数学代写|计算复杂度理论代写Computational complexity theory代考|FIT2014

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (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|计算复杂度理论代写Computational complexity theory代考|Random Oracles

Consider the class $\mathcal{C}$ of all subsets of ${0,1}^{*}$ and define subclasses $\mathcal{A}=\left{A \in \mathcal{C}: P^{A}=N P^{A}\right}$ and $B=\left{B \in \mathcal{C}: P^{B} \neq N P^{B}\right}$. One of the approaches to study the relativized $P=$ ? $N P$ question is to compare the two subclasses $\mathcal{A}$ and $\mathcal{B}$ to see which one is larger. In this subsection, we give a brief introduction to this study based on the probability theory on the space $\mathcal{C}$.

To define the notion of probability on the space $\mathcal{C}$, it is most convenient to identify each element $X \in \mathcal{C}$ with its characteristic sequence $\alpha_{X}=\chi_{X}(\lambda) \chi_{X}(0) \chi_{X}(1) \chi_{X}(00) \ldots$ (i.e., the $k$ th bit of $\alpha_{X}$ is 1 if and only if the $k$ th string of ${0,1}^{*}$, under the lexicographic ordering, is in $\left.X\right)$ and treat $\mathcal{C}$ as the set of all infinite binary sequences or, equivalently, the Cartesian product ${0,1}^{\infty}$. We can define a topology on $\mathcal{C}$ by letting the set ${0,1}$ have the discrete topology and forming the product topology on $\mathcal{C}$. This is the well-known Cantor space. We now define the uniform probability measure $\mu$ on $\mathcal{C}$ as the product measure of the simple equiprobable measure on ${0,1}$, that is, $\mu{0}=\mu{1}=1 / 2$. In other words, for any integer $n \geq 1$, the $n$th bit of a random sequence $\alpha \in \mathcal{C}$ has the equal probability to be 0 or 1 . If we identify each real number in $[0,1]$ with its binary expansion, then this measure $\mu$ is the Lebesgue measure on $[0,1] . .^{5}$

For any $u \in{0,1}^{}$, let $\Gamma_{u}$ be the set of all infinite binary sequences that begin with $u$, that is, $\Gamma_{u}^{u}=\left{u \beta: \beta \in{0,1}^{\infty}\right}$. Each set $\Gamma_{u}$ is called a cylinder. All cylinders $\Gamma_{u}, u \in{0,1}^{}$, together form a basis of open neighborhoods of the space $\mathcal{C}$ (under the product topology). It is clear that $\mu\left(\Gamma_{u}\right)=2^{-|u|}$ for all $u \in{0,1}^{}$. The smallest $\sigma$-field containing all $\Gamma_{u}$, for all $u \in{0,1}^{}$, is called the Borel field. ${ }^{6}$ Each set in the Borel field, called a Borel set, is measurable.

The question of which of the two subclasses $\mathcal{A}$ and $B$ is larger can be formulated, in this setting, as to whether $\mu(\mathcal{A})$ is greater than $\mu(\mathcal{B})$. In the following, we show that $\mu(\mathcal{A})=0$.

An important idea behind the proof is Kolmogorov’s zero-one law of tail events, which implies that if an oracle class is insensitive to a finite number of bit changes, then its probability is either zero or one. This property holds for the classes $\mathcal{A}$ and $B$ as well as most other interesting oracle classes.

数学代写|计算复杂度理论代写Computational complexity theory代考|Structure of Relativized NP

Although the meaning of relativized collapsing and relativized separation results is still not clear, many relativized results have been proven. These results show a variety of possible relations between the well-known complexity classes. In this section, we demonstrate some of these relativized results on complexity classes within $N P$ to show what the possible relations are between $P, N P, N P \cap \operatorname{co} N P$, and $U P$. The relativized results on the classes beyond the class $N P$, that is, those on $N P, P H$, and $P S P A C E$, will be discussed in later chapters.

The proofs of the following results are all done by the stageconstruction diagonalization. The proofs sometimes need to satisfy simultaneously two or more potentially conflicting requirements that make the proofs more involved.
Theorem 4.20 (a) $(\exists A) P^{A} \neq N P^{A} \cap \operatorname{coN} P^{A} \neq N P^{A}$.
(b) $(\exists B) P^{B} \neq N P^{B}=\cos P^{B}$.
(c) $(\exists C) P^{C}=N P^{C} \cap \operatorname{coN} P^{C} \neq N P^{C}$.
Proof. (a): This can be done by a standard diagonalization proof. We leave it as an exercise (Exercise 4.16(a)).
(b): Let $\left{M_{i}\right}$ be an effective enumeration of all polynomial-time oracle DTMs and $\left{N_{i}\right}$ an effective enumeration of all polynomial-time oracle NTMs. For any set $B$, let $K_{B}=\left{\left\langle i, x, 0^{j}\right\rangle: N_{i}^{B}\right.$ accepts $x$ in $j$ moves $}$. Then, by the relativized proof of Theorem $2.11, K_{B}$ is $\leq_{m}^{P}$-complete for the class $N P^{B}$. Let $L_{B}=\left{0^{n}:(\exists x)|x|=n, x \in B\right}$. Then, it is clear that $L_{B} \in N P^{B}$. We are going to construct a set $B$ to satisfy the following requirements:
$R_{0, t}$ : for each $x$ of length $t, x \notin K_{B} \Longleftrightarrow(\exists y,|y|=t) x y \in B$, $R_{1, i}:(\exists n) 0^{n} \in L_{B} \Longleftrightarrow M_{i}^{A}$ does not accept $0^{n} .$
Note that if requirements $R_{0, t}$ are satisfied for all $t \geq 1$, then $\overline{K_{B}} \in N P^{B}$ and, hence, $N P^{B}=\operatorname{coN} N P^{B}$, and if requirements $R_{1, i}$ are satisfied for all $i \geq 1$, then $L_{B} \notin P^{B}$ and, hence, $P^{B} \neq N P^{B}$.

