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Chap 2

Chap 2. Context-Free Languages. Context-free Grammars. is not regular Context-free grammar : eg. G 1 : A 0A1 substitution rules AB production rules B# A,B : variables 0,1,# : terminals. Derivation: generate 000#111 Parse tree:. Def : (context-free grammar)

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Chap 2

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  1. Chap 2 Context-Free Languages

  2. Context-free Grammars • is not regular • Context-free grammar : eg. G1: A0A1 substitution rules AB production rules B# A,B : variables 0,1,# : terminals

  3. Derivation: generate 000#111 • Parse tree:

  4. Def: (context-free grammar) A context-free grammar is a 4-tuple (V,,R,S) 1. V: a finite set called the variables 2. : a finite set, disjoint from V, called the terminals 3. R: a finite set of rules 4. S: start symbol A Language of the grammar is

  5. Eg : • Eg : rules

  6. Def: • A string w is derived ambiguously in context-free grammar G if it has 2 or more leftmost derivations • Grammar G is ambiguous if it generates some string ambiguously • A language that can only be generated by ambiguous grammars is called inherently ambiguous eg:

  7. Chomsky Normal Form Every rule is of the form a : terminal A,B,C : any variables B,C may not be the start variable allow

  8. Thm: Any context-free language is generated by a context-free grammar in Chomsky Normal Form

  9. Proof: 1. Add a new start symbol : original start symbol 2. Remove -rule A, A is NOT the start symbol • If is a rule, then add • If is a rule, then add Do so for each occurrence of A • If is a rule, we add unless it had been removed Repeat these steps until all  rules not involving the start variable

  10. u is a string of variables and terminals Proof cont.: 3. Handle all unit rules Remove a unit rule For , we add unless it was a removed unit rule. Repeat until all unit rules are removed 4. Replace with Ai : new variables ui : variable or terminal symbol If ui is terminal, add Uiui■

  11. Eg. Convert the following CFG to Chomsky Normal Form ◆ ◆Remove B→ ◆ Remove A→ S→ASA|aB A→B|S B→b| S0→S S→ASA|aB A→B|S B→b| S0→S S→ASA|aB|a A→B|S| B→b S0→S S→ASA|aB|a|SA|AS|S A→B|S B→b

  12. Eg conti. ◆Remove unit rule S→S◆ Remove S0→S ◆Remove A→B ◆ Remove A→S S0→S S→ASA|aB|a|SA|AS A→B|S B→b S0→ASA|aB|a|SA|AS S→ASA|aB|a|SA|AS A→B|S B→b S0→ASA|aB|a|SA|AS S→ASA|aB|a|SA|AS A→b|S B→b S0→ASA|aB|a|SA|AS S→ASA|aB|a|SA|AS A→b|ASA|aB|a|SA|AS B→b

  13. Eg conti. ◆ S0→AA1|UB|a|SA|AS S→AA1|UB|a|SA|AS A→b| AA1|UB|a|SA|AS B→b A1→SA U→a

  14. Pushdown Automata • Finite automata • Pushdown automata State control a a b b input State control a a b b x input y z stack ⋮

  15. Def: A pushdown automata is a 6-tuple(Q, , ,  , q0, F)where Q, ,  and F are all finite sets, and 1. Q: the set of states 2. : the input alphabet 3. : the stack alphabet 4. : Q× ×  → P(Q×), the transition function 5. q0∈Q is the start state 6. F⊆Q is the set of accept state

  16. A PDA M= (Q, , ,  , q0, F) accepts input w=w1w2 …wm, where each wi∈and sequences of states r0, r1,…, rm ∈Q and strings s0, s1,…, sm∈* exist that satisfy the next 3 conditions: (si : stack contents) 1. r0=q0 and s0= 2. For i=0,…,m-1, we have (ri+1,b) ∈ (ri, wi+1,a),where si=at and si+1 =bt for some a,b ∈ and t ∈* 3. rm∈ F ri ri+1 ⋯wi+1⋯⋯ ⋯wi+2⋯⋯ a t b t

  17. Eg: M1=(Q, , , , q1, F ), where Q={q1, q2, q3, q4} ={0,1} ={0,$} F={q1,q4}

  18. Eg conti. a,b→c : reading a from the input it may replace b on the top of the stack with c , → $ 0, → 0 q1 q2 1, 0→ q4 q3 , $→ 1, 0→

  19. Eg: Show a pushdown automaton that recognizes the language b, a→ c, → q1 q3 , $→ q4 , → , → $ q2 , → q5 q6 q7 , → , $→ b, → c, a→ a, → a

  20. Eg: Give a PDA recognizing the language wR means w written backwards , → $ 0, → 0 1, → 1 q1 q2 , → 0, 0→ 1, 1→ q4 q3 , $→

  21. Thm: A language is context-free iff some pushdown automaton recognizes it

  22. qstart Proof: (1) If a language is context free, then some PDA recognizes it , →S$ , A→w for rule A→w a, a→ for terminal a qloop , $→ qaccept q a,s→z  (q,a,s)=(r,xyz) q q1 ⇒ ,→y a,s→xyz q2 r r ,→x

