 5 -2 =M , where M = , gn−1 3 -1 fn−1 f2 fn n−2 =M = ... = M , gn gn−1 g2

then, we compute M n−2 : √ √ det (M − λI2 ) = λ2 −4λ + 1 = 0, λ1 = 2 − 3 and λ2 = 2 + 3, λ1 = λ2 then there is P invertible such that M = P DP −1 where λ1 0 D= , 0 λ2 P is the transformation matrix formed by eigenvectors

1√

P =

P −1

3+ 3 2

−1 =√ 3

1√

,

3− 3 2

√ 3− 3 2√ −3− 3 2

-1 1

,

M n−2 = P D n−2 P −1 where D

n−2

=

(2 −

√ 0

3)n−2 (2 +

0 √

3)n−2

,

from which we obtain M n−2 , then f2 fn n−2 =M , gn g2 hence the result. Let En be a maximal planar map shown in Fig. 5. If we delete the edge e = v1 v2 of En , we obtain the crystal planar map Cn with n vertices (see Fig. 9).

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Figure 9: The crystal planar map Cn

Theorem 5. (A crystal planar map) The complexity of the crystal planar map Cn (see Fig. 9) is given by the following formula: √ n−2 √ n−2 1 − (2 − 3) τ (Cn ) = √ (n − 2) (2 + 3) , n ≥ 3. 2 3 Proof: By Lemma 1, we have: τ (En ) = τ (En − e) + τ (En .e), since τ (En − e) = τ (Cn ) and τ (En .e) = τ (Fn ) (see Fig. 10):

Figure 10: The maps En , En − e and En .e then τ (Cn ) = τ (En ) − τ (Fn ). We replace by the value of τ (En ) from Theorem 3 and the value of τ (Fn−1 ) from Theorem 4, thus our result follows.

4

Conclusion

In this paper, we have interested by calculating the number of spanning trees of some particular planar maps, for that we have derived the explicit formula to calculate the number of spanning trees of the maximal planar map, then we have deduced the formula for the number of spanning trees of the crystal planar map.


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