### 数学代写|计算复杂度理论代写Computational complexity theory代考|Universal Turing Machine

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• Statistical Inference 统计推断
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## 数学代写|计算复杂度理论代写Computational complexity theory代考|Universal Turing Machine

One of the most important properties of a computation system like TMs is that there exists a universal machine that can simulate each machine from its code.

Let us first consider one-tape DTMs with the input alphabet ${0,1}$, the working alphabet ${0,1, \mathrm{~B}}$, the initial state $q_{0}$, and the final state set $\left{q_{1}\right}$, that is, DTMs defined by $\left(Q, q_{0},\left{q_{1}\right},{0,1},{0,1, \mathrm{~B}}, \delta\right)$. Such a TM can be determined by the definition of the transition function $\delta$ only, for $Q$ is assumed to be the set of states appearing in the definition of $\delta$. Let us use the notation $q_{i}, 0 \leq i \leq|Q|-1$, for a state in $Q, X_{j}, j=0,1,2$, for a tape symbol where $X_{0}=0, X_{1}=1$ and $X_{2}=\mathrm{B}$, and $D_{k}, k=0$, lor a moving direction for the tape head where $D_{0}=\mathrm{L}$ and $D_{1}=\mathrm{R}$. For each equation $\delta\left(q_{i}, X_{j}\right)=\left(q_{k}, X_{\ell}, D_{h}\right)$, we encode it by the following string in ${0,1}^{*}$ :
$$0^{i+1} 10^{j+1} 10^{k+1} 10^{\ell+1} 10^{h+1} .$$
Assume that there are $m$ equations in the definition of $\delta$. Let code $e_{i}$ be the code of the $i$ th equation. Then we combine the codes for equations together to get the following code for the TM:
$$\text { code }{1} 11 \text { code }{2} 11 \cdots 11 \text { code }_{m} \text {. }$$
Note that because different orderings of the equations give different codes, there are $m$ ! equivalent codes for a TM of $m$ equations.

The above coding system is a one-to-many mapping $\phi$ from TMs to ${0,1}^{}$. Each string $x$ in ${0,1}^{}$ encodes at most one TM $\phi^{-1}(x)$. Let us extend $\phi^{-1}$ into a function mapping each string $x \in{0,1}^{}$ to a TM $M$ by mapping each $x$ not encoding a TM to a fixed empty TM $M_{0}$ whose code is $\lambda$ and that rejects all strings. Call this mapping $t$. Observe that $t$ is a mapping from ${0,1}^{}$ to TMs with the following properties:
(i) For every $x \in \Sigma^{*}, t(x)$ represents a TM;
(ii) Every TM is represented by at least one $\iota(x)$; and
(iii) The transition function $\delta$ of the TM $t(x)$ can be easily decoded from the string $x$.

We say a coding system $t$ is an emumeration of one-tape DTMs if $t$ satisfies properties (i) and (ii). In addition, property (iii) means that this enumeration admits a universal Turing machine. In the following, we write $M_{x}$ to mean the TM $\iota(x)$. We assume that $\langle\cdot, \cdot\rangle$ is a pairing function on ${0,1}^{*}$ such that both the function and its inverse are computable in linear time, for instance, $\langle x, y\rangle=0^{|x|} 1 x y$.

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

Diagonalization is an important proof technique widely used in recursive function theory and complexity theory. One of the earliest applications of diagonalization is Cantor’s proof for the fact that the set of real numbers is not countable. We give a similar proof for the set of functions on ${0,1}^{*}$. A set $S$ is countable (or, enumerable) if there exists a one-one onto mapping from the set of natural numbers to $S$.

Proposition $1.20$ The set of functions from ${0,1}^{}$ to ${0,1}$ is not countable. Proof. Suppose, by way of contradiction, that such a set is countable, that is, it can be represented as $\left{f_{0}, f_{1}, f_{2}, \cdots\right}$. Let $a_{i}$ denote the $i$ th string in ${0,1}^{}$ under the lexicographic ordering. Then we can define a function $f$ as follows: For each $i \geq 0, f\left(a_{i}\right)=1$ if $f_{i}\left(a_{i}\right)=0$ and $f\left(a_{i}\right)=0$ if $f_{i}\left(a_{i}\right)=1$. Clearly, $f$ is a function from ${0,1}^{*}$ to ${0,1}$. However, it is not in the list $f_{0}, f_{1}, f_{2}, \cdots$, because it differs from each $f_{i}$ on at least one input string $a_{i}$. This establishes a contradiction.

An immediate consequence of Proposition $1.20$ is that there exists a noncomputable function from ${0,1}^{}$ to ${0,1}$, because we have just shown that the set of all TMs, and hence the set of all computable functions, is countable. In the following, we use diagonalization to construct directly an undecidable (i.e., nonrecursive) problem: the halting problem. The halting problem is the set $K=\left{x \in{0,1}^{}: M_{x}\right.$ halts on $\left.x\right}$, where $\left{M_{x}\right}$ is an enumeration of TMs.
Theorem $1.21 K$ is r.e. but not recursive.
Proof. The fact that $K$ is r.e. follows immediately from the existence of the universal TM $M_{u}$ (Proposition 1.17). To see that $K$ is not recursive, we note that the complement of a recursive set is also recursive and, hence, r.e. Thus, if $K$ were recursive, then $\bar{K}$ would be r.e. and there would be an integer $y$ such that $M_{y}$ halts on all $x \in \bar{K}$ and does not halt on any $x \in K$. Then, a contradiction could be found when we consider whether or not $y$ itself is in $K$ : if $y \in K$, then $M_{y}$ must not halt on $y$ and it follows from the definition of $K$ that $y \notin K$ and if $y \notin K$, then $M_{y}$ must halt on $y$ and it follows from the definition of $K$ that $y \in K$.

