自动人形会梦见 NP 完全问题吗？(笔记)

Posted on 2024-08-16 繁/简： set

计算理论 (Theory of Computation) 当之无愧是计算机科学王冠上的明珠；考虑到我贫瘠的智商，以后估计不会朝 TCS 方向来走；但对这些优雅的理论有一个最基本的了解应当是 CS 学生的素养。

讽刺的是，你坑近二十年前就没有开设这门课了 [1]，只得留待交换来上。老师是印度人，口音很重，我听完整节课才明白他口中的「GEO」其实是「zero」；但课还是讲得十分清楚，颇有 Ravi 之风。

This article is a self-administered course note.

It will NOT cover any exam or assignment related content.

Preliminaries

关于 formal proof 的一些 recap，一些自动机理论概念的 notations 与定义。

字母表 (alphabet) $\Sigma$, 符号/字符 (symbol), 字符串 (string) 与语言 (language) $L\subseteq \Sigma^{*}$。

关于 strings/languages 的可数性质：

Number of strings over any non-empty alphabet is countable. e.g., length then lexicographic order.
Number of languages over any non-empty alphabet is uncountable. Remember that language is just a subset of the set of all strings.

DFA, NFA and RegEx

关于 $\epsilon$-transition。不输入任何 symbol 的转移。$\epsilon$ closure $\text{Eclose}(q)$。

regular language is a set of strings, regular expression $M$ is a representation of regular language $L(M)$.

DFA & NFA: Equivalence

NFAs and DFAs are equivalent in terms of the languages they can recognize, even though DFAs may require more states (potentially exponentially more) than NFAs.

首先 DFA 一定是 NFA。

对于给定的 NFA $A=(Q,\Sigma, \delta,q_0,F)$ 构建等价的 DFA $A_D=(Q_D,\Sigma,\delta_D,\{q_0\},F_D)$。定义 $Q_D=\{S|S\subseteq Q\}$ (因此 $|Q_D|$ 是 $O(2^{|Q|})$ 级别的)，$F_D=\{S|S\subseteq Q \text{ and } S\cap F\neq \emptyset\}$，$\delta_D(S,a)=\bigcup_{q\in S}\delta(q,a)$。

Claim. If for all string $w$, $\hat\delta_D(\{q_0\},w)=\hat\delta(q_0,w)$, then $\hat\delta_D(\{q_0\},wa)=\hat\delta(q_0,wa)$ also holds.

Base case. $w=\epsilon$, $\hat\delta_D(\{q_0\},\epsilon)=\{q_0\}=\hat\delta(q_0,\epsilon)$ holds.

Induction case. \[ \begin{aligned} \hat\delta_D(\{q_0\},wa)&=\delta_D(\hat\delta_D(\{q_0\},w),a) \\ &=\bigcup_{q\in\hat\delta_D(\{q_0\},w)}\delta(q,a) \\ &=\bigcup_{q\in \hat\delta(q_0,w)} \delta(q,a) \\ &=\hat\delta(q_0,wa) \end{aligned} \] 因此我们构建出的 DFA $A_D$ 能够接受字符串 $w$，当且仅当 NFA $A$ 接受字符串 $w$。

DFA to RegEx

本质 DP。对于给定的 DFA $A=(Q,\Sigma,\delta,q_{start}, F)$，不妨定义 $Q=\{1,2,...,n\}$ 且 $q_{start}=1$。定义 $R_{i,j}^k$ 为从 $i$ 到 $j$，只能经过标号 $\leq k$ 状态的路径所对应的 RegEx。

Base case. 定义 $R_{i,j}^0$

If $i\neq j$. $R_{i,j}^0=a_1+a_2+...+a_m$ for all $\delta(i,a_r)=j$.
If $i=j$. $R_{i,j}^0=\epsilon+a_1+a_2+...+a_m$ for all $\delta(i,a_r)=i$.

Induction case. $R_{i,j}^{k+1}=R_{i,j}^k+R_{i,k+1}^k(R_{k+1,k+1}^k)^{*}R_{k+1,j}^k$

那么该 DFA 所对应的 regular expression 为 $L(A)=\sum_{j\in F} R_{1,j}^n$

RegEx to $\epsilon$-NFA

对于给定的 RegEx，构建 start state 与 final state 唯一的 $\epsilon$-NFA。

Base case.

$\emptyset$. $A=(\{q_0,q_f\}, \Sigma,\delta,q_0,\{q_f\})$ where $\delta$ is an empty transition.
$\epsilon$. $A=(\{q_0,q_f\}, \Sigma,\delta,q_0,\{q_f\})$ where $\delta(q_0,\epsilon)=q_f$.
$a$. $A=(\{q_0,q_f\}, \Sigma,\delta,q_0,\{q_f\})$ where $\delta(q_0,a)=q_f$.

Induction case. 对于 $r_1$ 对应的 $A_1=(Q_1,\Sigma,q_0^1,\delta_1,F_1)$ 与 $r_2$ 对应的 $A_2=(Q_2,\Sigma,q_0^2,\delta_2,F_2)$，分别为 $r_1+r_2$, $r_1\cdot r_2$, $r_1^*$ 构造新的 $\epsilon$-NFA $A=(\{q_0,q_f\}\cup Q_1\cup Q_2,\Sigma,\delta,q_0,\{q_f\})$。

$r_1+r_2$. 添加 $\delta(q_0,\epsilon)=\{q_0^1,q_0^2\}$, $\epsilon(q_f^1,\epsilon)=\{q_f\}$ for all $q_f^1\in F_1$, $\delta(q_f^2,\epsilon)=\{q_f\}$ for all $q_f^2\in F_2$。
$r_1\cdot r_2$. 添加 $\delta(q_0,\epsilon)=\{q_0^1\}$, $\delta(q_f^1,\epsilon)=\{q_0^2\}$ for all $q_f^1\in F_1$, $\delta(q_f^2,\epsilon)=\{q_f\}$ for all $q_f^2\in F_2$。
$r_1^*$. 添加 $\delta(q_0,\epsilon)=\{q_0^1, q_f\}$, $\delta(q_f^1,\epsilon)=\{q_0^1,q_f\}$ for all $q_f^1\in F_1$.

