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

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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

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

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MATLAB代写

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