### 数学代写|计算复杂度理论代写Computational complexity theory代考|COMP4500

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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

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

The polynomial-time hierarchy was formally defined by oracle TMs. As the oracles play a mysterious role in the computation of an oracle TM, it is relatively more difficult to analyze the computation of such machines. The characterization of Theorem $3.8$ provides a different view, and it has been found useful for many applications. In this section, we formalize this characterization as a computational model, called the alternating Turing machine (abbreviated as ATM), that can be used to define the complexity classes in the polynomial-time hierarchy without using the notion of oracles.

An ATM $M$ is an ordinary NTM with its states partitioned into two subsets, called the universal states and the existential states. An ATM operates exactly the same as an NTM, but the notion of its acceptance of an input is defined differently. Thus, the computation of an ATM $M$ is a computation tree of configurations with each pair $(\alpha, \beta)$ of parent and child configurations satisfying $\alpha \vdash_{M} \beta$. We say a configuration is a universal configuration if it is in a universal state, and it is an existential configuration if it is in an existential state.

To define the notion of an ATM accepting an input, we assign, inductively, the label ACCEPT to some of the nodes in this computation tree as follows: A leaf is labeled ACCEPT if and only if it is in the accepting state. An internal node in the universal state is labeled with ACCEPT if and only if all of its children are labeled with ACCEPT. An internal node in the existential state is labeled with ACCEPT if and only if at least one of its children is labeled with ACCEPT. We say an ATM M accepts an input $x$ if the root of the computation tree is labeled with ACCEPT using the above labeling system. Thus an NTM is just an ATM in which all states are classified as existential states.

When an NTM $M$ accepts an input $x$, an accepting computation path is a witness to this fact. Also, we define time $_{M}(x)$ to be the length of a shortest accepting path. For an ATM $M$, to demonstrate that it accepts an input $x$, we need to display the accepting computation subtree $T_{a c c}$ of the computation tree $T$ of $M(x)$ that has the following properties:
(i) The root of $T$ is in $T_{a c c}$.
(ii) If $u$ is an internal existential node in $T_{a c c}$, then exactly one child of $u$ in $T$ is in $T_{a c c}$.

## 数学代写|计算复杂度理论代写Computational complexity theory代考|PSPACE-Complete Problems

Our first PSPACE-complete problem is the space-bounded halting problem (SBHP).
SPACE Bounded Halting Problem (SBHP): Given a DTM $M$, an input $x$, and an integer $s$, written in the unary form $0^{s}$, determine whether $M$ accepts $x$ within space bound $s$.
Theorem 3.23 SBHP is $\leq_{m}^{P}$-complete for PSPACE.
Proof. The proof is similar to that of Theorem 2.11.
The existence of a PSPACE-complete set implies that if the polynomial-time hierarchy is properly infinite, then PSPACE properly contains $P H$.

Theorem 3.24 If $P H=P S P A C E$, then the polynomial-time hierarchy collapses to $\Sigma_{k}^{P}$ for some $k>0$.

Proof. If $P H=P S P A C E$, then SBHP $\in P H=\bigcup_{k \geq 0} \Sigma_{k}^{P}$ and, hence, SBHP $\in \Sigma_{k}^{P}$ for some $k \geq 0$. This implies that PSPACE $\subseteq \Sigma_{k}^{P}$, because $\Sigma_{k}^{P}$ is closed under the $\leq_{m}^{P}$-reducibility.

The first natural PSPACE-complete problem is a generalization of $S A T_{k}$. The inputs to this problem are Boolean formulas with quantifiers $(\exists x)$ and $(\forall x)$. An occurrence of a variable $v$ in a Boolean formula $F$ is a bounded variable if there is a quantifier $(\exists v)$ or $(\forall v)$ in $F$ such that this occurrence of $v$ is in the scope of the quantifier. A Boolean formula $F$ is called a quantified Boolean formula if every occurrence of every variable in $F$ is a bounded variable. For instance, $F=(\forall x)[(\forall y)[(\exists z)[x \bar{y} z+\bar{x} y \bar{z}] \rightarrow(\exists z)[(x \bar{z}+\bar{x} z)(y \bar{z}+\bar{y} z)]]]$ is a quantified Boolean formula. In the above, we used brackets […] to denote the scope of a quantifier and $\rightarrow$ to denote the Boolean operation $(a \rightarrow b)=(\bar{a}+b)$. Each quantified Boolean formula has a normal form in which all quantifiers occur before any occurrence of a Boolean variable, and the scope of each quantifier is the rest of the formula to its right. For instance, the normal form (with renaming) of the above formula $F$ is $(\forall x)(\forall y)(\forall z)(\exists w)[(x \bar{y} z+\bar{x} y \bar{z}) \rightarrow((x \bar{w}+\bar{x} w)(y \bar{w}+\bar{y} w))]$
QUANTIFIED BOOLEAN FORMULA (QBF): Given a quantified Boolean formula $F$, determine whether $F$ is true.

