Formal Languages and Compilers (形式语言和编译器) 的 自学笔记兼学习教程。
笔记作者介绍：大爽歌, b站小UP主 ，编程1对1辅导老师。

2.1 Context-Free Grammars

上下文无关语法

A context-free grammar (CFG) is a structure $G=(N,\sum ,P,S)$ where

• $N$: a finite set, the non-terminals.
  each element $v\in N$ is called a nonterminal characher on a variable.
  Each variable represents a different type of phrase or clause in the sentence. Variables are also sometimes called syntactic categories.
  Each variable defines a sub-language of the language defined by $G$.
  一个有限集合，非终结符。
  其元素为非终结字符或变量。
  每个变量代表句子中不同类型的短语或从句。
  变量有时也称为句法类别。
  每个变量定义了$G$定义的语言的一种子语言。

• $\sum$: a finite set of the terminals, disjoint form N.
  The set of terminals is the alphabet of the language defined by the grammar G.
  一个有限集合, 终结符，与$N$不相交。
  终结符集合是语法$G$定义的语言的字母表。

• $P$: $P\subseteq N\times (N\cup \sum )\ast$ is a finite set of productions.(a finite relation).
  The member of $P$ are called the rules or productions if the grammar.
  是生成式的有限集合。（关系的有限集合）。

  $P$ 的成员称为语法的规则或生成式。

• $S$: $S\in N$ is the start symbol(or start variable), used to represent the whole sentence(or program).
  $S$为起始变量（或起始符号），用于表示整个句子（或程序）。

Productions are denoted as follows:
Productions表示如下:

• A production $(A,w)\in P$ is written $A\to w$.
• Serveral productions $A \rightarrow w_1,…, A \rightarrow w_n $ are written $A\to w_1|...|w_n$
• The right-hand side may be empty: an $\epsilon$-production is written $A\to w$.

Example 1
能表示所有算术表达式的$G$
$G=(N,\sum ,P,E)$ for arithmetic expressions

• N = { E , T , F } N=\{E, T, F\} N={ E,T,F}
• $\sum =+,-,\ast ,/,(,),n$
• $E$： start symbol
• $P$:
  $E\to T|T+E|T-E$
  $T\to F|T\ast F|T/F$
  $F \rightarrow (E) | n $

其中n代表number
E代表Expression
T代表Term
F代表Factor

Example 2
回文字符(生成所有a和b构成的回文字符)
G 2 = ( { S } , { a , b } , P , S ) G_2 = (\{S\}, \{a,b\}, P, S) G2​=({ S},{ a,b},P,S)
P:

• $S\to aSa$
• $S\to bSb$
• $S\to a$
• $S\to b$
• $S\to \epsilon$

A typical derivation in this grammar is
$S\to aSa\to aaSaa\to aabSbaa\to aabbaa$

The language is context-free, however, it can be proved that it is not regular.

derivable

生成(导出，派生)

Consider a CFG $G=(N,\sum ,P,S)$

• Let $u,v,w\in (N\cup \sum {)}^{\ast }$ and $A\to w\in P$.
• Then $uAv$ yields uwv in one step, by replacing $A$ with $w$; this is denoted $uAv{\Rightarrow }_G^{1}uwv$.
• Alternatively, $uwv$ is derivable form $uAw$ in one step.
• Each of $u$, $v$, $w$ may be $\epsilon$.
• $A$ can be replaced with $w$ irrespective of the context $u$, $v$ in which A occurs.

Alternatively: 或者, 换言之

简单来讲，就是:
$uAv{\Rightarrow }_G^{1}uwv$ 表示：
$uwv$ is derivable form $uAw$ in 1 step.
$uwv$ 可由$uAw$ 一步生成(导出，派生).

进一步拓展
Derivability is a relation on $(N\cup \sum {)}^{\ast }$
Derivability: 可派生性

• Let $x_i\in (N\cup \sum {)}^{\ast }$ for each $ i\in \mathbb N$
• $x_n$ is derivable form $x_0$ in $n$ steps if $x_i{\Rightarrow }_G^1{x}_{i+1}$ for each $0\le i<n$; this is denoted $x_0 \Rightarrow_G^n x_n $
• $x{\Rightarrow }_G^{0}y$ if and only if $x=y$
• $y$ is derivable from $x$ if it is derivable in any number of steps.
• Alternatively, $x$ generates $y$ or $x$ yields $y$.
• $x{\Rightarrow }_G^{\ast }y$ if $x{\Rightarrow }_G^{n}y$ for some $ n \in \mathbb N$
• The relation ${\Rightarrow }_G^{\ast }$ is the reflexive-transitive closure of the relation ${\Rightarrow }_G^1$

