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

4.3. Main Results 79Theorem 4.3.31 [87, 89] Let C be a map of type C= C 1 | C 2 (v 1 and v 2 two vertices ofthe map C connected by an edge e) (see Figure 4.14), thenτ(C) = τ(C 1 ) × τ(C 2 ) − τ(C 1 − e) × τ(C 2 − e).Proof: We transform the map C as follows (see Figure 4.16):τ(C 1 | C 2 )= τ((C 1 − e) : C 2 )From Theorem 4.3.26, we haveFigure 4.16: The map C after the tranformationτ(C) = τ((C 1 − e).v 1 v 2 ) × τ(C 2 ) + τ(C 1 − e) × τ(C 2 .v 1 v 2 ).From Theorem 4.3.11, we have τ(C 2 .e) = τ(C 2 ) − τ(C 2 − e) and since τ((C 1 − e).v 1 v 2 ) =τ(C 1 .e), C 2 .v 1 v 2 = C 2 .e thenτ(C) = τ(C 1 .e) × τ(C 2 ) + τ(C 1 − e) × (τ(C 2 ) − τ(C 2 − e))We take τ(C 2 ) as a factor, then= τ(C 1 .e) × τ(C 2 ) + τ(C 1 − e) × τ(C 2 ) − τ(C 1 − e) × τ(C 2 − e).τ(C) = [τ(C 1 .e) + τ(C 1 − e)]τ(C 2 ) − τ(C 1 − e)τ(C 2 − e)We use Theorem 4.3.11, we obtain : τ(C) = τ(C 1 )τ(C 2 )−τ(C 1 −e)τ(C 2 −e).□Example 4.3.32 Here is an example of a map C of type C= C 1 | C 2 with calculation ofτ(C) (see Figure 4.17):Figure 4.17: An example of a map C of type C = C 1 | C 2we apply Theorem 4.3.31 in C 1 to calculate τ(C 1 ), we then obtain τ(C 1 )= 8, (see Figure4.18):