Lecture 4

Context-Free Grammar and Context-Free Language

Definition

A context-free grammar (CFG) \(G = (V, \Sigma, S, R)\), where

• \(V\) is a finite set of symbols.
• \(\Sigma \subseteq V\) is the set of terminals.
  • \(V - \Sigma\) is the set of non-terminals.
• \(S \in V - \Sigma\) is the start symbol.
• \(R \subseteq (V - \Sigma) \times V^*\) is a finite set of rules.
  • \((A, w)\) means \(A \rightarrow w\).

Definition

• \(\forall x, y, u \in U^*, A \in U - \Sigma\), if \((A, u) \in R\), then we have \(xAy \Rightarrow_G xuy\) and we say it derive in one step.

• \(\forall w, y, u \in U^*, A \in U - \Sigma\), if \(w = u\) or \(w \Rightarrow_G \cdots \Rightarrow_G u\), then we have \(w \Rightarrow^*_G u\) and we say it a derivation from \(w\) to \(u\) of length \(n\).

Definition

\(G\) generates \(w \in \Sigma^*\) if \(S \Rightarrow^*_G w\).

For \(L(G) = \{w \in \Sigma^* | G \text{ generates } w\}\)., we say \(G\) generates \(L(G)\).

Definition

We do a leftmost derivation of a string from a grammar by

1. Begin with a string consisting only of the start symbol.
2. Pick the leftmost variable occurrence in the current string.
3. Pick any production that has that variable on the left of the \(\rightarrow\). .
4. Replace the chosen occurrence of that variable by the right-hand side of the chosen production.
5. Repeat steps 2-4 until the string contains only terminals.

Similarly, we have rightmost derivation.

Sometimes, leftmost derivation is the same as rightmost derivation.

Example

For grammar \(S \rightarrow SS\), \(S \rightarrow (S)\), \(S \rightarrow e\), to get \(()()\) from \(S\),

• (leftmost) \(S \rightarrow SS \rightarrow (S)S \rightarrow ()S \rightarrow ()(S) \rightarrow ()()\)
• (rightmost) \(S \rightarrow SS \rightarrow S(S) \rightarrow S() \rightarrow (S)() \rightarrow ()()\)

But sometimes not.

Example

For grammar \(E \rightarrow E + E\), \(E \rightarrow E \times E\), \(E \rightarrow (E)\), \(E \rightarrow a\) to get \(a + a \times a\) from \(S\),

We call it an ambiguous expression. To find out the difference, we first introduce parse tree.

Parse Tree

Definition

Parse trees are a way of recording derivations that focus less upon the order in which derivation steps were applied and more on which productions were employed.

A parse tree for a string \(s\) in a languages described by a CFG is a tree in which

1. The root of the tree is labeled with \(S\).
2. Each internal node is labeled with a variable.
3. Each leaf node is labeled with a terminal.
4. The leaf nodes, read from left to right, provide the string \(s\).
5. Each internal node labeled with a variable \(A\) has children labeled with the symbols of \(\alpha\) (reading left to right through the children) only if \(A \rightarrow \alpha\) is a production in the grammar.

Example

For the two cases above, their corresponding parse trees are

graph TD;
    N1(("S"))
    N2(("S"))
    N3(("S"))
    N4(("("))
    N5(("S"))
    N6((")"))
    N7(("("))
    N8(("S"))
    N9((")"))
    N10(("e"))
    N11(("e"))
    N1 === N2
    N1 === N3
    N2 === N4
    N2 === N5
    N2 === N6
    N3 === N7
    N3 === N8
    N3 === N9
    N5 === N10
    N8 === N11

First Probability (multiply then add)Second Probability (add then muliply)

graph TD;
    N1(("E"))
    N2(("E"))
    N3(("+"))
    N4(("E"))
    N5(("a"))
    N6(("E"))
    N7(("×"))
    N8(("E"))
    N9(("a"))
    N10(("a"))
    N1 === N2
    N1 === N3
    N1 === N4
    N2 === N5
    N4 === N6
    N4 === N7
    N4 === N8
    N6 === N9
    N8 === N10

graph TD;
    N1(("E"))
    N2(("E"))
    N3(("×"))
    N4(("E"))
    N5(("E"))
    N6(("+"))
    N7(("E"))
    N8(("a"))
    N9(("a"))
    N10(("a"))
    N1 === N2
    N1 === N3
    N1 === N4
    N2 === N5
    N2 === N6
    N2 === N7
    N4 === N8
    N5 === N9
    N7 === N10

