组合数学 (Spring 2013)/Cayley's formula

Cayley's Formula

We now present a theorem of the number of labeled trees on a fixed number of vertices. It is due to Cayley in 1889. The theorem is often referred by the name Cayley's formula.

Cayley's formula for trees

There are [math]\displaystyle{ n^{n-2} }[/math] different trees on [math]\displaystyle{ n }[/math] distinct vertices.

The theorem has several proofs. Classical methods include the bijection which encodes a tree by a Prüfer code, through the Kirchhoff's matrix tree theorem, and by double counting.

Prüfer code

The Prüfer code encodes a labeled tree to a sequence of labels. This gives a bijections between trees and tuples.

In a tree, the vertices of degree 1 are called leaves. It is easy to see that:

• each tree has at least two leaves; and
• after removing a leaf (along with the edge adjacent to it) from a tree, the resulting graph is still a tree.

The following algorithm transforms a tree [math]\displaystyle{ T }[/math] of [math]\displaystyle{ n }[/math] vertices [math]\displaystyle{ 1,2,\ldots,n }[/math], to a tuple [math]\displaystyle{ (v_1,v_2,\ldots,v_{n-2})\in\{1,2,\ldots,n\}^{n-2} }[/math].

Prüfer code (encoder)

Input: A tree [math]\displaystyle{ T }[/math] of [math]\displaystyle{ n }[/math] distinct vertices, labeled by [math]\displaystyle{ 1,2,\ldots,n }[/math]. let [math]\displaystyle{ T_1=T }[/math]; for [math]\displaystyle{ i=1 }[/math] to [math]\displaystyle{ n-1 }[/math], do let [math]\displaystyle{ u_i }[/math] be the leaf in [math]\displaystyle{ T_i }[/math] with the smallest label, and [math]\displaystyle{ v_i }[/math] be its neighbor; let [math]\displaystyle{ T_{i+1} }[/math] be the new tree obtained from deleting the leaf [math]\displaystyle{ u_i }[/math] from [math]\displaystyle{ T_i }[/math]; end return [math]\displaystyle{ (v_1,v_2,\ldots,v_{n-2}) }[/math];

It is trivial to observe the following lemma:

Lemma 1

For each [math]\displaystyle{ 1\le i\le n-1 }[/math], [math]\displaystyle{ T_i }[/math] is a tree of [math]\displaystyle{ n-i+1 }[/math] vertices. In particular, the vertices of [math]\displaystyle{ T_i }[/math] are [math]\displaystyle{ u_i,u_{i+1},\ldots,u_{n-1},v_{n-1} }[/math], and the edges of [math]\displaystyle{ T_i }[/math] are precisely [math]\displaystyle{ \{u_j,v_j\} }[/math], [math]\displaystyle{ i\le j\le n-1 }[/math].

And there is a reason that we do not need to store [math]\displaystyle{ v_{n-1} }[/math] in the Prüfer code.

Lemma 2

It always holds that [math]\displaystyle{ v_{n-1}=n }[/math].

Proof. Every tree (of at least two vertices) has at least two leaves. The [math]\displaystyle{ u_i }[/math], [math]\displaystyle{ 1\le i\le n-1 }[/math], are the leaf of the smallest label in [math]\displaystyle{ T_i }[/math], which can never be [math]\displaystyle{ n }[/math], thus [math]\displaystyle{ n }[/math] is never deleted. [math]\displaystyle{ \square }[/math]

Lemma 1 and 2 together imply that given a Prüfer code [math]\displaystyle{ (v_1,v_2,\ldots,v_{n-2}) }[/math], the only remaining task to reconstruct the tree [math]\displaystyle{ T }[/math] is to figure out those [math]\displaystyle{ u_i }[/math], [math]\displaystyle{ 1\le i\le n-1 }[/math]. The following lemma state how to obtain [math]\displaystyle{ u_i }[/math], [math]\displaystyle{ 1\le i\le n-1 }[/math], from a Prüfer code [math]\displaystyle{ (v_1,v_2,\ldots,v_{n-2}) }[/math].

Lemma 3

For [math]\displaystyle{ i=1,2,\ldots,n-1 }[/math], [math]\displaystyle{ u_i }[/math] is the smallest element of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] not in [math]\displaystyle{ \{u_1,\ldots,u_{i-1}\}\cup\{v_i,\ldots,v_{n-1}\} }[/math].

