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## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|Time versus alternations: time-space tradeoffs for SAT

Despite the fact that SAT is widely believed to require exponential (or at least super-polynomial) time to solve, and to require linear (or at least super-logarithmic) space, we currently have no way to prove these conjectures. In fact, as far as we know, SAT may have both a linear time algorithm and a logarithmic space one. Nevertheless, we can prove that SAT does not have an algorithm that runs simultaneously in linear time and logarithmic space. In fact, we can prove the following stronger theorem:
THEOREM $5.13$ (??)
For every two functions $S, T: \mathbb{N} \rightarrow \mathbb{N}$, define $\operatorname{TISP}(T(n), S(n))$ to be the set of languages decided by a $T M M$ that on every input $x$ takes at most $O(T(|x|))$ steps and uses at most $O(S(|x|))$ cells of its read/write tapes. Then, SAT $\notin \mathbf{T I S P}\left(n^{1.1}, n^{0.1}\right)$.

The class $\operatorname{TISP}(T(n), S(n))$ is typically defined with respect to TM’s with RAM memory (i.e., TM’s that have random access to their tapes; such machines can be defined in a similar way to the definition of oracle TM’s in Section 3.5). Theorem $5.13$ and its proof carries over for that model as well. We also note that a stronger result is known for both models: for every $c<(\sqrt{5}+1) / 2$, there exists $d>0$ such that SAT $\notin \mathbf{T I S P}\left(n^c, n^d\right)$ and furthermore, $d$ approaches 1 from below as $c$ approaches 1 from above.
ProOF: We will show that
$$\boldsymbol{N T I M E}(n) \nsubseteq \mathbb{\operatorname { T I S }}\left(n^{1.2}, n^{0.2}\right)$$

## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|Defining the hierarchy via oracle machines

Recall the definition of oracle machines from Section 3.5. These are machines that are executed with access to a special tape they can use to make queries of the form “is $q \in O$ ” for some language $O$. For every $O \subseteq{0,1}^*$, oracle TM $M$ and input $x$, we denote by $M^O(x)$ the output of $M$ on $x$ with access to $O$ as an oracle. We have the following characterization of the polynomial hierarchy:
THEOREM $5.15$
For every $i \geq 2, \boldsymbol{\Sigma}i^p=\mathbf{N P}^{\boldsymbol{\Sigma}{i-1} \mathrm{SAT}}$, where the latter class denotes the set of languages decided by polynomial-time NDTMs with access to the oracle $\boldsymbol{\Sigma}_{i-1}$ SAT.

Proof: We showcase the idea by proving that $\boldsymbol{\Sigma}_2^p=\mathbf{N P}^{\mathrm{SAT}}$. Suppose that $L \in \boldsymbol{\Sigma}_2^p$. Then, there is a polynomial-time TM $M$ and a polynomial $q$ such that
$$x \in L \Leftrightarrow \exists u_1 \in{0,1}^{q(|x|)} \forall u_2 \in{0,1}^{q(|x|)} M\left(x, u_1, u_2\right)=1$$
yet for every fixed $u_1$ and $x$, the statement “for every $u_2, M\left(x, u_1, u_2\right)=1$ ” is the negation of an NP-statement and hence its truth can be determined using an oracle for SAT. We get that there is a simple NDTM $N$ that given oracle access for SAT can decide $L$ : on input $x$, non-deterministically guess $u_1$ and use the oracle to decide if $\forall u_2 M\left(x, u_1, u_2\right)=1$. We see that $x \in L$ iff there exists a choice $u_1$ that makes $N$ accept.

## 数学代写|计算复杂度理论代写Computational Complexity Theory代 考|Time versus alternations: time-space tradeoffs for SAT

ProOF: 我们将证明
$$\boldsymbol{N T I M E}(n) \nsubseteq \operatorname{TIS}\left(n^{1.2}, n^{0.2}\right)$$

## 数学代写|计算复杂度理论代写Computational Complexity Theory代 考|Defining the hierarchy via oracle machines

: $5.15$

$$x \in L \Leftrightarrow \exists u_1 \in 0,1^{q(|x|)} \forall u_2 \in 0,1^{q(|x|)} M\left(x, u_1, u_2\right)=1$$

## 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。

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## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|PSPACE completeness

As already indicated, we do not know if $\mathbf{P} \neq$ PSPACE, though we strongly believe that the answer is YES. Notice, $\mathbf{P}=$ PSPACE implies $\mathbf{P}=\mathbf{N P}$. Since complete problems can help capture the essence of a complexity class, we now present some complete problems for PSPACE.
DEFINITION $4.8$
A language $A$ is PSPACE-hard if for every $L \in$ PSPACE, $L \leq_p A$. If in addition $A \in$ PSPACE then $A$ is PSPACE-complete.