Minimization of DFA

Equivalence Classes

给定 regular language $L$。

$u\equiv_L w$ iff for all $x$, $ux\in L$ iff $wx\in L$.
$\equiv_L$ is an equivalence relation - it is reflexive, symmetric, and transitive.
通过 equivalence classes $\text{equiv}(w)$ 构造 minimal DFA $(Q, \Sigma, \delta,q_0,F)$
- $Q=\{\text{equiv}(w):w\in \Sigma^*\}$
- $q_0=\text{equiv}(\epsilon)$
- $F=\{\text{equiv}(w):w\in L\}$
- $\delta(\text{equiv}(w), a)=\text{equiv}(wa)$

该方法对给定的 regular language $L$ 构造一个 unique 且 minimal 的 DFA $A$ from scratch。

Distinguishable Pairs

给定 DFA $A=(Q,\Sigma,\delta,q_0,F)$。通过合并 non-distinguishable pairs 的方式找到其对应的 minimal DFA。其实，本质上就是找到一个将 $\equiv_A$ 划分为 finer equivalence classes $\equiv_L$ 的过程。

Base case. $\epsilon$ of length $0$ distinguishes the accepting and non-accepting states.
Induction case. suppose the algorithm finds all pairs of states distinguished using strings of length at most $k$, then, consider states $(p,q)$ distinguished using string $w=ax$ of length $k+1$.
If the pair $(\delta(p,a),\delta(q,a))$ are distinguishable, then $(p,q)$ are distinguishable.

画一张 $|Q|\times |Q|$ 的表格标记所有的 distinguishable pairs，并按照上述的长度归纳法把它们都找出来；剩下的 pairs 即是 non-distinguishable pairs，把它们合并即可。

Parallel Simulation of 2-DFAs

Given $A=(Q,\Sigma,\delta,q_0,F)$ and $A'=(Q’, \Sigma,\delta',q_0',F')$.

Construct $A''=(Q\times Q',\Sigma,\delta'',(q_0,q_0'),F'')$, where $\delta''((q,q'),a)=(\delta(q,a),\delta'(q',a))$. DFA $A''$ simulates the two DFAs $A$ and $A'$ in parallel.

If we take $F''=F\times F'$, then the new DFA $A''$ accepts $L(A)\cap L(A')$.
If we take $F''=F\times Q'\cup Q\times F'$, then the new DFA $A''$ accepts $L(A)\cup L(A')$.
Appropriate choice of $F''$ results in other boolean combination of languages accepted by $A$ and $A'$.

Properties of Regular Languages

Pumping Lemma

Let $L$ be a regular language. There exists a constant $n$ such that for every string in $L$ satisfying $|w|\geq n$, we can break $w$ into three strings $w=xyz$, such that,

$y\neq \epsilon$
$|xy|\leq n$
For all $k\geq 0$, the string $xy^kz$ is also in $L$

这里的 $n$ 实际上就是 $L$ 对应的 DFA 中 states 的数目，即 $n=|Q|$。只要 $L$ 中存在长度 $\geq n$ 的字符串，说明 DFA 中一定存在环 (鸽巢原理)；该环对应的就是 $y$。

Pumping lemma 一般用来证否，即，通过找到某个长度 $\geq n$ 的字符串 $xyz$，其扩增过后的 $xy^kz\notin L$，进而说明 $L$ 并非 regular language。

Properties

Closure Properties.

之前提到的，若 $L_1,L_2$ 是 regular language，so are $L_1\cup L_2,L_1\cdot L_2,L_1^*$
so are $L_1\cap L_2,L_1-L_2,\overline{L}=\Sigma^*-L, L^R$
$h$ is a homomorphism, if $L$ is regular, so is $h(L)$

Calculation Properties.

$M+N=N+M$, $L(M+N)=LM+LN$
$L+L=L$, $(L^*)^*=L^*$, $L^+=LL^*=L^*L$, $L^*=\epsilon+L^+$
$(L+M)^*=(L^*M^*)^*$

Homomorphism

$\Sigma$ and $\Gamma$ are two alphabets. Define homomorphism $h:\Sigma\to\Gamma^*$ such that,

$h(\epsilon)=\epsilon$
(extend to string) $h(aw)=h(a)\cdot h(w)$

If $L$ is regular then $h(L)=\{h(w)|w\in L\}$ is also regular.

新的证明模式：若 $M$ 是 $L$ 的 RegEx，定义 $R(M)$ 满足以下条件 (RegEx 的性质) 并说明 $R(M)$ 所对应的是 $h(L(M))$。即，通过说明 $h(L(M))$ 存在 RegEx 的形式证明 $h(L(M))$ 本身是 regular language。

$R(\emptyset)=\emptyset$
$R(\epsilon)=\epsilon$
$R(a)=h(a)$ for $a\in \Sigma$
$R(M+N)=R(M)+R(N)$
$R(M\cdot N)=R(M)\cdot R(N)$
$R(M^*)=(R(M))^*$

Base case. $L(R(\emptyset))=h(L(\emptyset))=\emptyset, L(R(\epsilon))=h(L(\epsilon))=\epsilon, L(R(a))=h(L(a))=h(a)$.

Induction case. 完整的证明应当包括 $M+N,M\cdot N,M^*$。这里以证明 $M+N$ 为例：若 $L(R(M))=h(L(M))$ 且 $L(R(N))=h(L(N))$，求证 $L(R(M+N))=h(L(M+N))$。 \[ \begin{aligned} L(R(M+N))&=L(R(M)+R(N)) \ \ \text{by definition} \\ &= L(R(M)) + L(R(N)) \\ &= h(L(M))+h(L(N)) \\ &= h(L(M+N)) \end{aligned} \] 注意把握好 regular language 与 RegEx 间的关系。这里就 implicitly 应用了 $L(M+N)=L(M)+L(N)$。

Context Free Grammars (CFG)

Context-Free Grammars (CFG) 是一种语法，通过 derivation 过程生成对应的 language。

CFG $G=(V,T,P,S)$.
$L(G)=\{w\in T^{*}|S\Rightarrow^*_G w \}$. 即，从 start symbol $S$ 开始，能够 derive 出字符串 $w\in$ terminals $T^*$.
If $S\Rightarrow^*_G \alpha$, $\alpha$ is called a sentential form (without any non-terminals).

Right-Linear Grammars

A CFG is right-linear if all the productions in it are of the form,

$A\to wB$ for $B\in V$ and $w\in T^*$, or
$A\to w$, for $w\in T^*$

右线性文法的两个重要性质：

Every regular language can be generated by some right-linear grammar.
Language generated by a right-linear grammar is regular.

证明 $1$：若 regular language $L$ 被 DFA $A=(Q,\Sigma,\delta,q_0,F)$ 识别，根据 $A$ 一定能构造出 right-linear CFG $G$.

$G=(Q,\Sigma,P,q_0)$
If $\delta(q,a)=p$, then there is a production in $P$ of the form $q\to ap$.
If $q\in F$, then $q\to \epsilon$.