For the latter case, to eliminate ambiguity, we can define another grammar,

\[ \begin{aligned} E &\rightarrow E + T \\ E &\rightarrow T \\ T &\rightarrow T \times F \\ T &\rightarrow F \\ F &\rightarrow (E) \\ F &\rightarrow a \\ \end{aligned} \]

Then we can build the unique parse tree of \(a + a \times a\).

graph TD;
    N1(("E"))
    N2(("E"))
    N3(("+"))
    N4(("T"))
    N5(("T"))
    N6(("T"))
    N7(("×"))
    N8(("F"))
    N9(("F"))
    N10(("F"))
    N11(("a"))
    N12(("a"))
    N13(("a"))

    N1 === N2
    N1 === N3
    N1 === N4
    N2 === N5
    N4 === N6
    N4 === N7
    N4 === N8
    N5 === N9
    N6 === N10
    N8 === N11
    N9 === N12
    N10 === N13

Definition

A CFG is in Chomsky normal form (CNF) if every rule of it is of one of the following forms.

• \(S \rightarrow e\).
• \(A \rightarrow BC\), where \(B, C \in V - \Sigma - \{S\}\).
• \(A \rightarrow a\), where \(a \in \Sigma\).

Property

Suppose \(G\) is a CFG in CNF, if \(G\) generates a string of length \(n \ge 1\), then the length of derivation is exactly \(2n-1\).

Theorem

Every CFG has an equivalent CFG in CNF.

Pushdown Automata (PDA)

Definition

PDA = NFA + stack

A PDA is a 6-tuple \(P = (K, \Gamma, \Sigma, \Delta, S, F)\),

• \(K\) is a finite set of states.
• \(\Gamma\) is the stack alphabet.
• \(\Sigma\) is the input alphabet.
• \(S \in k\) is initial state.
• \(F \subseteq k\) is the set of final states.

• \(\Delta\) is a transition relation. It describes in the current state and a string at the top of the stack, when read a symbol, what will the next state be, and then pop the top string, push another string onto the stack.

  \[ \Delta \subseteq (K \times (\Sigma \cup \{e\}) \times \Gamma^*) \times (K \times \Gamma^*) \]

Definition

• Yields in one step

  • \((p, x, \alpha) \vdash_P (q, y, \beta)\) if \(\exists ((p, a, \gamma), (q, \eta))\), such that \(x = ay\), \(\alpha = \gamma\tau\) and \(\beta = \gamma\tau\) for some \(\tau \in \Gamma^*\).

• Yields

  • \((p, x, \alpha) \vdash_P^* (q, y, \beta)\) if \((p, x, \alpha) = (q, y, \beta)\) or \((p, x, a) \vdash_P \cdots \vdash_P (q, y, \beta)\).

Definition

\(P\) accepts \(w \in \Sigma\) if \((s, w, e) \vdash_P^* (q, e, e)\) for some \(q \in F\).

For \(L(P) = \{ w \in \Sigma^*: P \text{ accepts } w \}\), we say \(P\) accepts \(L(P)\).

Example

Show that \(\{w \in \{0, 1\}^* | \text{ the number of 0 equals the number of 1}\}\) can be accepted by a PDA.

Proof

Construct a PDA \(P = (K, \Gamma, \Sigma, \Delta, S, F)\), where

• \(K = \{s, q, f\}\).
• \(\Gamma = \{0, 1, \$\}\), where \(\$\) means the bottom of the stack.
• \(\Sigma = \{0, 1\}\).
• \(S = s\).
• \(F = \{f\}\).

• transition relation

  \[ \begin{aligned} \Delta = \{ & ((s, e, e), (q, \$)), \\ & ((q, 0, 0), (q, 00)), \\ & ((q, 0, \$), (q, 0\$)), \\ & ((q, 0, 1), (q, e)), \\ & ((q, 1, 1), (q, 11)), \\ & ((q, 1, \$), (q, 1\$)), \\ & ((q, 1, 0), (q, e)), \\ & ((q, e, \$), (f, e)) \} \end{aligned} \]

I think \(\$ \in \Gamma\) here is to make \(q\) not the final state.

Considering that PDA is NFA, we can construct a better PDA, with

• \(K = \{s\}\).
• \(\Gamma = \{0, 1\}\).
• \(\Sigma = \{0, 1\}\).
• \(S = s\).
• \(F = \{s\}\).

• transition relation

  \[ \begin{aligned} \Delta = \{ & ((s, 0, 1), (s, e)), \\ & ((s, 0, e), (s, 0)), \\ & ((s, 1, 0), (s, e)), \\ & ((s, 1, e), (s, 1)) \} \end{aligned} \]

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最后更新: 2024.01.08 10:20:32 CST
创建日期: 2023.10.24 15:22:40 CST