Using our observations about polynomial-time reductions from Chapter ?? we see that if any PSPACE-complete language is in $\mathbf{P}$ then so is every other language in PSPACE. Viewed contrapostively, if PSPACE $\neq \mathbf{P}$ then a PSPACE-complete language is not in $\mathbf{P}$. Intuitively, a PSPACE-complete language is the “most difficult” problem of PSPACE. Just as NP trivially contains NP-complete problems, so does PSPACE. The following is one (Exercise 3):
SPACETM $=\left{\left\langle M, w, 1^n\right\rangle:\right.$ DTM $M$ accepts $w$ in space $\left.n\right}$.

## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|Savitch’s theore

The astute reader may notice that because the above proof uses the notion of a configuration graph and does not require this graph to have out-degree one, it actually yields a stronger statement: that TQBF is not just hard for PSPACE but in fact even for NPSPACE!. Since TQBF $\in$ PSPACE this implies that PSPACE = NSPACE, which is quite surprising since our intuition is that the corresponding classes for time ( $\mathbf{P}$ and $\mathbf{N P})$ are different. In fact, using the ideas of the above proof, one can obtain the following theorem:
THEOREM $4.12$ (SAVITCH [SAV70])
For any space-constructible $S: \mathbb{N} \rightarrow \mathbb{N}$ with $S(n) \geq \log n, \operatorname{NSPACE}(S(n)) \subseteq \mathbf{S P A C E}\left(S(n)^2\right)$
We remark that the running time of the algorithm obtained from this theorem can be as high as $2^{\Omega\left(s(n)^2\right)}$.
Proof: The proof closely follows the proof that TQBF is PSPACE-complete. Let $L \in \operatorname{NSPACE}(S(n))$ be a language decided by a TM $M$ such that for every $x \in{0,1}^n$, the configuration graph $G=G_{M, x}$ has at most $M=2^{O(S(n))}$ vertices. We describe a recursive procedure REACH? $(u, v, i)$ that returns “YES” if there is a path from $u$ to $v$ of length at most $2^i$ and “NO” otherwise. Note that REACH? $(s, t,\lceil\log M\rceil)$ is “YES” iff $t$ is reachable from $s$. Again, the main observation is that there is a path from $u$ to $v$ of length at most $2^i$ iff there’s a vertex $z$ with paths from $u$ to $z$ and from $z$ to $v$ of lengths at most $2^{i-1}$. Thus, on input $u, v, i$, REACH? will enumerate over all vertices $z$ (at a cost of $O(\log M)$ space) and output “YES” if it finds one $z$ such that REACH? $(u, z, i-1)=$ “YES” and REACH? $(z, v, i-1)=$ “YES”. Once again, the crucial observation is that although the algorithm makes $n$ recursive invocations, it can reuse the space in each of these invocations. Thus, if we let $s_{M, i}$ be the space complexity of REACH? $(u, v, i)$ on an $M$-vertex graph we get that $s_{M, i}=s_{M, i-1}+O(\log M)$ and thus $s_{M, \log M}=O\left(\log ^2 M\right)=O\left(S(n)^2\right)$.

## 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。

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## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|What if P = NP?

