Context Free Grammars

31 Jan 04:30 PM

Prof. Calvin

CSCI 5100

Sketch

• Context Free Grammars
  • Rules
  • Formal Definition
  • Examples
  • Ambiguity

Rules

Context Free Grammars

\[ \begin{align*} G_1& \\ &\left. \begin{aligned} &S \rightarrow 0S1 \\ &S \rightarrow R \\ &R \rightarrow \varepsilon \end{aligned} \right\} \quad \text{(Substitution) Rules} \end{align*} \]

• We use \(\varepsilon\) to denote the empty string.

Rules

• Term these “rules”
• They are of form:
  • Symbol
  • →
  • String of symbols

Terms

• Rule: Statements of form Variable → (string of symbols and terminals)
• Variable: Those symbols on the left-hand side (LHS) of a → in a rule
• Terminals: Those symbols which appear in only on the right-hand side (RHS)
• Starting variable: The topmost symbol.

3 Rules

• \(S \rightarrow 0S1\)
• \(S \rightarrow R\)
• \(R \rightarrow \varepsilon\)

2 Variables / 2 Terminals

• Variables
  • \(S\)
  • \(R\)
• Terminals
  • \(0\)
  • \(1\)

Generation

1. Write down the start variable.
2. Substitute the start variable according to any rule.
3. Substitute any variable for any rule until no variables remain.

• That is, only terminals.

4. Every possible string is the language \(L(G)\)

Example

\[ \begin{aligned} &S \rightarrow 0S1 \\ &S \rightarrow R \\ &R \rightarrow \varepsilon \end{aligned} \]

• \(S\)
• \(0S1\)
• \(00S11\)
• \(00R11\)
• \(0011\)

• \(S\)
• \(0S1\)
• \(0R1\)
• \(01\)

Graphically

• \(S\)
  
• \(0S1\)
• \(00S11\)
• \(00R11\)
• \(0011\)

Graphically

Example

\[ \begin{aligned} &S \rightarrow 0S1 \\ &S \rightarrow R \\ &R \rightarrow \varepsilon \end{aligned} \]

• \(L(G_1) = \{ 0^k1^k | k \in \mathbb{N} \}\)
• This is a language a CFG can do.
  • That a DFA/NFA/GNFA cannot.

Shorthand

\[ \begin{aligned} &S \rightarrow 0S1 \\ &S \rightarrow R \\ &R \rightarrow \varepsilon \end{aligned} \]

\[ \begin{aligned} &S \rightarrow 0S1\quad|\quad R\\ &R \rightarrow \varepsilon \end{aligned} \]

Formal Definition

Context Free Grammar

• A Context Free Grammar (CFG)
• A CFG is a 4-tuple \((V, \Sigma, R, S)\)
  • \(V\) : finite set of variables
  • \(\Sigma\) : finite set of terminal symbols
    • We note this is the alphabet, a la DFA
  • \(R\) finite set of rules
    • Of form \(V \rightarrow (V \cup \Sigma)*\)
  • \(S\) start variable

Substitutions

• \(\forall u,v \in (V \cup \Sigma)*\)
• \(u \Rightarrow v := \exists R : R(u) \rightarrow v\)
• \(u \overset{*}{\Rightarrow} v := \exists u_1, u_2, ... u_n : \underset{i \leq n}{\Rightarrow} u_i = v\)
  • \(\overset{*}{\Rightarrow}\) is the transitive closure of \(\Rightarrow\)
  • Term this a derivation of \(v\) from \(u\)
  • If \(u = S\) term this the derivation of \(v\)
• \(L(G) = \{ w | w \in \Sigma* \land S \overset{*}{\Rightarrow} w \}\)

Check in

• Which of the following is a CFG

\[ \begin{align*} C_1& \\ &\left. \begin{aligned} &B \rightarrow 0B1 \\ &B1 \rightarrow 1B \\ &0B \rightarrow 0B \end{aligned} \right. \end{align*} \]

• This has a context.

\[ \begin{align*} C_2& \\ &\left. \begin{aligned} &S \rightarrow 0S \quad | \quad S1 \\ &R \rightarrow RR \end{aligned} \right. \end{align*} \]

• This happens to be the empty language.

Example

Arithmetic

\[ \begin{align*} G_2& \\ &\left. \begin{aligned} &E \rightarrow E+T \quad | \quad T\\ &T \rightarrow T\times F \quad | \quad F \\ &F \rightarrow ( E ) \quad | \quad a \end{aligned} \right. \end{align*} \]

\[ \begin{aligned} &V = \{ E, T, F \}\\ &\Sigma = \{+, \times, (, ), a \}\\ &R = \text{As shown}\\ &S = E \end{aligned} \]

Graphically

• \(E\)
  
• \(E+T\)
• \(T+T\times F\)
• \(F+F\times a\)
• \(a+a\times a\)

Graphically

Takeaways

• This is a viable way to decribe a programming language.
  • In fact I wrote a series of grammars for my doctoral thesis (pg 18).
• Parse trees may encode e.g. precedence
  • \(\times\) over \(+\)
• If a string can be derived by different substitutions it is ambigious.

Ambiguity

Arithmetic

\[ \begin{align*} G_2& \\ &\left. \begin{aligned} &E \rightarrow E+T \quad | \quad T\\ &T \rightarrow T\times F \quad | \quad F \\ &F \rightarrow ( E ) \quad | \quad a \end{aligned} \right. \end{align*} \]

\[ \begin{align*} G_3& \\ &\left. \begin{aligned} &E \rightarrow E+E \quad | \quad E \times E \quad | \quad ( E ) \quad | \quad a\\ \end{aligned} \right. \end{align*} \]

• These represent the same language (\(L(G_2) = L(G_3)\))!
• But \(G_3\) is ambigious!

\(a + a \times a\)

Exercise

Python

\[ \begin{align*} G_3& \\ &\left. \begin{aligned} &E \rightarrow E+E \quad | \quad E \times E \quad | \quad ( E ) \quad | \quad a\\ \end{aligned} \right. \end{align*} \]

• Express CFG \(G_3\) in Python.

Solution

from itertools import count # infinite iterator
from itertools import combinations_with_replacement as cwr
# define variables, so we can use them
rules = (
  lambda s : s.replace("E", "E+E"),
  lambda s : s.replace("E", "ExE"), 
  lambda s : s.replace("E", "(E)"), 
  lambda s : s.replace("E", "a")
)
trans = lambda fs, s: fs[0](trans(fs[1:], s)) if len(fs) else s

G_2 = (trans(fs,'E') for n in count() for fs in cwr(rules, n))

next(G_2), next(G_2), next(G_2)

('E', 'E+E', 'ExE')

• This isn’t quite right. What’s wrong?

End of Day :)

• Good work!