reflexive-transitive closure: 自反传递闭包

举例说明
以前面例2中的CFG $G2$ 为例
derivation:
$aSa\to aba$
所以$aSa{\Rightarrow }_G^{1}aba$

$S\to aSa\to aaSaa\to aabSbaa\to aabbaa$

所以
$aaSaa{\Rightarrow }_G^{2}aabbaa$
$aSa{\Rightarrow }_G^{3}aabbaa$
$S{\Rightarrow }_G^{4}aabbaa$

Context-Free Languages

A context-free grammar generates a context-free language (CFL).

• A sentential form is any $x\in (N\cup \sum {)}^{\ast }$ deriable from the start symbol S, that is, $S{\Rightarrow }_G^{\ast }x$
• A sentence is a sentential form that consists only of terminal symbols: $x\in {\sum }^{\ast }$
• L ( G ) = { x ∈ ∑ ∗ ∣ S ⇒ G ∗ x } L(G) = \{x \in \sum^* | S \Rightarrow_G^* x \} L(G)={ x∈∑∗∣S⇒G∗​x} is the language generated by G.
• $A\subseteq {\sum }^{\ast }$ is context-free if $A=L(G)$
  for some CFG $G$.

A sentential form: 一个句子形式

简单来讲，CFG $G$ 生成的语言就是 CFL

Example 3
A = { a n b n ∣ n ∈ N } A = \{ a^nb^n | n \in \mathbb N \} A={ anbn∣n∈N} is context-free.

$A=L(G)$ for the CFG, G = ( { S } , { a , b } , P , S ) G=(\{S\}, \{a, b\}, P, S) G=({ S},{ a,b},P,S)
with production P:

• $S\to aSb$
• $S\to \epsilon$

Futher examples of CFLs:
A = { a i b j c k ∣ ( i = j    o r    j = k )    a n d    i , j , k ≥ 1 } A=\{a^ib^jc^k | (i=j\;or\; j=k)\; and\; i, j, k \geq 1 \} A={ aibjck∣(i=jorj=k)andi,j,k≥1} is generated by G = ( { S , T , U , A , C } , { a , b , c } , P , S ) G=(\{S, T, U, A, C\}, \{a, b, c\}, P, S) G=({ S,T,U,A,C},{ a,b,c},P,S) with production P:

• $S\to TC|AU$
• $T\to aTb|ab$
• $U\to bUc|bc$
• $A\to a|aA$
• $C\to c|cC$

The languages of balanced parentheses is generated by G = ( { S , T , U , A , C } , { ( , ) } , P , S ) G=(\{S, T, U, A, C\}, \{(, )\}, P, S) G=({ S,T,U,A,C},{(,)},P,S) with production P:

• $S\to (S)|SS|\epsilon$

balanced paratheses: 平衡括号

Regular Grammars

A regular grammar is a CFG $G=(N,\sum ,P,S)$ where for each $A\to w\in P$

• $w=\epsilon$, or
• $w\in \sum N$
• The right-hand side of each production is either $\epsilon$ or a terminal followed by a non-terminal.
  即每个产生式的右边要么是$\epsilon$，要么是一个终结符后面是一个非终结符(即以非终结符结尾)。

Example 4
An example of a regular grammar $G=(N,\sum ,P,S)$

• N = { S , A } N=\{S, A\} N={ S,A}
• ∑ = { a , b , c } \sum=\{a, b, c\} ∑={ a,b,c}
• P:
  $S\to aS|bA$
  $A\to cA|\epsilon$

This grammar describes the same language as the regular expression $a^{\ast }b{c}^{\ast }$, viz. the set of all strings consisting of arbitrarily many "a"s, followed by a single "b", followed by arbitrarily many "c"s.

