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

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CHAPTER 6.110COUNTING THE NUMBER OF SPANNING TREES IN THE STARFLOWER PLANAR MAPFigure 6.4: The star flower planar maps S 2,k and S 3,k6.4 An explicit formula for the number of spanningtrees in S n,kNow, we derive the explicit formula for τ(S n,k ) the number of spanning trees in the starflower planar map S n,k .Theorem 6.4.1 The number of spanning trees of the star flower planar map S n,k (seeFigure. 6.3) is given by the following formula:τ(S n,k ) = 2kn(k + 2) n−1 , n ≥ 2.Proof: We cut a triangle as shown in Figure. 6.3, then we obtain the star flower planarmap S n−1,k after cutting (see Figure. 6.5):Figure 6.5: The star flower planar map S n,k after cutting
6.4. An explicit formula for the number of spanning trees in S n,k 111and we apply Theorem 4.3.26, then we obtain:τ(S n,k ) = (k + 2) n−1 ∗ 2k + τ(S n−1,k ) ∗ (k + 2)τ(S n,k ) = 2k(k + 2) n−1 + (k + 2)τ(S n−1,k )(k + 2)τ(S n−1,k ) = 2k(k + 2) n−1 + (k + 2) 2 τ(S n−2,k )(k + 2) 2 τ(S n−2,k ) = 2k(k + 2) n−1 + (k + 2) 3 τ(S n−3,k )(k + 2) 3 τ(S n−3,k ) = 2k(k + 2) n−1 + (k + 2) 4 τ(S n−4,k ). . . . . . . . .(k + 2) n−4 τ(S 4,k ) = 2k(k + 2) n−1 + (k + 2) n−3 τ(S 3,k )(k + 2) n−3 τ(S 3,k ) = 2k(k + 2) n−1 + (k + 2) n−2 τ(S 2,k )(k + 2) n−2 τ(S 2,k ) = 2k(k + 2) n−1 + (k + 2) n−1 2kBy the sum of all the previous equations, we obtain:τ(S n,k ) = 2k(k + 2) n−1 (n − 1) + (k + 2) n −1 2kWe take 2k(k+2) n−1 as a factor, hence the result.□Particular cases:1) In the previous Theorem 6.4.1, if we take k = 1 (see the star flower planarmap S n,k shown in figure 6.3), then we obtain the star flower planar map S n,1 , which has2n vertices (n vertices of degree 4 and n of degree 2), 3n edges and n + 2 faces (n facesof degree 3, one face of degree n and the other face of degree 2n)(see Figure. 6.6):Figure 6.6: The star flower planar map S n,1Example 6.4.2 Here is an example of the star flower planar map S 2,1 with n = 2 (2triangles) and the star flower planar map S 3,1 with n = 3 (3 triangles) (see Figure. 6.7).
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1To my daughter Hanin.
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4my wife who has been very supporti
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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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Page 71 and 72: 4.3. Main Results 69one vertex (any
Page 73 and 74: 4.3. Main Results 71Lemma 4.3.8 Let
Page 75 and 76: 4.3. Main Results 73This recursion
Page 77 and 78: 4.3. Main Results 75Proof: It is th
Page 79 and 80: 4.3. Main Results 77Figure 4.12: An
Page 81 and 82: 4.3. Main Results 79Theorem 4.3.31
Page 83 and 84: 4.3. Main Results 81Property 4.3.33
Page 85: Part IIIUse of Derived Theoretical
Page 88 and 89: CHAPTER 5.86THE NUMBER OF SPANNING
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Page 138 and 139: 136 BIBLIOGRAPHY[13] N. Biggs, E. L
Page 140 and 141: 138 BIBLIOGRAPHY[46] P. Erdös, and
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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