Cayley's formula

Cayley's formula

2^{2-2}=1 tree with 2 vertices,3^{3-2}=3 trees with 3 vertices and 4^{4-2}=16trees with 4 vertices.

In mathematics, Cayley's formula is a result in graph theory named after Arthur Cayley. It states that if "n" is an integer bigger than 1, the number of trees on "n" labeled vertices is

:n^{n-2}.,

It is a particular case of Kirchhoff's theorem.

Proof of the formula

Let T_{n} be the set of trees possible on the vertex set {v_{1}, v_{2},ldots, v_{n}}. We seek to show that

:|T_{n}| = n^{n-2}.

We begin by proving a lemma:

Claim: Let d_{1}, d_{2}, ldots, d_{n} be positive integers such that sum_{i=1}^{n}d_{i} = 2n-2 . Let mathcal{A} be the set of trees on the vertex set {v_{1}, v_{2},ldots, v_{n}} such that the degree of v_{i} (denoted mbox{d}(v_{i})) is d_{i} for i = 1, 2,ldots, n. Then

:|mathcal{A}| = frac{(n-2)!}{(d_{1}-1)!(d_{2}-1)!cdots(d_{n}-1)!}.

Proof:We go by induction on n. For n=1 and n=2 the proposition holds trivially and is easy to verify. We move to the inductive step. Assume n>2 and that the claim holds for degree sequences on n-1 vertices. Since the d_{i} are all positive but their sum is less than 2n, exists kin{1,2,ldots,n} such that d_{k} = 1. Assume without loss of generality that d_{n} = 1.

For i = 1, 2, ldots, n-1 let mathcal{B}_{i} be the set of trees on the vertex set {v_{1}, v_{2}, ldots v_{n-1} } such that:

:d(v_{j}) = egin{cases} mbox{d}_{j}, & mbox{if }j eq i \ mbox{d}_{j}-1, & mbox{if }j=i end{cases}

where d(v) is the degree of v.

Trees in mathcal{B}_{i} correspond to trees in mathcal{A} with the edge {v_{i}, v_{n}}, and if mbox{d}_{i} = 1, then mathcal{B}_{i} = emptyset.

Since v_{n} must have been connected to some node in every tree in mathcal{A}, we have that

:|mathcal{A}| = sum_{i=1}^{n-1}|mathcal{B}_{i}|.

Further, for a given i we can apply either the inductive assumption (if mbox{d}_{i} > 1) or our previous note (if mbox{d}_{i} = 1, then mathcal{B}_{i} = emptyset) to find |mathcal{B}_{i}|:

:|mathcal{B}_{i}| = egin{cases} 0, & mbox{if }d_{i}=1 \ frac{(n-3)!}{(d_{1}-1)!cdots(d_{i}-2)!cdots(d_{n-1}-1)!}, & mbox{otherwise} end{cases} for i=1,2,ldots,n-1

Observing that

:frac{(n-3)!}{(d_{1}-1)!cdots(d_{i}-2)!cdots(d_{n-1}-1)!} = frac{(n-3)!(d_{i}-1)}{(d_{1}-1)!cdots(d_{n-1}-1)!}
it becomes clear that, in either case, |mathcal{B}_{i}| = frac{(n-3)!(d_{i}-1)}{(d_{1}-1)!cdots(d_{n-1}-1)!}.

So we have

::|mathcal{A}| = sum_{i=1}^{n-1}|mathcal{B}_{i}|

:=sum_{i=1}^{n-1}frac{(n-3)!(d_{i}-1)}{(d_{1}-1)!cdots(d_{n-1}-1)!}

:=frac{(n-3)!}{(d_{1}-1)!cdots(d_{n-1}-1)!}sum_{i=1}^{n-1}(d_{i}-1).

And since d_{n} = 1 and sum_{i=1}^{n}d_{i} = 2n-2, we have:

:|mathcal{A}| = frac{(n-3)!}{(d_{1}-1)!cdots(d_{n}-1)!}(n-2) = frac{(n-2)!}{(d_{1}-1)!cdots(d_{n}-1)!}

which proves the lemma.

We have shown that given a particular list of positive integers d_{1}, d_{2}, ldots, d_{n} such that the sum of these integers is 2n-2, we can find the number of trees with labeled vertices of these degrees. Since every tree on n vertices has n-1 edges, the sum of the degrees of the vertices in an n-vertex tree is always 2n-2. To count the total number of trees on n vertices, then, we simply sum over possible degree lists. Thus, we have:

:|T_{n}| = sum_{d_{1}+d_{2}+cdots+d_{n} = 2n-2} frac{(n-2)!}{(d_{1}-1)!cdots(d_{n}-1)!}.

If we reindex with k_{i}=d_{i}-1 for i=1,2,ldots,n, we have:

:|T_{n}| = sum_{k_{1}+k_{2}+cdots+k_{n} = n-2} frac{(n-2)!}{k_{1}!cdots k_{n}!}.

Finally, we can apply the multinomial theorem to find:

:|T_{n}| = n^{n-2}

as expected. Box

Note

Prüfer sequences yield one of the many alternate proofs of Cayley's formula.


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