Every regular language is generated by a regular grammar.
Proof:

• Consider the DFA $M=(Q,\sum ,\delta ,q_0,F)$.
• Construct the regular grammar $G=(Q,\sum ,P,q_0)$ with the following productions P:
  $q_i \rightarrow aq_j \quad if ; \delta(q_i, a) = q_j $
  $q_i \rightarrow \varepsilon \quad if ; q_i \in F $
• Then for each $w\in {\sum }^{\ast }$,

$$\begin{array}{rl}& w=a_1{a}_2...a_{n-1}{a}_n\in L(M)\\ \iff & \hat{\delta }(q_0,w)\in F\\ \iff & \hat{\delta }(q_{i-1},a_i)=q_{i}foreach1\le i\le nand{q}_n\in F\\ \iff & q_{i-1}\to a_i{q}_i\in Pforeach1\le i\le nand{q}_n\to \epsilon \in F\\ \iff & q_0\Rightarrow a_1{q}_1\Rightarrow a_1{a}_2{q}_2\Rightarrow ...\Rightarrow a_1{a}_2...a_{n-1}{a}_n{q}_n\Rightarrow a_1{a}_2...a_{n-1}{a}_n\\ \iff & w=a_1{a}_2...a_{n-1}{a}_n\in L(G)\end{array}$$

• Hence $L(M)=L(G)$

Chomsky Normal Form (CNF)

A CFG $G=(Q,\sum ,P,S)$ is in Chomsky normal form if every production has the form

• $A\to BC$, where $B,C\in N$, or
• $A\to a$, where $a\in N$.
• The right-hand side of each production is either two non-terminals or a terminal.

简单来讲，每个生成式结果为两个非终结符($Q$中的)或者一个终结符($\sum$中的)。

RHS: right-hand side
LHS: left-hand side

Example
G 1 = ( { S , A , B } , { a , b , c } , P , S ) G_1 = (\{S, A, B\}, \{a,b,c\}, P, S) G1​=({ S,A,B},{ a,b,c},P,S) with $P$:

• $S\to AB$
• $S\to c$
• $A\to a$
• $B\to b$

G 1 = ( { S , A , B } , { a , b , c } , P , S ) G_1 = (\{S, A, B\}, \{a,b,c\}, P, S) G1​=({ S,A,B},{ a,b,c},P,S) with $P$:

• $S\to aA$
• $A\to a$
• $B\to c$

$G_1$ is in CNF
$G_2$ is not in CNF

For every CFG G with $\epsilon \notin L(G)$ there is a CFG $G^{\prime }$ in Chomsky normal form with $L(G)=L(G^{\prime })$

1. Eliminate $\epsilon$-productions of the form $A\to \epsilon$.
2. Eliminate unit-productions of the form $A\to B$.
3. Eliminate non-generating non-terminals.
4. Eliminate non-reachable non-terminals.
5. Eliminate terminals form right-hand sides of length at least 2.
6. Eliminate right-hand sides of length at least 3.

Eliminate: 消除
non-generating: 非生成的
non-reachable: 不可到达

Example

https://www.javatpoint.com/automata-chomskys-normal-form

Convert the given CFG to CNF. Consider the given grammar $G_1$:

G 1 = ( { S , A , B } , { a , b , c } , P , S ) G_1 = (\{S, A, B\}, \{a,b,c\}, P, S) G1​=({ S,A,B},{ a,b,c},P,S) with $P$:

• $S\to TbT$
• $T\to aU$
• $T\to U$
• $T\to V$
• $U\to \epsilon$
• $V\to b$

2.2 The Cocke-Younger-Kasami Algorithm

Given a string $w\in {\sum }^{\ast }$ and a CFL $A$, is $ w \in A$?

分析

• This is the test for membership in a CFL.
• Checking all derivations does not work, since there might be infinitely many.
• It suffices to consider derivations that introduce up to $|w|$ non-terminals.
• This gives an upper bound on the length of derivations that need to be checked.
• The number of derivations might still be exponential in the length of $w$.

A technique to improve the running time is dynamic programming.

The CYK algorithm solves the membership problem $w\in L(G)$

Assume that $G$ is in chomsky normal form, for example:
$S\to BB|AS|a$
$A\to BC$
$B\to BS|B$
$C\to a$

• let $n$ be the length of $w$: for example, $n=6$ for $w=bbabab$.
• mark the positions that separate symbols in $w$:
• Let $w_{ij}$ be the substring of $w$ between positions $i$ and $j$: for example, $w_{25}=aba$ and $w_{06}=w$.
• N i j = { A ∈ N    ∣    A ⇒ ∗ w i j } N_{ij} = \{A \in N \;|\; A \Rightarrow ^* w_{ij}\} Nij​={ A∈N∣A⇒∗wij​} contains the non-terminals that generate $w_{ij}$
• The CYK algorithm calculates $N_{ij}$ for each $0\le i\le j\le n$
• Then $w\in L(G)$ if and only if $S\in N_{0n}$

The CYK algorithm fills a table with $N_{ij}$ in column $i$, row $j$.
The calculation proceeds in increasing order of substring length.

