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

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CHAPTER 6.112COUNTING THE NUMBER OF SPANNING TREES IN THE STARFLOWER PLANAR MAPFigure 6.7: The star flower planar maps S 2,1 and S 3,1Corollary 6.4.3 The <strong>number</strong> <strong>of</strong> <strong>spann<strong>in</strong>g</strong> <strong>trees</strong> <strong>of</strong> <strong>the</strong> star flower planar map S n,1 (seeFigure. 6.6) is given by <strong>the</strong> follow<strong>in</strong>g formula:τ(S n,1 ) = 2n(3) n−1 , n ≥ 2.Pro<strong>of</strong>: We cut a triangle as shown <strong>in</strong> Figure. 6.6, we <strong>the</strong>n obta<strong>in</strong> <strong>the</strong> star flower planarmap S n−1,1 after cutt<strong>in</strong>g (see Figure. 6.8):Figure 6.8: The star flower planar map S n,1 after cutt<strong>in</strong>gand we apply Theorem 4.3.26, <strong>the</strong>n we obta<strong>in</strong>:τ(S n,1 ) = (3) n−1 ∗ 2 + τ(S n−1,1 ) ∗ (3)τ(S n,1 ) = 2(3) n−1 + (3)τ(S n−1,1 )(3)τ(S n−1,1 ) = 2(3) n−1 + (3) 2 τ(S n−2,1 )(3) 2 τ(S n−2,1 ) = 2(3) n−1 + (3) 3 τ(S n−3,1 )(3) 3 τ(S n−3,1 ) = 2(3) n−1 + (3) 4 τ(S n−4,1 ). . . . . . . . .(3) n−4 τ(S 4,1 ) = 2(3) n−1 + (3) n−3 τ(S 3,1 )(3) n−3 τ(S 3,1 ) = 2(3) n−1 + (3) n−2 τ(S 2,1 )(3) n−2 τ(S 2,1 ) = 2(3) n−1 + (3) n−1 ∗ 2

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