## 统计代写|数据结构作业代写data structure代考|Trees

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

## 统计代写|数据结构作业代写data structure代考|What is a Tree

A tree is a data structure similar to a linked list but instead of each node pointing simply to the next node in a lincar fakhion, each node points to a number of nodes. Tree is an example of non-linear data structures $A$ tree structure is a way of representing the hierarchical nature of a strocture in a graphical form.

In trees ADT (Abstract Data Type), the order of the elenents is not important. If we need ordering information linear data structures like linked lists, stacks, queues, cte, cin be used.

## 统计代写|数据结构作业代写data structure代考|Binary Tree Traversals

In order to process trees, we necd a mechanism for traversing them, and that forms the subject of this section. The process of visiting all nodes of a tree is called tree traversal. Fach node is processed only once but it may be visited more than once. As we have already secn in linear data structures (like linked lists, stacks, queues, cte.), the clemeats are visated in sequential order. But, in trec structures there are nuany differeat ways.
Tree traversal is lake searching the tree, excepx that in traversal the goal is to move through the tree in a particular order. In addition, all nodes are processed in the traversal but sear ching stops when the required aode is foumd.

In preorder traversal, each node is processed before (pre) either of its subtrees. Thais is the simplest traversal to uaderstand. However, cven though each node is processed before the subtrees, it still requires that some information must be maintaincd while moving down the tree. In the example abowe, 1 is processed first, then the left subtree, and this is followed by the right subtree.

Therefore, processing nust return to the right subtree after fanishiug the processing of the left subtree. To nuove to the right subtree after processing the left subtrec, we must maintain the root information. The obvious ADT for such information is a stack. Because of its IFO structure, it is possible to get the information about the right subtrees back in the reverse order.
Preorder traversal is defined as follows:

• Visal the root.
• Traverse the left subtree un Prcordcr.
• Traverse the right subtree in Preorder.

## 统计代写|数据结构作业代写data structure代考|Minimum depth of a binary tree

Minimum depth of a binary tree: Given a binary tree, find its mimimum depth. The minimum depth of a binazy tree is the number of nodes along the shortest path from the root node down to the nearest leaf node. For example, naimun depth of the following binary tree is $\supsetneqq$.

Solution: The algorithm is similar to the algorithmo of finding depth (or height) of a binary trec, except here we are finding minimum depth. One simplest approach to solve this problem would be by usimg recursion. But the question is when do we stop it? We stop the recursave calls when it is a leaf rode or None.
Algorithm I ret root be the pointer to the root aode of a subtrec.

• If the root is equal to None, then the maimimum depeh of the bänary tree would be $0 .$
• If the root is a keaf node, then the minimum depth of the binary trec woudd be 1 .
• If the root is not a leaf node and if left subtree of the root is None, then find the maimimum depth in the right suberce. Otherwise, find the naimimum depth in the left subtree.
• If the root is not a leaf node and both left suberee and right subtree of the root are aot None, then recursively find the mainimum depth of left and right subtree. I ct at be leftSubtreeMinDepth and rightSubtreeMinDepth respectively.
• To get the maininum height of the binary ree rooted at root, we will take nuininaum of leftSubtreeMinDepth and rightSwhtreeMinDepth and 1 for the ront node.

## 统计代写|数据结构作业代写data structure代考|Binary Tree Traversals

• 维萨根。
• 遍历左子树 un Prcordcr。
• 在 Preorder 中遍历右子树。

## 统计代写|数据结构作业代写data structure代考|Minimum depth of a binary tree

• 如果根等于无，则二叉树的最大深度为0.
• 如果根是 keaf 节点，则二进制 trec 的最小深度将是 1 。
• 如果根不是叶子节点并且根的左子树是None，则在右子节点中找到最大深度。否则，在左子树中找到最大深度。
• 如果根不是叶子节点，并且根的左子树和右子树都不是None，则递归找到左子树和右子树的最大深度。我分别在 leftSubtreeMinDepth 和 rightSubtreeMinDepth 处。
• 为了得到以根为根的二叉树的主高度，我们将 leftSubtreeMinDepth 和 rightSwhtreeMinDepth 的 nuininaum 和 1 用于 ront 节点。

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|数据结构作业代写data structure代考|Queues

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

## 统计代写|数据结构作业代写data structure代考|What is a Queue

A queue is a data structure used for storing data (simalar to I inked lists and Stacks). In queuc, the order in which data arrives is important. In general, a queue is a liac of people or things waiting to be served in sequential order starting at the beginaing of the line or sequence.

Definition: A queue is an ordered list in which insertions are done at one cnd (rear) and deletions are done at other end (front). The first element to be inserted is the first one to be deleted. Hence, it is called First in First out (FIFO) or Last in I.ast out (LII.O) list.

Sinalar to Stacks, special namacs are given to the two chaages that can be maade to a queuc. When an clearcnt is iaserted in a queuc, the concept is called EnQueue, and when an clensent is removed from the queue, the concept is called DeQueue.

DeQueueing an empty queue is called underflow and EnQueuing an element in a full queue is called over flow. Generally, we treat them as exceptions. As an exanuple, consuder the snapshot of the quete.

## 统计代写|数据结构作业代写data structure代考|How are Queues Used

The concept of a queuc can be explained by observing a line at a reservation counter. When we enter the lias, we stand at the end of the line and the person who is at the front of the line is the one who will be served next. He will exit the queue and be served.
As this happens, the nexi person wall come at the head of the line, wall exit the queue and will be served. As cach person at the head of the line kecps cxiting the queue, we nocve towards the head of the line. Finally, we wall reach the head of the line and we will exalt the queue aad be served. This behavion is very useful in casc3 where there is a necd to haintain the order of arrival.

## 统计代写|数据结构作业代写data structure代考|Queues

As you can see, the producer and consumer do not necessarily alternate in execution. In this solution, we use the Quete. We use random_randint0 to nake production and consumption sonewhat varied.

The writeQ0 and readQ0 functions each have a specific purpose: to place an object in the queuc-ue are using the string ‘MONK’, for example-ind to consume a quetued object, respectively. Notice that we are producing one object and reading one object each tine.

The producer0 is going to nun as a single thread whose sole purpose is to produce an item for the queuc, wait for a bit, and then do it again, up to the specified number of times, chosen randomly per script execution. The consumer() will do likewise, with the exception of consuming an item, of course.

You will notice that the random number of seconds that the producer slecps is in general shorter than the amount of time the consumer sleeps. This is to discourage the consumer from trying to take itens from an empty queue. By giving the producer a shorter time period of waiting, it is more likely that there will already be an objeet for the consumer to consume by the time their tum rolk around again.
These are just setup lines to set the total number of threads that are to be spawned and executed.
$\mathrm{~ F u a d l y , ~ w e ~ h a v e ~ u m ~ a n a i n 0 ~ f u n c u i o n , ~ w h i c h ~ s h r u b l e l ~ l e o k ~ y u i t e ~ s a i}$ appropriate threads and send them on their way, finishing up when both threads have conchded execution.

We infer from this example that a program that has multiple tasks to perform can be organized to use separate threads for each of the tasks. This can result in a much cleaner program design than a single-threaded program that attempts to do all of the tasks.

We illustrated how a single-threaded process can limit an application’s performance. In particular, programs with independent, nondeterministic, and non-causal tasks that execute sequentially can be improved by division into separate tasks executed by individual threads. Not all applications will bencfit from multithreading due to overhead and the fact that the Python interpreter is a single-threaded application, but now you are nore cognizant of Python’s threading capabilities and can use this tool to your advantage when appropriate.