Example

wikipedia example

CFG G 2 = ( { S , A , B , C } , { a , b } , S , P ) G_2 = (\{S, A, B, C\}, \{a, b\}, S, P) G2​=({ S,A,B,C},{ a,b},S,P) with P:
$S\to AB|BC$
$A\to BA|a$
$B\to CC|b$
$C\to AB|a$

check $bbabaa\in L(G_2)$ ?

下来用CYK算法绘制表格

0. 初始表格

1. 处理表格中长度为1的。
   First come the substrings of length 1, that is, $j=i+1$.

• The 1-symbol substring $w_{i,i+1}$ can be generated form $A$ if $A\to w_{i,i+1}\in P$.
• For each production $A\to a$ where $a=w_{i,i+1}$, add A to the entry at column $i$, row $j$.
• These entries form the main diagonal of the table.

2. 处理表格中长度为2的。
   Then come the substrings of length 2, that is, $j=i+2$.

• The 2-symbol substring $w_{i,i+2}$ is broken up into two 1-symbol substrings $w_{i,i+1}$ and $w_{i+1,i+2}$.
• If $B\in N_{i,i+1}$ and $C\in N_{i+1,i+2}$ and $A\to BC\in P$, then add $A$ to $N_{i,i+2}$.
• These entries form the diagonal below the main diagonal.

• 计算$w_{02}$, 将其拆分成$w_{01}$ 和$w_{12}$.
  N 01 = { B } N_{01} = \{B\} N01​={ B}, N 12 = { B } N_{12} = \{B\} N12​={ B}, 没有$A\in P\to BB$, 所有不存在

• 计算$w_{13}$, 将其拆分成$w_{12}$ 和$w_{23}$.
  N 12 = { B } N_{12} = \{B\} N12​={ B}, N 23 = { A , C } N_{23} = \{A, C\} N23​={ A,C}, $A\to BA\in P$, $S\to BC\in P$， 所以 N 13 = { S , A } N_{13} = \{S, A\} N13​={ S,A}

• 计算$w_{24}$, 将其拆分成$w_{23}$ 和$w_{34}$.
  N 23 = { A , C } N_{23} = \{A, C\} N23​={ A,C}, N 34 = { B } N_{34} = \{B\} N34​={ B}, $S\to AB\in P$, $C\to AB\in P$，所以 N 24 = { S , C } N_{24} = \{S, C\} N24​={ S,C}

3. 处理表格中长度为3的。
   The 3-symbol substring $w_{i,i+3}$ can be broken up in two ways.

• Consider how to generate $w_{i,i+3}$ using a production $A\to BC$, that is, $A\Rightarrow BC{\Rightarrow }^{\ast }{w}_{i,i+3}$

• This follows from $B{\Rightarrow }^{\ast }{w}_{i,i+1}$ and $C{\Rightarrow }^{\ast }{w}_{i+1,i+3}$ or from $B{\Rightarrow }^{\ast }{w}_{i,i+2}$ and $C{\Rightarrow }^{\ast }{w}_{i+2,i+3}$

• If $B\in N_{i,i+1}$ and $C\in N_{i+1,i+3}$ and $A\to BC\in P$, then add $A$ to $N_{i,i+3}$.

• If $B\in N_{i,i+2}$ and $C\in N_{i+2,i+3}$ and $A\to BC\in P$, then add $A$ to $N_{i,i+3}$.

• $w_{03}$ 可以分成 $w_{01}$+ $w_{13}$, 计算出$A$
  或分成$w_{02}$+ $w_{23}$, $w_{02}$空，没有
• $w_{14}$ 可以分成 $w_{12}$+ $w_{24}$, 计算出$S$
  或分成$w_{13}$+ $w_{34}$, 计算出$S,C$

4. 处理表格中长度为4的。
   此时每个$w_{i,i+4}$有三种分法

• $w_{i,i+1}$ 和 $w_{i+1,i+4}$
• $w_{i,i+2}$ 和 $w_{i+2,i+4}$
• $w_{i,i+3}$ 和 $w_{i+3,i+4}$

5. 处理表格中长度为5的。

6. 处理表格中长度为6的。

最后
$S\in N_{06}$
所以$bbabaa$可以由该语法生成。