We construct set $B$ by a stage construction. At each stage $n$, we will construct finite sets $B_{n}$ and $B_{n}^{\prime}$ such that $B_{n-1} \subseteq B_{n}, B_{n-1}^{\prime} \subseteq B_{n}^{\prime}$, and $B_{n} \cap$ $B_{n}^{\prime}=\emptyset$ for all $n \geq 1$. Set $B$ is define as the union of all $B_{n}, n \geq 0$.

The requirement $R_{0, t}, t \geq 1$, is to be satisfied by direct diagonalization at stage $2 t$. The requirements $R_{1, i}, i \geq 1$, are to be satisfied by a delayed diagonalization in the odd stages. We always try to satisfy the requirement $R_{1, i}$ with the least $i$ such that $R_{1, i}$ is not yet satisfied. We cancel an integer $i$ when $R_{1, i}$ is satisfied. Before stage 1 , we assume that $B_{0}=B_{0}^{\prime}=\emptyset$ and that none of integers $i$ is cancelled.

数学代写|计算复杂度理论代写Computational complexity theory代考|Graphs and Decision Trees

We first review the notion of graphs and the Boolean function representations of graphs. ${ }^{1}$ A graph is an ordered pair of disjoint sets $(V, E)$ such that $E$ is a set of pairs of elements in $V$ and $V \neq \emptyset$. The elements in the set $V$ are called vertices and the elements in the set $E$ are called edges. Two vertices are adjacent if there is an edge between them. Two graphs are isomorphic if there exists a one-to-one correspondence between their vertex sets that preserves adjacency.

A path is an alternating sequence of distinct vertices and edges starting and ending with vertices such that every vertex is an end point of its neighboring edges. The length of a path is the number of edges appearing in the path. A graph is connected if every pair of vertices are joined by a path.

Let $V={1, \ldots, n}$ be the vertex set of a graph $G=(V, E)$. Then its adjacency matrix $\left[x_{i j}\right]$ is defined by
$$x_{i j}= \begin{cases}1 & \text { if }{i, j} \in E, \ 0 & \text { otherwise. }\end{cases}$$
Note that $x_{i j}=x_{j i}$ and $x_{i i}=0$. So, the graph $G$ can be determined by $n(n-$ 1)/ 2 independent variables $x_{i j}, 1 \leq i<j \leq n$. For any property $\mathcal{P}$ of graphs with $n$ vertices, we define a Boolean function $f_{p}$ over $n(n-1) / 2$ variables $x_{i j}, 1 \leq i<j \leq n$, as follows:
$$f_{p}\left(x_{12}, \ldots, x_{n-1, n}\right)= \begin{cases}1 & \begin{array}{l} \text { if the graph } G \text { represented by }\left[x_{i, j}\right] \text { has } \ \text { the property } \mathcal{P} \end{array} \ 0 & \text { otherwise. }\end{cases}$$
Then $\mathcal{P}$ can be determined by $f_{p}$. For example, the property of connectivity corresponds to the Boolean functions $f_{\text {con }}$ of $n(n-1) / 2$ variables such that $f_{c o n}\left(x_{12}, \ldots, x_{n-1, n}\right)=1$ if and only if the graph $G$ represented by $\left[x_{i j}\right]$ is connected.

Not every Boolean function of $n(n-1) / 2$ variables represents a graph property because a graph property should be invariant under graph isomorphism. A Boolean function $f$ of $n(n-1) / 2$ variables is called a graph property if for every permutation $\sigma$ on the vertex set ${1, \ldots, n}$,
$$f\left(x_{12}, \ldots, x_{n-1, n}\right)=f\left(x_{\sigma(1) \sigma(2)}, \ldots, x_{\sigma(n-1) \sigma(n)}\right) \text {. }$$

数学代写|计算复杂度理论代写Computational complexity theory代考|Structure of Relativized NP

(二)(∃乙)磷乙≠ñ磷乙=因⁡磷乙.
（C）(∃C)磷C=ñ磷C∩和⁡磷C≠ñ磷C.

(b): 让\left{M_{i}\right}\left{M_{i}\right}是所有多项式时间预言机 DTM 的有效枚举，并且\left{N_{i}\right}\left{N_{i}\right}所有多项式时间预言机 NTM 的有效枚举。对于任何集合乙， 让K_{B}=\left{\left\langle i, x, 0^{j}\right\rangle: N_{i}^{B}\right.$接受$x$在$j$移动$}K_{B}=\left{\left\langle i, x, 0^{j}\right\rangle: N_{i}^{B}\right.$接受$x$在$j$移动$}. 然后，由定理的相对化证明2.11,ķ乙是≤米磷- 完成课程ñ磷乙. 让L_{B}=\left{0^{n}:(\exists x)|x|=n, x \in B\right}L_{B}=\left{0^{n}:(\exists x)|x|=n, x \in B\right}. 那么，很明显大号乙∈ñ磷乙. 我们将构建一个集合乙满足以下要求：
R0,吨: 对于每个X长度吨,X∉ķ乙⟺(∃是,|是|=吨)X是∈乙, R1,一世:(∃n)0n∈大号乙⟺米一世一个不接受0n.

数学代写|计算复杂度理论代写Computational complexity theory代考|Graphs and Decision Trees

X一世j={1 如果 一世,j∈和, 0 否则。

Fp(X12,…,Xn−1,n)={1 如果图表 G 代表为 [X一世,j] 有   财产 磷 0 否则。

F(X12,…,Xn−1,n)=F(Xσ(1)σ(2),…,Xσ(n−1)σ(n)).

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。