通过一个简单的 induction by length 可以得到 $\hat\delta(q_0,w)\in F \iff q_0\Rightarrow^*_G w$.

证明 $2$：给定 right-linear CFG $G=(V,\Sigma,P,S)$，证明 $G$ derive 出的 language $L(G)$ 一定能被某 NFA $A$ 识别。

$A=(V,\Sigma,\delta,S,F)$
If $(A\to aB)\in P$, then $B\in \delta(A,a)$.
$F=\{A|(A\to\epsilon)\in P\}$.

相似的，可证得 $A\Rightarrow_G^* wB \iff B\in\hat\delta(A,w)$，那么有 $S\Rightarrow_G^* w\iff \hat\delta(S,w)\cap F\neq \emptyset$.

Ambiguous Grammars

在模糊文法中，同一个字符串可能有多种不同的 derivations。我们可以通过定义 operator precedence 等规则来消除文法中的 ambiguity。但也存在 inherently ambiguous languages，它对应的文法总是 ambiguous 的。

Push Down Automata (PDA)

下推自动机。带一个容量无限栈 (作为 memory) 的 $\epsilon$-NFA。

$P=(Q,\Sigma,\Gamma,\delta,q_0,Z_0,F)$
$\Sigma$: Alphabet set, $\Gamma$: Stack alphabet set. $Z_0$: the only initial symbol on the stack.
$(p,\gamma)\in \delta(q,a,X)$: in state $p$, reading $a$, with top of stack being $X$, the machine's new state is $p$, $X$ is popped and $\gamma$ is pushed. (if $\gamma=RS$, $S$ is pushed first, then $R$)
Instantaneous description $(q,w,a)$ denotes current state is $q$, input left to read is $w$, and $\alpha$ being the top of the stack. $(q,aw,X\alpha)\vdash (p, w,\beta\alpha)$ if $(p, \beta)\in \delta(q, a,X)$. Extend $\vdash$ to $\vdash^*$.

Equivalent PDAs

PDA 按照不同的 language acceptance 方式分为两种。

By final state: $\{w|(q_0,w,Z_0)\vdash^*(q_f,\epsilon,\alpha), \text{ for some }q_f\in F\}$.
By empty stack: $\{w|(q_0,w,Z_0)\vdash^*(q,\epsilon,\epsilon)\text{ for some }q\in Q\}$.

其实这两种 PDA 是等价的。

From empty-set to final-state PDA - initially put a special symbol onto the stack. (give control to empty-set PDA), if ever see the top of stack as that symbol (seize control), then go to final state.

From final-state to empty-set PDA - place a transition from final state to a special state which empties the stack.

for all $p\in F$ and $Z\in \Gamma\cup\{X_0\}$, $\delta_E(p,\epsilon,Z)$ contains $(p_f,\epsilon)$
(pop until empty) for all $Z\in \Gamma\cup \{X_0\}$, $\delta_E(p_f,\epsilon,Z)$ contains $(p_f,\epsilon)$

Equivalence of CFGs and PDAs

CFG 和 PDA 本质等价。

First show how to accept a CFG with an empty-stack PDA (easy).

Given $G=(V,T,P,S)$, construct $PDA=(\{q_0\},\Sigma,\Gamma,\delta,q_0,S,F)$, where

$\Sigma=T$
$\Gamma=V\cup T$
For all $a\in \Sigma$, $\delta(q_0,a,a)=\{(q_0,\epsilon)\}$ [terminal alphabet is eliminated directly]
For all $A\in V$, $\delta(q_0,\epsilon,A)=\{(q_0,\gamma):A\to \gamma \text{ in } P\}$ [non-terminal alphabet is substituted]

Next show each language accepted by an empty-set PDA can be accepted by a CFG (hard).

Given $PDA=(Q,\Sigma,\Gamma,\delta,q_0,Z_0,F)$, construct grammar $G=(V, \Sigma,R,S)$, where

$V=\{S\}\cup\{[qZp]:p,q\in Q,Z\in \Gamma\}$
For each $p\in Q$, $S\to [q_0Z_0p]$ [in state $q$, with top of the stack being $Z_0$, going to state $p$]
If $\delta(q,a,X)$ contains $(r, Y_1...Y_k)$, $[qXr_k]\to a[rY_1r_1][r_1Y_2r_2]...[r_{k-1}Y_kr_k]$, for all $r_1,r_2,...,r_k\in Q$. [produce $a$, but need to get rid of $Y_1...Y_k$ one by one. The actual discarding path is permutated.]
Notice if $k=0$, $[qXr]\to a$

According to the above definition, $[qXr]$ generates $w$ iff $(q,w,X)\vdash^*(r, \epsilon,\epsilon)$ (stack being emptied), if inducted on number of steps of PDA or derivation in the CFG.

Deterministic PDA (DPDA)

与原始定义的 NPDA 不同，DPDA,

there is at most one element in $\delta(q,a,Z)$
if $\delta(q,\epsilon,X)$ is non-empty, for all $a\in\Sigma$, $\delta(q,a,X)$ is empty.

三个性质：

不同于等价的 DFA/NFA，存在能被 NPDA 接受但无法被任何 DPDA 接受的语言。
(退化为 DFA/NFA) Every regular language can be accepted by a final-state DPDA.
Any language accepted by an empty-stack DPDA has prefix property: For every $x,y$ in the language, $x$ is not a prefix of $y$. e.g., language $\{a.aa\}$ is not accepted by any DPDA.

Properties of CFGs

Useful Symbols

三个定义。

Symbol $A$ is useful if it appears as $S\Rightarrow ^*_G \alpha A\beta \Rightarrow^*_G w$ for some $w\in T^*$.
Symbol $A$ is generating if $A\Rightarrow_G^* w$, for some $w\in T^*$.
Symbol $A$ is reachable if $S\Rightarrow^*_G \alpha A\beta$, for some $\alpha,\beta \in (V\cup T)^*$.

A symbol is useful only if it is reachable and generating (vice versa need not be the case).

因此，给定一个 $G=(V,T,P,S)$ which generates at least one string，通过以下两步可以创造一个等价的 (生成的 language 与 $G$ 相同) CFG，且它不包含任何 useless symbols。

eliminate all symbols (and productions involving these symbols) that are non-generating.
eliminate all symbols (and productions involving these symbols) that are non-reachable.

Chomsky Normal Form

Chomsky Normal Form: All productions are of the form $A\to BC$ or $A\to a$ (where $a\in T$ and $A,B,C\in V$).

All CFGs can be converted to Chomsky Normal Form.

