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Lecture 11 Context-Free Grammar

Lecture 11 Context-Free Grammar. Definition. A context-free grammar (CFG) G is a quadruple (V, Σ , R, S) where V : a set of non-terminal symbols Σ : a set of terminals ( V ∩ Σ = Ǿ ) R : a set of rules ( R: V → (V U Σ )* ) S : a start symbol. Example. V = {q, f,} Σ = {0, 1}

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Lecture 11 Context-Free Grammar

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  1. Lecture 11 Context-Free Grammar

  2. Definition • A context-free grammar (CFG) G is a quadruple (V, Σ, R, S) where • V: a set of non-terminal symbols • Σ: a set of terminals (V ∩ Σ = Ǿ) • R: a set of rules (R: V → (V UΣ)*) • S: a start symbol.

  3. Example • V = {q, f,} • Σ = {0, 1} • R = {q → 11q, q → 00f, f → 11f, f → ε } • S = q • (R= {q → 11q | 00f, f → 11f | ε })

  4. How do we use rules? • If A → B, then xAy xBy and we say that xAy derivates xBy. • If s ··· t, then we write s * t. • A string x in Σ* is generated by G=(V,Σ,R,S) if S * x. • L(G) = { x in Σ* | S * x}.

  5. Example • G = ({S}, {0,1}. {S → 0S1 | ε }, S) • ε in L(G) because S ε . • 01 in L(G) because S 0S1 01. • 0011 in L(G) because S 0S1 00S11 0011. • 0 1 in L(G) because S * 0 1 . • L(G) = {0 1 | n > 0} n n n n n n

  6. Context-Free Language (CFL) • A language L is context-free if there exists a CFG G such that L = L(G).

  7. Theorem • For every regular set, there exists a CFG G such that L=L(G). Proof. Let L=L(M) for a DFA M=(Q, Σ, δ, s, F). Construct a CFG G=(V, Σ, R, S) as follows. V = Q, Σ = Σ, R = { q → ap | δ(q,a) = p } U { f → ε | f in F}, S = s. x1 xn S x1q1 x1x2q2 ··· x1…xnf x1…xn f=qn s q1

  8. There is a path associated with x from initial state to a final state. x in L(M) S * x Therefore, L(M) = L(G).

  9. Corollary • Every regular language is a CFL. • The class of regular languages is a proper subclass of CFLs. CFL Regular Why, proper?

  10. Regular Grammar • Regular grammar is a CFG (V, Σ, R, S) such that every rule is in form V→ Σ*(V+ε) • Example G = ({S, A}, {0, 1}, {S →1A, A → 00}, S) • Remark: Every regular language can be generated by a regular grammar.

  11. Theorem • Every regular grammar generates a regular language. Proof. Consider a regular grammar G=(V, Σ, R, S). Construct a string-labeled digraph with vertex set V U {f} as follows: For each rule A → xB, x in Σ* and B in V, draw an edge A → B. x x For each rule A → x, x in Σ*, draw an edge A → f

  12. Example G = ({S,A}, {0,1}, {S→0S | 10A, A→00}, S) 0 10 00 f S A This string-labeled digraph with initial state S and a final state f is a state diagram of an NFA M. There is a path associated with x from S to f in M. S * x in Σ* Therefore, L(G) = L(M).

  13. Corollary A language L is regular if and only if L can be generated by a regular grammar.

  14. Right-Linear and Left-Linear • The regular grammar is also called a right-linear grammar. • A grammar G=(V, Σ, R, S) is left-linear if every rule is in form V → (V+ε)Σ*. (e.g., ({S,A}, {0, 1}, {S→A01, A→10}, S) Remark: Every language generated from a left-linear grammar is regular. Why?

  15. Why? • For left-linear grammar G = (V, Σ, R, S), construct G = (V, Σ, R , S) where R = {A→W | A→W in R}. • G is right-linear. Hence, L(G ) is regular. • Therefore, L(G) = L(G ) is regular. R R R R R R R R

  16. Example 1 G = ({S,A}, {0, 1}, {S→A01, A→10}, S) G =({S,A}, {0, 1}, {S →10A, A →01}, S) NFA accepts L(G ) L(G )={1001} L(G)={1001} R R 10 01 S A R

  17. Example 2 L(G) = 0*1 L(G ) = 10* NFA accepts 10* G = ({S,A}, {0,1}, {S →1A, A →0A|ε}, S) G = ({S,A}, {0,1}, {S →A1, A →A0|ε}, S) R 1 S A 0 R

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