## 统计代写|数据结构作业代写data structure代考|What is a Queue

Sinalar 到 Stacks，特殊的 namacs 被赋予了两个可以编入队列的 chaages。当一个 clearcnt 在队列中被插入时，这个概念被称为 EnQueue，当一个 clensent 从队列中被移除时，这个概念被称为 DeQueue。

## 统计代写|数据结构作业代写data structure代考|Queues

writeQ0 和 readQ0 函数各有一个特定目的：将对象放入队列中 – 使用字符串“MONK”，例如 -ind 分别使用队列中的对象。请注意，我们正在生成一个对象，并且每个齿读取一个对象。

producer0 将作为单个线程运行，其唯一目的是为队列生成一个项目，稍等片刻，然后再做一次，直到指定的次数，每次脚本执行随机选择。consumer() 也会做同样的事情，当然，除了消费一个项目。

F在一种dl是, 在和 H一种在和 在米 一种n一种一世n0 F在nC在一世这n, 在H一世CH sHr在bl和l l和这ķ 是在一世吨和 s一种一世适当的线程并在途中发送它们，当两个线程都已确定执行时完成。

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|数据结构作业代写data structure代考|Stacks

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

## 统计代写|数据结构作业代写data structure代考|What is a Stack

A stack is a simple data structure used for storing data (simailar to Lanked Lists). In a stack, the order in which the data arrives is important. A pile of plates in a cafeteria is a good example of a stack. The plates are added to the stack as they are cleaned and they are placed on the top. When a plate, is required it is taken from the top of the stack. The first plate placed on the stack is the last one to be used.

Definition: A stack is an ordered list in which insertion and deletion are done at one end, called top. The last clement inserted is the first one to be delcted. Hence, it is called the Last in First out (I.IFO) or Furst in Last out (FILO) list.

Special names are given to the two changes that can be made to a stack. When an clenuent is inserted in a stack, the concept is called push, and when an element is removed from the stack, the concept is called pop. Tryang to pop out an cmpty stack is called under flow and trying to push an elenueat in a full stack is called overflow. Geacrally, we treat then as exceptions. As an example, consider the snapshots of the stack.

## 统计代写|数据结构作业代写data structure代考|How Stacks are Used

Consider a working day in the office. Let us assume a developer is working on a long-term project. The manager then gives the developer a uew task which is maore important. The developer pusts the long term project aside and begins work on the asew tasla. The phone rings, and this is the highest pronory as it must be answered immediately. The developer pushes the present task into the peadiag tray and answers the phone.

When the call is complete the task that was abandoned to answer the phone is retrieved from the pendirg tray and work progresses. To take another call, it nay have to be handled in the sanve manner, but eventually the new task will be funished, and the developer can draw the loangtern project from the pending tray and continue with thaat.

## 统计代写|数据结构作业代写data structure代考|Dynamic Array Implementation

First, ket’s consider how we inplemented a simple array-based stack. We took one index variable top which points to the iadex of the most. recenthy inserted clement in the stack. To insert (or push) an element, we increment top index and then place the new element at that index.
Simalarly, to delete (or pop) an element we take the element at top index and then decrenent the top index. We represent an empty quete with top value equal to $-1$. The issue that still needs to be resolved is what we do when all the slots in the fixed size array stack are oceupicdi? First tryy What if we increment the size of the aray by 1 every time the stack is fulle?

• Pusha 0 ibcrease size of $\mathrm{S} |$ by 1
• Pop0: decrease size of Sll by 1
Issues with this approach?
This way of incrementing the array size is too expensive. Let us see the reason for this. For example, at $n=1$, to push an element create a new array of size 2 and copy all the old array elements to the new array, and at the end add the new element. At $n=2$, to push an element create a new array of size 3 and copy all the old array elements to the new array, and at the end add the new elenent.

Similarly, at $n=n-1$, if we want to push an element create a new array of size $n$ and copy all the old array elements to the new array and at the end add the new element. After $n$ push operations the total tine $T(n)$ (number of copy operations) is proportional to $1+2+\ldots+$ $n \approx \mathrm{O}\left(n^{2}\right)$.
Alternative Approach: Repeated Doubling
Let us improve the conplexity by using the array doubling technique. If the array is full, create a new array of twice the size, and copy the itens. With this approash, pushing $n$ items take time proporional to $n$ (not $n^{2}$ ).
For simplieity, let us aosune that initinlly we started waith $n=1$ and moved up to $n=32$. That means, we do the doubling at $1,2,4,8,16$. The other way of analyzing the same approach is: at $n=1$, if we want to add (push) an element, double the current size of the array and copy all the elements of the old array to the new array.

## 统计代写|数据结构作业代写data structure代考|Dynamic Array Implementation

• Pusha 0 ibcrease 大小小号|1
• Pop0：将 Sll 的大小减小 1
这种方法有问题吗？
这种增加数组大小的方法太昂贵了。让我们看看这其中的原因。例如，在n=1，要推送一个元素，创建一个大小为 2 的新数组，并将所有旧数组元素复制到新数组，最后添加新元素。在n=2，要推送一个元素，创建一个大小为 3 的新数组并将所有旧数组元素复制到新数组，最后添加新元素。

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

By the time we traverse the complete list (for creating the hash table), we can find the list length. I ct us say the list length is $M$. To find $n^{\text {th }}$ from the end of linked list, we can convent this to $(M-n+1)^{\text {th }}$ from the beginuing. Sance we already know the length of the list, it is just a matter of retuming $(M-n+1)^{\text {th }}$ key value from the hash table.
‘ Iine Complexity: I inve lor creating the hash table, $T(m)=()(m)$. Space Cionaplexity: Since we need to create a hash table of size $m$, O( $m)$.
Problem-4 Can we use Problem-3 approach for solving Problem-2 without creating the hash table?
Solution Yeg. If we observe the Problenh3 solution, what we are actually doing is finding the size of the linked list. That means we are using the hash table to find the size of the linked list. We can find the length of the linked list just by starting at the head node and traversing the list. So, we can find the length of the list without creating the hash table. After finding the length, compute $M-n+1$ and with one more scan we can get the $(M-n+1)^{\text {th }}$ node from the beginning. This solution needs two scans: one for finding the length of the list and the other for finding $(M-n+1)^{t h}$ node from the hegianing-
Time Complexity: Time for finding the length + Time for finding the $(M-n+1)^{\text {th }}$ node from the beginning. Therefore, $T(n=O(n)+$ $O(n) \approx O(n)$. Space Complexity: $O(1)$. Hence, no need to create the hash table.
Problem5 Can we solve Problen-2 in one scan?
Solution: Yes. Efficient Approach: Use two pointers $p N$ thNode and $p$ Temp. Initially, both point to head node of the list. $p N$ thNode starts moving only alter $p$ Temp has made $n$ mones. From there both mowe forward until $p$ Temp reaches the end of the list. As a result, $p N$ thNode points to $n^{\text {th }}$ node from the end of the linked list.
Notet At any poant of time both move one node at a time.

## 统计代写|数据结构作业代写data structure代考|Efficient Approach

Solution: Yes. Fincient Approed QMemoryleas Appronch): The space conplexity can be reduced to O(1) by considering two pointers at differeat speed – a slow pointer and a fast pointer. The slow poanter moves one step at a time while the fast pointer mowes two steps at a tine. This problem was solved by Floyd. The solution is named the Floyd cycle finding algorithm. It uses two pointers moving at different speeds to walk the linked list. If there is no cycle in the list, the fast poanter will eventually reach the ead and we can return false in this case. Now consider a cyclic list and imagiae the slow and fast pointers are two rumers racing around a circle track. Once they enter the loop they are expected to neet, which denotes that there is a loop.

