# 数学代写|组合学代写Combinatorics代考|CS 519

#### Doug I. Jones

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## 数学代写|组合学代写Combinatorics代考|Lattice Path Representation for Planar Trees

The construction of the word $w(T)$ of a given tree we gave in Section $3.1$ can be reversed and the tree $T$ can be reconstructed from $w(T)$. Consequently the problem of enumerating planar binary trees may thus be transformed into the problem of counting tree words. However, this leads to a new question:
What words are planar tree words?
We can give this question a beautiful answer by means of a graphical representation of planar trees by lattice paths. We recall that a lattice path is a polygonal path in the $x, y$-plane whose vertices are the lattice points $(i, j)$ with integer coordinates. Given a tree $T$ we construct a lattice path $P(T)$ using the successive letters of $w(T)$ as a guide in the following manner.

We start at the origin and proceed in steps obtained by replacing each letter in $w(T)$ by a vector according to the following rule
$$x_{i} \rightarrow(1, i-1) .$$
This is best illustrated by an example. Consider the tree $T$ in Figure 3.19.

It will be good to use the letter $n$ to denote the total number of nodes in a tree. This given, we notice two properties of $P(T)$ neither of which is accidental:
(a) the path ends at the point $(n,-1)$, (heren $=13)$
(b) all but the last edge of $P(T)$ lies “above” thex $-$ axis.
Indeed we have the following remarkable theorem.

## 数学代写|组合学代写Combinatorics代考|Combinatorial Enumeration of Binary Trees

For a binary tree $T$ the path $P(T)$ has a very simple nature. Indeed, it consists only of edges with slopes 1 or $-1$. For example, when $T$ is as in Figure 3.22,
we have
$$w(T)=x_{2} x_{2} x_{0} x_{2} x_{2} x_{0} x_{0} x_{0} x_{0}$$
This gives the path in Figure 3.23.
Note that the word in (3.23) has $I(T)(=4) x_{2}$ ‘s and $E(T)(=5) x_{0}$ ‘s. Let $R\left(x_{0}{ }^{5} x_{2}^{4}\right)$ denote the set of all words which are rearrangements of the 9 letters $x_{0}^{5} x_{2}^{4}$. Clearly, any word $w(T)$ which corresponds to a tree $T$ with 5 external and 4 internal nodes belongs to $R\left(x_{0}^{5} x_{2}^{4}\right) .$ Note that there are all together
$$\left(\begin{array}{l} 9 \ 5 \end{array}\right)$$
words in $R\left(x_{0}{ }^{5} x_{2}{ }^{4}\right)$. Indeed, to construct any one of them we need only choose the positions of the $x_{0}$ ‘s in the word. Note also that the path corresponding to any one of the words of $R\left(x_{0}{ }^{5} x_{2}{ }^{4}\right)$ will necessarily have property $a$ ) of (3.19). However, only a subset of $R\left(x_{0}^{5} x_{2}^{4}\right)$ will also have property (3.19) (b).

We aim to find out the nature of this subset. For this purpose let us look again at an example. Let us pick at random a word with $4 x_{2}$ ‘s and $5 x_{0}$ ‘s. For instance,
$$w=x_{2} x_{0} x_{0} x_{2} x_{2} x_{2} x_{0} x_{0} x_{0}$$
The path corresponding to this word is given in Figure 3.24.
Property (3.19) ( $b$ ) does not hold for this path and we must conclude that $w$ is not a tree word. However, let us extend this path by going back to the beginning of $w$ and replacing letters by vectors as before. The resulting path is, of course, a juxtaposition of two identical copies of the previous path, as indicated in Figure 3.25.

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|Lattice Path Representation for Planar Trees

$$x_{i} \rightarrow(1, i-1) .$$

(a) 路径在该点結束 $(n,-1)$, (男性 $=13)$
(b) 除了最后一条边之外的所有边 $P(T)$ 位于”上方” $\mathrm{x}$ 一轴。

## 数学代写|组合学代写Combinatorics代考|Combinatorial Enumeration of Binary Trees

$$w(T)=x_{2} x_{2} x_{0} x_{2} x_{2} x_{0} x_{0} x_{0} x_{0}$$

(95)

$$w=x_{2} x_{0} x_{0} x_{2} x_{2} x_{2} x_{0} x_{0} x_{0}$$

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## MATLAB代写

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