If $\mathbf{P}=\mathbf{N P}$ – specifically, if an $\mathbf{N P}$-complete problem like 3SAT had a very efficient algorithm running in say $O\left(n^2\right)$ time – then the world would be mostly a Utopia. Mathematicians could be replaced by efficient theorem-discovering programs (a fact pointed out in Kurt Gödel’s 1956 letter and discovered three decades later). In general for every search problem whose answer can be efficiently verified (or has a short certificate of correctness), we will be able to find the correct answer or the short certificate in polynomial time. AI software would be perfect since we could easily do exhaustive searches in a large tree of possibilities. Inventors and engineers would be greatly aided by software packages that can design the perfect part or gizmo for the job at hand. VLSI designers will be able to whip up optimum circuits, with minimum power requirements. Whenever a scientist has some experimental data, she would be able to automatically obtain the simplest theory (under any reasonable measure of simplicity we choose) that best explains these measurements; by the principle of Occam’s Razor the simplest explanation is likely to be the right one. Of course, in some cases it took scientists centuries to come up with the simplest theories explaining the known data. This approach can be used to solve also non-scientific problems: one could find the simplest theory that explains, say, the list of books from the New York Times’ bestseller list. (NB: All these applications will be a consequence of our study of the Polynomial Hierarchy in Chapter 5.)

Somewhat intriguingly, this Utopia would have no need for randomness. As we will later see, if $\mathbf{P}=\mathbf{N P}$ then randomized algorithms would buy essentially no efficiency gains over deterministic algorithms; see Chapter 7. (Philosophers should ponder this one.)

This Utopia would also come at one price: there would be no privacy in the digital domain. Any encryption scheme would have a trivial decoding algorithm. There would be no digital cash, no SSL, RSA or PGP (see Chapter 10). We would just have to learn to get along better without these, folks.

This utopian world may seem ridiculous, but the fact that we can’t rule it out shows how little we know about computation. Taking the half-full cup point of view, it shows how many wonderful things are still waiting to be discovered.

## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|What if NP = coNP?

If $\mathbf{N P}=\mathbf{c o N P}$, the consequences still seem dramatic. Mostly, they have to do with existence of short certificates for statements that do not seem to have any. To give an example, remember the NP-complete problem of finding whether or not a set of multivariate polynomials has a common root, in other words, deciding whether a system of equations of the following type has a solution:
\begin{aligned} f_1\left(x_1, \ldots, x_n\right) & =0 \ f_2\left(x_1, \ldots, x_n\right) & =0 \ \vdots & \ f_m\left(x_1, \ldots, x_n\right) & =0 \end{aligned}
where each $f_i$ is a quadratic polynomial.

If a solution exists, then that solution serves as a certificate to this effect (of course, we have to also show that the solution can be described using a polynomial number of bits, which we omit). The problem of deciding that the system does not have a solution is of course in coNP. Can we give a certificate to the effect that the system does not have a solution? Hilbert’s Nullstellensatz Theorem seems to do that: it says that the system is infeasible iff there is a sequence of polynomials $g_1, g_2, \ldots, g_m$ such that $\sum_i f_i g_i=1$, where 1 on the right hand side denotes the constant polynomial 1.

What is happening? Does the Nullstellensatz prove coNP $=\mathbf{N P}$ ? No, because the degrees of the $g_i$ ‘s – and hence the number of bits used to represent them- could be exponential in $n, m$. (And it is simple to construct $f_i$ ‘s for which this is necessary.)

## 数学代写|计算复杂度理论代写Computational Complexity Theory代考|What if NP = coNP?

$$f_1\left(x_1, \ldots, x_n\right)=0 f_2\left(x_1, \ldots, x_n\right) \quad=0 \vdots f_m\left(x_1, \ldots, x_n\right) \quad=0$$

## 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。

Posted on Categories:Computational Complexity, 数学代写, 计算复杂度

## avatest™帮您通过考试

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## 数学代写|计算复杂度代写Computational Complexity代考|Warmup: Expressiveness of boolean formulae

As a warmup for the proof of Lemma $2.12$ we show how to express various conditions using CNF formulae.
EXAMPLE $2.13$
The formula $(a \vee \bar{b}) \wedge(\bar{a} \vee b)$ is in CNF form. It is satisfied by only those values of $a, b$ that are equal. Thus, the formula
$$\left(x_1 \vee \bar{y}_1\right) \wedge\left(\bar{x}_1 \vee y_1\right) \wedge \cdots \wedge\left(x_n \vee \bar{y}_n\right) \wedge\left(\bar{x}_n \vee y_n\right)$$
is True if and only if the strings $x, y \in{0,1}^n$ are equal to one another.
Thus, though $=$ is not a standard boolean operator like $\vee$ or $\wedge$, we will use it as a convenient shorthand since the formula $\phi_1=\phi_2$ is equivalent to (in other words, has the same satisfying assignments as $)\left(\phi_1 \vee \overline{\phi_2}\right) \wedge\left(\phi_1 \vee \phi_2\right)$