Eliminate $\epsilon$ productions.
Eliminate unit productions.
Convert all productions into the following forms:
- productions of length $2$: $A\to BC$ where $B,C$ are non-terminals.
- productions of length $1$: $A\to a$ where $a$ is a terminal.

首先来看第一步：消除所有 $\epsilon$ productions。重复以下过程，直到 $G$ 中不存在任何 $\epsilon$ production：

Identify nullable state. e.g., If $A\to\epsilon$, then $A$ is nullable.
Remove $\epsilon$ production associated to the nullable state. e.g., remove $A\to \epsilon$.
For each production $B\to\alpha$, replace it with all possible $B\to\alpha'$ resulted in possibly deleting some of the nullable non-terminals. e.g., $S\to ASA|aB|a$ is replaced by $S\to ASA|SA|AS|S|aB|a$.

接下来是第二步：消除所有 unit productions。重复以下过程，直到 $G$ 中不存在任何 unit production：

Identify non-terminal pair. e.g., If $A\to B$, then $(A,B)$ is a pair.
Remove unit production associated to the pair. e.g., remove $A\to B$.
For each non-unit productions of the form $B\to \gamma$, add production $A\to \gamma$.

经以上两步处理，$G$ 中所有的 productions $A\to \alpha$ 都表现为以下两个形式。

$|\alpha|=1$，此时 $\alpha$ 一定是 terminal。
$|\alpha|=|\alpha_1\alpha_2...\alpha_n|=n\geq 2$，$\alpha_i$ 既可以是 terminal 又可以是 non-terminal。

此时我们应用第三步，将第二种形式再进一步化成 $A\to BC$ 且 $B,C$ 均为 non-terminal 的形式：Given production $A\to X_1X_2...X_k$ is changed to the following set of productions with new non-terminals $B_i,Z_i$:

$A\to Z_1B_2,$
$B_2\to Z_2B_3,...,$
$B_{k-1}\to Z_{k-1}B_k,$ (two non-terminals, the second form)
$Z_i\to X_i$ if $X_i\in T$ (one terminal, the first form)
$Z_i=T_i$ if $X_i\in V$

乔姆斯基范式下的 CFG $G$ 的优美性质：其 parse tree 是一颗二叉树！若该二叉树的最大深度为 $s$，$G$ 所生成语言的 size 是 $O(2^{s-1})$ 级别的。

Pumping Lemma for CFL

Let $L$ be a CFL (Context Free Language, languages generated by CFG). Then there exists a constant $n$ such that, if $z$ is any string in $L$ such that $|z|\geq n$, then we can write $z=uvwxy$ such that,

$|vwx|\leq n$
$vx\neq \epsilon$
For all $i\geq 0$, $uv^iwx^iy\in L$.

Proof (所有 pumping lemma 证明的重点都在于重复). Given Chomsky Normal Form grammar $G=(V,T,P,S)$ for $L-\{\epsilon\}$. Let $m=|V|$. Let $n=2^m$.

Suppose a string $z\in L$ of length at least $n=2^m$ is given. Consider the parse tree for $z$, it must have a path from the root to a leaf of length at least $m+1$ (完全二叉树深度最小，但仍有 $m+1$).

In this path, among the last $m+1$ non-terminals, there must be two non-terminals which are the same (by pigeonhole principle).

repetition: A\Rightarrow^* A — repetition: $A\Rightarrow^* A$

Then, $z=uvwxy$, where $S\Rightarrow ^*_G uAy \Rightarrow _G^* uvAxy \Rightarrow uvwxy$.

Thus, we have $A\Rightarrow ^*_G vAx,A\Rightarrow_G^* w$
Thus, $A\Rightarrow_G^*v^iAx^i\Rightarrow_G^* v^iwx^i$
Thus, $S\Rightarrow_G^* uAy\Rightarrow_G^*uv^iwx^iy$ for all $i$ [condition 3]

Since $A$ is a subtree of $S$, the length of $vwx$ is at most $2^m=n$ [condition 1] Since $G$ is in Chomsky Normal Form, it does not have any unit productions or $\epsilon$ productions. Therefore, as $A\Rightarrow_G^* vAx$, $vx\neq \epsilon$ [condition 2]

Closure Properties

Substitution (类似 homomorphism).

Consider mapping each terminal $a$ to a CFL $L_a$, $s(a)=L_a$. For a string $w$, define $s(w)$ as follows:

$s(\epsilon)=\{\epsilon\}$
$s(wa)=s(w)\times s(a)$, for $a\in\Sigma, w\in \Sigma^*$

If $L$ is CFL over $\Sigma$ and $s$ is a substitution on $\Sigma$ such that $s(a)=L_a$ is CFL for each $a\in\Sigma$. Then, $\cup_{w\in L}s(w)$ is a CFL (证明很简单，把原 $P$ 中所有的 $A\to a$ 替换成 $A\to S_a$：$S_a$ 是 $L_a$ 对应的 starting state).

Reversal.

If $L$ is CFL, then $L^R$ is CFL (将原 $P$ 中所有的 $A\to \alpha$ 替换成 $A\to\alpha^R$)

Union.

If $L$ is CFL and $R$ is regular, then $L\cap R$ is CFL.

证明：parallel simulation of PDA for $L$ and DFA for $R$.

For $Z\in \Gamma,p\in Q,q\in Q'$, $\delta''((p,q),\epsilon,Z)=\delta(p,\epsilon,Z)\times \{q\}$
For $a\in\Sigma,Z\in \Gamma,p\in Q,q\in Q'$: $\delta''((p,q),a,Z)=\delta(p,a,Z)\times \{\delta'(q,a)\}$

注意一定是 CFL $\cap$ RL，不能是 CFL $\cap$ CFL。例如，$L_1=\{a^nb^nc^m:m,n\geq 1\}$ 与 $L_2=\{a^mb^nc^n:m,n\geq 1\}$ 均是 CFL (忽略 $m$，判断 $n=n$)，但 $L_3=L_1\cap L_2=\{a^nb^nc^n:n\geq 1\}$ 不是 context free 的 (单栈 PDA 只能处理两个长度相等的符号，三个就不行了)。

Empty.

A CFL is $\emptyset$ if $S$ is a useless symbol. Otherwise it is non-empty.

CYK Algorithm

CYK 算法 (Cocke-Younger-Kasami algorithm) 用于 CFL 的 membership testing (判断某个字串 $w$ 是否属于给定的 CFL $L$)。它是一个 $O(n^3)$ 的 DP 算法。

将 $L$ 化为 Chomsky Normal Form。对于 $w=a_1...a_n$，定义 $X_{i,j}$ 为所有能生成 $a_ia_{i+1}...a_j$ 的 non-terminals。

Base case. $X_{i,i}$ is the set of non-terminals which generate $a_i$.
Induction step. $X_{i,j}$ contains all $A$ such that $A\to BC$ and $B\in X_{i,k}, C\in X_{k+1,j}$, for $i\leq k<j$.

