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

4.3. Main Results 81Property 4.3.33 Let C be a map of type C = C 1 ‡ C 2 , where ‡ a simple path thatcontains k + 1 vertices and k edges, then- C 1 and C 2 have k + 1 common vertices v 1 , v 2 ,..., v k , v k+1 , k common edges (simple pathp) and a common face (the external face).- V C =V C1 +V C2 -(k + 1), E C =E C1 +E C2 -k and F C =F C1 +F C2 -1.Theorem 4.3.34 (Generalization of Theorem 4.3.31) Let C be a map, v 1 and v k+1 twovertices of the map C connected by a simple path p = v 1 , v 2 , ..., v k , v k+1 that contains kedges; see Figure 4.20, thenτ(C) = τ(C 1 ) × τ(C 2 ) − k 2 τ(C 1 − p) × τ(C 2 − p).Proof:From Theorem 4.3.20, and we use Theorems 4.3.3 and 4.3.26, we have:τ(C) = τ(C.p) + kτ(C − p)= τ(C 1 .p)τ(C 2 .p) + k[τ(C 1 .p)τ(C 2 − p) + τ(C 1 − p)τ(C 2 .p)]= [τ(C 1 .p) + kτ(C 1 − p)]τ(C 2 .p) + kτ(C 1 .p)τ(C 2 − p)= τ(C 1 )[τ(C 2 ) − kτ(C 2 − p)] + kτ(C 1 .p)τ(C 2 − p)= τ(C 1 )τ(C 2 ) − kτ(C 1 )τ(C 2 − p) + kτ(C 1 .p)τ(C 2 − p)= τ(C 1 )τ(C 2 ) − k[τ(C 1 ) − τ(C 1 .p)]τ(C 2 − p)= τ(C 1 )τ(C 2 ) − k 2 τ(C 1 − p)τ(C 2 − p), hence the result.Example 4.3.35 Here is an example of a map C of type C= C 1 ‡ C 2 with calculation ofτ(C) (see Figure 4.21):□Figure 4.21: An example of a map C of type C = C 1 ‡ C 2we apply Theorem 4.3.34 in C 1 to calculate τ(C 1 ), we then obtain τ(C 1 ) = 16, (see Figure4.22):