Proof. Note that [math]\displaystyle{ u_1,u_2,\ldots,u_{n-1},v_{n-1} }[/math] is a sequence of distinct vertices, because [math]\displaystyle{ u_1,u_2,\ldots,u_{n-1} }[/math] are deleted one by one from the tree, and [math]\displaystyle{ v_{n-1}=n }[/math] is never deleted. Thus, each vertex [math]\displaystyle{ v }[/math] appears among [math]\displaystyle{ u_1,u_2,\ldots,u_{n-1},v_{n-1} }[/math] exactly once. And each vertex [math]\displaystyle{ v }[/math] appears for [math]\displaystyle{ deg(v) }[/math] times among the edges [math]\displaystyle{ \{u_i,v_i\} }[/math], [math]\displaystyle{ 1\le i\le n-1 }[/math], where [math]\displaystyle{ deg(v) }[/math] denotes the degree of vertex [math]\displaystyle{ v }[/math] in the original tree [math]\displaystyle{ T }[/math]. Therefore, each vertex [math]\displaystyle{ v }[/math] appears among [math]\displaystyle{ v_1,v_2,\ldots,v_{n-2} }[/math] for [math]\displaystyle{ deg(v)-1 }[/math] times. Similarly, each vertex [math]\displaystyle{ v }[/math] of [math]\displaystyle{ T_i }[/math] appears among [math]\displaystyle{ v_i,v_{i+1},\ldots,v_{n-2} }[/math] for [math]\displaystyle{ deg_i(v)-1 }[/math] times, where [math]\displaystyle{ deg_i(v) }[/math] is the degree of vertex [math]\displaystyle{ v }[/math] in the tree [math]\displaystyle{ T_i }[/math]. In particular, the leaves of [math]\displaystyle{ T_i }[/math] are not among [math]\displaystyle{ \{v_i,v_{i+1},\ldots,v_{n-2}\} }[/math]. Recall that the vertices of [math]\displaystyle{ T_i }[/math] are [math]\displaystyle{ u_i,u_{i+1},\ldots,u_{n-1},v_{n-1} }[/math]. Then the leaves of [math]\displaystyle{ T_i }[/math] are the elements of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] not in [math]\displaystyle{ \{u_1,\ldots,u_{i-1}\}\cup\{v_i,\ldots,v_{n-1}\} }[/math]. By definition of Prüfer code, [math]\displaystyle{ u_i }[/math] is the leaf in [math]\displaystyle{ T_i }[/math] of smallest label, hence the smallest element of [math]\displaystyle{ \{1,2,\ldots,n\} }[/math] not in [math]\displaystyle{ \{u_1,\ldots,u_{i-1}\}\cup\{v_i,\ldots,v_{n-1}\} }[/math]. [math]\displaystyle{ \square }[/math]

Applying Lemma 3, we have the following decoder for the Prüfer code:

Prüfer code (decoder)

Input: A tuple [math]\displaystyle{ (v_1,v_2,\ldots,v_{n-2})\in\{1,2,\ldots,n\}^{n-2} }[/math]. let [math]\displaystyle{ T }[/math] be empty graph, and [math]\displaystyle{ v_{n-1}=n }[/math]; for [math]\displaystyle{ i=1 }[/math] to [math]\displaystyle{ n-1 }[/math], do let [math]\displaystyle{ u_i }[/math] be the smallest label not in [math]\displaystyle{ \{u_1,\ldots,u_{i-1}\}\cup\{v_i,\ldots,v_{n-1}\} }[/math]; add an edge [math]\displaystyle{ \{u_i,v_i\} }[/math] to [math]\displaystyle{ T }[/math]; end return [math]\displaystyle{ T }[/math];

In other words, the encoding of trees to tuples by the Prüfer code is reversible, thus the mapping is injective (1-1). To see it is also surjective, we need to show that for every possible [math]\displaystyle{ (v_1,v_2,\ldots,v_{n-2})\in\{1,2,\ldots,n\}^{n-2} }[/math], the above decoder recovers a tree from it.

It is easy to see that the decoder always returns a graph of [math]\displaystyle{ n-1 }[/math] edges on the [math]\displaystyle{ n }[/math] vertices. The only thing remaining to verify is that the returned graph has no cycle in it, which can be easily proved by a timeline argument (left as an exercise).

Therefore, the Prüfer code establishes a bijection between the set of trees on [math]\displaystyle{ n }[/math] distinct vertices and the tuples from [math]\displaystyle{ \{1,2,\ldots,n\}^{n-2} }[/math]. This proves Cayley's formula.

Double counting

We now present a proof of the Cayley's formula by double counting, which is considered by the Proofs from THE BOOK "the most beautiful of them all".