This works becanse the only way a faster movang pointer would point to the sanae location as a slower moving pointer is if sonachow the entire list or a part of it is circular. Think of a tortoise and a hare rumaing on a track. The faster numang hare will catch up with the tortoise if they are ruming in a loop. As an exanple, consider the followanng exmple and trace out the Floyd algonthm. From the diagrans below we can see that after the final step they are meeting at sone point in the loop which nay not be the starting point of the loop.
Note: slowPtr (tortoise) moves one pointer at a tine and fastPtr (hare) mones two pointers at a tine.

## 统计代写|数据结构作业代写data structure代考|Algorithm

Create two stacks one for the first list and one for the second list.
Traverse the first list and push all the node addresses onto the first stack.
Traverse the second list and push all the node addresses onto the second stack.
Now both stacks contain the node address of the corresponding lists.
Now compare the top node address of both stacks.
If they are the same, take the top elements from both the slacks and keep them in some temporary variable (since both node addresses are node, it is enough if we use one temporary varable).
Continue this process until the top node addresses of the stacks are not the same.
This point is the one where the lists merge into a single list.
Return the value of the tenporary variable.

## 数据结构代写

‘ 线复杂性：我想创建哈希表，吨(米)=()(米). Space Cionaplexity：因为我们需要创建一个大小为米， 这（米).

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|数据结构作业代写data structure代考|What is a Linked List

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

## 统计代写|数据结构作业代写data structure代考|Issues with Linked Lists

There are a number of issues wàth linked lists. The main disadvautage of lanked lists is access time to individual elements. Array is randonaccess, which means it takes $\mathrm{O}(1)$ to access any clement in the array. I inked lists ake $O(n)$ for access to an element an the list in the worst case. Another advantage of arrays in access tine is spacial locality in menory. Arays are defined as contiguous blocks of nuenory, and so any array element will be physically near its neighbors. This greatly benefits from modern CTI caching methods.

Although the dynamic allocation of storage is a great advantage, the overhead with storing and retricving data can nake a big differeace. Sometimes limked lists are hard to manipulate. If the last item is deleted, the last but one must then have its pointer changed to hold a None reference. This requires that the list is traversed to find the last but one link, and its poanter set to a None relerence. Finally, linked lists waste menory in terms of extra reference points.

The advantage of a doubly linked list (also called two – way linked list) is that given a node in the list, we can navigate in both directions. A mode in a singly linked list cuamot be renoysd uakss we have the poiater to its predecessor. But in a doubly linked list, we can delete a mode even if we dou’t have the previous mode’s address (since each node has a left pointer pointiag to the previous node and can move backward).

• Each node requires an extra pointer, requiring more space.
• The insertion or deletion of a node takes a bit longer (more pointer operations).
Similar to a singly linked list, let us implement the operations of a doubly linked list. If you understand the singly linked list operations, then doubly linked list operations are obwious. Following is a type declaration for a doubly linked list of integers:

## 统计代写|数据结构作业代写data structure代考|Skip Lists

Binary trees can be used for represcuting abstract data types such as dictionaries and ordered lists. They work well when the elenucnts are inserted in a randon order. Sonve sequences of operations, such as inserting the elensents in order, produce degcnerate data stroctures that give very poor performance. If it were possable to randomaly permute the list of itens to be inserted, trees would work well waith high probability for any iaput sequence. In most cases queries must be answered oas-line, so randomly permutimg the anput is impractical. Balanced tree algorithas re-arrange the tree as operations are performed to maintain certain balance couditions and assure good performance.

Skip list is a data structure that can be used as an altemative to balanced biaary trees (refer to Trees chapter). As compared to a binary trec, skip lists allow quick search, insertion aad deletion of elements. This is achicved by using probabilistic balaucing rather than strictly enforce balancing. It is basically a linked list with additional poanters such that internediate modes can be skipped. It uses a randon mumber generator to maike some decisions.In an ordinary sorted linked list, search, insert, ausd delete are in $\mathrm{O}(\mathrm{n})$ because the list must be scanned node-by-node from the head to find the relevant node. If somehow, we could scan down the list in bagher steps (skip down, as it were), we would reduce the cost of scauniag. This is the fuadanacntal xlea behind Skip I.sts.

## 统计代写|数据结构作业代写data structure代考|Issues with Linked Lists

• 每个节点都需要一个额外的指针，需要更多空间。
• 节点的插入或删除需要更长的时间（更多的指针操作）。
类似于单链表，让我们实现双链表的操作。如果您了解单链表操作，那么双链表操作是显而易见的。以下是整数双向链表的类型声明：

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|数据结构作业代写data structure代考|Backtracking

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

## 统计代写|数据结构作业代写data structure代考|Solutions

Solution:
def appendAt Frout $(\mathrm{x}, 1 \mathrm{j})$
return $\mid \mathrm{x}+$ clemeat for clement in L]
chef bitStrings(n):
if $\mathrm{n}=0$ – retura $[1$
if $\mathrm{a}=1$ a retum $\left[00^{},{ }^{} 1\right]$
clset
retum (appeadAtFrout(0″, bitStriags $(\mathrm{a}-1))+$ append.MtFrout(“1”, bitStrings(1-1)))
print (batStrings(1))
Afermetínset
def batStriags(n):
if $\mathrm{n}=0$ = retum |
if $\mathrm{n}=1:$ retum [0″, “1”]
return | digit bitstring for digit in bitStrings(1)
for bitstring in bitStrings(n-1)|
print (bathtrings(1))
Let $T(n)$ be the running time of binary $(n)$. Assume function print $f$ takes time $O(1)$.
$$T(n)=\left{\begin{array}{lr} c_{1} & \text { if } n<0 \ 2 T(n-1)+d, \text { otherwise } \end{array}\right.$$
Using Suberaction and Conquer Master theorem we get: $T(n)=O\left(2^{n}\right)$. This means the algorithm for generating bit-strings is optimal. Problem-4 Generate all the strings of length $n$ drawu from $0 . . k-1$.
Solution: Let us assume we keep current $k$-ary string in an array $A[0 . . n-1]$. Call function $k$-string $(n, k)$ :
def range Tol ist(k):
result – II
def range Tol ist(k):
result –
for i in range(0,k):
result.append(str(i))
retura result
for $i$ in range(0,k):
result.append(str(i))
return result
def baseKStrings $(\mathrm{n}, \mathrm{k})$ :

## 统计代写|数据结构作业代写data structure代考|Finding the length

Problem-6 Finding the length of connected cells of 1s (resions) in a mntrix of Os and 1s: Given a matrix, each of which may be 1 or 0 . The filled cells that are connected form a region. Two cells are said to be connected if they are adjacent to each other horizontally, vertically or diagonally. There may be several regions in the matrix. How do you find the largest region (in terms of number of cells) in the matrix?

