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

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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
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• 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

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