enumeration of the number of spanning trees in some ... - Toubkal

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130 CHAPTER 7. MAXIMAL PLANAR MAPSProof: We put f n = τ(F n ) and g n = τ(G n ). f 2 = 2, g 2 = 1, in the map F n , we cut thefirst cycle (see Figure. 7.11), and we use Theorem 4.3.31 (the same goes for the mapG n ), then we obtain: τ(G n ) = 3τ(F n−1 ) − τ(G n−1 ), τ(F n ) = 2τ(G n ) − τ(F n−1 ) therefore,we have the following system:{fn = 2g n − f n−1g n = 3f n−1 − g n−1 with f 2 = 2 and g 2 = 1,we replace by the value of g n in the first equation, we get:{fn = 5f n−1 − 2g n−1g n = 3f n−1 − g n−1 with f 2 = 2 and g 2 = 1(fn)= Mg n)= Mg n(fn(fn−1), where M =g n−1)g n−1(fn−1( 5 -23 -1= ... = M n−2 (f2g 2),then, we compute M n−2 :det (M − λI 2 ) = λ 2 −4λ + 1 = 0, λ 1 = 2 − √ 3 and λ 2 = 2 + √ 3, λ 1 ≠ λ 2 then there isP invertible such that M = P DP −1 where( )λ1 0D =,0 λ 2P is the transformation matrix formed by eigenvectors( 1 1P =M n−2 = P D n−2 P −1 whereD n−2 =3+ √ 32(P −1 = √ −133− √ 33− √ 322-1−3− √ 312),)( √ )(2 − 3)n−200 (2 + √ 3) n−2 ,from which we obtain M n−2 , then(fng n)= M n−2 (f2g 2),,),hence the result.□
7.3. Formulae for the Number of Spanning Trees in a Maximal Planar Map 131Numerical results The Table 7.2 illustrates some of the values of the number of spanningtrees in the sequence of planar maps, the Fan planar maps τ(F n ) and τ(G n ), byusing the formula given in Theorem 7.3.5:n 2 3 4 5 6 7 8 9 10τ(F n ) 2 8 30 112 418 1560 5822 21728 81090τ(G n ) 1 5 19 71 265 989 3691 13775 51409Table 7.2: Some values of τ(F n ) and τ(G n )Let E n be a maximal planar map shown in Figure. 7.7. If we delete the edge e = v 1 v 2 ofE n , we obtain the crystal planar map C n with n vertices (see Figure. 7.12).Figure 7.12: The crystal planar map C nTheorem 7.3.6 (A crystal planar map) The complexity of the crystal planar map C n(see Figure. 7.12) is given by the following formula:τ(C n ) = 12 √ 3 (n − 2) ((2 + √ 3) n−2 − (2 − √ 3) n−2 ), n ≥ 3.Proof: By Theorem 4.3.11, we have: τ(E n ) = τ(E n −e)+τ(E n .e), since τ(E n −e) = τ(C n )and τ(E n .e) = τ(F n ) (see Figure. 7.13):then τ(C n ) = τ(E n ) − τ(F n ). We replace by the value of τ(E n ) from Theorem 7.3.3 andthe value of τ(F n−1 ) from Theorem 7.3.5, thus our result follows.□
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1To my daughter Hanin.
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6 CONTENTSII Theoretical Part 533 H
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8 CONTENTS
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12 LIST OF FIGURES7.11 The maps F n
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14 LIST OF TABLES
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LIST OF TABLES 17the n-Barrel chain
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Part IPreliminaries19
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22 CHAPTER 1. DEFINITIONS AND PROPE
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42 CHAPTER 2. SPANNING TREESFigure
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44 CHAPTER 2. SPANNING TREESThe num
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46 CHAPTER 2. SPANNING TREESExample
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48 CHAPTER 2. SPANNING TREESFor exa
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50 CHAPTER 2. SPANNING TREESIn the
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52 CHAPTER 2. SPANNING TREES
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Chapter 3How to count the number of
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3.3. Counting the number of spannin
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Chapter 4New methods to compute the
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4.3. Main Results 69one vertex (any
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4.3. Main Results 71Lemma 4.3.8 Let
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4.3. Main Results 73This recursion
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4.3. Main Results 75Proof: It is th
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4.3. Main Results 77Figure 4.12: An
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Page 138 and 139: 136 BIBLIOGRAPHY[13] N. Biggs, E. L
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Page 142 and 143: 140 BIBLIOGRAPHY[77] D. Lotfi, M. E
Page 144 and 145: 142 BIBLIOGRAPHY[107] R. Shrock and
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