In fact, CNF formulae of sufficient size can express every Boolean condition, as shown by the following simple claim: (this fact is sometimes known as universality of the operations AND, OR and NOT)
CLAIM $2.14$
For every Boolean function $f:{0,1}^{\ell} \rightarrow{0,1}$ there is an $\ell$-variable CNF formula $\varphi$ of size $\ell 2^{\ell}$ such that $\varphi(u)=f(u)$ for every $u \in{0,1}^{\ell}$, where the size of a CNF formula is defined to be the number of $\wedge / \vee$ symbols it contains.

## 数学代写|计算复杂度代写Computational Complexity代考|Proof of Lemma 2.12

Let $L$ be an NP language and let $M$ be the polynomial time TM such that that for every $x \in{0,1}^$, $x \in L \Leftrightarrow M(x, u)=1$ for some $u \in{0,1}^{p(|x|)}$, where $p: \mathbb{N} \rightarrow \mathbb{N}$ is some polynomial. We show $L$ is polynomial-time Karp reducible to SAT by describing a way to transform in polynomial-time every string $x \in{0,1}^$ into a CNF formula $\varphi_x$ such that $x \in L$ iff $\varphi_x$ is satisfiable.

How can we construct such a formula $\varphi_x$ ? By Claim $2.14$, the function that maps $u \in{0,1}^{p(|x|)}$ to $M(x, u)$ can be expressed as a CNF formula $\psi_x$ (i.e., $\psi_x(u)=M(x, u)$ for every $\left.u \in{0,1}^{p(|x|)}\right)$. Thus a string $u$ such that $M(x, u)=1$ exists if and only if $\psi_x$ is satisfiable. But this is not useful for us, since the size of the formula $\psi_x$ obtained from Claim $2.14$ can be as large as $p(|x|) 2^{p(|x|)}$. To get a smaller formula we use the fact that $M$ runs in polynomial time, and that each basic step of a Turing machine is highly local (in the sense that it examines and changes only a few bits of the machine’s tapes).

In the course of the proof we will make the following simplifying assumptions about the TM M: (1) $M$ only has two tapes: an input tape and a work/output tape and (2) $M$ is an oblivious TM in the sense that its head movement does not depend on the contents of its input tape. In particular, this means that $M$ ‘s computation takes the same time for all inputs of size $n$ and for each time step $i$ the location of $M$ ‘s heads at the $i^{\text {th }}$ step depends only on $i$ and $M$ ‘s input length.
We can make these assumptions without loss of generality because for every $T(n)$-time TM $M$ there exists a two-tape oblivious TM $\tilde{M}$ computing the same function in $O\left(T(n)^2\right)$ time (see Remark $1.10$ and Exercise 8 of Chapter 1). ${ }^4$ Thus in particular, if $L$ is in NP then there exists a two-tape oblivious polynomial-time TM $M$ and a polynomial $p$ such that $x \in L \Leftrightarrow \exists u \in$ ${0,1}^{p(|x|)}$ s.t. $M(x, u)=1$.

The advantage of assuming that $M$ is oblivious is that for any given input length, we can define functions inputpos $(i), \operatorname{prev}(i)$ where inputpos $(i)$ denotes the location of the input tape head at the $i^{\text {th }}$ step and prev $(i)$ denotes the last step before $i$ that $M$ visited the same location on its work tape, see Figure 2.3. ${ }^5$ These values can be computed in polynomial time by simulating $M$ on, say, the all-zeroes input.

## 数学代写|计算复杂度代写计算复杂度代考|预热:布尔公式的表达

EXAMPLE $2.13$

$$\left(x_1 \vee \bar{y}_1\right) \wedge\left(\bar{x}_1 \vee y_1\right) \wedge \cdots \wedge\left(x_n \vee \bar{y}_n\right) \wedge\left(\bar{x}_n \vee y_n\right)$$

CLAIM $2.14$

## 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。

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## 数学代写|计算复杂度代写Computational Complexity代考|Reducibility and NP-completeness

It turns out that the independent set problem is at least as hard as any other language in NP: if it has a polynomial-time algorithm then so do all the problems in NP. This fascinating property is called NP-hardness. Since most scientists conjecture that $\mathbf{N P} \neq \mathbf{P}$, the fact that a language is NP-hard can be viewed as evidence that it cannot be decided in polynomial time.

