# 英国补考|组合学代写Combinatorics代考|CS586

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## 英国补考|组合学代写Combinatorics代考|Planar Binary Trees

This is easily shown by induction of the number of internal nodes of $T$. Indeed, (3.2) is clearly true for the trivial tree $o$, since in that case
$$E(o)=1, \quad I(o)=0 .$$
Let us assume that (3.2) is true for all trees with fewer than $n$ internal nodes. Let $T$ be a tree with $n$ internal nodes and let $T_{1}$ and $T_{2}$ be the principal subtrees of $T$. Since $T_{1}$ and $T_{2}$ will have fewer than $n$ internal nodes, we shall have
$$E\left(T_{1}\right)=I\left(T_{1}\right)+1 \text { and } E\left(T_{2}\right)=I\left(T_{2}\right)+1 .$$
This gives
$$E(T)=E\left(T_{1} \times T_{2}\right)=E\left(T_{1}\right)+E\left(T_{2}\right)=I\left(T_{1}\right)+1+I\left(T_{2}\right)+1=I(T)+1 .$$
Thus (3.2) must hold for all binary trees with $n$ nodes as well, and the induction is complete.

This given, we shall enumerate binary trees by the number of internal nodes. More precisely, let $c_{n}$ denote the number of binary trees with $n$ internal nodes. Our goal is to obtain a formula which will enable us to calculate $c_{n}$ for any value of $n$.
To this end, let us denote by $C$ the formal sum of all binary trees as shown in Figure 3.14.

## 英国补考|组合学代写Combinatorics代考|Ternary Trees

In this case it is not immediately clear what the relationship is between the numbers of internal and external nodes. Thus we shall use two variables in our generating function and we shall set
$$C(x)=\sum_{T \in C} t^{E(T)} x^{I(T)}$$
This gives
$$C(x)=t+t^{3} x+3 t^{5} x^{2}+\cdots \cdot$$
It is good to think of this series as a formal power series in $x$ with coefficients that are polynomials in $t$. I et us organize the terms in (3.9) according to increasing powers of $x$ and write
$$C(x)=t+\sum_{n \geq 1} c_{n}(t) x^{n} .$$

Our goal is to give a formula for the polynomials $c_{n}(t)$.
To this end, we can follow the same method we used for binary trees up to a point. Indeed, we have for the same reason as before the formal series identity
$$C=o+C \times C \times C .$$
Making the replacement $T \rightarrow t^{E(T)} x^{I(T)}$ we obtain the equation
$$C(x)=t+x(C(x))^{3}$$
Thus we are to solve the third degree equation
$$x(C(x))^{3}-C(x)+t=0 .$$

# 组合学代考

## 英国补考|组合学代写Combinatorics代考|Planar Binary Trees

$$E(o)=1, \quad I(o)=0 .$$

$$E\left(T_{1}\right)=I\left(T_{1}\right)+1 \text { and } E\left(T_{2}\right)=I\left(T_{2}\right)+1$$

$$E(T)=E\left(T_{1} \times T_{2}\right)=E\left(T_{1}\right)+E\left(T_{2}\right)=I\left(T_{1}\right)+1+I\left(T_{2}\right)+1=I(T)+1 .$$

## 英国补考|组合学代写Combinatorics代考|Ternary Trees

$$C(x)=\sum_{T \in C} t^{E(T)} x^{I(T)}$$

$$C(x)=t+t^{3} x+3 t^{5} x^{2}+\cdots$$

$$C(x)=t+\sum_{n \geq 1} c_{n}(t) x^{n} .$$

$$C=o+C \times C \times C .$$

$$C(x)=t+x(C(x))^{3}$$

$$x(C(x))^{3}-C(x)+t=0 .$$

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