### 数学代写|计算复杂度理论代写Computational complexity theory代考|CATS 2013

statistics-lab™ 为您的留学生涯保驾护航 在代写计算复杂度理论Computational complexity theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写计算复杂度理论Computational complexity theory代写方面经验极为丰富，各种代写计算复杂度理论Computational complexity theory相关的作业也就用不着说。

• 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

## 数学代写|计算复杂度理论代写Computational complexity theory代考|Unrelativizable Proof Techniques

First, a common view is that the question of whether $P$ is equal to $N P$ is a difficult question in view of Theorem 4.14. As most proof techniques developed in recursion theory, including the basic diagonalization and simulation techniques, relativize, any attack to the $P=$ ? NP question must use a new, unrelativizable proof technique. Many more contradictory relativized results like Theorem $4.14$ (including some in Section 4.8) on the relations between complexity classes tend to support this viewpoint. On the other hand, some unrelativizable proof techniques do exist in complexity theory. For instance, we will apply an algebraic technique to collapse the complexity class PSPACE to a subclass $I P$ (see Chapter 10). As there exists an oracle $X$ that separates $P S P A C E^{X}$ from $I P^{X}$, this proof is indeed unrelativizable. Though this is a breakthrough in the theory of relativization, it seems still too early to tell whether such techniques are applicable to a wider class of questions.

## 数学代写|计算复杂度理论代写Computational complexity theory代考|Independence Results

One of the most interesting topics in set theory is the study of independence results. A statement $A$ is said to be independent of a theory $T$ if there exist two models $M_{1}$ and $M_{2}$ of $T$ such that $A$ is true in $M_{1}$ and false in $M_{2}$. If a statement $A$ is known to be independent of the theory $T$, then neither $A$ nor its negation $\neg A$ is provable in theory $T$. The phenomenon of contradictory relativized results looks like a mini-independent result: neither the statement $P=N P$ nor its negation $P \neq N P$ is provable by relativizable techniques. This observation raises the question of whether they are provable within a formal proof system. In the following, we present a simple argument showing that this is indeed possible.

To prepare for this result, we first briefly review the concept of a formal proof system. An axiomatizable theory is a triple $F=(\Sigma, W, T)$, where $\Sigma$ is a finite alphabet, $W \subseteq \Sigma^{*}$ is a recursive set of well-formed formulas, and $T \subseteq W$ is an r.e. set. The elements in $T$ are called theorems. If $T$ is recursive, we say the theory $F$ is decidable. We are only interested in a sound theory in which we can prove the basic properties of TMs. In other words, we assume that TMs form a submodel for $F$, all basic properties of TMs are provable in $F$, and all theorems in $F$ are true in the TM model. In the following, we let $\left{M_{i}\right}$ be a fixed enumeration of multi-tape DTMs.
Theorem 4.15 For any formal axiomatizable theory $F$ for which $T M s$ form a submodel, we can effectively find a DTM $M_{i}$ such that $L\left(M_{i}\right)=\emptyset$ and neither ” $P P^{L\left(M_{i}\right)}=N P^{L\left(M_{i}\right) “}$ nor ” $P^{L\left(M_{i}\right)} \neq N P^{L\left(M_{i}\right) “}$ is provable in $F$.

Proof. Let $A$ and $B$ be two recursive sets such that $P^{A}=N P^{A}$ and $P^{B} \neq$ $N P^{B}$. Define a TM $M$ such that $M$ accepts $(j, x)$ if and only if among the first $x$ proofs in $F$ there is a proof for the statement ” $P^{L\left(M_{j}\right)}=N P^{L\left(M_{j}\right)}$ ” and $x \in B$ or there is a proof for the statement ” $P^{L\left(M_{j}\right)} \neq N P^{L\left(M_{j}\right)}$ ” and $x \in A$. By the recursion theorem, there exists an index $j_{0}$ such that $M_{j_{0}}$ accepts $x$ if and only if $M$ accepts $\left(j_{0}, x\right)$. (See, e.g., Rogers (1967) for the recursion theorem.)

Now, if there is a proof for the statement ” $P^{L\left(M_{k b}\right)}=N P^{L\left(M_{k}\right)}$ ” in $F$, then for almost all $x, M$ accepts $\left(j_{0}, x\right)$ if and only if $x \in B$. That is, the set $L\left(M_{j_{0}}\right)$ differs from the set $B$ by only a finite set and, hence, $P^{B} \neq N P^{B}$ implies $P^{L\left(M_{f 0}\right)} \neq N P^{L\left(M_{j 0}\right)}$. Similarly, if there exists a proof for the statement ” $P^{L\left(M_{j 0}\right)} \neq N P^{L\left(M_{f 0}\right)}$ “, then $L\left(M_{j_{0}}\right)$ differs from $A$ by only a finite set theory $F$, we conclude that neither ” $P^{L\left(M_{f 0}\right)}=N P^{L\left(M_{j 0}\right) “}$ nor ” $P^{L\left(M_{f 0}\right)} \neq$ $N P^{L\left(M_{f 0}\right)}$, , is provable in $F$.

Furthermore, because neither ” $P^{L\left(M_{f 0}\right)}=N P^{L\left(M_{f 0}\right)}$ ” nor ” $P^{L\left(M_{f 0}\right)} \neq$ $N P^{L\left(M_{j_{0}}\right) \text { ” }}$ is provable in $F$, the machine $M_{j_{0}}$ does not accept any input $x$, that is, $L\left(M_{j_{0}}\right)=\emptyset$.

## 数学代写|计算复杂度理论代写Computational complexity theory代考|Positive Relativization

Still another viewpoint is that the formulation of the relativized complexity class $N P^{A}$ used in Theorem $4.14$ does not reflect correctly the concept of nondeterministic computation. Consider the set $L_{B}$ in the proof of Theorem 4.14. Note that although each computation path of the oracle NTM $M$ that accepts $L_{B}$ asks only one question to determine whether $0^{n}$ is in $B$, the whole computation tree of $M^{B}(x)$ makes an exponential number of queries. While it is recognized that this is the distinctive property of an NTM to make, in the whole computation tree, an exponential number of moves, the fact that $M$ can access an exponential amount of information about the oracle $B$ immediately makes the oracle NTMs much stronger than oracle DTMs. To make the relation between oracle NTMs and oracle DTMs close to that between regular NTMs and regular DTMs, we must not allow the oracle NTMs to make arbitrary queries. Instead, we would like to know whether an oracle NTM that is allowed to make, in the whole computation tree, only a polynomial number of queries is stronger than an oracle DTM. When we add these constraints to the oracle NTMs, it turns out that the relativized $P=$ ?NP question is equivalent to the unrelativized version. This result supports the viewpoint that the relativized separation of Theorem $4.14$ is due to the extra information that an oracle NTM can access, rather than the nondeterminism of the NTM and, hence, this kind of separation results bear no relation to the original unrelativized questions. This type of relativization is called positive relativization. We present a simple result of this type in the following.

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## MATLAB代写

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