How can we prove that a language $B$ is at least as hard as some other language $A$ ? The crucial tool we use is the notion of a reduction (see Figure 2.1):
Definition $2.7$ (REdUCTIONS, NP-HARDNESS AND NP-COMPLETENESS)
We say that a language $A \subseteq{0,1}^$ is polynomial-time Karp reducible to a language $B \subseteq{0,1}^$ (sometimes shortened to just “polynomial-time reducible” ${ }^2$ ) denoted by $A \leq p B$ if there is a polynomial-time computable function
that for every $x \in{0,1}^*, x \in A$ if and only if $f(x) \in B$. We say that $B$ is NP-hard if $A \leq p B$ for every $A \in$ NP. We say that $B$ is NP-complete if $B$ is NP-hard and $B \in \mathbf{N P}$.

## 数学代写|计算复杂度代写Computational Complexity代考|The Cook-Levin Theorem: Computation is Local

Around 1971, Cook and Levin independently discovered the notion of NP-completeness and gave examples of combinatorial NP-complete problems whose definition seems to have nothing to do with Turing machines. Soon after, Karp showed that NP-completeness occurs widely and many problems of practical interest are NP-complete. To date, thousands of computational problems in a variety of disciplines have been found to be NP-complete.

Some of the simplest examples of NP-complete problems come from propositional logic. A Boolean formula over the variables $u_1, \ldots, u_n$ consists of the variables and the logical operators AND $(\wedge)$, $\operatorname{NOT}(\neg)$ and OR $(\vee)$; see Appendix A for their definitions. For example, $(a \wedge b) \vee(a \wedge c) \vee(b \wedge c)$ is a Boolean formula that is TruE if and only if the majority of the variables $a, b, c$ are TrUE. If $\varphi$ is a Boolean formula over variables $u_1, \ldots, u_n$, and $z \in{0,1}^n$, then $\varphi(z)$ denotes the value of $\varphi$ when the variables of $\varphi$ are assigned the values $z$ (where we identify 1 with TRUE and 0 with FALSE). A formula $\varphi$ is satisfiable if there there exists some assignment $z$ such that $\varphi(z)$ is TruE. Otherwise, we say that $\varphi$ is unsatisfiable.

A Boolean formula over variables $u_1, \ldots, u_n$ is in CNF form (shorthand for Conjunctive Normal Form) if it is an AND of OR’s of variables or their negations. For example, the following is a 3CNF formula:
$$\left(u_1 \vee \bar{u}_2 \vee u_3\right) \wedge\left(u_2 \vee \bar{u}_3 \vee u_4\right) \wedge\left(\bar{u}_1 \vee u_3 \vee \bar{u}_4\right) .$$
where $\bar{u}$ denotes the negation of the variable $u$.

## 数学代写|计算复杂度代写计算复杂度代考| Cook-Levin定理:计算是局部的

np完全问题的一些最简单的例子来自命题逻辑。变量$u_1, \ldots, u_n$上的布尔公式由变量和逻辑运算符and $(\wedge)$, $\operatorname{NOT}(\neg)$和OR $(\vee)$组成;它们的定义见附录A。例如，$(a \wedge b) \vee(a \wedge c) \vee(b \wedge c)$是一个布尔公式，当且仅当$a, b, c$的大部分变量为TruE时，该公式为TruE。如果$\varphi$是一个关于变量$u_1, \ldots, u_n$和$z \in{0,1}^n$的布尔公式，那么当$\varphi$的变量被赋值$z$时，$\varphi(z)$表示$\varphi$的值(其中我们将1标识为TRUE, 0标识为FALSE)。如果存在一些赋值$z$，使得$\varphi(z)$为TruE，则公式$\varphi$是可满足的。否则，我们说$\varphi$是不能满足的。

$$\left(u_1 \vee \bar{u}_2 \vee u_3\right) \wedge\left(u_2 \vee \bar{u}_3 \vee u_4\right) \wedge\left(\bar{u}_1 \vee u_3 \vee \bar{u}_4\right) .$$

