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

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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|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

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## MATLAB代写

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