Sample Input? 11000 Sample Output: 5 01100 00101 10001 01011
Solution
$\operatorname{def} \operatorname{getval}(\boldsymbol{A}, \mathrm{i}, \mathrm{j}, \mathrm{L}, \mathrm{H})$ !
if $(i<0$ or $i>-$ L or j<0 or $j>-H):$
return 0
else:
return Alillil
def findMaxBlock $\left(A, r, c, L_{\text {. }} H\right.$, sizc):
global maxsize
global cntart
if $(\mathrm{r}>-\mathrm{L}$ or $c>-\mathrm{H})$ :
return
cntart|r||c|-1
size $+-1$
if (size $>$ maxsize):
maxsize – size

$\mathrm{~ d i r e c t i o n – [ | – 1 , 0 ] , | – 1 , – 1 | , | 0 , – 1 ] , [ 1 , – 1 | , [ 1 , 0 ] , { 1 , 1 ] , | 0 , 1 | , | – 1 , 1 | |}$
for i in range(0,7):
newi – $r+$ direction[i]|이
val-getval (A, newi, new], L., H)
if (val>0 and (cutarr|newi][new] $\mid=-0)$ ):
findMaxBlock( $A$, newi, newj, L, H, size)
cutiurr $|r||c|-0$
def getMaxOnes(A, rmax, colmax):
global maxsize
global size
global cntarr
for $i$ in range $(0$, rax $)$ :
for $j$ in range(0,colnax):
if $(A|\mathrm{~A}| \mathrm{bl}-1)$ :
findMavRlork ( $\Lambda, i, j$, rnax, colmax, 0 )
return maxsize
$\mathrm{~ z a r r – | { 1 , 1 , 0 , 0 , 0 ] , | 0 , 1 , 1 , 0 , 1 | , [ 0 , 0 , 0 , 1 , 1 ] , [ 1 , 0 , 0 , 1 , 1 ] , { 0 , 1 , 0 , 1 , 1 ] |}$
$\max =5$

## 统计代写|数据结构作业代写data structure代考|Path finding problem

If we have reached the destination point
retum an array containing only the position of the destination clse

1. Mewe in the forwards diroction and cherk if this learle to a solution
2. If option a does uot work, then mone down
3. If either work, add the current position to the solution obtained at cither 1 or 2
def pathFinder(Matrix, position, N):

if position = $(\mathrm{N}-1, \mathrm{~N}-1)$ :
return ${(N-1, N-1) \mid$
$x, y=$ positicm
if $\mathrm{x}+1<\mathrm{N}$ and Matrix $|\mathrm{x}+1||\mathrm{y}|-1$ :
$a$ – pathFinder(Matrix $(x+1, y), N)$
if a !-None:
retum $|(x, y)|+a$
if $\mathrm{y}+1<\mathrm{N}$ and Matrix $[\mathrm{x}|| \mathrm{y}+1 \mid-1$ :
$b=p a r h F$ iader(Matrix , $(x+y+1), N)$
if $b$ !- None:
retum $|(\mathrm{x}, \mathrm{y})|+\mathrm{b}$
Matrix – $\lfloor 11,1,1,1,0],{0,1,0,1,0],[0,1,0,1,0],{0,1,0,0,0], \mid 1,1,1,1,1] \mid$
prisut (pathFinder(Matrix, $(0,0), 5)$ )

## 统计代写|数据结构作业代写data structure代考|Solutions

def appendAt Frout $(\mathrm {x}, 1 \ mathrm {j})$ return $\ mid \ mathrm {x +$ clemeat for clement in L] chef bitStrings(n): if $\ mathrm {n } = 0$ – retura $[1$ if $\ mathrm {a} = 1$ a retum $\ left [00 ^ { }, {} ^ } 1 \ right]$返回 | 位串中的数字位串 (1) 位串中的位串 (n-1) | 打印（浴巾（1））(\ mathrm {x}, 1 \ mathrm {j}) \ mid \ mathrm {x} + clemeat for clement in L] \ mathrm {n} = 0 – retura [1 \ mathrm {a} = 1 a retum $\ left [00 { clset retum (appeadAtFrout (0 ″, bitStriags (\ mathrm {a} -1))) + append.MtFrout (“1”, bitStrings (1-1))) print (batStrings (1)) Afermetínset def batStriags (n): 如果\mathrm {n} = 0 = 返回 | 如果\ mathrm {n} = 1:$ retum [0 ″, “1”](x,1j)
∣x+n=0[1
a=1clsetretum(appeadAtFrout(0″,bitStriagsíappend.MtFrout(“1”,bitStrings(1−1)))print(batStrings(1))AfermetínsetdefbatStriags(n):if=retum|if令 T(n) 为二进制 (n) 的运行时间。假设函数 print f 花费时间 O(1)。
$$T(n)=\left{\begin{array}{lr} c_{1} & \text { 如果 } n<0 \ 2 T(n-1)+d, \text { 否则 } \end{数组}\对。$$

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|数据结构作业代写data structure代考|Format of a Recursive Function

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

## 统计代写|数据结构作业代写data structure代考|Format of a Recursive Function

A recursive function perfomus a task in part by callimg itself to perform the subtasks. At sonve point, the function encounters a subtask that it can perform without calling itself. This case, where the function does not recur, is called the base case. The former, where the fuaction calls itself to perform a subtask, is referred to as the recursive case. We can write all recursive functions using the format:
if(test for the base casc):
return some base case value
cliftest for aaother base case):
return sone oxher base case value

clse:
return (some work and then a recursive call)
As an example, consuder the factorial function: $n$ ! is the product of all integers between $n$ and 1 . The definition of recursive factorial looks like:
$$\begin{gathered} n !=1, \quad \text { if } n=0 \ n !=n=(n-1) ! \text { if } n=0 \end{gathered}$$
This definition can easily be converted to recursive implementation. Here the problem is determining the value of $n$ !, and the subproblem is determining the value of $(n-l)$. In the recursive case, when $n$ is greater than 1 , the function calls itself to determine the value of ( $n-l)$ ! and multiplies that with $n$.
In the base ease, when $n$ is 0 or 1 , the function simply retums 1 . This looks like the following:
$/ /$ calculates factorial of a positive integer
def factorial(n):
if $\mathrm{n}=0$ : retum 1
retum $n$ “factorial(n-1)
priat (fictomảal(6))

## 统计代写|数据结构作业代写data structure代考|Problems & Solutions

In this chapter we cover a few problems wath recursion and we will discuss the rest ia other chapters. By the time you complete reading the entire book, you will encouater many recursion problems.
Problem-1 Discuss Towers of Hanoi puzale.
Solution: The Towers of Hanoi is a mathematical puzzle. It consists of three rods (or pegs or towers) and a number of disks of different sizes which can slide onto any rod. The puzale starts with the disks on one rod in ascending order of size, the smallest at the top, thus making a conical shape. The objective of the puzzle is to move the entire stack to another rod, satisfying the following rules:

• Only one disk may be moved at a tine.
• Each move consists of taking the upper disk from one of the rods and sliding it onto another rod, on top of the other disks that may already be present on that rod.
• No disk may be placed on top of a smaller disk.
Algorithm:
• Mone the top $n-1$ disks from Source to Auxiliary tower,
• Move the $n^{\text {th }}$ disk from Source to Destination tower,
• Mone the $n-1$ disks from Auxiliary tower to Destination tower.
• Transfering the top $n-1$ disks from Source to Auxiliary tower can again be thought of as a fresh problem and can be solved in the sime manner. Once we solve Towers of Hanoi with three disks, we can solve it with any number of disks with the above ialgorithm.
def towersOHHanoi(numberOIDisks, startPeg-1, endPeg-3):
if numberOfDisks:
towersOHHanoi (numberOIDisks-1, startPeg, 6-startPeg-endPeg)
print (“Move disk \%d from peg \%id to peg \%d” \% (numberOfDisks, startPeg, endPeg))
towersOHHanoi (numberOIDisks-1, 6-stantPeg-endPeg, endPeg)
towersOHHanoi (numberOHDisks-1)

## 统计代写|数据结构作业代写data structure代考|What is Backtracking

Backtracking is an inproveruent of the brute force approach. It systematically searches for a solution to a problem anbong all available options. In backtracking, we stant with one possable option out of many avalable options and try to solve the problem if we are able to solve the problem with the selected move then we wall priat the solution clse we wall backurack and select some other option and try solve it. If thone if the options work out, we wall clain that there is no solution for the problem.