## 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。

Posted on Categories:Computational Complexity, 数学代写, 计算复杂度

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## 数学代写|计算复杂度代写Computational Complexity代考|Big-Oh notation

As mentioned above, we will typically measure the computational efficiency algorithm as the number of a basic operations it performs as a function of its input length. That is, the efficiency of an algorithm can be captured by a function $T$ from the set of natural numbers $\mathbb{N}$ to itself such that $T(n)$ is equal to the maximum number of basic operations that the algorithm performs on inputs of length $n$. However, this function is sometimes be overly dependant on the low-level details of our definition of a basic operation. For example, the addition algorithm will take about three times more operations if it uses addition of single digit binary (i.e., base 2) numbers as a basic operation, as opposed to decimal (i.e., base 10) numbers. To help us ignore these low level details and focus on the big picture, the following well known notation is very useful:

DEFINITION $1.2$ (BIG-OH NOTATION)
If $f, g$ are two functions from $\mathbb{N}$ to $\mathbb{N}$, then we (1) say that $f=O(g)$ if there exists a constant $c$ such that $f(n) \leq c \cdot g(n)$ for every sufficiently large $n$, (2) say that $f=\Omega(g)$ if $g=O(f)$, (3) say that $f=\Theta(g)$ is $f=O(g)$ and $g=O(f),(4)$ say that $f=o(g)$ if for every $\epsilon>0, f(n) \leq \epsilon \cdot g(n)$ for every sufficiently large $n$, and $(5)$ say that $f=\omega(g)$ if $g=o(f)$.

To emphasize the input parameter, we often write $f(n)=O(g(n))$ instead of $f=O(g)$, and use similar notation for $o, \Omega, \omega, \Theta$.

## 数学代写|计算复杂度代写Computational Complexity代考|Modeling computation and eﬃﬃﬃciency

We start with an informal description of computation. Let $f$ be a function that takes a string of bits (i.e., a member of the set ${0,1}^$ ) and outputs, say, either 0 or 1 . Informally speaking, an algorithm for computing $f$ is a set of mechanical rules, such that by following them we can compute $f(x)$ given any input $x \in{0,1}^$. The set of rules being followed is fixed (i.e., the same rules must work for all possible inputs) though each rule in this set may be applied arbitrarily many times. Each rule involves one or more of the following “elementary” operations:

1. Read a bit of the input.Read a bit (or possibly a symbol from a slightly larger alphabet, say a digit in the set ${0, \ldots, 9})$ from the “scratch pad” or working space we allow the algorithm to use.
2. Based on the values read,
3. Write a bit/symbol to the scratch pad.
4. Either stop and output 0 or 1 , or choose a new rule from the set that will be applied next.
5. Finally, the running time is the number of these basic operations performed. Below, we formalize all of these notions.

## 数学代写|计算复杂度代写计算复杂度代考|Big-Oh符号

.

$f, g$ 是两个函数 $\mathbb{N}$ 到 $\mathbb{N}$，那么我们(1)说 $f=O(g)$ 如果存在一个常数 $c$ 如此这般 $f(n) \leq c \cdot g(n)$ 对于每一个足够大的 $n$，(2)说 $f=\Omega(g)$ 如果 $g=O(f)$，(3)说 $f=\Theta(g)$ 是 $f=O(g)$ 和 $g=O(f),(4)$ 说出来 $f=o(g)$ 如果对于每一个 $\epsilon>0, f(n) \leq \epsilon \cdot g(n)$ 对于每一个足够大的 $n$，以及 $(5)$ 说出来 $f=\omega(g)$ 如果 $g=o(f)$.

## 数学代写|计算复杂度代写Computational Complexity代考|建模计算和效率

.计算复杂度代考|

1. 读取输入的一个bit。从我们允许算法使用的“草稿板”或工作空间中读取一点(或者可能是稍大一点的字母表中的一个符号，比如集合${0, \ldots, 9})$中的一个数字)。
2. 根据读取的值，
3. 向刮痧板上写入位/符号。
4. 要么停止并输出0或1，要么从接下来将应用的集合中选择一个新规则。
5. 最后，运行时间是执行这些基本操作的次数。下面，我们将这些概念形式化。

## 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。