## 数学代写|计算复杂度理论代写Computational complexity theory代考|EXP-Complete Problems

All complete problems studied so far are candidates for intractable problems, but their intractability still depends on the separation of the classes $N P, P S P A C E$, and $P H$ from the class $P$. Are there natural problems that are provably intractable in the sense that they can be proved not belonging to $P$ ? In this section, we present a few problems that are complete for $E X P$ and, hence, not in $P$.

Our first $E X P$-complete problem is the bounded halting problem on deterministic machines with the time bound encoded in binary form.
EXPONENTIAL-Time BoUnDEd HALTING PROBLEM (EXP-BHP): Given a DTM $M$, a string $x$, and an integer $n>0$, written in the binary form, determine whether $M(x)$ halts in $n$ moves.
Proposition 3.30 EXP-BHP is EXP-complete.
Proof. If $L$ is accepted by a DTM $M$ in time $2^{c n}$, then the function $f(x)=$ $\left\langle M, x, 2^{c|x|}\right\rangle$ is a polynomial-time reduction from $L$ to ExP-BHP.

We note that in the above problem, if the time bound $n$ is written in the unary form (as in the problem BHP), then the problem becomes polynomial-time solvable. Indeed, there is a simple translation of most $P$-complete problems ${ }^{1}$ to $E X P$-complete problems by more succinct encodings of the inputs. In the following, we demonstrate this idea on the famous $P$-complete problem, CIRCUIT VALUE Problem (CVP).

Let $C$ be a Boolean circuit ${ }^{2}$ satisfying the following property: $C$ has $n$ gates numbered from 1 to $n$; we let $C(i)$ denote the gate of $C$ numbered $i$. There are four types of gates in circuit $C$ : ZERO gates, ONE gates, AND gates, and OR gates. A ZERO (ONE) gate has no input and one output whose value is 0 (1, respectively). An AND (OR) gate has two inputs and one output whose value is the Boolean product (Boolean sum, respectively) of the two inputs. If the gate $i$ is an AND or OR gate, then its two inputs are the outputs of two gates whose numbers are lower than $i$. Note that this circuit $C$ does not have input gates and so it computes a unique Boolean value (the output of gate $n$ ). If the circuit is given explicitly, then its output value is computable in polynomial time. (In fact, it is $P$-complete; see Theorem 6.41). In the following, we consider the encoding of the circuit by a DTM. We say that a DTM $M$ generates a circuit $C$ of size $n$ in time $m$ if for all $i, 0 \leq i \leq n, M(i)$ outputs a triple $\langle b, j, k\rangle$ in $m$ moves, with $0 \leq b \leq 3$ and $1 \leq j, k<i$ if $b \leq 1$, such that
(i) If $b=0$, then $C(i)=C(j) \cdot C(k)$;
(ii) If $b=1$, then $C(i)=C(j)+C(k)$;
(iii) If $b=2$, then $C(i)=0$;
(iv) If $b=3$, then $C(i)=1$.

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

(i) 的根吨在吨一个CC.
(ii) 如果在是一个内部存在节点吨一个CC，那么恰好是的一个孩子在在吨在吨一个CC.

## 数学代写|计算复杂度理论代写Computational complexity theory代考|PSPACE-Complete Problems

SPACE Bounded Halting Problem (SBHP)：给定一个 DTM米, 一个输入X, 和一个整数s，写成一元形式0s, 判断是否米接受X在空间范围内s.

PSPACE 完全集的存在意味着如果多项式时间层次适当地无限，则 PSPACE 适当地包含磷H.

## 数学代写|计算复杂度理论代写Computational complexity theory代考|EXP-Complete Problems

(i) 如果b=0， 然后C(一世)=C(j)⋅C(ķ);
(ii) 如果b=1， 然后C(一世)=C(j)+C(ķ);
(iii) 如果b=2， 然后C(一世)=0;
(iv) 如果b=3， 然后C(一世)=1.

## 有限元方法代写

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## MATLAB代写

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