Backuracking is a form of recursion. The ustal scenario is that youn are faced with a number of optionas, and you must choose one of these. After you make your choice you will get a new set of options; just what set of options you get depends on what choice you nade. This procedure is repeated over and over until you reach a final state. If you made a good sequence of choices, your final state is a goal state; if you

didn’t, it isn’. Backtracking can be thotght of as a selective trec/graph traversal racthod. The tree is a way of representing sonse initial starting position (the root node) and a fiaal goal state (one of the leaves). Backtracking allows us to deal with satuations in which a raw brute-force approash woudd cxphode mto an impossible mumber of options to consaidcr. Backuncking is a sort of refincd brute force. At sach mode, we elimainate choices that are obwiously asot poscible and proceed to recursively checl only those that have potential.
$\mathrm{~ W h a t ‘ s ~ i n t e r e s t i n g ~ a h o u n t h a r k t r a r k i n g ~ i s ~ t h a t ~ w e ~ h a r k ~ m p ~ o n l y ~ a s ~ f a r ~ a s ~ a c e d e r t ~ t r ~ w e a r h ~ a ~ p o r e v i n u s ~ d o r i}$ alternative. In general, that will be at the most recent decision point. Eventually, more and nore of these decision points will have becn fully explored, and we will have to backtrack further asd further. If we backtrack all the way to our initial state and have explored all alternatives from there, we can conchude the particular problen is unsolvable. In such a case, we will have done all the work of the exhatstive recursion and known that there is no viable solution possible.

• Sonvetimes the best algorithm for a problem is to try all possibilities.
• This is always slow, but there are standard tools that can be uscd to help.
• Tools: algorithams for gcacrating basic objects, such as binary strings |2n possabilities for n-bit stringl. permatations
• Backtracking speeds the exhaustive search by pruaing.

## 统计代写|数据结构作业代写data structure代考|Format of a Recursive Function

if(test for the base casc):
return some base case value
cliftest for aaother base case):
return sone oxher base case value

clse:
return (一些工作，然后是递归调用)

n!=1, 如果 n=0 n!=n=(n−1)! 如果 n=0

//计算正整数
def factorial(n) 的阶乘：

1n“阶乘(n-1)
priat (fictomảal(6))

## 统计代写|数据结构作业代写data structure代考|Problems & Solutions

• 一个齿只能移动一个磁盘。
• 每次移动都包括从一根杆上取下上面的圆盘，然后将其滑到另一根杆上，在该杆上可能已经存在的其他圆盘的顶部。
• 任何磁盘都不能放在较小的磁盘上。
算法：
• 蒙顶n−1从源到辅助塔的磁盘，
• 移动nth 从源到目标塔的磁盘，
• 钱n−1从辅助塔到目标塔的磁盘。
• 转移顶部n−1从源到辅助塔的磁盘可以再次被认为是一个新问题，并且可以以同样的方式解决。一旦我们用三个圆盘解决了河内塔，我们就可以用上述算法用任意数量的圆盘来解决它。
def towersOHHanoi(numberOIDisks, startPeg-1, endPeg-3):
if numberOfDisks:
towersOHHanoi (numberOIDisks-1, startPeg, 6-startPeg-endPeg)
print (“将磁盘 \%d 从 peg \%id 移动到 peg \%d” \% (numberOfDisks, startPeg, endPeg))
towersOHHanoi (numberOIDisks-1, 6-stantPeg-endPeg, endPeg)
towersOHHanoi (numberOHDisks-1)

## 统计代写|数据结构作业代写data structure代考|What is Backtracking

Backuracking 是递归的一种形式。通常情况下，您面临许多选项，您必须选择其中之一。做出选择后，您将获得一组新选项；您获得的选项集取决于您做出的选择。一遍又一遍地重复此过程，直到达到最终状态。如果您做出了良好的选择顺序，那么您的最终状态就是目标状态；如果你

在H一种吨‘s 一世n吨和r和s吨一世nG 一种H这在n吨H一种rķ吨r一种rķ一世nG 一世s 吨H一种吨 在和 H一种rķ 米p 这nl是 一种s F一种r 一种s 一种C和d和r吨 吨r 在和一种rH 一种 p这r和在一世n在s d这r一世选择。一般来说，这将是最近的决策点。最终，越来越多的这些决策点将被充分探索，我们将不得不进一步回溯 asd。如果我们一直回溯到我们的初始状态并从那里探索了所有替代方案，我们可以推断特定问题是无法解决的。在这种情况下，我们将完成详尽递归的所有工作，并且知道没有可行的解决方案。

• Sonvetimes 解决问题的最佳算法是尝试所有可能性。
• 这总是很慢，但是有一些标准工具可以用来提供帮助。
• 工具：用于 gcacrating 基本对象的算法，例如二进制字符串 |n 位字符串 l 的 2n 种可能性。排列
• 回溯通过 pruaing 加速穷举搜索。

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|数据结构作业代写data structure代考|Divide and Conquer Master Theorem

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

## 统计代写|数据结构作业代写data structure代考|Problems & Solutions

For each of the following recurrences, give an expression for the runtime $T(n)$ if the recurrence can be solved with the Master Theorem. Otherwise, indicate that the Master Theorem does not apply.
Problem-1 $T(n)=3 T(n / 2)+n^{2}$
Solutions $T(n)=3 T(n / 2)+n^{2} \Rightarrow T(n)=\Theta\left(n^{2}\right)$ (Mnster Theorem Case 3.a)
Problem-2 $T(n)=4 T(n / 2)+n^{2}$
Solution: $T(n)=4 T(n / 2)+n^{2}=>T(n)=\theta\left(n^{2} \log n\right)$ (Master Theorem Case 2.a)
Problem-3 $T(n)=T(n / 2)+n^{2}$
Solution: $T(n)=T(n / 2)+n^{2}=>\theta\left(n^{2}\right)$ (Master Theorcm Casc 3.a)
Problem-4 $T(n)=2^{n} T(n / 2)+n^{n}$
Solution: $T(n)=2^{n} T(n / 2)+n^{n} \rightarrow$ Does not apply $(a$ is not constant)
Problems $T(n)=16 T(n / 4)+n$
Solution: $T(n)=16 T(n / 4)+n=T(n)=\Theta\left(n^{2}\right)$ (Master Theorem Case 1)
Problem-6 $T(n)=2 T(n / 2)+n \log n$
Solution: $T(n)=2 T(n / 2)+n \log n=>T(n)=\theta\left(n \log ^{2} n\right)$ (Master Theorem Case 2.a)
Problem-7 $T(n)=2 T(n / 2)+n / \log n$
Solution: $T(n)=2 T(n / 2)+n / \log n=>T(n)=\theta($ nloglog$n)$ (Master Theorem Case 2.b)
Problem-8 $T(n)=2 T(n / 4)+n^{0.51}$
Solution: $T(n)=2 T(n / 4)+n^{0.51}=>T(n)=\theta\left(n^{0.51}\right)$ (Mnster Theorcm Casc 3.b)
Problem-9 $T(n)=0.5 T(n / 2)+1 / n$
Solution: $T(n)=0.5 T(n / 2)+1 / n=>$ Does not apply $(a<1)$ Problem-10 $T(n)=6 T(n / 3)+n^{2} \log n$ Solution: $T(n)=6 T(n / 3)+n^{2} \log n=>T(n)=\Theta\left(n^{2} \log n\right)$ (Master Theorem Case 3.a)
Problem-11 $T(n)=64 T(n / 8)-n^{2} \log n$
Solution: $T(n)=64 T(n / 8)-n^{2} \log n=>$ Does not apply (function is not positive)