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

We study, in this section, the relationship between deterministic and nondeterministic complexity classes, as well as the relationship between time- and space-bounded complexity classes. We show several different simulations of nondeterministic machines by deterministic ones.
Theorem $1.28$ (a) For any fully space-constructible function $f(n) \geq n$,
$$D T I M E(f(n)) \subseteq N T I M E(f(n)) \subseteq D S P A C E(f(n)) .$$
(b) For any fully space-constructible function $f(n) \geq \log n$,
$$D S P A C E(f(n)) \subseteq N S P A C E(f(n)) \subseteq \bigcup_{c>0} D T I M E\left(2^{f(n)}\right) .$$
Proof. (a): The relation DTIME(f(n)) $\subseteq \operatorname{NTIME}(f(n))$ follows immediately from the fact that DTMs are just a subclass of NTMs. For the relation $N T I M E(f(n)) \subseteq D S P A C E(f(n))$, we recall the simulation of an NTM $M$ by a DTM $M_{1}$ as described in Theorem 1.9. Suppose that $M$ has time complexity bounded by $f(n)$; then $M_{1}$ needs to simulate $M$ for at most $f(n)$ moves. That is, we restrict $M_{1}$ to only execute the first $\sum_{j=1}^{f(n)} k^{i}$ stages such that the strings written in tape 2 are at most $f(n)$ symbols long. As $f(n)$ is fully space constructible, this restriction can be done by first marking off $f(n)$ squares on tape 2 . It is clear that such a restricted simulation works within space $f(n)$.
$\mathrm{~ ( b ) : ~ A g a i n , ~ D S P A C E ( f ( n ) ) ~}$ $N S P A C E(f(n)) \subseteq \bigcup_{c>0} D T I M E\left(2^{c f(n)}\right)$, assume that $M$ is an NTM with the space bound $f(n)$. We are going to construct a DTM $M_{1}$ to simulate $M$ in time $2^{c f(n)}$ for some $c>0$. As $M$ uses only space $f(n)$, there is a constant $c_{1}>0$ such that the shortest accepting computation for each $x \in L(M)$ is of length $\leq 2^{c_{1} f(|x|)}$. Thus, the machine $M_{1}$ needs only to simulate $M(x)$ for, at most, $2^{c_{1} f(n)}$ moves. However, $M$ is a nondeterministic machine and so its computation tree of depth $2^{c_{1} f(n)}$ could have $2^{20(\text { a }}$ ) naive simulation as (a) above takes too much time.

To reduce the deterministic simulation time, we notice that this computation tree, although of size $2^{2^{2 \varphi(m))}}$, has at most $2^{O(f(n))}$ different configurations: Each configuration is determined by at most $f(n)$ tape symbols on the work tape, one of $f(n)$ positions for the work tape head,one of $n$ positions for the input tape head, and one of $r$ positions for states, where $r$ is a constant. Thus, the total number of possible configurations of $M(x)$ is $2^{O(f(n))} \cdot f(n) \cdot n \cdot r=2^{O(f(n))}$. (Note that $f(n) \geq \log n$ implies $\left.n \leq 2^{f(n)}\right)$

## 计算复杂度理论代考

0一世+110j+110ķ+110ℓ+110H+1.

代码 111 代码 211⋯11 代码 米.

(i) 对于每个X∈Σ∗,吨(X)代表 TM；
(ii) 每个 TM 至少由一个代表我(X); (
iii) 过渡函数dTM的吨(X)可以很容易地从字符串中解码X.

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

D吨我米和(F(n))⊆ñ吨我米和(F(n))⊆D小号磷一个C和(F(n)).
(b) 对于任何完全空间可构造函数F(n)≥日志⁡n,

D小号磷一个C和(F(n))⊆ñ小号磷一个C和(F(n))⊆⋃C>0D吨我米和(2F(n)).

(b): 一个G一个一世n, D小号磷一个C和(F(n))  ñ小号磷一个C和(F(n))⊆⋃C>0D吨我米和(2CF(n))， 假使，假设米是一个有空间限制的 NTMF(n). 我们将构建一个 DTM米1模拟米及时2CF(n)对于一些C>0. 作为米仅使用空间F(n), 有一个常数C1>0这样每个的最短接受计算X∈大号(米)有长度≤2C1F(|X|). 因此，机米1只需要模拟米(X)因为，至多，2C1F(n)移动。然而，米是一个不确定的机器，因此它的深度计算树2C1F(n)本来可以220( 一个 ) 上面 (a) 的幼稚模拟需要太多时间。

## 有限元方法代写

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## MATLAB代写

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