$w=a_1...a_n$ is in the language iff $X_{1,n}$ contains $S$.

Turing Machine: Some Concepts

Turing Machine $M=(Q,\Sigma,\Gamma,\delta,q_0,B,F)$

$\Sigma$: input alphabet set. $\Gamma$: tape alphabet. $\Sigma\subseteq \Gamma$.
$\delta: Q\times \Gamma \to Q\times \Gamma\times \{L,R\}$ (state, read $\to$ new_state, write, move left/right)
$B$: blank symbol, $B\in \Gamma-\Sigma$
$F$: set of final states. $F\subseteq Q$

Instantaneous Description. $x_0x_1,...,x_{n-1}qx_nx_{n+1}...x_m$ (无限长纸带，读写头对应的状态)

TM accepts $x$ iff $q_0 x \vdash^* \alpha q_f\beta$ (只要在转移过程中能够到达 $q_f\in F$，纸带上的字符串 $x$ 即可视为被对应的图灵机接受；与 DFA/PDA 不同，不需要 consume $x$ 中的所有字符)。$L(M)=\{x|q_0 x \vdash^* \alpha q_f \beta, q_f\in F\}$.

Function computed by Turing Machine.

Machine halts on input $x$ iff $f(x)$ is defined.
$f(x)$ 是图灵机在输入 $x$ 上停机后纸带上的结果。
This reveals that machine may never halt.

Coding of Turing Machine. Since Turing machines can be represented by finite descriptions (such as code or a set of rules), it is possible to enumerate them in an ordered way.

Church-Turing Thesis. Whatever can be computed by an algorithmic device (in function computation sense, or language acceptance sense), can be done by a Turing Machine.

Semi-Infinite-Tape TM

想象一下把纸带对折，以最左边的 cell 为界，纸带被分为 upper track 与 lower track。显然，lower track 中的「往右」实际上对应着原纸带中的「往左」。

Turing machines with semi-infinite tape are equivalent to standard Turing machines.

对 standard TM 的 simulation 如下 (新增 $U,D$ 标记 upper track, lower/down track)：

Initialization. $\delta(q_S,(X,B))=((q_0,U),(X,*),S)$: From $q_S$, read $(X,B)$, goes to state $(q_0,U)$ [$q_0$ on upper track], write $X$ on upper track, $*$ on lower track, do not move.

Simulation. Below $X,Z\in \Gamma$ and $m\in\{L,R\}$.

If $\delta(q,X)=(q',Y,m)$ [不在分界线上]
- upper track. $\delta'((q,U),(X,Z))=((q',U),(Y,Z),m)$ [rewrite $X$ to $Y$ if on upper track]
- lower track. $\delta'((q,D),(Z,X))=((q',D),(Z,Y), \overline{m})$ [rewrite $X$ to $Y$ if on lower track]
If $\delta(q,X)=(q',Y,R)$ [lower track 上的分界线]
- $\delta'((q,U),(X,*))=((q',U),(Y,*),R)$
- $\delta'((q,D),(X,*))=((q',U),(Y,*),R)$
If $\delta(q,X)=(q',Y,L)$ [upper track 上的分界线]
- $\delta'((q,U),(X,*))=((q',D),(Y,*),L)$
- $\delta'((q,D),(X,*))=((q',D),(Y,*),L)$

Multi-Tapes TM

$k$ 个无限长纸带，$k$ 个读写头，$\delta:(Q\backslash F)\times \Gamma^k \to Q\times \Gamma^k \times \{L,R\}^k$.

This model intuitively seems much more powerful than the single-tape model, but any multi-tape machine—no matter how many tapes—can be simulated by a single-tape machine using quadratically more computation.

Nondeterministic TM

在非确定性图灵机中，$\delta(q,a)$ 成为一个集合 (a finite set of possibilities)。

修改 Instantaneous Description 的定义：For any current ID, there maybe finite number of possible next ID. $x$ is accepted by non-deterministic TM iff there exists an accepting state $q_f$ such that $q_0 x\vdash^* \alpha q_f \beta$.

与 DFA/NFA 相同，DTM simulates NTM；其通过 BFS 遍历 NTM 的 computation tree 直到找到对应的接受路径。因此一般而言，与 NTM 等价的 DTM 的计算复杂度是 exponential 级别的。

Recursive Enumerable Language

$L$ is recursively enumerable (RE), or computably enumerable (CE), if some TM accepts the language $L$.
A language $L$ is recursive, or decidable, if some TM accepts the language $L$, and halts on all the inputs.
A function $f$ is partial recursive, or partially computable, if some TM computes the function (if halts on all the inputs on which $f$ is defined, and it does not halt on inputs on which $f$ is not defined).
A function $f$ is recursive, or computable, if some Turing Machine computes the function, and $f$ is defined on all elements of $\Sigma^*$ (图灵机一定会停机).

Construct a Non-RE Language (Diagonal Language) $L_d=\{w_i:w_i\notin L(M_i)\}$. Here, all Turing Machines are listed in order $M_0,M_1,M_2,...$

Define $L_u$ (Universal Language) as $L_u=\{(M,w):M \text{ accepts } w\}$. Then, $\overline{L_u}$ is also non-RE. Proof: $\overline{L_u}=\{(M,w): M \text{ does not accept } w\}$, therefore if there is a TM $M$ accepting $\overline{L_u}$, we can construct a TM $M'$ accepting $L_d$ using $M$ as a subroutine.

Theorem 1. If $L$ is recursive, then $\overline{L}$ is recursive.

WLOG assume Turing Machine $M$ accepts $L$, and there is only one accepting state.
Create a new state $q_{new}$.
Construct $M'$ accepting $\overline{L}$ as follows: for any non-accepting state $q$ and any letter $a$, if $\delta(q,a)$ is not defined in $M$, then let $\delta(q,a)=q_{new}$. Other transitions in $M'$ are as in $M$.
Finally, let $q_{new}$ to be the only accepting state in $M'$.

Theorem 2. $L$ is recursive iff $L$ is RE and $\overline{L}$ is RE.

$\Rightarrow$ part: Since $L$ is recursive, $\overline{L}$ is also recursive (Theorem 1), therefore $L$ and $\overline{L}$ are both RE.