Problem-12 $T(n)=7 T(n / 3)+n^{2}$
Solution: $T(n)=7 T(n / 3)+n^{2}=>T(n)=\Theta\left(n^{2}\right)$ (Master Theorem Case 3.as)
Problem-18 $T(n)=4 T(n / 2)+\log n$
Solution: $T(n)=4 T(n / 2)+\log n=T(n)=\theta\left(n^{2}\right)$ (Master Theorem Case 1)
Problem-14 $T(n)=16 T(n / 4)+n !$
Solution: $T(n)=16 T(n / 4)+n ! \Rightarrow T(n)=\Theta(n !)$ (Master Theorem Case 3.a)
Problem-15 $T(n)=\sqrt{2} T(n / 2)+\log n$
Solution: $T(n)=\sqrt{2} T(n / 2)+\log n=>T(n)=\Theta(\sqrt{n})$ (Master Theorem Case 1)
Problem-16 $T(n)=3 T(n / 2)+n$
Solution: $T(n)=3 T(n / 2)+n=>T(n)=\Theta\left(n^{\log 3}\right)$ (Master Theorem Case 1)
Problem-17 $T(n)=3 T(n / 3)+\sqrt{n}$
Solutiont $T(n)=3 T(n / 3)+\sqrt{n} \Rightarrow T(n)=\Theta(n)$ (Master Theorem Case 1)
Problem-18 $T(n)=4 T(n / 2)+c n$
Solution: $T(n)=4 T(n / 2)+c n=T(n)=\Theta\left(n^{2}\right)$ (Master Theorem Case 1)
Problem-19 $\bar{T}(n)=3 \bar{T}(n / 4)+n \log n$
Solution: $T(n)=3 T(n / 4)+n \log n=>T(n)=\Theta(n \log n)$ (Master Theorem Case 3.a)
Problem-20 $T(n)=3 T(n / 3)+n / 2$
Solution: $T(n)=3 T(n / 3)+n / 2=>T(n)=\Theta(n \log n)$ (Master Theorem Case 2.a)

## 统计代写|数据结构作业代写data structure代考|Method of Guessing and Confirming

Now, let us discuss a method which can be used to solve any recurrence. The basic idea behaind this method is:
guess the answer; and then prove it correct by induction.
In other words, it addresses the question: What if the given recurrence doesn’t seem to match with any of these inaster theoreml methods? If we guess a solution and then try to verify our guess inductively, usually either the proof will succeed (in which case we are done), or the proof will fail (in which case the failure will help us refine our guess).
As iul cximule, comsides toe rexuncue T(n) $-\sqrt{n} T(\sqrt{n})+\mathrm{~ t . ~ T h i s ~ d o}$ observing the recurrence gives us the impression that it is similar to the divide and conquer method (dividing the problem into $\sqrt{n}$ subproblems each with size $\sqrt{n})$. As we can see, the size of the subproblems at the first level of recursion is $n$. So, let us guess that T(n) – O(nlogn), and then try to prowe that our guess is correct.
Let’s start by trying to prove an upper bound T $(n) \leq$ cnlogn:
\begin{aligned} T(n) &=\sqrt{n} T(\sqrt{n})+n \ & \leq \sqrt{n} \cdot c \sqrt{n} \log \sqrt{n}+n \ &=n \cdot c \log \sqrt{n}+n \ &=n \cdot c \cdot \frac{1}{2} \cdot \log n+n \ & \leq c n \log n \end{aligned}
The last inequality assumes only that $1 \leq c, \frac{1}{2}$. $\log n$. This is correct if $n$ is sufficicntly lange and for aary constant $c$, no maatter how small. From the abowe proof, we can see that our guess is correct for the upper bound. Now, let us prone the lower bound for this recurrence.

## 统计代写|数据结构作业代写data structure代考|Amortized Analysis

Amortized analysis refers to deternaining the time-iveraged runaing time for a sequence of operations. It is differeat froma average case analysis, because amortized analysis does not make any assmption about the distribution of the data values, whereas average case analysis assumes the data are not “bad” (c.g., sone sorting algonthms do well on average oxer all input orderangs but very badly on certain input orderings). That is, anbortized analysis is a worst-case analysis, but for a sequence of operations rather than for individual operations.

The nootination for anortizcd anahsis is to better undersiand the nunning tinve of certain techniqucs, where standard worstecase analysis provides an overly pessimistic bouad. Amortized analysas generally applies to a method that consists of a sequence of operations, where the vast majority of the operations are cheap, but some of the operations are expeasive. If we can show that the expensive operations are particularly rare, we can change them to the cheap operations, and only bound the cheap operations.

The general approach is to assign an autificial cost to each operation in the sequence, such that the total of the artificial costs for the sequence of operations bounds the total of the real costs for the scquence. Thas iutificial cost is called the anortized cost of an operation. To analyze the rumning time, the amortized cost thus is a correct way of understanding the overall ruming time – but uote that particular operations can still take louger so it is not a way of bounding the nuning time of any individual operation in the sequence.
Whes one event in a sequence affects the cost of later events:

• Oae particular task nay be expensive.
• But it may leave data structure in a state that the next few operations become easier.
Dermple: Let us consider an array of elenucuts from which we wat to find the $k^{\text {th }}$ smallest elenaent. We can solve this problem using sorting. After sorting the given array, we just need to retura the $k^{\text {th }}$ clement fron it. The cost of performing the sort (assunuing conparison-based sorting algorithm) is $\mathrm{O}\left(n \log n\right.$. If we perform $n$ such selections then the average cost of each selection is $\mathrm{O}(\mathrm{nlog} / \mathrm{n})=\mathrm{O}(\log )^{\mathrm{l}}$. This clearly indicates that sorting ouce is roducing the conplexity of subsequcat operations.

## 统计代写|数据结构作业代写data structure代考|Omega-Ω Notation

Ω示例
Exrmple-1 查找下限F(n)=5n2.

Eample-2 证明F(n)=100n+5≠Ω(n2).

100n+5≤100n+5n(∀n≥1)=105n Cn2≤105n⇒n(Cn−105)≤0

⇒矛盾：n不能小于常数
Example-32n=Ω(n),n2=Ω(n2),日志⁡n=Ω(日志⁡n).

