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

4.3. Main Results 77Figure 4.12: An example of maps C= C 1 : C 2 , C 1 , C 2 , C 1 .v 1 v 2 and C 2 .v 1 v 2Theorem 4.3.26 [87, 89] Let C= C 1 : C 2 be a map, v 1 and v 2 two vertices of the map Cwhich is formed by two maps C 1 and C 2 (see Figure 4.10), thenτ(C) = τ(C 1 ) × τ(C 2 .v 1 v 2 ) + τ(C 1 .v 1 v 2 ) × τ(C 2 ).Proof: Let T be the set of spanning trees of C, τ(C) = |T |.Let T 1 be the set of spanning trees of C which has a path from the vertex v 1 to the vertexv 2 in C 1 and let T 2 be the set of spanning trees of C which has a path from the vertex v 1to the vertex v 2 in C 2 .We have T 1 ∩ T 2 = ∅ and T 1 ∪ T 2 = T , then τ(C) = |T | = |T 1 | + |T 2 ||T 1 | is the number of spanning trees of C which has a path from v 1 to v 2 in C 1 and hasnot any path from v 1 to v 2 in C 2 .On the other hand, all spanning trees of C 1 have a path from v 1 to v 2 in C 1 , all spanningtrees of C 2 .v 1 v 2 does not have any path from v 1 to v 2 then |T 1 | = τ(C 1 ) × τ(C 2 .v 1 v 2 ) thesame for |T 2 | = τ(C 2 )×τ(C 1 .v 1 v 2 ) hence the result.□Example 4.3.27 In Example 4.3.25, we apply Theorem 4.3.26, we then obtain:τ(C) = τ(C 1 ) × τ(C 2 .v 1 v 2 ) + τ(C 1 .v 1 v 2 ) × τ(C 2 )= 3 × 4 + 5 × 4= 32.Remark 4.3.28 Let C be one of the following maps (see Figure 4.13), thenτ(C) = deg v = deg uor,τ(C)= The number of edges of the map = The number of faces of the map.