$\Leftarrow$ part: Suppose $M$ accepts $L$ and $M'$ accepts $\overline{L}$. Construct $M''$ as follows:

$M''(x)$ copies $x$ into two different tapes.
Then it runs parallel simulation - $M(x)$ on the first tape, $M'(x)$ on the second tape.
If at any point $M(x)$ accepts, then $M''$ accepts. If $M'(x)$ accepts, then $M''$ halts and rejects.

Universal Turing Machine

Universal Turing Machine (UTM) accepts Universal Language $L_u=\{(M,w):M\text{ accepts }w\}$.

先引入 Turing Machine 的 coding scheme:

Denote tape symbols as $X_1,X_2,...,X_s$ (e.g., $X_1$ for 0, $X_2$ for 1, $X_3$ for blank)
Denote directions as $L=D_1,R=D_2$
Encode a transition $\delta(q_i,X_j)=(q_k,X_l,D_m)$ as $0^i10^j10^k10^l10^m$
Encode the TM as $C_111C_211C_3...C_n$ where $C_i$ is the $i$-th transition.

为使输入字符串 $w$ 与 $M$ 的编码一致，同样将 $w$ 按照上述方式 (1 与 3) 编码：$10^1$ for 0, $10^2$ for 1, $10^3$ for blank。

UTM has the following tapes.

Input tape: $M\#w$
Working tape: initialized with encoded $w$. read-write head on this tape always points to $1$
State tape: initialized with $0$ (represent starting state $q_1$)
Scratch tape

UTM simulates $M(x)$ as follows.

Suppose the State tape has $0^i$.
Suppose the head on the Working tape is at the start of $10^j1$.
Search the code of $M$ on the Input tape of the form $0^i10^j10^k10^l10^m$, where on both sides we have either blanks or $11$.
Change the content of State tape to $0^k$ (done using copying).
Change the string from $10^j1$ to $10^k1$ on the Working tape. (mark the current $1$ to a special symbol $*$, copy the content into Scratch tape - instead of copying the portion $*0^j1$, write $*0^k1$, then copy the Scratch tape back to the Working tape)
Revert the special symbol $*$ to $1$, move the head on the Working tape to the previous or next $1$ depending on the value of $m$.

When $M$ halts, UTM can tell if it is in the single accepting state and so moves to an accepting state of its own.

Decidability

Reduction

$P_1$ reduces to $P_2$ ($P_1\leq _m P_2$) if some recursive function $f$ behaves as follows: $x\in P_1$ iff $f(x)\in P_2$.

Some properties:

If $P_2$ is recursive, then so is $P_1$
If $P_2$ is RE, then so is $P_1$
If $P_1$ is undecidable, then so is $P_2$
If $P_1$ is non-RE, then so is $P_2$

TMs accepting empty set

Let $L_e=\{M:L(M)=\emptyset\}$, $L_{ne}=\{M:L(M)\neq \emptyset\}$.

Theorem: $L_{ne}$ is RE (存在 TM 只接受 non-empty TMs).
Theorem: $L_e$ is not recursive.
Corollary: $L_e$ is not RE (不存在 TM 只接受 empty TMs).

Proof of $L_{ne}$ is RE: Construct $M'$ to accept $M$ that $L(M)\neq \emptyset$. \[ \begin{aligned} &\text{For } t=0\text{ to }\infty \\ &\text{For }i=0 \text{ to } t \\ & \ \ \ \ \ \ \ \text{If } M(w_i) \text{ accepts within } t \text{ steps, then accept. } \\ &\text{Endfor} \\ &\text{Endfor} \end{aligned} \] Step 上界 $t$ 的引入保证 $M'$ 在输入某个 $L(M)\neq \emptyset$ 的 $M$ 时不会陷入无限循环。

Proof of $L_e$ is not RE: Reduce $\overline{L_u}$ to $L_e$. Given the input $M\#w$, construct a non-input TM $M'$ (没有输入，或者说输入 $x$ 不影响输出) \[ \begin{aligned} & M'(x) \ \ (x \text{ is not involved}) \\ &\text{For }t=0 \text{ to } \infty \\ & \ \ \ \ \ \ \ \text{If } M(w) \text{ accepts within } t \text{ steps, then accept. } \\ &\text{Endfor} \\ &\text{End } M' \end{aligned} \] 注意 $M'$ 这一图灵机是否接受输入 $x$ 与 $x$ 本身无关；它仅与 $w\in L(M)$ 有关。

若 $w\in L(M)$，即 $M\# w\notin \overline{L_u}$，$M'$ 接受任意字符串，即 $M'\notin L_e$
若 $w\notin L(M)$，即 $M\#w\in \overline{L_u}$，$M'$ 不接受任意字符串，即 $M'\in L_e$

Rice's Theorem

Suppose $P$ is a non-trivial property about RE languages. Then $L_p$ is undecidable.

non-trivial property: at least one RE language satisfies $P$, and at least one RE language doesn't satisfy $P$.
$L_p=\{M|L(M) \text{ satisfies property } P\}$.

Proof.

Suppose $P$ is a non-trivial property about RE languages. WLOG assume $\emptyset$ satisfies $P$ (otherwise switch $P$ to $\overline{P}$). Suppose $L$ is an RE language that does not satisfy $P$. Let $M''$ be the machine which accepts $L$.

Define $f:f(M)=M'$ as follows. \[ \begin{aligned} &M'(x) \\ &\text{For }t=0 \text{ to } \infty \\ &\text{For }i=0 \text{ to } t \\ &\ \ \ \ \ \ \ \text{If } M(w_i) \text{ accepts within }t \text{ steps and } \\ & \ \ \ \ \ \ \ M''(x) \text{ accepts with in }t \text{ steps, then accept } x \\ &\text{EndFor} \\ &\text{EndFor} \end{aligned} \] We can see that $L_e\leq_m L_p$.

If $L(M)=\emptyset \in L_e$, then $L(M')=\emptyset \in L_p$.
If $L(M)\neq \emptyset \notin L_e$, then $L(M')=L\notin L_p$.

As $L_e$ is not recursive, we have that $L_p$ is also not recursive.

Post's Correspondence Problem

PCP: 给定两个字符串表 (list of strings) $A=w_1,w_2,...,w_k$ 与 $B=x_1,x_2,...,x_k$；波斯特对应问题的解是一个序列 $i_1,i_2,...,i_m \ (m>0)$ 使得 $w_{i_1}w_{i_2}...w_{i_m}=x_{i_1}x_{i_2}...x_{i_m}$

MPCP: 在 PCP 基础之上规定两个字串的首个字符串是一种特定组合。举例来说，规定必须使用 $w_1$ 与 $x_1$，MPCP 的解序列将使得 $w_1w_{i_1}w_{i_2}...w_{i_m}=x_1x_{i_1}x_{i_2}...x_{i_m}$

Argument 1. MPCP $\leq_m$ PCP.