## 统计代写|数据结构作业代写data structure代考|Theta- Notation

θ示例

∴n22−n2=θ(n2)和C1=1/5,C2=1和n0=2
Ermple 2 证明n≠θ(n2)

∴n≠θ(n2)
2ramp 8 证明6n3−θ(n2)

∴6n3≠θ(n2)
Jrample 4 证明n≠θ(日志⁡n)

## 统计代写|数据结构作业代写data structure代考|Master Theorem for Divide and Conquer Recurrences

1) 如果一种>bķ， 然后吨(n)=θ(n日志⁡Gb)
2) 如果一种=bķ

3) 如果一种<bķ

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|数据结构作业代写data structure代考|Omega-Ω Notation

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

## 统计代写|数据结构作业代写data structure代考|Omega-Ω Notation

Similar to the $\mathrm{O}$ discussion, this notation gives the tighter lower bound of the given algonthm and we represent it as $f(n)=\Omega(g(n))$. That nueans, at larger values of $n$, the tighter lower bound of $f(n)$ is $g(n)$. The $\Omega$ notation can be defined as $\Omega(g(n))={f(n)$ there exist positive constants $\mathrm{c}$ and $n_{0}$ such that $0 \leq c g(n) \leq f(n)$ for all $\left.\mathrm{n} \geq n_{0}\right} \cdot g(n)$ is an asymptotic tight lower bound for $f(n)$. Our objective is to give the largest rate of growth $g(n)$ which is less than or equal to the given algonthm’s rate of growth $f(n)$.
For eximule, if $f(n)-100 n^{3}+10 n+50, g(n)$ is $\Omega\left(n^{2}\right)$.

$\Omega$ Examples
Exrmple-1 Find lower bound for $f(n)=5 n^{2}$.
Solutiont $\exists c, n_{0}$ Such that: $0 \leq c n^{2} \leq 5 n^{2} \Rightarrow c n^{2} \leq 5 n^{2} \Rightarrow c=5$ and $n_{0}=1$ $\therefore 5 n^{2}=\Omega\left(n^{2}\right)$ with $c=5$ and $n_{0}-1$
Emample-2 Prove $f(n)=100 n+5 \neq \Omega\left(n^{2}\right)$.
Solutiont $\exists c, n_{0}$ Such that: $0 \leq{c n^{2}}^{2} \leq 100 n+5$
\begin{aligned} &100 n+5 \leq 100 n+5 n(\forall n \geq 1)=105 n \ &c n^{2} \leq 105 n \Rightarrow n(c n-105) \leq 0 \end{aligned}
Since $n$ is positive $\Rightarrow c n-105 \leq 0 \Rightarrow n \leq 105 / c$
$\Rightarrow$ Contradiction: $n$ cannot be smaller than a constant
Example-3 $2 n=\Omega(n), n^{2}=\Omega\left(n^{2}\right), \log n=\Omega(\log n)$.

## 统计代写|数据结构作业代写data structure代考|Theta- Notation

This notation decides whether the upper and lower bounds of a given function (ahorithm) are the same. The average ruming time of an algorithm is aluays between the lower bound and the upper bound. If the upper bound (O) and lower bound (S2) give the same result, then the $\Theta$ notation wall also have the sanve rate of growth. $A$ s an eximple, let us assume that $f(n)=10 n+n$ is the expression. Then, its tight upper bound $g(n)$ is $\mathrm{O}(n)$. The rate of growth in the best case is $g(n)=\mathrm{O}(n)$.

In this case, the rates of growth in the best case and worst case are the same. As a result, the average case will also be the same. For a given function (algorithm), if the rates of growth (bounda) for $\mathrm{O}$ and $\mathrm{\Omega}$ are not the same, then the rate of growth for the $\Theta$ cnse may not be the samae. In this case, we neod to consader all possible tinve complexities and tate the average of those (for exanmple, for a quick sont itrerige case, refer to the Sorting chapuca).
Now consuder the defuition of $\Theta$ lwotition. It is defined as $\theta(g(n))=f f(n)$ : there cxist positive constants $c_{1}$, $c_{2}$ and $n_{0}$ such that 0 s $c_{1} g(n) \leqslant f(n) \leqslant c_{2} g(n)$ for all $\left.n \approx n_{0}\right] \cdot g(n)$ is an isynptotic tight botmd for $f(n)$. $\Theta(g(n))$ is the set of functions with the sence order of growtha as $g(n)$.
$\Theta$ Examples
Example 1 Find $\Theta$ bound for $f(n)=\frac{n^{2}}{2}-\frac{n}{2}$
Solution: $\frac{n^{2}}{5} \leq \frac{n^{2}}{2}=\frac{n}{2} \leq n^{2}$, for all, $n \geq 2$
$\therefore \frac{n^{2}}{2}-\frac{n}{2}=\Theta\left(n^{2}\right)$ with $c_{1}=1 / 5, c_{2}=1$ and $n_{0}=2$
Ermple 2 Prove $n \neq \Theta\left(n^{2}\right)$

Solution: $c_{1} n^{2} \leq n \leq c_{2} n^{2} \Rightarrow$ only holds for: $n \leq 1 / c_{1}$
$$\therefore n \neq \theta\left(n^{2}\right)$$
2rample 8 Prove $6 n^{3}-\Theta\left(n^{2}\right)$
Solution: $c_{1} n^{2} \leq 6 n^{3} \leq c_{2} n^{2} \Rightarrow$ only holds for: $n \leq c_{2} / 6$
$$\therefore 6 n^{3} \neq \Theta\left(n^{2}\right)$$
Jrample 4 Prove $n \neq \Theta(\log n)$
Solution: $c_{1} \log n \leq n \leq c_{7} \log n \Rightarrow c_{7} \geq \frac{n}{\text { lugn }} \cdot \forall n \geq n_{n}$ – Impossible
Important Notes
For analysis (best case, worst case and average), we try to give the upper bound (O) and lower bound (S2) and average rumáng time ( $\Theta$ ). From the above examaples, it should also be clear that, for a given function (algonthm), getting the upper bound (O) and lower bound (S2) and averuge numing tine $(\Theta)$ nay not always be possible. For examuple, if we are discussing the best case of an algonthm, we try to give the upper bound $(O)$ and lower bound $(\Omega)$ and average nmaing time $(\Theta)$.

In the renaining chapters, we generally focus on the upper bound (O) because knowing the lower bound (S2) of an algorithn is of no practical importance, and we use the $\Theta$ notation if the upper boumd $(\mathrm{O})$ and lower bound $(\Omega)$ are the same.

## 统计代写|数据结构作业代写data structure代考|Master Theorem for Divide and Conquer Recurrences

All divide and conquer algorithms (Also discussed in detial in the Divide and Conquer chapter) divide the problem into sub-problens, cach of which is part of the original problem, and then perform sonve additional work to compute the final answer. As an example, a merge sort algorithm [for details, refer to Sorting chapter] opcrates ons two sub-problems, cach of vhich is half the size of the original, and then perorms $\mathrm{O}(\mathrm{n})$ additional work for merging. This gives the running time equation:
$$T(n)=2 T\left(\frac{n}{2}\right)+O(n)$$
The following theorem can be used to deternine the running time of divide and conquer algonthms. For a given program (algonithm), first we try to find the recurrence relation for the problem. If the recurrence is of the below form then we can directly give the answer without fully solving it. If the recurrence is of the form $T(n)=a T\left(\frac{n}{b}\right)+\Theta\left(n^{k} \log ^{p} n\right)$, where $a \geq 1, b>1, k \geq 0$ and $p$ is a real number, then:
1) If $a>b^{k}$, then $T(n)=\Theta\left(n^{\log g_{b}}\right)$
2) If $a=b^{k}$
a. If $p>-1$, then $T(n)=\theta\left(n^{\log {b}^{a}} \log ^{p+1} n\right)$ b. If $p=-1$, then $T(n)=\Theta\left(n^{\log } \operatorname{l} \log \log n\right)$ c. If $p<-1$, then $T(n)=\theta\left(n^{\log {b}^{a}}\right)$
3) If $a<b^{k}$
a. If $p \geq 0$, then $T(n)=\theta\left(n^{k} \log ^{p} n\right)$
b. If $p<0$, then $T(n)=O\left(n^{k}\right)$

## 统计代写|数据结构作业代写data structure代考|Omega-Ω Notation

Ω示例
Exrmple-1 查找下限F(n)=5n2.