  23. Proof conti: Q={ qstart,qloop,qaccept } ∪ E  (qstart, ,  )={ (qloop,S$) }  (qloop, , A )={ (qloop, w) | A→w is a rule in R }  (qloop, , $ )={ (qaccept, ) } eg: construct a PDA from the following CFG G S→aTb | b T→Ta| qstart , S→b , →T , →a , →S$ , T→a , →T qloop S→aTb→ab T→ , $→ ,S→b ,T→ a,a→ b,b→ a T b $ T b $ S $ b $ qaccept

  24. Proof conti: (2)If a pushdown automaton recognizes some language, then it is context free Say that P=(Q, , , , q0, {qaccept}) and construct G. The variables of G are {Apq : p,q∈Q} The start variable is • For each p,q,r,s ∈ Q , t∈, and a,b ∈ If (r,t) ∈ (p,a, ) and (q, ) ∈ (s,b,t) Put Apq→ aArsb in G • For each p,q,r ∈ Q , put Apq→ AprArq in G • For each p ∈ Q , put App→ in G

  25. Proof conti: Suppose P has the following 3 features: • It has a single accept state, qaccept • It empties its stack before accepting • Each transition either pushes a symbol onto the stack or pops one off the stack, but does not do both at the same time. Stack height Apq→AprArq Input string p r q state Generated by Arq Generated by Apr Stack height Apq→aArsb p q r s state a Generated by Ars b

  26. Proof conti: • Claim : (Page112 & 113) Apq generates x iff x can bring P from p with empty stack to q with empty stack. ■

  27. Non-Context-Free Languages • Thm:(Pumping lemma for context-free languages) If A is context free, then there is a number p, where if s∈A of length at least p, then s may be divided into 5 pieces s=uvxyz satisfying the conditions: 1.For each i≥0, uvixyiz∈A 2.|vy|>0 3.|vxy|≤p

  28. T T R R R R Proof: R u v x y z u v y z v x y R R R x v x y

  29. ≤b Proof conti: Let G be a CFG for CFL A b: the max number of symbols in the right-hand side of a rule (b≥2) Thus, if a parse tree has height h, then the string generated has length ≤ bh |V|: the number of variables in G Let p=b|V|+2 > b|V|+1 ⇒ Any string of length ≥ p must have a parse tree of height ≥ |V|+2

  30. Proof conti: s: a string in A with length ≥ p Let  be a parse tree of s Suppose  has the smallest number of nodes ⇒ has height ≥ |V|+2 (2) |v y|>0 (3) |v x y|≤ p if |v y|=0 ■ R ≥|V|+2 R R |v x y| ≤ b|V|+1 <b|V|+2 =p R R ⇒ R R ≤|V|+1 R  x  x v x y

  31. Eg: Use the pumping lemma to show that the language B={anbncn |n≥0} is not context free Proof: Assume B is a CFL and obtain a contradiction. Let p be the pumping length for B. Select s=apbpcp∈ B =uvxyz

  32. Proof conti: By condition (2) , v or y is not empty case 1:When v and y contain only one type of alphabet u v2 x y2 z a a a b a c b b b c c c

  33. Proof conti: case 2:When either v or y contain more than one type of symbol u v2 x y2 zwill destroy the order of anbncn (ab)2 ■

  34. Eg: Let C={aibjck : 0≤ i≤ j≤ k}. Use the pumping lemma to show that C is not a CFL Proof: p : pumping length Choose s = apbpcp = uvxyz

  35. Proof conti: case 1: When both v and y contain only one type of alphabet u v2 x y2 z a a a b a c b b b c ⇒⇒ There are fewer b than a c c ⇒⇒ There are fewer c than b and a

  36. Proof conti: case 2: When either v or y contain more than one type of symbol uv2xy2z will not contain the symbols in the correct order ■

  37. Eg: Use the pumping lemma to show that D={ww : w∈{0,1}* } is not a CFL Proof: p : pumping length Choose s = 0p1p0p1p = uvxyz ~~~ ≤ p

  38. Proof conti: 0p1111…1100….00001p straddle the midpoint uv0xy0z has the form 0p1i0j1p , where i and j cannot both be p v x y

  39. Using dynamic programming to test if w∈ L Let L be generated by CFG G in Chomsky normal form. For i≤ j, Table(i,j) contains the collection of variables that generates the substring wiwi+1…wj CFL

  40. D=“On input w=w1w2…wn (page 241) 1.If w= and S→ is a rule, accept 2.For i=1 to n 3 For each variable A 4 Test whether A→b is a rule, where b=wi 5 if so, place A in Table(i,i) 6.For l=2 to n 7. For i=1 to n-l+1 8. j=i+l-1 9. For k=i to j-1 10. For each rule A→BC 11. If Table(i, k) contains B and Table(k+1, j) contains C Put A in Table(i, j) 12.If S is in Table(1,n), ACCEPT; otherwise Reject. “

  41. Designing CFG: Eg: {0n1n : n≥0}∪{1n0n : n≥0} S2→1S20| S1→0S11| S→S1|S2 S1→0S11| S2→1S20|

  42. Eg: Convert any DFA into an equivalent CFG • Make a variable Ri for each state qi of the DFA • Add the rule Ri→aRj if (qi,a)=qj • Add the rule Ri→ if qi is an accept state • Make R0 the start variable

  43. Ambiguity: <EXPR>→<EXPR>+<EXPR>|<EXPR>x<EXPR>|(<EXPR>)|a a+a×a

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