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

2.3. Spanning Trees and Enumeration 45Remark 2.2.5 If two spanning trees in a graph have different edge (arc) sets, we considerthem as different trees, even though, they may be isomorphic.2.3 Spanning Trees and EnumerationThere are 2 (n 2) simple graphs with vertex set [n] = {1, ..., n}, since each pair may or maynot form an edge. How many of these are trees? In this section, we solve this countingproblem, count spanning trees in arbitrary graphs, and discuss several applications.2.3.1 Enumeration of TreesWith one or two vertices, only one tree can be formed. With three vertices there is stillonly one isomorphism class, but the adjacency matrix is determined by which vertex isthe center. Thus there are three trees with vertex set [3]. With vertex set [4], there arefour stars and 12 paths, yielding 16 trees. With vertex set [5], a careful study yields 125trees.Now we may see a pattern. With vertex set [n], there are n n−2 trees; this is Cayley’sFormula. Prüfer, Kirchhoff, Pólya, Renyi, and others found proofs of that whileJ.W. Moon [92] wrote a book about enumerating classes of trees. We present a bijectiveproof, establishing a one-to-one correspondence between the set of trees with vertex set[n] and a set of known size.Remark 2.3.1 In the mathematical discipline of graph theory, a graph labeling is theassignment of labels, traditionally represented by integers, to the edges or vertices, orboth of a graph.Theorem 2.3.2 (Cayley’s Formula [1889]). For a set S ⊆ N of size n, there are n n−2trees with vertex set S.Corollary 2.3.3 The number of labeled trees with n ≥ 2 vertices is the number ofspanning trees of a labeled K n , which is n n−2 .2.3.2 Spanning trees in graphsWe can interpret Cayley’s Formula in another way. Since the complete graph with vertexset [n] has all edges that can be used in forming trees with vertex set [n], the numberof trees with a specified vertex set of size n equals the number of spanning trees in acomplete graph on n vertices.We now consider the more general problem of computing the number of spanningtrees in any graph G. In general, G will not have as much symmetry as a completegraph, so it is not reasonable to expect as simple a formula as for K n , but we can hopefor an algorithm that provides a simple way to compute the answer for a given graph G.