Eample-2 证明F(n)=100n+5≠Ω(n2).

100n+5≤100n+5n(∀n≥1)=105n Cn2≤105n⇒n(Cn−105)≤0

⇒矛盾：n不能小于常数
Example-32n=Ω(n),n2=Ω(n2),日志⁡n=Ω(日志⁡n).

## 统计代写|数据结构作业代写data structure代考|Theta- Notation

θ示例

∴n22−n2=θ(n2)和C1=1/5,C2=1和n0=2
Ermple 2 证明n≠θ(n2)

∴n≠θ(n2)
2ramp 8 证明6n3−θ(n2)

∴6n3≠θ(n2)
Jrample 4 证明n≠θ(日志⁡n)

## 统计代写|数据结构作业代写data structure代考|Master Theorem for Divide and Conquer Recurrences

1) 如果一种>bķ， 然后吨(n)=θ(n日志⁡Gb)
2) 如果一种=bķ

3) 如果一种<bķ

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。

## 统计代写|数据结构作业代写data structure代考|System-defined data types

statistics-lab™ 为您的留学生涯保驾护航 在代写数据结构data structure方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写数据结构data structure方面经验极为丰富，各种代写数据结构data structure相关的作业也就用不着说。

• 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

## 统计代写|数据结构作业代写data structure代考|Abstract Data Types

Before definimg abstract data types, let us consider the different view of systen-defiacd data types. We all know that, by default, all primátive data types (int, float, etc.) support basic operations such as addition and subtraction. The systen provides the inplenentations for the primative data types. For user-defined data types we also need to define operations. The implenacntation for these operations can be done when we Want to actually use them. That mueans, in general, user defined data types are defined along with their operations.
To simplify the process of solving problens, we conbine the data structures with their operations and we call this Abstract Data Types (AD)’s). An Alyl consists of two parts

1. Declaration of data
2 Declaration of operations
Conmonly used ADTs include: Linked Lists, Stacks, Queues, Pronity Queues, Binary Trees, Dictionaries, Dijoint Sets (Union and Find), Hash Tables, Graphs, and many others. For cxample, stack uses L.IFO (Last-In-First-Out) mechanism while storing the dafa in data structurcs. The last element inserted into the stack is the first element that gets deleted. Common operations of it are: creating the stack, pushing an element outo the stack, popping an element from stack, finding the current top of the stack, finding number of elements in the stack, etc.
While defining the ADTs do not worry about the implementation details. They come into the picture only when we want to use them. Different kinds of ADT $\mathrm{s}$ are suiterl to different kinds of applications, and anme are highly sperialized to sperific task. Ry the end of this book, we will go through many of them and you will be in a position to relate the data structures to the kind of problens they solve.

## 统计代写|数据结构作业代写data structure代考|Types of Analysis

algorithn takes a bong tinc. We have ailready secn that ian ilgorithn can be represented in the form of in expression. That means we represent the stgorithin with multiple expressionst one for the case where it takes less tinio and another for the chse where it takes hwore tinie.

In general, the first case is called the best case and the second case is called the worst case for the algonthm. To analyze an algonthm, we need some kind of syatax, and that forms the base for asymptotic analysis/motation. There are three types of analysis:

• Worst crse
• Defines the imput for which the algorithm takes a long tinve (slowest time to complete). – Input is the one for which the algorithm runs the slowest.
• Bent case
• Defines the input for which the algorithm takes the least tinve (fastest time to complete).
• Input is the one for which the algorithm runs the fastest.
• Average case
• Prowides a prediction about the running time of the algorithm.
• Run the algorithm many times, using many different inputs that come from some distribution that generates these inputs, compute the total rmning tine (by adding the individual times), and divide by the number of trials. Assunes that the input is randon.
Lower Bound $<=$ Average Time $<=$ Upper Bound
For a given algorithm, we can represent the best, worst and average cases in the form of expressions. $A$ s an example, let $f(n)$ be the function which represents the given algorithm.
$$\begin{gathered} f(n)=n^{2}+500, \text { for worst case } \ f(n)=n+100 n+500 \text {, for best case } \end{gathered}$$
Similarly, for the average case. The expression defines the inputs with which the algorithm takes the average running time (or memory).

## 统计代写|数据结构作业代写data structure代考|Big-O Notation

This notafion gives the tight upper bouad of the given function. Gencrally, it is represented as $f(n)=()(g(n))$. That neans, at larger values of $n$, the upper bound of $f(n)$ is $g(n)$. For example, if $f(n)=n^{4}+100 n^{2}+10 n+50$ is the given algorithm, then $n^{4}$ is $g(n)$. That means $g(n)$ gives the maxinum rate of growth for $f(n)$ at larger values of $n$.

Let us see the $\mathrm{O}$-notation with a little more detail. $\mathrm{O}$-notation defined as $\mathrm{O}(g(n))=f f(n)$ : there exist positive constants $c$ and $n_{0}$ sasch that $0 \leq f(n) \leq c g(n)$ for all $\left.n \geq n_{0}\right} \cdot g(n)$ is an asynptotic tight upper bound for $f(n)$. Our objective is to give the snallest rate of growth $g(n)$ which is greater than or equal to the given algorithms” rate of growth $f(n)$.
Gearerally, we discard bower values of $n$. That maeans the rate of growth at lower values of $n$ is not important. In the figure, $n_{0}$ is the point from which we need to consuder the rate of growth for a given algorithm. Below $n_{0}$, the rate of growth could be different. $n_{0}$ is called threshold for the given function.

## 统计代写|数据结构作业代写data structure代考|Abstract Data Types

1. 数据
声明 2 操作
声明 常用的 ADT 包括：链表、堆栈、队列、Pronity 队列、二叉树、字典、双联集（并集和查找）、哈希表、图等等。例如，堆栈使用 L.IFO（后进先出）机制，同时将 dafa 存储在数据结构中。插入堆栈的最后一个元素是第一个被删除的元素。它的常见操作是：创建堆栈，将一个元素压出堆栈，从堆栈中弹出一个元素，查找当前堆栈顶部，查找堆栈中的元素数量等。

## 统计代写|数据结构作业代写data structure代考|Types of Analysis

• 最坏的情况
• 定义算法需要很长时间（完成时间最慢）的输入。– 输入是算法运行最慢的输入。
• 弯曲案例
• 定义算法花费最少的时间（最快完成时间）的输入。
• 输入是算法运行速度最快的输入。
• 平均情况
• 扩展关于算法运行时间的预测。
• 多次运行算法，使用来自生成这些输入的某个分布的许多不同输入，计算总 rmning tine（通过添加各个时间），然后除以试验次数。假设输入是随机的。
下限<=平均时间<=上限
对于给定的算法，我们可以用表达式的形式表示最好的、最坏的和平均的情况。一种举个例子，让F(n)是表示给定算法的函数。
F(n)=n2+500, 对于最坏的情况  F(n)=n+100n+500, 最好的情况
同样，对于平均情况。该表达式定义了算法采用平均运行时间（或内存）的输入。

## 统计代写|数据结构作业代写data structure代考|Big-O Notation

Gearerally，我们丢弃 Bower 值n. 这意味着在较低值下的增长率n并不重要。图中，n0是我们需要考虑给定算法的增长率的点。以下n0, 增长率可能不同。n0被称为给定函数的阈值。

## 广义线性模型代考

statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## 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。