Proof of Cayley's formula by double counting (Due to Pitman 1999) Let [math]\displaystyle{ T_n }[/math] be the number of different trees defined on [math]\displaystyle{ n }[/math] distinct vertices. A rooted tree is a tree with a special vertex. That is, one of the [math]\displaystyle{ n }[/math] vertices is marked as the "root" of the tree. A rooted tree defines a natural direction of all edges, such that an edge [math]\displaystyle{ uv }[/math] of the tree is directed from [math]\displaystyle{ u }[/math] to [math]\displaystyle{ v }[/math] if [math]\displaystyle{ u }[/math] is before [math]\displaystyle{ v }[/math] along the unique path from the root. We count the number of different sequences of directed edges that can be added to an empty graph on [math]\displaystyle{ n }[/math] vertices to form from it a rooted tree. We note that such a sequence can be formed in two ways: Starting with an unrooted tree, choose one of its vertices as root, and fix an total order of edges to specify the order in which the edges are added. Starting from an empty graph, add the edges one by one in steps. In the first method, we pick one of the [math]\displaystyle{ T_n }[/math] unrooted trees, choose one of the [math]\displaystyle{ n }[/math] vertices as the root, and pick one of the [math]\displaystyle{ (n-1)! }[/math] total orders of the [math]\displaystyle{ n-1 }[/math] edges. This gives us [math]\displaystyle{ T_nn(n-1)!=T_nn! }[/math] ways. In the second method, we consider the number of choices in one step, and multiply the numbers of choices in all steps. This is done as follows. Given a sequence of adding [math]\displaystyle{ n-1 }[/math] edges to an empty graph to form a rooted tree, we reverse this sequence and get a sequence of removing edges one by one from the final rooted tree until no edge left. We observe that: At first, we remove an edge from the rooted tree. Suppose that the root of the tree is [math]\displaystyle{ r }[/math], and the removed directed edge is [math]\displaystyle{ (u,v) }[/math]. After removing [math]\displaystyle{ (u,v) }[/math], the original rooted tree is disconnected into two rooted trees, one rooted at [math]\displaystyle{ r }[/math] and the other rooted at [math]\displaystyle{ v }[/math]. After removing [math]\displaystyle{ k-1 }[/math] edges, there are [math]\displaystyle{ k }[/math] rooted trees. In the [math]\displaystyle{ k }[/math]th step, a directed edge [math]\displaystyle{ (u,v) }[/math] in the current forest is removed and the tree containing [math]\displaystyle{ (u,v) }[/math] is disconnected into two trees, one rooted at the old root of that tree, and the other rooted at [math]\displaystyle{ v }[/math]. We now again reverse the above procedure, and consider the sequence of adding directed edges to an empty graph to form a rooted tree. At first, we have [math]\displaystyle{ n }[/math] rooted trees, each of 0 edge ([math]\displaystyle{ n }[/math] isolated vertices). After adding [math]\displaystyle{ n-k }[/math] edges, there are [math]\displaystyle{ k }[/math] rooted trees. Denoting the directed edge added next as [math]\displaystyle{ (u,v) }[/math]. As observed above, [math]\displaystyle{ u }[/math] can be any one of the [math]\displaystyle{ n }[/math] vertices; but [math]\displaystyle{ v }[/math] must be the root of one of the [math]\displaystyle{ k }[/math] trees, except the tree which contains [math]\displaystyle{ u }[/math]. There are [math]\displaystyle{ n(k-1) }[/math] choices of such [math]\displaystyle{ (u,v) }[/math]. Multiplying the numbers of choices in all steps, the number of sequences of adding directed edges to an empty graph to form a rooted tree is given by [math]\displaystyle{ \prod_{k=2}^nn(k-1)=n^{n-2}n! }[/math]. By the principle of double counting, counting the same thing by different methods yield the same result. [math]\displaystyle{ T_nn!=n^{n-2}n! }[/math], which gives that [math]\displaystyle{ T_n=n^{n-2} }[/math]. [math]\displaystyle{ \square }[/math]

Kirchhoff's Matrix-Tree Theorem

Given an undirected graph [math]\displaystyle{ G(V,E) }[/math], its adjacency matrix [math]\displaystyle{ A }[/math] is defined as

[math]\displaystyle{ A(i,j)=\begin{cases} 1 & \{i,j\}\in E,\\ 0 & \{i,j\}\not\in E. \end{cases} }[/math]

Let [math]\displaystyle{ D }[/math] be a diagonal matrix such that

[math]\displaystyle{ D(i,j)=\begin{cases} \text{deg}(i) & i=j,\\ 0 & i\neq j, \end{cases} }[/math]

where [math]\displaystyle{ \text{deg}(i) }[/math] denotes the degree of vertex [math]\displaystyle{ i }[/math].

The Laplacian matrix [math]\displaystyle{ L }[/math] of graph [math]\displaystyle{ G }[/math] is defined as [math]\displaystyle{ L=D-A }[/math], that is,

[math]\displaystyle{ L(i,j)=\begin{cases} \text{deg}(i) & i=j,\\ -1 & i\neq j, \{i,j\}\in E,\\ 0 & \text{otherwise}. \end{cases} }[/math]