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

CHAPTER 5.86THE NUMBER OF SPANNING TREES OF CERTAIN FAMILIES OFPLANAR MAPSFigure 5.1: Example of A map C which has two faces (Cycle)5.3 The case of two cyclesTheorem 5.3.1 Let C be the following map; see Figure 5.2 (It contains two cycles C 1with h 1 edges and C 2 with h 2 edges, and k is the length of the simple path p), thenτ(C) = τ(C 1 ) × τ(C 2 ) − k 2 .Figure 5.2: Case of two cyclesProof:From Theorem 4.3.20, we haveτ(C) = τ(C.p) + kτ(C − p) = (h 1 − k)(h 2 − k) + k[(h 1 − k) + (h 2 − k)]= h 1 h 2 − k 2 = τ(C 1 ) × τ(C 2 ) − k 2 .□Particular case:In the previous Theorem 5.3.1, if k = 1, then τ(C) = τ(C 1 ) × τ(C 2 ) − 1.Example 5.3.2 Consider the following simple map (see Figure. 5.3); using the previousTheorem 5.3.1, it is clear that the number of spanning trees of this map is 23.