对于任意 MPCP 问题，我们能够构造一个等价的 PCP 问题；若该 PCP 问题有解，相应的 MPCP 问题亦有解。

MPCP 问题：给定 $A=w_1,w_2,...,w_k$ 与 $B=x_1,x_2,...,x_k$ 要求起始字符串组合为 $w_i$ 与 $x_j$。按照如下方式构造 PCP 问题的字符串表 $A'=w_0',w_1',...,w_{k+1}'$ 与 $B=x_0',x_1',...,x_{k+1}'$。

$w_i'=$ inserting $*$ after each symbol in $w_i$
$x_i'=$ inserting $*$ before each symbol in $x_i$
$w_0'=*w_i',x_0'=x_j'$ (灵魂)
$w'_{k+1}=\$$, $x_{k+1}'=*\$$

不得不说真的很妙。如果在 $A',B'$ 上 PCP 有解，首先可以观察到的是解序列一定以 $0$ 开头，以 $k+1$ 结尾：这是由 $*$ 与 $\$$ 的巧妙安排决定的。若解序列为 $0,i_1,i_2,...,i_m,k+1$，那么对应的 MPCP 问题的解序列为 $i,i_1,i_2,...,i_m$ (for $A$) 与 $j,i_1,i_2,...,i_m$ (for $B$)。

Argument 2. $L_u\leq_m$ MPCP.

给定输入 $M\#w$，我们的目的是构造一个等价的 MPCP 问题：若该 MPCP 问题有解，$M$ 接受 $w$。这就将 $L_u$ 归约到了 MPCP 问题上。构造方式 (真的很妙啊！)

若该 MPCP 有解，我们可以看到每个由 $\#$ 包围起来的 section 都是某一步的 instantaneous description；并且直观上来说，是 $A$ 形成的字符串在「追赶」$B$ 形成的字符串 (毕竟一开始是 $\#$ v.s. $\#q_0w\#$)。

什么时候算「追上」了呢？就是 $A$ 的 ID 到达 accepting state 的时候。此时 $A$ 中的 $XqY,Xq, qX$ 都换一个 $B$ 中的 $q$，两者之间的差距越来越小，直到 $A$ 中被消除到只剩一个 $q$；此时 $q\#\#$ 换一个 $B$ 中的 $\#$，两字符串完全一致。

MPCP 中这一「追赶」的过程对应的正是 $M$ 在输入 $w$ 上的转移过程。具体来说，$A$ 是前一步，$B$ 是后一步；如果 $A$ 最终能追赶上 $B$ (MPCP 有解)，那么 $M$ 也能够到达 accepting state ($M$ accepts $w$)。

Undecidable CFGs

PCP 问题又可视作一个歧义语法的判断问题。给出字符串表 $A=w_1,...,w_k$ 与 $B=x_1,...,x_k$，我们再添加一个 index list $I=\{a_1,a_2,...,a_k\}$；其中 $I\cap \Sigma=\emptyset$。定义以下语法 $G$： \[ \begin{aligned} &S\to A \\ &S\to B\\ &A\to w_iAa_i, \text{ for }1\leq i\leq k \\ &A\to w_ia_i,\text{ for }1\leq i\leq k \\ &B\to x_iBa_i, \text{ for }1\leq i\leq k\\ &B\to x_ia_i, \text{ for }1\leq i\leq k \end{aligned} \] 该语法又可分为两个子语法 $G_A$ (仅与 $A$ 有关的部分)，$G_B$ (仅与 $B$ 有关的部分)；由此可以得出一系列结论链条：$L(G_A)\cap L(G_B)\neq \emptyset$ iff $G$ is ambiguous iff PCP has a solution.

很好理解：若 $L(G_A)\cap L(G_B)$ 交集不为空，那么其中任意一个字符串的 index 部分都表示一个 PCP 问题的解；且由于 $G$ 在 $G_A$ 与 $G_B$ 方向都能生成相同的字符串，说明 $G$ 是歧义的。由于 PCP 问题的不可判定性，我们进一步证明了 $G$ 的歧义与否与 $L(G_A)\cap L(G_B)$ 非空与否均是不可判定的。

可以证明 $L(G_A),L(G_B), \overline{L(G_A)}, \overline{L(G_B)}$ 均为 CFG。注意，general CFG $G$ 取反并不一定是 CFG，是这里的特殊 CFG $G_A,G_B$ 取反后仍为 CFG。

从 PCP 问题的不可判定开始，我们又可以引入一系列与 CFG 相关的不可判定问题：

给出 CFGs $G_1$ 与 $G_2$，判定 $L(G_1)\cap L(G_2)=\emptyset$
- 无法判定！e.g., 取 $G_1=G_A,G_2=G_B$ 即可。
给出 CFGs $G_1$ 与 $G_2$，判定 $L(G_1)=L(G_2)$
- 无法判定！e.g., 取 $L(G_1)=\overline{L(G_A)}\cup\overline{L(G_B)}=\overline{L(G_A)\cap L(G_B)}$；而 $L(G_2)$ 则是 $(\Sigma\cup I)^*$ (全集)。
给出 CFG $G$ 与 RE $R$，判定 $L(G)=L(R)$
- 无法判定！与问题 2 类似。
Given CFGs $G,G_1,G_2$ 与 RE $R$，$L(G)=T^*,L(G_2)\subseteq L(G_1),L(R)\subseteq L(G)$ 均不可判定。

NP Problems

Complexity

TM $M$ 的时间复杂度：

If $M$ is deterministic TM, $Time_M(x)$ is the number of steps used by $M$ on input $x$ before halting (if $M$ does not halt on $x$, $Time_M(x)=\infty$)
If $M$ is non-deterministic TM, $Time_M(x)$ denotes the maximum time on any path, even non-accepting.
If $M$ is $T(n)$ time bounded, $Time_M(x)\leq T(n)$ for all $|x|=n$.

TM $M$ 的空间复杂度：

$Space_M(x)$ is the maximum number of cells touched by $M$ on input $x$. Usually we only consider the cells on work tapes (excluding input & output tape). If $M$ does not halt on $x$, $Space_M(x)=\infty$.
If $M$ is $S(n)$ space bounded, $Space_M(x)\leq S(n)$.

