# 数学代写|组合学代写Combinatorics代考|Systems of Recursions

#### Doug I. Jones

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|组合学代写Combinatorics代考|Systems of Recursions

So far we’ve dealt with only one recursion at a time. Now we look at ways in which systems of recursions arise and adapt our methods for a single recursion to solving simple systems of recursions. The adaptation is straightforward-allow there to be more than one recursion. As usual, examples are the best way to see what this is all about.

Example 11.1 Counting Batcher sort comparators Let’s study the Batcher sort. As with our study of merge sorting in Example 10.4 (p. 278), we’ll limit the number of things being sorted to a power of 2 for simplicity. We want to determine $b_k$, the number of comparators in a Batcher sort for $2^k$ things. We’ll rewrite the Batcher sorting algorithm in Section 8.3.2 to focus on the number of things being sorted and number of comparators. The comments indicate the contributions to the recursion.
$\begin{array}{ll}\text { BSORT }\left(2^k \text { things) }\right. & / * \text { uses } b_k \text { comparators } * / \ \text { If } k=0 \text {, Return } & / * b_0=0 * / \ \text { BSORT }\left(2^{k-1} \text { things) }\right. & / * b_k=b_{k-1} * / \ \text { BSORT }\left(2^{k-1} \text { things) }\right. & / * \quad+b_{k-1} * / \ \text { BMERGE }\left(2^k \text { things }\right) & / * \quad+m_k * / \ \text { Return }\end{array}$
End
BMERGE $\left(2^k\right.$ things $)$
If $k=0$, Return
End if
If $k=1$,
one Comparator and Return
BMERGE2 ( $2^k$ things)
$2^{k-1}-1$ Comparators
Return
End
BMERGE2 $\left(2^k\right.$ things)
BMERGE ( $2^{k-1}$ things)
BMERGE ( $2^{k-1}$ things)
Return
/* uses $m_k$ comparators $* /$
$/ * m_0=0 * /$
$/ * m_1=1 * /$
/* $m_k=t_k * /$
/* $\quad+2^{k-1}-1$ / / uses $t_k$ comparators $* /$
$/ * t_k=m_{k-1} * /$
$/ * \quad+m_{k-1} * /$
End

## 数学代写|组合学代写Combinatorics代考|Exponential Generating Functions

When we use ordinary generating functions, the parts we are counting are “unlabeled.” It may appear at first sight that this was not so in all the applications of ordinary generating functions. For instance, we had sequences of zeroes and ones. Isn’t this labeling the positions in a sequence? No, it’s dividing them into two classes. If they were labeled, we would require that each entry in the sequence be different; that is, each label would be used just once. Well, then, what about placing balls into labeled boxes? Yes the boxes are all different, but the parts we are counting in our generating functions are the unlabeled balls, not the boxes. The boxes simply help to form the structure.

In this section, we’ll use exponential generating functions to count structures with labeled parts. What we’ve said is rather vague and may have left you confused. We need to be more precise, so we’ll look at a particular example and then explain the general framework that it fits into.

Recall the problem of counting unlabeled full binary RP-trees by number of leaves. We said that any such tree could be thought of as either a single vertex OR an ordered pair of trees. Let’s look at the construction of this ordered pair a bit more closely. If the final tree is to have $n$ leaves, we first choose some number $k$ of leaves and construct a tree with that many leaves, then we construct a tree with $n-k$ leaves as the second member of the ordered pair. Thus there is a three step procedure:

1. Determine the number of leaves for the first tree (and hence also the second), say $k$.
2. Construct the first tree so that it contains $k$ leaves.
3. Construct the second tree so that it contains $n-k$ leaves.
Now let’s look at what happens if the leaves are to be labeled; i.e., there is a bijection from the $n$ leaves to some set $N$ of $n$ labels. (Usually we have $N=n$, but this need not be so.) In this case, we must replace our first step by a somewhat more complicated step and modify the other two steps in an obvious manner:
$1^{\prime}$. Determine a subset $K$ of $N$ which will become the labels of the leaves of the first tree.
$2^{\prime}$. Construct the first tree so that its leaves use $K$ for labels.
$3^{\prime}$. Construct the second tree so that its leaves use $N-K$ for labels.

# 组合学代考

## 数学代写|组合学代写Combinatorics代考|Systems of Recursions

$\begin{array}{ll}\text { BSORT }\left(2^k \text { things) }\right. & / * \text { uses } b_k \text { comparators } * / \ \text { If } k=0 \text {, Return } & / * b_0=0 * / \ \text { BSORT }\left(2^{k-1} \text { things) }\right. & / * b_k=b_{k-1} * / \ \text { BSORT }\left(2^{k-1} \text { things) }\right. & / * \quad+b_{k-1} * / \ \text { BMERGE }\left(2^k \text { things }\right) & / * \quad+m_k * / \ \text { Return }\end{array}$

BMERGE $\left(2^k\right.$ things $)$

BMERGE2 ($2^k$ things)
$2^{k-1}-1$比较对象

BMERGE2 $\left(2^k\right.$ things)
BMERGE ($2^{k-1}$ things)
BMERGE ($2^{k-1}$ things)

/使用$m_k$比较器$/$
$/ * m_0=0 * /$
$/ * m_1=1 * /$
/* $m_k=t_k * /$
/* $\quad+2^{k-1}-1$ / /使用$t_k$比较器$* /$
$/ * t_k=m_{k-1} * /$
$/ * \quad+m_{k-1} * /$

## 数学代写|组合学代写Combinatorics代考|Exponential Generating Functions

$1^{\prime}$。确定$N$的一个子集$K$，它将成为第一个树的叶子的标签。
$2^{\prime}$。构造第一个树，使它的叶子使用$K$作为标签。
$3^{\prime}$。构造第二个树，使它的叶子使用$N-K$作为标签。

## 有限元方法代写

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## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

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