[Concrete Mathematics Generating Function] Proof of the "generating function" of Cayley's formula

Cayley formula: k=n^{n-2}  The cayley formula gives the number of minimum spanning trees with n vertices.


The work of this article: a small supplementary interpretation of the simple to extreme discrete mathematics course ppt


This question gives a way to prove the generating function of the Cayley formula. In this method, by constructing a generative function, the form of the generative function of all spanning trees of n vertices is obtained by mathematical induction, and thus the number of spanning trees of n vertices is deduced.

1. In order to describe each spanning tree, we define an item

\prod _{i=1}^{n}x_{i}^{deg(i)-1}

explain:

        For a vertex  x_{i} whose degree in the spanning tree is deg(i),  x_{i}^{deg(i)-1} each vertex is described by .

        This item is: multiply the description information of each vertex of the spanning tree to obtain the description item of the entire spanning tree

2. In order to describe all spanning trees of n vertices, we construct a generating function F(n)

        F(n)=\sum (\prod _{i=1}^{n}x_{i}^{deg(i)-1})

        This generating function accumulates all spanning tree information. Then one of the multiplicative term coefficients represents the existence of a spanning tree. Number of spanning trees = sum of coefficients.

3. First, define the head of the logical chain of induction:

When n F(n)= (x_{1}+...+x_{n} )^{n-2} = 3, the formula is consistent , and there are 3 spanning trees at this time
       That is, F(3)=x1 + x2 + x3, the sum of various coefficients = the number of spanning trees, and there are 3 spanning trees at this time.

 4. Carry out logical recursion: We know that F(n-1) satisfies the formula, and prove that F(n) also satisfies the formula

         (The original ppt description is like this)

        

· To describe each spanning tree, we define a term   \prod _{i=1}^{n}x_{i}^{deg(i)-1}

·If node n is a leaf node, then deg(n)-1=0, then there is no x_{n}item in this item.

We can split the spanning tree into two sets:

Set 1: Node n is a leaf node

Set 2: Node n is a non-leaf node

 We define the description items of the spanning tree in set 2, that is, the sum of the items of the spanning tree where n is a leaf node, as G(n)

G(n)=\sum _{deg(n)=1}(\prod _{i=1}^{n}x_{i}^{deg(i)-1})

 Double counting, you can get another representation of G(n):

 In this type of spanning tree, node n is connected to the spanning tree set F(n-1) of n-1 nodes with an incoming edge.

Whereas joinpoints x_{1} ,x_{2} ,x_{3}....x_{n-1}have a choice. Each choice makes x_{i}the power of the term +1

but G(n)=F(n-1)*(x_{1}+x_{2}+...+x_{n-1})

Substituting into the recursive formula of F(n-1), we get

        G(n)=(x_{1}+x_{2}+...+x_{n-1})^{n-2}

 This "observation" is proven.

5. Prove that F(n) is established by establishing G(n).

         G(n) is the spanning tree that we established through F(n-1), and the obtained point n is a leaf node

        If we prove that G(n)==F(n) describes the spanning tree where n is a leaf node , it means that we can prove F(n) from F(n-1).

----- "Refer to the coefficient of each item in G(n) == in F(n), n is the coefficient of the item of the leaf node spanning tree. (a bit around)

 Evidence method: Violent demolition

                                G(n)=(x_{1}+x_{2}+...+x_{n-1})^{n-2}        

                                F(n)= (x_{1}+...+x_{n} )^{n-2}, the term excluding xn is exactly G(n)

Obviously not written here. Certificate completed.


Summarize:

        This method aims to prove the cayley formula, so there is prior knowledge, and mathematical induction can be used.

        Each item represents a spanning tree by constructing a generative function. Then by mathematical induction, it is proved that the generating function and the cayley formula are equivalent.

        The multiplicative definition of the generating function has little to do with the form of the Cayley formula. It just acts as a bridge: through the meaning of its definition, the equality relationship between F(n-1) and G(n) can be obtained, so that G(n) can successfully deduce F(n).

        The above is my understanding of the ppt in the picture, please point out if there is any mistake.

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Origin blog.csdn.net/m0_67441224/article/details/129104657