为什么在计算复杂度时可以忽略常数 $c$？可以证明图灵机总是可以进行 Tape Compression (空间上的常数省略) 与 Linear Speedup (时间上的常数省略)。

Tape Compression. WLOG fix $c>0$. If a language $L$ is accepted by a machine $M$, with $k$ tapes, that is $S(n)$ space bounded, then $L$ is accepted by a machine $M'$, with $k$ tapes, that is $\lceil c S(n)\rceil$ space bounded.
Linear Speedup. WLOG fix $c>0$. Suppose $L$ is accepted by a machine $M$, with $k\geq 2$ tapes, that is $T(n)$ time bounded, where $\lim_{n\to\infty}T(n)/n=\infty$. Then $L$ is also accepted by a machine $M'$ that is $\lceil cT(n)\rceil$ time bounded.

几大 complexity class 之间的关系。

$\text{DTIME}(S(n))\subseteq \text{DSPACE}(S(n))$
If $L$ is in $\text{DSPACE}(S(n))$, and $S(n)\geq \log n$, then there exists a constant $c$, which depends on $L$, such that $L$ is in $\text{DTIME}(c^{S(n)})$.
If $L$ is in $\text{NTIME}(T(n))$ then there exists a constant $c$, which depends on $L$. such that $L$ is in $\text{DTIME}(c^{T(n)})$.

NP

P & NP.

$P=\{L|\text{ some }poly(n) \text{ bounded deterministic DTM accepts }L\}$
$LP=\{L|\text{ some }poly(n)\text{ bounded NDTM accepts }L\}$

Proposition. If $L\in NP$, then there exists a deterministic polynomial time computable predicate $P(x,y)$, and a polynomial $q(\cdot)$ such that: \[ x\in L \iff \exists y:|y|\leq q(|x|) \text{ such that } P(x,y)=1 \] Proof. If $L\in NP$, there exists a $q(n)$ time bounded NDTM $N$ accepting $L$. WLOG assume that $N$ has exactly two choices in each state ($|\delta(q,\alpha)|=2$), and $y$ is a 01 string.

We verify $P(x,y)$ as follows.

Let $y=y_1y_2...y_m$. If $m>q(|x|)$, then reject.
Otherwise simulate $N$, where at step $i$, choose the next state based on whether $y_i$ is $0$ or $1$ (we can say $y$ guide through $x$'s matching in $N$)
If in the end $N$ stays in an accepting state ($N$ accepts in the above simulation), $P(x,y)=1$.

根据以上的 simulation，如果 $x\in L$，那么一定存在某个 $y$ 满足 $|y|\leq q(|x|)$，使得 $P(x,y)=1$。我们将这样的 $y$ 称为 $x\in L$ 的一个 certificate 或 proof。据此，有人把 NP 问题定义为 a class of languages for which "proofs" can be easily (in polynomial time) verified.

NP-Completeness

定义 reduction $L_1\leq_m^p L_2$: there exists a poly-time computable function $f$ such that $x\in L_1\iff f(x)\in L_2$. The relation $\leq_m^p$ is reflexive and transitive.

我们说 $L$ 是 NP-complete 的当前仅当：

$L\in NP$
$\forall L'\in NP, L'\leq_m^p L$

NP-hard 问题仅仅满足 (2)，不要求其是 NP 问题。

著名的 NP-complete 问题：Satisfiability (e.g., 3-SAT), 3D Matching (三分图匹配), Vertex Cover, MAX-CUT (最大割问题), K-Clique, Independent Set, Hamiltonian Circuit, Partition, Set Cover, Traveling Salesman Problem (权重和 $\leq B$ 的哈密顿回路)。

$P=NP$ 问题：

If one of the NP-complete problem is solvable in poly-time ($\exists L\in NPC\land L\in P$), then all the problems in NP are solvable in poly-time ($P=NP$).
If $P\neq NP$, then none of the NP-complete problems are solvable in poly-time.

将 3-SAT 问题归约到 Vertex Cover 问题：若有 $n$ 个变量，$m$ 个 3-SAT clauses，构造出一个点数为 $2n+3m$ 的图，分别是 $n$ 个 $x_i$，$n$ 个 $\lnot x_i$，$m$ 个三角形，每个三角形对应一个 clause 中的三个 literals。连边策略很显然：该图的 Vertex Cover 解对应的是原 3-SAT 的一个解。

独立集 (Independent Set) 问题：独立集中的点两两不相连。给出 $G=(V,E)$，如果 $G$ 有一个大小为 $k$ 的 vertex cover，那么它有一个大小为 $|V|-k$ 的独立集，且 $G'=(V,E^c)$ 有一个大小为 $|V|-k$ 的 clique。

那么有 $V'$ 是 $G$ 的 vertex cover $\iff$ $V-V'$ 是 $G$ 的独立集 $\iff$ $V-V'$ 是 $G'$ 的 clique。

Reference

This article is a self-administered course note.

References in the article are from corresponding course materials if not specified.

Course info: CS3231. Professor: Sanjay Jain.

Course textbook: Introduction to Automata Theory, Languages and Computation (3rd Edition), Hopcraft, Motwani & Ullman, (2014), Pearson

Course website: https://www.comp.nus.edu.sg/~sanjay/cs3231.html

[1] 引自杨老师，摘录如下：「据考证，本科生课程 CSIS0293/COMP3293 Introduction to Theory of Computation 最后一次开设是在 Autumn 2005，研究生课程 CSIS9601/COMP9601 Theory of Computation and Algorithm Design 最后一次开设是在 Autumn 2016。」

-----------------------------------そして、次の曲が始まるのです。-----------------------------------

Preliminaries

DFA, NFA and RegEx

DFA & NFA: Equivalence

DFA to RegEx

RegEx to \(\epsilon\)-NFA

Minimization of DFA

Equivalence Classes

Distinguishable Pairs

Parallel Simulation of 2-DFAs

Properties of Regular Languages

Pumping Lemma

Properties

Homomorphism

Context Free Grammars (CFG)

Right-Linear Grammars

Ambiguous Grammars

Push Down Automata (PDA)

Equivalent PDAs

Equivalence of CFGs and PDAs

Deterministic PDA (DPDA)

Properties of CFGs

Useful Symbols

Chomsky Normal Form

Pumping Lemma for CFL

Closure Properties

CYK Algorithm

Turing Machine: Some Concepts

Semi-Infinite-Tape TM

Multi-Tapes TM

Nondeterministic TM

Recursive Enumerable Language

Universal Turing Machine

Decidability

Reduction

TMs accepting empty set

Rice's Theorem

Post's Correspondence Problem

Undecidable CFGs

NP Problems

Complexity

NP

NP-Completeness

Reference